The online Taylor series calculator is used to solve the Taylor series of the given function around the center point. Our Taylor calculator provides step by step solution for a given function. This Taylor series expansion calculator is also used to specify the order of the Taylor polynomial.
Follow the below steps to find the Taylor series of functions.
“In mathematics, Taylor series is an expression of a function for which the differentiation of all orders exists at a point “a” in the domain of “f” in the form of the power series.”
The Taylor series of a function is infinite of terms that are expressed in terms of the derivatives of the function at a single point.
(Source: Wikipedia)
The formula for Taylor series expansion is:
\(F\left(x\right)=\sum _{n=0}^{\infty }\left(\frac{f^n\left(a\right)}{n!}\left(x-a\right)^n\right)\)
In the formula of Taylor series, \(f^n\left(a\right)\) is the nth order of the given function, “a” is a particular point or center point of the function, and “n” is the order.
The Taylor series can be finite or infinite depending on the order of the expression. This Taylor polynomial calculator works according to the above expansion formula.
Here is an example solved by our Taylor expansion calculator.
Example
Find Taylor series of sinx up to order four and the center point is 3.
Solution
Step 1: Identify the given terms.
f(x) = sin(x)
n = 4
a = 3
Step 2: Now write the Taylor series expansion formula for n=4 & a=3.
\(F\left(x\right)=\sum _{n=0}^4\left(\frac{f^n\left(a\right)}{n!}\left(x-a\right)^n\right)\)
\( F\left(x\right)=\frac{f\left(a\right)}{0!}\left(x-3\right)^0+\frac{f\:'\left(a\right)}{1!}\left(x-3\right)^1+\frac{f\:''\left(a\right)}{2!}\left(x-3\right)^2+\frac{f\:'''\left(a\right)}{3!}\left(x-3\right)^3+\frac{f^{iv}\left(a\right)}{4!}\left(x-3\right)^4\) …(1)
Step 3: Now calculate the derivative of sinx up to order four.
\(f\left(a\right)=sin\left(a\right)\)
\(f'\left(a\right)=cos\left(a\right)\)
\(f''\left(a\right)=-sin\left(a\right)\)
\(f'''\left(a\right)=-cos\left(a\right)\)
\(f^{iv}\left(a\right)=sin\left(a\right)\)
Step 4: Now expand the above formula up to n=4.
For n = 0
\(\frac{sin\left(3\right)}{0!}\left(x-3\right)^0=sin\left(3\right)\)
For n = 1
\(\frac{cos\left(3\right)}{1!}\left(x-3\right)^1=\left(x-3\right)cos\left(3\right)\)
For n = 2
\(\frac{-\sin \left(3\right)}{2!}\left(x-3\right)^2=-\frac{1}{2}\left(x-3\right)^2\sin \left(3\right)\)
For n = 3
\(\frac{-cos\left(3\right)}{3!}\left(x-3\right)^3=-\frac{1}{6}\left(x-3\right)^3cos\left(3\right)\)
For n = 4
\(\frac{sin\left(3\right)}{4!}\left(x-3\right)^4=\frac{1}{24}\left(x-3\right)^4sin\left(3\right)\)
Step 5: Now put the above calculated values in (1).
\(F\left(x\right)=sin\left(3\right)+\left(x-3\right)cos\left(3\right)-\frac{1}{2}\left(x-3\right)^2sin\left(3\right)-\frac{1}{6}\left(x-3\right)^3cos\left(3\right)+\frac{1}{24}\left(x-3\right)^4sin\left(3\right)\)
Some of the functions solved by this Taylor approximation calculator are given in the below table.
Taylor series for | Output |
e^x | \(\sum _{n=0}^{\infty }\left(\frac{x^n}{n!}\right)\) |
cosx | \(\sum _{n=0}^{\infty \:}\left(-1\right)^n\frac{x^{2n}}{\left(2n\right)!}\) |
ln(1+x) | \(\sum _{n=1}^{\infty \:}\left(-1\right)^{n+1}\frac{x^n}{n}\) |
1/(1+x) | \(1-x+x^2-x^3+x^4+\ldots \) |
1/(1-x) | \(\sum _{n=0}^{\infty \:}x^n\) |
Taylor series | Encyclopædia Britannica, inc. (n.d.)
Example of Taylor series | Tutorial.math.lamar.edu (n.d.)