Taylor Series of (\pi - x)^-2 around a = 0

[olyviab]

Homework Statement

Write the Taylor series of the function f(x) = (\pi -x)^-2 around a = 0

Homework Equations

(\pi - x)^-2 = f(a) + f'(a)(x-a) + [f''(a)(x-a)^2]/(2!) +...+ [f^n(a)(x-a)^n]/(n!)

The Attempt at a Solution

This is what i have and i am not sure i am showing it correctly or compleatly.

f(x) = (\pi -x)^-2
f'(x) = 2(\pi - x)^-3
f''(x) = 6(\pi - x)^-4
f'''(x) = 24(\pi - x)^-5

(\pi - x)^-2 = f(a) + [f'(a)(x-a)^2]/(2!) +...+ [f^n(a)(x-a)^n]/(n!)

= (\pi-0)^-2 + [(2(\pi-0)^-3)(x-0)^2]/(2!) + [(6(\pi-0)^-4)(x-0)^3]/(3!) + ...

= \pi^-2 + 2(\pi^-3)x +[6(\pi^-4)(x^2)]/(2!) + [24(\pi^-5)(x^3)]/(3!) + ...

 

[mrbohn1]

You seem to have written the series wrong: where is the f(a) term? And why isn't the f'(a) term multiplied by (x-a)? Your derivatives are correct, now you just need to substitute into the (correct) series that you wrote in the "relevant equations" section...

 

[olyviab]

You seem to have written the series wrong: where is the f(a) term? And why isn't the f'(a) term multiplied by (x-a)? Your derivatives are correct, now you just need to substitute into the (correct) series that you wrote in the "relevant equations" section...

I looked over my work again and found where i messed up, thanks.
the final answer i get is:

(\pi-x)^-2 = \pi^-2 + 2(\pi^-3)(x) + 3(\pi^-4)(x^2) +...