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# Taylor Series Formula

Taylor series of a function is an infinite sum of terms, that is expressed in terms of the function's derivatives at any single point, where each following term has a larger exponent like x, x^{2}, x^{3}, etc. Taylor series formula thus helps in the mathematical representation of the Taylor series. Let us study the Taylor series formula using a few solved examples at the end of the page.

## What Is Taylor Series Formula?

The Taylor series formula helps to expand a function around a value of the variable using the derivatives of the function. It can be represented as,

f(x) = f(a) + f'(a) (x − a) + [ \(\frac{f''(a)}{2!}\) (x − a)^{2}] + [\( \frac{f'''(a)}{3!}\) (x − a)^{3}] + ….. + [ \(\frac{f^{(n)}(a)}{n!}\) (x − a)^{n}]

OR

\( f(x) = \sum_{n = 0}^\infty \frac{ f^{(n)} a }{ n! } \times (x - a)^n \)

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

## Taylor Series Formula Proof

**Taylor's Series Theorem Statement:**

Assume that if \(f(x)\) be a real or composite function, which is a differentiable function of a neighbourhood number that is also real or composite. Then, the Taylor series describes the following power series:

\(f(x)=f(a) \frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{p}(a)}{2 !}(x-a)^{2}+\frac{f^{(i)}(a)}{3 !}(x-a)^{3}+\ldots\)

In terms of sigma notation, the Taylor series can be written as

\(\sum_{n=0}^{\infty} \frac{f^{n}(a)}{n !}(x-a)^{n}$\)

Where,

\(\mathrm{f}^{(n)}(\mathrm{a})=\mathrm{n}^{\text {th }}\) derivative of \(\mathrm{f}\)

\(\mathrm{n} !\) = factorial of \(\mathrm{n}\)

**Proof:**

We know that the power series can be defined as

\(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+\ldots\)

When \(x=0\)

\(f(x)=a_{0}\)

So, differentiate the given function, it becomes,

\(f^{\prime}(x)=a_{1}+2 a_{2} x+3 a_{2} x^{2}+4 a_{4} x^{3}+\ldots\)

Again, when you substitute \(x=0\), we get

\(\mathrm{f}^{\prime}(0)=\mathrm{a}_{1}\)

So, differentiate it again, we get

\(f^{\prime \prime}(x)=2 a_{2}+6 a_{3} x+12 a_{4} x^{2}+\ldots\)

Now, substitute \(x=0\) in second-order differentiation, we get

\(\mathrm{f}^{\prime \prime}(0)=2 \mathrm{a}_{2}\)

Therefore, \(\left[\mathrm{f}^{\prime \prime}(0) / 2 !\right]=\mathrm{a}_{2}\)

By generalising the equation, we get

\(\mathrm{f}^{\mathrm{n}}(0) / \mathrm{n} !=\mathrm{a}_{\mathrm{n}}\)

Now substitute the values in the power series we get,

\(f(x)=f(0)+f^{\prime}(0) x+\frac{f^{\prime \prime}(0)}{2 !} x^{2}+\frac{f^{\prime \prime}(0)}{3 !} x^{3}+\ldots\)

Generalise \(\mathrm{f}\) in more general form, it becomes

\(f(x)=b+b_{1}(x-a)+b_{2}(x-a)^{2}+b_{3}(x-a)^{3}+\ldots\)

Now, \(x=a\), we get

\(b_{n}=f^{n}(a) / n !\)

Now, substitute \(b_{n}\) in a generalised form

\(f(x)=f(a) \frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\frac{f^{(3)}(a)}{3 !}(x-a)^{3}+\ldots\)

Hence, the Taylor series is proved.

Let us see the applications of the Taylor series formula in the following section.

## Examples Using Taylor Series Formula

**Example 1:** Find the expansion for the function, f(x) = 2x - 2x^{2} centered at a = -3 using the Taylor series formula.

**Solution: **

To find: Taylor series for the given function

Given:

Function, f(x) = 2x - 2x^{2}

Center at a = -3

\(P_n\)(x) = f(a) + f′(a)(x − a) + f′′(a)/2! × (x − a)^{2 }+ f′′′(a)/3! × (x − a)^{3 }+ f^{(4)}(a)/4! × (x − a)^{4 }+ ... + f^{(n) }(a)/n! × (x − a)^{n}

**Function and its derivatives.**

f(x) = 2x − 2x^{2}

f′(x) = 2− 4x

f′′(x) = −4

f′′′(x) = 0

Since a = −3 and n = 3, the required expansion is:

f(x) = f(−3) + f′(−3)(x − (−3)) + f′′(−3)/2! × (x − (−3))^{2 }+ f′′′(−3)/3! × (x − (−3))^{3 }

f(x) = f(−3) + f′(−3)(x + 3) + f′′(−3)/2! × (x + 3)^{2 }+ f′′′(−3)/3! × (x + 3)^{3}

We evaluate the function and its derivatives at x = a = −3:

f(−3) = 2(−3) − 2(−3)^{2 }= -24

f′(−3) = 2 - 4(-3)^{ }= 14

f′′(−3) = −4

f′′′(−3) = 0 and all the derivatives from here onwards are zeros.

