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H: Indefinite Integration : $\int \frac{dx}{(x+\sqrt{(x^2-1)})^2}$
Problem :
Solve : $\int \frac{dx}{(x+\sqrt{(x^2-1)})^2}$......(i)
I tried :
Let $x =\sec\theta$ therefore , (i) will become after some simplification
$$\int \frac{\sin\theta}{(1+\sin\theta)^2}d\theta$$
but i think its wrong method of approaching please suggest further....thanks..
AI: HINT: $$\frac1{x+\sqrt{x^2-1}}=x-\sqrt{x^2-1}$$
So, $$\frac1{(x+\sqrt{x^2-1})^2}=(x-\sqrt{x^2-1})^2=x^2+x^2-1-2x\sqrt{x^2-1}=2x^2-1-2x\sqrt{x^2-1}$$
Put $x^2-1=y$ for the last part
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H: Determine depth of a partially filled hemisphere
Recently came across a question in a Year 9 math book of which there was no "working out" supplied and offers now description on how they obtained the answer.
The question goes like this:
A bowl is in the shape of a hemisphere with radius 10cm. The surface of the water in the container has a radius of 7cm. How deeps is the water?
The supplied answer is 2.86 cm but I am lost for how they go it. Does anyone have any pointers or an answer?
AI: Draw a vertical cross-section of the bowl, passing through the centre $O$ of the hemisphere.
Let $N$ be the nearest point to $O$ on the surface of the water, and let $F$ be the point on the surface of the water which is furthest from $O$.
Then $NOF$ is a right triangle, with right angle at $N$.
By the Pythagorean Theorem, we have $(ON)^2+7^2=10^2$, so $ON=\sqrt{51}\approx 7.14$. The depth of the water is therefore $10-\sqrt{51}$, approximately $2.85857$.
|
H: How many different numbers are composed by n repeated digits?
For example, there are 3 digits: 1, 1, 4 and they compose 3 different numbers: 114, 141, 411.
My questions is: given n repeated digits: 1 * n1, 2 * n2, 3 * n3, ..., 9 * n9, in which ni >= 0 and n1 + n2 + ... + n9 = n, how many different numbers are composed by the n digits?
AI: Usually, the easiest way to answer these questions is to think as follows:
1) Suppose that we can tell ALL of the objects apart. (For instance, say we numbers the 1's as $1_1,1_2,\ldots,1_{n_1}$, etc.). How many ways could these be arranged? In this case, that is just $n!$, of course.
2) Obviously this was an overcounting. But by how much? In other words, how many rearrangements of the "labeled" objects should all count as the same "unlabeled" object? Given a sequence, you can rearrange the 1's in any way you like, as long as the positions that have 1's don't change; since there are $n_1$ 1's, this can be done in $n_1!$ ways. Similarly for the rest.
So, overall, there are
$$
\frac{n!}{n_1!n_2!\cdots n_9!}=\binom{n}{n_1,\ldots,n_9},
$$
where this last is the multinomial coefficient.
|
H: Find the Dirichlet inverse of the identity function
I thought I got this one but now I'm having doubts. I have that if $f$ is a function such that $(f\ast\text{id})(n)=(\text{id}\ast f)(n)=\iota$ then $f$ is multiplicative since it is the inverse of a multiplicative function, and so it suffices to examine its values at prime factors. So letting $n=\prod_{i=1}^{k}p_{i}^{\alpha_{i}}$
we require $$\left(f\ast\text{id}\right)(n) = \sum_{ab=n}f\left(a\right)b
= \sum_{i=1}^{k}f\left(p_{i}^{\alpha_{i}}\right)\prod_{j\ne i}p_{j}^{\alpha_{j}}
= 0
$$
And since $\prod_{j\ne i}p_{j}^{\alpha_{j}}\ne0$
for all $i$, we need $f\left(p_{i}^{\alpha_{i}}\right)=0$ for all primes $p_i$. Then for $n=1$ we require $\left(f\ast\text{id}\right)\left(1\right) = f(1)1
= 1
$ so $f(1)=1$. This implies that $f(n)=\cases{1 \ n=1 \\ 0 \ n\ne 1}$, or, in other words, $f=\iota$.
But I am doubting this now, since $\iota$ is supposed to be the identity with respect to the Dirichlet convolution, so that $\iota\ast \text{id}$ should be $\text{id}$. But I can't see where I've made a mistake. Can anyone help?
AI: It is not true that the only divisors of a composite are prime powers - generally divisors of a composite number will themselves be products of more than one prime - so your expression after the $\sum_{ab=n}$ sum is not correct. Just because it is sufficient to look at prime powers does not mean one can suppose no other kinds of factors of a composite number exist!
You want to "localize": just look at the case where $n=$ prime power. Observe
$$\delta_{r}=\iota(p^r)=(f*{\rm id})(p^r)=\sum_{l=0}^rf(p^l)p^{r-l}$$
for all primes $p$ and integers $r\ge0$. In particular,
$$\begin{array}{lll} f(1)=1 & \implies & f(1)=1 \\
f(p)+pf(1)=0 & \implies & f(p)=-p \\
f(p^l)+\cdots+p^{l-1}f(p)+p^lf(1)=0 & \implies & ?~~?~~?~~?~~? \end{array}$$
the third line is the same for $l\ge2$. Prove it using induction. Can you now write down the general form for $f$ at all integer arguments? Hint: consider the $\mu(\cdot)$ function, squarefreeness and signs.
|
H: Prove that $\log _5 7 < \sqrt 2.$
Prove that $\log _5 7 < \sqrt 2.$
Trial : Here $\log _5 7 < \sqrt 2 \implies 5^\sqrt 2 <7.$ But I don't know how to prove this. Please help.
AI: Observe that:
$$ \begin{align*}
\log_5 7 &= \dfrac{3}{3}\log_5 7 \\
&= \dfrac{1}{3}\log_5 7^3 \\
&= \dfrac{1}{3}\log_5 343 \\
&< \dfrac{1}{3}\log_5 625\\
&= \dfrac{1}{3}\log_5 5^4\\
&= \dfrac{1}{3}(4)\\
&= \sqrt{\dfrac{16}{9}}\\
&< \sqrt{\dfrac{18}{9}}\\
&= \sqrt{2}\\
\end{align*} $$
as desired.
|
H: Getting a values from nodes
The goal: get horizontal values of vertical level N where level 1 is pinacle node (1).
Example: level 4 as input should produce: | 1 | 3 | 3 | 1 |
Note: the sum of two adjacent nodes of level N is the value of node of level N + 1 in between those nodes of level N (marked with circles on the image).
I have accomplished this task with the help of loops and arrays:
function getValues($level) {
$level = $level === 0 ? 1 : abs((int) $level);
$data = array(1 => array(1 => 1)
, 2 => array(1 => 1, 2 => 1));
for ($i = 3; $i <= $level; $i++) {
for ($ii = 1; $ii <= $i; $ii++) {
$val = $ii === 1 || $ii === $i
? 1 : $data[$i - 1][$ii - 1] + $data[$i - 1][$ii];
$data[$i][$ii] = $val;
}
}
return $data[$level];
}
echo '| '.implode(' | ', getValues(7))." |\n";
It works pretty good, but...
Question: how can I do the same with the help of Math instead of using control structures of programming language? Probably I need logarithms?
AI: You're trying to get the $N$th row of Pascal's triangle.
The $N$th row can be calculated with the formulae:
$$
\mathbf{C}_k^n= {n \choose k} = \frac{n!}{k!(n-k)!}
$$
where, $n$ is the row and $k$ is the vertical position.
|
H: Why does $\lim_{n\to \infty}\sqrt{n}(1-x^2)^n=0$ if $0<|x|\leq 1$?
I was working on constructing an approximation of the identity, and one point in my construction requires me to show that there is a sufficiently large $n$ such that given $\epsilon>0$ and some $0<\delta<1$, then $\sqrt{n}(1-x^2)^n<\epsilon$ if $\delta\leq|x|\leq 1$.
I know that $\lim_{n\to\infty}\sqrt{n}(1-x^2)^n=0$ since $(1-x^2)$ is a positive number less than $1$, and this goes to $0$ much faster than $\sqrt{n}$ goes to $\infty$. So such a large $n$ does exist.
I'm happy to leave it at that, but how could you more rigorously convince someone that the limit does in fact go to $0$?
AI: Deal with $|x|=1$ separately, it is very easy. Now suppose that $|x|\lt 1$.
Let $1-x^2=\dfrac{1}{1+a}$. Note that since $0\lt 1-x^2\lt 1$, $a$ is positive. By the Binomial Theorem, or the easily proved (by induction) Bernoulli Inequality, we have $(1+a)^n \ge 1+an$.
Thus
$$\sqrt{n} (1-x^2)^n \le \frac{\sqrt{n}}{1+an}.$$
Now for any positive $\epsilon$, it is relatively easy to come up with an $N$ such that
$$\frac{\sqrt{n}}{1+an}\lt \epsilon$$
if $n\gt N$. For example, we can take as $N$ any integer $\ge \left(\dfrac{1}{a\epsilon}\right)^2$.
|
H: Does the integrability of $\log(f(x))$ imply $f(x)$ is bounded?
Let $f(x):(a,b)\rightarrow \mathbb{R}$. Suppose $\int_a^b\log{f(x)}\,\mathrm{d}x<\infty$, can we claim that $0<f(x)<M<\infty$ a.e.. Why and why not?
AI: Hint: Consider $f\colon (0,1)\to \Bbb R, x\mapsto \dfrac{1}{x}$.
|
H: How to find this mass?
Let $R$ be the region in the first quadrant of the plane bounded by the lemniscates of the following equations:
$\rho^2=4\cos(2\theta)$,
$\rho^2=9\cos(2\theta)$,
$\rho^2=4\sin(2\theta)$, and
$\rho^2=9\sin(2\theta)$.
Find the mass of this portion of plane if the density at the point $P(\rho,\theta)$ is $\delta(\rho, \theta)=16\rho$.
AI: The mass integral will involve the double integral for area in polar coordinates with an embedded density function:
$$m \ = \ \int_{\theta_1}^{\theta_2} \int_{\rho_{inner}}^{\rho_{outer}}\ \delta (\rho , \theta) \ \ \rho \ d\rho \ \ d\theta \ . $$
Since the density function is $ \ \delta (\rho , \theta) \ = \ 16 \ \rho \ , $ this integral becomes
$$m \ = \ \int_{\theta_1}^{\theta_2} \int_{\rho_{inner}}^{\rho_{outer}}\ 16\rho^2 \ \ d\rho \ \ d\theta \ . $$
Some of the hard work is going to be finding the limits of integration. In the diagram below, the lemniscates (which, in full, look like figure-eights or "infinity signs") intersect to form a region with four vertices. The angles are found from solving the various combinations of $ \ a \cos 2 \theta \ = \ b \sin 2 \theta \ $ ; two vertices are on the ray $ \ \tan 2 \theta = 1 \ \Rightarrow \ \theta = \frac{\pi}{8} \ $ , one at $ \ \tan 2 \alpha = \frac{4}{9} \ , $ and the fourth at $ \ \tan 2 \beta = \frac{9}{4} \ . $ Despite appearances, the region is not symmetrical about $ \ \theta = \frac{\pi}{8} \ , $ so we need to split the integral into
$$m \ = \ \int_{\alpha}^{\pi / 8} \int_{\rho_{inner}}^{\rho_{outer}}\ 16\rho^2 \ \ d\rho \ \ d\theta \ \ + \ \ \int^{\beta}_{\pi / 8} \ \int_{\rho_{inner}}^{\rho_{outer}} \ 16\rho^2 \ \ d\rho \ \ d\theta \ . $$
The red curves are the cosine-lemniscates and the blue ones, the sine-lemniscates. For the first integral, the radial integral runs from $ \ 2 \ \sqrt{\cos 2 \theta} \ $ to $ \ 3 \ \sqrt{\sin 2 \theta} \ $ , while for the second, the radial integration goes from $ \ 2 \ \sqrt{\sin 2 \theta} \ $ to $ \ 3 \ \sqrt{\cos 2 \theta} \ . $ So what you need to carry out is
$$m \ = \ \int_{\alpha}^{\pi / 8} \int_{2 \sqrt{\cos 2 \theta}}^{3 \sqrt{\sin 2 \theta}}\ 16\rho^2 \ \ d\rho \ \ d\theta \ \ + \ \ \int^{\beta}_{\pi / 8} \ \int_{2 \sqrt{\sin 2 \theta}}^{3 \sqrt{\cos 2 \theta}} \ 16\rho^2 \ \ d\rho \ \ d\theta \ . $$
This will be a bit unpleasant since the radicals won't just "square away" and none of the angular limits of integration have "nice" values...
|
H: holomorphic on right half plane
Could any one tell me how to solve this one?
Let $f$ be holomorphic function in the right half plane, with $|f(z)|<1$ for all $z\in \{z:Re(z)>0\}$, $f(1)=0.$ Find out largest possible value of $|f(2)|.$
AI: Let $\phi(z) = \frac{1-z}{1+z}$. $\phi$ is a Möbius transformation that maps $U$, the open unit disk, into the right half plane. Let $\tilde{f} = f \circ \phi$. Then $\tilde{f}$ maps $U$ into itself, and $\tilde{f}(0) = f(1) = 0$. Hence the Schwartz lemma tells us that $|\tilde{f}(z)| \le |z|$ for $z \in U$. We have $\phi(-\frac{1}{3}) = 2$, hence this tells us that $|f(2)| = |\tilde{f}(-\frac{1}{3})| \le \frac{1}{3}$.
If we let $f(z) = \phi(z)$, we get $f(2) = -\frac{1}{3}$, hence the maximum is attained.
|
H: a codeword over $\operatorname{GF}(4)$ -> two codewords over $\operatorname{GF}(2)$ using MAGMA
A codeword $X$ over $\operatorname{GF}(4)$ is given. How can I write it as $X= A+wB$ using MAGMA? where $A$ and $B$ are over $\operatorname{GF}(2)$ and $w^2 + w =1$.
Is there an easy way, or do I have to write some for loops and if statements?
AI: Probably there is an easier way, but the following function should do the job:
function f4tof2(c)
n := NumberOfColumns(c);
V := VectorSpace(GF(2),n);
ets := [ElementToSequence(c[i]) : i in [1..n]];
return [V![ets[i][1] : i in [1..n]],V![ets[i][2] : i in [1..n]]];
end function;
|
H: Lie algebra isomorphism between $\mathfrak{sl}(2,{\bf C})$ and $\mathfrak{so}(3,\Bbb C)$
I think that this is an exercise. I can not find a solution.
We can define Lie bracket multiplication on $\mathbb{C}^3$ : $$ x\wedge y $$ where $x=(x_1,x_2, x_3)$, $y=
(y_1,y_2,y_3)$, and $\wedge $ is the wedge product we know.
Consider the Lie algebra $\mathfrak {sl}(2,\mathbb{C})= \{ X\in M_2(\mathbb{C}) \mid\ {\rm Trace} (X) =0\}$
and $$
e= \left(
\begin{array}{cc}
0 & 1 \\
0 & 0 \\
\end{array}
\right),\ f= \left(
\begin{array}{cc}
0 & 0 \\
1 & 0 \\
\end{array}
\right),\ h= \left(
\begin{array}{cc}
1 & 0 \\
0 & -1 \\
\end{array}
\right). $$ Note that $$ [e,f]=h,\ [e,h]=-2e,\
[f,h]=2f.$$
Here, the problem is to find an explicit isomorphism between $\mathfrak {sl}(2,\mathbb{ C})$ and $\mathbb{C}^3$.
Thank you in advance.
AI: Some hints: you have introduced a basis $\{e,f,h\}$ for the Lie algebra $sl(2,\mathbb C)$; all you need is a suitable basis $\{e_i\}$ in $\mathbb C^{3}$ and an isomorphism $\phi:\mathbb C^{3}\rightarrow sl(2,\mathbb C)$ of vector spaces s.t. $\phi(e_i\wedge e_j)=[\phi(e_i),\phi(e_j)]$. Can you find such basis? Try with the simplest one...
Then you should define the isomorphism $\phi$ simply as $\phi(e_i)=...$ (choose the right element of the basis for $sl(2,\mathbb C)$: you need to preserve compatibility with brackets) and extend it $\mathbb C$-linearly.
|
H: How to determine the no. of integral partitions into $k$ parts?
I wanted to know, if I was to partition $500$ into positive $k$ integers, not necessarily distinct under the following constraints
1.k is +ve.
2.all k parts need not be distinct.
3.the first integer subtracted from the last integer in the k parts is smaller than or equal to 1.( a(k)-a(1)<=1)
what would be a formula for the number of possibilities?
Any help appreciated.
Thanks.
AI: You simply have this expression
$$a_1+a_2+ \dots a_k=500$$
Now it's just a Stars and Bars problem. Assuming $a_i$'s are all non-negative integers as Alex says. You have infinitely many ways to partition if you let negative integers.
|
H: How to justify $\frac14(n^2(n+1)^2)+\frac14(4(n+1)^3) = \frac14(n+1)^2(n^2+4(n+1))$?
How does my lecturer go from :
$$ \frac {n^2(n+1)^2} {4} + \frac {4(n+1)^3} {4} \text{ to } \frac {(n+1)^2} {4} \times [n^2+4(n+1)] $$
I can understand that $$ \frac {n^2(n+1)^2} {4} = \frac {(n+1)^2} {4}\times n^2 $$
But I'm not sure how he can legally multiply $n^2+4(n+1) \text{ by } \frac {(n+1)^2} {4} $.
Thanks for reading my question.
AI: In $\frac{4(n+1)^3}{4}$ you can extract $4(n+1)$, i.e.
$$\frac{4(n+1)^3}{4}=4(n+1)*\frac{(n+1)^2}{4}; $$
collecting all terms in your expression, we arrive at
$$\frac{n^2(n+1)^2}{4}+\frac{4(n+1)^3}{4}=n^2*\frac{(n+1)^2}{4}+4(n+1)*\frac{(n+1)^2}{4} =\frac{(n+1)^2}{4}*[n^2+4(n+1)],$$
as desired.
|
H: $\log(0,05)$ is minus, but $\log(0,04999\ldots)$ is plus?
