Q. 3 - Calculus

Question

If $\mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\mathrm{\infty }} {\mathrm{a}}_{\mathrm{k}}{\left(\mathrm{x}-{\mathrm{x}}_0\right)}^{\mathrm{k}}$ , what is $\mathrm{f}\left({\mathrm{x}}_0\right)?$What is $\mathrm{f}\text{'}\left({\mathrm{x}}_0\right)?$ What is $\mathrm{f}"\left({\mathrm{x}}_0\right)?$What is ${\mathrm{f}}^{\left(\mathrm{k}\right)}\left({\mathrm{x}}_0\right)?$

Step-by-Step Solution

Answer

The value of  ${\mathrm{f}}^{\left(\mathrm{k}\right)}\left({\mathrm{x}}_0\right)=\mathrm{k}! {\mathrm{a}}_{\mathrm{k}}$

Step 1. Given information.

Consider the power series of the function is $\mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\mathrm{\infty }} {\mathrm{a}}_{\mathrm{k}}{\left(\mathrm{x}-{\mathrm{x}}_0\right)}^{\mathrm{k}}$

Step 2. Finding the power series

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

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

So, 

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

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

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

Step 3. Finding the first derivative of the power series

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

therefore,

$\begin{array}{r}{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left[\sum _{\mathrm{k}=0}^{\mathrm{\infty }} {\mathrm{a}}_{\mathrm{k}}{\left(\mathrm{x}-{\mathrm{x}}_0\right)}^{\mathrm{k}}\right]\\ =\sum _{\mathrm{k}=0}^{\mathrm{\infty }} {\mathrm{a}}_{\mathrm{k}}\frac{\mathrm{d}}{\mathrm{dx}}{\left(\mathrm{x}-{\mathrm{x}}_0\right)}^{\mathrm{k}}\\ =\sum _{\mathrm{k}=0}^{\mathrm{\infty }} {\mathrm{a}}_{\mathrm{k}}\mathrm{k}{\left(\mathrm{x}-{\mathrm{x}}_0\right)}^{\mathrm{k}-1}\end{array}$

Changing the index of the function, 

So, 

${\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\mathrm{\infty }} {\mathrm{a}}_{\mathrm{k}+1}(\mathrm{k}+1){\left(\mathrm{x}-{\mathrm{x}}_0\right)}^{\mathrm{k}}$

That is implies that , 

$$\begin{gathered} {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\mathrm{x}} {\mathrm{a}}_{\mathrm{k}+1}(\mathrm{k}+1){\left(\mathrm{x}-{\mathrm{x}}_0\right)}^{\mathrm{k}} \\ ={\mathrm{a}}_1\cdot 1\cdot {\left(\mathrm{x}-{\mathrm{x}}_0\right)}^0+{\mathrm{a}}_2\cdot 2\cdot {\left(\mathrm{x}-{\mathrm{x}}_0\right)}^1+{\mathrm{a}}_3\cdot 3\cdot {\left(\mathrm{x}-{\mathrm{x}}_0\right)}^1+\dots \end{gathered}$$

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

Step 4. Finding the seocnd derivative of the power series

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

Therefore,

$\begin{array}{r}{\mathrm{f}}^{\mathrm{n}}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left[\sum _{\mathrm{k}=0}^{\mathrm{x}} {\mathrm{a}}_{\mathrm{k}+1}(\mathrm{k}+1){\left(\mathrm{x}-{\mathrm{x}}_0\right)}^{\mathrm{k}}\right]\\ =\sum _{\mathrm{k}=0}^{\mathrm{\infty }} {\mathrm{a}}_{\mathrm{k}+1}(\mathrm{k}+1)\frac{\mathrm{d}}{\mathrm{dx}}{\left(\mathrm{x}-{\mathrm{x}}_0\right)}^{\mathrm{k}}\\ =\sum _{\mathrm{k}=0}^{\mathrm{\infty }} {\mathrm{a}}_{\mathrm{k}+1}(\mathrm{k}+1)\mathrm{k}{\left(\mathrm{x}-{\mathrm{x}}_0\right)}^{\mathrm{k}-1}\end{array}$

Changing the index of the function, 

So, 

${\mathrm{f}}^{\prime \prime }\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\mathrm{\infty }} {\mathrm{a}}_{\mathrm{k}+2}(\mathrm{k}+2)(\mathrm{k}+1){\left(\mathrm{x}-{\mathrm{x}}_0\right)}^{\mathrm{k}}$

That is implies that , 

$\boxed{{\mathrm{f}}^{\prime \prime }\left({\mathrm{x}}_0\right)=2{\mathrm{a}}_2}$

Step 5. Finding Final Derivative

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 function  ${\mathrm{f}}^{\left(\mathrm{k}\right)}\left({\mathrm{x}}_0\right)=\mathrm{k}! {\mathrm{a}}_{\mathrm{k}}$