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, x², x³, 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 - 2x² centered at a = -3 using the Taylor series formula.

Solution:  

To find: Taylor series for the given function

Given: 

Function, f(x) = 2x - 2x²

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 − 2x²

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 − 2x²  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 − x²/2! + x⁴/4! − ...

Answer: Taylor series expansion for given function, cos(x) = 1 − x²/2! + x⁴/4! − ...

Example 3: Find the Taylor Series for f(x) = x³ - 10x² + 6 at x=3.
Solution: First, let us find the derivatives of the given function.

f(x) =  x³ − 10x² + 6 ⇒ f(3) = -57

f’(x) = 3x² − 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