How is this calculated, and why is this?
We're calculating fixed-rate mortgage, with following formular:
$$
n = 1-\frac{\log(\frac{L\cdot x}{y})}{\log(1+x)}
$$
Where: $L$ is the loan size, $x$ is the interest, $y$ is the amount to pay back per month, and $n$ is the number of months it takes to pay back the loan.
AI: Of the many ways $\log$ can be defined, one is $\log x = \int_1^x\!\frac{1}{t}\,dt$ (I'm assuming you mean natural log, but if not, everything can be easily adapted, since $\log_n x = \log x/\log n$). Letting $x > 1$ and interpreting $\int_1^x\!\frac{1}{t}\,dt$ as the area under the graph of $1/x$ from $1$ to $x$, we see that since $1/x$ is positive for $x > 0$, the integral will also be positive for $x > 1$. However, if $x\in(0,1)$, we use the properties of the integral to see that
$$
\int_1^x\!\frac{1}{t}\,dt = -\int_x^1\!\frac{1}{t}\,dt.
$$
The integral $\int_x^1\!\frac{1}{t}\,dt$ is greater than $0$, because again we can interpret it as the area under $1/x$ from $x$ to $1$. This implies that $\log x < 0$ whenever $x\in(0,1)$. So, $\log(.05)$ and $\log(.049999\dots)$ are less than $0$.
I also feel the need to remark that $.05$ and $.04999\dots$ are the same number, so you shouldn't be getting a positive answer for one's log and a negative answer for the other's log, you should be getting the same answer. (You probably computed $\log(.49999\ldots9)$ for finitely many $9$'s, but as I explained above, this should also be negative.) The reason that these two numbers are the same is exactly the same as the reason $.99999\ldots = 1$, and we already have plenty of threads on that, so you can check one of those out if you are in doubt of this fact.
As for how $\log$ is calculated, it's usually tough to compute $\log x$ by hand for some random $x$, but you can approximate it by integral approximation methods, or expand $\log$ in a Taylor series near $x$, and use only finitely many terms of the series to get an approximation (this is likely what your calculator does).
|
H: adding infinitely many equations side by side in a recurrence relation
we are given that $x+\beta y_{n+1}=k_n+y_n$ for all $n\in\mathbb{N}\cup\{0\}$, where $\beta\in(0,1)$, $y_0=0$, and $k_n$ is 6 whenever $n$ is even and 4 whenever odd. Being the naive mathematician I do the following
$x+\beta y_1=6$
$\beta x+\beta^2y_2=4\beta + \beta y_1$
$\beta^2 x + \beta^3 y_3 = 6\beta^2 + \beta^2 y_2$
etc.
Then I add the countably infinitely many equations side by side and cancel terms.
$x+\beta y_1+\beta x+\beta^2y_2+\beta^2x+\beta^3y_3+\cdots= 6+4\beta+\beta y_1 +6\beta^2+\beta^2 y_2+\cdots$
The remaining part of the algebra is not of interest. I was wondering under what conditions what I'm doing is formally correct. Under what circumstances it would fail, in the sense that I cannot add them all side by side and simplify. My intuition tells me that we would need $\lim_{n\rightarrow\infty} \beta^ny_n = 0$ but if people here can help me out that will be very much appreciated. Also, is there a textbook I can read more on this topic?
AI: You can't literally add infinitely many things, but you can take an infinite series, assuming it converges. In this case what you want to look at is
a partial sum.
$$(x + \beta y_1) + (\beta x + \beta^2 y_2) + \ldots + (\beta^{n-1} x + \beta^{n} y_{n}) = 6 + (4 \beta + \beta y_1) + \ldots + (k_{n-1} \beta^{n-1} + \beta^{n-1} y_{n-1})$$
which you can then solve for $\beta^n y_n$.
|
H: Hilbert's finiteness theorem over arbitrary fields; reductive groups
As a generalization of the finiteness result Hilbert proved in his 1890 paper, one usually formulates the following nowadays:
Let $G\to\operatorname{GL}(V)$ be a rational representation of a linearly reductive group $G$ on a finite-dimensional $k$-vector space $V$. Then $k[V]^G$ is a finitely generated $k$-algebra.
The sources I looked at so far assume either $k=\mathbb{C}$, $k$ algebraically closed, or $k$ infinite. In what generality does this result hold, regarding $k$?
From the non-constructive proof of the result, I can't really see any reliance on $k$ being algebraically closed; just the existence of a Reynolds operator is needed. But $k$ algebraically closed or infinite is often assumed when reductive algebraic groups are defined. Any reference for this result with $k$ arbitrary or whatever the maximum generality is is greatly appreciated!
Edit: If I remember correctly, Emmy Noether proved the case $k$ arbitrary and $G$ a finite subgroup of $\operatorname{GL}_n(k)$ acting on the polynomial ring $k[x_1,\dots,x_n]$. The only "problem" there could be is that in general $k[V]$ is not a polynomial ring if $k$ is a finite field.
AI: In Ideals, Varieties, and Algorithms, by Cox, Little, and O'Shea, the theorem is shown for finite groups over any characteristic 0 field. The requirement that the field has characteristic 0 is clearly necessary if we want the Reynold's operator to be defined for arbitrary groups. The book doesn't really cover enough algebra to go into full generality for the linearly reductive case, so I'm not sure how this result carries over to that.
|
H: Open and closed set on a given metric
I'm stuck on this one, could you please give me a tip or two:
Let $d$ be a metric in $X$ such that $d : X^2 \rightarrow \mathbb{R} : d(x, y) = d_{1}(x,y) + d_{2}(x,y)$ for $x,y \in X$. Let $d_{1}$ be a discrete metric. Show that in the metric space $(X,d)$ for any $x\in X$ the open metric ball $K(x, \frac{1}{2})$ of radius $\frac{1}{2}$ and centered in $x$ is a one-element set $\left\{x\right\}$. Also prove that any $A \subset X$ is both open and closed in the meaning of metric $d$.
Thanks in advance!
AI: The latter is simple enough, given the former, since if $\{x\}$ is open for all $x\in X$, then every subset of $X$ is open (as a union of singletons). It then follows that every subset of $X$ is closed. (Why?)
As for the former, take $x\in X$, and suppose that $y\in X$ such that $d(x,y)<\frac12.$ Since $d=d_1+d_2,$ then $d_1(x,y)<\frac12$--supposing that $d_2$ is a (pseudo)metric, anyway, so that $d_2(x,y)\ge0.$ By definition of the standard discrete metric, though, $$d(x,y)=\begin{cases}0 & x=y\\1 & x\ne y.\end{cases}$$ What can we conclude?
More generally, suppose that $d_1,d_2$ are any pseudometrics on $X$, and put $d=d_1+d_2$. It can readily be seen that $d$ is a pseudometric on $X$ as well. Now, for $x,y\in X$, we have $d_1(x,y)\le d_1(x,y)+d_2(x,y)=d(x,y)$ and likewise $d_2(x,y)\le d(x,y).$ Since $d$-distances are not smaller than $d_1$-distances or $d_2$-distances, then $d$-balls are not larger than $d_1$-balls or $d_2$-balls with the same center and radius.
In fact, we say even more than that: for any $x\in X$ and any $r>0,$ we find that the open $d$-ball of radius $r$ centered at $x$ is contained in the corresponding open $d_1$- and $d_2$-balls. Indeed, if $d(x,y)<r,$ then $d_1(x,y)\le d(x,y)<r,$ so the open $d$-ball of radius $r$ centered at $x$ is contained in the open $d_1$-ball of radius $r$ centered at $x$ (similar for $d_2$).
For example, define $d_1,d_2:\Bbb R^2\to\Bbb R$ by $d_1(v,w)=|v_1-w_1|$ and $d_2(v,w)=|v_2-w_2|$, where $v=\langle v_1,v_2\rangle$ and $w=\langle w_1,w_2\rangle$. Visually, $d_1$ is the pseudometric on $\Bbb R^2$ that only thinks about how much two points in the plane differ horizontally; $d_2$, only vertically. Neither of these is a metric--$d_1$ notices no difference between any two points on a vertical line, and $d_2$ notices no difference between any two points on a horizontal line. However, if we put $d=d_1+d_2,$ then $d$ is a metric sometimes known as the "taxicab metric."
The open $d_1$-ball of radius $1$ about the origin (for example) is the region of the plane lying strictly between the lines $x=-1$ and $x=1$. The open $d_2$-ball of radius $1$ about the origin is the region lying between $y=-1$ and $y=1$. The intersection of these two "balls" is the open square with vertices $\langle 1,\pm 1\rangle$ and $\langle-1,\pm 1\rangle$ (which, incidentally, is precisely the open ball of radius $1$ about the origin using the "chess king metric"), while the open $d$-ball about the origin of radius $1$ is the open square (or diamond, if you want to look at it that way) with vertices $\langle 0,\pm 1\rangle$ and $\langle\pm 1,0\rangle$ (a proper subset of the intersection of the corresponding $d_1$- and $d_2$-balls). In this case, we see that adding the two pseudometrics together actually loses us a bunch of points from corresponding balls of a given radius, since $d$ thinks that $d_1$- and $d_2$-balls are unbounded!
For a related example (though not precisely the same thing), suppose you're given a metric $d_1$ on $X,$ and a positive constant $c>1.$ Let $d=c\cdot d_1$. Then $d$ is a metric on $X,$ and the open $d$-ball of radius $r$ about $x$ is the open $d_1$-ball of radius $\frac{r}{c}$ (which is less than $r$) about $x$. Intuitively (from the "point of view" of $d_1$), $d$ has poor depth perception, and always sees objects as being farther apart than they "really" are but at least $d$ describes balls in the "right" way, so when $d$ is asked to describe a ball of radius $r$, he will describe a ball, but because $d$ thinks of the "actual" distance $r$ as $cr$, then the ball $d$ describes will be of "actual" radius $\frac rc.$
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H: Tangent map $F_{*}: T_{p}(M) \to T_{F(p)}(N)$ are linear tranformations.
How to show that tangent map $F_{*}: T_{p}(M) \to T_{F(p)}(N)$ are linear tranformations?
I know that it suffices to show that
$$F_{*}(ax_{u}+bx_{v})=aF_{*}(x_{u})+bF_{*}(x_{v})$$
where $x$ is a patch in $M$.
By definition,
$$F_{*}(ax_{u}+bx_{v})=\frac{d}{dt}F(\alpha(t))_{t=0}$$
where $\alpha'(0)=ax_{u}+bx_{v}$.
I also know that if $y=F(\alpha(t))$, then $y'=F_{*}(\alpha'(t))$.
We can express $\alpha(t) =x(a_{1}(t),a_{2}(t))$, then $$\alpha'(t) =a_{1}'(t)[x_{u}(a_{1}(t),a_{2}(t))]+a_{2}'[x_{v}(a_{1}(t),a_{2}(t))]$$
For $t=0$, $\alpha'(0) =a_{1}'(0)[x_{u}(a_{1}(0),a_{2}(0))]+a_{2}'[x_{v}(a_{1}(0),a_{2}(0))]=ax_{u}+bx_{v}$.
Thus, I get $a_{1}(0)=u$, $a_{2}(0)=v$, $a_{1}'(0)=a$, $a_{2}'(0)=b$.
Now, we can say that
$$F_{*}(ax_{u}+bx_{y})=\frac{d}{dt}F(x(a_{1}(t),x(a_{2}(t)))_{t=0}$$
$$=a_{1}'(0)\frac{\partial}{\partial{u}}F(x(a_{1}(t),x(a_{2}(t)))+a_{2}'(0)\frac{\partial}{\partial{v}}F(x(a_{1}(t),x(a_{2}(t))$$
If $\frac{\partial}{\partial{u}}F(x)=F(x_{u})$
, done. but is it right?
Thanks in advance.
AI: To see that $F_*$ is linear, let us describe its image in the natural basis of $T_{F(p)}N$ coming from the choice of a coordinate patch. Let $\tilde{x}$ be a coordinate patch around $F(p)$, and let $\phi,\psi$ be the (smooth) functions on the domain of $x$ such that
$$F(x(u,v)) = \tilde{x}(\phi(u,v),\psi(u,v)).$$
Also, let $\tilde{\alpha} = F \circ \alpha$ and write $\tilde{\alpha}(t) = \tilde{x}(\tilde{a}_1(t),\tilde{a}_2(t))$ where $\tilde{a}_1(t) = \phi(a_1(t),a_2(t))$, and $\tilde{a}_2(t) = \psi(a_1(t),a_2(t))$. Then
$$F_*(\alpha'(0)) = \tilde{\alpha}'(0) = \tilde{a}_1'(0) \tilde{x}_u + \tilde{a}_2'(0) \tilde{x}_v \\ =(a\phi_u + b\phi_v) \tilde{x}_u + (a\psi_u+b\psi_v)\tilde{x}_v.$$
As an added bonus, this gives you a matrix representation of $F_*$ with respect to the bases given by $x$ and $\tilde{x}$.
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H: Function Projection: Orthogonal Polynomials
I am currently reading a paper called "Numerical Quadrature" by Timothy J. Giese (2008) which describes the numerical quadrature technique in detail. At one point (just before equation 19) it states that:
$"\dots$, we begin by projecting $f(x)$ into a set of orthogonal polynomials $[p_0(x),\dots,p_n(x)]$
$$f(x) = w(x)\sum_j p_j(x) \int f(x') p_j(x') dx'",$$
where $w(x)$ is the weighting function.
My question is, what is the intuition behind the truth of this equality?
AI: After looking at the paper, I think that the author is simply abusing notation here. Simply by virtue of words "by projecting...", the author means that $f(x)$ is to be approximated by the expression on the right-hand-side. The point is that $f(x)$ can be some arbitrary smooth function, and we have a quadrature that works on polynomials with weight $w(x)$, so we wish to express $f(x)$ approximately in terms of a polynomial multiplied by $w(x)$.
In particular, suppose $p_0, \ldots, p_n$ are orthonormal polynomials with respect to the weight $w(x)$; that is, we have inner product $\langle f,g \rangle = \int f(x) g(x) w(x) \, dx$. Then the orthogonal projection of $f(x)/w(x)$ onto that polynomial space is given by
\begin{align}
\mathrm{Proj} (f(x)/w(x)) & = \sum_{j=0}^n \langle f/w, p_j \rangle p_j(x) \\
& = \sum_j p_j(x) \int [f(x')/w(x')] p_j(x') w(x') \, dx' \\
& = \sum_j p_j(x) \int f(x') p_j(x') \, dx'
\end{align}
Now, by virtue of being an orthogonal projection, this means that so long as $f(x)$ and its higher derivatives are well-behaved, $f(x)/w(x)$ is "close" to the above expression (in an appropriate norm). Therefore, we have, informally,
$$f(x)/w(x) \approx \sum_j p_j(x) \int f(x') p_j(x') \, dx' $$
Multiplying by $w(x)$ on both sides yields the "equality" (which is really an approximation) that the author derived.
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H: Trying to understand the standard deviation's formula
I have just started learning standard deviation and I'm trying to understand the formula.
$$ s = \sqrt{ \frac{\sum(x-\bar x)^2}{n-1} } $$
Can anyone explain to me the square (and square root) part? If they are using the square & square root to prevent having negative value, why not just use | | for absolute value?
And I have another question. Why is the $-1$ sometimes omitted?
$$ \sigma = \sqrt{\frac{\sum(x-\bar x)^2}n} $$
Should I -1 or just omit it? I googled the standard deviation formula and some show -1 while some don't.
Thank you so much.
AI: It's not just to prevent negative values: standard deviation has many nice
mathematical properties that your alternative proposal would not. In particular, the absolute value function is not differentiable at $0$.
The reason for the $-1$ is to make $s^2$ an unbiased estimator of the variance.
See e.g. http://en.wikipedia.org/wiki/Sample_standard_deviation#Sample_standard_deviation
Whether you should use it or not may depend on what you're using it for.
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H: Check for directional derivative and show that $f(1,1,1) > f(0,0,0)$ when given the partial derivatives
I'm kinda lost in this exercise
Let $f:\mathbb{R}^3\rightarrow \mathbb{R}$ be of class $C^1$ and $\forall x=(x_1,x_2,x_3)\in\mathbb R^{3}$
$$\frac{\partial f}{\partial x_{1}}(x)=x_{2}, \frac{\partial f}{\partial x_{2}}(x)=x_{1}, \frac{\partial f}{\partial x_{3}}(x) > -1$$
Check if $f$ has a directional derivative in point $(1,2,3)$ and direction $v = (2,1,0)$. If so, what's the value of it? Also show that $f(1,1,1) > f(0,0,0)$.
Alright, when it comes to checking the directional derivative we can (since the function is of class $C^1$ that it is differentiable in all $x$ from the interior of its domain $\Omega$. Therefore it is differentiable in $(1,2,3)$ so it has a directional derivative which equals : $$\frac{\partial f}{\partial x_{1}}(x)(v_{1}) + \frac{\partial f}{\partial x_{2}}(x)(v_{2})+ \frac{\partial f}{\partial x_{3}}(x)(v_{3})$$ but how am I supposed to count it if we don't know what the $\frac{\partial f}{\partial z}$ is? (EDIT: Ok, $v_{3}$ is $0$ so this one doesnt matter :)) Also, how should I do the 2nd part of this exercise?
