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H: contradicting identity theorem?
the identity theorem for holomorphic functions states: given functions $f$ and $g$ holomorphic on a connected open set $D$, if $f = g$ on some open subset of $D$, then $f = g$ on $D$
Let $f(z) = \sin z + \cos z$,
$g(z) = \cos z$
then $(f-g)(0) = 0$
$\implies$ by the above theorem : $f(z) = g(z)$ on $\left|z\right|<1$
$\implies$ $\sin z + \cos z = \cos z$
$\implies \sin z =$ identically zero in $\left|z\right|<1$ which is not true
Where could I be making a mistake ?
AI: To conclude that $f=g$ on $D$, you must show that $f=g$ on some open subset of $D$. You did not show that, you just noticed that $f(0)=g(0)$, but $\{0\}$ is not an open subset of $D$. |
H: Solving recurrence relation, $a_n=6a_{n-1} - 5a_{n-2} + 1$
I'm trying to solve this recurrence relation:
$$
a_n = \begin{cases}
0 & \mbox{for } n = 0 \\
5 & \mbox{for } n = 1 \\
6a_{n-1} - 5a_{n-2} + 1 & \mbox{for } n > 1
\end{cases}
$$
I calculated generator function as:
$$
A = \frac{31x - 24x^2}{1 - 6x + 5x^2} + \frac{x^3}{(1-x)(1-6x+5x^2)} =
\frac{31x - 24x^2}{(x-1)(x-5)} + \frac{x^3}{(1-x)(x-1)(x-5)}
$$
(I'm not sure if that's right)
and its partial fractions decomposition looks like:
$$
A = \left(\frac{-7}{4} \cdot \frac{1}{x-1} - \frac{445}{4} \cdot \frac{1}{x-5}\right) +
\left( \frac{39}{16} \cdot \frac{1}{x-5} + \frac{3}{4} \cdot \frac{1}{(x-1)^2} - \frac{375}{16} \cdot \frac{1}{x-5} \right)
$$
(again - I'm not sure if it's ok)
I'm stuck here... From solutions I know that I should get:
$$
a_n = \frac{-21}{16} - \frac{1}{4}n + \frac{21}{16}5^n
$$
but I have no idea how it's solved... I hope somebody can help me (I spend more than 3h trying to solve this myself...)
AI: I did not check your work, so I’ll outline what you need to do to finish. You have something of the form
$$A(x)=\frac{a}{1-x}+\frac{b}{5-x}+\frac{c}{(1-x)^2}=\frac{a}{1-x}+\frac{b/5}{1-\frac{x}5}+\frac{c}{(1-x)^2}\;.$$
Expand these three terms into power series:
$$\begin{align*}
A(x)&=a\sum_{n\ge 0}x^n+\frac{b}5\sum_{n\ge 0}\left(\frac{x}5\right)^n+c\sum_{n\ge 0}(n+1)x^n\\\\
&=\sum_{n\ge 0}\left(a+\frac{b}{5^{n+1}}+c(n+1)\right)x^n\;.
\end{align*}$$
Now you can read off the coefficient of $x^n$.
Added: I’ve now had a chance to check your work, and it appears to be a bit off. I’ll use my preferred approach, which begins by assuming that $a_n=0$ for all $n<0$. Then the recurrence can be written
$$a_n=6a_{n-1}-5a_{n-2}+1-[n=0]+4[n=1]\;,$$
for all $n\ge 0$, where the last two terms contain Iverson brackets and are added to make the recurrence give the correct values to $a_0$ and $a_1$. Now multiply through by $x^n$ and sum over $n\ge 0$:
$$\begin{align*}
\sum_{n\ge 0}a_nx^n&=\sum_{n\ge 0}\Big(6a_{n-1}-5a_{n-2}+1-[n=0]+4[n=1]\Big)x^n\\\\
&=6x\sum_{n\ge 0}a_{n-1}x^{n-1}-5x^2\sum_{n\ge 0}a_{n-2}x^{n-2}+\sum_{n\ge 0}x^n-1+4x\;.
\end{align*}$$
Since $A(x)=\sum_{n\ge 0}a_nx^n=\sum_{n\ge 0}a_{n-1}x^{n-1}=\sum_{n\ge 0}a_{n-2}x^{n-2}$ (remember the blanket assumption that $a_n=0$ for $n<0$), we have
$$A(x)=6xA(x)-5x^2A(x)+\frac1{1-x}-1+4x\;,$$
and hence
$$A(x)=\frac1{(1-x)^2(5-x)}+\frac{4x-1}{(1-x)(5-x)}\;.$$
I’ll leave the partial fraction decomposition to you, at least for now. |
H: Isomorphism $\left(\mathbb{C}^{n}\setminus\{0\}\right)/\mathbb{Z}$ with $S^{1} \times S^{2n-1}$
I have to prove that there is this isomorphism:
$$\frac{\mathbb{C}^{n}{\setminus\{0\}}}{ \mathbb{Z}} \simeq S^{1} \times S^{2n-1},$$
where there is this equivalence relation in the left side:
$(w_1, \dots, w_n) \sim (z_1, \dots, z_n)$ if exists $\lambda \in \mathbb{C}$ with $|\lambda| \ge 1$ and a fixed $d \in \mathbb{N}$ such that
$$(w_1, \dots, w_n) \sim (\lambda^d z_1, \dots,\lambda^d z_n).$$
AI: Consider the map $$\phi:z\in\mathbb C^n\setminus\{0\}\mapsto\Bigl(\frac{z}{|z|},\log|z|\Bigr)\in S^{2n-1}\times\mathbb R.$$
Show that it is an homeomorphism.
Next, define on the codomain of $\phi$ an equivalence relation $\equiv$ such that $\phi$ induces an homeomorphism from $(\mathbb C^n\setminus\{0\})/\sim$ to $(S^{2n-1}\times\mathbb R)/\equiv$.
Figure out what $(S^{2n-1}\times\mathbb R)/\equiv$ is. |
H: Quick way to solve computational congruences
The specific problem at hand is $$34x \equiv 60 \bmod{98}$$
I reduced to get $$17x \equiv 30 \bmod{49}$$
and from this I have
$$17x \equiv 30 \bmod{7}$$
which is easy to solve and yields $x \equiv 3 \bmod{7}$. How can I use this fact to help me solve the more complex equation? I know the solution will be of the form $7k + 3$, but how can I utilize this fact?
AI: You can use your fact directly: plug in $7k+3$ for $x$ and continue solving!
More precisely, you can take $x=7k+3$ as the definition of the new variable $k$, and your calculation shows that $k$ is an integer. So you can continue solving congruences to determine the complete solution for $k$, which in turn gives you the complete solution for $x$. |
H: Expected value of function of negative binomial
With $X$ representing the total number of trials, and m the fixed number of successes. The pdf is then
$f(x|p)=$${x-1}\choose{m-1}$$p^m (1-p)^{x-m} \ \ \ \ x \ge m$
As a step in something else I'm trying to find $E[\frac{1}{X-1}]$.
Here's what I've got up to now:
$E[\frac{1}{X-1}]=\sum_{x=m}^{\infty}\frac{1}{x-1}$${x-1}\choose{m-1}$$p^m (1-p)^{x-m} \ \ \ \ \ \ \ \ $ #lotus
$E[\frac{1}{X-1}]=(1-p)\sum_{x=m}^{\infty}(x-m)$${x-2}\choose{m-1}$$p^m (1-p)^{x-1-m} \ \ \ \ \ \ \ \ $ #algebra
$E[\frac{1}{X-1}]=(1-p)\sum_{y=m-1}^{\infty}(y+1-m)$${y-1}\choose{m-1}$$p^m (1-p)^{y-m} \ \ \ \ \ \ \ \ $ #let y = x -1
Note the first term of the sum is zero to change the starting index.
$E[\frac{1}{X-1}]=(1-p) \left[ \sum_{y=m}^{\infty}y P(y|p)+ (1-m)\sum_{x=m}^{\infty} P(y|p) \right] \ \ \ \ \ \ \ \ $ #splitting up the sum
$E[\frac{1}{X-1}]=(1-p)\left[ \frac{m}{p} + (1-m) \right] \ \ \ \ \ \ \ \ $ #left = expectation of negative binomial, right = pdf sums to one
Is this correct? Is there a neater way to do it?
AI: $$
\begin{align}
&\sum_{x=m}^\infty\frac1{x-1}\binom{x-1}{m-1}p^m(1-p)^{x-m}\\
&=\sum_{x=m}^\infty\frac1{m-1}\binom{x-2}{m-2}p^m(1-p)^{x-m}\tag{1}\\
&=\frac{p^m}{m-1}\sum_{x=m}^\infty\binom{x-2}{x-m}(1-p)^{x-m}\tag{2}\\
&=\frac{p^m}{m-1}\sum_{x=m}^\infty\binom{-m+1}{x-m}(-1)^{x-m}(1-p)^{x-m}\tag{3}\\
&=\frac{p^m}{m-1}\sum_{x=0}^\infty\binom{-m+1}{x}(-1)^x(1-p)^x\tag{4}\\
&=\frac{p^m}{m-1}p^{-m+1}\tag{5}\\[6pt]
&=\frac{p}{m-1}\tag{6}
\end{align}
$$
$(1): \binom{x-1}{m-1}=\frac{x-1}{m-1}\binom{x-2}{m-2}$
$(2): \binom{x-2}{m-2}=\binom{x-2}{x-m}$ and bring constants out front
$(3): \binom{x-2}{x-m}=\binom{-m+1}{x-m}(-1)^{x-m}$
$(4): x\mapsto x+m$
$(5):$ binomial theorem
$(6):$ algebra |
H: Expected value of $|H-T|$ in $n$ coin flips
Let $H_n$ be the number of heads in $n$ coin flips.
Let $T_n$ be the number of tails in $n$ coins flips.
Is there a good way to calculate $E_n = E[|H-T|]$ that isn't brute force computation, i.e. directly evaluating $$E[|H-T|] = \frac{1}{2^n} \sum {n \choose r} |n-2r|. $$
I don't see any approach that can simplify this calculation.
AI: $$\sum_r \dbinom{n}r \vert n-2r \vert = \sum_{r < n/2} \dbinom{n}r (n-2r) + \sum_{r>n/2} \dbinom{n}r (2r-n) = 2\sum_{r>n/2} \dbinom{n}r r - \sum_{r<n/2}\dbinom{n}r r$$
I trust you can take it from here. You may need to use the identities
$$r \binom{n}r = n\dbinom{n-1}{r-1}$$
$$\sum_{r=0}^{k-1} \dbinom{2k}r = \dfrac{4^k - \dbinom{2k}k}2$$
$$\sum_{r=0}^{k} \dbinom{2k+1}r = 2^{2k}$$ |
H: Partitions of $n$: proving $p(n+2)+ p(n) \geq 2p(n+1)$
For $n \geq 2$ give an alternative description of $p(n) - p(n-1)$ as the number of partitions of $n$ which have a certain property.
I have done that part, it is fine. I have not included it here as I am not accustomed with LaTeX but can put it up if wanted. But then it says
Hence or otherwise prove $p(n+2)+ p(n) \geq 2p(n+1)$.
This is the part I do not know how to do. Any help appreciated thank you
AI: Hint: The stated inequality is equivalent to
$$p(n+2)-p(n+1)\ge p(n+1)-p(n)$$ |
H: A question regarding random variables
Consider the average $Y$ of $n$ independent random variables, each uniformly distributed on $[a,b]$. Does that mean $Y= \sum_{i=1}^{n}Y_{i}$ or $Y= \dfrac{\sum_{i=1}^{n}Y_{i}}{n}$?
Thank you for your time.
AI: $\text{The sample $\color{red}{\text{mean}}/\text{average}$ always $\color{blue}{\text{means}}$}$ $$Y = \dfrac{\sum_{i=1}^n Y_i}n$$
$\text{irrespective of the distribution.}$ |
H: Notation minimum of a column vector
I'd like to know the notation to express the minimum of a column vector.
Is this notation correct?
\begin{equation}
\min
\left[\matrix{
\left|b_{n}-b_{n+1}\right| \cr
\left|\left(b_{n}+360\right)-b_{n+1}\right| \cr
\left|b_{n}-\left(b_{n+1}+360\right)\right| \cr
}\right]
\end{equation}
If not, how would i have to write it down?
AI: You shouldn't need it to be a column vector unless you refer to it elsewhere as being a column vector, in which case I'd give it a name $\mathbf v$ and write $\min_i |v_i|$ or whatever.
If this is the only place it is used, just use the normal $\min\{\cdot,\cdot,\cdot\}$.
It's all personal taste really, I just instinctively feel the above are more natural. |
H: The last 2 digits of $7^{7^{7^7}}$
What is the calculation way to find out the last $2$ digits of $7^{7^{7^7}}$? WolframAlpha shows $...43$.
AI: $7^2=49=50-1$
$\implies 7^4=(50-1)^2=50^2-2\cdot50\cdot1+1\equiv1\pmod {100}$
Alternatively, $100=4\cdot25, 7^2\equiv1\pmod 4$
and $7^2\equiv-1\pmod{25}\implies 7^4\equiv(-1)^2\pmod{25}\equiv1$
$\implies 7^{\text{lcm}(2,4)}\equiv1\pmod {4\cdot25}$ as $(25,4)=1\implies 7^4\equiv1\pmod {100}$
So, we need to determine $7^{7^7}\pmod 4$
As $7^7$ is odd, and $7\equiv-1\pmod4\implies 7^{7^7}\equiv-1\pmod4\equiv3$
$\implies 7^{7^7}=4n+3$ for some integer $n>0$
So, $7^{7^{7^7}}=7^{4n+3}\equiv (7^4)^n\cdot7^3\pmod{100}\equiv1^n\cdot343$ |
H: Lie subalgebra, Lie subgroup and membership
Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$ and let $H$ be a connected Lie subgroup with Lie algebra $\mathfrak{h}$.
We have that $X \in \mathfrak{h} $ iff $exp(tX) \in H \ \ \ \forall t \in \mathbb{R} $ .
I have seen a proof of this but I don't understand why we need that $\forall t \in \mathbb{R}$.
An answer to this question shows that it is equivalent to ask $\forall t \in \mathbb{R}$ or just in $\forall t \in \, ]-t_0,t_0[$ for some $t_0$. Therefore I'm not worried about "for all $t$" versus "for $t$ small enough", but about "for just one $t$" versus "for more than one $t$" (of course $t\neq 0$).
I'd like to see an example in which you have some vector $X \in \mathfrak{g}$ such that $exp(X) \in H$ but $X \not \in \mathfrak{h}$. Is this possible?
AI: Take $G=\Bbb C^\times$, $H=\{1\}$ so $\mathfrak h=\{0\}$ and $X=2\pi\mathbf i\notin\mathfrak h$ but $\exp(X)\in H$. |
H: bifurcation value
I tried to understand how to locate the bifurcation value for the one-parameter family. From my understanding the bifurcation value is the maximum or minimum point of a parabola, so I set the differential equals to zero to find the equilibrium points. From two equilibrium point, I'll find the midpoint and try to find the output value that corresponding to that midpoint. For example: I got this differential equation
$dy/dt=y^2-ay+1$
I let $dy/dt=0$ then solve for $y$. I got $y= (a+\sqrt{a^2-4})/2$ and $y= (a-\sqrt{a^2-4})/2$ . I find the midpoint of these two value and got $y=a/2$. Plug back to the differential equation, solve for $a$ and got $a=2$ and $a=-2$. What do I do from here if I want to use the values of $a$ to draw the phase line ?
AI: 1.You're using a somewhat circuitous way to find bifurcation points. Most of the work is computing the equilibria of this differential equation,
$$
y = \frac{a \pm \sqrt{a^2- 4}}{2} ~~~~~ (*)
$$
Inspecting these forms, the parameter values $a = \pm 2$ correspond to the points at which equilibria are 'created' or 'destroyed', and so correspond to parameter values for which the qualitative aspects of the system change drastically. More specific to this one-dimensional problem, the phase line changes as $a$ crosses $\pm 2$ and remains the same elsewhere.
