Taylor Series

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. Let us discuss the definition, formula and derivations of Taylor’s series in detail.

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

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


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, 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!)+ …..

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