Study the uniform convergence of $f_n(x)=\cos^nx\sin^{2n}x$ on $\mathbb{R}$

My attempt:

Obviuosly $f_n(x)$ converges pointwise to $0,$ so I would like to proof uniform convergence to $0.$

I calculated the derivative $f_n(x):$ $n\cos^{n-1}(x)\sin^{2n-1}(x)(3\cos^2(x)-1)$

It's roots are: $0,\frac{\pi}{2},\arccos(\frac{1}{\sqrt{3}})$

$\arccos(\frac{1}{\sqrt{3}})$ turns out to be maxima point, placing it in $f_n,$ by WolframAlpha we get $2^n3^{-3n/2}\rightarrow0$ as $n\to \infty$.

Therefore $f_n(x) \xrightarrow{u}0.$

My questions are:

$1)$ How that result was obtained by WolframAlpha?

$2)$ Is there a simpler way to prove it, such as bounding from above by something that converges to $0?$

Any help appreciated.


Since: $$\sin x\cos x=\frac{1}{2}\sin 2x$$ you have $$\cos ^{n}x\sin^{2n}x =\frac{1}{2^n}\sin^n 2x\sin^n x$$

So $|f_n(x)|$ is bounded by...

It's very hard to say how WA calculates anything, but if $\cos x=\frac1{\sqrt{3}}=3^{-1/2}$ then $\sin x=\pm \frac{\sqrt{2}}{\sqrt{3}}=\pm 2^{1/2}{3^{-1/2}}$.


$f_n $ is even and $2\pi $ periodic. So we take only $[0,\pi] $.

the roots of $f'_n (x)=0$ are

$x_1=0 , f_n (x_1)=0$

$x_2=\pi/2 ; f_n (x_2)=0$

$x_3=\pi ; f_n (x_3)=0$

$\cos (x_4)=1/\sqrt {3 } ;$ $$ f_n (x_4)=(1/3)^\frac n2 (1-1/3)^n=$$

$$=3^{-n/2-n}2^n $$ $$=(\frac {2}{3\sqrt {3}})^n=\max_{[0,\pi]} |f| =\max_{\Bbb R} |f| $$ $$\to 0$$

  • $\begingroup$ I found the same bound. $\endgroup$ – Gribouillis Aug 20 '17 at 13:36

$f_1(x) = \cos(x)\sin^2(x)$ has a max and a min, since it's contunious and periodic. More specifically, $|f_1(x)| = |\cos(x)\sin^2(x)|$ has a max, let's say $g$. Then $$|f_n(x)| = |\cos(x)\sin^2(x)|^n\leq g^n$$ for all $x$. If $g<1$, then $g^n\to 0$.

It might be worth noting, if you do this by hand and want exact results, that $|\sin(\arccos(x))| = \sqrt{1-x^2}$. If you don't want exact convergence results, and you rather just want to prove convergence, it's really pretty clear that $g<1$ just from inspecting the sine and cosine function.

WolframAlpha probably calculated $f_n(\arccos(1/\sqrt3))$ the hard way, but relatively few have actual insight into the inner workings of WA, so it's hard to tell.


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