Thanks in advance! :)
AI: The directional derivative of $f$ at a point $(x,y,z)$ in the direction of a vector $v=(v_x,v_y,v_z)$ is
$$
\frac{\partial f}{\partial x}(x,y,z)\cdot v_x+\frac{\partial f}{\partial y}(x,y,z)\cdot v_y+\frac{\partial f}{\partial z}(x,y,z)\cdot v_z.
$$
In the specific case asked in the question, this is
$$
2\cdot2+1\cdot1+(?)\cdot0=5.
$$
Using the path $t\mapsto(t,t,t)$ from $(0,0,0)$ to $(1,1,1)$, one gets
$$
f(1,1,1)-f(0,0,0)=\int_0^1\left(\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}\right)(t,t,t)\,\mathrm dt\gt\int_0^1(t+t-1)\,\mathrm dt=0.
$$
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H: Expected value: Showing $[\Bbb E(X)-\Bbb E(Y)]^2 \geq 2 \cdot \Bbb{Cov}(X,Y)$
The original question is to show that for any Random variables $X,Y$ and $0\leq p \leq 1$
$$p\Bbb V(X)+(1-p)\Bbb V(Y) + p(1-p)[\Bbb E(X)-\Bbb E(Y)]^2 \geq p^2 \Bbb V(X)+(1-p)^2 \Bbb V(Y) +2p(1-p) \Bbb{Cov}(X,Y)$$
I tried to focus on showing $[\Bbb E(X)-\Bbb E(Y)]^2 \geq 2\Bbb{Cov}(X,Y)$, but it didnt work for me..
Any ideas ?
thanks
AI: Substracting the RHS from the LHS and simplifying by the nonnegative factor $p(1-p)$, one sees that the original question is solved as soon as
$$
\Bbb V(X)+\Bbb V(Y)+[\Bbb E(X)-\Bbb E(Y)]^2 -2\Bbb{Cov}(X,Y)\geqslant0.
$$
The fact that
$$
\Bbb V(X)+\Bbb V(Y)-2\Bbb{Cov}(X,Y)=\Bbb V(X-Y)\geqslant0,
$$
concludes the proof.
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H: Asymptotic behaviour of $1- \left( \frac{\Gamma(n+\frac{1}{2})}{\sqrt{n} \Gamma(n)} \right) ^2$
I know that $$\lim_{n\rightarrow \infty}\frac{\Gamma(n+\frac{1}{2})}{\sqrt{n} \Gamma(n)}=1,$$ but I'm interested in the exact behaviour of
$$a_n =1- \left( \frac{\Gamma(n+\frac{1}{2})}{\sqrt{n} \Gamma(n)} \right) ^2$$
particularily compared to $$b_n = \frac{1}{4n}$$
I haven't studied asymptotics yet, so I have no idea how to approach this, but I need this particular result in a statistics problem I'm working on.
AI: The following code in Mathematica
Series[1 - (Gamma[x + 1/2]/(Sqrt[x] Gamma[x]))^2, {x, Infinity, 6}]
gives
$$
\frac{1}{4 x}-\frac{1}{32 x^2}-\frac{1}{128 x^3}+\frac{5}{2048 x^4}+\frac{23}{8192 x^5}-\frac{53}{65536 x^6}+O\left[\frac{1}{x}\right]^7
$$
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H: Summing a series - Calculus 1.
I'm learning Calculus 1 at the collage,
and the semester's end is close, which bring with it the exams period.
So I pretty much understand all the topics,
except for a series summing.
I don't know why, but I just don't get it. I Googled a lot, but nothing. so I hope to find my luck here.
For example, I have the following series:
${\sum_{n=1}^{\infty}} ({1\over 2^{n-1}})$.
I need a detailed explanation, how to sum this series. Thanks in advance..!
AI: In mathematics, a geometric series is a series with a constant ratio between successive terms. For example, the series
$1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+ \ldots$
is geometric, because each successive term can be obtained by multiplying the previous term by $\frac {1}{2} $
In general the sum of a geometric series with the first term 'a' and common ratio 'r' is given by :. $$ S=\frac{a (1-r^n)}{1-r} $$
Suppose, $|r|<1 $ ie. $-1<r<1$ and $n$ tends to $\infty$, we can say that $r^n$ tends to $0$ (why?)
Hence, the sum becomes $$ S=\frac {a}{1-r} $$
Here, the first term is 1. the common ratio is $\frac {1
}{2} $ (Observe)
substituting, we get, $$ S=\frac {1}{1-\frac{1}{2}} $$
so, $S=2$
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H: Quick method /Birds eye view to determine the value
Is there any way to guess the answer without doing elaborating calculations?
AI: Corrected after a comment of André Nicolas:
We have $4^5=1024>1016$. This shows that $y$ can only be $1$, $2$ or $3$. Now if $y=3$, then since $3^2\times 3^5>1016$, the only possible values of $x$ are $1$ and $2$. Similarly, if $y=2$, the only possible values of $x$ are $1,2,3,4,5$. Computing the products $xy$ and comparing them with the values proposed in the question, out of these seven options there remain only two: $(x,y)=(2,2)$ and $(x,y)=(5,2)$. Checking for $z^3$, out of these two possibilities there remains only one: $(x,y,z)=(5,2,6)$.
Now if $y=1$, then $xy=x$, and all we have to check is: which of the six values among $1016-(xy)^2$ are cubes. This gives one more triple $(x,y,z)=(4,1,10)$.
So the possible values of $xy$ (among indicated) are $4$ and $10$.
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H: Finding the number of each type of fruit bought
A man has $\$100$. If we have that:
$1$ apple costs $\$1$
$1$ oranges cost $\$0.05$
$1$ watermelon costs $\$5$
and the man buys exactly $100$ pieces of fruit with exactly $\$100$ (He must buy at least 1 fruit from each category), then how many of each fruit did he buy? Could you please explain the method used? Thanks :)
AI: Given your edit, we must be able to purchase individual oranges, for otherwise the $100$ fruits for $\$100$ situation is impossible. (Why?)
Suppose $x,y,z$ are (respectively) the numbers of apples, oranges, and watermelons bought. Observe that one orange costs $\$0.05$ (why?). The amount he paid for apples was the cost per apple times the number of apples--that is, $1.00\cdot x=x$ dollars--likewise, he paid $0.05y$ dollars for the oranges and $5z$ dollars for the watermelons. Hence, since he paid $\$100$ in total, then $$100=x+.05y+5z\tag{$\heartsuit$}$$ On the other hand, he bought $100$ fruits, so we know that $$100=x+y+z\tag{$\clubsuit$}$$ Subtracting $(\heartsuit)-(\clubsuit)$ gives us $$0=-.95y+4z,$$ whence multiplication by $20$ yields $$0=-19y+80z$$ and so $$19y=80z.$$ Now, we know that $y,z$ are integers, so since $19$ is prime and is not a factor of $80,$ then $19$ is a factor of $z$. In particular, we know that $z$ is positive, so $z=19k$ for some integer $k\ge1,$ whence $y=80k.$ Now, putting those back into $(\clubsuit)$ gives us $$100=x+80k+19k=x+99k.\tag{$\clubsuit'$}$$ Since $x,k$ are positive integers, then it follows by $(\clubsuit')$ that $x=k=1,$ so he bought an apple, $80$ oranges, and $19$ watermelons.
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H: Please the Inequality in the proof of The Isoperimetric Inequality
From A proof of the Isoperimetric Inequality, can you please explain the starred inequality $$A + \pi r^2 = \int_{\gamma} x\,dy + \int_C -y\,dx = \int^l_0 x(s)y_s(s)\,ds - \int^l_0 \overline{y}(s)x_s(s)\,ds = $$
$$ = \int^l_0 ( x(s)y_s(s) - \overline{y}(s)x_s(s)) \,ds \le \int^l_0 \sqrt{ (x(s)y_s(s) - \overline{y}(s)x_s(s))^2} \,ds \stackrel{*}{\le}$$
$$ \stackrel{*}{\le} \int^l_0 \sqrt{ (x^2(s) + \overline{y}^2(s))} \,ds = lr$$
Where the starred inequality follows from the fact that:
$$(x y_s - \overline{y} x_s)^2 = [(x, - \overline{y}) \cdot (y_s, x_s)]^2 \le (x^2 + \overline{y}^2) \cdot (y^2_s + x^2_s) = x^2 + \overline{y}^2 $$
This is my question: How do you justify this inequality?
$$(x y_s - \overline{y} x_s)^2 \le (x^2 + \overline{y}^2) \cdot (y^2_s + x^2_s)$$
If you expand the left and right hand side you get
$$ x^2 y_s^2 -2xx_s \overline{y}y_s + \overline{y}^2 x_s^2 \le x^2y_s^2 +x^2 x_s^2 + \overline{y}^2 y_s^2 +\overline{y}^2 x_s^2 $$
So why is
$$ -2xx_s \overline{y}y_s \le x^2 x_s^2 + \overline{y}^2 y_s^2 $$
Thanks in advance!
AI: Hint. Expand $(xx_s + \bar yy_s)^2$.
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H: Prove $f\colon X/{\sim} \to Y \text{ is continuous} \iff \pi\circ f\colon X \to Y \text{ is continuous}$
I need to show that $$f\colon X/{\sim} \to Y \text{ is continuous} \iff \pi\circ f\colon X \to Y \text{ is continuous}$$
where $X/{\sim}$ is a quotient topology and $\pi$ is the quotient map.
I understand the proof for $$f\colon X/{\sim} \to Y \text{ is continuous} \iff f\circ \pi\colon X \to Y \text{ is continuous}$$ which is quite simple. But I can't see how to break this one down.
AI: A general hint: by definition of the quotient topology, the projection $\pi$ is continuous. Question: is the composition of continuous functions a continuous function?
In your case I do not agree with notation, as $\pi$ is a quotient map, i.e.
$\pi: X\rightarrow X\setminus\sim$ or $\pi: Y\rightarrow Y\setminus\sim$ . In either the first or second case, I can not compose $\pi$ with $f: X\setminus\sim\rightarrow Y$ and arrive at the map $\pi\circ f: X\rightarrow Y$.
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H: Application of Cauchy theorem to prove normality of a subgroup
Let $G$ be a group $o(G)=pq$, where $p,q$ are both distinct prime numbers. Let $H<G$ be a subgroup of $G$ and $o(H)=p$. I want to show that $H$ is normal in $G$. My argument goes as follows. First we observe that based on Cauchy theorem for groups there exist $a,b \in G$ such that $o(a)=p$ and $o(b)=q$. Now we know that $o(ab)=LCM(o(a), o(b))=pq$, so $G$ is cyclic and $ab$ generates $G$. If $G$ is cyclic then $G$ is abelian and as every subgroup of an abelian group is normal it follows that $H$ is normal in $G$. Can someone verify if this argument is valid?
AI: Your argument can't possibly be right since what you try to prove is, in general, false. For example, take
$$G=S_3\;,\;\;|G|=2\cdot 3\;,\;\;H=\{\;(1)\,,\,(12)\;\}\le G\;\;\text{ but }\;\;H\rlap{\;\;/}\lhd G$$
What is true is that if $\,|G|=pq\;,\;p>q\,$ primes and $\;p\neq 1\pmod q\;$ then any Sylow $\,q$-subgroup of $\,G\,$ is normal.
And your proof begins with problem at $\,ord(ab)=LCM(ord(a)\,,\,ord(b))\,$ , since this isn't true in general (it though is true if $\,ab=ba\,$ and $\;gcd(ord(a)\,,\,ord(b))=1\;$ ), and thus you can not deduce $\,G\,$ is cyclic, which again is false in the general case.
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H: How do I show that an angle is a certain value in a triangle with two sides given?
In the following example I am told that angle x is 60° and that I have to prove it is (without a calculator). What is the simplest way of showing that it is true?
AI: $12-5\sqrt{3}=\sqrt{3}*(4\sqrt{3}-5)$
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H: Joining finite sequences
How do I describe the joining of two finite sequences in mathematical notation? For example, suppose the following:
$$
A=(a_i)_{i=1,2}=(4,2)\\
B=(b_i)_{i=1,2}=(9,5)\\
C=(c_i)_{i=1,...,4}=(4,2,9,5)
$$
Sequence $C$ can be considered sequence $A$ with sequence $B$ attached to the end. How do I describe sequence $C$ in terms of sequence $A$ and $B$?
AI: Those are sets, not sequences. If you ment sets, then it is $C = A \cup B$, the union of the two sets.
If you ment finite sequences you could write \begin{equation} c_i = \begin{cases} a_i & i=1,2 \\ b_{i-2} & i=3,4 \end{cases} \end{equation}
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H: How to write two for loops in math notation?
I have a vector of numbers that looks like this:
[1, 2, 3, 4, 5]
For every number the vector, I would like to multiply each number by every other number and find the sum:
1*1 + 1*2 + 1*3 + 1*4 + 1*5 + 2*1 + 2*2 + 2*3 + 2*4 + 2*5 + ... + 5*5
How can I write this in math notation?
AI: $\sum_{i=1}^5 \sum_{j=1}^5 ij$.
However, you can also give the array $v$ values $v_i$ other than the index $i$. Then you would write:
$\sum_{i=1}^5 \sum_{j=1}^5 v_iv_j$.
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H: Finding all the integer solutions for :$y^2=x^6+17$
Assume that $x,y$ are integers .How to find the solutions for:
$$y^2=x^6+17$$
AI: $y^2-x^6=17\Rightarrow (y-x^3)(y+x^3)=17$.And we know $17$ is a prime.
So the only possibility is
$y+x^3=17$ and $y-x^3=1$ .
or $y+x^3=-17$ and $y-x^3=-1$
or $y+x^3=1$ and $y-x^3=17$
(As $x,y$ are integers so $(y-x^3)(y+x^3)=17$ implies that the only factors of $17 $ are $y+x^3$ and $y-x^3$ )
$\Rightarrow y=9,x=2$
or $y=9,x=-2$
or $x=-9,y=2$
or $x=-9,y=-2$
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H: Contractive mapping on compact space
A contractive mapping on $M$ is a function $f$ from the metric space $(M,d)$ into itself satisfying $$d(f(x),f(y))<d(x,y)$$ whenever $x,y\in M$ with $x\ne y$. Prove that if $f$ is a contractive mapping on a compact metric space $M$, there exists a unique point $x\in M$ with $f(x)=x$.
Consider the function $g(x)=d(f(x),x)$. Note that
\begin{align*}d(f(x),x)-d(f(y),y)&\leq (d(x,y)+d(y,f(x))-(d(y,f(x)-d(f(x),f(y)) \\
&=d(x,y)+d(f(x),f(y)) \\
&<2d(x,y)
\end{align*}
And similarly for $d(f(y),y)-d(f(x),x)$, so we have $|d(f(x),x)-d(f(y),y)|<2d(x,y)$. This means $g(x)$ is continuous.
So $g(x)$ is a real-valued continuous function on a compact space, so must be bounded and has a minimum $c$, i.e. there exists $x_0$ such that $d(f(x_0),x_0)$ equals the minimum $c$. What then?
AI: If $f(x_0) \not= x_0$, then $c > 0$ and $d(f(f(x_0)),f(x_0)) < d(f(x_0),x_0) = c$, which is a contradiction. Thus $f(x_0) = x_0$. Uniqueness is pretty easy to show.
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H: Principal ideal and free module
Let $R$ be a commutative ring and $I$ be an ideal of $R$.
Is it true that $I$ is a principal ideal if and only if $I$ is a free $R$-module?
AI: Definitely not. Any proper principal ideal in a finite commutative ring is a counterexample.
On the other hand, a commutative ring is a principal ideal domain if and only if all of its nonzero ideals are free modules with unique rank. This is a result on "free ideal rings" (FIRs) studied by P.M. Cohn.
As you mentioned, it is easy to show that every principal ideal of a domain is isomorphic to $R$. (One way is to notice that $xR\cong R/ann(x)$, and $ann(x)=0$ if $x$ is nonzero.)
Now suppose $J\neq 0$ is a free principal ideal of a commutative domain. Then $J\cong \oplus_{i\in I} R$ for some copies of $R$. In particular, $J$ this says (through the isomorphism) that $J$ has submodules corresponding to the copies of $R$, and so they are also ideals of $R$.
Suppose for a moment more than one copy of $R$ is used. Since the sum is direct, these copies have pairwise intersection zero. However, nonzero ideals of a domain always have nonzero intersection! To avoid this contradiction the sum can only have one term, hence $J\cong R$. Being isomorphic to a cyclic module, $J$ is itself cyclic (so it is a principal ideal).
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H: Equations on generic algebraic stuctures.
Is there a general theory that studies the solution of equations on structures with a binary operation like, for example, Magmas, Quasigroups, Semigroups, Monoids, Loops and Groups from the most general point of view?
If there isn't a single theory, then is there a specific "theory of equations" for every structure as above?
AI: If there's no known axioms for a magma, and you don't have an explicit representation of the magma, there's literally no way to solve for an equation. Using no axioms means not being able to restrict the problem to a class of algebras, either, unless you're studying a process that generates algebras with interesting solution properties that don't reduce to axioms on the class. The question is not whether it's possible to research a thing, but what reason there is to suspect there's any metastructure to solutions of equations that doesn't depend on the algebras themselves having structure. What are the semantics of the questions, the motivation?
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H: double comb space is not contractible
I'm trying to show that the double comb space is not contractible.
Intuitively I can see why this is true, but I can't seem to formalize a prof.
I try to do the following:
Let $D$ be the double comb space
Suppose $H:D\times I \rightarrow D$ so that $H(x,0)=x$ and $H(x,1)=x_0$ where $x_0\in D$
$D$ is path-connected so we can assume $x_0=(0,0)$.
Now we need to somehow show that H is not continuous, I'm just not sure how.
I have a feeling that I'm not understanding some basic idea here, and that this prof should be quite simple, but I'm not sure what it might be.
Thanks.
AI: Have you tried using uniform continuity? $D \times I$ is a compact metric space so the function $H$, if it is continuous, must also be uniformly continuous. This could have implications for how $H$ maps two sets $x \times I$ and $y \times I$ when $x,y \in D$ are very close to $x_0$ but $x$ is to the left and $y$ is to the right of $x_0$.
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H: Let $f \colon \Bbb C \to \Bbb C$ be a complex valued function given by $f(z)=u(x,y)+iv(x,y).$
I am stuck on the following question :
MY ATTEMPT:
By Cauchy Riemann equation ,we have $u_x=v_y,u_y=-v_x.$ Now $v(x,y)=3xy^2 \implies v_x=3y^2 \implies -u_y=3y^2 \implies u=-y^3+ \phi(x) $. Now,I am not sure which way to go? Can someone give some explanation about which way to go in order to pick the correct option?
