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Question Number 92923 by Ar Brandon last updated on 09/May/20

$$\mathrm{Solve}$$$$1+\frac{\mathrm{x}}{2!}+\frac{{\mathrm{x}}^2}{4!}+\frac{{\mathrm{x}}^3}{6!}+\cdot \cdot \cdot =0$$

Answered by mr W last updated on 09/May/20

$$ifx>0,LHS>0\Rightarrow nosolution$$$$i.e.x<0$$$$letx=-t^2$$$$1+\frac{\mathrm{x}}{2!}+\frac{{\mathrm{x}}^2}{4!}+\frac{{\mathrm{x}}^3}{6!}+\cdot \cdot \cdot$$$$=1-\frac{t^2}{2!}+\frac{t^4}{4!}-\frac{t^6}{6!}+....=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{t}^{2n}}{\left(2n\right)!}$$$$=\cos t=0$$$$\Rightarrow t=2k\pi \pm \frac{\pi }{2}$$$$\Rightarrow x=-{(2k\pi \pm \frac{\pi }{2})}^{2}withk\in \mathbb{Z}$$

Commented by Ar Brandon last updated on 09/May/20

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