The Taylor series formula is a representation of a function as an infinite sum of terms that are calculated from the values of the function’s derivatives at a single point. The concept of a Taylor series was formulated by the Scottish mathematician James Gregory and formally introduced by the English mathematician Brook Taylor in 1715.

A function can be approximated by using a finite number of terms of its Taylor series. Taylor’s theorem gives quantitative estimates on the error introduced by the use of such an approximation. The polynomial formed by taking some initial terms of the Taylor series is called a Taylor polynomial.

The Taylor series of a function is the limit of that function’s Taylor polynomials as the degree increases, provided that the limit exists. A function may not be equal to its Taylor series, even if its Taylor series converges at every point. A function that is equal to its Taylor series in an open interval (or a disc in the complex plane) is known as an analytic function in that interval.

## Formula for Taylor Series

The Taylor series of a real or complex-valued function **f(x)** that is infinitely differentiable at a real or complex number “a” is the power series.

f(x) = f(a) (x − a) + [f'(a)/2!(x−a)^{2}] + [f'(a)/3!(x − a)^{3}] + ….. + [f'(a)/n!(x − a)^{n}] |

Or, you can make this formula simple and more compact by using the sigma notation.

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

**Also Check:** Taylor Series Calculator

### Solved Examples Using Taylor Series Formula

**Example: **Find the Taylor series with center $x_{0}=0 for the hyperbolic cosine function *f(x)* = *cosh x* by using the fact that *cosh x* is the derivative of the hyperbolic sine function *sinh x*, which has as its Taylor series expansion.

$\large sinh: x=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{\left(2n+1\right)!}$

**Solution: **

(Note: If you remember the Taylor expansions for *sin x* and *cos x*, you get an indication, why their hyperbolic counterparts might deserve the names “*sine*” and “*cosine*”).

Since,

$\large sinh: x=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{\left(2n+1\right)!}=x+\frac{x^{3}}{3!}+\frac{x^{5}}{5!}+…$

Its derivative *cosh x* has the Taylor expansion.

$\large cosh : x=\sum_{n=0}^{\infty}\frac{\left ( 2n+1 \right )x^{2n}}{\left ( 2n+1 \right )!}=\sum_{n=0}^{\infty}\frac{x^{2n}}{\left ( 2n \right )!}$

The last step follows from the fact that.

$\large \frac{2n+1}{\left ( 2n+1 \right )!}=\frac{2n+1}{1\cdot2\cdot\cdot\cdot\left (2n\right)\cdot\left (2n+1\right)}=\frac{1}{\left (2n\right)!}$

**Note:** The hyperbolic functions differ from their trigonometric counterparts in that they do not sport alternating signs.