Math, asked by abhi0207, 11 months ago

expand sec x by maclaurins series

Answers

Answered by saounksh
32

Answer:

sec(x) = 1 + x²/2 + 5x⁴/24 + .........

Step-by-step explanation:

Concept

Maclaurins series of a function f(x) is a special case of Taylor Series Expansion. In maclaurin series, the function is expanded around a = 0. Maclaurin series of f(x) is given by

f(x) = f(0) + f´(0)x + f´´(0)x²/2! + f´´´(0)x³/3! + .......

Calculations

It is given that

f(x) = sec(x)

⇒ f(0) = sec(0) = 1

f´(x) = sec(x)tan(x)

⇒ f´(0) = sec(0)tan(0) = 1*0 = 0

f´´(x) = sec³(x) + sec(x)tan²(x)

⇒ f´´(0) = sec³(0) + sec(0)tan²(0) = 1 + 1*0

⇒ f´´(0) = 1

f´´´(x) = 3sec²(x).sec(x)tan(x) + sec(x).2tan(x)sec²(x) + sec(x)tan(x).tan²(x)

f´´´(x) = 5sec³(x)tan(x) + sec(x)tan³(x)

⇒ f´´´(0) = 5sec³(0)tan(0) + sec(0)tan³(0)

⇒ f´´´(0) = 0

f´´´´(x) = 5sec³(x)sec²(x) + 5.3sec²(x).sec(x)tan(x).tan(x) + sec(x).3tan²(x)sec²(x) + sec(x)tan(x).tan³(x)

f´´´´(x) = 5sec⁵(x) + 18sec³(x)tan²(x) + sec(x)tan³(x)

⇒ f´´´´(0) = 5sec⁵(0) + 18sec³(0)tan²(0) + sec(0)tan³(0)

⇒ f´´´´(0) = 5

Therefore, Maclaurin Series Expansion of sec(x) is

sec(x) = 1 + 0.x + x²/2! + 0.x³/3! + 5.x⁴/4! + ........

= 1 + x²/2 + 5x⁴/24 + .........

Answered by lakshmilakku
2

Answer:

sec(x) = 1 + x²/2 + 5x⁴/24 + .........

Step-by-step explanation:

Maclaurin's Theorem: What is it?

image outcome

Unless one of the functions is identically zero, the sum of an even and an odd function is neither even nor odd.

How to Expand ?

As can be seen, when all derivatives are evaluated at x = 0, they all result in the value 1. Additionally, f(0)=1, therefore we can infer that the expansion of the Maclaurin Series will be as follows: e x  1 + x + 1 2 x 2 + 1 6 x 3 + 1 24 x 4 + 1 120 x 5 +...

f(x) = sec(x)

⇒ f(0) = sec(0) = 1

f´(x) = sec(x)tan(x)

⇒ f´(0) = sec(0)tan(0) = 1*0 = 0

f´´(x) = sec³(x) + sec(x)tan²(x)

⇒ f´´(0) = sec³(0) + sec(0)tan²(0) = 1 + 1*0

⇒ f´´(0) = 1

f´´´(x) = 3sec²(x).sec(x)tan(x) + sec(x).2tan(x)sec²(x) + sec(x)tan(x).tan²(x)

f´´´(x) = 5sec³(x)tan(x) + sec(x)tan³(x)

⇒ f´´´(0) = 5sec³(0)tan(0) + sec(0)tan³(0)

⇒ f´´´(0) = 0

f´´´´(x) = 5sec³(x)sec²(x) + 5.3sec²(x).sec(x)tan(x).tan(x) + sec(x).3tan²(x)sec²(x) + sec(x)tan(x).tan³(x)

f´´´´(x) = 5sec⁵(x) + 18sec³(x)tan²(x) + sec(x)tan³(x)

⇒ f´´´´(0) = 5sec⁵(0) + 18sec³(0)tan²(0) + sec(0)tan³(0)

⇒ f´´´´(0) = 5

Therefore, Maclaurin Series Expansion of sec(x) is

sec(x) = 1 + 0.x + x²/2! + 0.x³/3! + 5.x⁴/4! + ........

= 1 + x²/2 + 5x⁴/24 + .........

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