Find limit involving square of sine

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  • #1
songoku
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Homework Statement:
Find
$$\lim_{n \rightarrow \infty} \sin^{2} (\pi \sqrt{n^2+n})$$
Relevant Equations:
Limit
$$\lim_{n \rightarrow \infty} \sin^{2} (\pi \sqrt{n^2+n})$$
$$=\lim_{n \rightarrow \infty} \sin^{2} (\pi \sqrt{n^2+n}-n\pi)$$
$$=\lim_{n \rightarrow \infty} \sin^{2} (\pi \sqrt{n^2+n}-n\pi)$$
$$=\lim_{n \rightarrow \infty} \sin^{2} (\pi (\sqrt{n^2+n}-n))$$
$$=\lim_{n \rightarrow \infty} \sin^{2} \left(\pi \left((\sqrt{n^2+n}-n) . \frac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n}\right)\right)$$
$$=\sin^{2} \left(\pi \lim_{n \rightarrow \infty} \left(\frac{n}{\sqrt{n^2+n}+n}\right)\right)$$
$$=\sin^{2} \left(\pi \left(\frac{1}{2}\right)\right)$$
$$=1$$

But if I imagine the graph of ##\sin^{2} (\pi \sqrt{n^2+n})##, it will oscillate between 0 and 1 so when ##n \rightarrow \infty##, the limit would be undefined.

Where is the mistake in my reasoning?

Thanks
 

Answers and Replies

  • #2
PeroK
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But if I imagine the graph of ##\sin^{2} (\pi \sqrt{n^2+n})##, it will oscillate between 0 and 1
Why?
 
  • #3
Office_Shredder
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n is supposed to be an integer right? Consider what you did vs the same limit but n can be any real number. That should resolve your confusion hopefully.
 
  • #4
songoku
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Why?
n is supposed to be an integer right? Consider what you did vs the same limit but n can be any real number. That should resolve your confusion hopefully.
I understand.

Thank you very much PeroK and Office_Shredder
 
  • #5
Hall
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Let me attempt to clarify the doubt of Songoku concretely and demonstrate that the sequence ##s_n = \sin^2 \left( \pi \sqrt{n^2 +n} \right)## converges, instead of it characteristic oscillation. So, pal, we're given
$$
\sin^2 \left( \pi \sqrt{n^2 +n} \right)
$$
Write ## a_n = \sqrt{n^2 +n}##. We will use, gently, (no exploitation), the fact increment in ##n## is by 1 and not less than that is possible. We will analyse the difference between consecutive ##a_n## as ##n## gets large,
$$
a_{n+1} - a_n = \sqrt{ (n+1)^ + n+1} - \sqrt{ n^2 + n}
$$
After rationalisation we will find
## a_{n+1} - a_n = \dfrac{ 2n+2}{ \sqrt{n^2+3n +2} + \sqrt{n^2+n} }##
We want to know what happens to this difference when ##n## gets very large, so,
$$
\lim_{n \to \infty} (a_{n+1} - a_n) = \lim_{n \to \infty} \dfrac{ 2n+2}{ \sqrt{n^2+3n +2} + \sqrt{n^2+n} } = 1
$$

Fix a very large ##n## and call it ##N##, such that error in ##a_{N+1} -a_N \approx 1## is beyond four decimal places, and thus, by definition of limit all subsequent ##n## will have the same property.

##\sin^2 \left( \pi a_N \right)##
## \sin^2 \left( \pi a_{N+1} \right) = \sin^2 \left( \pi a_{N}+1 \right)= \left( \sin (\pi a_N) \cos \pi + sin\pi \cos (\pi a_N) \right)^2= \sin^2 \left( \pi a_N \right)##

For any natural number ##k##,
##\sin^2 \left( \pi a_{N+k} \right) = \sin^2 \left( \pi (a_N + k) \right) = \left( sin (\pi a_N) \cos k\pi + \sin \pi \cos (\pi a_N) \right)^2 = \sin^2 \left( \pi a_N \right)##

Thus, the sequence ## \sin^2 \left( \pi \sqrt{n^2 +n} \right)## converges.
 
  • #6
SammyS
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Homework Statement:: Find
$$\lim_{n \rightarrow \infty} \sin^{2} (\pi \sqrt{n^2+n})$$
Relevant Equations:: Limit


But if I imagine the graph of ##\sin^{2} (\pi \sqrt{n^2+n})##, it will oscillate between 0 and 1 so when ##n \rightarrow \infty##, the limit would be undefined.

Where is the mistake in my reasoning?

Thanks
Essentially, what you have shown is that ##\displaystyle \lim_{n\to \infty} \left(\sqrt{n^2+n\,} -n \right) = \frac 1 2 ## .

