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

Commented bytabata last updated on 24/Jul/21

$${Msr}\:{olaf}\:{help}\:{me}\:{please}\:? \\$$

Answered by Olaf_Thorendsen last updated on 24/Jul/21

$${h}\left({z}\right)\:=\:\frac{\mathrm{1}}{{z}^{\mathrm{2}} \mathrm{sinh}{z}}\:=\:\frac{{f}\left({z}\right)}{{z}^{\mathrm{2}} },\:\mid{z}\mid\:<\:\mathrm{R} \\$$ $$\mathrm{The}\:\mathrm{zeros}\:\mathrm{of}\:\mathrm{sinh}\left({z}\right)\:\mathrm{are}\:\left\{{k}\pi{i}\right\}_{{k}\in\mathbb{Z}} \:\:\mathrm{so} \\$$ $$\mathrm{the}\:\mathrm{highest}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{for}\:\mathrm{R}\:\mathrm{is}\:\pi. \\$$ $$\\$$ $$\mathrm{For}\:\mid{z}\mid<\pi,\:\mathrm{we}\:\mathrm{have} \\$$ $${f}\left({z}\right)\:=\:\frac{\mathrm{1}}{\mathrm{sinh}{z}}\:=\:\frac{\mathrm{1}}{{z}+\frac{{z}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{z}^{\mathrm{5}} }{\mathrm{5}!}+\frac{{z}^{\mathrm{7}} }{\mathrm{7}!}...} \\$$ $${f}\left({z}\right)\:=\:\frac{\mathrm{1}}{{z}}.\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{{z}^{\mathrm{2}} }{\mathrm{6}}+\frac{{z}^{\mathrm{4}} }{\mathrm{120}}+\frac{{z}^{\mathrm{6}} }{\mathrm{5040}}...\right)} \\$$ $${f}\left({z}\right)\:=\:\frac{\mathrm{1}}{{z}}\left(\mathrm{1}−\left(\frac{{z}^{\mathrm{2}} }{\mathrm{6}}+\frac{{z}^{\mathrm{4}} }{\mathrm{120}}+\frac{{z}^{\mathrm{6}} }{\mathrm{5040}}+...\right)+\left(\frac{{z}^{\mathrm{2}} }{\mathrm{6}}+\frac{{z}^{\mathrm{4}} }{\mathrm{120}}+\frac{{z}^{\mathrm{6}} }{\mathrm{5040}}+...\right)^{\mathrm{2}} +...\right) \\$$ $${f}\left({z}\right)\:=\:\frac{\mathrm{1}}{{z}}\left(\mathrm{1}−\frac{{z}^{\mathrm{2}} }{\mathrm{6}}−\frac{{z}^{\mathrm{4}} }{\mathrm{120}}−\frac{{z}^{\mathrm{6}} }{\mathrm{5040}}+\frac{{z}^{\mathrm{4}} }{\mathrm{36}}+\frac{{z}^{\mathrm{6}} }{\mathrm{360}}...\right) \\$$ $${f}\left({z}\right)\:=\:\frac{\mathrm{1}}{{z}}\left(\mathrm{1}−\frac{{z}^{\mathrm{2}} }{\mathrm{6}}+\frac{\mathrm{7}{z}^{\mathrm{4}} }{\mathrm{360}}+\frac{\mathrm{13}{z}^{\mathrm{6}} }{\mathrm{5040}}+...\right) \\$$ $$\Rightarrow\:{h}\left({z}\right)\:=\:\frac{\mathrm{1}}{{z}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{6}{z}}+\frac{\mathrm{7}{z}}{\mathrm{360}}+\frac{\mathrm{13}{z}^{\mathrm{3}} }{\mathrm{5040}}+... \\$$

Commented bytabata last updated on 24/Jul/21

$${thank}\:{you}\:{msr}\:{olaf}\: \\$$

Commented bytabata last updated on 24/Jul/21

$${sir}\:{can}\:{how}\:{exactly}\:\left(\frac{{z}^{\mathrm{2}} }{\mathrm{6}}+\frac{{z}^{\mathrm{4}} }{\mathrm{120}}+\frac{{z}^{\mathrm{6}} }{\mathrm{5040}}+...\right)^{\mathrm{2}} {how}\:{me}\:{solution}\:{please}? \\$$

Commented bytabata last updated on 24/Jul/21

$${and}\:{sir}\:{how}\:\frac{\mathrm{1}}{{z}}\left(\mathrm{1}−\frac{{z}^{\mathrm{2}} }{\mathrm{6}}+\frac{\mathrm{7}{z}^{\mathrm{4}} }{\mathrm{360}}+\frac{\mathrm{13}{z}^{\mathrm{6}} }{\mathrm{5040}}+...\right)=\frac{\mathrm{1}}{{z}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{6}{z}}+\frac{\mathrm{7}{z}}{\mathrm{360}}+\frac{\mathrm{13}{z}^{\mathrm{3}} }{\mathrm{5040}} \\$$ $$\\$$ $${i}\:{want}\:{clearly}\:{that}\:{because}\:{its}\:{very}\:{hard}\:? \\$$

Commented byOlaf_Thorendsen last updated on 24/Jul/21

$$\mathrm{sinh}{x}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\$$ $$\frac{\mathrm{1}}{\mathrm{1}+{u}}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {u}^{{n}} \\$$

Commented bymr W last updated on 24/Jul/21

$${please}\:{recheck}\:{sir}: \\$$ $${i}\:{think}\:{it}\:{should}\:{be} \\$$ $${f}\left({z}\right)\:=\:\frac{\mathrm{1}}{{z}}\left(\mathrm{1}−\frac{{z}^{\mathrm{2}} }{\mathrm{6}}−\frac{{z}^{\mathrm{4}} }{\mathrm{120}}−\frac{{z}^{\mathrm{6}} }{\mathrm{5040}}+\frac{{z}^{\mathrm{4}} }{\mathrm{36}}+\frac{{z}^{\mathrm{6}} }{\mathrm{360}}−\frac{{z}^{\mathrm{6}} }{\mathrm{216}}...\right) \\$$ $${f}\left({z}\right)\:=\:\frac{\mathrm{1}}{{z}}\left(\mathrm{1}−\frac{{z}^{\mathrm{2}} }{\mathrm{6}}+\frac{\mathrm{7}{z}^{\mathrm{4}} }{\mathrm{360}}−\frac{\mathrm{31}{z}^{\mathrm{6}} }{\mathrm{15120}}+...\right) \\$$ $$\Rightarrow\:{h}\left({z}\right)\:=\:\frac{\mathrm{1}}{{z}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{6}{z}}+\frac{\mathrm{7}{z}}{\mathrm{360}}−\frac{\mathrm{31}{z}^{\mathrm{3}} }{\mathrm{15120}}+... \\$$

Commented byOlaf_Thorendsen last updated on 24/Jul/21

$$\mathrm{You}\:\mathrm{are}\:\mathrm{right}\:\mathrm{mister}.\:\mathrm{I}\:\mathrm{forgot}\:\mathrm{one}\:\mathrm{term}. \\$$