Question

Find the nth Maclaurin polynomial for the function. f(x) = sec(x), n = 2 P_2(x) =

Answer 1

`\mathbf{P_2(x) = 1+(x^2)/(2)}`

Explanation:

Given that:

f(x) = sec (x) , n = 2

Where are to find P_2(x)

Suppose ; f(x) = sec (x) , n = 2

then

`f(0) = sec (0) = 1`

`f'(x) = sec (x)* tan (x)|_(x=0) = 0`

`f''(x) = sec (x)*tan ^2(x)+ sec (x) * sec^2(x)`

`f''(x) = sec (x)*tan ^2(x)|_(x=0) + sec^3(x)`

`f''(x) = 0 + sec^3(0)`

`f''(x) = 1`

`f(x) = f(0) + (f'(0)x)/(1!)+ (f''(0)x^2)/(2!)+ (f'''(0)x^3)/(3!)+...`

`f(x) = 1 + (0)/(1!)x+ (x^2)/(2!)+...`

`f(x) = 1 + (x^2)/(2)+...`

since order n =2

`\mathbf{P_2(x) = 1+(x^2)/(2)}`