In Mathematics, the Taylor series is the most famous series that is utilized in several mathematical as well as practical problems. The Taylor theorem expresses a function in the form of the sum of infinite terms. These terms are determined from the derivative of a given function for a particular point. The standard definition of an algebraic function is provided using an algebraic equation. Likewise, transcendental functions are defined for a property that holds for them. A function may be well described by its Taylor series too. This series can also be used to determine various functions in plenty 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 + 2a3x2 + 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, the Taylor series is proved.

## Taylor Series in Several Variables

The Taylor series is also represented in the form of functions of several variables. The general form of the Taylor series in several variables is

$T(x_{1}, x_{2}, x_{3},…x_{m})=f(a_{1}, a_{2}, a_{3},…a_{m})+\sum_{j=1}^{m}\frac{\partial f(a_{1}, a_{2}, a_{3},…a_{m})}{\partial x_{j}}(x_{j}-a_{j})+\frac{1}{2!\sum_{j=1}^{m}}$

## Maclaurin Series Expansion

If the Taylor Series is centred at 0, then the series is known as the Maclaurin series. It means that,

If a= 0 in the Taylor series, then we get

$f(x)= f(0)+f'(0)x+\frac{f”(0)}{2!}x^{2}+\frac{f”‘(0)}{3!}x^{3}+….$ is known as the Maclaurin series.

## Applications of Taylor Series

The uses of the Taylor series are:

• Taylor series is used to evaluate the value of a whole function in each point if the functional values and derivatives are identified at a single point.
• The representation of Taylor series reduces many mathematical proofs.
• The sum of partial series can be used as an approximation of the whole series.
• Multivariate Taylor series is used in many optimization techniques.
• This series is used in the power flow analysis of electrical power systems.

### Problems on Taylor’s Theorem

Question: Determine the Taylor series at x=0 for f(x) = ex

Solution: Given: f(x) = ex

Differentiate the given equation,

f’(x) = ex

f’’(x) =ex

f’’’(x) = ex

At x=0, we get

f’(0) = e0 =1

f’’(0) = e0=1

f’’’(0) = e0 = 1

When Taylor series at x= 0, then the Maclaurin series is

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

ex = 1+ x(1) + (x2/2!)(1) + (x3/3!)(1) + …..

Therefore, ex = 1+ x + (x2/2!) + (x3/3!)+ …..