Answer:
[tex]\mathbf{P_2(x) = 1+\dfrac{x^2}{2}}[/tex]
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
[tex]f(0) = sec (0) = 1[/tex]
[tex]f'(x) = sec (x)* tan (x)|_{x=0} = 0[/tex]
[tex]f''(x) = sec (x)*tan ^2(x)+ sec (x) * sec^2(x)[/tex]
[tex]f''(x) = sec (x)*tan ^2(x)|_{x=0} + sec^3(x)[/tex]
[tex]f''(x) = 0 + sec^3(0)[/tex]
[tex]f''(x) = 1[/tex]
[tex]f(x) = f(0) + \dfrac{f'(0)x}{1!}+ \dfrac{f''(0)x^2}{2!}+ \dfrac{f'''(0)x^3}{3!}+...[/tex]
[tex]f(x) = 1 + \dfrac{0}{1!}x+ \dfrac{x^2}{2!}+...[/tex]
[tex]f(x) = 1 + \dfrac{x^2}{2}+...[/tex]
since order n =2
[tex]\mathbf{P_2(x) = 1+\dfrac{x^2}{2}}[/tex]
