Find-a-series-for-x-2-tanh-xpi-tan-xpi-

Question Number 136132 by frc2crc last updated on 19/Mar/21

F i n d a s e r i e s f o r \frac{x^{2}}{\tanh (x π) \tan (x π)}

Answered by Dwaipayan Shikari last updated on 19/Mar/21

s i n h (π x) = π x \underset{n = 1}{\prod^{\infty}} (1 + \frac{x^{2}}{n^{2}})

\Rightarrow \frac{1}{t a n h (π x)} = \frac{d}{d x} (l o g (π x) + \underset{n = 1}{\sum^{\infty}} l o g (1 + \frac{x^{2}}{n^{2}}))

\Rightarrow \frac{1}{t a n h (π x)} = \frac{1}{x} + 2 \underset{n = 1}{\sum^{\infty}} \frac{x}{n^{2} + x^{2}}

S i m i l a r l y \frac{1}{t a n (π x)} = \frac{1}{x} - 2 \underset{n = 1}{\sum^{\infty}} \frac{x}{n^{2} - x^{2}}

\frac{x^{2}}{t a n h (π x) t a n (π x)} = 1 + 2 (\underset{n = 1}{\sum^{\infty}} \frac{x^{2}}{n^{2} + x^{2}} - \underset{n = 1}{\sum^{\infty}} \frac{x^{2}}{n^{2} - x^{2}}) - 4 x^{4} (\underset{n = 1}{\sum^{\infty}} \frac{1}{(n^{2} + x^{2})} \underset{n = 1}{\sum^{\infty}} \frac{1}{(n^{2} - x^{2})})

Commented by frc2crc last updated on 19/Mar/21

i w o u l d h a v e a s u m l i k e

\underset{n = 1}{\sum^{\infty}} a_{n} {(x π)}^{2 n - 1}

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