Previous in Relation and Functions Next in Relation and Functions
Question Number 40114 by maxmathsup by imad last updated on 15/Jul/18
$${let}\:{g}\left({x}\right)=\:{cos}\left({x}+\mathrm{1}\right) \\ $$$${developp}\:{g}\:{at}\:{integr}\:{serie} \\ $$
Commented by prof Abdo imad last updated on 17/Jul/18
$${we}\:{have}\:{g}\left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{g}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \\ $$$${but}\:{g}^{\left({n}\right)} \left({x}\right)={cos}\left({x}+\mathrm{1}+\frac{{n}\pi}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${g}^{\left({n}\right)} \left(\mathrm{0}\right)\:={cos}\left(\mathrm{1}+\frac{{n}\pi}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${g}\left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{cos}\left(\mathrm{1}+\frac{{n}\pi}{\mathrm{2}}\right)\frac{{x}^{{n}} }{{n}!} \\ $$