What you're doing, effectively, is solving for the parameter value(s) when these equilibria coincide; maybe in a more complicated system, it might be useful to pursue this method, but here you ought to be able to read off this behavior directly from $(*)$.
2.The phase line will depend on the value of $a$, and so a complete answer would include a phase line for each of the 5 cases: $a < -2, a = -2, -2 < a < 2, a = 2, a > 2$.
For each of $a < -2, a > 2$, there are two equilibria. I can tell you that in each case, one of them is a sink and other a source; you have to figure out which is which by writing
$$
\dot{y} = (y - \lambda_1) ( y - \lambda_2)
$$
where $\lambda_i$ are the equilibria given by $(*)$, and figuring out the sign of $\dot{y}$ for each of $y \in (-\infty, \lambda_1), (\lambda_1, \lambda_2), (\lambda_2, \infty)$.
For $-2< a < 2$ there are no equilibria, and the sign of $\dot{y}$ is positive for all $y$. What phase line corresponds to this situation? (Hint: it's pretty simple).
For $a = \pm 2$, there is a single equilibrium at $y = \pm 1$ that is neither a sink nor a source, because the sign of $\dot{y} = (y \mp 1)^2$ is always positive. |
H: $\int_\Omega fd\mu_n\to\int_\Omega fd\mu,\ \forall\ f\in C_0(\Omega)$ implies $\mu_n(\Omega)\to \mu(\Omega)$?
Let $\Omega\subset\mathbb{R}^N$ be a bounded open smooth domain and $C_0(\Omega)$ the set of bounded continuous functions with compact support. It is know that $C_0(\Omega)^\star =M(\Omega)$, where $M(\Omega)$ denotes the set of Radon measures (regular, bounded and infinitely additive). Suppose that $\mu_n\to \mu$ ($\mu_n,\mu\geq 0$) in the weak star sense, i.e. $$\int_\Omega fd\mu_n\to\int_\Omega fd\mu,\ \forall\ f\in C_0(\Omega)$$
Is it true that $\mu_n(\Omega)\to \mu(\Omega)$?
Thank you
AI: If $\Omega:=(0,1)$ and $\mu_n:=\delta_{n^{-1}}$, then we have the assumption with $\mu=0$, but not $μ_n(\Omega)=1\to \mu(\Omega)=0$.
One can generalize this taking $x\in \overline{\Omega}\setminus\Omega$ and $(x_n,n\in\Bbb N)\subset \Omega$ converging to $x$. |
H: How to calculate $\lim_{x\to 1^+} \log (x)^{\log(x)}$?
How to calculate $\lim_{x\to 1^+} \log (x)^{\log(x)}$ ?
i know that its "1", but why?
How can i calculate this?
Thank you very very much =)
AI: Substituting $t = \log x$ you obtain
$$
\lim_{x \to 1^+} \log(x)^{\log(x)} = \lim_{t \to 0^+} t^t = \lim_{t \to 0^+} e^{t \log(t)} = e^{\lim_{t \to 0^+} t \log(t)}.
$$
To compute $\lim_{t \to 0^+} t \log(t)$ use, for example, de l'Hospital rule. |
H: Disc with two points identified
Is a disc $D^2$ with two points on the boundary identified, same as $D^2 \vee D^2$ ? They both have boundary $S^1 \vee S^1$. I am confused because an exercise in Hatcher seems to ask the same question about the two spaces (i.e. to prove that there doesn't exist a retraction from each of this space to $S^1 \vee S^1$).
AI: The two spaces are not even homotopy equivalent. I'll describe the intuition for the homotopy type of each space.
For $D^2$ with two (distinct) points on the boundary identified, visualize it as the cylinder $S^1 \times [0,1]$ with one part of the wall "pinched" to a point. It's easy to see this is homotopy equivalent to $S^1$.
$D^2 \vee D^2$ is contractible since we can just contract each copy of $D^2$ to the basepoint. |
H: Homeomorphism between open unit ball and $\mathbb R^n$
Let $B=\{x\in\mathbb R^n : ||x||<1\}$ the open unit ball with the subapce topology of $\mathbb R^n$. I want to show that $B^n\cong\mathbb R^n$ with the map $F(x)=\tan(\frac{\pi ||x||}{2})\frac{x}{||x||}$ for $x\not=0$ and $F(0)=0$
Well $\frac{x}{||x||}$ is more or less the direction of the vector $x$, the norm is a number out of $[0,1)$ and the scaled tangent is a Homeomorphism from $[0,1)$ to the positive real numbers.
I want to show that $F(x)$ is contionous, bijective and there is an inverse map.
Injectivity: Let $F(x)=F(y)\leftrightarrow ||F(x)||=||F(y)||\leftrightarrow \tan(\frac{\pi ||x||}{2})||\frac{x}{||x||}||=\tan(\frac{\pi ||y||}{2})||\frac{y}{||y||}||$
I know that the tangent is injective but I do not see how $x=y$ can be followed.
Surjectivity: Let $y\in\mathbb R^n$, now I need to show $\exists x\in B^n: F(x)=y$, i.e $\tan(\frac{\pi ||x||}{2})\frac{x}{||x||}=y$ I do not see how surjectivity can be followed form this.
Inverse map: I know that the tangent has an inverse map, namely the arctangent, but how can this be used here?
Continuity: Well I know that the tangent is only contin. on $\mathbb R\setminus\{(n+\frac{1}{2})\pi: n\in \mathbb Z\}$
AI: Injectivity
$F(x)=F(y)$, with $x\ne0$ and $y\ne0$ means
$$
\tan\biggl(\frac{\pi \|x\|}{2}\biggr)\frac{x}{\|x\|}=
\tan\biggl(\frac{\pi \|y\|}{2}\biggr)\frac{y}{\|y\|}
$$
So we can assume $x\ne0$ and $y\ne0$. Then, taking norms,
$$
\tan\biggl(\frac{\pi \|x\|}{2}\biggr)=\tan\biggl(\frac{\pi \|y\|}{2}\biggr)
$$
because $x/\|x\|$ has norm $1$ and the same for $y/\|y\|$; moreover $\tan t>0$ if $t>0$. By the injectivity of the tangent you get
$$
\frac{\pi \|x\|}{2}=\frac{\pi \|y\|}{2}
$$
so $\|x\|=\|y\|$ and, finally $x=y$.
It's clear that $F(x)=0$ if and only if $x=0$, so the proof is complete.
Surjectivity
$0=F(0)$, so we only need to show that, for $z\ne0$, we can find $x$ with $F(x)=z$. We should have
$$
z=\tan\biggl(\frac{\pi \|x\|}{2}\biggr)\frac{x}{\|x\|}
$$
so
$$
\|z\|=\tan\biggl(\frac{\pi \|x\|}{2}\biggr)
$$
hence
$$
\frac{\pi\|x\|}{2}=\arctan\|z\|
$$
that is,
$$
\|x\|=\frac{2\arctan\|z\|}{\pi}.
$$
Thus the candidate is
$$
x=\frac{2\arctan\|z\|}{\pi\|z\|}z
$$
Verify it's the right one.
Continuity
The continuity of $F$ is obvious in the points different from $0$, because it's obtained by continuous functions. Since your domain consists of vectors with norm $<1$, there's no problem with the tangent function, because you consider only arguments in $(0,\pi/2)$. The continuity of the inverse is obvious as well
outside $0$
Are $F$ and its inverse continuous at $0$?
A well known fact is that $\lim_{x\to0}F(x)=0$ if and only if $\lim_{x\to0}\|F(x)\|=0$; now
$$
\|F(x)\|=\biggl\|\tan\biggl(\frac{\pi\|x\|}{2}\biggr)\frac{x}{\|x\|}\biggr\|
=\tan\biggl(\frac{\pi\|x\|}{2}\biggr)
$$
that obviously satisfies the requested property. The same for the continuity at $0$ of $F^{-1}$. |
H: how to calculate the real probability
Say I have a 50% probability of winning a game if I play against person 1, and a 50% probability of winning a game against person 2. I will play with both people, one after another. Before the matches, what was the probability that I would win at least 1 match? It has to be something between 90-99% I think. Any suggestions? Thank you!
AI: Note: as Austin Mohr's helpful follow-up comment explains, the answer relies on the fact that the likelihood of a win and a loss are the same: we have $50\%$ chance of a win $= 50\%$ chance of a loss in each game, so every outcome is equally likely.
W: win... L: lose
Four possible outcomes: (Game 1 vs. person 1) followed by (Game 2 vs. person 2)
W W <--
W L <--
L W <--
L L
In $3$ of $4$ outcomes, you win one or both (at least one) of the two games. In only one outcome, will you lose both (and hence not win at least once).
Therefore: The probability of winning at least one game: $P(\text{win at least one game}) = \dfrac 34 = 75\%$ |
H: Is $y=|x^3|$ a smooth function?
Is this a smooth function? $y=|x^3|$
The graph of this function has no sharp cuts or corners, so I think it is a smooth function but someone told me that it's not.
AI: You can check that
$$
f'(x)=3x^2\operatorname{sign}(x)\\
f''(x)=6x\operatorname{sign}(x)\\
f'''(x)=6\operatorname{sign}(x)
$$
Since $f'''$ is discontinuous at zero then $f\notin C^2(\mathbb{R})$. Hence $f$ is not smooth. |
H: Wrong Reasoning about the problem of breaking a stick in $2$ points and build a triangle with the $3$ parts.
For homework I was asked to solve this classical problem "If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4" and ok, it must result $1/4$. But I can't figure out why. This is my reasoning (obviously wrong somewhere!). Let $X,Y$ be the points (picked uniformy at random on the stick)
There are $3$ case:
1) $0<X<Y<1$ with probability $\frac{1}{2}$ to happen
2) $0<Y<X<1$ with probability $\frac{1}{2}$ to happen
3) $0<Y=X<1$ which has probability $0$ to happen.
Consider the first case: ($X$ indicate now the length of the segment $[0,X]$)
the triangle inequalities leads to:
$X<(Y-X)+(1-Y) = 1-X \longrightarrow X<\frac{1}{2}$
$Y-X < (X) + (1 - Y) \longrightarrow Y-X < \frac{1}{2}$
$1-Y < Y \longrightarrow Y > \frac{1}{2}$
which leads to a probability of "success" of $\frac{1}{8}$
and the "global" probability of this event is $\mathbb{P}(\text{case} \ 1)*\mathbb{P}(\text{ I can build a triangle in this case})=$ $\frac{1}{16}$.
for the second case is the same.
the third case has probability $0$, so it doesn't give any contribution to the probability. According to my reasoning the global probability is $\frac{1}{8} \neq \frac{1}{4}$.
I can't find why my reasoning is wrong. I don't want a solution of the problem, but WHERE and WHY I've made an error. I know there are a lot of questions about this problem, but I need a correction in this reasoning not a complete solution, and I didn't found an answer in the other questions :) Thank you in advance.
AI: There's nothing wrong with your (corrected) calculation. The triangular region specified by the three inequalities $x<\frac 12$, $y<x+\frac 12$, and $y>\frac 12$, does indeed have area $\frac{1}{8}$. However, the total area under consideration is the portion of the $[0,1]$ square subject to $x<y$, which has area $\frac 12$. Hence the conditional probability is their ratio, or $\frac 14$.
You've made things more complicated by splitting into cases this way. Looking at the $[0,1]$ square, there are two triangular regions where the stick-breaking leads to a triangle; their total area is $\frac 14$. |
H: finding the probability density function of $ dY_t = - Y_t X_t dW_t$
Could someone point me to where I can learn how to derive the stationary distribution for the martingale $Y_t$ which itself has stochastic volatility drive by $X_t$:
\begin{align}
dY_t &= - Y_t\ X_t\ dW^Y_t \\
dX_t &= - \kappa X_t dt + \sigma dW^X_t
\end{align}
where the brownians $W_t^Y$ and $W_t^X$ are uncorrelated. Many thanks in advance !
AI: I am not sure I can give a reference for this particular question, but it can be done easily enough.
If the two Brownian motions are independent you can find the probability distribution for $Y_t$ conditioned on a path of $X_t$. You can write
$$ \log Y_t = \log Y_0 - \int_0^t X_t dW^Y_t - \frac12 \int_0^t X_t^2\,dt. $$
The point is that $\int_0^t X_t^2\,dt = Z_t$ is just a random variable and, conditioned on a particular path of $X_t$, the first integral is an Ito integral of a deterministic function, so
$$ \log Y_t = \log Y_0 - N(0, Z_t) - \frac12 Z_t.$$
So the distribution of $\log Y_t$ conditioned on the value of $Z_t=\int_0^t X_t^2\,dt$ is a normal distribution, as above, and the distributions of $X_t$ and $Z_t$ you can calculate independently of this question.
The particular $X_t$ that you wrote down is an Ornstein-Uhlenbeck process. See Wikipedia, for example, which also gives the pdfs for $X_t$, $X_t^2$, etc., from which you can find the distribution of $Z_t$. |
H: Equivalent norms and isometries
Let $X$ be a vector space, $\|\cdot\|_1$ and $\|\cdot\|_2$ two equivalent norms on $X$. Under what further assumptions can we prove there is an isometry between $(X,\|\cdot\|_1)$ and $(X,\|\cdot\|_2)$?
In particular, are $(\mathbb{R}^2,\|\cdot\|_1)$ and $(\mathbb{R}^2,\|\cdot\|_2)$ isometric (where the two norms are the maximum norm and the usual euclidean norm respectively)?
AI: Normally, one proves that two normed linear spaces are isometric in a constructive™ way, by exhibiting an isometry between them. One common exception: since it's widely known that any two separable Hilbert spaces are isometrically isomorphic to each other, an explicit isometry is unnecessary in this case.
One elementary way to show that two spaces $X,Y$ are not isometrically isomorphic is to find a subset of $X$ that cannot be isometrically embedded into $Y$. For example, consider the subset $$A=\{(\pm 1,0), (0,\pm 1)\}\subset (\mathbb R^2,\|\cdot \|_1)$$
Any two points of $A$ are at distance $2$ from each other. Such a configuration is impossible in $(\mathbb R^2,\|\cdot \|_2)$. Indeed, three points at distance $2$ from one another must lie at the vertices of an equilateral triangle, and there is nowhere to put the fourth one. |
H: Without calculating limit directly show that it is equal to zero
$$\lim_{n\rightarrow\infty}\left(\frac{n+1}{n}\right)^{n^2}\frac{1}{3^n}=0$$
I am not really sure what it means by "without calculating limit" and I don't really have ideas how to do it.
AI: $\left(1+\dfrac 1 n\right)^n < e \,\,$so $\left(1+\dfrac 1 n\right)^{n^2} < e^n$
so $\left(1+\dfrac 1 n\right)^{n^2} \dfrac 1{3^n} < \dfrac{e^n}{3^n}
= \left(\dfrac e 3 \right)^n \to 0$. |
H: Inconsistent System of Linear Equations
Let $A ∈ M_{n\times n}(F)$. Suppose that the system of linear equations $AX = B$
has more than one solution. Prove that there is a column $C ∈ F^n$ such that
the system of linear equations $AX = C$ is inconsistent.
I thought it'd be clever to use this (underlined by my pencil):
AI: Hints: What does the premise imply about the kernel of $A$? Also:
$$\dim(\text{Im } A)+\dim(\ker A)=n.$$ |
H: Generating primes from other primes
For a natural number $n$ let $M$ be an $n$ by $n$ matrix w/$0$'s on diagonal and natural numbers off diagonal and let $p_1, p_2, \dots, p_n$ be a set of prime numbers.
Note then, that
$$p=\sum_{i=1}^n (\pm) \prod_{j=1}^n p_i^{m_{ij}}$$
is not divisible by any of the primes $p_1, p_2, \dots, p_n$.
Questions: Is there a finite set of primes that would generate
(a) all co-primes to the set,
(b) all other primes
in this fashion (varying $M$ and signs).
