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Question Number 147948 by tabata last updated on 24/Jul/21

Commented by tabata last updated on 24/Jul/21

$M s r o l a f h e l p m e p l e a s e ?$
	Answered by Olaf_Thorendsen last updated on 24/Jul/21
	$h(z) = (1/(z^2 sinhz)) = ((f(z))/z^2 ), ∣z∣ < R The zeros of sinh(z) are {kπi}_(k∈Z) so the highest possible value for R is π. For ∣z∣<π, we have f(z) = (1/(sinhz)) = (1/(z+(z^3 /(3!))+(z^5 /(5!))+(z^7 /(7!))...)) f(z) = (1/z).(1/(1+((z^2 /6)+(z^4 /(120))+(z^6 /(5040))...))) f(z) = (1/z)(1−((z^2 /6)+(z^4 /(120))+(z^6 /(5040))+...)+((z^2 /6)+(z^4 /(120))+(z^6 /(5040))+...)^2 +...) f(z) = (1/z)(1−(z^2 /6)−(z^4 /(120))−(z^6 /(5040))+(z^4 /(36))+(z^6 /(360))...) f(z) = (1/z)(1−(z^2 /6)+((7z^4 )/(360))+((13z^6 )/(5040))+...) ⇒ h(z) = (1/z^3 )−(1/(6z))+((7z)/(360))+((13z^3 )/(5040))+...$
	$h (z) = \frac{1}{z^{2} \sinh z} = \frac{f (z)}{z^{2}}, ∣ z ∣ < R$ $The zeros of \sinh (z) are {k π i}_{k \in Z} so$ $the highest possible value for R is π .$ $For ∣ z ∣< π, we have$ $f (z) = \frac{1}{\sinh z} = \frac{1}{z + \frac{z^{3}}{3!} + \frac{z^{5}}{5!} + \frac{z^{7}}{7!} . . .}$ $f (z) = \frac{1}{z} . \frac{1}{1 + (\frac{z^{2}}{6} + \frac{z^{4}}{120} + \frac{z^{6}}{5040} . . .)}$ $f (z) = \frac{1}{z} (1 - (\frac{z^{2}}{6} + \frac{z^{4}}{120} + \frac{z^{6}}{5040} + . . .) + {(\frac{z^{2}}{6} + \frac{z^{4}}{120} + \frac{z^{6}}{5040} + . . .)}^{2} + . . .)$ $f (z) = \frac{1}{z} (1 - \frac{z^{2}}{6} - \frac{z^{4}}{120} - \frac{z^{6}}{5040} + \frac{z^{4}}{36} + \frac{z^{6}}{360} . . .)$ $f (z) = \frac{1}{z} (1 - \frac{z^{2}}{6} + \frac{7 z^{4}}{360} + \frac{13 z^{6}}{5040} + . . .)$ $\Rightarrow h (z) = \frac{1}{z^{3}} - \frac{1}{6 z} + \frac{7 z}{360} + \frac{13 z^{3}}{5040} + . . .$
	Commented by tabata last updated on 24/Jul/21

	$t h a n k y o u m s r o l a f$
	Commented by tabata last updated on 24/Jul/21

	$s i r c a n h o w e x a c t l y {(\frac{z^{2}}{6} + \frac{z^{4}}{120} + \frac{z^{6}}{5040} + . . .)}^{2} h o w m e s o l u t i o n p l e a s e ?$
	Commented by tabata last updated on 24/Jul/21

	$a n d s i r h o w \frac{1}{z} (1 - \frac{z^{2}}{6} + \frac{7 z^{4}}{360} + \frac{13 z^{6}}{5040} + . . .) = \frac{1}{z^{3}} - \frac{1}{6 z} + \frac{7 z}{360} + \frac{13 z^{3}}{5040}$ $i w a n t c l e a r l y t h a t b e c a u s e i t s v e r y h a r d ?$
	Commented by Olaf_Thorendsen last updated on 24/Jul/21

	$\sinh x = \underset{n = 0}{\sum^{\infty}} \frac{x^{2 n + 1}}{(2 n + 1)!}$ $\frac{1}{1 + u} = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} u^{n}$
	Commented by mr W last updated on 24/Jul/21

	$p l e a s e r e c h e c k s i r :$ $i t h i n k i t s h o u l d b e$ $f (z) = \frac{1}{z} (1 - \frac{z^{2}}{6} - \frac{z^{4}}{120} - \frac{z^{6}}{5040} + \frac{z^{4}}{36} + \frac{z^{6}}{360} - \frac{z^{6}}{216} . . .)$ $f (z) = \frac{1}{z} (1 - \frac{z^{2}}{6} + \frac{7 z^{4}}{360} - \frac{31 z^{6}}{15120} + . . .)$ $\Rightarrow h (z) = \frac{1}{z^{3}} - \frac{1}{6 z} + \frac{7 z}{360} - \frac{31 z^{3}}{15120} + . . .$
	Commented by Olaf_Thorendsen last updated on 24/Jul/21

	$You are right mister . I forgot one term .$
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