anonymous
  • anonymous
A limit (challenge) problem for smart people. In the question n is an integer Condition:- Try to solve it in not more than 3-4 steps.
Mathematics
chestercat
  • chestercat
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ganeshie8
  • ganeshie8
Is \(n\) an integer ?
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
Yes n is an integer. :)

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dan815
  • dan815
0
anonymous
  • anonymous
no
dan815
  • dan815
infinity
dan815
  • dan815
1
anonymous
  • anonymous
no
dan815
  • dan815
pi
dan815
  • dan815
e,
anonymous
  • anonymous
u need to show your method , don't just give out answers like a idiot
dan815
  • dan815
lulz :)
dan815
  • dan815
maybe complex numbers i dont like the root in there
anonymous
  • anonymous
no real sollution :)
dan815
  • dan815
you can get real solution from that
dan815
  • dan815
is it indeterminate?
dan815
  • dan815
is the answer even converging
anonymous
  • anonymous
no there is a answer
dan815
  • dan815
okay then ill work on it : )
anonymous
  • anonymous
It dosen't have stupid answers like it does not exist etc
TrojanPoem
  • TrojanPoem
No.name is right. It got answer.
anonymous
  • anonymous
You got the answer @TrojanPoem
anonymous
  • anonymous
?
TrojanPoem
  • TrojanPoem
Nearly , but more than 3 steps.
anonymous
  • anonymous
its ok tell your answer
ganeshie8
  • ganeshie8
im getting pi but im still messing wid it
anonymous
  • anonymous
@ganeshie8 you are correct !
dan815
  • dan815
psh i already guessed pi lol
anonymous
  • anonymous
Ok , before i post my sollution , how did u get it @ganeshie8
ganeshie8
  • ganeshie8
bit embarrassing i kno |dw:1434985625298:dw|
TrojanPoem
  • TrojanPoem
haha @ganeshie8 Like ++
anonymous
  • anonymous
You made my day @ganeshie8 :)
ganeshie8
  • ganeshie8
I'm glad you find it was entertaining :) please provide a hint as im kinda stuck..
anonymous
  • anonymous
no problem , never mind Well the hint is itself the whole sollution , so let me post the sollution itself ok?
dan815
  • dan815
no odont
ganeshie8
  • ganeshie8
^ give us some time
anonymous
  • anonymous
Yeah don't give up so easily , u will get it!
anonymous
  • anonymous
After giving hint it is easy , so i will not give any hint :)
anonymous
  • anonymous
Ok a small hint
ganeshie8
  • ganeshie8
nope wait almost there
anonymous
  • anonymous
ok ok
anonymous
  • anonymous
Don't even think of binomial expansions , you will get the answer no doubt , but that's a waste of time and space
ganeshie8
  • ganeshie8
\[\large{\begin{align}&\lim\limits_{n\to\infty}~n*\sin(2\pi\sqrt{n^2+1})\\~\\ &=\lim\limits_{n\to\infty}~n*\sin(2\pi(\sqrt{n^2+1}-n)+2n\pi)\\~\\ &=\lim\limits_{n\to\infty}~n*\sin(2\pi(\sqrt{n^2+1}-n))\\~\\ &=\lim\limits_{n\to\infty}~n*\sin\left(\frac{2\pi}{\sqrt{n^2+1}+n}\right)\\~\\ \end{align}}\]
ganeshie8
  • ganeshie8
next i want to use sinx/x limit but im gonna need more paper, one sec..
dan815
  • dan815
its so weird i cant believe the 1 is skewing it form 2pi to pi
dan815
  • dan815
for a big number sqrt(1+n^2) is pretty much sqrt(n^2)
dan815
  • dan815
so i said it was just n*sin(2pi*n) and thats inf * 0 so from lohopitals ud get 2pi
dan815
  • dan815
but that +1 in there changed it to pi O_O still working
ganeshie8
  • ganeshie8
I just figured out a really neat way to solve this w/o using lhopital xD
dan815
  • dan815
are you squaring and sqrting the whole limit
dan815
  • dan815
or k^2=1+n^2!
