A Taylor series is a series expansion of a function at a single point. When the Taylor series is centered with zero, then that particular series is called as

A Taylor series, which is an expansion of a real function** f(x)** at single point **x = a** is given as the formula stated below :

or you can make this formula simple and more compact by using the sigma notion

here **n!** represents the factorial of '**n'** and **ƒ (n)(a)** represents the nth derivative of '**ƒ' **calculate at a single point '**a'**.

or you can make this formula simple and more compact by using the sigma notion

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Below are the problems on Taylor series :

### Solved Examples

**Question 1: **Find the Taylor Series for the function f(x) = e^{x} at about x = 0 ?

** Solution: **

Given function: f(x) = e^{x}

f^{(n)}(0) = e^{0} = 1; n = 0, 1, 2, 3....

f^{(n)}(x) = e^{x} ; n = 0, 1, 2, 3....

Taylor series for the function f(x) at about x = a is given by $\sum_{n = 0}^{\infty}$ $\frac{f^{n}a}{n!}$ $(x - a)^{n}$

So, for the function f(x) = e^{x} = $\sum_{n = 0}^{\infty}$ $\frac{x^{n}}{n!}$

**Question 2: **Find the taylor series expansion for the function f(x) = e^{-x} at about x = -3 ?

** Solution: **

Given function: f(x) = e^{x}

f^{(n)}(-3) = e^{-3} = 1; n = 0, 1, 2, 3....

f^{(n)}(x) = e^{x} ; n = 0, 1, 2, 3....

Taylor series for the function f(x) at about x = a is given by $\sum_{n = 0}^{\infty}$ $\frac{f^{n}a}{n!}$ $(x - a)^{n}$

So, for the function f(x) = e^{x} = $\sum_{n = 0}^{\infty}$ $\frac{e^{-3}}{n!}$ $(x + 3)^{n}$

Given function: f(x) = e

f

f

Taylor series for the function f(x) at about x = a is given by $\sum_{n = 0}^{\infty}$ $\frac{f^{n}a}{n!}$ $(x - a)^{n}$

So, for the function f(x) = e

Given function: f(x) = e

f

f

Taylor series for the function f(x) at about x = a is given by $\sum_{n = 0}^{\infty}$ $\frac{f^{n}a}{n!}$ $(x - a)^{n}$

So, for the function f(x) = e