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help-1
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331,558,346,333,224,960
jan Niku
12/23/2021 08:46:34
so it doesnt have to converge to 0
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b1ff0db61ea64ab092075179366b4182
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jan Niku
12/23/2021 08:46:45
well look just because it has that (-1)^n+1 term doesnt make it alternating
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jan Niku
12/23/2021 08:46:57
the 2^n drags it up enough
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Dzoni
12/23/2021 08:47:08
Okay, but how do i know if (-1)^inf is 1 or -1?
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jan Niku
12/23/2021 08:47:20
its not going to matter for large n
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b1ff0db61ea64ab092075179366b4182
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Dzoni
12/23/2021 08:47:28
Since 2^n grows much faster?
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b1ff0db61ea64ab092075179366b4182
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jan Niku
12/23/2021 08:47:31
yea
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jan Niku
12/23/2021 08:47:36
a million plus or minus one
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b1ff0db61ea64ab092075179366b4182
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jan Niku
12/23/2021 08:47:40
i mean were still in handwavey land
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jan Niku
12/23/2021 08:47:45
actually dealing with it analytically is another thing
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help-1
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Dzoni
12/23/2021 08:47:46
So essentially for large n i have (2^n)^1/n ?
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b1ff0db61ea64ab092075179366b4182
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jan Niku
12/23/2021 08:47:50
but just in figuring out what to show here
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jan Niku
12/23/2021 08:47:56
well
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jan Niku
12/23/2021 08:48:18
$( \frac{ 2^n }{ n } ) ^{\sfrac{1}{n}}$
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TeXit
12/23/2021 08:48:21
**jan Niku**
https://cdn.discordapp.c…346333224964.png
331558346333224964.png
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Dzoni
12/23/2021 08:48:42
I mean that n doesn't matter too, right? The same reason why (-1) doesn't matter
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Dzoni
12/23/2021 08:48:48
2^n grows much faster than n i think
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jan Niku
12/23/2021 08:48:59
well it becomes $\frac{2}{\sqrt[n]{n}}$
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TeXit
12/23/2021 08:49:02
**jan Niku**
https://cdn.discordapp.c…346333224964.png
331558346333224964.png
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b1ff0db61ea64ab092075179366b4182
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jan Niku
12/23/2021 08:49:19
,w graph 2/( xth root x)
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b1ff0db61ea64ab092075179366b4182
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TeXit
12/23/2021 08:49:25
Results provided by WolframAlpha
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jan Niku
12/23/2021 08:49:28
:sully:
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Dzoni
12/23/2021 08:50:09
I am not sure why you didn't remove that / n since it grows much slower than 2^n. I always thought exponential grows much faster than polynomial
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jan Niku
12/23/2021 08:50:19
dont remove it
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Dzoni
12/23/2021 08:50:21
And that we would have (2^n)^1/n then which would be 1
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jan Niku
12/23/2021 08:50:22
use exponent rules
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jan Niku
12/23/2021 08:50:27
were already one step into hand wavey
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jan Niku
12/23/2021 08:50:34
everything we remove we have to go back and show is insignificant
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jan Niku
12/23/2021 08:50:39
once we figure out what result we want to show
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Ann
12/23/2021 08:50:46
2
0
1
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jan Niku
12/23/2021 08:50:47
anyways the limit doesnt look like 1
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jan Niku
12/23/2021 08:50:48
i think its 2
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Dzoni
12/23/2021 08:52:01
Oh, its actually not inf^0 but rather 2^(n*1/n) which would be 2^1 = 2.
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jan Niku
12/23/2021 08:52:08
👀
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Dzoni
12/23/2021 08:52:10
So both ways work
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jan Niku
12/23/2021 08:52:21
im not sure how you get that
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Dzoni
12/23/2021 08:52:44
Well by removing the /n because i think its insignificant 😄
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jan Niku
12/23/2021 08:52:50
its not
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jan Niku
12/23/2021 08:53:09
i mean idk im a dumbie right but to me
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jan Niku
12/23/2021 08:53:18
you dont usually remove pieces that have n in them
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Dzoni
12/23/2021 08:53:25
I don't understand how it isn't insignificant. I mean even if you leave it we would also get 2 at the end right?
