91Ó°ÊÓ

Consider the power series of the function is$\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{k}=0}{\overset{\mathrm{∞}}{∑}} {\mathrm{a}}_{\mathrm{k}}{\left(\mathrm{x}−{\mathrm{x}}_0\right)}^{\mathrm{k}}$

The power series can also be written as $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{a}}_0+{\mathrm{a}}_1{\left(\mathrm{x}−{\mathrm{x}}_0\right)}^1+{\mathrm{a}}_2{\left(\mathrm{x}−{\mathrm{x}}_0\right)}^2+…$

Therefore, $\mathrm{f}\left({\mathrm{x}}_0\right)$can be found by putting $\mathrm{x}={\mathrm{x}}_0$

So,

$\mathrm{f}\left({\mathrm{x}}_0\right)={\mathrm{a}}_0+{\mathrm{a}}_1{\left({\mathrm{x}}_0−{\mathrm{x}}_0\right)}^1+{\mathrm{a}}_2{\left({\mathrm{x}}_0−{\mathrm{x}}_0\right)}^2+…$

$={\mathrm{a}}_0+{\mathrm{a}}_1(0{)}^1+{\mathrm{a}}_2(0{)}^2+…$

It implies that can written as$\boxed{\mathrm{f}\left({\mathrm{x}}_0\right)={\mathrm{a}}_0}$

Determine the function's derivative is $\mathrm{f}\left(\mathrm{x}\right)$to find $\mathrm{f}\text{'}\left(\mathrm{x}\right)$

therefore,

$\begin{array}{r}{\mathrm{f}}^{\mathrm{′}}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left[\underset{\mathrm{k}=0}{\overset{\mathrm{∞}}{∑}} {\mathrm{a}}_{\mathrm{k}}{\left(\mathrm{x}−{\mathrm{x}}_0\right)}^{\mathrm{k}}\right]\\ =\underset{\mathrm{k}=0}{\overset{\mathrm{∞}}{∑}} {\mathrm{a}}_{\mathrm{k}}\frac{\mathrm{d}}{\mathrm{dx}}{\left(\mathrm{x}−{\mathrm{x}}_0\right)}^{\mathrm{k}}\\ =\underset{\mathrm{k}=0}{\overset{\mathrm{∞}}{∑}} {\mathrm{a}}_{\mathrm{k}}\mathrm{k}{\left(\mathrm{x}−{\mathrm{x}}_0\right)}^{\mathrm{k}−1}\end{array}$

Changing the index of the function,

So,

${\mathrm{f}}^{\mathrm{′}}\left(\mathrm{x}\right)=\underset{\mathrm{k}=0}{\overset{\mathrm{∞}}{∑}} {\mathrm{a}}_{\mathrm{k}+1}(\mathrm{k}+1){\left(\mathrm{x}−{\mathrm{x}}_0\right)}^{\mathrm{k}}$

That is implies that ,

$$\begin{gathered} {\mathrm{f}}^{\mathrm{′}}\left(\mathrm{x}\right)=\underset{\mathrm{k}=0}{\overset{\mathrm{x}}{∑}} {\mathrm{a}}_{\mathrm{k}+1}(\mathrm{k}+1){\left(\mathrm{x}−{\mathrm{x}}_0\right)}^{\mathrm{k}} \\ ={\mathrm{a}}_1â‹\dots 1â‹\dots {\left(\mathrm{x}−{\mathrm{x}}_0\right)}^0+{\mathrm{a}}_2â‹\dots 2â‹\dots {\left(\mathrm{x}−{\mathrm{x}}_0\right)}^1+{\mathrm{a}}_3â‹\dots 3â‹\dots {\left(\mathrm{x}−{\mathrm{x}}_0\right)}^1+… \end{gathered}$$

The derivative of the function of$\boxed{{\mathrm{f}}^{\mathrm{′}}\left({\mathrm{x}}_0\right)={\mathrm{a}}_1}$

Determine the function's derivative is $\mathrm{f}\left(\mathrm{x}\right)$to find $\mathrm{f}"\left(\mathrm{x}\right)$

Therefore,

$\begin{array}{r}{\mathrm{f}}^{\mathrm{n}}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left[\underset{\mathrm{k}=0}{\overset{\mathrm{x}}{∑}} {\mathrm{a}}_{\mathrm{k}+1}(\mathrm{k}+1){\left(\mathrm{x}−{\mathrm{x}}_0\right)}^{\mathrm{k}}\right]\\ =\underset{\mathrm{k}=0}{\overset{\mathrm{∞}}{∑}} {\mathrm{a}}_{\mathrm{k}+1}(\mathrm{k}+1)\frac{\mathrm{d}}{\mathrm{dx}}{\left(\mathrm{x}−{\mathrm{x}}_0\right)}^{\mathrm{k}}\\ =\underset{\mathrm{k}=0}{\overset{\mathrm{∞}}{∑}} {\mathrm{a}}_{\mathrm{k}+1}(\mathrm{k}+1)\mathrm{k}{\left(\mathrm{x}−{\mathrm{x}}_0\right)}^{\mathrm{k}−1}\end{array}$

Changing the index of the function,

So,

${\mathrm{f}}^{′â\P IJ}\left(\mathrm{x}\right)=\underset{\mathrm{k}=0}{\overset{\mathrm{∞}}{∑}} {\mathrm{a}}_{\mathrm{k}+2}(\mathrm{k}+2)(\mathrm{k}+1){\left(\mathrm{x}−{\mathrm{x}}_0\right)}^{\mathrm{k}}$

That is implies that ,

role="math" localid="1650121702962" $\boxed{{\mathrm{f}}^{′â\P IJ}\left({\mathrm{x}}_0\right)=2{\mathrm{a}}_2}$

Finally, ${\mathrm{f}}^{\left(\mathrm{k}\right)}\left({\mathrm{x}}_0\right)$can be calculated by differentiating $\mathrm{f}\left(\mathrm{x}\right)$, k number of times and then substituting $\mathrm{x}={\mathrm{x}}_0$

So, the Power series of the functionrole="math" localid="1650121953688" ${\mathrm{f}}^{\left(\mathrm{k}\right)}\left({\mathrm{x}}_0\right)=\mathrm{k}!{\mathrm{a}}_{\mathrm{k}}$