Taylor series expansion for the given function:

\(P_3\)(x) = -24 + 14(x + 3) - 4/2! (x + 3)^{2 }- 0/3! (x + 3)^{3}

\(P_3\)(x) = -24 + 14(x + 3) - 2(x + 3)^{2}

**Answer:**Taylor series expansion around a = −3 for the function f(x) = 2x − 2x^{2} is -24 + 14(x + 3) - 2(x + 3)^{2}.

**Example 2:** Find the Taylor series expansion for function, f(x) = cos x, centred at x = 0.

**Solution:**

To find: Taylor series expansion

Given:

Function, f(x) = Cos x

Using Taylor series formula,

f(x) = f(a) + f′(a)(x − a) + f′′(a)/2! × (x − a)^{2 }+ f′′′(a)/3! × (x − a)^{3 }+ f^{(4)}(a)/4! × (x − a)^{4 }+ ... + f^{(n) }(a)/n! × (x − a)^{n}

We evaluate the function and its derivatives:

f(x) = cos(x)

f'(x) = −sin(x)

f''(x) = −cos(x)

f'''(x) = sin(x)

Thus,

cos(x) = cos(a) − sin(a)/1! (x - a) − cos(a)/2! (x - a)^{2} + sin(a)/3! (x - a)^{3} + ...

Now, put a = 0.

cos(x) = 1 − 0/1! (x - 0) − 1/2! (x - 0)^{2} + 0/3! (x - 0)^{3} + 1/4! (x - 0)^{4} + ...

cos(x) = 1 − x^{2}/2! + x^{4}/4! − ...

**Answer: **Taylor series expansion for given function, cos(x) = 1 − x^{2}/2! + x^{4}/4! − ...

**Example 3: Find the Taylor Series for f(x) = x ^{3 }- 10x^{2} + 6 at x=3.**

**Solution:**First, let us find the derivatives of the given function.

f(x) = x^{3} − 10x^{2} + 6 ⇒ f(3) = -57

f’(x) = 3x^{2} − 20x ⇒ f’(3) = 33

f’’(x) = 6x – 20 ⇒ f’’(3) = -2

f’’’(x) = 6 ⇒ f’’’(3) = 6

f’’’’(x) = 0

Thus, the required series is:

\(\begin{aligned}

x^{3}-10 x^{2}+6 &=\sum_{n=0}^{\infty} \frac{f^{(n)}(3)}{n !}(x-3)^{n} \\

&=f(3)+f^{\prime}(3)(x-3)+\frac{f^{\prime \prime}(3)}{2 !}(x-3)^{2}+\frac{f^{\prime \prime \prime}(3)}{3 !}(x-3)^{3}+0 \\

&=-57-33(x-3)-(x-3)^{2}+(x-3)^{3}

\end{aligned}\)

**Answer: Taylor series expansion for given function is = − 57 − 33(x−3) − (x−3) ^{2 }+ (x−3)^{3}**

## FAQs on Taylor Series Formula

### What Is Taylor Series Formula Used For?

A Taylor series is useful in computer science, calculus, chemistry, physics, etc. Taylor series is used to create an estimation of what a function looks like.

### What Is Taylor Series Formula?

The Taylor series formula helps to expand a function around a value of the variable using the derivatives of the function. It can be represented as,

f(x) = f(a) + f'(a) (x − a) + [ \(\frac{f''(a)}{2!}\) (x − a)^{2}] + [\( \frac{f'''(a)}{3!}\) (x − a)^{3}] + ….. + [ \(\frac{f^{(n)}(a)}{n!}\) (x − a)^{n}]

OR

\( f(x) = \sum_{n = 0}^\infty \frac{ f^{(n)} a }{ n! } \times (x - a)^n \)

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

### Who Invented Taylor Series Formula?

The concept of the Taylor series was given by the Scottish mathematician James Gregory and later it was formally introduced by the English mathematician Brook Taylor in 1715.

### Are Taylor Series Formula and Maclaurin Series Formula the Same?

Taylor series is the representation of a function as an infinite sum of terms that are worked out from the values of the function's derivatives at a single point. On the other hand, a Maclaurin series gives the expansion of the Taylor series of a function about zero.

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