So, as ##n## approaches ##\infty##, ##\displaystyle \sqrt{n^2+n\,}## approaches ##\displaystyle n+ \frac 1 2 ## .

What can you say regarding ##\displaystyle \ \sin\left( \pi \left(n+ \frac 1 2 \right)\right) ##, if ##n## is even?

... if ##n## is odd?
 
  • #7
Office_Shredder
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Let me attempt to clarify the doubt of Songoku concretely and demonstrate that the sequence ##s_n = \sin^2 \left( \pi \sqrt{n^2 +n} \right)## converges, instead of it characteristic oscillation. So, pal, we're given
$$
\sin^2 \left( \pi \sqrt{n^2 +n} \right)
$$
Write ## a_n = \sqrt{n^2 +n}##. We will use, gently, (no exploitation), the fact increment in ##n## is by 1 and not less than that is possible. We will analyse the difference between consecutive ##a_n## as ##n## gets large,
$$
a_{n+1} - a_n = \sqrt{ (n+1)^ + n+1} - \sqrt{ n^2 + n}
$$
After rationalisation we will find
## a_{n+1} - a_n = \dfrac{ 2n+2}{ \sqrt{n^2+3n +2} + \sqrt{n^2+n} }##
We want to know what happens to this difference when ##n## gets very large, so,
$$
\lim_{n \to \infty} (a_{n+1} - a_n) = \lim_{n \to \infty} \dfrac{ 2n+2}{ \sqrt{n^2+3n +2} + \sqrt{n^2+n} } = 1
$$

Fix a very large ##n## and call it ##N##, such that error in ##a_{N+1} -a_N \approx 1## is beyond four decimal places, and thus, by definition of limit all subsequent ##n## will have the same property.

##\sin^2 \left( \pi a_N \right)##
## \sin^2 \left( \pi a_{N+1} \right) = \sin^2 \left( \pi a_{N}+1 \right)= \left( \sin (\pi a_N) \cos \pi + sin\pi \cos (\pi a_N) \right)^2= \sin^2 \left( \pi a_N \right)##

For any natural number ##k##,
##\sin^2 \left( \pi a_{N+k} \right) = \sin^2 \left( \pi (a_N + k) \right) = \left( sin (\pi a_N) \cos k\pi + \sin \pi \cos (\pi a_N) \right)^2 = \sin^2 \left( \pi a_N \right)##

Thus, the sequence ## \sin^2 \left( \pi \sqrt{n^2 +n} \right)## converges.
Amazingly, I think this is not a proof. If ##a_{n+1}=a_n+1+1/n## is defined recursively ,(with ##a_0=0## say) then the difference converges to 1, but I think this sequence doesn't converge if you plug it in instead of ##\sqrt{n^2+n}##
 
  • #8
Hall
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Amazingly, I think this is not a proof. If ##a_{n+1}=a_n+1+1/n## is defined recursively ,(with ##a_0=0## say) then the difference converges to 1, but I think this sequence doesn't converge if you plug it in instead of ##\sqrt{n^2+n}##
I didn't understand much of that.

What I did is the following:
## a_n = \sqrt{n^2 +n}##
##s_n = a_{n+1} - a_n##
##\lim s_n = 1##.
 
  • #9
Office_Shredder
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I didn't understand much of that.

What I did is the following:
## a_n = \sqrt{n^2 +n}##
##s_n = a_{n+1} - a_n##
##\lim s_n = 1##.
Suppose instead you had
##a_n = n + \sum_{k=1}^n 1/k##
##s_n = 1+1/n## converges to 1.

But $$\lim_{n\to \infty} \sin^2(\pi a_n)$$
Does not exist.
 
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  • #10
Hall
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69
Suppose instead you had
##a_n = n + \sum_{k=1}^n 1/k##
##s_n = 1+1/n## converges to 1.

But $$\lim_{n\to \infty} \sin^2(\pi a_n)$$
Does not exist.
I realised that in post #5 in fact I proved that the sequence converges to ##\sin^2{\pi a_N}##, so if we were to change ##N## our point of convergence will change too. Upon analysing it further and contrasting it with your objection, I found that upon increasing ##N## indefinitely the difference between ##\sin^2{\pi a_N}## and ##\sin^2{\pi a_{N+k} }## goes to zero, that is the points of convergence itself converge and hence the original sequence converges; in your case we can still apply the steps of post #5 to prove that ##\sin^2{\pi a_N}## is the convergence point, but by changing ##N## the difference between convergence points doesn't reach zero. Well, all this analysis is nothing but basically Cauchy criterion.

"You're not as great as not to make any mistakes, but be as great as to accept them."

I will put myself in the second category. Thanks Office_Shredder.
 

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