# How to Calculate Taylor Series?

The Taylor Series, which is sometimes called the McLaurin Series on special circumstances, is a popular power series for functions. The Taylor series of a function is simply a representation of the function infinite sums of terms which are expressed in the form of a derivative at a sine point.

This power series is very easy to determine as your only need to find the n + 1 terms to get the nth polynomial. For example, if you want to find the derivative of a function polynomial degree of 5, you will need to calculate the Taylor series of that function as the 6th sum of terms (n + 1).

The Taylor series polynomials are an approximation of a function whose accuracy gets better as n increases on a constant point of a.  when a = 0, then what you get will be the McLaurin Series.

This article will discuss the formula of the Taylor Series and how to accurately use it to calculate the approximations of functions such that the derivate becomes a very good presentation of that function.

## Taylor Series Formula

The Taylor Series of Function f(x), which is infinitely differentiable at a centered real or complex number denoted by  a is the power series;

$F(x)=&space;\frac{f(a)}{0!}(x-a)^{0}&space;+&space;\frac{f^{/}(a)}{1!}&space;(x-a)^{2}+&space;\frac{f^{//}(a)}{2!}(x-a)^{2}+\frac{f^{///}(a)}{3!}&space;(x-a)^{3}+...$

The above series can be written as

Where  $f^{n}(a)$ represents the nth derivative that is evaluated at the point of $a$

If $a=0$, then the Taylor series becomes a special power series known as the McLaurin Series.

The two formulas of the Taylor Series are the same. However, the first is expanded expansion of the former. It is the preferred form when calculating this power series as it helps students better substitute and et their answers.

## How to Calculate the Taylor Series?

You have to know how to calculate derivatives using the product rule, quotient rule, and Chain Rule. Understanding the concept of factorial is also very important.

Write out the formula. This is extremely important as it will help you get the answer more quickly

Determine the respective  $f(a),&space;f^{/}&space;(a),f^{//}&space;(a),f^{///}&space;(a),$  and f///a and substitute into the expanded formula

Simplify as much as possible to arrive at the Taylor Series

## Example 1

Find the Taylor Series for $e^{x}$  with a centered point of $a=0$

Solution

The Taylor series formula is

The second step is to expand the formula

$e^{x}=\frac{f^{0}(a)}{0!)}(x-a)^{0}+\frac{f^{/}(a)}{1!)}(x-a)^{1}+\frac{f^{//}(a)}{2!)}(x-a)^{2}+&space;\frac{f^{///}(a)}{3!)}(x-a)^{3}+...$

The third step is to determine faand the respective derivatives $f^{/}(a),f^{//}(a),&space;f^{///}(a),$

Clearly;

Since $f(x)&space;=&space;e^{x},&space;f(a)&space;=e^{a}&space;where&space;x&space;=&space;a$

Since $f^{/}(x)&space;=&space;e^{x},&space;f^{/}(a)&space;=e^{a}$

Since $f^{//}(x)&space;=&space;e^{x},&space;f^{//}(a)&space;=e^{a}$

Since $f^{///}(x)&space;=&space;e^{x},&space;f^{///}(a)&space;=e^{a}$

This is true because the derivative of an exponential with a coefficient of 1 and a positive x will always be the same.

## Example 2

Substitute the derived values into the expanded formula

## Conclusion

Generally, when a=0, the Taylor series is much easier to calculate because a is naturally eliminated from the formula leading to a more straightforward answer. You can take on more examples of your choice with the same formula, and you will arrive at an accurate answer.

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