Taylor and Maclaurin series: Derivation and calculations

 

Introduction to Taylor and Maclaurin series: Derivation and calculations

 

In mathematics, the Taylor series and Maclaurin series are the fundamental concepts of calculus that are used to represent functions as an infinite sum of terms. The Taylor series is a generalization of the Maclaurin series, which is a specific type of Taylor series.

The purpose of this post is to provide the basic concept of the Taylor series and Maclaurin series with the help of solved examples.

What is the Taylor series?

In calculus, the Taylor series is a well-known technique that is used to represent algebraic and trigonometric functions as an infinite sum of terms. In this technique, each term of the Taylor series is a differential of the function calculated at a particular point.

The general expression of finding the Taylor series of the given function at a point u = c is:

F(u) = f(c) + [f’(c) (u – c)]/ 1! + [f’’(c) (u – c)²]/ 2! + [f’’’(c) (u – c)³] / 3! + … + [f ⁿ (c) (u – c) ⁿ]/ n!

Here,

·       f'(c), f''(c), f'''(c) represent the first, second, and third derivatives of f(u) evaluated at u = c respectively.

·       The terms (u – c), (u – c)², (u – c)³, etc. are known as the powers of (u – c).

The summation form of the Taylor formula is:

F(u) = n=0Σ∞ [f n(c) (u – c) n/ n!]

Where ·       F(u) is the Taylor expression of the given trigonometric, algebraic, and quadric functions. ·       “c” is the particular point of the function. ·       n=0Σ∞ is the sum of the terms. ·       “n” is the total number of differential to be taken for expansion. ·       “fn(c)” = the result of the differentiated function at a particular point.

How to calculate the Taylor series?

The calculations of the taylor series can be done easily by taking the differentials of the given function and evaluating them at the particular point. But the process of calculating the Taylor series is rather lengthy, so you can take help from a Taylor series calculator to avoid lengthy calculations.

Below is a sloved example to learn how to evcaluate it manually.

Example

Find the infinite sum of f(u) = sin(u) – 5u⁴ + 2u⁵, if the particular point is u = 2 and the order of the function is 6.

Solution

Step 1: First of all, take the first 6 differentials of the given function with respect to “u”.

Number of derivatives	Symbol	Results
First derivative of sin(u) – 5u4 + 2u5	f’(u)	= cos(u) – 20u3 + 10u4
Second derivative of sin(u) – 5u4 + 2u5	f’’(u)	= -sin(u) – 60u2 + 40u3
Third derivative of sin(u) – 5u4 + 2u5	f’’’(u)	= -cos(u) – 120u + 120u2
Fourth derivative of sin(u) – 5u4 + 2u5	fiv(u)	= sin(u) – 120 + 240u
Fifth derivative of sin(u) – 5u4 + 2u5	fv(u)	= cos(u) + 240
Sixth derivative of sin(u) – 5u4 + 2u5	fvi(u)	= -sin(u)

Step 2: Now substitute the value of u = 2 to the above differential functions.

Differential functions	At u = 2	Results
f(u) = sin(u) – 5u4 + 2u5	f(2)	= sin(2) – 5(2)4 + 2(2)5 = sin(2) – 5(16) + 2(32) = sin(2) – 80 + 64 = sin(2) – 16
f’(u) = cos(u) – 20u3 + 10u4	f’(2)	= cos(2) – 20(2)3 + 10(2)4 = cos(2) – 20(8) + 10(16) = cos(2) – 160 + 160 = cos(2)
f’’(u) = -sin(u) – 60u2 + 40u3	f’’(2)	= -sin(2) – 60(2)2 + 40(2)3 = -sin(2) – 60(4) + 40(8) = -sin(2) – 240 + 320 = -sin(2) + 80
f’’’(u) = -cos(u) – 120u + 120u2	f’’’(2)	= -cos(2) – 120(2) + 120(2)2 = -cos(2) – 120(2) + 120(4) = -cos(2) – 240 + 480 = -cos(2) + 240
fiv(u) = sin(u) – 120 + 240u	fiv(2)	= sin(2) – 120 + 240(2) = sin(2) – 120 + 480 = sin(2) + 360
fv(u) = cos(u) + 240	fv(2)	= cos(2) + 240
fvi(u) = -sin(u)	fvi(2)	= -sin(2)

Step 3: Substitute the differential values at u = 2 to the general formula of the Taylor series.

F(u) = _n=0Σ^∞ [f ⁿ(c) (u – c) ⁿ/ n!]

Put n = 6 to the above expression and write the series.

F(u) = _n=0Σ⁶ [f ⁿ(c) (u – c) ⁿ/ n!]

F(u) = f(c) + [f’(c) (u – c)]/ 1! + [f’’(c) (u – c)²]/ 2! + [f’’’(c) (u – c)³] / 3! + [f’’’’(c) (u – c)⁴] / 4! + [f^v(c) (u – c)⁵] / 5! + [f^vi(c) (u – c)⁶] / 6!

At u = 2 expressions will be:

F(u) = f(2) + [f’(2) (u – 2)]/ 1! + [f’’(2) (u – 2)²]/ 2! + [f’’’(2) (u – 2)³] / 3! + [f’’’’(2) (u – 2)⁴] / 4! + [f^v(2) (u – 2)⁵] / 5! + [f^vi(2) (u – 2)⁶] / 6!

