Question

Use differentiation to find a power series representation for the following function.

\[\frac{1}{(x+6)^2}\]

Answer 1

In any case, let's see what we can do.
\[f'(x)=\frac{-2}{(x+6)^3}\]
\[f''(x)=\frac{6}{(x+6)^4}\]
\[f'''(x)=\frac{-24}{(x+6)^5}\]
Can you see the pattern for the nth derivative?
Also, where should the power series be centered?

Answer 2

It should be centered at 0 and the nth derivative im not to sure on

Answer 3

just taking a stab at the nth derivative.. is it \[\frac{(n+2)!}{(x+6)^{n+3}}\]

Answer 4

\[f^{(n)}(x)=\frac{(-1)^n (n+1)!}{(x+6)^{n+2}}\]
Thus, \[c_n=\frac{f^{(n)}(0)}{n!}=\frac{(-1)^n n}{6^{n+2}}\]
and its Taylor series is \[\sum_{n=0}^{\infty} \frac{(-1)^nn}{6^{n+2}}x\]

Answer 5

ohh i started at n=0 oops

Answer 6

Alternatively, you can see that \[\frac{1}{(x+6)^2}={dy \over dx}\left (\frac{-1}{x+6}\right)\]. \[\frac{-1}{x+6}=\frac{1}{6-x}\].
Do you know how to get the power series of 1/(6-x)?

Answer 7

Err, I meant 1/(-6-x), not 1/(6-x). :))

Answer 8

not quite

Answer 9

Anyway, the power series of -1/(x+6) is:

\[\begin{align}
\frac{-1}{x+6} & =\frac{-1}{6} \cdot \frac{1}{\frac{x}{6}+1} \\
& = {-1 \over 6} \sum_{n=0}^{\infty} \left ( {-x \over 6} \right )^{n} \\
& = {-1 \over 6} \sum_{n=0}^{\infty} {(-1)^n x^n \over 6^n}
\end{align}\]

Answer 10

That is equal to \[\sum_{n=0}^{\infty} {(-1)^{n+1} x^n \over6^{n+1}}\]
then differentiate this with respect to x.

Answer 11

Sorry for the late reply and any errors you see. I'm very tired.

Answer 12

Its ok, thank you

Answer 13

And as I typed my previous reply, I saw an error in my first answer. n should be n+1 and x should be x^n. Anyway, the second solution should produce the same answer as the first.