Q. 5 - Calculus

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 $f\left(x\right) and k!=k·(k-1)·(k-2)....2·1$

Step-by-Step Solution

Answer

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

Step 1: 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 

Step 2: Identify the differentiation rules needed

Examine the function to determine which differentiation rules apply: power rule, product rule, quotient rule, chain rule, or special function derivatives.

Step 3: Apply the differentiation rules

Differentiate each term of the function systematically, applying the chain rule for composite functions.

Step 4: Simplify the derivative

Combine like terms, factor where appropriate, and write the derivative in its simplest form.

Step 5: State the final answer

Write the final derivative clearly.

Step 6: Conclude with the answer

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