Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f_n \in C([0,+\infty))$ be defined by $f_n(t):=\sin{\sqrt{t+4n^2\pi^2}}$, for $n \in \mathbb N$ and $t \ge 0$.

  1. Prove that $f_n$ converges pointwise to $f \in C([0,+\infty))$ and determine $f$;

  2. study the uniform convergence of the sequence on bounded intervals and on $[0;+\infty)$.

Well, I've got some problems and I need your kind help.

First of all, I've noted that $f_n(0)=0$ for every $n \in \mathbb N$. My guess is that the pointwise limit is $f \equiv 0$. But how can I prove it rigorously? I think that one should note that $4n^2\pi^2=(2n\pi)^2$ so there must be something related to periodicity of the function $\sin(\cdot)$...

Thanks in advance.

share|improve this question
add comment

2 Answers 2

up vote 3 down vote accepted

$\sin(\sqrt{t+4n^2\pi^2})=\sin\left(2n\pi\sqrt{1+\frac{t}{4n^2\pi^2}}\right)\approx \sin\left(2n\pi+\frac{t}{4n\pi}\right)=\sin\left(\frac{t}{4n\pi}\right)\approx\frac{t}{4n\pi}\rightarrow 0$

so $f(t)=0$.

On any interval, the convergence is bounded as $|\sin(x)|\le|x|$.

share|improve this answer
add comment

Hint $$\sin \left( {\sqrt {t + 4{n^2}{\pi ^2}} } \right) = \sin \left( {2n\pi \sqrt {\frac{{t + 4{n^2}{\pi ^2}}}{{4{n^2}\pi }}} } \right) = \sin \left( {2n\pi \sqrt {\frac{t}{{4{n^2}\pi }} + 1} } \right)$$

and

$$\sqrt {1 + x} = 1 + \frac{x}{2} + o\left( x \right)$$ for $x\to 0$

Of course $\sin x =x+o(x)$ for $x\to 0$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.