Q. 70 - Calculus

Question

Let $f$ be a function with an nth-order derivative at a point $x_o$ and let $P_n\left(x\right)=\sum _{k=0}^n\frac{f^{\left(k\right)}\left(x_0\right)}{k!}{\left(x-x_0\right)}^k$. Prove that $f^{\left(k\right)}\left(x_0\right)=P_n^{\left(k\right)}\left(x_0\right)$ for every non-negative integer $k\le n$.

Step-by-Step Solution

Answer

The equation $P_n^{\left(k\right)}\left(x_0\right)=f^{\left(k\right)}\left(x_0\right)$ is true.

Step 1. Given information

An function is given as $P_n\left(x\right)=\sum _{k=0}^n\frac{f^{\left(k\right)}\left(x_0\right)}{k!}{\left(x-x_0\right)}^k$

Step 2. Verification

First we have to find the derivative $P_n\left(x\right)$,

$$\begin{gathered} P_n^{\text{'}}\left(x\right)=\frac{d}{dx}\left[\sum _{k=0}^n\frac{f^{\left(k\right)}\left(x_0\right)}{k!}{\left(x-x_0\right)}^k\right] \\ =\sum _{k=0}^n\frac{f^{\left(k\right)}\left(x_0\right)}{k!}\frac{d}{dx}{\left(x-x_0\right)}^k \\ =\sum _{k=0}^n\frac{f^{\left(k\right)}\left(x_0\right)}{k!}\times k{\left(x-x_0\right)}^{k-1} \\ =\sum _{k=0}^n\frac{f^{\left(k\right)}\left(x_0\right)}{(k-1)!}{\left(x-x_0\right)}^{k-1} \end{gathered}$$

Similarly,

$$\begin{gathered} P_n^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}\left[\sum _{k=0}^n\frac{f^{\left(k\right)}\left(x_0\right)}{(k-1)!}{\left(x-x_0\right)}^{k-1}\right] \\ =\sum _{k=0}^n\frac{f^{\left(k\right)}\left(x_0\right)}{(k-2)!}{\left(x-x_0\right)}^{k-2} \end{gathered}$$

Implies that $P_n^{\left(k\right)}\left(x_0\right)=f^{\left(k\right)}\left(x_0\right)$

Hence proved.