AI: If this function is holomorphic, than $u(x,y)$ and $v(x,y)$ must be harmonically adjoint. That means, that $v(x,y)$ must satisfy
$$\Delta v(x,y)=0; $$
We found the values of $\frac{\partial^2 v}{\partial x^2}$, $\frac{\partial^2 v}{\partial y^2}$
$$\frac{\partial^2 v}{\partial x^2}=0; \frac{\partial^2 v}{\partial y^2}=6x $$
So, the next equality must be true for all $(x,y)\in\mathbb{C}$
$$6x=0 $$
But this is true only for $x=0.$ So, this function can not be holomorphic for any $u(x,y)$
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H: The Mystery of the Integrating Factor
The other day I was taught that to solve equations of the form
$$
f'+pf=q,
$$
where $f=f(t)$, $p=p(t)$ and $q=q(t)$, I need to use a function, or integrating factor, $\mu=\mu(t)$ such that
$$
\mu'=\mu p.
$$
However, how do I know that such function $\mu$ exists at all? In other words, can I always find $\mu$?
AI: Subject to some assumptions, you can figure out what $\mu$ must look like. Consider that you have
$$\frac{d\mu}{d x} = \mu \cdot p(x)$$
Separating variables and integrating:
$$\int \frac{d\mu}{\mu} = \int p(x) dx$$
which implies that
$$\log \mu = \int p(x) dx$$
or
$$\mu(t) = \exp \left( \int_{t_0}^t p(x) dx \right)$$
So as long as $p(x)$ has a first integral, the function $\mu(x)$ exists.
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H: continuity and differentiability of two variables.
Consider the map $f:\mathbb{R}^2 \to \mathbb{R}^2$ defined by
$$f(x,y) = (3x-2y+y^2 , 4x+5y+y^2)$$
then which of the following is true?
$f$ is discontinuous at (0,0)
$f$ is continuous at (0,0) and all directional derivatives exists at (0,0)
$f$ is differentiable at (0,0) but the derivative $Df(0,0)$ is not invertible
$f$ is differentiable at (0,0) and the derivative $Df(0,0)$ is invertible
AI: All coordinates are polynomial functions in $x,y$ .So f is continuous.
$D_vf(0)=f(tv)-f(0)=((3tv_1-2tv_2+t^2v_2^2),(4tv_1+5tv_2+t^2v_2^2))$
So $\displaystyle \frac{f(tv)-f(0)}{t}=\frac{((3tv_1-2tv_2+t^2v_2^2),(4tv_1+5tv_2+t^2v_2^2))}{t}=((3v_1-2v_2+tv_2^2),(4v_1+5v_2+tv_2^2))$
$\Rightarrow \displaystyle \lim_{t\to 0}\frac{f(tv)-f(0)}{t}=\lim_{t\to 0}((3v_1-2v_2+tv_2^2),(4v_1+5v_2+tv_2^2))=(3v_1-2v_2,4v_1+5v_2)$
So the directional derivative exists.
Now the derivative will be
$\begin{pmatrix}
3&-2\\
4&5
\end {pmatrix}$ (First column is made up of $D_{(1,0)}f(0)$ and the 2nd column is made up of $D_{(0,1)}f(0)$)which is invertible.
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H: Derivative of a quotient
I am trying to find
$$y = \frac{x^2 + 4x + 3}{ \sqrt{x}}$$
I am reviewing this so I am suppose to do it without the quotient rule, just using what I know about 16 years of algebra and the power rule, difference rule and such. I cannot get it, or anything close to the answer.
I attempted to find the answer using the quotient rule and it was still wrong, I am not sure why I am following the rules and this does not work. There is probably some trick or some exception at play here that I am not familiar with.
AI: Hint:$$y = \frac{x^2 + 4x + 3}{ \sqrt{x}}=x^{3/2}+4x^{1/2}+3x^{-1/2}$$
and we know for all $r\in R$: $$\frac{d}{dx}x^r=rx^{r-1}$$
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H: Is there a term for "is mapped to by an isomorphism"?
In any context where isomorphisms are defined.
For example, if $G$ and $H$ are two isomorphic groups, then there exists an isomorphism mapping their identity elements together. That is to say, their identity elements are _____, where _____ is the word I'm looking for.
Example two: a graph is vertex transitive if and only if its vertices are pairwise _____.
AI: In general, if $G$, $H$ are groups, there does not exist an isomorphism between them.
However, maybe you are looking for the word characteristic.
A characteristic thingie is such that it is unchanged under automorphisms.
For example
the identity element of a group,
the center of a group,
the set of elements of even order
are characteristic (elements, subgroups, subsets) of $G$.
On the other hand, e.g. $p$-Sylow groups are in general not characteristic (because there can be several, i.e. there is not "the" Sylow group).
A graph is vertex transitive if and only if its vertices are pairwise in the orbit of one another under the operation of the automorphism group(!?)
In a way, looking at the operation of the automorphism group on (elements, subgroups, ...) of $G$, characteristic (elements, subgroups, ...) are those having one-point orbits.
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H: prove that $ \lim_{x \to 0} \frac{e^{1/x}}{x}$ does not exist.
By Substitution of y = $\frac {1}{x}$ i have managed to show that
$\lim_{x \to 0^+}\frac{e^{1/x}}{x} = \infty $
but i can't find a way to show that
$\lim_{x \to 0^-}\frac{e^{1/x}}{x} = 0 $
I've tried L'Hospital rule but ended with
$\lim_{x \to 0^-}-\frac{e^{1/x}}{x^2}$ = -"$\frac {0}{0}$" again.
I found by deriving the function that in a small left environment of 0 the function is monotonously increasing.
AI: HINT:
$$\lim_{x \to 0^-}\frac{e^{1/x}}{x} = \lim_{x \to 0^+}-\frac{e^{-1/x}}{x} = \lim_{y \to +\infty}-\frac{e^{-y}}{1/y} = \lim_{y \to +\infty}-\frac{y}{e^{y}}.$$
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H: Characterization of Subset Sum via Linear Programming
I have a sample subset sum problem.
Given numbers $x_1, x_2... x_N$ and a target value to sum to $x_S$
Minimize $x_S - x_1y_1 - x_2y_2 - x_3y_3 ... x_Ny_N$
such that
0 <= $y_1$ <= 1
0 <= $y_2$ <= 1 ...
0 <= $y_N$ <= 1
$x_S - x_1y_1 - x_2y_2 - x_3y_3.... - x_ny_n >= 0 $
$x_S - x_1y_1 - x_2y_2 - x_3y_3.... - x_ny_n <= 0 $
The set of vertices of the polyhedron will have coordinates equal to the $x_i$ meaning that usual simplex should hit the required values... By adding the third and fourth inequalities the only things that remain are those edges which satisfy the required constraints (only solution edges/vertices are now present)
Now if there is a way to find a trivial solution (some combination of edges) if there exists an algorithm that lets you traverse the figure and find its vertices... those correspond to solution of the problem.
Is this a good approach? I can always cut out the last (4th) inequality and then use simplex to handle it.
AI: Not sure this is helpful if no exact solution exists. In your case, the solver will report the problem is infeasible. Also, you would be much better off specifying an equality constraint instead of two inequality constraints, but the solver should be robust to handle these in preprocessing automatically.
I would instead try to maximize $\sum x_i y_i$ subject to $\sum x_i y_i \leq x_S$ and $y_i \in [0,1]$. (If you want the integer version, then $y_i \in \{0,1\}$.
Here in case the equality cannot be achieved, you will at least report the closest available solution.
It also sounds like this is a flavor of the classic Knapsack problem (my formulation makes it apparent). The usefullness of having a standard problem is that there are available heuristics and approximation algorithms...
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H: Finding the Exact Value
How would one go about finding the exact value of $\theta$ in the following:$\sqrt{3}\tan \theta -1 =0$? I am unsure of how to begin this question. Any help would be appreciated!
AI: Hint: $\sqrt{3}\tan \theta -1 =0\implies \sqrt{3}\tan \theta =1\implies \tan \theta=1/\sqrt{3}$
Recall from SOH CAH TOA, that $\tan \theta=\text{opposite}/\text{adjacent}$, so in this case $1$ is the opposite side and $\sqrt{3}$ is the adjacent side.
The above is a special triangle. Since you know that $\tan \theta=1/\sqrt{3}$, it follows that the angle $\theta=\pi/6$. However since $\tan \theta$ is a trigonometric periodic function, that value will come up again, and again. (Just like $\sin0=\sin2\pi).$
$\tan \theta$ has a period of $\pi.$
So to be fully accurate, $\theta=\pi/6 +n\pi, n \in \mathbb{Z}$
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H: Need help with integration by parts
I absolutely despise integration by parts, because it never seems to work for me. Here's an example:
$$ \int 4x \sin(2x) \, \mathrm{d}x $$
What I did:
$$ \int 4x \sin(2x) \, \mathrm{d}x = -2x \cos(2x) - \int - 4\cos(2x) \, \mathrm{d}x$$
Here I'm already stuck. I know about the ILATE/LIATE/whatever but it didn't work out for me either. What should I do at this juncture?
AI: You've done the hard part of the integration by parts right; you're just off by a constant multiple. If we let $u = 4x,\; dv=\sin(2x)dx \implies du = 4\,dx,\;v=\frac{-\cos(2x)}{2}$.
When we put this together:
$$\begin{align}uv - \int v\,du &= -2x\cos(2x) + \int\left(\frac{\cos(2x)}{2}\right)4\,dx\\
&=-2x\cos(2x) + \int2\cos(2x)\,dx
\end{align}$$
So, that's basically where you're at (except for the $2$ instead of the $4$). At this point, we can use $u$ substitution.
Hint:
Let $u = 2x$.
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H: On irreducible polynom $\frac{X^p-1}{X-1}$
Let $m(X) := \frac{X^p-1}{X-1}$ and $n(X) := m(X^p)$. I have shown that $m$ is irr. over $\mathbb Q$. Now I want to show that this is true for $n(X)$, too. I know that
$$
n(X+1)((X+1)^p-1)= (X+1)^{p^2}-1
$$
Working modulo $p$ I get that
$$
n(X+1)X^p \equiv X^{p^2} \mod p
$$ which means that $p$ divives each coefficient of $n(X+1)$ and not the leading coefficient. But I do not know how to show that $p^2 \nmid a_0$ where $a_0$ is the constant term of $n(X+1)$.
Pleas help :)
AI: Turning my comment into an answer: write $m(X) = \sum_{i=0}^{p-1}X^i = 1+X+X^2+\ldots+X^{p-1}$. Then $n(X+1) = m\left((X+1)^p\right) = 1+(X+1)^p+(X+1)^{2p}+\ldots+(X+1)^{p(p-1)}$. Now just use the binomial theorem to find the constant term of each of the terms of this sum.
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H: Galois group of $\mathbb{Q}(\sqrt[3]{2},\omega,\sqrt{-1})$ over $\mathbb{Q}$
How to find the Galois group $Gal(E|Q)$, where $E=\mathbb{Q}(\sqrt[3]{2},\omega,\sqrt{-1})$? I know that $\text{Gal}(\mathbb{Q}(\sqrt[3]{2},\omega)|\mathbb{Q})=D_3$, where $D_3$ is the dihedral group, $\omega^3=1$. Does it implies that
$$\text{Gal}(\mathbb{Q}(\sqrt[3]{2},\omega,\sqrt{-1})|\mathbb{Q})=D_3\times \mathbb{Z}_2 \quad \mbox{?}$$
Thanks for your attention!
AI: Yes in this case, but no in general. What you know is $\text{Gal}(\mathbb{Q}(\sqrt[3]{2},\omega,\sqrt{-1})|\mathbb{Q})$ has $D_3$ as a subgroup of index two. If $G$ is a group which has a subgroup $H$ of index two, it is not in general the case that $G$ is isomorphic to $H\times \mathbb{Z}_2$.
However, it is the case that the only group of order 12 having $D_3$ as a subgroup of index two is $D_3\times\mathbb{Z}_2$. Thus the Galois group in this case must be isomorphic to $D_3\times\mathbb{Z}_2$ (or equivalently $D_6$).
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H: Nullity of Kernel, Range of transpose
Define the linear transformation $T$ by $T(x) = Ax$, where
$A=\left(\begin{matrix} \frac{9}{10} & \frac{3}{10}\\ \frac{3}{10} & \frac{1}{10} \end{matrix}\right)$.
Find (a) $\ker(T)$, (b) $\text{nullity}(T)$, (c) $\text{range}(T)$ and (d) $\text{rk}(T)$.
Apparantly, $\ker(T) = \{(t,-3t) : t \in\mathbb{R}\}$ and $\text{range}(T)= \{(3t,t) :t \in\mathbb{R}\}$. Can anyone explain how to get those answers?
How would this change if we use the following matrix instead of $A$?
$B=\left(\begin{matrix}5 & -1\\ 1 &1 \\ 1 & -1\end{matrix}\right)$;
($\text{range}(T) = \{(4s,4t,s-t) :s,t \in\mathbb{R}\}$)
AI: Let's look at $A=\left(\begin{matrix} \frac{9}{10} & \frac{3}{10}\\ \frac{3}{10} & \frac{1}{10} \end{matrix}\right)$. If we envision multiplying a vector $(v_1, v_2)$ by $A$, then we see that we get $(\frac{9}{10} v_1+\frac{3}{10}v_2, \frac{3}{10}v_1+\frac{1}{10}v_2)$. Now, if $(v_1, v_2)$ is in the kernel of $A$ precisely when $A$ multiplied by $(v_1, v_2)$ is the zero vector. So this is true when $\frac{9}{10} v_1+\frac{3}{10}v_2=0$ and so $3v_1+v_2=0$. The vectors that satisfy this are precisely those of the form $\{(t, -3t): t \in \mathbb{R} \}$. Notice also that any vector $v$ of this form will satisfy $Av=(0,0)$ and so this set of vectors is the kernel.
The range of $A$ consists of the vectors $v$ so that there exists some vector $w$ with $Aw=v$. Try using a similar approach to figure out the form of any vector in the range.
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H: Roots of $x^{2}+e^{0.1x}-1$
I saw an exercise that asks to prove that $f(x):=x^{2}+e^{0.1x}-1$
have a root $r<0$.
The solution stated that $f''(x)=2+(0.1)^{2}e^{0.1x}>0$ hence there
is a maximum of two roots, since $0$ is a root and since $$f'(0)=0.1>0$$
there is a root $r<0$.
I know that a convex function have at most $2$ roots, but I don't
understand why $f'(0)>0$ imply that there is a second root and that
it is negative.
I would appreciate it if someone could explain the reasoning
AI: The derivative is positive at $0$ (which is a root). This means the function is increasing there. Further, the second derivative is always positive, so the function will always be increasing for positive $x$. Thus if the second root exists, then it must be negative.
Unfortunately, simply knowing the derivative is positive at $0$ and that the second derivative is always positive does not tell us that there is a second root. But we can quickly verify that there is a second root, by noting that $f(-10) > 0$ trivially and seeing that the function is continuous. Now, since the derivative is positive at zero, $f$ will be negative in some neighborhood of $0$. By continuity, $f$ will be zero between $-10$ and $0$.
And that's the best you can do.
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H: Probability, combinations with repetition
A store sells n different kinds of fruits. A boy buys k fruits. Find the probability that he buys all the kinds of fruits.
Give me a hint, please.
Thank you.
AI: I assume that you mean for the store to have an unlimited number of each different kind of fruit. Are you also assuming that any multiset of fruit is equally likely? That is, the probability of choosing 1 apple, 3 oranges, and 2 bananas is the same as the probability of choosing 6 apples?
This being the case, think of it this way: you must choose at least one of each of the $n$ fruits. So, to count the number of ways to choose all $k$ fruits - such that each fruit is represented at least once - you really need to count in how many ways you can choose the $k-n$ "extra" pieces.
See if you can make that work.
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H: Prove that this graph doesn't exist
A group of 3141 students gather together. Some of them have 13 friends in this group, some have 33 friends, and the rest has 37 friends. Prove using graph theory that this group does not exist. Assume that if A is friends with B, then B is friends with A.
AI: This seems like homework; so, let me just give you a hint.
Say that you have $a$ people with 13 friends, $b$ with 33, and the remaining $3141-a-b$ have 37. In the graph, this corresponds to having $a$ vertices of degree 13, $b$ vertices of degree $33$, and the rest of degree 37.
So, the total degree of the graph is
$$
13a+33b+37(3141-a-b).
$$
Simplify this, and see if you can prove that no matter how you choose $a$ and $b$, the resulting total degree will always be odd. Why is this impossible in a graph?
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H: Some questions about Goldbach's conjecture
I was thinking about the usage of Dirichlet's theorem in proving some facts about the Goldbach's conjecture.
I will start with an example. Using Dirichlet's theorem, we know that there are infinitely many primes in the form of, let's say, $6k+1$. This means there are infinitely many primes in the form of $6k+4-3$. Let's say $p=6k+4-3$ and so $p+3=6k+4$. Now tell me if my reasoning is not true. From this, can we say there are infinitely many numbers in the form of $6k+4$ which the Goldbach's conjecture is true about them? I may assume this is true for the next things im gonna say.
Now, we are going to prove something more general, by using one special case of the Dirichlet's theorem: $p=2ak+(b-q)$ where $gcd(2a,b-q)=1$ and so $2ak+b=p+q$. If we find a number $a$ and a prime number $q$ so that $gcd(2a,2-q)=1$, $gcd(2a,4-q)=1$, ... and $gcd(2a,2a-q)=1$ ,we may say there are infinitely many numbers in the form of $2ak+2$ which the Goldbach's conjecture is true about them, same for the forms of $2ak+4$, $2ak+6$, ... and $2ak+2a$. From this we can say Goldbach's conjecture is true for infinitely many numbers in any even form. Can you help me find such a numbers (that $q$ and $a$ i mean)? What can it lead to?
AI: You are correct that this approach will show that there are an infinite number of even numbers in each of these series that can be expressed as the sum of two primes. $a=2,q=3$ seems to work for what you want.