ganeshie8
  • ganeshie8
nvm its not working, was trying to interpret the given expression as a variation of area of regular polygon formula... but no success
ganeshie8
  • ganeshie8
as \(n\to\infty\), the regular polygon becomes an unit circle evaluating the limit to \(\pi\) thought it would be that simple but looks tricky to conclude like that
dan815
  • dan815
ya not sure what im doing wrong, i keep getting 2pi again
dan815
  • dan815
look at my work, can u spot something wrong?
ganeshie8
  • ganeshie8
okie
dan815
  • dan815
|dw:1434988291816:dw|
dan815
  • dan815
first is that step legal
dan815
  • dan815
can i just square and sqrt the lmit like that, now this lets me use lhopitals rule a couple times with no problems
dan815
  • dan815
because k has remained an integer
ganeshie8
  • ganeshie8
that is okay because sqrt(x) is a continuous function \[\lim \sqrt{f(x)} = \sqrt{\lim f(x)}\] But aren't you assuming \(\sqrt{1+n^2}\) is an integer which is incorrect ? let me go read it agian..
dan815
  • dan815
oh true okay throw that out then
anonymous
  • anonymous
Does this count? Step 1: Make a conjecture about the limit value (\(\pi\)). Step 2: Prove it.
ganeshie8
  • ganeshie8
i smell epsilon delta but it smells good to me :)
anonymous
  • anonymous
I was actually thinking monotone convergence, but showing either condition might be a bit tricky. As for an \(\epsilon\) proof, not sure how much analysis would be required relative to monotone convergence. We have to find some \(N\) for which \(n>N\) yields \[\left|n\sin\left(2\pi\sqrt{1+n^2}\right)-\pi\right|<\epsilon\]
anonymous
  • anonymous
(It should go without saying that \(\epsilon>0\))
anonymous
  • anonymous
My point of view is: When a function has the limit where n tends to infinity, then the result from sine will eventually tend to undefinition, meaning the function becomes undefined.
ganeshie8
  • ganeshie8
\[\large{\begin{align}&\lim\limits_{n\to\infty}~n*\sin(2\pi\sqrt{n^2+1})\\~\\ &=\lim\limits_{n\to\infty}~n*\sin(2\pi(\sqrt{n^2+1}-n)+2n\pi)\\~\\ &=\lim\limits_{n\to\infty}~n*\sin(2\pi(\sqrt{n^2+1}-n))\\~\\ &=\lim\limits_{n\to\infty}~n*\sin\left(\frac{2\pi}{\sqrt{n^2+1}+n}\right)\\~\\ &=\lim\limits_{n\to\infty}~n*\left(\frac{2\pi}{\sqrt{n^2+1}+n} +\mathcal{O}(1/n^2)\right) ~~~\color{gray}{(\text{taylor})}\\~\\ &= \lim\limits_{n\to\infty}~n*\left(\frac{2\pi}{\sqrt{n^2+1}+n} \right) + 0 \\~\\ &=\pi \end{align}}\]
xapproachesinfinity
  • xapproachesinfinity
can you do it without taylor?
ganeshie8
  • ganeshie8
yeah just use \(\sin x\approx x\) ;p
ganeshie8
  • ganeshie8
(jk)
ParthKohli
  • ParthKohli
Amazing!
anonymous
  • anonymous
Well shall i give my sollution?
ganeshie8
  • ganeshie8
Yes waiting
anonymous
  • anonymous
1 Attachment
ganeshie8
  • ganeshie8
wow thats exact same as my solution
anonymous
  • anonymous
I cannot see your sollution , it is in latex form , so i am not able to understand it. nice if u solved it
ganeshie8
  • ganeshie8
what do you mean it is in latex form
anonymous
  • anonymous
Like the only the code is visible not the output
ganeshie8
  • ganeshie8
Ahh ok some problem with rendering here it is |dw:1435050449486:dw|
anonymous
  • anonymous
nice but what is taylor
ganeshie8
  • ganeshie8
https://www.khanacademy.org/math/integral-calculus/sequences_series_approx_calc/maclaurin_taylor/v/maclauren-and-taylor-series-intuition
anonymous
  • anonymous
thnx
ganeshie8
  • ganeshie8
Here taylor series of most common functions |dw:1435050691842:dw|
anonymous
  • anonymous
Oh i know these expansions

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