# 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.