The final result is:

F(u) = (sin(2) – 16) + [cos(2) (u – 2)]/ 1! + [(-sin(2) + 80) (u – 2)²]/ 2! + [(-cos(2) + 240) (u – 2)³] / 3! + [(sin(2) + 360) (u – 2)⁴] / 4! + [(cos(2) + 240) (u – 2)⁵] / 5! + [-sin(2) (u – 2)⁶] / 6!

F(u) = (sin(2) – 16) + [cos(2) (u – 2)] + [(-sin(2) + 80)/2 (u – 2)²] + [(-cos(2) + 240)/6 (u – 2)³] + [(sin(2) + 360)/24 (u – 2)⁴] + [(cos(2) + 240)/120 (u – 2)⁵] + [-sin(2) (u – 2)⁶] /720

F(u) = (sin(2) – 16) + [cos(2) (u – 2)] + [(-sin(2)/2 + 40) (u – 2)²] + [(-cos(2)/6 + 40) (u – 2)³] + [(sin(2)/24 + 15) (u – 2)⁴] + [(cos(2)/12- + 2) (u – 2)⁵] + [-sin(2) (u – 2)⁶] /720

What is the Maclaurin Series?

In calculus, the Maclaurin series is part of the Taylor series. In simple words, the Maclaurin series is a special form of the Taylor series in which the central point is u = 0. In this technique, the terms (x – c), (x – c)², (x – c)³, etc. will be written as (x – 0), (x – 0)², (x – 0)³, etc. [x, x², x³, etc].

The formula for the Maclaurin series of a function f(u) is:

f(u) = f(0) + f'(0)x + (f''(0)/2!)x² + (f'''(0)/3!)x³ + ... + (fⁿ(0)/n!) xⁿ

The summation form of the Maclaurin formula is:

F(u) = _n=0Σ^∞ [f ⁿ(0) (u) ⁿ/ n!]

How to calculate Maclaurin Series?

Example

Find the infinite sum of f(u) = cos(u) 5u⁵ + 12u⁶, if the particular point is u = 0 and the order of the function is 4.

Solution

Step 1: First of all, take the first 4 differentials of the given function with respect to “u”.

Number of derivatives	Symbol	Results
1st derivative of cos(u) + 5u5 + 12u6	f’(u)	= -sin(u) + 25u4 + 72u5
2nd derivative of cos(u) + 5u5 + 12u6	f’’(u)	= - cos(u) +100u3 + 360u4
3rd derivative of cos(u) + 5u5 + 12u6	f’’’(u)	= sin(u) + 300u2 + 1440u3
4th derivative of cos(u) + 5u5 + 12u6	fiv(u)	= cos(u) + 600u + 4320u2

Step 2: Now substitute the value of u = 1 to the above differential functions.

Differential functions	At u = 1	Results
f(u) = cos(u) + 5u5 + 12u6	f(0)	= cos(0) + 5(0)5 + 12(0)6 = cos(0) + 5(0) + 12(0) = 1 + 0 + 0 = 1
f’(u) = -sin(u) + 25u4 + 72u5	f’(1)	= -sin(0) + 25(0)4 + 72(0)5 = -sin(0) + 25(0) + 72(0) = 0 + 0 + 0 = 0
f’’(u) = - cos(u) +100u3 + 360u4	f’’(1)	= -cos(0) + 100(0)3 + 360(0)4 = -cos(0) + 100(0) + 360(0) = -1 + 0 + 0 = -1
f’’’(u) = sin(u) + 300u2 + 1440u3	f’’’(1)	= sin(0) + 300(0)2 + 1440(0)3 = sin(0) + 300(0) + 1440(0) = 0 + 0 + 0 = 0
f’’’’(u) = cos(u) + 600u + 4320u2	fiv(1)	= cos(0) + 600(0) + 4320(0)2 = cos(0) + 600(0) + 4320(0) = 1 + 0 + 0 = 1

Step 3: Substitute the differential values at u = 0 to the general formula of the Taylor series.

F(u) = _n=0Σ^∞ [f ⁿ(0) (u – 0) ⁿ/ n!]

Put n = 4 to the above expression and write the series.

F(u) = _n=0Σ⁴ [f ⁿ(0) (u – 0)ⁿ/ n!]

F(u) = f(0) + [f’(0) (u – 0)]/ 1! + [f’’(0) (u – 0)²]/ 2! + [f’’’(0) (u – 0)³] / 3! + [f’’’’(0) (u – 0)⁴] / 4!

F(u) = f(0) + [f’(0) (u)]/ 1! + [f’’(0) (u)²]/ 2! + [f’’’(0) (u)³] / 3! + [f’’’’(0) (u)⁴] / 4!

The final result is:

F(u) = 1 + [(0) (u)]/ 1! + [(-1) (u)²]/ 2! + [(0) (u)³] / 3! + [(1) (u)⁴] / 4!

F(u) = 1 + [(0) (u)] + [(-1) (u)²]/ 2 + [(0) (u)³] / 6 + [(1) (u)⁴] / 24

F(u) = 1 + 0 - u²/ 2 + 0 + u⁴ / 24

F(u) = 1 - u²/ 2 + u⁴ / 24

Conclusion

Now you can take help from this post to understand the terms Taylor and Maclaurin series. In this post, we have discussed the basics of the Taylor and Maclaurin series with the help of solved examples to understand them more precisely.