I want to show the uniform convergence of $f_n(x)=n\sin (\sqrt{4\pi^2n^2+x^2})$ on $[0,a],a>0$.
I tried it as follows:
$f_n(x)=n\sin (\sqrt{4\pi^2n^2+x^2}-2n\pi)=n\sin\left(\frac{x^2}{\sqrt{4\pi^2n^2+x^2}+2n\pi}\right)\sim n\left(\frac{x^2}{\sqrt{4\pi^2n^2+x^2}+2n\pi}\right)\to \frac{x^2}{4\pi}$.
It follows that $f_n$ converges pointwise to $f:x\mapsto \frac{x^2}{4\pi}$.
To show uniform convergence, I want to show that the sequence $y_n:=\sup_{x\in [0,a]}|f_n(x)-f(x)|$ converges to $0$.
$\begin{align} f_n(x)&=n\left((\sqrt{4\pi^2n^2+x^2}-2n\pi)-(\sqrt{4\pi^2n^2+x^2}-2n\pi)^3 \frac 1{3!}\right)+o(1/n)\\ &=\left(\frac{x^2}{\sqrt{4\pi^2+(x/n)^2}+2\pi}\right)-\left(\frac{x^2}{\sqrt{4\pi^2+(x/n)^2}+2\pi}\right)^3\frac{1}{n^23!}+o(1/n) \end{align}$
$\begin{align}|f_n(x)-f(x)|&\le \left|\left(\frac{x^2}{\sqrt{4\pi^2+(x/n)^2}+2\pi}\right)-\frac {x^2}{4\pi}\right|+\left(\frac{x^2}{\sqrt{4\pi^2+(x/n)^2}+2\pi}\right)^3\frac{1}{n^23!}+o(1/n)\\ &\le a^2\left(\frac 1{4\pi}-\frac 1{\sqrt{4\pi^2+(x/n)^2}+2\pi}\right)+o(1/n)\\ &\le a^2\left( \frac 1{4\pi}-\frac 1{\sqrt{4\pi^2+(a/n)^2}+2\pi}\right)+o(1/n)\end{align}$
It follows that $y_n\to 0$.
Is my proof correct? Thanks.
Note:
$(1): f_n(x)=n\sin (\sqrt{4\pi^2n^2+x^2}-2n\pi)=n\sin t= n(t-\frac {t^3}{3!}) +nR(t),$ where $t=\sqrt{4\pi^2n^2+x^2}-2n\pi=\frac{x^2}{\sqrt{4\pi^2n^2+x^2}+2n\pi}\le \frac {a^2}{4n\pi}$ and $R(t)$ is the remainder term.
$(2): |R(t)|\le t^5\sum_{k=1}^\infty \frac{t^{2k-2}}{(2k+3)!}\le \frac{a^{10}}{{4\pi}^{5}n^5}\color{blue}{\sum_{k=1}^\infty \frac{a^{4k-4}}{(4n\pi)^{2k-2}(2k+3)!}}=M\frac{a^{10}}{{4\pi}^{5}n^5}$.
The blue colored series converges and is hence bounded by some $M>0$.
It follows that $|n R(t)|\le M\frac{a^{10}}{{4\pi}^{5}n^4}$, whence $R(t)=o(1/n)$.
I carried out the program of $a-b=\frac{a^2-b^2}{a+b}$ twice to good effect along with some trigonometric identities.
