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Taylor Series: What is it and how to calculate it?

In calculus, the Taylor series is widely used to find the power series of the one variable functions with respect to a specific point. It can also take the order of the function that tells how many derivatives of the function must be involved in the series.

The derivative is a branch of calculus that differentiate the function according to the independent variable. In this post, we will learn the definition, formula, examples, and solutions of the Taylor series.

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What is the Taylor series?

In calculus, a power series that gives the expansion of a function f(y) in the region of a point “b” provided that in a region the function is continuous and all the derivatives exist is known as a Taylor series.

It can also have defined as a function of the endless sum of the terms that are expressed in terms of the differential of a function at a center point. It is usually used to expand a function in series by taking the specific point and the nth order.

The function must be continuous and the derivative of that function must exist. If the differentiation of the function is not existing, then the series of Taylor does not exist for that function.

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Taylor series Formula

Here is a general formula of the Taylor series:

F(y) = f(b) + f’(b) (y – b) / 1! + f’’(b) (y – b)²/ 2! + f’’’(b) (y – b)³/ 3! + … + fⁿ(b) (y – b)ⁿ/ n!

The formula of the Taylor series in the form of summation is given below.

fⁿ(b) is the n^th differential of the function at a specific point b.
n is the order of the differential or total numbers.
b is the specific point.
F(y) is the series of the given function.

To get the result of the Taylor series according to the above formula, use a Taylor series calculator.

How to solve the problems of the Taylor series?

The problems of the Taylor series can be calculated easily with the help of its formula. For solving the problems of the Taylor series, the knowledge of finding derivatives is compulsory. Here are a few examples of the Taylor series.

Example 1

Evaluate the Taylor series of y * cos(y) around the center point 5 and the order is 6.

Solution

Step 1: First of all, take the given information into the problem.

f(y) = ycos(y)

order of the function= n = 5

center point = b = 2

Step 2: Now take the formula of the Taylor series and expand it according to the given data.

According to the given order, this equation becomes.

F(y) = f(b) + f’(b) (y – b)/1! + f’’(b) (y – b)²/2! + f’’’(b) (y – b)³/3! + f’’’’(b) (y – b)⁴/4! + f’’’’’(b) (y – b)⁵/5! + f’’’’’’(b) (y – b)⁶/6! … (1)

Step 3: Now evaluate the first six derivatives of ycos(y) with respect to “y”

f(y) = ycos(y)

f’(y) = cos(y) – ysin(y)

f’’(y) = -sin(y) – (sin(y) + ycos(y)) = -2sin(y) – ycos(y)

f’’’(y) = -2cos(y) – cos(y) – (-ysin(y)) = -2cos(y) – cos(y) + ysin(y) = -3cosy + ysin(y)

f’’’’(y) = -3(-sin(y)) + sin(y) + ycos(y) = 3sin(y) + sin(y) + ycos(y) = 4sin(y) + ycos(y)

f’’’’’(y) = 4cos(y) + cos(y) + (-ysin(y)) = 5cos(y) – ysin(y)

f’’’’’’(y) = -5sin(y) – (sin(y) + ycos(y)) = -5sin(y) – sin(y) – ycos(y) = -6sin(y) – ycos(y)

For n = 0

f(b) (y – b)⁰/ 0! = f(b) = bcos(b)

For n = 1

f’(b) (y – b)¹/ 1! = f’(b) (y – b) = (cos(b) – bsin(b)) (y – b)

For n = 2

f’’(b) (y – b)²/ 2! = f’’(b) (y – b)²/2 = ½ (-2sin(b) – bcos(b)) (y – b)²

For n = 3

f’’’(b) (y – b)³/3! = f’’’(b) (y – b)³/6 = 1 /6 (-3cos(b) – bsin(b)) (y – b)³

For n = 4

f’’’’(b) (y – b)⁴/4! = f’’’’(b) (y – b)⁴/24 = 1 /24 (4sin(b) + bcos(b)) (y – b)⁴

For n = 5

f’’’’’(b) (y – b)⁵/5! = f’’’’’(b) (y – b)⁵/120 = 1 /120 (5cos(b) – bsin(b)) (y – b)⁵

