Question

Suppose f is a polynomial of degree n and let k be some integer with $0\le k\le n$. Prove that if f(x) is of the form $f\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+....+a_{1}x+a_0$

Then $a_k=\frac{f^k\left(0\right)}{k!}$where $f^k\left(x\right)$is the k-th derivative of$f\left(x\right)andk!=k·(k-1)·(k-2)....2·1$

Short answer

We use Principal of mathematical induction to prove$a_k=\frac{f^k\left(0\right)}{k!}where0\le k\le n$

Step-by-step solution

Given information

We are given a polynomial function $f\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+....+a_k{x}^k+.....+a_{1}x+a_0$

We use mathematical induction to prove

When n=0

$f\left(x\right)=a_0$

Hence LHS= RHS

Now consider that statement is true for k where $0\le k\le n$

Hence the polynomial become

And

$f^k\left(x\right)={}_k{}^{n}c{a}_n{x}^{n-k}+....+(k+1)!a_{k+1}x+k!a_k$ (1)

At x=0

$$\begin{gathered} f^k\left(x\right)=k!a_k \\ {a}_k=\frac{f^k\left(x\right)}{k!} \end{gathered}$$

Now we prove that the statement is true for k+1

To prove it we differentiate equation 1 again

We get,

$f^{k+1}\left(x\right)={}_{k+1}{}^{n}c{a}_n{x}^{n-k-1}+.....+(k+1)!a_{k+1}$

At x=0

$$\begin{gathered} f^{k+1}\left(0\right)=(k+1)!a_{k+1} \\ {a}_{k+1}=\frac{f^{k+1}\left(0\right)}{(k+1)!} \end{gathered}$$

Hence proved