expand sec x by maclaurins series
Answers
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 + .........
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 + .........