Example: $p_1 = 2$, $p_2=3$,
$$ 2+3 = -2^2 + 3^2 =5, $$
$$ 2^2+3 = 2^4 - 3^2 = 7, $$
$$ 2^3+3 = -2^4+ 3^3 = 11, $$
$$ 2^2+ 3^2 = 13,$$
$$\dots,$$
$$ ? =37. $$
What restriction on $M$ and signs would produce only primes for a set $p_1, p_2, \dots, p_n$?
AI: Intuitively, the answer to the first two questions is no and the last is that there is no restriction on $M$ and signs that will produce only primes. The fundamental reason is that powers get very far apart when the exponent is large.
Pick your set $\{p_1,p_2,p_3,\ldots p_n\}$. It is finite and has a maximum. Now pick a prime and a composite that are about $10^{100^n}$ times the largest one. It is quite unlikely that you can express them this way. We could calculate how many powers of the primes are less that this and find it is small compared to the number of numbers to express. This will not be rigorous-there could be some cancellation between larger powers of the primes that brings it back into range.
For the last, note that $2^5+3=35$. You just need to find one prime that $2$ is a primitive root of and you will be able to find a multiple of that prime. |
H: Limit as N goes to Infinity
Consider this limit: $$\lim_{n\rightarrow\infty} \left( 1+\frac{1}{n} \right) ^{n^2} = x$$
I thought the way to solve this for $x$ was to reduce it using the fact that as $n \rightarrow \infty$, $\frac{1}{n} \rightarrow 0$:
$$\therefore \lim_{n\rightarrow\infty}(1+0)^{n^2} = x$$
Apparently this is wrong! Why is it wrong?
AI: But as $n\rightarrow \infty $ that exponent gets ever larger. A smaller argument, yes, but raised to an ever larger power. Can't assume that the whole thing is going to go to one. Try a few examples on your calculator to see what happens.
For a little more detail, if $n$ is some big number, when you multiply everything out, the first two terms (look up the binomial expansion) will be
$1+ n^2 * \displaystyle\frac{1}{n} = 1+n \rightarrow \infty$
None of the remaining terms will be negative, so you know this is a bottom limit on what you are going to get. |
H: Closed form for $\sum_{n=1}^\infty\frac{1}{2^n\left(1+\sqrt[2^n]{2}\right)}$
Here is another infinite sum I need you help with:
$$\sum_{n=1}^\infty\frac{1}{2^n\left(1+\sqrt[2^n]{2}\right)}.$$
I was told it could be represented in terms of elementary functions and integers.
AI: Note that
$$\frac{2^{-n}}{2^{2^{-n}}-1}-\frac{2^{-(n-1)}}{2^{2^{-(n-1)}}-1} = \frac{2^{-n}}{2^{2^{-n}}+1} $$
Thus we have a telescoping sum. However, note that
$$\lim_{n \to \infty} \frac{2^{-n}}{2^{2^{-n}}-1} = \frac{1}{\log{2}}$$
Therefore the sum is
$$a_1-a_0 + a_2-a_1 + a_3-a_2 + \ldots + \frac{1}{\log{2}} = \frac{1}{\log{2}}- a_0$$
where
$$a_n = \frac{1}{2^n \left ( 2^{2^{-n}}-1\right)}$$
or
$$\sum_{n=1}^{\infty} \frac{1}{2^n \left ( 1+ \sqrt[2^n]{2}\right)}= \frac{1}{\log{2}}-1$$ |
H: Calculating $-\frac{4}{x^2}+\frac{1}{(x-1)^2}$
Hello everyone I have the following question.
I have the following fraction$$f(x)=-\frac{4}{x^2}+\frac{1}{(x-1)^2}$$
But how would I reduce it? I know I have to multiply the opposite numerator by denominator and I got:
\begin{gather*}
(x-1)^2(4)=4x^2-8x+4\\
1(x^2)=x^2
\end{gather*}
So I got:$$f(x)=-\frac{4x^2-8x+4+x^2}{x^2(x-1)^2}$$
But this is incorrect. What am I doing wrong?
AI: You are close.
Your mistake was applying the "-"
to both terms instead of just one.
You wrote
$f(x)=-\frac{4x^2-8x+4+x^2}{x(x-1)^2}$.
What you should have is
$f(x)=\frac{-(4x^2-8x+4)+x^2}{x(x-1)^2}
=\frac{-4x^2+8x-4+x^2}{x(x-1)^2}
=\frac{-3x^2+8x-4}{x(x-1)^2}
$. |
H: Why is the free pro-c-group on an infinite set not the pro-c-completion of the free group?
The set-up is the following: $\mathfrak c$ is a collection of finite groups closed under subgroups, homomorphic images, and extensions. For any group $G$, the pro-$\mathfrak c$-completion $G(\mathfrak c)$ is defined as the topological group $\varprojlim G/U$ where $U$ ranges over all normal subgroups of $G$ such that $G/U\in\mathfrak c$, with $G/U$ treated as discrete topological groups. Note that we have a group homomorphism $\eta\colon G\to G(\mathfrak c)$.
If I understand correctly, the pro-$\mathfrak c$-completion is the left adjoint to the forgetul functor from pro-$\mathfrak c$-groups to groups, as it satisfies the universal property that any group homomorphism $f\colon G\to H$ where $H$ is pro-$\mathfrak c$-group factors as $f=h\circ\eta$ where $h\colon G(\mathfrak c)\to H$ is a continuous group homomorphism (so a morphism of pro-$\mathfrak c$-groups).
It seems to me that the above implies that the free pro-$\mathfrak c$-group on a set $S$ should be the pro-$\mathfrak c$-completion of the free group on $S$. I am reading, however, that this is the case only when $S$ is finite, but not when $S$ is infinite. What am I missing?
AI: Adjoint functors compose: if $F_1, F_2$ are composable functors with left adjoints $G_1, G_2$, then $F_1 \circ F_2$ has left adjoint $G_2 \circ G_1$, and this is straightforward to prove. In this situation $F_1$ is the forgetful functor from groups to sets and $F_2$ is the forgetful functor from pro-c-groups to groups. Hence $G_1$ is the free group functor and $G_2$ is the pro-c-completion.
So you are right if you interpret "free pro-c-group on an infinite set" in the obvious way. However, a cursory google search found Wilson's Profinite Groups which gives a different universal property from the obvious one. |
H: Regular expressions, is it always true that (r+s)*=r*+s*?
I'm really confused about this, can some one please help me understand this better.
If r and s are regular expressions then is it always true that (r+s)=r+s*?
Are r and s sets and does the plus mean the union?
AI: Let $R$ be the set of words described by the regular expression $r$, and let $S$ be the set of words described by the regular expression $s$. Then the regular expression $(r+s)^*$ describes the words in the set $(R\cup S)^*$, and $r^*+s^*$ describes the words in the set $R^*\cup S^*$. Thus, the question boils down to asking whether it’s necessarily true that $(R\cup S)^*=R^*\cup S^*$. (Looking at some very simple regular expressions should give you the answer pretty easily.) |
H: Commuting square of functors
Let $\mathcal{E}$ be a complete and cocomplete category. Given a functor $i: \mathcal{C} \to \mathcal{D}$ between small categories, there is a triple of adjoint functors between their respective categories of presheaves with values in $\mathcal{E}$:
$$Lan_i =: i_!: [\mathcal{C}, \mathcal{E}] \to [\mathcal{D}, \mathcal{E}]$$
$$(-)\circ i =: i^\ast: [\mathcal{D}, \mathcal{E}] \to [\mathcal{C}, \mathcal{E}]$$
$$Ran_i =: i_\ast: [\mathcal{C}, \mathcal{E}] \to [\mathcal{D}, \mathcal{E}]$$
Suppose we are further given functors $p: \mathcal{C} \to \mathcal{C}$ and $q: \mathcal{D} \to \mathcal{D}$ such that $i \circ p = q \circ i$. It is then obvious that $p^\ast \circ i^\ast\simeq i^\ast \circ q^\ast$. Using the unit $\mathrm{id}_\mathcal{C} \to i^\ast \circ i_!$ and counit $i_! \circ i^\ast \to \mathrm{id}_\mathcal{D}$ of the first adjunction, we obtain a natural transformation of functors $\mathcal{C} \to \mathcal{D}$:
$$i_! \circ p^\ast \to i_! \circ p^\ast \circ i^\ast \circ i_! \simeq i_! \circ i^\ast \circ q^\ast \circ i_! \to q^\ast \circ i_!$$
Is this a natural isomorphism in general? What about if $i$ is fully faithful (in which case the unit $\mathrm{id}_\mathcal{C} \to i^\ast \circ i_!$ is a natural isomorphism)?
AI: You are asking about the Beck–Chevalley conditions for Kan extensions, which is a fairly non-trivial question. Here is one generic situation where things work. Let $u : \mathcal{A} \to \mathcal{C}$ and $v : \mathcal{B} \to \mathcal{C}$ be functors between small categories, let $(u \downarrow v)$ be the comma category, let $p : (u \downarrow v) \to \mathcal{A}$ and $q : (u \downarrow v) \to \mathcal{B}$ be the projections, and let $\theta : u p \Rightarrow v q$ be the canonical natural transformation. Then, the Beck–Chevalley transformations
$$q_! v^* \Rightarrow q_! p^* u^* u_! \Rightarrow q_! q^* v^* u_! \Rightarrow v^* u_!$$
$$u^* v_* \Rightarrow p_* p^* u^* v_* \Rightarrow p_* q^* v^* v_* \Rightarrow p_* q^*$$
are natural isomorphisms. This is Proposition 4.1.19 in my notes.
More generally, there is a notion of exact square, and there is an exact square condition characterising fully faithful functors. |
H: How to maximize $n!\sum^n_{k = 0}\frac{a^k(1+(-1)^{n-k})}{k!(n-k)!} \pmod{a^2}$ for a given $a$?
Short Version of the Question:
How do I maximize the value of $n!\sum^n_{k = 0}\frac{a^k(1+(-1)^{n-k})}{k!(n-k)!} \pmod{a^2}$ for a given $a$?
Long Version of the Question:
I'm currently attempting a Project Euler question using as much mathematics as possible before resorting to coding.
The problem itself is as follows:
Let r be the remainder when (a-1)^n + (a+1)^n is divided by a^2.
For example, if a = 7 and n = 3, then r = 42: 63^3 + 83^3 = 728 42 mod 49. And as n varies, so too will r, but for a = 7 it turns out that rmax = 42.
For 3 < a < 1000, find the sum of all rmax.
My first attempt at this problem was to change $(a-1)^n + (a + 1)^n$ to make it easier for coding purposes. And so I used the binomial theorem to come up with the following:
$(a-1)^n + (a + 1)^n$ = $n!\sum^n_{k = 0}\frac{a^k(1+(-1)^{n-k})}{k!(n-k)!}$
Obviously what I want to do now, is to maximize this new equation with the modulo. However, I cannot come up with how to do this. It may very well be that the original statement would be easier to simplify!
Nonetheless, I've come to ask: how can I maximize the value of $n!\sum^n_{k = 0}\frac{a^k(1+(-1)^{n-k})}{k!(n-k)!} \pmod{a^2}$ for a given $a$?
On this point, I would rather a hint, as opposed to a full answer. That would ruin the fun!
Note: I have a limited knowledge of calculus, and have yet to find the derivative of a modulo statement, or a summation. Would this need to be learned?
AI: Hint: If you expand $(a-1)^n$, you will find that every term but two are multiples of $a^2$. Similarly for $(a+1)^n$. Hence you can have a much simpler expression, with no $\Sigma$, $\mod{a^2}$. |
H: Comparison test for series $\sum_{n=1}^{\infty}\frac{n}{n^3 - 2n + 1}$
I am trying to prove the convergence of the series $\sum_{n=1}^{\infty}\frac{n}{n^3 -2n +1}$ with the simple comparison test. I know it can be done with other tests but this question came up in my homework for the comparison test before the other tests were discussed so I'm being stubborn and want to use that test.
I know it converges but I haven't been able to find a rigorous justification. This is my reasoning so far.
$\sum_{n=1}^{\infty}\frac{n}{n^3 -2n +1}$ converges because $\sum_{n=1}^{\infty}\frac{n}{n^3}$ = $\sum_{n=1}^{\infty}\frac{1}{n^2}$ and I know the latter converges by the p-series criterion.
To use the comparison test as my justification I must show $\frac{n}{n^3−2n+1} \leq \frac{n}{n^3} \ \ \forall \ n$. This would be obvious if the denominator on the left were greater than the denominator on the right but that can't be true as on the left I am subtracting from $n^3$ and on the right $n^3$ is unchanged.
So I switch to proving $\frac{n}{n^3−2n+1} \leq \frac{1}{n^2}$ and then I'm stuck.
If the numerators were both 1 then $n^3−2n+1$ is eventually going to be larger than $n^2$ even if we're subtracting a little bit from $n^3$.
That is, at n=1 we'll get $1^3−2+1=0 \leq 1$ but thereafter $n^3− 2n +1 \geq n^2$ is always true.
But the statement of the comparison test says $a_n \leq b_n \forall n$ where $\sum b_n$ is known to converge.
So that's where I am so far. The problem comes down to showing the necessary inequality holds and is it enough the it holds "eventually".
Any ideas?
AI: How about the Limit Comparison Test? From Wikipedia:
Suppose that we have two series $\sum_n a_n$ and $\sum_n b_n$ with $a_n, b_n > 0$ for all $n$.
Then if $\lim_{n\to\infty}\frac{a_n}{b_n} = c$ with $0 < c < \infty$, then either both series converge or both series diverge.
For your problem, use:
$$a_n=\frac{n}{n^3 -2n +1}\\
b_n = \frac {1}{n^2}$$
Then:
$$\lim_{n\to\infty} \frac {a_n}{b_n} = \lim_{n\to \infty}\frac {\frac{n}{n^3 -2n +1}} {\frac 1 {n^2}} = \lim_{n\to \infty} {\frac{n^3}{n^3 -2n +1}} = 1\\$$
The last limit is a consequence of L'Hôpital's rule (see below).
Thus, because $0 < 1 < \infty$, and $\sum \frac 1 {n^2}$ converges (P-series), the given series must also converge.
The L'Hôpital's rule derivation is:
$\lim_{n\to\infty} \frac{n^3}{n^3 - 2n + 1} = \frac \infty \infty$ so we can take the derivative of the numerator and denominator.
$\lim_{n\to\infty} \frac{3n^2}{3n^2 - 2} = \frac \infty \infty$ so we can do it again.
$\lim_{n\to\infty} \frac{6n}{6n} = \frac \infty \infty$ so we do it once more.
$\lim_{n\to\infty} \frac{6}{6} = 1$ and we're done. |
H: Proving existence of unique polynomial satisfying some integral
I came across a problem that I was having some trouble with.
Fix a positive integer $n$. Let $f(x) \in C(R)$ be a (real valued) continuous function.
Show that there exists a unique polynomial $q(x) \in P_n(R)$, such that for every
polynomial $p(x) \in P_n(x)$ we have $\int^1_0 p(x)f(x)dx=\int^1_0 p(x)q(x)dx$.
I know that typically to solve uniqueness questions, you need to assume another $g(x)$ exists and then manipulate the resulting equation(s) to show that $g(x)=q(x)$. However, it's not so clear to me here how I should go about this here. I think that the $p(x)$ might be confusing me in this question.
If anyone can give me some pointers, it would be greatly appreciated. Thanks!
