Taylor Series | Engineering Mathematics - Engineering Mathematics PDF Download

Taylor series is the polynomial or a function of an infinite sum of terms. Each successive term will have a larger exponent or higher degree than the preceding term.
Taylor Series | Engineering Mathematics - Engineering Mathematics
The above Taylor series expansion is given for a real values function f(x) where f’(a), f’’(a), f’’’(a), etc., denotes the derivative of the function at point a. If the value of point ‘a’ is zero, then the Taylor series is also called the Maclaurin series.

Taylor’s Series Theorem

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 :
Taylor Series | Engineering Mathematics - Engineering Mathematics
In terms of sigma notation, the Taylor series can be written as
Taylor Series | Engineering Mathematics - Engineering Mathematics
Where
f(n) (a) = nth derivative of f
n! = factorial of n.

Proof

We know that the power series can be defined as
Taylor Series | Engineering Mathematics - Engineering Mathematics
When x = 0,
f(x)= a0 
So, differentiate the given function, it becomes,
f’(x) = a1+ 2a2x + 3a3x+ 4a4x+….
Again, when you substitute x = 0, we get
f’(0) =a1
So, differentiate it again, we get
f”(x) = 2a2 + 6a3x +12a4x2 + …
Now, substitute x=0 in second-order differentiation, we get

f”(0) = 2a2
Therefore, [f”(0)/2!] = a2
By generalising the equation, we get
f n (0) / n! = an
Now substitute the values in the power series we get,
Taylor Series | Engineering Mathematics - Engineering Mathematics
Generalise f in more general form, it becomes
f(x) = b + b(x-a) + b2( x-a)2 + b3 (x-a)3+ ….
Now, x = a, we get
bn = fn(a) / n!
Now, substitute bn in a generalised form
Taylor Series | Engineering Mathematics - Engineering Mathematics
Hence, the Taylor series is proved.

Taylor Series of Sin x

f(x) = sin x
f’(x) = cos x
f’’(x) = -sin x
f’’’(x) = -cos x
f’’’’(x) = sin x
And so on
The taylor series for Sin x at x = 0, is given by:
Taylor Series | Engineering Mathematics - Engineering Mathematics

Taylor Series in Several Variables

The Taylor series is also represented in the form of functions of several variables. The general form of the Taylor series in several variables is
Taylor Series | Engineering Mathematics - Engineering Mathematics
Maclaurin Series

If the Taylor Series is centred at 0, then the series is known as the Maclaurin series. It means that,
If a= 0 in the Taylor series, then we get;
Taylor Series | Engineering Mathematics - Engineering Mathematics
This is known as the Maclaurin series.
Example: Maclaurin series of 1/(1-x) is given by:
1+x+x2+x3+x4+…, 

Applications of Taylor Series

The uses of the Taylor series are:

  • Taylor series is used to evaluate the value of a whole function in each point if the functional values and derivatives are identified at a single point.
  • The representation of Taylor series reduces many mathematical proofs.
  • The sum of partial series can be used as an approximation of the whole series.
  • Multivariate Taylor series is used in many optimization techniques.
  • This series is used in the power flow analysis of electrical power systems.

Problems and Solutions

Question 1: Determine the Taylor series at x=0 for f(x) = ex 

Given: f(x) = ex
Differentiate the given equation,
f’(x) = ex
f’’(x) =ex
f’’’(x) = ex
At x=0, we get
f’(0) = e=1
f’’(0) = e0=1
f’’’(0) = e0 = 1
When Taylor series at x= 0, then the Maclaurin series is
Taylor Series | Engineering Mathematics - Engineering Mathematics
ex = 1+ x(1) + (x2/2!)(1) + (x3/3!)(1) + ….. 
Therefore, ex = 1+ x + (x2/2!) + (x3/3!)+ …..

Question 2: EValuate the Taylor Series for  f ( x ) = cos ( x )  for  x = 0.

We need to take the derivatives of the cos x and evaluate them at  x = 0.
f(x) = cos x  ⇒ f(0) = 1
f’(x) = -sin x ⇒ f’(0) = 0
f’’(x) = -cos x ⇒ f’’(0) = -1
f’’’(x) = sin x ⇒ f’’’(0) = 0
f’’’’(x) = cos x ⇒ f’’’’(4) = 1
f(5)(x) = -sin x ⇒ f(5) (0) = 0
f(6) (x) = -cos x ⇒ f(6)(0) = -1
Therefore, according to the Taylor series expansion;
Taylor Series | Engineering Mathematics - Engineering Mathematics
Taylor Series | Engineering Mathematics - Engineering Mathematics

Question 3: Evaluate the Taylor Series for  f (x) = x3 − 10x2 + 6  at  x = 3. 

First, we will 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 
Therefore, the required series is:
Taylor Series | Engineering Mathematics - Engineering Mathematics

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