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jan Niku
12/23/2021 08:53:27
because the interaction of pieces with n determine the growth of this thing
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jan Niku
12/23/2021 08:53:48
you want them as simple as possible before you 'remove' them because then you can clearly reason to yourself why they are insignificant
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b1ff0db61ea64ab092075179366b4182
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jan Niku
12/23/2021 08:53:57
but i am also dumb :thonk:
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Dzoni
12/23/2021 08:54:03
Well, my reasoning was that 2^n grows much faster than n
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jan Niku
12/23/2021 08:54:04
i would go this way
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jan Niku
12/23/2021 08:54:10
sure
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Dzoni
12/23/2021 08:54:14
It was clear enough for me, but if its wrong i won't remove it
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jan Niku
12/23/2021 08:54:16
but look how complicated that interaction is
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jan Niku
12/23/2021 08:54:30
idk its up to you i guess im being a stickler i dont understand your reasoning but im also tired
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jan Niku
12/23/2021 08:54:40
i feel like this is much clearer
https://cdn.discordapp.c…346333224964.png
331558346333224964.png
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jan Niku
12/23/2021 08:54:58
since then all that remains is to convince yourself that the limit of nth root of n is 1
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jan Niku
12/23/2021 08:55:06
but either way
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b1ff0db61ea64ab092075179366b4182
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jan Niku
12/23/2021 08:55:12
now that you believe the limit is 2
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jan Niku
12/23/2021 08:55:17
all thats left to do is go back and show it
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jan Niku
12/23/2021 08:55:21
if thats part of the exercise at least
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Dzoni
12/23/2021 08:55:47
It's not a part of it
1
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jan Niku
12/23/2021 08:55:51
oh
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jan Niku
12/23/2021 08:55:53
:hype:
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Dzoni
12/23/2021 08:55:59
I just needed to see what that converges to
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Dzoni
12/23/2021 08:56:12
So i could finally calculate my R 🙂
1
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Dzoni
12/23/2021 08:56:24
Thank you so much man!
1
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jan Niku
12/23/2021 08:56:26
np
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Dzoni
12/23/2021 08:56:33
.close
1
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918a1813fd1b4c1197f831ca8e316c4b
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dldh06
01/28/2022 22:57:22
Not a bad deal on yogurt
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Random guy
01/28/2022 23:31:57
how to prove that this improper integral converges?
https://cdn.discordapp.c…022-01-29_10.png
2022-01-29_10.png
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Random guy
01/28/2022 23:32:27
i tried using the comparison test but im not sure what functions i should use
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Kaynex
01/28/2022 23:34:10
It's definitely smaller than 9x/(x²)⁴
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Random guy
01/28/2022 23:35:14
but how to prove that it converges
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icebear
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Geopchad
01/28/2022 23:36:57
the function that kaynex gives you converges--and if the integral is SMALLER than that, then it also converges
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Kaynex
01/28/2022 23:36:57
What does it mean for an integral to converge?
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Random guy
01/28/2022 23:37:56
doesnt \int _{0}^{\infty }\dfrac{9x}{\left( x^{2}\right) ^{4}}dx diverge?
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Random guy
01/28/2022 23:37:59
$\int _{0}^{\infty }\dfrac{9x}{\left( x^{2}\right) ^{4}}dx$
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TeXit
01/28/2022 23:38:03
**Random guy**
https://cdn.discordapp.c…964059701258.png
673062964059701258.png
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Geopchad
01/28/2022 23:38:56
no?
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Geopchad
01/28/2022 23:38:59
oh wait
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Kaynex
01/28/2022 23:39:08
Oh kek
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Bungo
01/28/2022 23:39:11
It diverges due to the asymptote at the origin being too "thick"
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Geopchad
01/28/2022 23:39:14
yeah it does lmao
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Bungo
01/28/2022 23:39:25
The tail behavior at infinity is OK though
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Random guy
01/28/2022 23:39:40
:bleakkekw:
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Kaynex
01/28/2022 23:39:42
It converges into infinity, which is what we want to capture
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Random guy
01/28/2022 23:40:06
then what function shall i consider
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Kaynex
01/28/2022 23:40:08
Ignore the 0 part it never mattered anyway
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help-1
e4ea2888718446c69540ee50f21127b3
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Random guy
01/28/2022 23:40:29
bruh
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Kaynex
01/28/2022 23:40:36
bruh
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Geopchad
01/28/2022 23:42:04
are you allowed to just
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Geopchad
01/28/2022 23:42:05
find it?
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Geopchad
01/28/2022 23:42:10
and show that the limit goes to infinity
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Random guy
01/28/2022 23:42:33
im pretty sure this limit converges
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help-1
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Random guy
01/28/2022 23:42:49
i mean integral
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Random guy
01/28/2022 23:42:51
lol
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Bungo
01/28/2022 23:43:32
The denominator satisfies $(x^2 + 4)^4 \geq 4^4$, so the integrand is less than $9x / 4^4$, which you can use as a comparison function on, say, the interval $[0,1]$.
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TeXit
01/28/2022 23:43:35
**OurBelovedBungo**
https://cdn.discordapp.c…030513352705.png
761155030513352705.png
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Random guy
01/28/2022 23:44:06
oh
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Random guy
01/28/2022 23:44:07
i see
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help-1
e4ea2888718446c69540ee50f21127b3
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Bungo
01/28/2022 23:44:59
Then on $[1,\infty)$ use the fact that the denominator also satisfies $(x^2 + 4)^4 \geq x^8$, so the function is also bounded by $9x / x^8$, which is integrable on $[1,\infty)$.
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TeXit
01/28/2022 23:45:03
**OurBelovedBungo**
https://cdn.discordapp.c…030513352705.png
761155030513352705.png
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help-1
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Random guy
01/28/2022 23:45:28
thanks for your help
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Random guy
01/28/2022 23:45:52
i think i understand now
1
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