Taylor Series

Introduction

Taylor series or Taylor expansion of a function is a finite sum of terms that are expressed in terms of the functions derivatives at a single point The polynomial or function of an infinite sum of terms is the Taylor series. The exponent or degree of each succeeding term will be greater than the exponent or degree of the one before it.

$$\mathrm{f(a)\:+\:\frac{f'(a)}{1!}(x\:-\:a)\:+\:\frac{f"(a)}{2!}(x\:-\:a)^{2}\:+\:\frac{f'''(a)}{3!}(x\:-\:a)^{3}\:+\:.......}$$

For a real value function f(x), where f'(a), f"(a), f"'(a), etc., stands for the derivative of the function at point a, the aforementioned Taylor series expansion is provided. The Taylor series is also known as the Maclaurin series if the value of point "a" is zero.

What is a Taylor series?

Assume that f(x) is a real or composite function and that it is a differentiable function of a real or composite neighborhood number. The following power series is then described by the Taylor series −

$\mathrm{f(x)\:=\:f(a)\:+\:\frac{f'(as)}{1!}(x\:-\:a)\:+\:\frac{f"(a)}{2!}(x\:-\:a)^{2}\:+\:\frac{f'''(a)}{3!}(x\:-\:a)^{3}\:+\:.......}$

Here,

• f(x) = Real or complex-valued function, that is infinitely differentiable at a real or complex number “a” is the power series

• n = Total number of terms in the series

The Taylor series can be represented using sigma notation as

$$\mathrm{f(x)\:=\:\displaystyle\sum\limits_{n=0}^\infty\:\frac{f^{n}(a)}{n!}\:(x\:-\:a)^{n}}$$

Where

$\mathrm{f^{n}(a)\:=\:n^{th}}$

n! = factorial of n.

How to find a Taylor Series?

If we want to find the Taylor series of function f(x), the following steps can be followed to find a Taylor series.

Step 1 − Determine f(x)’s initial few derivatives.

The formula shows f(a). This is f(x) at x = a. Next, we observe f' (a). This is the f(x) first derivative calculated for x = a. Like this, we will calculate f "(x), f "(x), and so on as per need.

Step 2 − Calculate the derivatives of the function when x = a

Substitute a for x in each of the outcomes from the preceding step. We obtain f(a) using the initial derivative − $\mathrm{f'(a)\:,\:f''(a)\:,\:f'''(a).........}$

Step 3 − Complete the Taylor series expression's right side.

From this, we will now construct the Taylor series −

$$\mathrm{f(x)\:=\:f(a)\:+\:\frac{f'(as)}{1!}(x\:-\:a)\:+\:\frac{f"(a)}{2!}(x\:-\:a)^{2}\:+\:\frac{f'''(a)}{3!}(x\:-\:a)^{3}\:+\:.......}$$

Step 4 − Write the result using a summation. Hence, you get the final Taylor series.

Find the Taylor series of some common functions

Let’s find the Taylor series for some common functions.

• $\mathrm{\sin\:x}$

Taylor series for x=0

$\mathrm{f(x)\:=\:\tan\:x}$

Step 1 − Determine f(x)’s initial few derivatives.

$\mathrm{f'(x)\:=\:\sec^{2}\:x}$

$\mathrm{f''(x)\:=\:2\sec^{2}(x)\:\tan(x)}$

$\mathrm{f'''(x)\:=\:-4\sec^{4}\:+\:6\sec^{4}(x)}$

$\mathrm{f''''(x)\:=\:-8\sec^{2}(x)\tan(x)\:+\:24\sec^{2}\tan(x)}$

Step 2 − Calculate the derivatives of the function when $\mathrm{x\:=\:a\:=\:0}$

$\mathrm{f(0)\:=\:\tan\:0\:=\:0}$

$\mathrm{f'(0)\:=\:\sec^{2}\:0\:=\:1}$

$\mathrm{f(x)\:=\:f(0)\:=\:\frac{f'(0)}{1!}}$

$\mathrm{f'''(x)\:=\:-\cos\:0\:=\:-1}$

$\mathrm{f''''(x)\:=\:\sin\:0\:=\:0}$

Step 3 − Complete the Taylor series expression's right side.

$\mathrm{f(x)\:=\:f(0)\:+\:\frac{f'(0)}{1!}(x\:-\:0)\:+\:\frac{f"(0)}{2!}(x\:-\:0)^{2}\:+\:\frac{f'''(0)}{3!}(x\:-\:0)^{3}\:+\:\:\frac{f'''(0)}{4!}(x\:-\:0)^{4}......}$

$\mathrm{f(x)\:=\:\frac{1}{1!}(x)\:-\:-\:\frac{1}{3!}(x)^{3}\:+\:\frac{1}{5!}(x)^{5}\:-\:\frac{1}{7!}(x)^{7}\:+\:......}$

$\mathrm{\sin\:x\:=\:\displaystyle\sum\limits_{n=0}^\infty \frac{(-1)^{n}}{(2n\:+\:1)!}(x)^{2n\:+\:1}\:=\:\frac{1}{1!}(x)\:-\:\frac{1}{3!}(x)^{3}\:+\:\frac{1}{5!}(x)^{5}\:-\:\frac{1}{7!}(x)^{7}}$

• $\mathrm{\tan\:x}$

Taylor series for x=0

$\mathrm{f(x)\:=\:\tan\:x}$

Step 1 − Determine f(x)’s initial few derivatives.

