- Examine whether the sequence of functions $f_n:[0, 1] \rightarrow \mathbb{R}$ defined as
$$f_n(x)=\sqrt{n} x^2 \left( 1 - x^2 \right)^n$$
converges uniformly on $[0, 1]$ or not. If so, find the limit. - Examine whether the series of functions defined as $\displaystyle \sum_{n=1}^{\infty} \frac{\sin (n^2 x)}{n (x^2+n)}$ converges uniformly on $\mathbb{R}$.
Uniform convergence
Uniform convergence
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
- Grigorios Kostakos
- Founder
- Posts: 461
- Joined: Mon Nov 09, 2015 1:36 am
- Location: Ioannina, Greece
Re: Uniform convergence
i. For $x\in(0,1)$ we have that \begin{align*}
\displaystyle\mathop{\lim}\limits_{n\to+\infty}f_n(x)&=\mathop{\lim}\limits_{n\to+\infty}\sqrt{n} \,x^2 \,( 1 - x^2)^n\\
&=x^2\mathop{\lim}\limits_{n\to+\infty}\frac{\sqrt{n}}{( 1 - x^2)^{-n}}\\
&\stackrel{\frac{\infty}{\infty}}{=}x^2\mathop{\lim}\limits_{n\to+\infty}\frac{\frac{1}{2\sqrt{n}}}{-n\,( 1 - x^2)^{-n-1}}\\
&=-x^2\mathop{\lim}\limits_{n\to+\infty}\frac{1}{2\sqrt{n}\,( 1 - x^2)^{-n-1}}\\
&=-x^2\cdot0\\
&=0\,,
\end{align*} and because $f_n(0)=f_n(1)=0$, we have that $f_n(x)\xrightarrow{p.w.}0\,, \; x\in[0,1]\,.$
$f'_n(x)=-2x\,\sqrt {n}\,( 1-x^2)^{n-1}\,((n+1)\,x^2-1)$. It's easy to check that, for $n\in\mathbb{N}$, the functions $f_n(x)\,,\; x\in[0,1]\,,$ reach their total maximum value at $x=\frac{1}{\sqrt{n+1}}$ which is $\frac{\sqrt{n}}{n+1}\,\big(1-\frac{1}{n+1}\big)^n$. But then \begin{align*}
\mathop{\lim}\limits_{n\to+\infty}\sup\big\{|f_n(x)-0|\; |\; x\in[0,1]\big\}&=\mathop{\lim}\limits_{n\to+\infty}\frac{\sqrt{n}}{n+1}\Big(1-\frac{1}{n+1}\Big)^n\\ &=\mathop{\lim}\limits_{n\to+\infty}\frac{n\sqrt{n}}{(n+1)^2}\,\mathop{\lim}\limits_{n\to+\infty}\Big(1-\frac{1}{n+1}\Big)^{n+1}\\
&=0\cdot{\rm{e}}^{-1}\\
&=0\,.\end{align*} So, the sequence $\{f_n\}_{n\in\mathbb{N}}$ converges uniformly to zero function.
[/centre]
ii.Because for $x\in\mathbb{R}\,,\; n\in\mathbb{N}$ we have that $\Big|\frac{\sin (n^2 x)}{n (x^2+n)}\Big|\leqslant\frac{1}{n^2}$ and the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges, by the Weierstass criterion, we conclude that the series of functions $\sum_{n=1}^{\infty} \frac{\sin (n^2 x)}{n (x^2+n)}$ converges uniformly on $\mathbb{R}$.
\displaystyle\mathop{\lim}\limits_{n\to+\infty}f_n(x)&=\mathop{\lim}\limits_{n\to+\infty}\sqrt{n} \,x^2 \,( 1 - x^2)^n\\
&=x^2\mathop{\lim}\limits_{n\to+\infty}\frac{\sqrt{n}}{( 1 - x^2)^{-n}}\\
&\stackrel{\frac{\infty}{\infty}}{=}x^2\mathop{\lim}\limits_{n\to+\infty}\frac{\frac{1}{2\sqrt{n}}}{-n\,( 1 - x^2)^{-n-1}}\\
&=-x^2\mathop{\lim}\limits_{n\to+\infty}\frac{1}{2\sqrt{n}\,( 1 - x^2)^{-n-1}}\\
&=-x^2\cdot0\\
&=0\,,
\end{align*} and because $f_n(0)=f_n(1)=0$, we have that $f_n(x)\xrightarrow{p.w.}0\,, \; x\in[0,1]\,.$
$f'_n(x)=-2x\,\sqrt {n}\,( 1-x^2)^{n-1}\,((n+1)\,x^2-1)$. It's easy to check that, for $n\in\mathbb{N}$, the functions $f_n(x)\,,\; x\in[0,1]\,,$ reach their total maximum value at $x=\frac{1}{\sqrt{n+1}}$ which is $\frac{\sqrt{n}}{n+1}\,\big(1-\frac{1}{n+1}\big)^n$. But then \begin{align*}
\mathop{\lim}\limits_{n\to+\infty}\sup\big\{|f_n(x)-0|\; |\; x\in[0,1]\big\}&=\mathop{\lim}\limits_{n\to+\infty}\frac{\sqrt{n}}{n+1}\Big(1-\frac{1}{n+1}\Big)^n\\ &=\mathop{\lim}\limits_{n\to+\infty}\frac{n\sqrt{n}}{(n+1)^2}\,\mathop{\lim}\limits_{n\to+\infty}\Big(1-\frac{1}{n+1}\Big)^{n+1}\\
&=0\cdot{\rm{e}}^{-1}\\
&=0\,.\end{align*} So, the sequence $\{f_n\}_{n\in\mathbb{N}}$ converges uniformly to zero function.
[/centre]
ii.Because for $x\in\mathbb{R}\,,\; n\in\mathbb{N}$ we have that $\Big|\frac{\sin (n^2 x)}{n (x^2+n)}\Big|\leqslant\frac{1}{n^2}$ and the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges, by the Weierstass criterion, we conclude that the series of functions $\sum_{n=1}^{\infty} \frac{\sin (n^2 x)}{n (x^2+n)}$ converges uniformly on $\mathbb{R}$.
Grigorios Kostakos
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