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anonymous
 one year ago
A limit (challenge) problem for smart people.
In the question n is an integer
Condition:
Try to solve it in not more than 34 steps.
anonymous
 one year ago
A limit (challenge) problem for smart people. In the question n is an integer Condition: Try to solve it in not more than 34 steps.

This Question is Closed

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Is \(n\) an integer ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes n is an integer. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0u need to show your method , don't just give out answers like a idiot

dan815
 one year ago
Best ResponseYou've already chosen the best response.1maybe complex numbers i dont like the root in there

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no real sollution :)

dan815
 one year ago
Best ResponseYou've already chosen the best response.1you can get real solution from that

dan815
 one year ago
Best ResponseYou've already chosen the best response.1is the answer even converging

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no there is a answer

dan815
 one year ago
Best ResponseYou've already chosen the best response.1okay then ill work on it : )

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It dosen't have stupid answers like it does not exist etc

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0No.name is right. It got answer.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You got the answer @TrojanPoem

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Nearly , but more than 3 steps.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its ok tell your answer

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3im getting pi but im still messing wid it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 you are correct !

dan815
 one year ago
Best ResponseYou've already chosen the best response.1psh i already guessed pi lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok , before i post my sollution , how did u get it @ganeshie8

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3bit embarrassing i kno dw:1434985625298:dw

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0haha @ganeshie8 Like ++

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You made my day @ganeshie8 :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3I'm glad you find it was entertaining :) please provide a hint as im kinda stuck..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no problem , never mind Well the hint is itself the whole sollution , so let me post the sollution itself ok?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah don't give up so easily , u will get it!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0After giving hint it is easy , so i will not give any hint :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3nope wait almost there

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Don't even think of binomial expansions , you will get the answer no doubt , but that's a waste of time and space

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\[\large{\begin{align}&\lim\limits_{n\to\infty}~n*\sin(2\pi\sqrt{n^2+1})\\~\\ &=\lim\limits_{n\to\infty}~n*\sin(2\pi(\sqrt{n^2+1}n)+2n\pi)\\~\\ &=\lim\limits_{n\to\infty}~n*\sin(2\pi(\sqrt{n^2+1}n))\\~\\ &=\lim\limits_{n\to\infty}~n*\sin\left(\frac{2\pi}{\sqrt{n^2+1}+n}\right)\\~\\ \end{align}}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3next i want to use sinx/x limit but im gonna need more paper, one sec..

dan815
 one year ago
Best ResponseYou've already chosen the best response.1its so weird i cant believe the 1 is skewing it form 2pi to pi

dan815
 one year ago
Best ResponseYou've already chosen the best response.1for a big number sqrt(1+n^2) is pretty much sqrt(n^2)

dan815
 one year ago
Best ResponseYou've already chosen the best response.1so i said it was just n*sin(2pi*n) and thats inf * 0 so from lohopitals ud get 2pi

dan815
 one year ago
Best ResponseYou've already chosen the best response.1but that +1 in there changed it to pi O_O still working

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3I just figured out a really neat way to solve this w/o using lhopital xD

dan815
 one year ago
Best ResponseYou've already chosen the best response.1are you squaring and sqrting the whole limit

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3nvm its not working, was trying to interpret the given expression as a variation of area of regular polygon formula... but no success

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3as \(n\to\infty\), 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
 one year ago
Best ResponseYou've already chosen the best response.1ya not sure what im doing wrong, i keep getting 2pi again

dan815
 one year ago
Best ResponseYou've already chosen the best response.1look at my work, can u spot something wrong?

dan815
 one year ago
Best ResponseYou've already chosen the best response.1first is that step legal

dan815
 one year ago
Best ResponseYou've already chosen the best response.1can i just square and sqrt the lmit like that, now this lets me use lhopitals rule a couple times with no problems

dan815
 one year ago
Best ResponseYou've already chosen the best response.1because k has remained an integer

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3that 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
 one year ago
Best ResponseYou've already chosen the best response.1oh true okay throw that out then

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Does this count? Step 1: Make a conjecture about the limit value (\(\pi\)). Step 2: Prove it.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3i smell epsilon delta but it smells good to me :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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 \[\leftn\sin\left(2\pi\sqrt{1+n^2}\right)\pi\right<\epsilon\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(It should go without saying that \(\epsilon>0\))

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0My 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
 one year ago
Best ResponseYou've already chosen the best response.3\[\large{\begin{align}&\lim\limits_{n\to\infty}~n*\sin(2\pi\sqrt{n^2+1})\\~\\ &=\lim\limits_{n\to\infty}~n*\sin(2\pi(\sqrt{n^2+1}n)+2n\pi)\\~\\ &=\lim\limits_{n\to\infty}~n*\sin(2\pi(\sqrt{n^2+1}n))\\~\\ &=\lim\limits_{n\to\infty}~n*\sin\left(\frac{2\pi}{\sqrt{n^2+1}+n}\right)\\~\\ &=\lim\limits_{n\to\infty}~n*\left(\frac{2\pi}{\sqrt{n^2+1}+n} +\mathcal{O}(1/n^2)\right) ~~~\color{gray}{(\text{taylor})}\\~\\ &= \lim\limits_{n\to\infty}~n*\left(\frac{2\pi}{\sqrt{n^2+1}+n} \right) + 0 \\~\\ &=\pi \end{align}}\]

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0can you do it without taylor?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3yeah just use \(\sin x\approx x\) ;p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well shall i give my sollution?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3wow thats exact same as my solution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I cannot see your sollution , it is in latex form , so i am not able to understand it. nice if u solved it

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3what do you mean it is in latex form

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Like the only the code is visible not the output

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Ahh ok some problem with rendering here it is dw:1435050449486:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0nice but what is taylor

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Here taylor series of most common functions dw:1435050691842:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh i know these expansions
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