Introduction to Taylor series explained with examples

Introduction to Taylor series explained with examples: In calculus, the Taylor series is frequently used to expand the given function up to the nth order around the center point. The function can be a trigonometric or algebraic expression of one variable. Taylor series finds the limits of the functions at a specific point.

In this type of power series, the derivatives are used to expand the functions. In this post, we will learn the definition, formula, and examples of the Taylor series.

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Introduction to Taylor series explained with examples
Introduction to Taylor series explained with examples

Introduction to Taylor series explained with examples

Taylor series

According to Britannica, the Taylor series is:

In mathematics, the expression of a function (f) for which the differential of all orders exists at a point “a” in the domain of f in the form of power series. It expands the given expression in the infinite sum of the terms up to the nth order derivative at the center point (a).

To expand the functions using the Taylor series keep one thing in mind that the given should be continuous e.g., the derivatives of the function must exist. If the function is discontinuous the Taylor series can’t exist.

You can use a Taylor expansion calculator to get the series expansions of the given function around a center point up to the nth term.

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The formula of the Taylor series

The general formula of the Taylor series is:

F(y) = f(a) + 1/1! [f’(a) (y – a)] + ½! [f’’(a) (y – a)2] + 1/3! [f’’’(a) (y – a)3 + … + 1/n! [f n (a) (y – a) n]

The summation form of the above equation is

F(y) = f n(a) (y – a) n/ n!

In the formula of the Taylor series, F(y) is the Taylor series of the given function, a is the center point, n is the order of derivative, and f(a) is the given function around the center point a.

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Examples of the Taylor series

The Taylor series can be expanded easily by applying the formula using order and the center point of the function. Before solving the problems of this series you must have a sound knowledge of the derivative and how to find the derivative of the functions.

Example 1: For trigonometric function

Calculate the Taylor series of the given trigonometric function whose center point is 7 and the order is 6, ycos(y) + 23.

Solution

Step 1: Take the given data from the problem.

f(y) = ycos(y) + 23

nth term of the function= n = 6

center point = a = 3

Step 2: Take the formula of the Taylor series to expand the given trigonometric function.

F(y) = f n(a) (y – a) n/ n!

Put n = 6 in the above summation form and write the series.

F(y) = f(a) + 1/1! [f’(a) (y – a)] + ½! [f’’(a) (y – a)2] + 1/3! [f’’’(a) (y – a)3] + 1/4! [f’’’’(a) (y – a)4] + 1/5! [f’’’’’(a) (y – a)5] + 1/6! [f’’’’’(a) (y – a)6] … equation (1)

Step 4: Now find the first six derivatives of ycos(y) + 23.

f(y) = ycos(y) + 23

f’(y) = d/dx [ycos(y) + 23] = cos(y) + (y(-sin(y)) + 0 = cos(y) – ysin(y)  

f’’(y) = d/dx [cos(y) – ysin(y)] = -sin(y) – sin(y) – ycos(y) = -2sin(y) – ycos(y)

f’’’(y) = d/dx [-2sin(y) – ycos(y)] = -2cos(y) – cos(y) + ysin(y) = -3cos(y) + ysin(y)

f iv(y) = d/dx [-3cos(y) + ysin(y)] = – (-3sin(y)) + sin(y) + ycos(y) = 4sin(y) + ycos(y)

f v(y) = d/dx [4sin(y) + ycos(y)] = 4cos(y) + cos(y) – ysin(y) = 5cos(y) – ysin(y)

f vi(y) = d/dx [5cos(y) – ysin(y)] = -5sin(y) – sin(y) –ycos(y) = -6sin(y) – ysin(y)

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Step 5: Put n = 0, 1, 2, 3, 4, 5, 6 in the formula of Taylor series to get the result of series at a.

For n = 0

f(a) (y – a)0/ 0!  = f(a) = acos(a) + 23

For n = 1

f’(a) (y – a)1/ 1!  = f’(a) (y – a) = (cos(a) – asin(a)) (y – a)

For n = 2

f’’(a) (y – a)2/ 2!  = f’’(a) (y – a)2/2 = 1/2 (-2sin(a) – acos(a)) (y – a)2

For n = 3

f’’’(a) (y – a)3/3!  = f’’’(a) (y – a)3/6 = 1 /6 (-3cos(a) + asin(a)) (y – a)3

For n = 4

f’’’’(a) (y – a)4/4!  = f’’’’(a) (y – a)4/24 = 1/24 (4sin(a) + acos(a)) (y – a)4

For n = 5

f’’’’’(a) (y – a)5/5! = f’’’’’(a) (y – a)5/120   = 1 /120 (5cos(a) – asin(a)) (y – a)5

For n = 6

f’’’’’’(a) (y – a)6/6! = f’’’’’’(a) (y – a)6/720   = 1/720 (-6sin(a) – asin(a)) (y – a)6

Step 6: Substitute all the values in equation (1).

F(y) = f(a) + 1/1! [f’(a) (y – a)] + ½! [f’’(a) (y – a)2] + 1/3! [f’’’(a) (y – a)3] + 1/4! [f’’’’(a) (y – a)4] + 1/5! [f’’’’’(a) (y – a)5] + 1/6! [f’’’’’(a) (y – a)6] … equation (1)

F(y) = (acos(a) + 23) + (cos(a) – asin(a)) (y – a) + 1/2 (-2sin(a) – acos(a)) (y – a)2 + 1 /6 (-3cos(a) + asin(a)) (y – a)3 + 1/24 (4sin(a) + acos(a)) (y – a)4 + 1/120 (5cos(a) – asin(a)) (y – a)5 + 1/720 (-6sin(a) – asin(a)) (y – a)6

Step 7: Put a = 3 in the above expression.

F(y) = (3cos(3) + 23) + (cos(3) – 3sin(3)) (y – 3) + 1/2 (-2sin(3) – 3cos(3)) (y – 3)2 + 1 /6 (-3cos(3) + 3sin(3)) (y – 3)3 + 1/24 (4sin(3) + 3cos(3)) (y – 3)4 + 1/120 (5cos(3) – 3sin(3)) (y – 3)5 + 1/720 (-6sin(3) – 3sin(3)) (y – 3)6

Hence, the given function ycos(y) + 23 is expanded in a power series.

You can also use an online Taylor series calculator to get the result in a fraction of seconds without going through such a large calculation. Follow the following steps to get the result using this tool.

Step 1: Input the one variable function.

Step 2: Enter the corresponding variable, center point (a), and the order (n) in the required input boxes.

Step 3: Click the calculate button.

Step 4: The result will show below the calculate button in a fraction of seconds.

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FAQs

What is the Taylor series of sin(x) at a=0?

The Taylor series of sin(x) is

x – x3/6 + x5/120 – x7/5040 + …

What is the Taylor series of sin(x) at a=0?

The Taylor series of cos(x) at a=0 is

1 – x2/2 + x4/24 – x6/720 + …

What is the Maclaurin series?

The Taylor series at a = 0 is said to be the McLaurin series.

Summary

In the above post, we have learned the definition and formula of the Taylor series along with the example. Now you can expand any function just by learning the above mentioned example.

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