AI: Here's how you can claim uniqueness: Fix $f\in C(\Bbb{R})$, and suppose that both $g$ and $q$ satisfy the condition in question. This means we have
$$
\int_0^1\!p(x)f(x)\,dx = \int_0^1\!p(x)g(x)\,dx = \int_0^1\!p(x)q(x)\,dx
$$
for all $p\in P_n(\Bbb R)$. In particular, this property holds for $g(x) - q(x)$ (which is again in $P_n(\Bbb R)$). Then we have
$$
\int_0^1\!\left(g(x) - q(x)\right)g(x)\,dx = \int_0^1\!\left(g(x) - q(x)\right)q(x)\,dx
$$
so that
\begin{align*}
0 &= \int_0^1\!\left(g(x) - q(x)\right)g(x)\,dx - \int_0^1\!\left(g(x) - q(x)\right)q(x)\,dx\\
&= \int_0^1\!\left(g(x) - q(x)\right)g(x) - \left(g(x) - q(x)\right)q(x)\,dx\\
&= \int_0^1\!\left(g(x) - q(x)\right)\left(g(x) - q(x)\right)\,dx\\
&= \int_0^1\!\left(g(x) - q(x)\right)^2\,dx.
\end{align*}
However, $\left(g(x) - q(x)\right)^2$ is continuous and $\left(g(x) - q(x)\right)^2\geq 0$ for all $x\in [0,1]$, so that $\left(g(x) - q(x)\right)^2 = 0$ for all $x\in [0,1]$ (because if $h$ is continuous and nonnegative on $I = [a,b]$ and $\int_a^b\!h = 0$, then $h\equiv 0$ on $[a,b]$). This implies that $g(x) = q(x)$ on $[0,1]$, and since they are polynomials, they must then be the same polynomial.
Alternatively, we can view this same proof in terms of linear algebra language:
$\langle p,q\rangle = \int_0^1 pq$ defines an inner product on $P_n(\Bbb R)$. In the same manner as before, if $g,q$ both satisfy your desired condition, then
\begin{align*}
\langle q - g,q\rangle &= \langle q - g,g\rangle\\
\implies 0 &= \langle q - g,q\rangle - \langle q - g,g\rangle\\
&= \langle q - g,q - g\rangle\\
&= \left|\left| q - g\right|\right|^2,
\end{align*}
which implies that $q - g = 0$, or $q = g$. (Added because you tagged this "linear algebra.") |
H: Integrating $\int^{\infty}_0 e^{-x^2}\,dx$ using Feynman's parametrization trick
I stumbled upon this short article on last weekend, it introduces an integral trick that exploits differentiation under the integral sign. On its last page, the author, Mr. Anonymous, left several exercises without any hints, one of them is to evaluate the Gaussian integral
$$
\int^\infty_0 e^{-x^2} \,dx= \frac{\sqrt{\pi}}{2}
$$
using this parametrization trick. I had been evaluating it through trial and error using different paramatrizations, but no luck so far.
Here are what I have tried so far:
A first instinct would be do something like:$$
I(b) = \int^\infty_0 e^{-f(b)x^2}\,dx
$$
for some permissible function $f(\cdot)$, differentiating it will lead to a simple solvable ode:
$$
\frac{I'(b)}{I(b)} = -\frac{f'(b)}{2f(b)}
$$
which gives:
$$
I(b) = \frac{C}{\sqrt{f(b)}}.
$$
However, finding this constant $C$ basically is equivalent to evaluating the original integral, we are stuck here without leaving this parametrization trick framework.
A second try involves an exercise on the same page:
$$
I(b) = \int^\infty_0 e^{-\frac{b^2}{x^2}-x^2}dx.
$$
Taking derivative and rescaling the integral using change of variable we have:
$$
I'(b) = -2I(b).
$$
This gives us another impossible to solve constant $C$ in:
$$
I(b) = C e^{-2b}
$$
without leaving this framework yet again.
The third try is trying modify Américo Tavares's answer in this MSE question:
$$
I(b) = \int^\infty_0 be^{-b^2x^2}\,dx.
$$
It is easy to show that:
$$
I'(b) = \int^\infty_0 e^{-b^2x^2}\,dx - \int^\infty_0 2b^2 x^2 e^{-b^2x^2}\,dx = 0
$$
by an integration by parts identity:
$$
\int^\infty_0 x^2 e^{- c x^2}\,dx = \frac{1}{2c}\int^\infty_0 e^{- c x^2}\,dx .
$$
Then $I(b) = C$, ouch, stuck again at this constant.
Notice in that Proving $\displaystyle\int_{0}^{\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}$ question, Bryan Yocks's answer is somewhat similar to the idea of parametrization, however he has to introduce another parametric integration to produce a definite integral leading to $\arctan$.
Is there such a one shot parametrization trick solution like the author Anonymous claimed to be "creative parameterizations and a dose of differentiation under the integral"?
AI: Just basically independently reinvented Bryan Yock's solution as a more 'pure' version of Feynman.
Let $$I(b) = \int_0^\infty \frac {e^{-x^2}}{1+(x/b)^2} \mathrm d x = \int_0^\infty \frac{e^{-b^2y^2}}{1+y^2} b\,\mathrm dy$$ so that $I(0)=0$, $I'(0)= \pi/2$ and $I(\infty)$ is the thing we want to evaluate.
Now note that rather than differentiating directly, it's convenient to multiply by some stuff first to save ourselves some trouble. Specifically, note
$$\left(\frac 1 b e^{-b^2}I\right)' = -2b \int_0^\infty e^{-b^2(1+y^2)} \mathrm d y = -2 e^{-b^2} I(\infty)$$
Then usually at this point we would solve the differential equation for all $b$, and use the known information at the origin to infer the information at infinity. Not so easy here because the indefinite integral of $e^{-x^2}$ isn't known. But we don't actually need the solution in between; we only need to relate information at the origin and infinity. Therefore, we can connect these points by simply integrating the equation definitely; applying $\int_0^\infty \mathrm d b$ we obtain
$$-I'(0)= -2 I(\infty)^2 \quad \implies \quad I(\infty) = \frac{\sqrt \pi} 2$$ |
H: Recursion relation for Euler numbers
I am trying to solve the following:
The Euler numbers $E_n$ are defined by the power series expansion
$$\frac{1}{\cos z}=\sum_{n=0}^\infty \frac{E_n}{n!}z^n\text{ for }|z|<\pi/2$$
(a) Show that $E_n=0$ when $n$ is an odd integer.
(b) Establish a recursion relation for the sequence.
Through research it seems that part a seems well established but I couldn't find a proof.
Part (b) I found a formula, but couldn't find how to establish it. Any help appreciated.
AI: For part (a): since $\cos(z)$ is an even function (that is, $\cos(z) = \cos(-z)$), then it follows that $\dfrac{1}{\cos(z)}$ is also even (why?). You should be able to convince yourself that the Taylor series of an even function $f_e(z)$ can't have any odd terms (try differentiating both sides of the identity $f_e(-z) = f_e(z)$ at $z=0$ repeatedly using the chain rule).
For part (b): You know that $\cos(z)\times\dfrac{1}{\cos(z)} = 1$ identically; expand both terms on the LHS in terms of their Taylor series, multiply them out using the usual method for multiplying series, and then set the left and right sides of the equation to be equal as Taylor series — that is to say, look at the terms in $z^n$ on both sides (note that on the RHS, that term is $1$ if $n=0$ and $0$ otherwise). You should find that this gives you a recursive formula for the Euler numbers. |
H: Calculus II, Curve length question.
Find the length of the curve
$x= \int_0^y\sqrt{\sec ^4(3 t)-1}dt, \quad 0\le y\le 9$
A bit stumped, without the 'y' in the upper limit it'd make a lot more sense to me.
Advice or solutions with explanation would be very appreciated.
AI: $$\frac{dx}{dy} = \sqrt{\sec^4{3 y}-1}$$
Arc length is then
$$\begin{align}\int_0^9 dy \sqrt{1+\left ( \frac{dx}{dy} \right )^2} &= \int_0^9 dy\, \sec^2{3 y} \\ &= \frac13 \tan{27} \end{align}$$ |
H: Annulus Theorem
I'm trying to read Rolfsen's "Knots and Links" and I'm a little discouraged that I can't do one of the first and seemingly more important exercises. The question is
Use the Schoenflies theorem (every topological imbedding $S^1 \rightarrow \mathbb{R}^2$ is the boundary of a 2-disk) to prove the annulus theorem: Given two disjoint imbeddings of $S^1$ in $\mathbb{R}^2$, where one is "inside" the other (I.e. one is in the bounded component of the complement of the other), then the space "between" the two curves is homemorphic to $S^1 \times [0,1]$.
There is a hint that says "connect the two curves with arcs". Unfortunately I can't say much about "what I've tried." At this point in the book, we know only that all simple closed curves in $\mathbb{R}^2$ or $S^2$ are "equivalent" in the sense that there exists a homeomorphism of the ambient space taking one to the other. This allows us to assume one of the curves is the standardly imbedded $S^1$. After that I'm totally unsure of what to do. I would appreciate any hints or advice. Thank you for reading my question.
AI: I haven't thought this through carefully, but just following the hint:
choose a point $a$ on the outer cirlce and a point $b$ on the inner circle. Join $a$ and $b$ with an arc that doesn't cross the two circles. Now "thicken up" this arc very slightly (and thicken up each of $a$ and $b$ to a very short piece of circular arc on their respective circles) so that now we have to the two circles with a very thing rectangle sitting between them. (I think that making this "thickened" picture rigorous is probably part of what is involved in translating the hint into a full argument.)
Now the consider the boundary of the union of the two circles and this rectangle. If you are picturing what I want you to picture, you will see that it is a circle. (Just to be sure: if you take an annulus and snip out a small rectangle joining the inner and outer annulus, you get a kind of "sliced" annulus which is homeomorphic to a disk with circle as boundary. But you can verify that the boundary of the region I am describing is a circle without knowing that the region inbetween the two circles is an annulus just follow the outer circle around from the top left corner of the rectangle all the way around to the top right corner, then move along the right boundary of the rectangle to its bottom right corner, now go back around the inner circle to the bottom left corner of the rectangle, and now go back up the left edge of the rectangle to get back where you started: you've described a circle.) So it bounds a disk.
So the region between the two circles, with a thin rectangle removed, is a disk. Gluing back in the rectangle should make into an annulus, as required. |
H: Arc length parameter s
Consider the metric $$ds^2 = \frac{dx^2+dy^2}{y^2}.$$
Assume $R>0, a\in\mathbb{R}$. Consider the curve $$\gamma(\theta)=(a+R\sin\theta,R\cos\theta)$$
for $-\frac{\pi}{2}\leq\theta\leq \frac{\pi}{2}$. Find the arclength parameter s. How would I go about doing this? Thank you in advance!
AI: Since for $y>0$,
$$ds^2=\frac{dx^2+dy^2}{y^2} \iff ds=\frac{\sqrt{(dx/d\phi)^2+(dy/d\phi)^2}}{y}d\phi,$$
the arclength parameter is
$$s(\theta)=\int_0^\theta ds=\int_0^\theta \frac{\sqrt{\left(\frac{d}{d\phi}(a+R\sin\phi)\right)^2+\left(\frac{d}{d\phi}(R\cos\phi)\right)^2}}{R\cos\phi}d\phi.$$
This boils down to finding the antiderivative of $\sec\phi$. |
H: Perfect squares using 20 1's, 20 2's and 20 3's.
How many perfect squares can be formed using 20 1's, 20 2's and 20 3's. This is a recent exam question, which I had no clue how to solve?
There is some kind of trick here, since time allotted to solve it was just 4 minutes.
AI: Hint: The resulting number is divisible by $3$ but not by $9$, since the digit sum is $120$. |
H: Linearly independent subset?
If $u,v,w$ and $z$ are distinct elements in $R$, then
{$(1,u,u^2,u^3), (1,v,v^2,v^3), (1,w,w^2,w^3), (1,z,z^2,z^3)$} is a linearly independent subset of $R^4$.
Is that true or false?
I even can't start the first step..
AI: If they are linearly dependent, then the equation
$$
\underbrace{\pmatrix{1&u&u^2&u^3\\ 1&v&v^2&v^3\\ 1&w&w^2&w^3\\ 1&z&z^2&z^3}}_{A}
\pmatrix{a\\ b\\ c\\ d}=\pmatrix{0\\ 0\\ 0\\ 0}
$$
has a nontrivial solution $(a,b,c,d)$, i.e. there is a nonzero real polynomial $a+bx+cx^2+dx^3$ of degree $\le3$ that has at least four distinct roots $u,v,w,z$. This is impossible.
As pointed out by the others, The matrix $A$ in the above displayed equation is called a Vandermonde matrix. As we have demostrated, showing that it is invertible is easy (and the above proof works for any field, not just $\mathbb{R}$). The tricky part, though, is to find its determinant or inverse. |
H: Is it a standard to say that $a \oplus a_{\small 1}=0$ or $a \veebar a_{\small 1}=0$?
I am trying to express the following:
$a$ or $a_{\small 1}=0$ but only one of them equals zero.
so if $a=0$ then $a_{\small 1}\neq 0$ and if $a\neq 0$ then $a_{\small 1}=0$.
And I'm looking for a standard way to say that, from my little background in logic, I remember the logical operator xor which truth table fit my purposes. So is it ok to use any of the following notations?
$$a \oplus a_{\small 1}=0$$
$$a \veebar a_{\small 1}=0$$
AI: Both symbols are variously used to denote XOR
But it would need to be expressed thusly:
$$(a = 0) \oplus (a_1 = 0)$$
or
$$(a = 0) \veebar (a_1 = 0)$$
And NOT, e.g., $$a \oplus a_1 = 0$$ |
H: Curvature of a non-compact complete surface
Assume $\Sigma$ is a non compact, complete surface. Assume the integral $$\int_{\Sigma}K$$ is convergent, where K is the Gauss curvature of $\Sigma$. Is it always true that $$\frac{1}{2\pi}\int_{\Sigma}K$$ is an integer?
AI: No. Start with a one-nappe circular cone. Cut off the tip and replace with a subset of a sphere, resulting in what is now a $C^1$ infinite surface. Smooth near the circle where they join, so the result is at least $C^2.$ The result has zero curvature except near the spherical part. Since we did not specify the central angle of the cone, we get whatever we like in the way of total Gauss curvature, up to an effective maximum given by a hemisphere and an infinite cylinder. |
H: Prove that $n^2 \leq \frac{c^n+c^{-n}-2}{c+c^{-1}-2}$.
Let $c \not= 1$ be a real positive number, and let $n$ be a positive integer.
Prove that $$n^2 \leq \frac{c^n+c^{-n}-2}{c+c^{-1}-2}.$$
My initial thought was to try and induct on $n$, but the negative powers annoyed be badly.
I still think that induction is the way to go, but maybe it is something more clever than I am doing. Although perhaps induction is unhelpful, and I simply need to use inequalities to do this, like AM-GM and Cauchy-Schwarz. But even there I get very stuck. I have a feeling that there is some neat thing I should do to make this a lot simpler.
Any help would be really appreciated. Thanks in advance.
AI: Hint: the RHS is $$\frac{-(1-c^n)(1-c^{-n})}{-(1-c)(1-c^{-1})}$$
To clarify my bonus hint, we may rearrange the AM-HM inequality to state $$(a_1+a_2+\cdots+a_n)(\frac 1a_1+\frac 1a_2+\cdots+\frac 1a_n)\ge n^2$$ |
H: Inverse and derivative of a function
Find an example of an inverse function f(x) such that its derivative is the same as its inverse.
I tried many different functions but non of them worked.
AI: We find a function defined for all positive $x$, with the right properties.
There may be no general theory, so let's fool around and look for a function of the shape $f(x)= cx^a$, for constants $a$ and $c$.
Then the derivative of $f$ is given by $f'(x)=acx^{a-1}$. The inverse function is given by $f^{-1}(x)=\frac{x^{1/a}}{c^{1/a}}$. So we need
$$acx^{a-1}=\frac{x^{1/a}}{c^{1/a}}.$$
For the the identity above to hold, we want $a-1=\frac{1}{a}$. Rewrite as $a^2-a-1=0$, which has $a=\frac{1+\sqrt{5}}{2}$ as a root. Nice number!
We also want $ca=\frac{1}{c^{1/a}}$. There is such a $c$. It is given by
$$c=\left(\frac{\sqrt{5}-1}{2}\right)^{1/\sqrt{5}}.$$ |
H: Homomorphism from $\mathbb{Z}/n\mathbb{Z}$
Does there exist a non-zero homomorphism from $\mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Z}$ ? If yes, state the mapping. How is this map exactly?
