# Taylor Series Calculator

Enter the function in the input box, select variable, enter points, enter the order, and hit the "calculate" button to get the result using this Taylor series calculator.

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Taylor series calculator with steps is a tool used to find the Taylor series of one-variable functions. This Taylor series solver calculates the Taylor series around the center point of the function.

The order of the Taylor polynomial can be specified by using our Taylor series expansion calculator.

## What is the Taylor series?

Taylor series is a form of power series that gives the expansion of a function f(x) in the region of a point provided that in the region the function is continuous and all its differentials exist.

The order of the function tells how many derivatives of the function have to be taken to make a series.

### The formula of the Taylor series

The formula used to expand the Taylor series is given below.
$F\left(x\right)=\sum _{n=0}^{\infty }\left(\frac{f^n\left(a\right)}{n!}\left(x-a\right)^n\right)$

• $f^n\left(a\right)$ is nth order of the function.
• “n” is the total number.
• “a” is the center point of the function.

## How to calculate the Taylor series?

Following is an example of the Taylor series solved by our Taylor polynomial calculator.

Example

Find the Taylor series of cos(x) having 5 as a center point and the order is 4.

Solution

Step 1: Write the given terms.

f(x) = cos(x)
a = 5
n = 4

Step 2: Take the Taylor expansion formula for n=4 & a=5.

$F\left(x\right)=\sum _{n=0}^4\left(\frac{f^n\left(a\right)}{n!}\left(x-a\right)^n\right)$
$F\left(x\right)=\frac{f\left(a\right)}{0!}\left(x-5\right)^0+\frac{f'\left(a\right)}{1!}\left(x-5\right)^1+\frac{f''\left(a\right)}{2!}\left(x-5\right)^2+\frac{f'''\left(a\right)}{3!}\left(x-5\right)^3+\frac{f^{iv}\left(a\right)}{4!}\left(x-5\right)^4$ …(1)

Step 3: Now find the first four derivatives of cos(x) at x=a.

f(a) = cos(a)
f’(a) = -sin(a)
f’’(a) = -cos(a)
f’’’(a) = -(-sin(a)) = sin(a)
f’’’’(a)= cos(a)

Step 4: Put the values of n up to 4.

For n = 0

$\frac{cos\left(5\right)}{0!}\left(x-5\right)^0=cos\left(5\right)$

For n = 1

$\frac{-sin\left(5\right)}{1!}\left(x-5\right)^1=-\left(x-5\right)sin\left(5\right)$

For n = 2

$\frac{-\cos \left(5\right)}{2!}\left(x-5\right)^2=-\frac{1}{2}\left(x-5\right)^2\cos \left(5\right)$

For n = 3

$\frac{sin\left(5\right)}{3!}\left(x-5\right)^3=\frac{1}{6}\left(x-5\right)^3sin\left(5\right)$

For n = 4

$\frac{cos\left(5\right)}{4!}\left(x-5\right)^4=\frac{1}{24}\left(x-5\right)^4cos\left(5\right)$

Step 5: Now place the values in (1).

$F\left(x\right)=cos\left(5\right)-\left(x-5\right)sin\left(5\right)-\frac{1}{2}\left(x-5\right)^2cos\left(5\right)+\frac{1}{6}\left(x-5\right)^3sin\left(5\right)+\frac{1}{24}\left(x-5\right)^4cos\left(5\right)$

## FAQs

### What is the Taylor series of e^x?

The taylor series of e^x is $\sum _{n=0}^{\infty }\frac{x^n}{n!}$

### What is the Taylor series of e^2x?

The Taylor series of e^2x center at 0 is $1+2x+2x^2+\frac{4}{3}x^3+\frac{2}{3}x^4+\ldots$

### What is the Taylor series of tanx?

The Taylor series of tanx is $x+\frac{1}{3}x^3+\frac{2}{15}x^5+\frac{17}{315}x^7+\frac{62}{2835}x^9+\ldots$