The Taylor series calculator uses partial sums of an infinite number of terms to approximate the given values. The tool finds the Taylor series of one variable functions around the center of the function.

So, get to know the series that come from the given value derivatives at a single point.

## What is the Taylor Series?

**“Taylor series is an infinite sum given values that are demonstrated as the function derivative at a single point”.**

## Taylor Series Formula:

Around the given points, the Taylor series calculator calculates the series expansion of given functions. To determine this for a polynomial with an infinite number of terms. Look at the formula below:

$$ f (x) = \sum\limits_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!}(x - a)^{k} $$

**Where:**

- f(x) _ Function for the nth-order derivative
- a _ Central term of the function
- n _ Order that indicates the total numbers
- k! _
__factorial__

## How To Calculate the Taylor Series? - Example

Evaluate the Taylor series of (x^5)^(1/3) up to order n = 5 and a = 1

**Given Data:**

3√x^5

n = 5

a = 1

### Solution:

As we already know the __Maclaurin equation__ to find the Taylor series of polynomials is:

$$ f (x) = \sum\limits_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!}(x - a)^{k} $$

$$ f (x) ≈ P(x) = \sum\limits_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!}(x - a)^{k} = \sum\limits_{k=0}^{4} \frac{f^{(k)}(a)}{k!}(x - a)^{k} $$

So, what we need to do to get the required polynomial is to evaluate the derivatives, calculate them at a given point, and substitute the results into the formula.

$$ f^{(0)}(x) = f(x)= \sqrt[3]{x^{5}} $$

Calculate the point: $$ f(1) = 1 $$

#### Step # 1:

$$ f^{(1)}(x) = \left(f^{(0)}(x)\right)^{'}= \left(\sqrt[3]{x^{5}}\right)^{'} = \frac{5 \sqrt[3]{x^{5}}}{3 x} $$

Calculate the 1 derivative at the given point:

$$ \left(f(1)\right)^{'} = \frac{5}{3} $$

#### Step # 2:

Find the 2 derivatives:

$$ f^{(2)}(x) = \left(f^{(1)}(x)\right)^{'}= \left(\frac{5 \sqrt[3]{x^{5}}}{3 x}\right)^{'} = \frac{10 \sqrt[3]{x^{5}}}{9 x^{2}} $$

Calculate the 2 derivative at the given point:

$$ \left(f(1)\right)^{''} = \frac{10}{9} $$

#### Step # 3:

Find the 3 derivatives:

$$ f^{(3)}(x) = \left(f^{(2)}(x)\right)^{'}= \left(\frac{10 \sqrt[3]{x^{5}}}{9 x^{2}}\right)^{'} = - \frac{10 \sqrt[3]{x^{5}}}{27 x^{3}} $$

Now, calculate the 3 derivative at the given point:

$$ \left(f(1)\right)^{'''} = - \frac{10}{27} $$

#### Step # 4:

Find the 4 derivatives:

$$ f^{(4)}(x) = \left(f^{(3)}(x)\right)^{'}= \left(- \frac{10 \sqrt[3]{x^{5}}}{27 x^{3}}\right)^{'} = \frac{40 \sqrt[3]{x^{5}}}{81 x^{4}} $$

Calculate the 4 derivative at the given point:

$$ \left(f(1)\right)^{''''} = \frac{40}{81} $$

#### Step # 5:

Now use the calculated value to obtain the polynomial:

$$ f(x) ≈ \frac{1}{0!}(x- (1))^{0} + \frac{\frac{5}{3}}{1!}(x- (1))^{1} + \frac{\frac{10}{9}}{2!}(x- (1))^{2} + \frac{- \frac{10}{27}}{3!}(x- (1))^{3} + \frac{\frac{40}{81}}{4!}(x- (1))^{4} $$

After simplification, we finally get the final answer:

$$ f(x)≈P(x)=1+\frac{5 x}{3} - \frac{5}{3}+\frac{5 \left(x - 1\right)^{2}}{9}- \frac{5 \left(x - 1\right)^{3}}{81}+\frac{5 \left(x - 1\right)^{4}}{243} $$

## How To Use This Calculator?

The Taylor series expansion calculator operates well if you provide the following inputs.

### What To Do?

- Put the function in the designated field of the tool
- Enter a point to find the Taylor series
- Insert the value of n for approximation
- Put the optional value to determine the error at that point
- Tap on
**“Calculate”**

### What To Get?

- Taylor series after simplification
- The function and derivatives of a given point
- Finds the series of given functions for order n
- Step-by-step results after the simplify

## References:

**Wikipedia: **Taylor series, Definition, Example, Approximation error, and convergence, List of Maclaurin series of some common functions, Calculation of Taylor series.