Question Number 186689 by SANOGO last updated on 08/Feb/23
$$\underset{{n}={o}} {\overset{+{oo}} {\sum}}\:\frac{{x}^{\mathrm{3}{n}} }{\left(\mathrm{2}{n}\right)!}\:\:=\:\:\:? \\ $$
Commented by mr W last updated on 08/Feb/23
$${why}\:{not}\:\infty\:{instead}\:{of}\:{oo}? \\ $$$${why}\:{not}\:\mathrm{0}\:{instead}\:{of}\:{o}? \\ $$
Answered by mr W last updated on 08/Feb/23
$${e}^{{x}} =\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}+…\:\:\:\:\:\left({i}\right) \\ $$$${e}^{−{x}} =\mathrm{1}−{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}−\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}+…\:\:\:\:\:\left({ii}\right) \\ $$$$\frac{\left({i}\right)+\left({ii}\right)}{\mathrm{2}}: \\ $$$$\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}}=\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}+…=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!} \\ $$$${replace}\:{x}\:{with}\:{x}^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\frac{{e}^{\frac{\mathrm{3}{x}}{\mathrm{2}}} +{e}^{−\frac{\mathrm{3}{x}}{\mathrm{2}}} }{\mathrm{2}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{3}{n}} }{\left(\mathrm{2}{n}\right)!} \\ $$$${or} \\ $$$$\mathrm{cosh}\:\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{3}{n}} }{\left(\mathrm{2}{n}\right)!}\:\:\checkmark \\ $$