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How to find all the solutions of the following equation?

$1+\frac{x}{2!}+\frac{{{x}^{2}}}{4!}+\frac{{{x}^{3}}}{6!}+\frac{{{x}^{4}}}{8!}+...=0$

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3 Answers 3

up vote 3 down vote accepted

$$ \cosh x = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots $$

and

$$ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \dots $$

This means that for $x > 0$, $$\cosh \sqrt x = 1 + \frac{x}{2!} + \frac{x^2}{4!} + \dots $$

which clearly has no roots. For $x < 0$, $$ \cos\sqrt{-x} = 1 + \frac{x}{2!} + \frac{x^2}{4!} + \dots $$

So you are after negative values of $x$ such that $\cos \sqrt{-x} = 0$

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So the solutions set would be $\left\{ \left. -{{\pi }^{2}}\left( k+\frac{1}{2} \right) \right|k\ge 0 \right\}$. Thank you very much for the help! –  Sleepingip Mar 15 '13 at 17:20
    
Not quite, it would be $\{ -( \pi(k + \frac{1}{2}))^2 | k \geq 0 \}$ –  muzzlator Mar 15 '13 at 17:22
    
Oh, sorry! Typing error there. Thanks for the correction. –  Sleepingip Mar 15 '13 at 17:26

Notice that for $x \ge 0$, $\cosh \sqrt{x}= 1+\frac{x}{2!}+\frac{{{x}^{2}}}{4!}+\frac{{{x}^{3}}}{6!}+\frac{{{x}^{4}}}{8!}+...$, and for $x<0$, $\cos \sqrt{|x|}= 1-\frac{|x|}{2!}+\frac{{{|x|}^{2}}}{4!}-\frac{{{|x|}^{3}}}{6!}+\frac{{{|x|}^{4}}}{8!}-...$

Since $\cosh y >0$ for all $y$, we need only look for negative solutions to $\cos \sqrt{|x|}= 0$.

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And the difference is... –  copper.hat Mar 15 '13 at 17:11

Hint: Write down the series expansion for $e^t$, also for $e^{-t}$.

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will we need to prove convergence of the LHS for this equation? –  Vincent Tjeng Mar 15 '13 at 16:57
    
My guess is that since it is well-known that the series for $e^t$ converges for all $t$, you can freely manipulate without detailed step by step justification. –  André Nicolas Mar 15 '13 at 17:05

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