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H: Abstract Algebra and Parallel Computing
I've recently been learning a bit about parallel computation. Two things I recently learned about are Reduce and Scan.
Where Reduce is defined as 2-Tuple of a Set of elements and a binary operation that is associative.
Scan is defined as a 2-Tuple of a set of elements and a binary operation that is associative and that this pair has an identity.
So a Scan is essentially a Group. Now I haven't looked too much further into this, but I was wondering if any of you might know if this idea that parallel computation works on sets of information with binary (or really any associative function) operations and applied principles of abstract algebra to it?
AI: Reduce qualifies as a semigroup, and Scan is a monoid (a monoid is a semi-group with an identity).
Neither is a group, which is, essentially, a monoid in which the inverse of every element in the "monoid" is also in the "monoid".
Take a peek at the linked entries in Wikipedia to get a feel for the properties shared by semigroups, and those shared by monoids.
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H: Convergeance and Lebesgue Integral exercise
Can you help me formally prove that $f_n:\mathbb R \to \mathbb R$,$f_n=n\chi_{[\frac{1}{n},\frac{2}{n}]}$ $\forall n\in\mathbb N$ converges pointwise to $f\equiv 0$ and that every $f_n$ is Lebesgue integrable?
It is not a homework,I am solving exercises for my tomorrow's measure theory exam.I thank you all in advance !
AI: Fix any $x\in\mathbb{R}$. If $x> 0$, then there exists $N$ such that for all $n\geq N$ $x\notin[\frac{1}{n}, \frac{2}{n}]$ (because $\frac{2}{n} < x$) and thus $f_n(x)=0$. . And if $x \leq 0$, $f_n(x)=0$ for all $n$ anyway, as the support of $f_n$ is in $(0,\infty)$.
So for all $x\in\mathbb{R}$, $f_n(x) \xrightarrow[n\to\infty]{}0$; hence $(f_n)_n$ converges pointwise to $\bar{0}$ on $\mathbb{R}$.
Further, for any fixed $n$, $f_n$ is a simple function, and is thus Lebesgue integrable because $\mathbf{1}_{[\frac{1}{n}, \frac{2}{n}]}$ is measurable and has finite measure.
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H: At least two participants in a meeting received the same number of phone numbers
The question is: there are $n$ participants in a meeting ($n \geq 2$). During this meeting, people exchanged phone numbers with each other. Prove that at least $2$ participants received the same number of phone numbers.
I was thinking of solving it using contradiction. So somehow prove that at least $2$ participants doesn't receive the same number of phone numbers, or no one received the same number of phone numbers. Would that work?
AI: How many numbers can a person get? Anywhere from $0$ to $n-1$ numbers.
If they all get a different number of numbers, then somebody must get $0$ numbers and somebody must get $n-1$ numbers. (Why?) Is that possible?
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H: Area between a curve and a horizontal line
I had 2 make 2 exercises which I found very similar but they have a crucial difference which puzzles me:
1 Calculate the area of the plane $V$, between the graph of the function $f(x) = \dfrac{8x}{x^2+4}$ and the line $y= 1\dfrac{3}{5}$.
2 Calculate the area of the plane $V$, between the graph of the function $f(x) = \dfrac{1}{x^2-4x+5}$, the y-axis and the line $y=\dfrac{1}{2}$.
1 $\displaystyle\int_0^4 (f(x)-1\dfrac{3}{5}) \, \mathrm{dx}$
2 $\displaystyle\int_0^1 (\dfrac{1}{2} - f(x)) \, \mathrm{dx}$
What is the difference between these cases which causes the different calculation? I am puzzled.
AI: Your standard formula for calculating areas between curves has the form:
$$
\int_a^b (f(x)-g(x)) dx
$$
where $f(x)$ is the upper curve and $g(x)$ is the lower curve. This makes the integral positive, since we want area. So, draw a picture. For the first question, we see that for $x\in [1,4]$, we have $\dfrac{8x}{x^2+4} \ge 1\dfrac{3}{5}$. (For example, pick a sample point and note that $f(2)=\dfrac{8(2)}{2^2+4}=2 \ge 8/5$). Thus, $f(x)$ is the upper curve, which yields the integral:
$$
\int_1^4 \left(\dfrac{8x}{x^2+4}-1\dfrac{3}{5}\right) dx
$$
On the other hand, for the second question, we see that for $x\in [0,1]$, we have $\dfrac{1}{x^2-4x+5} \le \dfrac{1}{2}$. (For example, pick a sample point and note that $f(0.5)=\dfrac{1}{0.5^2-4(0.5)+5}=4/13 \le 1/2$). Thus, $f(x)$ is the lower curve, which yields the integral:
$$
\int_0^1 \left(\dfrac{1}{2} - \dfrac{1}{x^2-4x+5}\right) dx
$$
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H: Do negative dimensions make sense?
Some time ago I read in a popular physics book that in M-theory, there are some "things" which can be said to have dimension $-1$.
Probably, the author was vastly exaggerating, but this left me wondering:
Are there mathematical theories which contain a notion that can be regarded as some sort of generalization of the classical notion of dimensionality and which allows negative values?
For example the Hausdorff dimension can assume, as far as I know, only nonnegative real values. I think that stable homotopy groups can be defined for arbitrary integer dimensions, but this doesn't really count, since one is not dealing with negative dimensional objects per se.
AI: One way to define the dimension of a finite-dimensional vector space naturally extends to the definition of the Euler characteristic of a bounded chain complex of finite-dimensional vector spaces, and these can be negative. Somewhat relatedly, there is a notion of super vector space which also have a notion of dimension which can be negative. These explain, in some sense, the formula
$$\left( {n \choose d} \right) = (-1)^d {-n \choose d}$$
if you think of ${-n \choose d}$ as the dimension of the exterior power $\Lambda^d(V)$ where $\dim V = -n$. See this blog post for details.
Thinking in terms of negative dimensions also suggests some interesting dualities between Lie groups; for example, I think an inner product on a negative-dimensional vector space is a symplectic form, so in some sense the orthogonal groups of negative-dimensional vector spaces are the symplectic groups, or something like that. See, for example, this paper.
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H: Constrained ( Variable Length ) Permutation Calculation.
I am writing some tracking software, but I think this is pretty purely a math question. I don't need to know the math to accomplish this with my code, but I like math and I want to learn! Thus please keep in mind, I fully expect to have to look up definitions of math terms in your answers :-)
I am trying to determine how many possible workflows will exist in my new system. There are 6 "starting" posibilites, 0 or 1 "middle" possibilities, and 5 "ending" possibilities. So, some workflows are 2 items long, some are 3 items long. None of these possibilities are the same. The 6 starting, can only be options for the starting item. The middle only has 1 possibility, and it is either present or not. The 5 end possibilites can only be options for the ending position, wether it's the 2nd or 3rd item in the particular instance.
4 instance examples:
1starting - 1ending
1starting - 1middle - 1ending
1starting - 2ending
1starting - 1middle - 2ending
etc...
How many workflows are possible? How would this be calculated / what would the formula look like? Thanks!
AI: Let $S_1,S_2,...,S_6$ be the $6$ possible "starting" possibilities, let $M_0,M_1$ be the $2$ possible "middle" possibilities, and let $E_1,E_2,...,E_5$ be the $5$ possible "ending" possibilities. Note that there are $6$ ways to choose the first option, $2$ was to choose the second option, and $5$ ways to choose the last option. Thus, by the Fundamental Counting Principle, the total number of possibilities is:
$$
6 \cdot 2 \cdot 5 = 60
$$
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H: Sum of positive i.i.d. random variables.
Let $X_1, X_2, \ldots, $ be i.i.d. random variables with $X_1 > 0$. Let $S_n = \sum_{m=1}^n X_m$. Can we conclude $[\sup_{n \ge 1} S_n = \infty$] almost surely?
Assuming the statement is true, by Kolmogorov's 0-1 law, $[\sup_{n \ge 1} S_n = \infty]$ has probability either $0$ or $1$, so it is enough to show that this event has positive probability. But how can we prove this without knowing the distribution of $X_1$?
AI: If $X_1\gt0$ almost surely, then there exists $x\gt0$ such that $P(X_1\geqslant x)\gt0$. Thus, $S_n\geqslant xN_n$ where $N_n=\#\{k\mid1\leqslant k\leqslant n, X_k\geqslant x\}$. The series $\sum\limits_nP(X_n\geqslant x)=\sum\limits_nP(X_1\geqslant x)$ diverges hence, by Borel-Cantelli, $X_k\geqslant x$ for infinitely many $k$ almost surely, that is, $N_n\to\infty$ almost surely. In other words, $P(S_n\to\infty)=1$.
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H: Calculate Exact Value of $\sin\theta, \cos\theta$ and $\tan\theta$
Having trouble getting a start on this problem, any help would be appreciated!
Given point $P = (-3,5)$ is on the terminal arm of angle $\theta$ in standard position. Calculate the exact value of $\sin\theta, \cos\theta$, and $\tan\theta$.
AI: Hints:
What quadrant is the point $\;P = ({\bf x, y}) = (-3, 5)$ located?
Draw the right triangle that point $P$ makes with the $x$ axis - the length of the hypotenuse of the right triangle will equal $\;{\bf h} = \sqrt{(-3)^2 + 5^2} = \sqrt{34}$
Use SOH CAH TOA to unpack the definitions of $\tan \theta, \;\sin\theta,\;\cos\theta$:
$$\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac yx = \quad?\;$$
$$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac yh = \quad?\;$$
$$\cos \theta= \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{h} = \quad?$$
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H: Real integral $ \int_{-\infty}^{\infty} \frac{dx}{1+x^2} $ with the help of complex friends
I have to solve the integral $$ \int_{-\infty}^{\infty} \frac{dx}{1+x^2} $$ by doing this:
Given a rectangle that is defined by the points $ r+i, -r+i,-r-i,r-i$, $r>0$ and $\gamma_r$ is a closed positively oriented curve around the boundary of this rectangle. Then I should proof that $$ \lim_{r \rightarrow \infty} \int_{\gamma_r} \frac{1}{z} dz=2\pi i.$$
And by using this, I am supposed to evaluate $$ \int_{-\infty}^{\infty} \frac{dx}{1+x^2}. $$
There are two things that look difficult to me, the first one is: Why is there a limit for the first integral? Cause by using Cauchy's integral theorem this should be the same as the integral around the circle $$C: \gamma(t)=|r+i|e^{it}$$ and therefore
$$ \int_0^{2\pi} \frac{1}{|r+i|e^{it}}^{i|r+i|e^{it}} dt=2 \pi i.$$
So there is no limit necessary, which makes me thinking that this way should be wrong.
The second thing I do not understand is, how both integrals are related to each other?
AI: Let us write down the integrals over the four sides of the rectangle:
\begin{align}
&I_1=\int_{r-i}^{r+i}\frac{dz}{z}=\int_{-1}^{1}\frac{idt}{r+it},\\
&I_2=\int_{r+i}^{-r+i}\frac{dz}{z}=-\int_{-r}^{r}\frac{dt}{t+i},\\
&I_3=\int_{-r+i}^{-r-i}\frac{dz}{z}=-\int_{-1}^{1}\frac{idt}{-r+it},\\
&I_4=\int_{-r-i}^{r-i}\frac{dz}{z}=\int_{-r}^{r}\frac{dt}{t-i}.
\end{align}
Next consider the sum
$$I_2+I_4=\int_{-r}^{r}\left(\frac{1}{t-i}-\frac{1}{t+i}\right)dt=2i\int_{-r}^r\frac{dt}{1+t^2}.$$
The limit of this sum as $r\rightarrow\infty$ is proportional to the integral we want to find. On the other hand, the limit of $I_1$ and $I_3$ is zero (they are both $O(1/r)$).
Hence we can write
$$\lim_{r\rightarrow\infty}(I_1+I_2+I_3+I_4)=2i\int_{-\infty}^{\infty}\frac{dt}{t^2+1}=2\pi i,$$
which gives the answer.
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H: Defining the winding number for a general curve
In the Complex Analysis text by Ahlfors, he says that we can define the winding number $n(\gamma,a)$ for any continuous, closed curve $\gamma$ which doesn't pass through the point $a$ (differentiability is not required).
Divide $\gamma$ into subarcs $\gamma_1,\dots,\gamma_n$, such that each subarc lies in disk which doesn't include $a$.
Form the line segments $\sigma_1,\dots,\sigma_n$ between the endpoints (with direction as the curve's).
Lastly, define $n(\gamma,a)$ to be $n(\sigma,a)$ where $\sigma=\sigma_1+\dots+\sigma_n$.
I want to show that this number is independent of the choice of the subarcs. That is if $\delta_1,\dots,\delta_m$ is another admissible partition, with line segments $\tau_1,\dots,\tau_m$ than $n(\sigma,a)=n(\tau,a)$.
I've been thinking about this for a couple of days now, and any help will be appreciated.
AI: Hint: If you have two partitions $(\sigma_k)$ and $(\tau_l)$, you can always find a common refined partition $(\rho_m)$, that is a partition such that each $\sigma_k$ and each $\tau_l$ is a $\rho_m$. Now if you prove that $n(\sigma,a)=n(\rho,a)$ and $n(\tau,a)=n(\rho,a)$, you're done; in other words, you only need to consider the case where one of the two partitions is a refinement of the other, and I believe that case should be much easier.
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H: Vector spaces and subspaces.
a) Find an abelian group V and a field F for which V is a vctor space over F in at least two different ways, that is, there are two different definitions of scalar multiplication making V a vector spae over F.
We can choose both V and F to be $\Bbb{Q}$ and define the two different scalar multiplications by $aq := aq$ and $aq := a^{-1}q$ for all $a,q \in \Bbb{Q}$. Is this correct?
b) Find a vector space V over F and a subset S of V that is (1) a subspace of V and (2) a vector space using operations that differ from those of V.
I find this a bit surprising, because the textbook (Advanced Linear Algebra by Roman) says "A subspace of a vector space V is a subset S of V that is a vector space in its own right under the operations obtained by restricting the operations of V to S...". However, if S has a different operation that V does, then it won't fit this definition, right?
Thank you in advance
AI: Your answer to (a) doesn't work because $a^{-1}$ doesn't exist when $a=0$. Even apart from this difficulty, your proposed multiplication won't satisfy the distributive law (with respect to scalar addition). There won't be any example over the field of rational numbers; I suggest trying the field of complex numbers.
For (b), I think the intention is that $S$ should be a subspace of $V$ using (as the definition requires) the restrictions to $S$ of the operations of $V$, and should also be a vector space with some other operations. In other words, it's similar to question (a) in that the same set is a vector space for two different choices of operations.
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H: How to integrate an exponential raise to the inverse sine?
Find the $\space \displaystyle\int e^{\sin^{-1}x}~\mathrm dx$ .
I started by making a substitution. Let $u=\sin^{-1}x$, and so one can conclued that:
$\begin{align}1)&\mathrm du=\displaystyle\frac{1}{\sqrt{1-x^2}}\mathrm dx\\2)&x=\sin u
\end{align}$
So, the integral stays:
$\begin{align}\int e^{\sin^{-1}x}\frac{\sqrt{1-x^2}}{\sqrt{1-x^2}} \mathrm dx &=\int e^u\sqrt{1-\sin^2u}~\mathrm du=\int e^u\sqrt{\cos^2u}~\mathrm du=\int e^u\cos u~\mathrm du\end{align}$
Now, I tryed integration by parts but I could't managed. In Wolfram there is a complicated formula that I never heard about. Is there an intuitive way to finish this integral? Thanks.
AI: Two consecutive integrations by parts should give you an equation satisfied by the primitive. More specifically,
$$ \int_{-\infty}^a e^u\cos u\,\mathrm du = e^a\cos a + \int_{-\infty}^a e^u\sin u\,\mathrm du.$$
A second integration by parts gives an equation for $\int_{-\infty}^a e^u\cos u\,\mathrm du$.
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H: Cardinal arithmetic questions
I have problem to solve:
Let $a,b$ and $c$ be cardinal numbers. Prove that $a+b=b$, $b \le c$ implies $a+c=c$.
And trying to prove this I got couple questions:
For infinite cardinal $c$, is $c+c=c$ always? And how to prove it? (Is $c \cdot c=c$?)
From $a+b=b$ can I conclude that $a \le b$ since $a \ge b$ and 1. will imply $a+b=a$? I know that statement that every two cardinals are comparable leads to AC (we can use it when solving problems)
(I apologize if my questions are too simple, I tried to google it, but Google is not math-friendly tonight for me =D)
P.S. Is there solution without using AC?
AI: Since $b\le c$, then there is a $d$ such that $c=b+d$. Then
$$a+c=a+(b+d)=(a+b)+d=b+d=c.$$
Note that this proof does not use the axiom of choice.
If you assume the axiom of choice, then $c+c=c$ for all infinite cardinals $c$. Without choice, it is consistent that this fails in general (see the comments here), though of course it holds in particular cases (for any well-ordered $c$, or for $c=\mathfrak c:=|\mathbb R|$, for example).
That $c\cdot c=c$ for all infinite $c$ is equivalent to choice. To prove that choice gives us this, it suffices to argue that $c\cdot c=c$ whenever $c$ is well-orderable. Typically, one uses some kind of pairing function to achieve this. See for example here. The point is that pairing functions give us explicit bijections between $\kappa\times \kappa$ and $\kappa$ for any infinite well-ordered cardinal $\kappa$. Since $\kappa\le\kappa+\kappa\le \kappa\cdot \kappa$, it follows from this that $\kappa+\kappa=\kappa$ as well.
To see that $c\cdot c=c$ for all infinite $c$ gives us choice, see here.
And yes, if $a+b=b$, then $a\le b$. On the other hand, without assuming choice, just knowing that $a\le b$ does not suffice to conclude that $a+b=b$. For example, if $b$ is infinite and Dedekind finite, then $1<b<b+1$.
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H: Closed subsets of $\mathbb{R}$ characterization
I remember the characterization of open subsets of $\mathbb{R}$ as a countable union of disjoint open intervals. I was thinking about whether this allows us to characterize closed subsets as a countable union of disjoint closed intervals. As we know, closed subsets are just complements of open subsets.
