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Question Number 210441 by universe last updated on 09/Aug/24

Answered by aleks041103 last updated on 09/Aug/24

$$\left(i\right)$$$$foroddn,thefunction{sin}^n\left(nx\right)issimilar$$$$tosin\left(x\right)withthefactthattheprimitivefunctionoscilates$$$$between0andthemaximalpossiblevalue$$$${\int }_0^{\pi /n}{sin}^n\left(nx\right)dx=\frac{1}{n}{\int }_0^{\pi }{sin}^n\left(x\right)dx⩽$$$$⩽\frac{1}{n}{\int }_0^{\pi }sin\left(x\right)dx=\frac{2}{n}$$$$$$$$s_n=F_n\left(1\right)-F_n\left(0\right)$$$$foroddnwehave{F}_n\left(x\right)\in [0,2/n]$$$$then{s}_n⩽\frac{2}{n},oddn.$$

Answered by aleks041103 last updated on 09/Aug/24

$$\left(ii\right)forevenn:$$$${\int }_0^1{sin}^n\left(nx\right)dx=\frac{1}{n}{\int }_0^n{sin}^n\left(x\right)dx=$$$$=\frac{1}{n}[\underset{k=0}{\sum ^{\lfloor n/\pi \rfloor -1}}{\int }_{\pi k}^{\pi (k+1)}{sin}^n\left(x\right)dx+{\int }_{\pi \lfloor n/\pi \rfloor }^n{sin}^n\left(x\right)dx]=$$$$=\frac{1}{n}(\lfloor \frac{n}{\pi }\rfloor +c){\int }_0^{\pi }{sin}^n\left(x\right)dx$$$$where0<c<1.$$$$$$$$nowlet$$$$I_n={\int }_0^{\pi }{sin}^{2n}\left(x\right)dx=$$$$={\int }_0^{\pi }{sin}^{2n-2}\left(x\right)dx-{\int }_0^{\pi }cos\left(x\right)d\left(\frac{{sin}^{2n-1}\left(x\right)}{2n-1}\right)dx=$$$$=I_{n-1}-{\left[\frac{{sin}^{2n-1}\left(x\right)cos\left(x\right)}{2n-1}\right]}_0^{\pi }-\frac{1}{2n-1}{\int }_0^{\pi }{sin}^{2n}\left(x\right)dx=$$$$=I_{n-1}-\frac{1}{2n-1}{I}_n$$$$\Rightarrow I_{n-1}=\frac{2n}{2n-1}{I}_n$$$$\Rightarrow I_n=\frac{2n-1}{2n}{I}_{n-1}$$$$\Rightarrow I_n=\frac{(2n-1)(2n-3)...3.1}{\left(2n\right)(2n-2)...4.2}{I}_0$$$$I_0={\int }_0^{\pi }dx=\pi$$$$\Rightarrow I_n=\frac{(2n-1)!!}{\left(2n\right)!!}\pi$$$$\left(2n\right)!!=2^{n}n!$$$$(2n-1)!!=\frac{\left(2n\right)!}{\left(2n\right)!!}=\frac{\left(2n\right)!}{2^{n}n!}$$$$\Rightarrow I_n=\frac{\left(2n\right)!}{4^n{(n!)}^2}\pi$$$$\Rightarrow s_{2n}=\frac{1}{2n}(\lfloor \frac{2n}{\pi }\rfloor +c_n)I_n=\frac{\pi }{2n}(\lfloor \frac{2n}{\pi }\rfloor +c_n)\frac{\left(2n\right)!}{4^n{(n!)}^2}$$$$Stirling:$$$$x!\sim \sqrt{2\pi x}{\left(\frac{x}{e}\right)}^x$$$$\Rightarrow \left(2n\right)!\sim \sqrt{4\pi n}{\left(\frac{2n}{e}\right)}^{2n}=\sqrt{4\pi n}{4}^n{\left(\frac{n}{e}\right)}^{2n}$$$$\Rightarrow {(n!)}^2\sim 2\pi n{\left(\frac{n}{e}\right)}^{2n}$$$$\Rightarrow s_{2n}=\frac{\pi }{2n}(\lfloor \frac{2n}{\pi }\rfloor +c_n)\frac{\left(2n\right)!}{4^n{(n!)}^2}\sim \frac{\pi }{2n}(\lfloor \frac{2n}{\pi }\rfloor +c_n)\frac{\sqrt{4\pi n}}{2\pi n}$$$$\Rightarrow s_{2n}\sim \frac{\pi }{2n}(\lfloor \frac{2n}{\pi }\rfloor +c_n)\frac{1}{\sqrt{\pi n}}$$$$itiseasytoseethat\frac{\pi }{2n}\lfloor \frac{2n}{\pi }\rfloor \sim 1$$$$\Rightarrow s_{2n}\sim \frac{1}{\sqrt{\pi n}}$$$$\Rightarrow s_{2n}\to 0$$$$$$$$from\left(i\right)wehave0<s_{2n+1}<\frac{2}{2n+1}$$$$\Rightarrow s_{2n+1}\to 0$$$$$$$$\Rightarrow s_n\to 0$$$$\Rightarrow theonlylimitingpointis0.$$

Commented by hardmath last updated on 11/Aug/24

$$$$Dear professor, your solutions are perfect

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