Chapter 2: Q. 5 (page 200)

Suppose f is a polynomial of degree n and let k be some integer with $0 \leq k \leq n$ . Prove that if f(x) is of the form $f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + . . . . + a_{1} x + a_{0}$
Then $a_{k} = \frac{f^{k} (0)}{k!}$ where $f^{k} (x)$ is the k-th derivative of $f (x) a n d k! = k \cdot (k - 1) \cdot (k - 2) . . . . 2 \cdot 1$

Short Answer

Expert verified

We use Principal of mathematical induction to prove $a_{k} = \frac{f^{k} (0)}{k!} w h e r e 0 \leq k \leq n$

Step by step solution

Given information

We are given a polynomial function $f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + . . . . + a_{k} x^{k} + . . . . . + a_{1} x + a_{0}$

We use mathematical induction to prove

When n=0

$f (x) = a_{0}$

Hence LHS= RHS

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

Hence the polynomial become

And

$f^{k} (x) = {}_{k}^{n}c a_{n} x^{n - k} + . . . . + (k + 1)! a_{k + 1} x + k! a_{k}$ (1)

At x=0

$f^{k} (x) = k! a_{k} a_{k} = \frac{f^{k} (x)}{k!}$

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

To prove it we differentiate equation 1 again

We get,

$f^{k + 1} (x) = {}_{k + 1}^{n}c a_{n} x^{n - k - 1} + . . . . . + (k + 1)! a_{k + 1}$

At x=0

$f^{k + 1} (0) = (k + 1)! a_{k + 1} a_{k + 1} = \frac{f^{k + 1} (0)}{(k + 1)!}$

Hence proved

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Suppose f is a polynomial of degree n and let k be some integer with 0≤k≤n. Prove that if f(x) is of the form f(x)=anxn+an-1xn-1+....+a1x+a0Then ak=fk(0)k!where fk(x)is the k-th derivative off(x)andk!=k·(k-1)·(k-2)....2·1

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