AI: Suppose there is a homomorphism $f: \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}$. What could $f(1)$ be? Let's call it $a$. Then $f(2) = f(1+1) = f(1)+f(1) = a+a = 2a$, and likewise $f(3) = 3a$, $\dots,$ $f(\underbrace{1+\cdots+1}_{n\text{ times}}) = na$. But $\underbrace{1+\cdots+1}_{n\text{ times}} = 0$, and $f(0) = 0$. So what can $a$ be?
Once you've worked out which value(s) of $a$ is/are allowed, can you check that the associated map $f$ is a homomorphism? (We've already done this, really, but make sure you're sure of that.) |
H: Systems of Differential Equations and higher order Differential Equations.
I've seen how one can transform a higher order ordinary differential equation into a system of first-order differential equations, but I haven't been able to find the converse. Is it true that one can transform any system into a higher-order differential equation? If so, is there a general method to do so?
AI: If I am understanding your question, you just would reverse the process on the last equation from the system.
An $n^{th}$ order differential equation can be converted into an $n$-dimensional system of first order equations.
There are various reasons for doing this, one being that a first order system is
much easier to solve numerically (using computer software) and most differential equations you encounter in “real life” (physics, engineering etc) don’t have nice exact solutions.
If the equation is of order $n$ and the unknown function is $y$, then set:
$$x_1 = y, x_2 = y', \ldots , x_n = y^{n−1}.$$
Note (and then note again) that we only go up to the $(n − 1)^{st}$ derivative in this process. Lets do an example in both directions (practice some where a known system has been converted to such a system and make sure you can work backward).
Forward Approach
$$\tag 1 y^{(4)} - 3y' y'' + \sin(t y'') -7ty^2 = e^t$$
Let: $x_1 = y, x_2 = y', x_3 = y'', x_4 = y'''$ and substitute into $(1)$, yielding:
$x_1' = y' = x_2$
$x_2' = y'' = x_3$
$x_3' = y''' = x_4$
$x_4' = y^{(4)} = 3yy''-\sin(ty'')+7ty^2+e^t = 3x_2x_4-\sin(tx_3)+tx_1^2+e^t$
Backward Approach
Looking at the last equation from the system, we let: $y = x_1, y' = x_2, y''=x_3, y'''=x_4$ and substitute into the system's last equation above, yielding:
$y^{(4)} = 3y'y''-\sin(ty'')+7ty^2+e^t$ |
H: Trigonometric equality $x = 99 \sin (\pi x)$
Find the number of real solutions of $\displaystyle x = 99 \sin (\pi x)$.
I am getting stuck in some trigonometric relations.
AI: There is an obvious root at $x=0$. Since $\sin(-\pi x)=-\sin(\pi x)$, there are just as many negative solutions as there are positive solutions.
So we count the positive solutions, double the result, and add $1$. Now we show how to count the positive solutions. We will use a picture to do the count.
Rewrite the equation as
$$\sin(\pi x)=\frac{x}{99}.$$
For $x\gt 99$, the right-hand side is $\gt 1$, while the left is always $\le 1$. So we will not have to look for solutions beyond $x=99$.
To count the positive solutions $\lt 99$, draw the graph of $y=\sin(\pi x)$. We need to go a moderately long way. Note that $\sin(\pi x)$ has period $2$. So when we go up to $x=99$, there will be $49$ full periods, plus a half period.
Let's take a look at $y=\frac{x}{99}$. This is a line with very shallow slope. It enters the first period of $y=\sin(\pi x)$ and leaves it, giving $2$ intersections, one in the interval $(0,\pi/2)$, the other in the interval $(\pi/2,\pi)$. It enters and leaves the second period ($2$ more solutions). Continue. You will have to think a little about what happens in the final half-period.
Remark: A graphing calculator may help, but you will need to play with the viewing window. |
H: Schema of separation and set of all sets
The schema of separation states (this is probably simplified) states that if $P$ is a property and $X$ is a set, then there exists a set $Y = \{ x \in X : P(x)\}$.
The notes I'm reading say that from this we can conclude that the set of all sets doesn't exist, by applying the schema of separation to the property $x \notin x$.
I'm not seeing how applying the schema of separation to that property shows it. Can someone clarify this?
AI: Let $V$ be the collection of all sets, and assume that $V$ was a set, let $W=\{x\in V\mid x\notin x\}$, then by the separation schema $W$ is a set.
Now ask yourself, $W\in W$ or $W\notin W$?
If $W\in W$ then $W$ satisfies $W\notin W$, which is a contradiction to the fact that $W\in W$.
If $W\notin W$ then $W$ does not satisfy $W\notin W$, which is again a contradiction!
Therefore $W$ cannot be a set, but this can only mean that $V$ cannot be a set to begin with.
This is also known as Russell's paradox. |
H: Number of real roots of $\sum_{k=0}^{n}\frac{x^{k}}{k!}=0$
Prove the following, without induction. Is it possible?
The equation $\sum_{k=0}^{n}\frac{x^{k}}{k!}=0$ has no real root if $n$ is even.
And if $n$ is odd, it has only one real root.
I also tried searching the proofs many times with search key words "number of real root" or "exponential function" or "$\sum_{k=0}^{n}\frac{x^{k}}{k!}$".
But failed. What is the way to search the related topics?
AI: Note that, $f'(x)+\frac{x^k}{k!}=f(x).$ If $k$ is even and $x_0$ is the smallest root of $f,$ then $f'(x_0)<0.$ So immediately after $x_0$ $f(x)<0$ and the function decreases. Thus, in order for the function $f$ to have its second zero which it must have as a polynomial of even degree (the multiplicity of each root is clearly one) it has to increase. However, if the function $f$ increases at a point then $f(x)>0$ (follows from our identity). This shows that our assumption is wrong. So, $f$ does not have a root. Once even case is established, odd case follows immediately. |
H: power of a number in a factorial
what is the largest power of 24 in 150! ?
HINT : answer is 48
I need to know the method for solving such questions when the highest power of the number to be found is non-prime ..
AI: You have to compute the highest power of each prime factor
of the number you want
(for 24, these are 2 and 3)
and then for each of these,
compute how many times
that power of the primes divide the factorial
(for 24 it is $2^3$ and $3^1$),
and choose the smaller.
For your case,
2 goes 75+37+18+9+4+2+1=146,
so $2^3$ goes $[146/3] = 48$.
3 goes 50+16+5+1 = 73.
The smaller of these is 48. |
H: How to calculate $\sum_{i=2}^n {\frac 1{\log_2 i}}$
How to calculate
$$\sum_{i=2}^n {\frac 1{\log_2 i}}$$
AI: If you meant $$\sum_{i=2}^n \dfrac1{\log_2(i)},$$ then there is no closed form; though you can compute its asymptotic using Euler-Maclaurin formula:
$$\sum_{k=2}^{n} \dfrac1{\log_2(k)} \sim \log 2\times\int_2^n \dfrac{dx}{\log(x)} \sim \dfrac{n}{\log_2(n)}$$
However, if you meant $$\sum_{i=2}^n \dfrac1{\log_2(n)}$$ note that the term inside the summation is independent of $i$ and hence equals (exactly) $$\dfrac{n-1}{\log_2(n)}$$ |
H: A basic doubt on the sojourn time of a CTMC
By using the memoryless property, I find the CDF of sojourn time of a CTMC as follows :
$$F_X(t) = 1-e^{-F_X'(0)t}$$
I am slightly confused about the intuitive meaning of the term $F_X'(0)$. How this term represents the rate of leaving that state.
AI: These are the basics of the exponential distributions, probably explained on the WP page.
If $T$ is exponential with parameter $a$, then $F_T(t)=1-\mathrm e^{-at}$ for every $t\geqslant0$ hence $a=F_T'(0^+)$ and $F_T(t)=1-\mathrm e^{-F_T'(0^+)t}$.
Furthermore, for every $t\geqslant0$, $P[t\leqslant T\leqslant t+s\mid T\geqslant t]=1-\mathrm e^{-as}\sim as$ when $s\to0$ hence, indeed, $a$ represents the rate of $T$ happening at $t$. Exponential distributions are characterized by the fact that this rate does not depend on $t\geqslant0$. |
H: Making a cube root function analytic on $\mathbb{C}\backslash [1,3]$
I am still not convinced by the post that the function$$\sqrt[3]{(z-1)(z-2)(z-3)}$$ can be defined so it is analytic on $\mathbb{C}\backslash [1,3]$. We define for each $z\in \mathbb{C}\backslash (-\infty,3]$ the function $$f(z)=\int_4^z \frac{((z-1)(z-2)(z-3))'}{(z-1)(z-2)(z-3)}\,dz +\ln 6$$ and it claims that function $$\exp\left(\frac{1}{3}f(z)\right)$$ is continuous on $(-\infty,1)$. This is the part I don't understand. How do we prove that it is continuous on $(-\infty,1)$. I computed the integral from 4 to $-4$ along the upper semicircle of $|z|=4$, and the integral from 4 to $-4$ along the lower semicircle of $|z|=4$ and they are not equal. So I don't see how the function can be continuous at $-4$, for example.
AI: Given that
$$\frac{((z-1)(z-2)(z-3))'}{(z-1)(z-2)(z-3)}=\frac1{z-1}+\frac1{z-2}+\frac1{z-3}$$ you see that all the three residues of the integrand are equal to one, so their sum is $3$. The "upper" and "lower" integrals thus differ by $6\pi i$.
After multiplication by $1/3$ they differ by $2\pi i$, and after $\exp$ they agree. IOW $f(z)$ won't be continuous, but $\exp(f(z)/3)$ still is. |
H: Proving an equality
Let $f(n) = n^ {\log n}$. Let $p(n)$ and $q(n) \geq n$ be polynomials. I want to show that for
$n$ sufficiently large $f (n)$ satisfies
$$p(n) < f (n) < 2^{q(n)}$$
starting from the above inequality doesn't yield any satisfying result.
AI: Since $p(n)$ is a polynomial,
$p(n) < c n^d$ where $d$ is the degree of $p$
and $c$ is the sum of the absolute value of its coefficients.
So $p(n) < n^{d+1}$ for $n > c$.
Therefore,
if $\log n > d+1$,
$p(n) < f(n)$.
For the upper bound,
$f(n) = n^{\log n}
= e^{\log^2 n}
= 2^{(\log^2 n)/\log 2}
$,
so any polynomial $q$ such that
$q(n) > (\log^2 n)/\log 2$
will work.
Since $(\log^2 n)/n \to 0$ as $n \to \infty$,
any polynomial will work. |
H: Reducibility of $x^{2n} + x^{2n-2} + \cdots + x^{2} + 1$
Just for fun I am experimenting with irreducibility of certain polynomials over the integers. Since $x^4+x^2+1=(x^2-x+1)(x^2+x+1)$, I thought perhaps $x^6+x^4+x^2+1$ is also reducible. Indeed:
$$x^6+x^4+x^2+1=(x^2+1)(x^4+1)$$
Let $f_n(x)=x^{2n}+x^{2n-2}+\cdots + x^2+1$. Using Macaulay2 (powerful software package) I checked that:
$$f_4(x) = (x^4-x^3+x^2-x+1)(x^4+x^3+x^2+x+1)$$
We get that
$$ f_5(x)=(x^2+1)(x^2-x+1)(x^2+x+1)(x^4-x^2+1)$$
The polynomial is reducible for $n=6, 7, 8, 9$ as far as I checked. I suspect that $f_n(x)$ is reducible over integers for all $n\ge 2$. Is this true?
Thanks!
AI: Note that $f_n(x)(x^2-1) = x^{2n+2} - 1 = (x^{n+1}-1)(x^{n+1}+1)$. Now use unique factorization... |
H: Infinitely many primes of the form $4n+3$
I've found at least 3 other posts$^*$ regarding this theorem, but the posts don't address the issues that I have.
Below is a proof that for infinitely many primes of the form $4n+3$, there's a few questions I have in the proof which I'll mark accordingly.
Proof: Suppose there were only finitely many primes $p_1,\dots, p_k$, which are of the form $4n+3$. Let $N = 4p_1\cdots p_k - 1$. This number is of the form $4n+3$ and is also not prime as it is larger than all the possible primes of the same form. Therefore, it is divisible by a prime $ \color{green}{ \text{(How did they get to this conclusion?)}}$. However, none of the $p_1,\dots, p_k$ divide $N$. So every prime which divides $N$ must be of the form $4n+1$ $ \color{green}{ \text{(Why must it be of this form?)}}$. But notice any two numbers of the form $4n+1$ form a product of the same form, which contradicts the definition of $N$. Contradiction. $\square$
Then as a follow-up question, the text asks "Why does a proof of this flavor fail for primes of the form $4n+1$? $ \color{green}{ \text{(This is my last question.)}}$
$^*$One involves congruences, which I haven't learned yet. The other is a solution-verification type question. The last one makes use of a lemma that is actually one of my questions, but wasn't a question in that post.
AI: Every number $n>1$ is divisible by some prime $p$ (which includes the case $n=p$). Assume otherwise and let $n$ be the smallest such number. As this $n$ is not prime, it has a nontrivivial divisor $d$ with $1<d<n$. By minimality of $n$, $d$ is divisible by some prime $p$. But then $p$ also divides $n$.
All numbers are of the form $4n$, $4n+1$, $4n+2$, or $4n+3$.
This is also true for primes $p$, but $p=4n$ is not possible and $p=2n$ only for $p=2$. Here, we have excluded $p=2$ as well as $p=4n+3$ by construction, which leaves only primes $p=4n+1$.
This proof fails for $p=4n+1$ because a number of the form $4n+1$ may well be the product of two numbers of the form $4n-1$. For example $3\cdot 7=21$. Therefore the step that at least one divisor must be of form $4n+1$ fails. |
H: How do you graph $x + y + z = 1$ without using graphing devices?
How can I graph $x + y + z = 1$ without using graphing devices?
I equal $z = 0$ to find the graph on the xy plane. So I got a line, $y = 1-x$
But when I equal 0 for either the $x$ or the $y,$ I get $z = 1-y$ or $z = 1-x$ , and those are two different lines from different angles. Different graphing websites were telling me different answers...
Please don't show some crazy and complicated methods to graph this. I just want simple steps just as plugging $x,y,z$ as zeros and etc.
AI: I hope you don't mind if I post here a silence solution. I think it is telling you everything, however, other nice answers give you the theoretical points. :-) |
H: Differentiation Operator a Contraction Mapping
Let $C^{\infty}[a,b]$ be the space of all infinitely differentiable functions on [a,b] with norm $$ || f || = \max _{[0,1]} | f(x) | , f \in C^{\infty}[a,b]$$
Is the differentiation operator $\frac{d}{dx}$ a contraction mapping on $C^{\infty}[a,b]$?
I'm confused. Operators are not until later in the textbook, and the contraction mappings I've worked with are using the Mean Value Theorem which are referencing a derivative? Ah!
AI: It's not a contraction.
Let $f(x)=\sin nx$, $g(x)=\cos nx$ ($n$ is a number we need to find). Then $f,g\in C^{\infty}[a,b]$. And
\begin{align*}
\left\|f-g\right\|=&\underset{[a,b]}{\max}|\sin nx-\cos nx|=\sqrt{2}\underset{[a,b]}{\max}|\sin(nx-\frac{\pi}{4})|
\\
\left\|\frac{df}{dx}-\frac{dg}{dx}\right\|=&\left\|n\cos nx+n\sin nx\right\|=n\underset{[a,b]}{\max}|\cos nx+\sin nx|=\sqrt{2}n\underset{[a,b]}{\max}|\sin(nx+\frac{\pi}{4})|
\end{align*}
If we take $n$ large enough that $n\ge\frac{2\pi}{b-a}$, then
\begin{align*}
\left\|f-g\right\|=\sqrt{2},\qquad\left\|\frac{df}{dx}-\frac{dg}{dx}\right\|=\sqrt{2}n.
