Solve-1-x-2-x-2-4-x-3-6-0-

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

Solve

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

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

i f x > 0, L H S > 0 \Rightarrow n o s o l u t i o n

i . e . x < 0

l e t x = - t^{2}

1 + \frac{x}{2!} + \frac{x^{2}}{4!} + \frac{x^{3}}{6!} + \cdot \cdot \cdot

= 1 - \frac{t^{2}}{2!} + \frac{t^{4}}{4!} - \frac{t^{6}}{6!} + \dots . = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} t^{2 n}}{(2 n)!}

= \cos t = 0

\Rightarrow t = 2 k π \pm \frac{π}{2}

\Rightarrow x = - {(2 k π \pm \frac{π}{2})}^{2} w i t h k \in Z

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

��yes !