By using free Taylor Series Calculator, you can easily find the approximate value of the integration function. Just provide the function, expansion order and expansion variable in the specified input fields and press on the calculate button to check the result of integration function immediately.

**Taylor Series Calculator: **If you are searching for a best tool that finds the integration function value using Taylor Series, then you are at the correct place. You will get the handy Taylor Series Calculator tool that provides you the answer within no time. Along with the exact result, you will also find the lengthy explanation to solve a function using Taylor Series. Below provided step by step procedure is helpful for you in solving the function easily.

Follow these simple steps to solve any function integration using taylor series. These steps are useful for you to get a clear idea on the concept.

- Let us consider any function to find the integration, upper and lower limits of integration.
- Taylor Series formula for a function f is f(x) = f(a) + f'(a)(x-a) + f''(a)/2 (x-a)
^{2}+ f'''(a)/3! (x-a)^{3}+ . . . + f^{(n)}(a)/n! (x-a)^{n} - Where, a is the centre.
- First of all find the derivative of given function and evaluate the derivative at the given point.
- Substitute the obtained values in the above formula to get a polynomial.
- Finally, simplify your polynomial expression to get the final answer.

**Example**

**Question: Calculate Taylor Series of sin(x) up to n=5, centre is 1?**

**Solution:**

Given function is f(x) = sin(x)

n = 0 to 5, centre = 1

Taylor Series formula is f(x) = f(a) + f^{1}(a)(x-a) + f^{2}(a)/2 (x-a)^{2} + f^{3}(a)/3! (x-a)^{3} + . . . + f^{(n)}(a)/n! (x-a)^{n}

By using the above values f(x) will become

f(x) = f(0) + f^{1}(1)(x-1) + f^{2}(2)/2 (x-2)^{2} + f^{3}(3)/3! (x-1)^{3} + f^{4}(4)/4! (x-1)^{4} + f^{5}(5)/5! (x-1)^{5}

So, calculate the derivative and evaluate them at the given point to get the result into the given formula.

f^{0}(x) = f(x) = sin(x)

f(0) = sin(0) = 0

First derivative of f(x) is f^{1}(x) = [f^{0}(x)]'

= [sin(x)]'

= cos(x)

[f(0)]' = cos(0) = 1

Second Derivative: f^{2}(x) = [f^{1}(x)]' = [cos(x)]' = -sin(x)

(f(0))′′= 0

Evaluate third derivative (f(0))''' = -cos(0) = -1

Fourth derivative f^{4}(x) = [f^{3}(x)]' = [-cos(x)]' = sin(x)

(f(0))'''' = sin(0) = 0

Fifth derivative: f^{5}(x) = [f^{4}(x)]' = [sin(x)]' = cos(x)

(f(0))''''' = cos(0) = 1

Substitute these values in the taylor formula

f(x) ≈ 0 + 1 (x-1) + 0/2 (x-2)^{2} + (-1)/6 (x-1)^{3} + 0/4! (x-1)^{4} + 1/5! (x-1)^{5}

≈ (x-1) - 1/6 (x-1)^{3} + 1/120 (x-1)^{5}

sin(x) ≈ (x-1) - 1/6 (x-1)^{3} + 1/120 (x-1)^{5}

Study more mathematical concepts from our site Onlinecalculator.guru to make your calculations easier.

**1. How do you use Taylor Series Calculator to solve the functions?**

The step by step process to use this calculator is provided here. Enter the function, limit and center in the respective input fields. Hit on the calculate button to get the series in the output window.

**2. Solve x ^{2}+2x with centre 3 using taylor series?**

Given fuunction is x^{2}+2x

Taylor Series formula is f(x) = f(a) + f'(a)(x-a) + f''(a)/2 (x-a)^{2} + f'''(a)/6 (x-a)^{3}

f(3) = (3)^{2}+2(3) = 15

f'(x) = 2x+2

f'(3) = 6+2=8

f''(x) = 2

f'''(x) = 0

Replace the obtained values in the formula

f(x) = 15+ 8(x-3) + 2(x-3)^{2} + 0(x-3)^{3}

x^{2}+2x = 15 + 8(x-3) + 2(x-3)^{2}

**3. What is meant by Taylor Series?**

Taylor Series is defined as the expression of the function as the infinite series, in which the terms are expressed for the value of functions derivative at a single point. It is used to draw the conclusion of what a function looks like.

**4. What is Taylor series method?**

Taylor series method is one of the earliest analytic numeric algorithms to find approximate solution of initial value problems for ordinary differential equations.