\end{align*}
If we also make $n\ge1$,
$$\left\|\frac{df}{dx}-\frac{dg}{dx}\right\|\ge\left\|f-g\right\|.$$
Therefore the operator $d/dx$ is not contraction. |
H: Finitely additive measure on $\mathbb R$
Suppose $\mathcal B$ is the Borel $\sigma$-algebra on $\mathbb R$. Let $\mu : \mathcal B \rightarrow [0, \infty ]$ be a finitely additive(but not necessarily countably additive), translation-invariant, ''measure'', with the property that $\mu(K) < \infty$ for every compact $K \subset \mathbb R $.
Let $P(\mathbb R)$ denote the power set of $\mathbb R$.
Does there exist a finitely additive, translation-invariant "measure", $\tilde \mu : P(\mathbb R) \rightarrow [0, \infty ]$, extending $\mu$ ?
In particular, let $\mu$ be the Lebesgue measure on the Borel $\sigma$-algebra. Can something be said about this special case at least?
AI: I don't know about the general case. But here Tao states that the Lebesgue case is true, using the Hahn-Banach theorem. |
H: Find the greatest integer $k$ for which $1991^k$ divides $1990^{{1991}^{1992}}+1992^{{1991}^{1990}}$
Find the greatest integer $k$ for which $1991^k$ divides $$1990^{{1991}^{1992}}+1992^{{1991}^{1990}}$$
It is easy to see that $k \geq 1$ as $1990 \equiv -1$ and $1992 \equiv 1 \pmod{1991}$
Also, I thought that perhaps as $1991$ is the product of two distinct primes, it would be worth looking at small values of $(pq)^k||(pq-1)^{{pq}^{pq+1}}+(pq+1)^{{pq}^{pq-1}}$ for primes $p$ and $q$. Any help would be really appreciated.
AI: I shall prove a more general result: Let $n>1$ be an odd positive integer. Then $n^n \|[(n-1)^{n^{n+1}}+(n+1)^{n^{n-1}}]$.
Proof:
\begin{align}
(n-1)^{n^2}+(n+1)=\sum_{i=0}^{n^2}{\binom{n^2}{i}(-1)^{n^2-i}n^i}+(n+1) & \equiv \binom{n^2}{1}n-1+(n+1) \pmod{n^2}\\
& \equiv n \pmod{n^2}
\end{align}
Now applying Lifting the Exponent Lemma on each prime factor of $n$, we have $n^n \|[(n-1)^{n^{n+1}}+(n+1)^{n^{n-1}}]$.
For your special case, $k=1991$. |
H: Integral of fractional expression $\int^3_0 \frac{dx}{1+\sqrt{x+1}}$
I want to solve this integral and think about call $\sqrt{x+1} = t \rightarrow t^2 = x+1$
$$\int^3_0 \frac{dx}{1+\sqrt{x+1}}$$
Now the integral is : $$\int^3_0 \frac{2tdt}{1+t}$$ now I need your suggestions.
Thanks.
AI: The substitution is a good idea. The details are not quite right. We get $dx=2t\,dt$, and therefore our definite integral is equal to
$$\int_{t=1}^2 \frac{2t}{1+t}\,dt.$$
Now can you take over? It may be useful to note that $\frac{2t}{1+t}=2-\frac{2}{1+t}$. |
H: Trace of an Inverse Matrix
I want to know if there is a way to simplify, or a closed form solution of $tr(\Sigma^{-1})$ where $\Sigma$ is a symmetric positive definite matrix.
AI: Let $A$ be symmetric positive definite matrix hence $\exists$ a diagonal matrix $D$ whose diagonal entries are nonzero and $A=P D P^{-1}$ so $A^{-1} = P D^{-1} P^{-1}$ and $Tr(A^{-1})= Tr(D^{-1})$. Now $D$ being diagonal matrix with non zero diagonal entries $D^{-1}$ has diagonal entries reciprocal of the diagonal entries of $D$ so $Tr(D^{-1})$ is sum of the inverses of the diagonal entries of $D$. |
H: If $A$ is invertible show that $\det(A) \neq 0$
If $A \in M_{n\times n}(F)$ is not invertible then the rank of $A$ is less than $n$,
thus $\det(A) =0$.
I proved that way, but looks like too simple so I think maybe there is a trick that I missed.
Or I proved it right?
Modification:
($\Rightarrow$) If $A$ is invertible, $AB=I$, $\det(AB)=\det(A)\det(B)=1$ so $\det(A) \neq 0$
($\Leftarrow$) If $\det(A) \neq 0$ , by Cramer's rule, $Ax=b$ has a unique solution and it means $A$ is invertible ($x=A^{-1}b$)
Now I think I complete the proof.
AI: Your proof of the $\Rightarrow$ implication is right. For the other one, do you know about the Adjugate Matrix? To each $A\in M_{n\times n}(F)$ we can associate a matrix $adj(A)$ defined as the tranpose of the cofactor matrix of $A$ such that one always has $A\cdot adj(A)=\det (A) I_{n\times n}$. If $\det(A)\neq 0$ then $\det(A)$ is invertible and therefore you get $A^{-1}=(\det(A))^{-1}adj(A)$.
Notice that the result can be actually strengthened: given a unitary commutative ring $R$, a matrix $A\in M_{n\times n}(R)$ is invertibile if and only if its determinant is invertible in $R$. The proof of this fact goes exactly as the former one just by noticing that what we really used previously was the fact that any non zero element in a field is invertible, as I underlined. |
H: Integrate over the region bounded by two regions. Using Polar coordinates.
Using Polar Coordinates integrate over the region bounded by the two circles:
$$x^2+y^2=4$$
$$x^2+y^2=1$$
Evaluate the integral of $\int\int3x+8y^2 dx$
So what I did was said that as $x^2+y^2=4$ and $x^2+y^2=1$
That $1 \le r \le 2$. And as there is a symmetry in the four quadrants
$0 \le \theta \le \frac{\pi}{2}$
which gave me $\int_0^\frac{\pi}{2}\int_1^2 3r^2\cos(\theta) +8r^3\sin^2(\theta) ~dr d\theta$
The answer it gives in the book is $30\pi$.
I'm getting
$28 +30\pi$
AI: There is less symmetry than you think because of the $x$. Well, there is symmetry there too, but it is cancellation symmetry: the contribution of $x$ to the integral is $0$. You can either note that, or integrate from $0$ to $2\pi$, or integrate from $0$ to $\pi$ and double.
Note that we need to replace $dx\,dy$ by $r\,dr\,d\theta$. |
H: Find limits for functions of natural numbers
So I am dealing with some problems about random graphs where we find limits of functions of natural numbers. A simple example can be the limit of $\ln(n)$ as $n \rightarrow \infty$, where $n$ is the number of vertices of a graph. My concern is how do we intepret this type of limit? Can we view it as a special type of limits in the calculus setting, where we have the variable $x \in \mathbb{R}$? In particular, can we still apply tools from calculus (say the l'Hospital's rule) to find limits of $f(n)$, where $n \in \mathbb{N}$?
AI: Well, if we know that $f(x)$ converges as $x \to \infty$, then it doesn't matter how we get there (we can take $ x_n = n$). But if we aren't sure there can be problems. For example $\lim_{n \to \infty} \sin{n\pi x} = 0$, so if we know convergence "in $\mathbb{N}$" it doesn't necessarily buy us anything in the real sense. You aren't sampling all the possible sequences, in a sense.
So yes, if we know a limit exists, we can apply calculus tools. But if we aren't sure, we might get gobbledygook. |
H: Easy question about $H_0^1$ space
I have some trouble with proper understanding of $H_0^1(0,1)$ space. Consider the following space $$H_D = \{u\in H^1(0,1): u(0) = u(1) = 0\}.$$ What can we say about the connection between $H_D$ and $H^1_0(0,1)$. Is $H_D$ in $H^1_0(0,1)$? In literature stays, that functions in $H_0^1(0,l)$ are interpreted as $$''u = 0\; \text{on}\; \partial\Omega''.$$
AI: Do you know anything about the trace operator? There is a theorem that says that if $U$ is a bounded domain and $\partial U$ is $C^1$ and $u \in W^{1,p}$, then
$$
u \in W_0^{1,p}(U) \iff Tu=0 \text{ on } \partial U,
$$
where $T$ is the trace operator. You want $p = 2$ here. |
H: How to express the sum of a set?
Suppose I have a set of numbers. How can I express in set-theory terms the sum of the elements in that set?
AI: There is no set-theoretic notation for that since a sum of numbers (presumably real or complex numbers, but more generally could be elements in a ring) is not a set-theoretic notion.
If the numbers you have are indexed by some set, that is if you have a function $f:I\to \mathbb R$ (just taking the reals here to simplify things), then (if $I$ is finite), the sum of the multi(!)set $\{f(i)\mid i\in I\}$ is commonly denoted by $\sum_{i\in I}f(i)$. If $I$ is infinite then some limiting notion is required to make sense of the sum. |
H: Probability that subsets intersect
Given a set $N$ I would like to calculate the probability that two arbitrarily chosen and equally likely subsets $K\subseteq N$ and $J\subseteq N$ both of fixed size intersect. Let's say $n=\#N$, $k=\#K$, and $j=\#J$ denote the number of elements of the respective sets. The probability should be called $\rho(p,j,k)$ and so far I think I have two formulas which do the job.
The first is recursively defined as:
$$ \rho(p,j,k)=\begin{cases}\frac{k}{n}+\frac{n-k}{n}\cdot\rho(j-1,k,n) & \text{if $j$>0},\\ 0 & \text{otherwise}.\end{cases} $$
The idea is that taking the first element of $J$ there is a chance of $\frac{k}{n}$ that it will be an element of $K$. If not, i.e. with a probability of $1-\frac{k}{n}=\frac{n-k}{n}$ there might be the probability that the next element will be in $K$ and so on.
I have an explicit formula as well:
$$ \rho(p,j,k)=\frac{k}{n}+\frac{k(k-1)}{n^2}+\cdots+\frac{k(k-1)\cdots(k-j+1)}{n^j}$$
The idea here is that either one element of $J$ is in $K$ (Probability: $\frac{k}{n}$) or two elements of $J$ are in $K$ (Probability: $\frac{k(k-1)}{n^2}$) and so on...
Now I have two questions:
First, I think I need to assume that $j\le k$ and I would like to not put any restrictions on the sizes of the sets.
Second, the problem seems very general so I wonder whether there is some known obvious result to it which I am overlooking.
AI: I suppose that $k=\#K$ and $j=\#J$ are given. The number of all $J$'s is $\binom{n}{j}$, the number of those $J$'s that don't intersect $K$ is $\binom{n-k}{j}$, so the probability that $K$ and $J$ don't intersect is $\binom{n-k}{j}/\binom{n}{j}$.
edit: oops, you actually know the answer - but I'll leave it here anyway, just in case someone finds it useful |
H: Uniqueness of "Punctured" Tubular Neighborhoods (?)
Here is a question that has been haunting me for a while: Let $\mathbb{R}^{n-1} \times [0, \infty)$ be the upper half space of $\mathbb{R}^n$ and suppose we have a smooth homeomorphism (not a diffeo)
$f : \mathbb{R}^{n-1} \times [0, \infty) \rightarrow \mathbb{R}^{n-1} \times [0, \infty)$
such that $f$ is the identity on $\mathbb{R}^{n-1} \times \{0\}$ and $f$ restricts to a diffeomorphism from the punctured (at the origin) half-space to itself. In other words, $f$ is the identity on the boundary and it would be a diffeomorphism from the half-space to itself except that the derivative is singular at the origin.
Question : Is the restriction of $f$ to the punctured half space smoothly isotopic to the identity?
A little background: If $Df$ were nonsingular at 0 this would just be an application of the uniqueness of collars, which is a special case of the uniqueness of tubular neighborhoods. That proof uses the morse lemma to write the coordinates of the imbedding as a linear combination of coordinate functions. The coefficients are given by smooth functions whose values at zero are the entries of the Jacobian of the imbedding at the origin. The isotopy one constructs takes the imbedding to the linear map $Df_0$. The issue here is of course that this linear map is degenerate. But perhaps we can construct a different isotopy if we remove the bad point?
This is mainly motivated by my (borderline obsessive and unhealthy) thinking about this question: Different definitions of handle attachment
Thanks for reading my question and I hope It's not too cranky.
AI: I don't really see the connection to your handle attachment question -- it looks to me like you're approaching it as a more general problem than it actually is.
But this question has a reasonable answer. Think of your half-space as sitting inside the Euclidean space $\mathbb R^n$. Then you can intersect the half-space with spheres of radius $r$ centred around the origin. This converts your punctured half-space (via a diffeomorphism) to:
$$ D^{n-1} \times \mathbb R $$
So you are studying the group of diffeomorphisms of this manifold where the diffeomorphism restricts to the identity on the boundary. I think in full generality this space is not connected. Aside from dimension $4$ I think this space is known to have the homotopy-type of $\Omega Diff(D^n)$, where $Diff(D^n)$ is the group of diffeomorphisms of the $n$-disc $D^n$ which restrict to the identity on the boundary.
The basic idea for why you'd expect something like this is that a diffeomorphism of $D^{n-1} \times \mathbb R$ restricts to an embedding of $D^{n-1} \times \{0\}$ into $D^{n-1} \times \mathbb R$. That map is a fibre bundle. There's a standard sequence of relations between this embedding space and $Diff(D^n)$, it's outlined at the end of Hatcher's paper on the Smale conjecture. Needless to say, the spaces $Diff(D^n)$ for $n$ large tend not to be contractible, and the fundamental groups are also known to be non-trivial in a large number of cases. Partial computations of the fundamental groups were done by Milnor and Kervaire. |
H: $\int^1_0 \frac{xdx}{x^2+2x+1}$
I need some suggestion how to solve this integral.
$$\int^1_0 \frac{xdx}{x^2+2x+1}$$
I think about to do the following step :
$$\frac{1}{2}\int^1_0\frac{2x+2-2dx}{x^2+2x+1}$$$$ t=x^2+2x+1 \rightarrow 2x+2dx=dt$$
then the integral will be : $$\frac{1}{2}\int^1_0 \frac{-2dt}{t}$$
its a correct way to solve it?
Thanks!
AI: Hint:
Notice that $x^2+2x+1=(x+1)^2$
$\int^1_0 \dfrac{xdx}{x^2+2x+1}=\int_0^1\dfrac{A}{x+1}+\int_0^1\dfrac{B}{(x+1)^2}$
What is wrong with your approach?
After you have $dt=2(x+1)dx$, your expression will be $\dfrac{1}{2} \int_0^1 \dfrac{2(\sqrt{t}-1)}{2\sqrt{t} \cdot t} dt$ |
H: If an integer $n$ is such that $7n$ is the form $a^2 + 3b^2$, prove that $n$ is also of that form.
If an integer $n$ is such that $7n$ is the form $a^2 + 3b^2$, prove that $n$ is also of that form.
I thought that looking at quad residues mod $7$ might??? help. But that didn't take me anywhere so apart from that I'm at a loss. I would appreciate any help. Thanks in advance.
AI: $\text{Make use of the identity }$$$\color{red}{(x^2+3y^2)(u^2+3v^2) = \underbrace{(xu\pm3yv)^2 + 3(xv\mp yu)^2}_{a^2 + 3b^2}}$$
$\text{Now we have}$
$$7 = 2^2 + 3 \times 1^2$$
$\text{i.e., }x=2 \text{ and }y=1$.
$\text{Use this to obtain $u$ and $v$.}$
$\textbf{EDIT}$:
$\text{Here is the full solution. The main claim is that if $7 \vert (a^2+3b^2)$, then}$
Either $7 \vert (a+2b)$ and $7 \vert (2a-3b)$
Or $7 \vert (a-2b)$ and $7 \vert (2a+3b)$
This can be checked by listing out the different possibilities for $a$ and $b$.