I had an idea of starting with an open interval at $(a,b)$, then looking at the next open interval to the right at $(c,d)$, and concluding that $[b,c]$ belongs to the closed subset. But this idea can break down, for example, if the open subset contains the intervals $(-1,0), (1/2,1), (1/8,1/4), (1/32,1/16), \ldots$. In this case, there is no next open interval to the right of $(-1,0)$. How can we modify this idea to characterize closed subsets?
AI: It’s not true that all closed sets are countable unions of disjoint closed intervals: the middle-thirds Cantor set is a counterexample. It is closed and nowhere dense in $\Bbb R$, so it contains no non-trivial closed interval. That is, the only closed intervals that it contains are degenerate ones of the form $[x,x]=\{x\}$. And since it is uncountable, it isn’t a countable union even of these degenerate closed intervals.
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H: Derivative of $ y = (2x - 3)^4 \cdot (x^2 + x + 1)^5$
$$ y = (2x - 3)^4 \cdot (x^2 + x + 1)^5$$
I know that it should be the chain rule and product rule used together to get the answer
$$ y = \frac{dx}{dy}((2x - 3)^4) \cdot (x^2 + x + 1)^5 + \frac{dx}{dy}(x^2 + x + 1)^5 \cdot (2x - 3)^4 $$
this gives me something ridiculous like this
$$8(2x-3)^3 \cdot (x^2 + x + 1)^5 + (x^2 + x + 1)^4 \cdot (2x+1) (2x-3)^4$$
This is wrong and I keep getting it, I don't know how to simplify it without expanding everything.
The book Houdini's out $(2x -3)^3 (x^2 + x + 1)^4 (28x^2 - 12x - 7)$
AI: You're derivative is close: $$\frac d{dx}\left(x^2 + x + 4)^5\right) = \color{blue}{\bf 5}(x^2 + x + + 4)^4\cdot (2x + 1)$$ This gives us:
$$f'(x) = 8(2x-3)^3 \cdot (x^2 + x + 1)^5 + \color{blue}{\bf 5}(x^2 + x + 1)^4 (2x+1) (2x-3)^4$$
Then we can factor out common factors of each term of the sum:
$$ = (2x - 3)^3(x^2 + x + 1)^4\left(8(x^2 + x + 1) + 5(2x+1)(2x - 3)\right)$$
And then expand the factors where needed, and combine like terms in the right-most factor:
$$ = (2x - 3)^3(x^2 + x + 1)^4\left(8x^2 + 8x + 8 + 5(4x^2 - 4x - 3)\right)$$
$$\bf = (2x - 3)^3(x^2 + x + 1)^4(28x^2 - 12x - 7)$$
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H: How can evaluate $\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}$
I don't know if I apply for this case sin (a-b), or if it is the case of another type of resolution, someone with some idea without using derivation or L'Hôpital's rule? Thank you.
$$\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}$$
AI: Using the identity
$$
\sin(A)-\sin(B)=2\sin\left(\frac{A-B}{2}\right)\cos\left(\frac{A+B}{2}\right)
$$
we get
$$
\begin{align}
\lim_{x\to0}\frac{\sin\left(x^2+\frac1x\right)-\sin\left(\frac1x\right)}{x}
&=\lim_{x\to0}\frac{2\sin\left(\frac{x^2}{2}\right)\cos\left(\frac{x^2}{2}+\frac1x\right)}{x}\\
&=\lim_{x\to0}x\frac{\sin\left(\frac{x^2}{2}\right)\cos\left(\frac{x^2}{2}+\frac1x\right)}{\frac{x^2}{2}}\\
&=\lim_{x\to0}x\cos\left(\frac{x^2}{2}+\frac1x\right)\lim_{x\to0}\frac{\sin\left(\frac{x^2}{2}\right)}{\frac{x^2}{2}}\\[12pt]
&=0\cdot1
\end{align}
$$
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H: I know the basic definition of continuity. But here, the definition is applied for a ball.
I am studying the topology of $\Bbb R^n$ from W. R . Wade's Introduction to analysis book.
I know the basic definition of continuity. But here, the definition is applied for a "ball".
I dont understand the section which i have underlined with a blue pencil.
How do we rewrite the definition for a ball?
Please can you explain this to me in a more clear way?
Thank you.
AI: The open ball $B_\delta(a)$ is defined to be the set of all points whose distance from $a$ is less than $\delta$:
$$B_\delta(a)=\{x:d(x,a)<\delta\}\;,$$
where $d(x,a)$ is the distance from $x$ to $a$. In the real line we normally define distance using the absolute value function: $d(x,a)=|x-a|$. Thus, $x\in B_\delta(a)$ just means that $|x-a|<\delta$. Similarly, $f(x)\in B_\epsilon(f(a))$ just means that $|f(x)-f(a)|<\epsilon$. Here’s the $\epsilon$-$\delta$ definition:
The function $f:E\to\Bbb R$ is continuous as $a\in E$ if and only if given $\epsilon>0$ there is a $\delta>0$ such that $|x-a|<\delta$ and $x\in E$ imply that $|f(x)-f(a)|<\epsilon$.
If we make the substitutions that I indicated, it becomes:
The function $f:E\to\Bbb R$ is continuous as $a\in E$ if and only if given $\epsilon>0$ there is a $\delta>0$ such that $x\in E\cap B_\delta(a)$ implies that $f(x)\in B_\epsilon(f(a))$.
Now recall that $f[A]$ just means $\{f(x):x\in A\}$. In that second version of the definition we’re saying that $f$ is continuous at $a$ if
$$\{f(x):x\in E\cap B_\delta(a)\}\subseteq B_\epsilon(f(a))\;,$$
i.e., if
$$f[E\cap B_\delta(a)]\subseteq B_\epsilon(f(a))\;.$$
Thus, we can further rewrite it thus:
The function $f:E\to\Bbb R$ is continuous as $a\in E$ if and only if given $\epsilon>0$ there is a $\delta>0$ such that $f[E\cap B_\delta(a)]\subseteq B_\epsilon(f(a))$.
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H: Prove that it is possible to divide integral area into two equal parts
Assuming $f$ is locally integrable on interval $<a,b>$, I'd like to show that it is always possible to divide it into two equal parts in terms of enclosed areas.
In other words, I'd like to show there exists $x \in [a,b]$ with $\int_a^x f(t)\ dt = \int_x^b f(t)\ d(t)$.
I got stuck composing a help function $h(x) = \int_a^x f(t)\ dt - \int_x^b f(t)\ d(t)$. I thought it might be useful, but I don't really know how could apply the mean value theorem to show that it is possible to find such an $x$ that $g(x)=0$.
P.S.: I also think it only works on $[a,b]$.
AI: The function $h(x)$ you defined is continuous on $[a,b]$. At $a$, it is $-\int_a^b f(t)\ dt$. At $b$, it is $\int_a^b f(t)\ dt$. So, if the integral over $[a,b]$ is not zero, you're done by the intermediate value theorem.
If the integral is zero over $[a,b]$, the point $x=a$ will work.
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H: What is the equation representing a constant elasticity of 1?
I'm reading the chapter in my textbook about the price elasticity of demand, and it was pointed out that most demand curves do not represent a constant elasticity of demand - even linear curves like $f(x)=x$ is not constant elasticity although it has constant slope. It then points out three curves that have constant elasticity all throughout the curve.
I'm just curious, from a mathematical standpoint, what function represents the rounded curve? It's definitely something akin to an exponential function, but I'm not quite sure how to calculate it. I played around on google and found that $f(x)=e^{-x}$ looks something like it, but it's not quite the right shape. The straight lines are simple and straightforward - they have a slope of 0 and a slope of infinity, respectively. But what about the round curve?
FWIW, the equation I have to calculate elasticity is:
$$\eta=\frac{\text{Percentage change in quantity demanded}}{\text{Percentage change in price}}$$
Would I have to use integration to find this out?
AI: To expand on ronno's comment, recall that you can use the price elasticity of demand to determine whether you should raise or lower your prices (If elasticity of demand is below 0, higher prices yield greater profit and vice versa) until the elasticity is exactly 1, and then you are maximizing revenue. Thus, if the elasticity is 1 throughout the entire curve, then you should be making the same revenue regardless of what point on the curve you "choose" by setting your price. Revenue being price times quantity, it follows that...
pQ = c for some constant c.
I've only actually taken one semester of microeconomics, so I'm not sure why you would write c^2 instead of just c. Perhaps someone else can explain that.
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H: computation of $\int_0^T \frac{\sin (t)}{t} dt$
I need approaches to solve the problem of the following type
$\int_{0}^{T} \frac{\sin (t)}{t} dt $. Is there a closed form solution available.
AI: The integral
$$
\int_0^T\frac{\sin(t)}{t}\,\mathrm{d}t
$$
is called the Sine Integral. It has no closed form in terms of elementary functions.
However, its limit as $T\to\infty$ is
$$
\int_0^\infty\frac{\sin(t)}{t}\,\mathrm{d}t=\frac\pi2
$$
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H: What does an expression $[x^n](1-x)^{-1}(1-x^2)^{-1}(1-x^3)^{-1}(1-x^4)^{-1}...$ mean?
I came across the function that describes number of partitions of $n$ (I mean partitions like $5=4+1=3+2=3+1+1$ and so on.
There was defined a Cartesian product: $$\{0,1,1+1,1+1+1,...\}\times\{0,2,2+2,2+2+2,...\}\times\{0,3,3+3,...\}\times...$$ so that each partition is an element of finite sum (sum over components is finite) of this product. For example $1+2+2=(1,2+2,0,0,...)$.
And then followed something I didn't understand: author introduced polynomials in the following way:
$$\begin{align}1:1+x+x^2+x^3+...&=(1-x)^{-1}\\2:1+x^2+x^4+x^6+...&=(1-x^2)^{-1}\\3:1+x^3+x^6+x^9+...&=(1-x^3)^{-1}\end{align}\\...$$
(powers are corresponding elements of product), then multiplied all the sums and stated, that number of partitions of $n$ is:
$$[x^n](1-x)^{-1}(1-x^2)^{-1}(1-x^3)^{-1}...$$
Later he showed an example, but it was not numeric – $x$ was left in the expression.
So how do I understand this?
Thanks in advance!
AI: Often the notation
$$
[x^n](a_0+a_1x+a_2x^2+a_3x^3+\dots+a_nx^n+\dots)
$$
is used to mean the coefficient of $x^n$ in the given polynomial or power series ($a_n$ in the series I've given).
The formulas presented by the author are not polynomials, but power series. Polynomials are finite versions of power series. That is,
$$
\frac1{1-x}=1+x+x^2+x^3+x^4+\dots
$$
never terminates; there is no highest power of $x$.
For example,
$$
\left[x^6\right]\frac1{1-x}\frac1{1-x^2}\frac1{1-x^3}\frac1{1-x^4}\cdots=11
$$
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H: Galois group of a reducible polynomial over a arbitrary field
How to proceed to determine the Galois group of a reducible polynomial over a field $F$. As an example I tried to compute the Galois group of $f(X)=X^4+4\in\mathbb{Q}[X]$; one can check that $f(X)=(X^2-2X+2)(X^2+2X+2)$ and the polynomials $h(X)=X^2-2X+2$ and $g(X)=X^2+2X+2$ are irreducible (Eisenstein). Computing the the discriminant $d_g=d_h=-4$, then $\Delta_g,\Delta_h\not\in\mathbb{Q}$. Therefore, $G_h=S_2=G_g$. The roots of $h(X)$ are $\{1+i,1-i\}$ and the roots of $g(X)$ are $\{-1+i,-1-i\}$. As all the roots are in the same field, namely $\mathbb{Q}(i)$, it follows that $G_f=S_2$ and thus a cyclic group of order $2$ in $S_4$. Is that right? If not all the roots were not in the same field, the Galois group of a reducible polynomial $f(X)=h_n(X)\dots h_0(X)$ would be the direct sum of the Galois group of each $h_i(X)$?
AI: In this case, you are correct. In general, Galois groups of different polynomials don't always combine as easily as you have stated. Consider $(x^3-2)(x^3-5)$. The roots of the first factor are $ \sqrt[3]{2}, \pm \sqrt[3]{2}z$ (where $z$ is a primitive third root of unity) and those of the second are $ \sqrt[3]{5}, \pm \sqrt[3]{5}z$. Now, each has Galois group $S_3$, but the splitting field of either polynomial contains $z$, so that together, the splitting field only has dimension 18 as opposed to 36 if the Galois group were $S_3 \times S_3$.
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H: What is convection-dominated pde problems?
Can you explain for me what is convection-dominated problems? Definition and examples if possible.
Why don't we can apply standard discretization methods (finite difference, finite element, finite volume methods) for convection-dominated equation?
What is the importance of global Péclet number in this problem?
AI: Suppose we are modeling a quantity $u$, say the concentration of a chemical, driven by a flow in some fluid in a region $\Omega$ with no source (meaning we are not adding more chemical into the fluid after starting the timer). Then the convection-diffusion pde to describe the phenomenon is:
$$
\frac{\partial u}{\partial t} = \nabla \cdot (D \nabla u - \vec{b}u ).\tag{1}
$$
This comes from linear hyperbolic conservation law (in the differential form):
$$
\frac{\partial u}{\partial t} = -\nabla \cdot \vec{F},
$$
where $\vec{F}$ is the flux vector, and
$$
\int_{\partial \Omega}\vec{F}\cdot n\,dS
$$
measure the amount of chemical flows out from the domain of interest $\Omega$, and $\vec{F} = -D \nabla u + \vec{b}u$. $D$ is the diffusion constant, it describes how the quantity diffuses, $\vec{b}$ is a flow field, it may carry $u$ around. If we assume $D$ is a constant and $\vec{b}$ is a divergence free flow (incompressible), then (1) is:
$$
\frac{\partial u}{\partial t} = \underbrace{D\Delta u}_{\text{Diffusion term}} - \underbrace{\vec{b}\cdot \nabla u}_{\text{Convection term}}. \tag{2}
$$
Convection-dominance just means, in the convection-diffusion equation (2):
$$D\ll \|\vec{b}\|.$$
The diffusion term is very small relative to the convection term.
Intuitively, diffusion means "smooth", while convection could possibly contain non-smoothness due to its derivation from an integral conservation law. For example, if we make the diffusion term $D\sim 0$ almost gone, the solution may looks like the following in a 2D problem:
where the solution's discontinuity "flows" with the field $\vec{b}$ on the right. If the diffusion constant is huge, i.e., not convection-dominated, then the steep slope will be smoothened as the quantity being transported along with the flow $\vec{b}$.
As I said in the comments, traditional numerical methods relies on the smoothness of the solution. Solution like above can't be solved by tradition numerical schemes like you listed. That's why people using Discontinuous Galerkin finite element to numerically solve this type of problems, while we don't impose the continuity conditions of the numerical solutions, in the meantime the conservation can be guaranteed numerically.
For the last question, please just read 2.1.4 here on page 33.
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H: Nitpicky Sylow Subgroup Question
Would we call the trivial subgroup of a finite group $G$ a Sylow-$p$ subgroup if $p \nmid |G|$? Or do we just only look at Sylow-$p$ subgroups as being at least the size $p$ (knowing that a Sylow-$p$ subgroup is a subgroup of $G$ with order $p^k$ where $k$ is the largest power of $p$ that has $p^k \mid |G|$)?
AI: I believe it is customary among most authors to take only primes that divide the group's order. Now, if you're doing some calculations over a large or infinite number of primes then you could always define a Sylow subgroup of a prime not dividing the group's order as the trivial subgroup.
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H: Is this fact about matrices and linear systems true?
Let $A$ be a $m$-by-$n$ matrix and $B=A^TA$. If the columns of $A$ are linearly independent, then $Bx=0$ has a unique solution.
If is true, can you help me prove it? If is false, could you give a counterexample?
Thanks.
AI: Yes. If $Bx=0$, then $x^T B x = (Ax)^T (Ax) = \|Ax\|^2 = 0$. Since the columns of $A$ are linearly independent, $Ax=0$ iff $x=0$. Hence $x=0$ and the solution to $Bx=0$ is not just unique, it is zero.
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H: prove that $\ln (x^2+ cos^2x)$ is uniformly continuous.
I know that the direction of the proof is to show that the derivative of the function is bounded, and hence meets the Lipschitz condition. So I differentiated it:
$$f'(x) = \frac{2x-\sin2x}{x^2 + \cos^2x}$$
But I am stuck here. I can't manage to show that it is bounded.
EDIT:
The domain is $\Bbb R$
AI: Show that the limits as $|x| \to \infty$ are both $0$. So there is some interval $[-a, a]$ where $f'(x)< 1$ for $x \not \in [-a,a]$. Then show that $f'(x)$ is continuous on $[-a,a]$, and therefore takes a maximum.
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H: Is a function $f:X\to Y$ continuous if and only if its graph on each connnected component of $X$ is connected?
I was thinking about this question today. Is the following true:
Let $X$ be a topological space with connected components $\{C_i\}_{i\in I}$. Let $Y$ be a topological space and let $f:X\rightarrow Y$ be a function. Then $f$ is continuous iff $$\forall i\in I, \;\{(x,f(x))\mid x\in C_i\}\text{ is a connected subspace of }X\times Y$$
The forward direction is easy, as $\{(x,f(x))\mid x\in C_i\}$ is the continuous image of the connected subspace $C_i$ (the mapping would send $x$ to $(x,f(x))$).
I am not sure if the backward direction is true for all topological spaces.
Thank you
AI: Let $f:\Bbb R \to \Bbb R,\ x\mapsto\sin(1/x)$ if $x>0$ and $0$ otherwise. Then the graph is connected but $f$ is not continuous.
For an example where the implication works, see If the graph $G(f)$ of $f : [a, b] \rightarrow \mathbb{R}$ is path-connected, then $f$ is continuous.
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H: How to understand $\delta$ and $\varepsilon$ in real analysis?