For $x\gt0$, the Mean Value Theorem says $$ \begin{align} f_0(x)&=1-\cos(x)\ge0\tag{1a}\\[9pt] f_1(x)&=x-\sin(x)\ge0\tag{1b}\\[3pt] f_2(x)&=-1+\frac{x^2}2+\cos(x)\ge0\tag{1c}\\ f_3(x)&=-x+\frac{x^3}6+\sin(x)\ge0\tag{1d} \end{align} $$ Explanation:
$\text{(1a)}$: $-1\le\cos(x)\le1$
$\text{(1b)}$: for some $c\in(0,x)$, $f_1(x)-f_1(0)=(x-0)f_0(c)\ge0$
$\text{(1c)}$: for some $c\in(0,x)$, $f_2(x)-f_2(0)=(x-0)f_1(c)\ge0$
$\text{(1d)}$: for some $c\in(0,x)$, $f_3(x)-f_3(0)=(x-0)f_2(c)\ge0$
Thus, for $x\ge0$, $\text{(1b)}$ and $\text{(1d)}$ say $$ 0\le x-\sin(x)\le\frac{x^3}6\tag2 $$ Furthermore, for $n\ge\frac{x^2}{2\pi^2}$, $\frac{x^2}{4\pi n}\le\frac\pi2$ and therefore, $$ \begin{align} &\sin\left(\frac{x^2}{4\pi n}\right)-\sin\left(\sqrt{4\pi^2n^2+x^2}\right)\tag{3a}\\ &=\sin\left(\frac{x^2}{4\pi n}\right)-\sin\left(\sqrt{4\pi^2n^2+x^2}-2\pi n\right)\tag{3b}\\ &=\sin\left(\frac{x^2}{4\pi n}\right)-\sin\left(\frac{x^2}{\sqrt{4\pi^2n^2+x^2}+2\pi n}\right)\tag{3c}\\ &=\scriptsize2\sin\left(\frac{x^2\left(\sqrt{4\pi^2n^2+x^2}-2\pi n\right)}{2\cdot4\pi n\left(\sqrt{4\pi^2n^2+x^2}+2\pi n\right)}\right)\cos\left(\frac{x^2\left(\sqrt{4\pi^2n^2+x^2}+6\pi n\right)}{2\cdot4\pi n\left(\sqrt{4\pi^2n^2+x^2}+2\pi n\right)}\right)\tag{3d}\\ &=\scriptsize2\sin\left(\frac{x^4}{2\cdot4\pi n\left(\sqrt{4\pi^2n^2+x^2}+2\pi n\right)^2}\right)\cos\left(\frac{x^2\left(\sqrt{4\pi^2n^2+x^2}+6\pi n\right)}{2\cdot4\pi n\left(\sqrt{4\pi^2n^2+x^2}+2\pi n\right)}\right)\tag{3e}\\ &\le\frac{x^4}{64\pi^3n^3}\tag{3f} \end{align} $$ Explanation:
$\text{(3b):}$ $\sin(x)$ has period $2\pi$
$\text{(3c):}$ $a-b=\frac{a^2-b^2}{a+b}$
$\text{(3d):}$ $\sin(a)-\sin(b)=2\sin\left(\frac{a-b}2\right)\cos\left(\frac{a+b}2\right)$
$\text{(3e):}$ $a-b=\frac{a^2-b^2}{a+b}$
$\text{(3f):}$ $\sin(x)\le x$ and $\cos(x)\le1$
Note that since $\frac{x^2}{4\pi n}\le\frac\pi2$, $\text{(3c)}$ shows that $$ \sin\left(\frac{x^2}{4\pi n}\right)-\sin\left(\sqrt{4\pi^2n^2+x^2}\right)\ge0\tag4 $$ Thus, for $n\ge\frac{x^2}{2\pi^2}$, $$ \begin{align} \frac{x^2}{4\pi n}-\sin\left(\sqrt{4\pi^2n^2+x^2}\right) &=\frac{x^2}{4\pi n}-\sin\left(\frac{x^2}{4\pi n}\right)\\ &+\sin\left(\frac{x^2}{4\pi n}\right)-\sin\left(\sqrt{4\pi^2n^2+x^2}\right)\tag{5a}\\ &\le\frac16\left(\frac{x^2}{4\pi n}\right)^3+\frac{x^4}{64\pi^3n^3}\tag{5b}\\ &=\frac{6x^4+x^6}{384\pi^3n^3}\tag{5c} \end{align} $$ Explanation:
$\text{(5a):}$ break the sum into two pieces
$\text{(5b):}$ apply $(2)$ and $(3)$
$\text{(5c):}$ simplify
$(2)$ and $(4)$ show that $\text{(5a)}$ is $\ge0$. Therefore, if we multiply $(5)$ by $n$, we get $$ 0\le\frac{x^2}{4\pi}-n\sin\left(\sqrt{4\pi^2n^2+x^2}\right) \le\frac{6x^4+x^6}{384\pi^3n^2}\tag6 $$