Question

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

Answer 1

Answer:

$\mathbf{P_2(x) = 1+\dfrac{x^2}{2}}$

Step-by-step explanation:

Given that:

f(x) = sec (x)  , n = 2

Where are to find P_2(x)

Suppose ; f(x) = sec (x)  , n = 2

then

$f(0) = sec (0) = 1$

$f'(x) = sec (x)* tan (x)|_{x=0} = 0$

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

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

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

$f''(x) = 1$

$f(x) = f(0) + \dfrac{f'(0)x}{1!}+ \dfrac{f''(0)x^2}{2!}+ \dfrac{f'''(0)x^3}{3!}+...$

$f(x) = 1 + \dfrac{0}{1!}x+ \dfrac{x^2}{2!}+...$

$f(x) = 1 + \dfrac{x^2}{2}+...$

since order n =2

$\mathbf{P_2(x) = 1+\dfrac{x^2}{2}}$