I'm a bit confused as to how to interpret $\delta$ and $\varepsilon$ mean in real analysis. My textbook gives an example demonstrating that $\frac{1}{x^2}$ is not uniformly continuous on $(0,1)$.
Definition of not uniformly continuous: $(\exists \varepsilon >0)(\forall \delta > 0)(\exists x)(\exists y)$ such that $|x-y| < \delta \implies |f(x) - f(y)| \geq \varepsilon$.
Let $\varepsilon = 1$ and $y = x + \frac{\delta}{2}$ and show that $\left\vert f(x) - f(x + \frac{\delta}{2}) \right\vert \geq 1.$
With some algebra we see that $\displaystyle 1 \leq \frac{\delta (2x+\frac{\delta}{2})}{2x^2(x+\frac{\delta}{2})^2}$.
Now my book makes the claim that it suffices to prove that the function is not uniformly continuous for $\delta < \frac{1}{2}$. My book picks $x = \delta$ and then $\displaystyle \frac{\delta (2\delta+\frac{\delta}{2})}{2\delta^2(\delta+\frac{\delta}{2})^2} = \frac{5}{9\delta^2} > 1$.
So I see that this shows that if $0 < \delta < \frac{1}{2}$ then $|f(\delta) - f(\delta + \frac{\delta}{2})| > 1$. But the inequality $\displaystyle \frac{5}{9\delta^2} > 1$ is not true if $\delta = 3/4$ and by definition it needs to be true for all $\delta > 0$.
So in general should I think of $\delta$ and $\varepsilon$ as positive numbers very very close to $0$? And why does it suffice to take $\delta < 1/2$?
AI: The variables $x$ and $y$ are bounded by $\exists$ in the written definition.
So, if there exist such $x$ and $y$ for some $\delta_0<1/2$, then the same $x$ and $y$ will be good for any $\delta>\delta_0$, as both $|x-y|<\delta$ and $|f(x)-f(y)|\ge\varepsilon$ will hold.
By the way, the implication part ($P\implies Q$) rather goes to 'and' ($P\land\,\lnot Q$) under negation:
$$ (\exists \varepsilon >0)(\forall \delta > 0)(\exists x)(\exists y): \
(|x-y| < \delta)\, \land \,(|f(x) - f(y)| \geq \varepsilon)\,. $$
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H: prove $A \sim B\implies 2^A \sim 2^B$.
I want to prove that if $A \sim B$ then $2^A \sim 2^B$.
$A\sim B$: There is a bijection from $A$ to $B$
thanks.
AI: HINT: Suppose that $f:A\to B$ is a bijection, and consider the map
$$F:2^A\to 2^B:X\mapsto f[X]=\{f(a):a\in X\}\;.$$
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H: Related rate volume increasing
I am trying to find out how fast the water level is rising in a tub that is filling at a rate of $.7 \text{ft}^3 / \text{min}$
I am not too sure how to do this, I am given that the base of the bathtub is $18 \text{ft}^2$ and that it is rectangular.
I know that $$b * h = v$$ but I am not so sure how to use this.
I know that the derivitive of the volume is $.7$
and that the base is a constant so it goes to zero, that leaves me with
$$h = .7$$
So does that mean the height increases $.7$ feet a minute? That is the wrong answer but I don't see why. My logic seems correct and the formulas are right as well.
AI: $$\frac{dV}{dt} = \text{area} \cdot \frac{dh}{dt}$$
Solve for $dh/dt$.
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H: Why it has Trace?
The operator tr : $T_1^1 (V ) \rightarrow R$ is just the trace of $F$ when it is considered as an endomorphism of $V$ . Since the trace of an endomorphism is basis-independent.
I am very confused with this statement because to my understanding, $T_1^1$ is $V^* \otimes V$. How can it be a matrix and have trace?
AI: When $V$ is finite-dimensional, there is a canonical isomorphism $V^\ast \otimes V \cong L(V)$ given by sending the $(1,1)$-tensor $a \otimes b$ to the linear transformation $v \mapsto a(v)b$ on $V$. Thus, the trace of a $(1,1)$-tensor can be defined as the trace of the corresponding linear transformation, which in turn can be computed as the trace of the matrix of that linear transformation with respect to any basis of $V$.
An equivalent way to view the trace on $T^1_1(V)$ is directly as the map $T^1_1(V) = V^\ast \otimes V \to F$ defined by $a \otimes b \mapsto a(b)$.
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H: $\mathbb{A}^2\backslash\{(0,0)\}$ is not affine variety
In our lecture notes we have this example, with the proof why $X = \Bbb{A}^2\setminus \{(0,0)\}$ is not an affine variety:
Let $i:X\hookrightarrow \mathbb{A}^2$ be an inclusion map. We show, that any regular function on $X$ extends uniquely to a regular function on $\mathbb{A}^2$, so that $i^*:k[x,y]\hookrightarrow \mathscr{O}(X)$ is an isomorphism.
Then we have, if $X$ was affine, $i$ is an isomorphism, which is not the case.
So we want to see that $i^*$ is an isomorphism. Therefore note that $\mathscr{O}(X)$ is a subring of the function field $k(x,y)$, that contains $k[x,y]$.
Take $P=(a,b)\in X$. Then $\mathscr{O}_{X,P}=\mathscr{O}_{\mathbb{A}^2,P}=k[x,y]_{\mathfrak{m}_P}$ whith $\mathfrak{m}_P=(x-a,y-b)$ is maximal ideal of $k[x,y]$.
-> Here is my first questions. How did we get this continued equality? I don't see how this is connected... (my commutative algebra knowledge has some gaps.)
But accepting that, we have $\mathscr{O}(X)\subset\bigcap\limits_{P\not=(0,0)}k[\mathfrak{m}_P]$. And this intersection lies inside of k(x,y).
Now we take $f=\frac{g}{h}\in K(x,y)$ with $g,h\in k[x,y]$ coprime.
Then $f\in k[x,y]_{\mathfrak{m}_P}$ iff $h(P)\not=0$. (Use that $k[x,y]$ is a UFD).
-> First, how did we see that $k[x,y]$ is UFD? And then, how did we conclude out of this the above property?
If $h$ is not a unit, then there are infinitely many $P\in \mathbb{A}^2$ such that $h(P)=0$ and thus also such points in $X$, and so $f\not\in\mathscr{O}(X)$.
So we have that $f\in\mathscr{O}(X)$ if $h$ is a unit. So $f\in k[x,y]$.
Thus $k[x,y]=\mathscr{O}(X)$ and hence $X$ not affine.
-> Now the last question, why does follow that $X$ is not affine?
I hope my questions are not too stupid, and that someone feels like helping!
All the best, Luca :)
AI: Let me try to answer your questions.
The reason why $\mathcal{O}_{P,X} = \mathcal{O}_{P,\Bbb{A}^2}$ is because for any point $P$ in an affine variety $Y$ and $U \subseteq Y$ open, we have that $\mathcal{O}_{P,U} \cong \mathcal{O}_{P,Y}$. This should essentially be immediate (what is the definition of the local ring at a point?) The second equality comes from Theorem 1.3.2 (c) of Hartshorne.
That $k[x,y]$ is a UFD is a standard fact in basic abstract algebra. More generally for any UFD $R$, the polynomial ring in $n$ number of variables over $R$ is also a UFD. For your second question, suppose I take a quotient of two polynomials $f = g/h$ with $h$ and $g$ coprime. Suppose that there is another representation $g'/h'$ of $f$ with $g'$ and $h'$ coprime. Then it will follow that $h'g =g'h$ which means to say that $h| h'g$. By unique factorization, it follows that every prime factor of $h$ divides $h'$ and in fact $h|
h'$ since $h$ divides $h'g$ but not $g$.
Similarly we conclude that $h' | h$ and so upto multiplication by a scalar, $h' = h$. Mutadis mutandis the same argument also shows that $g' = g$ and so the representation of $f$ as a quotient of two polynomials is unique upto multiplication by a scalar. Thus whenever we choose an element $f$ of $k(x,y)$, essentially to ask questions about whether a denominator vanishes, it is enough to work with just any representative of $f$ as a quotient of two polynomials with no common factor.
Thus if $f \in k[x,y]_{\mathfrak{m}_P}$ then $f$ can be written as the quotient of two polynomials $g/h$ with $h(P) \neq 0$ (definition of the localization at $\mathfrak{m}_P$). Conversely if $f = g/h$ with $h(P) \neq 0$, then for any other representative $g'/h'$ of $f$, $h'(P) \neq 0$ too for $h'$ is a non-zero scalar multiple of $h$. Thus $f \in k[x,y]_{\mathfrak{m}_P}$.
Once you know that $k[x,y] = \mathcal{O}(X)$ it is clear that $i^\ast$ is an isomorphism. But now their is an equivalence of categories between affine varieties over $k$ and finitely generated integral domains over $k$, and so if $X$ is affine the map $i$ has to be an isomorphism too. But this is impossible because it is not even surjective! Thus $X$ cannot be an affine variety.
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H: Find the inverse Laplace transformation of $\dfrac{s+1}{(s^2 + 1)(s^2 +4s+13)}$
My question is : find the function $f(t)$ that has the following Laplace transform
$$ F(s) = \frac{s+1}{(s^2 + 1)(s^2 +4s+13)} $$
thanks
AI: I am going to evaluate this using residues. If you have no idea of what these are, then I will just give you an easy-to-understand intermediate result: if $f(s) = p(s)/q(s)$ and $q$ has a zero at $s_0$, then the residue of $f$ at the pole $s=s_0$ is
$$\frac{p(s_0)}{q'(s_0)}$$
The inverse LT of the given
$$\hat{f}(s) = \frac{s+1}{(s^2+1)(s^2+4 s+13)}$$
is simply the sum of the residues of $\hat{f}(s) e^{s t}$ at the poles. The poles here are at $s_1=i$, $s_2=-i$, $s_3=-2+i 3$, and $s_4=-2-i 3$. I will allow you to apply the above formula to compute the residues of $\hat{f}(s) e^{s t}$ and add them up. I get for the final result,
$$f(t) = - e^{-2 t}\left[ \frac{1}{15} \sin{3 t}+\frac{1}{20} \cos{3 t}\right] + \frac{1}{20} (\cos{t}+2 \sin{t})$$
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H: can all triangle numbers that are squares be expressed as sum of squares
I'm not sure if this is just a subset of Which integers can be expressed as a sum of squares of two coprime integers? which in turn points to http://en.wikipedia.org/wiki/Brahmagupta%E2%80%93Fibonacci_identity, but if so, I'm not seeing it.
Basically: looking at the numbers between 0 and 1000, if (but not iff) n is a square, then the nth triangle number (i.e. $$\frac{n(n+1)}{2}$$) can be expressed as the sum of two perfect squares . Does this hold for all squares, and can someone point me in the direction of why this is? (For what it's worth, I is an engineer but haven't really touched number theoryish stuff since college.)
(Also saw Prove that there are infinitely many natural numbers $n$, such that $n(n+1)$ can be expressed as sum of two positive squares in two distinct ways. which proves there are infinitely many for the similar $n(n+1)$ case but not that all squares will work unless I missed some aspect)
(Brute force approach to checking numbers, because brute force always works)
import math
maxsquare = 1001
squares = [i*i for i in xrange(maxsquare)]
for j in xrange(int(math.sqrt(maxsquare))):
i = j * j
n = i * (i+1) / 2
for s in squares:
f = n - s
if f in squares:
print "%d^2 + %d^2 = %d * %d+1 / 2" % (int(math.sqrt(f)),int(math.sqrt(s)),i,i)
break
else:
print "Could not find anything for i = %d" % i
AI: The answer is yes, it is true for all $n$.
Let me reformulate your question. You want to know if, whenever $n$ is a perfect square, it is the case that $\frac12 n(n+1)$ is a sum of two squares. If $n$ is a perfect square, then it has an integer square root, which we can call $m$, and write $n = m^2$. Then your question becomes whether $\frac12 m^2(m^2+1)$ is a sum of two squares for all integers $m$.
There is a very useful theorem about sums of two squares which says that a number $N$ is a sum of two squares if and only if every prime of the form $4k+3$ (such as $3, 7, 11, 19,$ etc.) appears in the prime factorization of $N$ an even number of times.
Now $m^2(m^2 + 1) = {\left(m^2\right)}^2 + m^2$ is obviously a sum of two squares. And dividing $m^2(m^2+1)$ by 2 can't possibly affect the number of times any prime of the form $4k+3$ appears in its factorization, so $\frac12 m^2(m^2+1)$ is also a sum of two squares.
(Addendum: Erick Wong points out below that we do not even need to invoke the Fermat $4k+3$ theorem.)
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H: Why are projective spaces over a ring of different dimensions non-isomorphic?
Let $A$ be a nonzero commutative ring with unit. Define $\mathbb P_A^n$ to be the scheme $\operatorname {Proj} A[T_0,\dots,T_n]$, where the grading on the polynomial ring is by degree. Why is it true that $\mathbb P_A^n\not\cong \mathbb P_A^m$ when $n\neq m$? (I admit I am just guessing here. It seems like it should be true. But I do not know why.)
I am still learning the basics of scheme theory and would prefer as simple an answer as possible. Can this be proved directly from the definitions, without much machinery?
My motivation for asking this question is to show that the only $n$ for which $\mathbb P_A^n$ is affine is $n=0$. I can show that for any $\mathbb P_A^n$, the sheaf of global sections is $A$. So it is easy to see that if $\mathbb P_A^n$ is affine, the only possibility is that it is isomorphic to $\operatorname{Spec} A$. One can directly verify that they are isomorphic for $n=0$, but I do not know how to rule out larger $n$. Hence this question.
AI: There is a certain sense in which the dimension tells them apart. This can be made precise by a cohomological argument: Thm. 5.1 of Chapter 3 of Hartshorne implies that there is a coherent sheaf on $\mathbb{P}^n$ with a non-zero cohomology group in dimension $n$. Meanwhile, if $m<n$, then $\mathbb{P}^m$ has a covering by $m+1$ affine schemes, all of whose intersections are also affine, and this implies that no coherent sheaf on $\mathbb{P}^m$ has cohomology in any dimension bigger than $m$ (start with Thm 4.5 of Hartshorne).
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H: Let $f$ be analytic on $D=\{z\in\mathbb C:|z|<1\}.$ Then $g(z)=\overline{f(\bar z)}$ is analytic on $D.$
Let $f$ be analytic on $D=\{z\in\mathbb C:|z|<1\}.$ Then $g(z)=\overline{f(\bar z)}$ is analytic on $D$.
My attempt: Let $f$ be analytic on $D,~f(x,y)=u(x,y)+iv(x,y)$ where $u,v:D\to\mathbb{R}$ are the real and imaginary parts of $f$. Then $g:D\to\mathbb{C}:(x,y)\mapsto u(x,-y)-iv(x,-y).$ Let
$$g(x,y)=p(x,y)+iq(x,y)$$
where $p,q:D\to\mathbb R$ are real and imaginary parts of $g$ on $D$. Then
$$p:D\to\mathbb R:(x,y)\mapsto u(x,-y)
\\q:D\to\mathbb R:(x,y)\mapsto -v(x,-y).$$
Let $(x_0,y_0)\in D.$ Then
$$\begin{array}{ll}p_x(x_0,y_0)& = \lim_{x\to x_0}\frac{p(x,y_0)-p(x_0,y_0)}{x-x_0}\\ & =\lim_{x\to x_0}\frac{u(x,-y_0)-u(x_0,-y_0)}{x-x_0}\\ & =u_x(x_0,-y_0)\end{array}$$
$$\begin{array}{ll}q_y(x_0,y_0)& =\lim_{y\to y_0}\frac{q(x_0,y)-q(x_0,y_0)}{y-y_0}\\ & =\lim_{(-y)\to (-y_0)}\frac{v(x_0,-y)-v(x_0,-y_0)}{(-y)-(-y_0)}\\ & =v_y(x_0,-y_0)\end{array}$$
$u_x=v_y$ at $(x_0,-y_0)\implies p_x=q_y$ at $(x_0,y_0)$
Is it okay now?
AI: Here is a more direct approach (which I think is a little clearer):
We have $\frac{g(z+h)-g(z)}{h} = \frac{\overline{f}(\overline{z+h})-\overline{f}(\overline{z})}{h} = \overline{ \left( \frac{f(\overline{z+h})-f(\overline{z})}{\overline{h}} \right) } = \overline{ \left( \frac{f(\overline{z}+\overline{h})-f(\overline{z})}{\overline{h}} \right) }$.
Since conjugation is continuous, we have
\begin{eqnarray}
g'(z) &=& \lim_{h \to 0 }\frac{g(z+h)-g(z)}{h} \\
&=& \lim_{h \to 0 }\ \overline{ \left( \frac{f(\overline{z}+\overline{h})-f(\overline{z})}{\overline{h}} \right) } \\
&=& \overline{ \left( \lim_{h \to 0 } \frac{f(\overline{z}+\overline{h})-f(\overline{z})}{\overline{h}} \right) } \\
&=& \overline{ \left( \lim_{h \to 0 } \frac{f(\overline{z}+h)-f(\overline{z})}{h} \right) } \\
&=& \overline{ f'(\overline{z}) }
\end{eqnarray}
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H: Shortest Distance From a Point to a Plane
My professor did this question (Finding the shortest distance from a point to a plane) in class, but before doing the example he showed us a formula for calculating it, but didn't really explain how he got the formula. I was hoping someone could tell me how or why the formula works.
The question was as follows: Find the formula for the shortest distance from a point $P_0(x_0,y_0,z_0)$ to a plane $Ax+By+Cz+D=0$.
So the shortest distance would be a straight line from the point to the plane, which means that straight line would have to be in the direction of a normal vector to the plane. Let me denote the normal by $\vec{N}=[A,B ,C]^T$.