$\mathrm{f'(x)\:=\:\sec^{2}\:x}$

$\mathrm{f''(x)\:=\:2\sec^{2}(x)\:\tan\:(x)}$

$\mathrm{f''(x)\:=\:-4\sec^{2}(x)\:+\:6\sec^{4}(x)}$

$\mathrm{f''''(x)\:=\:-8\sec^{2}(x)\tan(x)\:+\:24\sec^{4}(x)\:\tan(x)}$

Step 2: Calculate the derivatives of the function when x = a = 0.

$\mathrm{f'(0)\:=\:\tan\:0\:=\:0}$

$\mathrm{f'(0)\:=\:\sec^{0}\:=\:1}$

$\mathrm{f''(0)\:=\:2\sec^{2}(0)\:\tan(0)\:=\:0}$

$\mathrm{f'''(0)\:=\:-4\sec^{2}(0)\tan(0)\:+\:6\sec^{4}(0)\:=\:2}$

$\mathrm{f''''(0)\:=\:-8\sec^{2}\:\tan(0)\:+\:24\sec^{4}\:\tan(0)\:=\:0}$

Step 3 − Complete the Taylor series expression's right side

$\mathrm{f(x)\:=\:f(0)\:+\:\frac{f'(0)}{1!}(x\:-\:0)\:+\:\frac{f"(0)}{2!}(x\:-\:0)^{2}\:+\:\frac{f'''(0)}{3!}(x\:-\:0)^{3}\:+\:\:\frac{f'''(0)}{4!}(x\:-\:0)^{4}......}$

$$\mathrm{f(x)\:=\:\frac{1}{1!}(x)\:+\:\frac{2}{2\times\:3}(x)^{3}\:+\:\frac{2\times\:8}{2\times\:3\times\:4\times\:5}(x)^{5}}$$

$$\mathrm{f(x)\:=\\frac{1}{1}(x)\:+\:\frac{1}{3}(x)^{3}\:+\:\frac{2}{15}(x)^{5}\:+\:......}$$

$$\mathrm{\tan\:x\:=\:\displaystyle\sum\limits_{n=0}^\infty \frac{f^{n}(a)}{n!}(x)^{n}\:=\:\frac{1}{1}(x)\:+\:\frac{1}{3}(x)^{3}\:+\:\frac{2}{15}(x)^{5}\:+\:.....}$$

• $\mathrm{e^{\bigwedge}x}$

Taylor series for x=0

$\mathrm{f(x)\:=\:e^{x}}$

Step 1 − Determine f(x)’s initial few derivatives.

$\mathrm{}$

Step 2 − Calculate the derivatives of the function when x = 0.

$\mathrm{f(0)\:=\:f'(0)\:=\:f'(0)\:=\:f''(0)\:=\:f'''(0)\:=\:f''''(0)\:=\:e^{0}\:=\:1}$

Step 3 − Complete the Taylor series expression's right side.

$\mathrm{f(x)\:=\:f(0)\:+\:\frac{f'(0)}{1!}(x\:-\:0)\:+\:\frac{f''(0)}{2!}(x\:-\:0)^{2}\:+\:\frac{f'''(0)}{3!}(x\:-\:0)^{3}\:+\:\frac{f''''(0)}{4!}(x\:-\:0)^{4}\:+\:.....}$

$\mathrm{\:\:\:\:\:\:\:\:\:\:\:\:\:\:f(x)\:=\:1\:+\:\frac{1}{1!}(x)\:+\:\frac{1}{2!}(x)^{2}\:+\:\frac{1}{3!}(x)^{3}\:+\:\frac{1}{4!}(x)^{4}\:+\:.....}$

$\mathrm{\:\:\:\:\:\:\:\:\:\:\:\:\:\:e^{x}\:=\:\displaystyle\sum\limits_{n=0}^\infty \frac{1}{n!}(x)^{n}\:=\:1\:+\:\frac{1}{1!}(x)\:+\:\frac{1}{2!}(x)^{2}\:+\:\frac{1}{3!}(x)^{3}\:+\:\frac{1}{4!}(x)^{4}\:+\:.....}$

Properties of Taylor Series

Property 1

Only odd powers of x are found in the Taylor series for an odd function, and only even powers of x are found in the Taylor series for an even function

Property 2

By substituting an existing series, a new series can be made.

Property 3

• Multiple existing series can be combined into one new series by adding, subtracting, multiplying, or dividing them.

• When it comes to division, we must employ the long division or synthetic division method. As long as there is no division by zero, the result is accurate.

