Taylor series calculator
Taylor series calculator is used to find the Taylor series of a function around n and center point a. Our Taylor series solver takes one variable function to calculate the Taylor series. It can also calculates the order of Taylor polynomial.
How does Taylor series expansion calculator work?
Use the below steps to calculate the Taylor series of any function.
- Enter the given function into the input box.
- Write the corresponding variable of the function.
- Enter the center point and the order of the function.
- Press the calculate button to get the result.
- Hit the clear key to enter a new function.
What is the Taylor series?
A power series that gives the expansion of a function f(x) in the region of center point a provided that in the region function is continuous, all its derivatives exist, and series converges to the function is known as a Taylor series.
The formula of the Taylor series
The general equation of the Taylor series used to epand the function is given below.
\(F\left(x\right)=\sum _{n=0}^{\infty }\left(\frac{f^n\left(a\right)}{n!}\left(x-a\right)^n\right)\)
In the equation of Taylor series, \(f^n\left(a\right)\) is nth derivative of the function, “a” is the center point of the function, and “n” is the total number.
Our Taylor series approximation calculator follows the above formula to calculate the Taylor series of the function.
How to calculate the Taylor series?
Below is an example of the Taylor series solved by the Taylor series calculator.
Example
Calculate the Taylor series of e^x having 4 as a center point and the order is 3.
Solution
Step 1: Take the given data.
\( f\left(x\right)=e^x\)
\( a=4\)
\( n=3\)
Step 2: Now write the equation of Taylor series.
\(F\left(x\right)=\sum _{n=0}^3\left(\frac{f^n\left(a\right)}{n!}\left(x-a\right)^n\right)\)
\( F\left(x\right)=\frac{f\left(a\right)}{0!}\left(x-4\right)^0+\frac{f'\left(a\right)}{1!}\left(x-4\right)^1+\frac{f''\left(a\right)}{2!}\left(x-4\right)^2+\frac{f'''\left(a\right)}{3!}\left(x-4\right)^3\) …(1)
Step 3: Calculate the first three derivatives of given function at x=a.
\( f\left(a\right)=e^a\)
\( f\:'\left(a\right)=e^a\)
\( f''\left(a\right)=-cos\left(a\right)\)
\( f\:'''\left(a\right)=e^a\)
Step 4: Now put n=0, 1, 2, 3 in the equation of the Taylor series.
For n = 0
\(\frac{e^4}{0!}\left(x-4\right)^0=e^4\)
For n = 1
\(\frac{e^4}{1!}\left(x-4\right)^1=e^4\left(x-4\right)\)
For n = 2
\(\frac{e^4}{2!}\left(x-4\right)^2=\frac{e^4}{2}\left(x-4\right)^2\)
For n = 3
\(\frac{e^4}{3!}\left(x-4\right)^3=\frac{e^4}{6}\left(x-4\right)^3\)
Step 5: Put the calculated values in the equation (1).
\( F\left(x\right)=e^4+e^4\left(x-4\right)+\frac{e^4}{2}\left(x-4\right)^2+\frac{e^4}{6}\left(x-4\right)^3\)
FAQs
What is the Taylor series of sinx at x=0?
The Taylor series of sinx at x=0 is
\( \frac{1}{1!}x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\frac{1}{7!}x^7+\frac{1}{9!}x^9+\ldots\)
What is the Taylor series of sin(y^2) at x=1?
The taylor series of sin(y^2) at x=1 is
\(\sin \left(1\right)+2\cos \left(1\right)\left(y-1\right)+\left(-2\sin \left(1\right)+\cos \left(1\right)\right)\left(y-1\right)^2+\frac{-4\cos \left(1\right)-6\sin \left(1\right)}{3}\left(y-1\right)^3+\frac{2\sin \left(1\right)-24\cos \left(1\right)}{12}\left(y-1\right)^4+\ldots\)