 snowlovelydayM

2021-01-19

Taylor series
a. Use the definition of a Taylor series to find the first four nonzero terms of the Taylor series for the given function centered at a.
b. Write the power series using summation notation.
$f\left(x\right)={2}^{x},a=1$ pierretteA

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Given :
The function is $f\left(x\right)={2}^{x}$ and center at a=1
The definition of a Taylor series :

a)To use the definition of a Taylor series to find the first four nonzero terms of the Taylor series for the given function centered at a
Find derivative of the function .
$f\left(x\right)={2}^{x}$
${f}^{\prime }\left(x\right)={2}^{x}\mathrm{ln}\left(2\right)$
${f}^{″}\left(x\right)={2}^{x}{\mathrm{ln}}^{2}\left(2\right)$
${f}^{‴}\left(x\right)={2}^{x}{\mathrm{ln}}^{3}\left(2\right)$
${f}^{4}\left(x\right)={2}^{x}{\mathrm{ln}}^{4}\left(2\right)$
To find derivative f(x) at x=1
$f\left(1\right)=2$
${f}^{\prime }\left(1\right)=2\mathrm{ln}\left(2\right)$
${f}^{″}\left(1\right)=2{\mathrm{ln}}^{2}\left(2\right)$
${f}^{‴}\left(1\right)=2{\mathrm{ln}}^{3}\left(2\right)$
${f}^{4}\left(1\right)=2{\mathrm{ln}}^{4}\left(2\right)$
To find the first four nonzero terms of the Taylor series .
${c}_{0}=f\left(1\right)=2,{c}_{1}=\frac{{f}^{\prime }\left(1\right)}{1!}=\frac{2\mathrm{ln}\left(2\right)}{1!},{c}_{2}=\frac{{f}^{″}\left(1\right)}{2!}=\frac{2{\mathrm{ln}}^{2}\left(2\right)}{2!},{c}_{3}=\frac{{f}^{‴}\left(1\right)}{3!}=\frac{2{\mathrm{ln}}^{3}\left(2\right)}{3!},{c}_{4}=\frac{{f}^{4}\left(1\right)}{4!}=\frac{2{\mathrm{ln}}^{4}\left(2\right)}{4!}$
The in general ${c}_{k}=\frac{2{\mathrm{ln}}^{k}\left(2\right)}{k!}$
b) To write the power series using summation notation .
By using definition of a Taylor series and part a.
The series for centered at a=1 is
$\sum _{k=0}^{\mathrm{\infty }}{c}_{k}\left(x-a{\right)}^{k}=\sum _{k=0}^{\mathrm{\infty }}\frac{2{\mathrm{ln}}^{k}\left(2\right)}{k!}\left(x-1{\right)}^{k}$ Jeffrey Jordon

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