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Question Number 132153 by Ñï= last updated on 11/Feb/21

Commented by Ar Brandon last updated on 11/Feb/21

No username ?! ����

Answered by Olaf last updated on 11/Feb/21

$$\mathrm{S}\left(x\right)=\underset{n=0}{\sum ^{\mathrm{\infty }}}(2n-1)!!\frac{x^n}{n!}$$$$\mathrm{Let}f\left(x\right)={(1-2x)}^{-\frac{1}{2}}$$$$f^{\prime }\left(x\right)={(1-2x)}^{-\frac{3}{2}}$$$$f^{″}\left(x\right)=3{(1-2x)}^{-\frac{5}{2}}$$$$f^{\left(3\right)}\left(x\right)=3.5{(1-2x)}^{-\frac{7}{2}}$$$$f^{\left(4\right)}\left(x\right)=3.5.7{(1-2x)}^{-\frac{9}{2}}$$$$...$$$$f^{\left(n\right)}\left(x\right)=3.5.7...(2n-1){(1-2x)}^{-\frac{(2n+1)}{2}}$$$$f^{\left(n\right)}\left(0\right)=3.5.7...(2n-1)=\underset{k=1}{\prod ^n}(2k-1)=(2n-1)!!$$$$f\left(x\right)=\underset{n=0}{\sum ^{\mathrm{\infty }}}{f}^{\left(n\right)}\left(0\right)\frac{x^n}{n!}=\mathrm{S}\left(x\right)$$

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