Question Number 132153 by Ñï= last updated on 11/Feb/21
Commented by Ar Brandon last updated on 11/Feb/21
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Answered by Olaf last updated on 11/Feb/21
$$\mathrm{S}\left({x}\right)\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{2}{n}−\mathrm{1}\right)!!\frac{{x}^{{n}} }{{n}!} \\ $$$$\mathrm{Let}\:{f}\left({x}\right)\:=\:\left(\mathrm{1}−\mathrm{2}{x}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${f}'\left({x}\right)\:=\:\left(\mathrm{1}−\mathrm{2}{x}\right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${f}''\left({x}\right)\:=\:\mathrm{3}\left(\mathrm{1}−\mathrm{2}{x}\right)^{−\frac{\mathrm{5}}{\mathrm{2}}} \\ $$$${f}^{\left(\mathrm{3}\right)} \left({x}\right)\:=\:\mathrm{3}.\mathrm{5}\left(\mathrm{1}−\mathrm{2}{x}\right)^{−\frac{\mathrm{7}}{\mathrm{2}}} \\ $$$${f}^{\left(\mathrm{4}\right)} \left({x}\right)\:=\:\mathrm{3}.\mathrm{5}.\mathrm{7}\left(\mathrm{1}−\mathrm{2}{x}\right)^{−\frac{\mathrm{9}}{\mathrm{2}}} \\ $$$$… \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\:\mathrm{3}.\mathrm{5}.\mathrm{7}…\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{1}−\mathrm{2}{x}\right)^{−\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{2}}} \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\:\mathrm{3}.\mathrm{5}.\mathrm{7}…\left(\mathrm{2}{n}−\mathrm{1}\right)\:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{2}{k}−\mathrm{1}\right)\:=\:\left(\mathrm{2}{n}−\mathrm{1}\right)!! \\ $$$${f}\left({x}\right)\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{f}^{\left({n}\right)} \left(\mathrm{0}\right)\frac{{x}^{{n}} }{{n}!}\:=\:\mathrm{S}\left({x}\right) \\ $$