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**Table of Content**

The Taylor series calculator calculates all coefficients of a Taylor series expansion of a function centered at point **“n”**. Also, you can set point **“n” **as zero (**0**) to get the Maclaurin series representation.

In mathematics, the Taylor series is defined as the representation of a given function. It is an infinite series that represents the value of the derivative of a function at a certain point.

There is a special case of the Taylor and Maclaurin series calculator. This series helps to reduce the number of mathematics proofs and is used for power flow analysis.

The formula used by the Taylor series formula calculator for calculating a series for a function is given as:

$$ F (x) = âˆ‘^ âˆž_{n=0} f^k (a) / k! (x â€“ a) ^k $$

Where

f^(n)(a) = nth order derivative of function f(x), as evaluated at x = a

n =Â Â where the series is centered.

The series will be most precise near the centering point.

A Taylor expansion may be infinite, but we can select to make our series or function as little or long terms as we want. We can set the maximum n value to make it an n-order series.

Calculate Taylor expansion of (x^2+4)^{1/2} up to n = 4?

Given function f(m)= (x^2+4)^{1/2} and order point n = 1 to 4

Maclaurin equation for the function is:

$$ f(y)=âˆ‘k=0^âˆž f (k) (a)/ k! (x â€“ a)^k $$

$$ f(y)â‰ˆ P (x) = âˆ‘_k=0^4 f^(k) (a) / k! (x â€“ a)^k = âˆ‘k=0^âˆž f (k) (a)/ k! (x â€“ a)^k $$

$$ F^0(y) = f (y) = \sqrt {x^2 + 4 } $$

$$f (1) = \sqrt {5} $$

Take the first derivative \( f^1(y) = [f^0(y)]’ \)

$$ [\sqrt {x^2 + 4 }]’ = \frac {x} { \sqrt {x^2 + 4 }} $$

$$(f (1))’ = \frac { \sqrt { 5}} {5}$$

$$f^2 (y) = [f^1 (y)]’ = \frac {x} { \sqrt {x^2 + 4 }} = 4 / (x^2 + 4) ^{3/2} $$

Calculate the second derivative at a given point:

F (1)â€™â€™ = 4\sqrt{5} / 25

$$ f^3(y) = [f^2(y)]’ = (4/ (x^2 + 4) ^{3/2}) = – 12x / (x^2 + 4) ^{5/2} $$

Calculate the third derivative of \( (f (0))”’ = – 12 \sqrt {5} / 125 \)

$$f^4 (y) = [f^3 (y)]’ = [- 12x / (x^2 + 4) ^{5/2}]’ = 48x^2 â€“ 48 / \sqrt {x^2 + 4} (x^6 + 12x^4 + 48x^2 + 64) $$

Then, find the fourth derivative of the function (f(0))”” Â = 0

$$ f(y) â‰ˆ \sqrt {5}/0! (x â€“ 1)^0 + \frac { \sqrt{5} / 5} {1!} (x â€“ 1)^1 + â€¦ + 0 / 4! (x â€“ 1)^4 $$

After simplification, we get the final results:

$$ f(y) â‰ˆ P(x) = \sqrt {5} + \sqrt {5} (x-1) / 5 + 2 \sqrt {5} (x-1)^2 / 25 â€“ 2 \sqrt {5} (x – 1)^3 / 125Â $$

All the steps are specifically solved and represented in the Taylor series calculator calculations to make them easy to understand.

- Firstly, substitute a function with respect to a specific variable.
- Now, enter a particular point to evaluate the Taylor series of functions around this point.
- Then, add the order n for approximation.
- By using this Taylor series error calculator, find the series and determine the error at the given point. (optional)
- Click the calculate button for further solutions.

- The sum of the Taylor series calculator with steps shows the series after simplification.
- It computes the series of entered functions around the given order number n.
- The third-degree Taylor polynomial calculator takes the derivative for getting the polynomials and puts the results into the Taylor series formula.
- It displays the results after the simplification of polynomials.

From the source of Wikipedia: Analytic functions, Approximation error, convergence, Generalization, List of Maclaurin series of some common functions, Exponential function, Natural logarithm, Geometric series, Binomial series.