For n = 6

f’’’’’’(b) (y – b)⁶/6! = f’’’’’’(b) (y – b)⁶/720 = 1 /720 (-6sin(b) – bcos(b)) (y – b)⁶

Step 5: Now Put the calculated values in (1)

F(y) = bcos(b) + (cos(b) – bsin(b)) (y – b) + ½ (-2sin(b) – bcos(b)) (y – b)²+ 1 /6 (-3cos(b) – bsin(b)) (y – b)³+ 1 /24 (4sin(b) + bcos(b)) (y – b)⁴+ 1 /120 (5cos(b) – bsin(b)) (y – b)⁵ + 1 /720 (-6sin(b) – bcos(b)) (y – b)⁶

Step 6: Now replace b = 5 in the above series.

F(y) = 5cos(5) + (cos(5) – 5sin(5)) (y – 5) + ½ (-2sin(5) – 5cos(5)) (y – 5)²+ 1 /6 (-3cos(5) – 5sin(5)) (y – 5)³+ 1 /24 (4sin(5) + 5cos(5)) (y – 5)⁴+ 1 /120 (5cos(5) – 5sin(5)) (y – 5)⁵ + 1 /720 (-6sin(5) – 5cos(5)) (y – 5)⁶

This is the required expansion of the cos(x) by using the Taylor series

Example 2

Evaluate the Taylor series of y^4 around the center point 20 and the order is 5.

Solution

Step 1: First of all, take the given information into the problem.

f(y) = y^4

order of the function= n = 5

center point = b = 20

Step 2: Now take the formula of the Taylor series and expand it according to the given data.

According to the given order, this equation becomes.

F(y) = f(b) + f’(b) (y – b)/1! + f’’(b) (y – b)²/2! + f’’’(b) (y – b)³/3! + f’’’’(b) (y – b)⁴/4! + f’’’’’(b) (y – b)⁵/5! … (1)

Step 3: Now evaluate the first six derivatives of y^4 with respect to “y”

f(y) = y^4

f’(y) = 4y^3

f’’(y) = 12y^2

f’’’(y) = 24y

f’’’’(y) = 24

f’’’’’(y) = 0

For n = 0

f(b) (y – b)⁰/ 0! = f(b) = y^4 = b^4

For n = 1

f’(b) (y – b)¹/ 1! = f’(b) (y – b) = 4y^3 (y – b) = 4b^3 (y – b)

For n = 2

f’’(b) (y – b)²/ 2! = f’’(b) (y – b)²/2 = ½ 12y^2 (y – b)² = 12b^2 (y – b)²

For n = 3

f’’’(b) (y – b)³/3! = f’’’(b) (y – b)³/6 = 1 /6 24y (y – b)³ = 4b (y – b)³

For n = 4

f’’’’(b) (y – b)⁴/4! = f’’’’(b) (y – b)⁴/24 = 1 /24 (24) (y – b)⁴ = (y – b)⁴

For n = 5

f’’’’’(b) (y – b)⁵/5! = f’’’’’(b) (y – b)⁵/120 = 1/120 (0) (y – b)⁵ = 0

Step 5: Now Put the calculated values in (1)

F(y) = b^4 + 4b^3 (y – b) + 12b^2 (y – b)²+ 4b (y – b)³ + (y – b)⁴ + 0

Step 6: Now replace b = 2 in the above series.

F(y) = 2^4 + 4(2^3) (y – 2) + 12(2^2) (y – 2)²+ 4(2) (y – 2)³ + (y – 2)⁴ + 0

F(y) = 16 + 4(8) (y – 2) + 12(4) (y – 2)²+ 4(2) (y – 2)³ + (y – 2)⁴

F(y) = 16 + 24 (y – 2) + 48 (y – 2)²+ 8 (y – 2)³ + (y – 2)⁴

Conclusion

In this post, we have covered all the basics of the Taylor series along with definition, formula, and solved examples. Now the confusion about the Taylor series can be cleared by learning this post.