What is the Maclaurin series of

${z}^{3}\mathrm{sin}({z}^{2})$

${z}^{3}\mathrm{sin}({z}^{2})$

Ryder Ferguson
2022-10-14
Answered

What is the Maclaurin series of

${z}^{3}\mathrm{sin}({z}^{2})$

${z}^{3}\mathrm{sin}({z}^{2})$

You can still ask an expert for help

Kenley Rasmussen

Answered 2022-10-15
Author has **13** answers

Recall that

$\mathrm{sin}(z)=\sum _{n=0}^{\mathrm{\infty}}\frac{(-1{)}^{n}{z}^{2n+1}}{(2n+1)!}$

Thus,

$\mathrm{sin}({z}^{2})=\sum _{n=0}^{\mathrm{\infty}}\frac{(-1{)}^{n}{z}^{4n+2}}{(2n+1)!}$

and finally

${z}^{3}\mathrm{sin}({z}^{2})=\sum _{n=0}^{\mathrm{\infty}}\frac{(-1{)}^{n}{z}^{4n+5}}{(2n+1)!}$

$\mathrm{sin}(z)=\sum _{n=0}^{\mathrm{\infty}}\frac{(-1{)}^{n}{z}^{2n+1}}{(2n+1)!}$

Thus,

$\mathrm{sin}({z}^{2})=\sum _{n=0}^{\mathrm{\infty}}\frac{(-1{)}^{n}{z}^{4n+2}}{(2n+1)!}$

and finally

${z}^{3}\mathrm{sin}({z}^{2})=\sum _{n=0}^{\mathrm{\infty}}\frac{(-1{)}^{n}{z}^{4n+5}}{(2n+1)!}$

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