Calculating Taylor series.

Question

I want to calculate the Taylor McLaurin series of order 4 of this polynomial:

$$\frac{1}{9-x^2}$$

Calculating all the derivatives and evaluating at $x_0=0$ (since I am calculating McLaurin series)

$$P(x)=\frac{1}{9}+\frac{x^2}{9^2}+\frac{x^4}{9^3}$$

Now I am asked to calculate $f^8(0)$, the eight derivative of $\frac{1}{9-x^2}$. I know I am not supposed to derivate the polynomial 8 times, instead I can use the McLaurin polyonimal I got. How can I use the McLaurin polynomial to calculate $f^8(0)$?.

Finally, I have to calculate the Taylor polynomial of order 4 of $\frac{1}{x^2+4x-5}$. Here I am also supposed to use the two first results to conclude how the Taylor polyonimial is. I realized that $\frac{1}{x^2+4x-5}=\frac{1}{(x-1)(x+5)}$

Answer 1

They want you to write

$$\frac{1}{9-x^2} = \frac{1}{9}\frac{1}{1-\left(\frac{x}{3}\right)^2} $$

and then realize that the last fraction is the sum of a geometric series. The above equals

$$\frac{1}{9} \left(1 + \left(\frac{x}{3}\right)^2 +\left(\frac{x}{3}\right)^4 + \cdots \right).$$

There was no need to take any derivatives.

So now it's easy to answer your two questions. The coefficient on $x^8$ is

$$\frac{1}{3^{10}} = \frac{f^{(8)}(0)}{8!}. $$

The relevant fact you need for the second question is

$$\frac{1}{x^2+4x-5} = \frac{1}{(x+2)^2 -9}.$$