# Showing that $\sin(\sqrt{4 \pi^{2}n^{2} + x})$ converges uniformly on $[0,1]$

Suppose we are considering the sequence of functions $f_{n}(x)=\sin(\sqrt{4 \pi^{2}n^{2} + x})$ and I am having trouble showing that that $f_{n}$ converges uniformly on the interval $[0,1]$.

An idea, I've tried is to consider the Taylor series:

$$\sin(\sqrt{4 \pi^{2}n^{2} + x}) = (\sqrt{4 \pi^{2}n^{2} + x})- \frac{(\sqrt{4 \pi^{2}n^{2} + x})^{3}}{6} + O((\sqrt{4 \pi^{2}n^{2} + x})^{5})$$

but I haven't gotten anything useful as of yet.

• What tools do you have to show a sequence of functions converges uniformly on an interval? Why do you think the Taylor series would be helpful? – DaveNine Jun 27 '18 at 16:16
• I was hoping to show something like for each $\epsilon>0$, we can find an integer $N$ such that if $m,n \geq N$ and any $x$, we have $|f_{n}(x)-f_{m}(x)| \leq \epsilon$. I think this is a problem which only requires elementary analysis, but if there is a better way to do it using other techniques, I'd love to see it. – user135520 Jun 27 '18 at 16:24

Fix $x\in[0,1]$. The Mean Value Theorem shows that $$|\sqrt{4\pi^2n^2+x}-2\pi n|<\frac x{4\pi n}\le\frac1{4\pi n}.$$ So $2\pi n \le \sqrt{4\pi^2n^2+x} < 2\pi n +1/(4\pi n)$. Then apply MVT again to show$$|\sin(\sqrt{4\pi^2n^2+x})| =|\sin(\sqrt{4\pi^2n^2+x})-\sin(2\pi n)|<\frac1{4\pi n}.$$
• I think your argument also works to show that $f_{n}$ does not uniformly converge on $[0,\infty)$. Because we have that $|f_{n}(x)| = |f'_{n}(x_{0})| |x|$ for some $x_{0} \in [0, x)$ and we can take $x$ to be very large. – user135520 Jun 27 '18 at 20:27
• @user135520 Could be. It's obvious that the convergence is not uniform on $[0,\infty)$; for every $n$ there exists $x$ with $\sqrt{4\pi^2n^2+x}=(2n+1)\pi$.. – David C. Ullrich Jun 28 '18 at 0:18
• @user135520 Or rather $(2n+1/2)\pi$. – David C. Ullrich Jun 28 '18 at 0:26
Your sequence converges uniformly to $0$.
uniformly in $x \in [0,1]$.