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Question Number 117311 by eric last updated on 10/Oct/20

Answered by Dwaipayan Shikari last updated on 10/Oct/20

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{2n+1}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+.....$$$${tan}^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+....$$$${tan}^{-1}1=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+....$$$$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+....=\frac{\pi }{4}$$

Commented by TANMAY PANACEA last updated on 10/Oct/20

$$plsprove{tan}^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+..$$$$pls$$

Commented by Dwaipayan Shikari last updated on 10/Oct/20

$$f\left(x\right)={tan}^{-1}x$$$$f^{\prime }\left(x\right)=\frac{1}{1+x^2}={(1+x^2)}^{-1}=1-x^2+x^4-x^6+...$$$$f\left(x\right)=\int 1-x^2+x^4-x^6+..$$$$f\left(x\right)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+....+C$$$$f\left(0\right)={tan}^{-1}\left(0\right)=0$$$$C=0$$$$f\left(x\right)={tan}^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+....$$$$$$

Commented by TANMAY PANACEA last updated on 10/Oct/20

$$excellent$$

Answered by mathmax by abdo last updated on 10/Oct/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{2\mathrm{n}+1}{\mathrm{x}}^{2\mathrm{n}+1}\mathrm{with}\mid \mathrm{x}\mid ⩽1\mathrm{we}\mathrm{have}$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-{\mathrm{x}}^2)}^{\mathrm{n}}=\frac{1}{1+{\mathrm{x}}^2}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=\int \frac{\mathrm{dx}}{1+{\mathrm{x}}^2}+\mathrm{c}=\mathrm{arctanx}+\mathrm{c}$$$$\mathrm{f}\left(0\right)=0=\mathrm{c}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\mathrm{arctanx}\mathrm{and}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{2\mathrm{n}+1}=\mathrm{f}\left(1\right)=\arctan \left(1\right)=\frac{\pi }{4}$$

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