Property 4

By differentiating or integrating a function's Taylor series term by term, a function can be either integrated or differentiated.

Solved examples

1) Find the Taylor series for $$

Answer − We must calculate the cos x derivatives and evaluate them at x = 0.

$\mathrm{f(x)\:=\:\cos\:x\:\Longrightarrow\:f(0)\:=\:1}$

$\mathrm{f'(x)\:=\:-\sin\:x\:\Longrightarrow\:f'(0)\:=\:0}$

$\mathrm{f''(x)\:=\:-\sin\:x\:\Longrightarrow\:f''(0)\:=\:-1}$

$\mathrm{f'''(x)\:=\:-\cos\:x\:\Longrightarrow\:f'''()}\:=\:0$

$\mathrm{f''''(x)\:=\:cos\:x\:\Longrightarrow\:f''''(0)\:=\:1}$

$\mathrm{f{(5)}(x)\:=\:-\:\sin\:x\:\Longrightarrow\:f^{(5)}(0)\:=\:0}$

$\mathrm{f^{(6)}(x)\:=\:-\:\cos\:x\:\Longrightarrow\:f^{(6)}(0)\:=\:-1}$

Complete the Taylor series expression's right side.

$\mathrm{f(x)\:=\:f(0)\:+\:\frac{f'(0)}{1!}(x\:-\:0)\:+\:\frac{f''(0)}{2!}(x\:-\:0)^{2}\:+\:\frac{f'''(0)}{3!}(x\:-\:0)^{3}\:+\:\frac{f''''(0)}{4!}(x\:-\:0)^{4}\:+\:.....}$

$\mathrm{\:\:\:\:\:\:=\:1\:+\:\frac{0}{1!}(x\:-\:0)\:+\:\frac{1}{2!}(x\:-\:0)^{2}\:+\:\frac{0}{3!}(x\:-\:0)^{3}\:+\:\frac{1}{4!}(x\:-\:0)^{4}\:+\:.....}$

$\mathrm{\:\:\:\:\:\:=\:1\:-\:\frac{1}{2!}(x)^{2}\:+\:\frac{1}{4!}(x)^{4}\:+\:......}$

$$\mathrm{\cos\:x\:=\:\displaystyle\sum\limits_{n=0}^\infty \frac{f^{n}(a)}{n!}(x)^{n}\:=\:1\:-\:\frac{1}{2!}(x)^{2}\:+\:\frac{1}{4!}(x)^{4}\:-\:......}$$

$$\mathrm{\cos\:x\:=\:\displaystyle\sum\limits_{n=0}^\infty \frac{(-1)^{n}}{(2n)!}(x)^{2n}}$$

Conclusion

• The polynomial or function of an infinite sum of terms is the Taylor series. The exponent or degree of each succeeding term will be greater than the exponent or degree of the one before it.

• The following power series is then described by the Taylor series −

$$\mathrm{f(x)\:=\:f(a)\:+\:\frac{f'(a)}{1!}(x\:-\:a)\:+\:\frac{f"(a)}{2!}(x\:-\:a)^{2}\:+\:\frac{f'''(a)}{3!}(x\:-\:a)^{3}\:+\:.......}$$

• The Taylor series can be represented using sigma notation as

$$\mathrm{f(x)\:=\:\displaystyle\sum\limits_{n=0}^\infty \frac{f^{n}(a)}{n!}\:(x\:-\:a)^{n}}$$

• If we want to find the Taylor series of function f(x), the following steps can be followed to find a Taylor series.

  • Determine f(x)’s initial few derivatives

  • Calculate the derivatives of the function when x = a.

  • Complete the Taylor series expression's right side.

  • Write the result using a summation. Hence, you get the final Taylor series.

FAQs

1. What is the Taylor series?

The polynomial or function of an infinite sum of terms is the Taylor series. The exponent or degree of each succeeding term will be greater than the exponent or degree of the one before it.

2. What is the Taylor Series Formula?

$\mathrm{f(x)\:=\:f(a)\:+\:\frac{f'(a)}{1!}(x\:-\:a)\:+\:\frac{f"(a)}{2!}(x\:-\:a)^{2}\:+\:\frac{f'''(a)}{3!}(x\:-\:a)^{3}\:+\:.......}$

OR

$$\mathrm{f(x)\:=\:\displaystyle\sum\limits_{n=0}^\infty \frac{f^{n}(a)}{n!}\:(x\:-\:a)^{n}}$$

3. Are the formulas for the Taylor and Maclaurin series equivalent?

A function is represented as an infinite sum of terms in a Taylor series, which is calculated from the values of the function's derivatives at a single point. A Maclaurin series, on the other hand, provides the Taylor series expansion of a function at zero.

4. Who Created the Formula for the Taylor Series?

The English mathematician Brook Taylor first explicitly developed the Taylor series in 1715, after the Scottish mathematician James Gregory first proposed the idea.

5. Can we combine multiple series into one series?

Multiple existing series can be combined into one new series by adding, subtracting, multiplying, or dividing them.