Next the professor said that $||\vec{P_{1}P_{0}}||\cos \theta=\frac{||\vec{P_{1}P{0}||||\vec{N}||\cos \theta}}{||\vec{N}||}=\frac{(\vec{P_1}P_{0}\cdot\vec{N})}{||\vec{N}||}=\frac{Ax_{0}+By_{0}+Cz_{0}+D.}{\sqrt{A^2+B^2+C^2}}$
I understood how he manipulated it to go to the last formula, however I guess I am having trouble understanding how/where the $\cos \theta$ in this formula: $||\vec{P_{1}P_{0}}||\cos \theta$ came from.
If it is for the simple reason that $\text{adjacent}=\text{hypotenuse} \cos \theta$, then I don't understand how the $||\vec{P_{1}P_{0}}|$ is the hypotenuse.
Can anyone shed any light on any of this?
AI: It is a projection.
Make a vector with the starting point as the given point, and the ending point as any point one the plane. We here call is $v$. The distance is exactly the projection of $v$ on an unit vector $n$ perpendicular to the plane. And $n$ can be easily calculated according to the plane.
And this is the result.
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H: $360\times 60$ nautical mile is not equal with $6400$ km of earth radius
One degree on a great circle of earth equals to $60$ Nautical Miles. Hence:
$\ 360 \times 60 = 21,600$ M (Nautical Mile)
$\ 21,600M \times 1852$ Meters $= 40,003,200$ Meters $= 40,003.2$ Kilometers
While the radius of earth is $6400$ Kilometers. What is problem with my calculation?
UPDATE: The question is wrong in nature. I confused the radius of earth with circumference of the earth!
AI: You need to divide by $2\pi$ to get the radius. You have calculated the circumference.
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H: Understanding equivalent definitions of left cosets
I understand the standard definition of left coset, but the one I do not understand (or see why it is advantageous) is the definition that follows:
Let $H\leq G$. Then a left coset of $H$ is a nonempty set $S\subset G$ such that $x^{-1}y \in H$ for any $x,y\in S$, and for any fixed $x\in S$, the map $y\mapsto x^{-1}y $ is a surjection from $S$ to $H$.
I see how it works, but it doesn't seem to say much... What am I not seeing?
AI: I'm not exactly sure what you mean by the canonical definition, but one standard definition of a coset is a translate $xH$ of $H$, for some $x\in G$. Another definition is that a coset is an equivalence class of the equivalence relation $\sim$ given by $x\sim y$ if $x^{-1}y\in H$. This is, in spirit, close to the definition you quote.
The equivalence class definition is useful in a couple of ways.
First, it's good to know that $\sim$ really is an equivalence relation, because that tells you immediately that the cosets partition $G$.
Second, it gives you a way of checking when two elements $x$ and $y$ represent the same coset: simply check if $x^{-1}y\in H$. This is similar to the idea of checking when two numbers $a$ and $b$ have the same remainder mod $m$: check if $m$ divides $a-b$.
The definition you quoted connects the two standard definitions -- the definition in terms of translates and the definition in terms of equivalence classes. Indeed, the fact that the transformation $y\mapsto x^{-1}y$ from $S$ to $H$ is surjective tells us that if we translate $S$ by $x^{-1}$ we get $H$ back.
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H: Prove that $\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b} \ge a+b+c$
If $a,b,c$ are positive , show that $$\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b} \ge a+b+c$$
Trial: Here I proceed in this way $$\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b} \ge \dfrac{2bc}{b+c}+\dfrac{2ca}{c+a}+\dfrac{2ab}{a+b}$$ then how I proceed. Please help.
AI: It suffices to show, by symmetry, that
$$\frac{a^2+b^2}{a+b}\ge \frac{a+b}{2}.$$
This is equivalent to
$$2a^2+2b^2\ge a^2+b^2+2ab\iff a^2+b^2\ge 2ab.$$
The last inequality is just the standard AM-GM inequality.
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H: What does the notation min max mean?
Min clearly means minimum and max maximum, but I am confused about a question that says "With $x, y, z$ being positive numbers, let $xyz=1$, use the AM-GM inequality to show that min max $[x+y,$ $x+z,$ $y+z]=2$ What does this mean? (I am not looking for the answer this particular question, but just what "min max" means.
AI: The meaning will depend on context. Here it means that for each triple $\langle x,y,z\rangle$ such that $xyz=1$ we find the maximum of $x+y,x+z$, and $y+z$, and then we find the smallest of those maxima: it’s
$$\min\Big\{\max\{x+y,x+z,y+z\}:xyz=1\Big\}\;.$$
In general it will be something similar: you’ll be finding the minimum of some set of maxima.
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H: Is every almost complex structure tame up to sign?
Let $V$ be a real vector space equipped with a symplectic form $\omega : V\times V \to \mathbb{R}$. An almost complex structure $J$ is compatible with $\omega$ if $\omega(J(u), J(v)) = \omega(u, v)$ for all $u, v \in V$. In addition, $J$ is said to tame $\omega$ if $J$ is compatible with $\omega$ and $\omega(u, J(u)) > 0$ for all $u \in V\setminus\{0\}$.
Given a compatible almost complex structure $J$ which tames $\omega$, we have another compatible almost complex structure given by $-J$ and satisfies $\omega(u, -J(u)) < 0$ for all $u \in V\setminus\{0\}$.
In both of these cases, the resulting bilinear form $\omega(\cdot, J\cdot)$ is definite; is this always the case? Alternatively phrased,
Given an almost complex structure $J$ which is compatible with the symplectic form $\omega$, must one of $J$ or $-J$ tame $\omega$?
AI: What if $J(e_1)=e_2$, $J(e_3)=e_4$, and $\omega=dx_1\wedge dx_3+dx_2\wedge dx_4$?
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H: Related rates with a cone
I am trying to figure out the rate the water level increases in a conical tank that is 3 m height, 2 m radius at top and water flows in at $2\text{m} ^3 / \text{minute}$
I know that
$$(1/3) \pi r^2 h = V$$
$$4 \pi = V$$
or at 2 seconds the volume of the water is $8.37758$
so now I have
$$(1/3) \pi r^2 h * dh/dt = V * dV/dt$$
which gives an incorrect answer, I am not sure if I use t = 2 volume and height or what.
AI: By similar triangles, observe that:
$$
\dfrac{h}{3}=\dfrac{r}{2} \iff r=\dfrac{2h}{3}
$$
Hence, substituting into the formula for the volume of a cone will help us to avoid product rule:
$$
V=\dfrac{1}{3}\pi \left(\dfrac{2h}{3}\right)^2h = \dfrac{4\pi}{27}h^3
$$
Differentiating each side with respect to $t$ yields:
$$
\dfrac{dV}{dt} = \dfrac{4\pi}{27}(3h^2)\dfrac{dh}{dt} = \dfrac{4\pi}{9}h^2\dfrac{dh}{dt} \iff \boxed{\dfrac{dh}{dt} = \dfrac{9}{4\pi h^2}\dfrac{dV}{dt}}
$$
Since we know that $dV/dt = 2$, we simply need to plug in the height at the specific instant in time that the question is asking for into the final equation, and we are done. If the radius is given instead, simply multiply it by $3/2$ to convert to a height.
EDIT: Alright, here's my MS Paint skills. Hopefully this diagram explains where the similar triangles came from:
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H: Continuous function from non-compact space onto compact space
Give an example of metric spaces $M_1$ and $M_2$ and a continuous function $f$ from $M_1$ onto $M_2$ such that $M_2$ is compact, but $M_1$ is not compact.
So there must exist a sequence $x_1,x_2,\ldots$ with no convergent subsequence in $M_1$, but any sequence in $M_2$ must have a convergent subsequence. In particular, $f(x_1),f(x_2),\ldots$ must have a subsequence that converges to $f(a)$ for some $a\in M_1$.
AI: HINT: A one-point space is compact and metrizable. Constant functions are continuous.
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H: Proving a lemma - show the span of a union of subsets is still in the span
This is part of proving a larger theorem but I suspect my prof has a typo in here (I emailed him about it to be sure)
The lemma is written as follows:
Let $V$ be a vector space. Let {$z, x_1, x_2, ..., x_n$} be a subset of $V$. Show that if $z \in\ span(${$x_1, x_2,..., x_n$}$)$, then $span(${$z, x_1, x_2,..., x_r$}$)=span(${$ x_1, x_2,..., x_r$})
I feel like this should be really simple and I saw a proof that you can take out a vector from a subset and not change the span, but I am unsure of the reverse -- assuming that is what this lemma is about. (To me, the "if" implies something besides just unifying the two sets should follow).
Anyhow, the proof should, I think, start with that we can modify a subset (let's call it S) without affecting the span if we start like this:
$\exists x \in S$ such that $ x \in span(S-${$x$})
then you build up a linearly independent subset somehow. The proof that you can take vectors out of the subset says that since $x\in span(${$x$}$)\ \exists\ \lambda_1, \lambda_2,..., \lambda_n \ \in\ K$ such that $x=\sum_{i=1}^n\lambda_i x_i $
since we know $ span(S) \supset span(S-${$x$}) we just need to show $ span(S) \subseteq span(S-${$x$})
But honestly I am not sure I understand what's happening here well enough to prove the above lemma. (This class moves fast enough that we're essentially memorizing proofs rather than re-deriving them I guess).
I am really starting to hate linear algebra. :-(
(Edited to fix U symbol and make it a "is a member of" symbol)
AI: We want to show $\operatorname{span}\{z, x_1, \dots, x_n\} = \operatorname{span}\{x_1, \dots, x_n\}$. In general, to show $X = Y$ where $X, Y$ are sets, we want to show that $X \subseteq Y$ and $Y \subseteq X$.
So suppose $v \in \operatorname{span}\{x_1, \dots, x_n\}$. Then, we can find scalars $c_1, \dots, c_n$ such that $$v = c_1x_1 + \dots + c_nx_n$$ so clearly, $v \in \operatorname{span}\{z, x_1, \dots, x_n\}$. This proves $$\operatorname{span}\{x_1, \dots, x_n\} \subseteq \operatorname{span}\{z, x_1, \dots, x_n\}$$
Now let $v \in \operatorname{span}\{z, x_1, \dots, x_n\}$. Again, by definition, there are scalars $c_1, \dots, c_{n+1}$ such that $v = c_1x_1 + \dots + c_nx_n + c_{n+1}z$. But hold on, $z \in \operatorname{span}\{x_1, \dots, x_n\}$, right? This means there are scalars $a_1, \dots, a_n$ such that $z = a_1x_1 + \dots + a_nx_n$. Hence,
$$v = c_1x_1 + \dots + c_nx_n + c_{n+1}(a_1x_1 + \dots + a_nx_n)$$
$$= (c_1 + c_{n+1}a_1)x_1 + \dots + (c_n+c_{n+1}a_n)x_n$$
and so we conclude that $v \in \operatorname{span}\{x_1, \dots, x_n\}$. Therefore,
$$\operatorname{span}\{z, x_1, \dots, x_n\} = \operatorname{span}\{x_1, \dots x_n\}$$
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H: The definition of tensor
$(E_1, \ldots, E_n)$ are basis for $V$. But this statement makes no sense to me. How could it define $F$ with $F^{j_1 \ldots j_l}_{i_1 \ldots i_k} E_{j_1} \ldots$, but then define $F^{j_1 \ldots j_l}_{i_1 \ldots i_k}$ with $F$? Totally lost...
AI: One presumably has that $\{\varphi^i\}$ is the dual basis of $V^\ast$ defined by $\{E_j\}$, so that
$$
\varphi^i(E_j) = \delta^i_j.
$$
Hence, $\{E_{j_1} \otimes \cdots \otimes E_{j_l} \otimes \varphi^{i_1} \otimes \cdots \varphi^{i_k}\}$ is a basis of $T^k_l(V) = V^{\otimes l} \otimes (V^\ast)^{\otimes k}$, so that any $F \in T^k_l(V)$ can be written as
$$
F = F^{j_1\cdots j_l}_{i_1 \cdots i_k} E_{j_1} \otimes \cdots \otimes E_{j_l} \otimes \varphi^{i_1} \otimes \cdots \varphi^{i_k}
$$
for unique constants $F^{j_1\cdots j_l}_{i_1 \cdots i_k}$.
In light of the canonical isomorphism $V^{\ast\ast} \cong V$, you can identify $T^k_l(V) = V^{\otimes l} \otimes (V^\ast)^{\otimes k}$ as the vector space of all multilinear maps $(V^\ast)^l \times V^k \to F$ via
$$
(w_1 \otimes \cdots w_k \otimes \psi_1 \otimes \cdots \otimes \psi_l)(f_1,\dotsc,f_k,v_1,\dotsc,v_l) := f_1(w_1) \cdots f_k(w_k) \psi_1(v_1) \cdots \psi_l(v_l)
$$
for $w_m \in V$, $\psi_n \in V^\ast$. Given this, then, you have that
$$
(E_{j_1} \otimes \cdots \otimes E_{j_l} \otimes \varphi^{i_1} \otimes \cdots \varphi^{i_k})(\varphi^{j_1^\prime},\dotsc,\varphi^{j_l^\prime}, E_{i_1^\prime}, \dotsc, E_{i_k^\prime}) = \delta^{j_1^\prime}_{j_1} \cdots \delta^{j^\prime_l}_{j_l}\delta^{i_1}_{i^\prime_1} \cdots \delta^{i_k}_{i^\prime_k}.
$$
Thus, for our $F \in T^k_l(V)$,
$$
F(\varphi^{j_1^\prime},\dotsc,\varphi^{j_l^\prime}, E_{i_1^\prime}, \dotsc, E_{i_k^\prime}) = F^{j_1\cdots j_l}_{i_1 \cdots i_k} E_{j_1} \otimes \cdots \otimes E_{j_l} \otimes \varphi^{i_1} \otimes \cdots \varphi^{i_k}(\varphi^{j_1^\prime},\dotsc,\varphi^{j_l^\prime}, E_{i_1^\prime}, \dotsc, E_{i_k^\prime}) = F^{j_1\cdots j_l}_{i_1 \cdots i_k}\delta^{j_1^\prime}_{j_1} \cdots \delta^{j^\prime_l}_{j_l}\delta^{i_1}_{i^\prime_1} \cdots \delta^{i_k}_{i^\prime_k} = F^{j_1^\prime\cdots j_l^\prime}_{i_1^\prime \cdots i_k^\prime}.
$$
This links to @anon's answer as follows, if you replace "inner product" with "pairing between a vector space and its dual":
Just as for $T^k_l(V)$, $\{\varphi^{j_1} \otimes \cdots \otimes \varphi^{j_l} \otimes E_{i_1} \otimes \cdots \otimes E_{i_k}\}$ can therefore be identified as a basis of $(V^\ast)^{\otimes l} \otimes V^{\otimes k} \cong V^{\otimes k} \otimes (V^\ast)^{\otimes l} =: T^l_k(V)$.
The canonical isomorphism $V^{\ast\ast} \cong V$ induces a canonical isomorphism
$$
T^k_l(V) := V^{\otimes l} \otimes (V^\ast)^{\otimes k} \cong (V^{\ast\ast})^{\otimes l} \otimes (V^\ast)^{\otimes k} \cong ((V^\ast)^{\otimes l} \otimes V^{\otimes k})^\ast \cong T^l_k(V)^\ast.
$$
Under this isomorphism, the basis $\{E_{j_1} \otimes \cdots \otimes E_{j_l} \otimes \varphi^{i_1} \otimes \cdots \varphi^{i_k}\}$ of $T^k_l(V)$ can be identified as the dual basis of $T^k_l(V) \cong T^l_k(V)^\ast$ dual to the basis $\{\varphi^{j_1} \otimes \cdots \otimes \varphi^{j_l} \otimes E_{i_1} \otimes \cdots \otimes E_{i_k}\}$ of $(V^\ast)^{\otimes l} \otimes V^{\otimes k} \cong T^l_k(V)$. Thus,
$$
(E_{j_1} \otimes \cdots \otimes E_{j_l} \otimes \varphi^{i_1} \otimes \cdots \varphi^{i_k})(\varphi^{j_1^\prime} \otimes \cdots \otimes \varphi^{j_l^\prime} \otimes E_{i_1^\prime} \otimes \cdots \otimes E_{i_k^\prime}) = \delta^{j_1^\prime}_{j_1} \cdots \delta^{j^\prime_l}_{j_l}\delta^{i_1}_{i^\prime_1} \cdots \delta^{i_k}_{i^\prime_k}.
$$
In general, if $W$ is a finite dimensional vector space with basis $\{e_j\}$, and if $\{\varepsilon^j\}$ is the dual basis of $W^\ast$ defined by $\{e_j\}$, i.e., $\varepsilon^j(e_k) = \delta^j_k$, then for any $f \in W^\ast$, writing $f = f_j \varepsilon^j$, we have that $f(e_k) = f_j \varepsilon^j(e_k) = f_j \delta^j_k = f_k$.
Putting everything together, we therefore have that $F \in T^k_l(V)$ can be written as
$$
F = F^{j_1\cdots j_l}_{i_1 \cdots i_k} E_{j_1} \otimes \cdots \otimes E_{j_l} \otimes \varphi^{i_1} \otimes \cdots \varphi^{i_k}
$$
for unique constants $F^{j_1\cdots j_l}_{i_1 \cdots i_k}$, so that
$$
F(\varphi^{j_1^\prime} \otimes \cdots \otimes \varphi^{j_l^\prime} \otimes E_{i_1^\prime} \otimes \cdots \otimes E_{i_k^\prime})\\ = F^{j_1\cdots j_l}_{i_1 \cdots i_k} E_{j_1} \otimes \cdots \otimes E_{j_l} \otimes \varphi^{i_1} \otimes \cdots \varphi^{i_k}(\varphi^{j_1^\prime} \otimes \cdots \otimes \varphi^{j_l^\prime} \otimes E_{i_1^\prime} \otimes \cdots \otimes E_{i_k^\prime})\\ = F^{j_1\cdots j_l}_{i_1 \cdots i_k}\delta^{j_1^\prime}_{j_1} \cdots \delta^{j^\prime_l}_{j_l}\delta^{i_1}_{i^\prime_1} \cdots \delta^{i_k}_{i^\prime_k} = F^{j_1^\prime\cdots j_l^\prime}_{i_1^\prime \cdots i_k^\prime}.
$$
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