If $7 \vert (a+2b)$ and $7 \vert (2a-3b)$, then set $u = \dfrac{2a-3b}7$ and $v = \dfrac{a+2b}7$.
Else, if $7 \vert (a-2b)$ and $7 \vert (2a+3b)$, then set $u = \dfrac{3b-2a}7$ and $v = \dfrac{a+2b}7$.
This guarantess that $u$ and $v$ are integers. Hence, $\text{$n$ can be written as $u^2+3v^2$.}$ |
H: Is uncountable subset of separable space separable?
I have to prove that any uncountable $B\subseteq \mathbb{R}$, where $(\mathbb{R},\epsilon^1)$ is euclidean topology and topology on B is relative, is separable. And I know it's true because every subset of separable metric space is separable.
But what if we are given separable space $(X,\tau)$, $X$ uncountable, and $A \subseteq X$ uncountable subset with relative topology. Is $(A,\tau_A)$ separable and if it is, how to prove it?
AI: Consider the Niemytzki (or Moore) plane. This space is separable (the family of points with both coordinates rational is dense), but the $x$-axis $A = \{ \langle x , 0 \rangle : x \in \mathbb{R} \}$ is an uncountable closed discrete subset (and so $A$ with the subspace topology is discrete, and is therefore not separable). |
H: Showing that the segment joining $0$ to $z$ is perpendicular to the segment joining $0$ and $w$ iff $Re[z\bar{w}]=0$
I'm reading Beardon's Algebra and Geometry.
Suppose that $zw\neq0$. Show that the segment joining $0$ to $z$ is perpendicular to the segment joining $0$ to $w$ if and only if $Re[z\bar{w}]=0$.
From here I first expanded $z\bar{w}$:
$$i \left(a_2 b_1-a_1 b_2\right)+a_1 a_2+b_1 b_2$$
Then I obtained the real part of $z\bar{w}$:
$$a_1 a_2+b_1 b_2$$
From here, I was kinda stuck on how to proceed in the demonstration but then I decided to proceed with some of the most obvious examples for getting complex numbers that wouldfit perpendicularity: $(0+7i)$ and $(7+0i)$ for which $0 \cdot 7+7\cdot0=0$ and we have one case of perpendicularity. This first case would be:
$$((a_1=0)\oplus (a_2=0))\wedge ((b_1=0)\oplus (b_2=0))$$
With this reasoning in mind, I thought that for all the other cases of perpendicularity, we must find any numbers that satisfy $a_1 a_2=n$ and $b_1 b_2=-n$.
Is my reasoning correct? Is there something I should add?
AI: Two lines are perpendicular iff the scalar product of some vectors in their direction is zero. Apply this to the vectors $(a_1,b_1)$ and $(a_2,b_2)$. |
H: Do Boolean rings always have a unit element?
Let $(B, +, \cdot)$ be a non-trivial ring with the property that every $x \in B$ satisfies $x \cdot x = x$. How does one prove that such a ring $(B, +, \cdot)$ must have a unit element $1_B$? (Or, in case this is not true in general, what is a counterexample?)
BTW, I'm looking for an elementary proof, not requiring anything more than the definition of a ring, the definition of $(B, +, \cdot)$, and, if necessary, the easily shown facts that $x + x = 0$ and $x\cdot y = y\cdot x,\,\forall\, x,y \in B$.
AI: An example: the family of all finite subsets of a given infinite set. |
H: Can you use modulus to make 0 > 2?
I wanted to create a rock-paper-scissors game that didn't use a lot of conditionals, and I was wondering if there were any mathematical way of representing the cycle of rock-paper-scissors. So Rock beats Paper beats Scissors beats Rock, or Rock > Paper > Scissors > Rock. If you assigned numbers to these, it wouldn't work: 0 > 1 > 2 > 0. But I was wondering if there were some way you could use modulus function to make this work? I understand that if you had a series of integers and modded them by 3 you would get a repeating cycle of 0, 1, 2, e.g. 60 % 3 = 0; 61 % 3 = 1; 62 % 3 = 2, 63 % 3 = 0, etc.
Do you think there could be any way to make this programmable using mods? I'm interested to know if there's an algorithm or some way to use this to create a cycle where 0>1>2>0. Just a warning, though, I don't study maths or know a lot about complex maths, so if you know of a way to do this, I'd really appreciate it if you tried not to overwhelm me with maths terms!
AI: Codify Rock by 0, Scissors by 1, and Paper by 2. Notice that, modulo $3$, option $k$ beats option $k+1$ and is beaten by option $k-1$. So, if $x$ and $y$ are the choices of player 1 and 2, respectively, then compute $k=x-y$ modulo $3$. Then player 2 wins iff $k=1$, player 1 wins iff $k=-1$, and they tie iff $k=0$. This easily generalizes to cycles of any length. Thus, after correctly codifying the options by numbers (actually elements of the group $\mathbb Z_n$), all you need to do is compute the difference, compute modulo $n$, check whether the result is $0$ (resulting in a tie), or is it in the range 1 to floor(n/2) (player 2 wins), or else player 1 wins (some care needed according to the parity of $n$ and the rules of the game). |
H: Trigonometry Addition Thereom
Using the expansion of
a. $\sin(+)$, prove that $\sin75°=\sqrt 6+\sqrt{24}$
b. $\sin(+)$, prove that $\tan75°=2+\sqrt 3$
Where to start? draw up triangle of sin 75? find other values? help please.
AI: I suggest, you start by listing all angles of which you know exact values for $\sin$ and $\cos$ (should include $0°$, $30°$, $45°$, $60°$, $90°$ at least), then see how you can express $75°$. |
H: Understanding big O notation
I'm not a mathematician by any stretch and I'm trying to translate some maths terms into simple maths terms. Please don't laugh, I do consider this complicated!
The equations in question are
O(n) and O(n ^ 2)
Now, I have read up on Wiki about this but it has been written (IMO) for people who already understand it!
I believe n ^ 2 translates to the power of, in this case it is also the equivalent of squaring it (i.e. n * n).
However, I can't get my head around O in terms of what it is describing. Wiki says it's the limiting behaviour. So, does this mean O is more of a description than a function or command? In my understanding, the following 2 equations are the same
O(n^2)
n^2
AI: Quickly, $O(n^2)$ is any function $f=f(n)$ such that $$\left| \frac{f(n)}{n^2} \right|$$ remains bounded as $n \to +\infty$. It may be $n^2$ itself, but it may also be $n$, or $\sin \cos n$, etc. |
H: Free online mathematical software
What are the best free user-friendly alternatives to Mathematica and Maple available online?
I used Magma online calculator a few times for computational algebra issues, and was very much satisfied, even though the calculation time there was limited to $60$ seconds.
Very basic computations can be carried out with Wolfram Alpha. What if one is interested in integer relation detection or integration involving special functions, asymptotic analysis etc?
Thank you in advance.
Added: It would be nice to provide links in the answers so that the page becomes easily usable. I would also very much appreciate short summary on what a particular software is suitable/not suitable for. For example, Magma is in my opinion useless for doing the least numerics.
AI: I propose Sage. In my opinion it is the best free open-source mathematics software system. |
H: Showing that $\mathbb{P}[X\geq a]\leq \exp[-ta]\mathbb{E}[\exp[tX]]$
The problem is to show that $\mathbb{P}[X\geq a]\leq \exp[-ta]\mathbb{E}(\exp[tX])$ given $\exp(tX)<\infty$ for $t\in \mathbb{R}$ where $X$ is a random variable.
Then to show that $\mathbb{P}[X\geq a]\leq \inf\limits_{t\in\mathbb{R}} \exp[-ta]\mathbb{E}(\exp[tX])$. But I imagine this shouldn't be too difficult after the first part.
I tried to use Jensen's inequality.
$\psi(X)=\exp(tX)$ is a convex function so
$\psi(\mathbb E (X)) \leq\mathbb E (\psi (X)) \equiv\mathbb E( \exp[tX])$
So I am left with $\mathbb P [X\geq a]\exp[ta]$ on the LHS which seems to have no relation to $\exp(\mathbb E[tX])$ i.e. $\psi(\mathbb E(tX))$
AI: Hint: The inequality is equivalent to showing that:
$$
{\rm E}\left[1_{\{X\geq a\}}e^{ta}\right]\leq {\rm E}\left[e^{tX}\right].
$$ |
H: Quadratic form positive semidefinite if limits in every direction are nonnegative?
Let
$$q(x_1,\ldots,x_n) = \sum_{i,j=1}^{n}a_{ij}{x_i}{x_j}, \quad a_{ij}\in \mathbb{R}.$$
be a quadratic form with real coefficients.
Suppose that the limit is nonnegative in every direction. That is, for any unit vector $u$,
$$ \lim_{t \rightarrow \infty} q(tu) \geq 0.$$
Is $q$ necessarily positive semidefinite?
If the condition is strengthened to
$$ \lim_{t \rightarrow \infty} q(tu) > 0,$$
is it necessary positive definite?
AI: Hint: if the coefficients are constant, then you don't need the limit. Bilinearity allows to evacuate $t^2>0$, so all you get is
$\forall u\in\mathbb R^n\, q(u) = \langle Au,u\rangle\ge 0$ where $A$ is a square symetric matrix. |
H: Limit of a continued fraction
Given the continued fraction:
$$f(x,N)=\left[2,3,4,...N,x\right]$$
$$f(x,N)=\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{4+\cfrac{1}{...+\cfrac{1}{x}}}}}$$
is it possible to find an expression for the integral:
$$g(x,N)=\int f(x,N)dx$$
as function of $N$ and $x$?
Thanks.
AI: UPDATE
Let's observe that for :
$$0+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{4+\cfrac{1}{...+\cfrac{1}{N+\cfrac{1}x}}}}}}$$
the $n$-th convergent is known (from MathWorld, see too OEIS A052119) using Bessel functions :
\begin{align}
A_n&=\frac{I_n(-2)K_1(2)-I_1(-2)K_n(2)}{I_2(-2)K_1(2)-I_1(-2)K_2(2)}\\
&=2\left(I_n(-2)K_1(2)+I_1(2)K_n(2)\right)\\
&=2\left((-1)^n I_n(2)K_1(2)+I_1(2)K_n(2)\right)\\
\\
B_n&=\frac{I_n(-2)K_0(2)-I_0(-2)K_n(2)}{I_1(-2)K_0(2)-I_0(-2)K_1(2)}\\
&=2\left(-I_n(-2)K_0(2)+I_0(2)K_n(2)\right)\\
&=2\left((-1)^{n-1} I_n(2)K_0(2)+I_0(2)K_n(2)\right)\\
\end{align}
(there was a sign error in MathWorld in the above expression, see too OEIS entries A001053 and A001040)
This means that we have (needing the $N+1$-th convergent for the $N$ at the end) :
$$[0;1,2,\dotsc,N]=\frac{A_{N+1}}{B_{N+1}}$$
From this we may get the next convergent using Lord_Farin's helpful link :
$$[0;1,2,\dotsc,N,x]=\frac{xA_{N+1}+A_{N}}{xB_{N+1}+B_{N}}$$
To get your continued fraction you only need to revert this and subtract $1$ obtaining :
$$[0;2,\dotsc,N,x]=\frac{xB_{N+1}+B_{N}}{xA_{N+1}+A_{N}}-1$$
Allowing the computation of :
\begin{align}
g(x,N)&=\int f(x,N)\;dx\\
&=\int \frac{xB_{N+1}+B_{N}}{xA_{N+1}+A_{N}}-1\;dx\\
&=\int \frac{B_{N}A_{N+1}-A_{N}B_{N+1}}{A_{N+1}(xA_{N+1}+A_{N})}+\frac {B_{N+1}}{A_{N+1}}-1\;dx\\
&=\frac{B_{N}A_{N+1}-A_{N}B_{N+1}}{A_{N+1}^2}\log(A_{N+1}\;x+A_{N})+\frac {B_{N+1}-A_{N+1}}{A_{N+1}}x\\
\\
g(x,N)&=\boxed{\displaystyle (-1)^{N+1}\frac{\log(A_{N+1}\;x+A_{N})}{A_{N+1}^2}+\frac {B_{N+1}-A_{N+1}}{A_{N+1}}x}\quad\text{(after simplification)}\\
\end{align}
I'll let you reverify all this, |
H: Integral of $\int(4-2x)^\frac{1}{3}dx$
I solved this integral then I did $\frac{d}{dx}$ of $F(x)$ and saw that it's not the same, so I did something wrong in my integration process.
$$\int(4-2x)^\frac{1}{3}dx$$
What I did is $$F(x) =\frac{(4-2x)^\frac{4}{3}}{-2\times\frac{4}{3}} + C$$
Thanks!
AI: Hint:
use this formula :$\displaystyle \int (ax+b)^n\,dx=\frac{(ax+b)^{n+1}}{a(n+1)}$
in your question $\displaystyle a=-2 , b=4,n=\frac{1}{3}$
so answer will be $\displaystyle \frac{3\cdot(4-2x)^\frac{4}{3}}{-8}$ |
H: Binary vs. Ternary Goldbach Conjecture
Is there an "understandable" explanation of why the ternary Goldbach conjecture is tractable with current methods, while the binary Goldbach conjecture seems to be out of scope with current techniques? As a non-expert it seems to me that these conjectures should be equally hard to prove, which must of course not be true since these statements are not equivalent, but they are of the same "nature" and I don't see why it is such a difference that a number is the sum of two or of three primes...
AI: The "current methods" have almost always been refinements of Hardy-Littlewood circle method. Terry Tao has a blog post where he describes how the circle method applies to the problem, and why experts think this method alone will not yield the even Goldbach conjecture. |
H: Why do the French count so strangely?
Today I've heard a talk about division rules. The lecturer stated that base 12 has a lot of division rules and was therefore commonly used in trade.
English and German name their numbers like they count (with 11 and 12 as exception), but not French:
# | English | German | French
-----------------------------------------------
0 | zero | null | zero
1 | one | eins | un
2 | two | zwei | deux
3 | three | drei | trois
4 | four | vier | quatre
5 | five | fünf | cinq
6 | six | sechs | six
7 | seven | sieben | sept
8 | eight | acht | huit
9 | nine | neun | neuf
10 | ten | zehn | dix
11 | eleven | elf | onze
12 | twelve | zwölf | douze
13 | thir|teen | drei|zehn | treize
14 | four|teen | vier|zehn | quatorze
15 | fif|teen | fünf|zehn | quinze
16 | six|teen | sech|zehn | seize
17 |seven|teen | sieb|zehn | dix-sept
18 and 19 are "regular"
20 | twenty | zwanzig | vingt
21 |twenty-one | ein|und|zwanzig | vingt et un
22 |twenty-two | zwei|und|zwanzig | vingt-deux
23 - 69 are "regular"
70 | seven|ty | sieb|zig | soixante-dix = 60 + 10
....
80 | eigh|ty | acht|zig | quatre-vingts = 4*20 ?!?!
81 |eighty-one | ein|und|achtzig | quatre-vingt-un = 4*20 + 1
...
So my question is:
Why do French count so strangely after 79?
(Are there other languages that count similar? What's the historic / mathematical reason for this system?)
Related Questions
Why is a full turn of the circle 360°?
What could be better than base 10?
AI: Many languages have (at least relicts of) non-decimal counting, very often vigesimal (because we have 20 fingers plus toes), but also many other systems. I recommend an old Gutenberg project of mine, The Number Concept
Note for example that the Danish word for 55 is femoghalvtreds "five more than half the third twenty-block" |
H: Binomial Coefficients Combinatorics
For a positive integers n, prove that
$$\displaystyle\sum\limits_{v=0}^n \frac{(2n)!}{(v!)^2 ((n-v)!)^2} = \binom{2n}{n}^2.$$
If somebody could please help me with this question, I would greatly appreciated it.
AI: A combinatorial proof. The right-hand side corresponds to choosing two (possibly overlapping) sets of size $n$ from $2n$ elements.
