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Question Number 195126 by Erico last updated on 25/Jul/23

$$\mathrm{Soit}{f}_n\left(x\right)=2^{n+1}\left[\frac{\frac{1}{2^n}cotan\left(\frac{x}{2^n}\right)-cotanx}{sin\left(\frac{x}{2^n}\right)}\right]$$$$\text{Double subscripts: use braces to clarify}$$

Answered by MM42 last updated on 25/Jul/23

$$f_n\left(x\right)=\frac{2sinxcos\left(\frac{x}{2^n}\right)-2^{n+1}sin\left(\frac{x}{2^n}\right)cosx}{{sin}^2\left(\frac{x}{2^n}\right)sinx}$$$$=\frac{sin(x+\frac{x}{2^n})+sin(x-\frac{x}{2^n})-2^{n}sin(\frac{x}{2^n}+x)-2^{n}sin(\frac{x}{2^n}-x)}{{sin}^2\left(\frac{x}{2^n}\right)sinx}$$$$=\frac{(1-2^n)sin\left(\frac{1+2^n}{2^n}\right)x-(1+2^n)sin\left(\frac{1-2^n}{2^n}\right)}{{sin}^2\left(\frac{x}{2^n}\right)sinx}$$$$=2^n\times \frac{\frac{1-2^n}{2^n}sin\left(\frac{1+2^n}{2^n}\right)x-\frac{1+2^n}{2^n}sin\left(\frac{1-2^n}{2^n}\right)}{{sin}^2\left(\frac{x}{2^n}\right)sinx}$$$$Tip:ifx\to 0\Rightarrow asinbx-bsinax\sim \frac{ab(a-b)(a+b)}{6}{x}^3$$$$\Rightarrow {lim}_{x\to 0}{f}_n\left(x\right)={lim}_{x\to 0}{2}^n\times \frac{\left(\frac{1-2^n}{2^n}\right)\left(\frac{1+2^n}{2^n}\right)(\frac{1-2^n}{2^n}-\frac{1+2^n}{2^n})(\frac{1-2^n}{2^n}+\frac{1+2^n}{2^n})\frac{x^3}{6}}{{\left(\frac{x}{2^n}\right)}^2\times x}$$$$=\frac{(1-2^{2n})(-2^{n+1})\times 2^{3n}\left(2\right)}{2^{4n}\times 6}=$$$$=\frac{2^{2n+1}-2}{3}✓$$$$$$$${lim}_{n\to \mathrm{\infty }}\frac{f_n\left(x\right)}{2^{2n+2}}=\frac{1}{6}✓$$$$$$

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