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Question Number 34696 by abdo imad last updated on 10/May/18

$$letf\left(x\right)=\sqrt{2-x}$$$$developpfatintegrserieandgivetheradiusof$$$$convergence.$$

Commented by math khazana by abdo last updated on 11/May/18

$$f\left(x\right)=\sqrt{2}\sqrt{1-\frac{x}{2}}=\sqrt{2}{(1-\frac{x}{2})}^{\frac{1}{2}}butweknowthat$$$${(1+u)}^{\alpha }=1+\alpha u\frac{\alpha (\alpha -1)}{2!}{u}^2+...$$$$=1+\sum _{n=1}^{\mathrm{\infty }}\frac{\alpha (\alpha -1)...(\alpha -n+1)}{n!}{u}^n$$$$and{(1-u)}^n=1+\sum _{n=1}^{\mathrm{\infty }}\alpha (\alpha -1)...(\alpha -n+1)\frac{{(-1)}^n}{n!}{u}^n$$$${(1-\frac{x}{2})}^{\frac{1}{2}}=1+\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2}(\frac{1}{2}-1)...(\frac{1}{2}-n+1)\frac{{(-1)}^n}{n!}\frac{x^n}{2^n}$$$$=1+\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n!2^{n+1}}(-\frac{1}{2})...(-\frac{3}{2})...(-\frac{2n-3}{2})x^n$$$$=1+\sum _{n=1}^{\mathrm{\infty }}\frac{(-1)}{n!2^{n+1}}.\frac{1.3.5.....(2n-3)}{2^{n-1}}{x}^n$$$$=1+\sum _{n=1}^{\mathrm{\infty }}\frac{-1}{n!2^{2n}}(1.3.5.....(2n-3))x^n\Rightarrow$$$$f\left(x\right)=\sqrt{2}-\sqrt{2}\sum _{n=1}^{\mathrm{\infty }}\frac{1.3.5....(2n-3)}{n!2^{2n}}{x}^n.$$$$$$

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