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If $a_i$ are positive real roots of transcendental equation$$\left( {\cos x} \right)\left( {\cosh x} \right) + 1 = 0.$$And we have $0<a_1<a_2<a_3<\cdots$, prove: $$\sum_{i=1}^{\infty}{a_{i}^{-6}\left(\frac{\sin a_i-\sinh a_i}{\cos a_i+\cosh a_i}\right)^2}=\frac{1}{80}.$$

First, we have$${\left( {\sinh {a_i}} \right)^2} = {\left( {\cosh {a_i}} \right)^2} - 1 = \frac{1}{{{{\cos }^2}{a_i}}} - 1 = {\tan ^2}{a_i} \Rightarrow \sinh {a_i} = \left| {\tan {a_i}} \right|.$$

If $\sinh {a_i} = \tan {a_i}$, we obtain$${\left( {\frac{{\sin {a_i} - \sinh {a_i}}}{{\cos {a_i} + \cosh {a_i}}}} \right)^2} = {\left( {\frac{{\sin {a_i} - \tan {a_i}}}{{\cos {a_i} - \frac{1}{{\cos {a_i}}}}}} \right)^2} = {\tan ^2}\frac{{{a_i}}}{2}.$$

If $\sinh {a_i} = -\tan {a_i}$, we obtain$${\left( {\frac{{\sin {a_i} - \sinh {a_i}}}{{\cos {a_i} + \cosh {a_i}}}} \right)^2} = {\left( {\frac{{\sin {a_i} + \tan {a_i}}}{{\cos {a_i} - \frac{1}{{\cos {a_i}}}}}} \right)^2} = {\cot ^2}\frac{{{a_i}}}{2}.$$ In fact, both cases will happen.

Eufisky
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    The roots are approximately $a_k\approx(2k-1)\frac\pi2+(-1)^k\operatorname{sech}\left((2k-1)\frac\pi2\right)$. That is, the sign of $\tan(a_k)$ is $(-1)^k$. – robjohn Nov 16 '17 at 15:22

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