Now we observe that:
$$\dfrac{2n!}{v!^2(n-v)!^2} = \binom{2n}{n} \binom n v \binom n{n-v}$$
If we divide the set of $2n$ elements in two fixed sets of $n$ elements, then picking $n$ from the original set amounts to picking $v$ from the first set, and $n-v$ from the second, for $v =0 \ldots n$.
Thus we find that
$$\sum_{v=0}^n \binom nv \binom n{n-v} = \binom{2n}n$$
which finally leads to the desired
$$\sum_{v=0}^n \frac{2n!}{v!^2(n-v)!^2} = \sum_{v=0}^n\binom{2n}{n} \binom n v \binom n{n-v} = \binom{2n}n \sum_{v=0}^n \binom nv\binom n{n-v} = \binom{2n}n^2$$ |
H: Integral of $\int^1_0 \frac{dx}{1+e^{2x}}$
I am trying to solve this integral and I need your suggestions.
I think about taking $1+e^{2x}$ and setting it as $t$, but I don't know how to continue now.
$$\int^1_0 \frac{dx}{1+e^{2x}}$$
Thanks!
AI: With the change of variable $u=e^{x}$, you get
$$
\int_{[0,1]}\frac{dx}{1+e^{2x}} = \int_{[1,e]}\frac{1}{u(1+u^2)}du
$$ |
H: In how many ways can we distribute $m$ balls to $n$ baskets when $n > m$
Given $m$ balls and $n$ baskets, in how many ways can we distribute the $m$ balls to the $n$ baskets when given that $n > m$ (the number of baskets is greater than the number of balls)?
So I said, that it does not differ if $n>m$ the solution is still ${n+m-1 \choose n-1}$, is that correct? I merely considered it as the number of solutions $x_1 + x_2 + ... + x_n = m$ when $0 \leq x_i$. Is that correct?
Also, as an addition:
In addition to the question, what if there can be only a maximum of one ball in each basket?
So I said, it is still the same $x_1 + x_2 + ... + x_n = m$ but this time $0 \leq x_i \leq 1$ and it is still the same solution. Is that correct?
AI: We assume the usual convention: the balls are indistinguishable, and the baskets are distinguishable.
For your first question, yes, it is precisely the formula you are familiar with.
When we are restricted to at most one ball per basket, then yes, we are counting the solutions of $x_1+x_2+\cdots+x_n=m$ with the restriction that for any $i$ we have $x_i=0$ or $x_i=1$.
But to me the easiest way to think of it is that we are choosing the $m$ baskets that will be lucky and get a ball. Or if you wish choosing the $i$ such that $x_i=1$. Or else think of it as forming an $n$-letter "word" made up of $m$ $1$'s and the rest $0$'s.
Whichever way one views it, this can be done in $\dbinom{n}{m}$ ways. |
H: Trigonometry Addition Thereom With Only one exact value?
Use the expression of $\sin(A+B)$ to evaluate $\sin 195$.
Do I use one exact value like $45+150$ or $60$ or is there another way?
AI: Here you go: $\sin 195° = -\sin 15° = -\sin(45-30)°$. After that you use standard formulas. |
H: hyperbola: equation for tangent lines and normal lines
Find the equations for
(a) the tangent lines, and
(b) the normal lines,
to the hyperbola
$y^2/4 - x^2/2 = 1$ when $x = 4$.
AI: If $x=4, \frac{y^2}4=1+\frac{4^2}2=9\implies y=\pm6$
Using Article 305 of this, the tangent of $$\frac{y^2}4-\frac{x^2}2=1$$
at $(h,k)$ is $$ \frac{y\cdot k}4-\frac{x\cdot h}2=1$$
Do you know how to find the perpendicular of a given line from a given point $(4,\pm 6)$? |
H: Scale free property of Pareto distribution
I am trying to show that the Pareto distribution is scale free, defined as:
p(bx) = g(b)p(x)
I get to this stage:
x dp/dx = p'(1)/p(1) p(x)
I have a rough solution which simply jumps to the next step:
ln(p(x)) = p(1)/p'(1) ln(x) + ln(p(1))
Can anyone explain?
AI: From the given equation, we have:
$$x \frac{\mathrm dp}{\mathrm dx} = \frac{p'(1)}{p(1)}p(x) \iff \frac{p'(x)}{p(x)} = \frac{p'(1)}{p(1)}\cdot \frac1x$$
and by the chain rule $\dfrac{\mathrm d(\log p(x))}{\mathrm dx} = \dfrac{p'(x)}{p(x)}$.
Integrating both sides with respect to $x$ gives the desired expression
$$\log p(x) = \dfrac{p'(1)}{p(1)}\log x+\log p(1)$$
because at $x = 1$ both expressions agree (thus verifying the choice of the constant of integration). |
H: How to find the minimum of $x+y^2+z^3$?
let $x,y,z>0$, and $x+3y+z=9$, find the minimum of
$$x+y^2+z^3$$
I think this problem is very interesting. I have found this
when $$x=\dfrac{9}{2}-\dfrac{1}{\sqrt{3}},y=\dfrac{3}{2},z=\dfrac{1}{\sqrt{3}}$$
I belive this inequality have $AM-GM$ methods,becasue I have see this same problem can use $AM-GM$ methods,and I think this methods is very very nice.
if $a,b,c>0$,and $ a+b^2+c^3=\dfrac{325}{9}$, prove that
$$a^2+b^3+c^4\ge\dfrac{2807}{27}$$
my methods: let $a=x,b=y,c=z$,and then
$$x+y^2+z^3=\dfrac{325}{9}$$
use $AM-GM$,we have
$$a^2+x^2\ge 2ax$$
$$b^3+b^3+y^3\ge 3yb^2$$
$$c^4+c^4+c^4+z^4\ge 4zc^3$$
then we have
$$a^2+b^3+c^4+x^2+\dfrac{y^3}{2}+\dfrac{z^4}{3}\ge 2ax+\dfrac{3}{2}b^2+\dfrac{4}{3}zc^3$$
AI: Since $x+y^2+a^2+z^3+b^3+b^3\ge x+2ay+3b^2z$, now let $2a=3, 3b^2=1$. You can determine $ y=a=\frac{3}{2}, z=b=\frac{1}{\sqrt{3}}$ and $x=9-3y-z$ if equality holds.
You do get an AM-GM approach. |
H: How to show $T$ is diagonalizable?
Let $T\colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be linear with distinct eigenvalues $\lambda_1, \lambda_2, \lambda_3$. Show that $T$ is diagonalizable.
It seems as if this is a very simple question but I'm a bit confused about how to start!
AI: You need to prove that $\mathbb{R}^3$ has a basis of eigenvectors. The key is that if you have vectors $\vec{v}_1, \vec{v}_2, \vec{v}_3$ that are eigenvectors for distinct eigenvalues, then the vectors must be linearly independent.
To prove this latter fact, assume that
$$c_1 \vec{v}_1 + c_2 \vec{v}_2 + c_3 \vec{v}_3 = \vec{0}.$$
Apply $(T - \lambda_1 I)(T -\lambda_2 I) = (T -\lambda_2 I)(T - \lambda_1 I)$ to both sides to find that $c_3 = 0$. (Here $I$ is the identity operator.) Note that you must use $\lambda_3 \neq \lambda_1, \lambda_2$ here. Similar operators can be applied to show that $c_1$ and $c_2$ must equal $0$ as well.
In this case, the argument gives you three linearly independent eigenvectors in $\mathbb{R}^3$, which are therefore a basis. |
H: On the no trivial $3$-tuples $(p, q, \alpha) \in \mathbb{N}^3$ such that $\sum_{k = 1}^{n}k^p =\Big [\sum_{k=1}^{n}k^q\Big]^\alpha $.
It is well known that $\sum_{k = 1}^{n}k^3 =\Big [\sum_{k=1}^{n}k^1\Big]^2$. My question is very simple.
There are $3$-tuples $(p, q, \alpha) \in
\mathbb{N}\times\mathbb{N}\times\mathbb{N}$, in addition to $(3,1,2)$,
such that $\alpha\geq 2$ and $$\sum_{k = 1}^{n}k^{\,p} =\Big [\sum_{k=1}^{n}k^{\,q}\Big]^\alpha, \quad \forall n \in \mathbb{N}{\;\Large ?}$$
AI: [The following claims should be obvious, and will not be proven.]
Claim: The degree of $ \sum_{i=1}^n i^r$ is $r+1$.
Claim: The leading coefficient of $ \sum_{i=1}^n i^r$ is $\frac{1}{r+1} $.
The first claim gives us $p+1 = \alpha (q+1) $. The second claim gives us $\frac{1}{p+1} = \left( \frac{ 1}{q+1} \right)^\alpha$.
Multiplying these two equations, we get $(q+1) = \alpha^\frac{1}{\alpha-1}$
Claim: $1 \leq \alpha^{\frac{1}{\alpha-1}} \leq 2$, with equality if and only if $\alpha = 1, 2$.
With this claim, the RHS is never an integer except for $\alpha = 1, 2 $.
If $\alpha = 1$, then $p+1 = q+1, p=q$.
If $\alpha = 2$, this forces $q+1=2$ or $q=1$, and that $p+1 = 2(q+1) $ so $p=3$.
Finally, verify that $(p,p,1) $ and $(3,1,2)$ are solutions. |
H: Finding the subgroup of $(\mathbb Z_{56},+)$ which is isomorphic with $(\mathbb Z_{14},+)$ by GAP
I am sorting some easy questions for the students in Group Theory I. One of them is:
Is $(\mathbb Z_{14},+)$ isomorphic to a subgroup of $(\mathbb Z_{35},+)$? What about $(\mathbb Z_{56},+)$?
I know the first claim is false because if it is true then I have $14\nmid 35$ which is a contradiction. For the second one, I see the function $f:(\mathbb Z_{14},+)\to(\mathbb Z_{56},+), f(g)=4g$ is a nice one-one homomorphism.
My problem is to find $f(\mathbb Z_{14})$ by using GAP. I know how to define Cyclic Groups or Free Groups in GAP's environment also. Thanks for the ANY help.
AI: gap> G:=CyclicGroup(56);
<pc group of size 56 with 4 generators>
gap> S:=AllSubgroups(G); # available since GAP 4.5, mainly for teaching purposes
[ Group([ <identity> of ... ]), Group([ f3*f4^3 ]), Group([ f2*f3*f4 ]),
Group([ f4 ]), Group([ f1*f2*f3 ]), Group([ f3 ]), Group([ f2 ]),
Group([ f1, f2, f3, f4 ]) ]
gap> H:=Filtered(S,x->Size(x)=14);
[ Group([ f3 ]) ] |
H: The Subspace $M=\{f\in C[0,1]:f(0)=0\}$
Let $C[0,1]$ with the supremum norm. It's easy to see that $M=\{f\in C[0,1]:f(0)=0\}$ is a closed subspace and so $C[0,1]/M$ is a Banach space. But I'm having trouble in finding a Banach space isometric to $C[0,1]/M$.
AI: For a given $a\in[0,1]$ consider linear bounded functional
$$
\operatorname{ev}_a:C[0,1]\to\mathbb{C}:f\mapsto f(a)
$$
Obviously $M=\operatorname{Ker}(\operatorname{ev}_0)$, i.e. $M$ is a kernel of bounded linear functional so $C([0,1])/M\cong \mathbb{C}$. |
H: $\mathbb{Q}/\mathbb{Z}$ has cyclic subgroup of every positive integer $n$?
I would like to know whether $(\mathbb{Q}/\mathbb{Z},+)$ has
$1$. Cyclic subgroup of every positive integer $n$?
$2$. Yes, unique one.
$3$. Yes, but not necessarily unique one.
$4$. Does not have cyclic subgroup of every positive integer $n$.
What I know about the given group is infinite group but every element has finite order. Please help how to proceed.
AI: Under the identification
$$
\frac{\Bbb R}{\Bbb Z}\stackrel{\sim}\longrightarrow S^1\simeq
\{\text{$z\in\Bbb C$ such that $||z||=1 $}\},\quad
r+\Bbb Z\mapsto e^{2\pi i r}
$$
the group ${\Bbb Q}/{\Bbb Z}$ is identified to the group $\mu(\Bbb C)$ of the complex roots of 1. Thus, a primitive $n$-th root $\zeta_n$ will correspond to an element in ${\Bbb Q}/{\Bbb Z}$ of exact order $n$.
Moreover, we know that every finite subgroup of the multiplicative group $\Bbb C^\times$ is cyclic. Thus there's just one subgroup of $\mu(\Bbb C)$ of order $n$, and same is true for ${\Bbb Q}/{\Bbb Z}$. |
H: $\sum_{n=0}^{\infty}3^{-n} (z-1)^{2n}$ converges when
$\sum_{n=0}^{\infty}3^{-n} (z-1)^{2n}$ converges when,
$1.|z|\le 3$
$2. |z|<\sqrt{3}$
$3.|z-1|<\sqrt{3}$
$4.|z-1|\le \sqrt{3}$
The radius of convergence can be found by applying the root test to the terms of the series. The root test uses the number $$C=\lim\sup|3^{-n}(z-1)^{2n}|^{1\over n}$$ for convergence $C<1$ and I get $|z-1|<\sqrt{3}$, Is it okay?
AI: Do the root test.
Set $$\limsup_{n\to\infty} \left(3^{-n} (z -1)^{2n}\right)^{1/n} < 1.$$
Cancelling exponents leaves
$$|z - 1|^2/3 < 1$$
or $$|z - 1| < \sqrt{3}.$$
Note that this fails to converge at both endpoints. It is, in fact, a geometric series. You are lookin' good. |
H: Prove using an example that there is no plane on $\mathbb{R}^3$ that contains every group of 4 points
Well, this is a homewrok question (which I know I should not be asking, but I cannot find an answer to this anywhere):
The exercise is as follows:
i) Find the equation of the plane of $\mathbb{R}^3$ that contains the following three points:
(0,0,1),(-1,3,5) and (1,-1,0)
ii) Prove using an example that there is no plane on $\mathbb{R}^3$ that contains every group of 4 points
So I have managed to solve (i) using a lot of Google and a third party tool (my notes are not all that good), but I cannot find (ii) anywhere. Can someone prove this providing an example and explaining it properly so that I can understand the reason this is impossible?
Thanks in advance!
P.S: This is an exercise from a course on Matrixes and how to solve Systems using matrices, in case anyone wants to know.
AI: From the phrasing of the question, it sounds like you understand that there is only one plane through those three points in part i). Knowing that, then we can say this:
Hint: If you find a point $P$ that does not satisfy the equation you found in i), then that point, along with the original three points, is an example of a set of four points that doesn't fit in a single plane. |
H: Sequence of the ratio of two successive terms of a sequence
If $(a_n)_{n\in N}$ is a strictly decreasing sequence of real number converging to $0$ and s.t. $\forall n\in N$, $0<a_n<1$, does the following limit:
$$
\lim_{n}\frac{a_{n+1}}{a_n}
$$
exists?
When does the limit is strictly greater than $0$ (it can be zero if for example the sequence is defined as $a_n=\frac{1}{n!}$)?
AI: For the first question (existence of a limit), if you set $u_n\stackrel{\rm{}def}{=}\ln \frac{a_{n+1}}{a_n}$, what can you say about the sequence $(u_n)_{n\in\mathbb{N}}$? |
H: Relation between continuity and connection between topologies
If $X$ is a set and $\tau_1,\tau_2$ two topologies on $X$. What does it mean to put the continuity of the identic map on $X$ (i.e $id_X(x)=x\forall x\in X$) in a relation to the comparability fo two topologies (in our case $\tau_1,\tau_2$)
AI: If the identity $(X,\tau_1)\to(X,\tau_2)$ is continuous, then the preimage of each open set in the codomain, that is a set in $\tau_2$, is an open set in the domain, i.e. an element of $\tau_1$. So $\tau_2$ is coarser than $\tau_1$.
If you know how to characterize continuity locally, then this means that for every $x\in X$ and every neighborhood of $x$ w.r.t. $\tau_2$ there is a neighborhood of $x$ w.r.t. $\tau_1$ contained in the given neighborhood, which means that the local basis is finer. |
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