Explain Taylor series

The Taylor theorem expresses a function in the form of the sum of infinite terms. These terms are circumscribed from the derivative of a given function for a particular point. The standard definition of an algebraic function is presented using an algebraic equation. A function may be well illustrated by its Taylor series too. This series can also be used to determine various functions in lots of areas of mathematics

Taylor’s Series Theorem

Assume that if f(x) be a real or composite function, which is a differentiable function of a neighbourhood number that is also real or composite. Then, the Taylor series describes the following power series :

\(f(x) =f(a)\frac{f'(a)}{1!}(x-a)+\frac{f”(a)}{2!}(x-a)^{2}+\frac{f^{(3)}(a)}{3!}(x-a)^{3}+….\)

In terms of sigma notation, the Taylor series can be written as

\(\sum_{n=0}^{\infty }\frac{f^{n}(a)}{n!}(x-a)^{n}\)

Where f(n) (a) = nth derivative of f n!

= factorial of n.

Taylor Series Formula and Proof

We know that the power series can be defined as

\(f(x)= \sum_{n=0}^{\infty }a_{n}x^{n}=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+…\)

When x = 0, f(x)= a0 So, differentiate the given function, it becomes, f’(x) = a1+ 2a2x + 3a3x2 + 4a4x3 +….

Again, when you substitute x = 0, we get

f’(0) =a1 So, differentiate it again, we get f”(x) = 2a2 + 6a3x +12a4x2 + …

Now, substitute x=0 in second-order differentiation, we get f”(0) = 2a2

Therefore, [f”(0)/2!] = a2

By generalising the equation, we get

f n (0) / n! = an

Now substitute the values in the power series we get,

\(f(x)= f(0)+f'(0)x+\frac{f”(0)}{2!}x^{2}+\frac{f”‘(0)}{3!}x^{3}+….\)

Generalise f in more general form, it becomes

f(x) = b + b1 (x-a) + b2( x-a)2 + b3 (x-a)3+ ….

Now, x = a, we get bn = fn(a) / n!

Now, substitute bn in a generalised form

\(f(x) =f(a)\frac{f'(a)}{1!}(x-a)+\frac{f”(a)}{2!}(x-a)^{2}+\frac{f^{(3)}(a)}{3!}(x-a)^{3}+….\)

Hence proved.

Leave a Comment

Your email address will not be published. Required fields are marked *


Free Class