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.
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.
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 :
In terms of sigma notation, the Taylor series can be written as
Where
f(n) (a) = nth derivative of f
n! = factorial of n.
We know that the power series can be defined as
When x = 0,
f(x)= a0
So, differentiate the given function, it becomes,
f’(x) = a1+ 2a2x + 3a3x2 + 4a4x3 +….
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,
Generalise f in more general form, it becomes
f(x) = b + b1 (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
Hence, the Taylor series is proved.
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:
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;
This is known as the Maclaurin series.
Example: Maclaurin series of 1/(1-x) is given by:
1+x+x2+x3+x4+…,
The uses of the Taylor series are:
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) = e0 =1
f’’(0) = e0=1
f’’’(0) = e0 = 1
When Taylor series at x= 0, then the Maclaurin series is
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;
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:
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