# Thread: Expand a function into a series of powers?

1. ## Expand a function into a series of powers?

How would I expand the function $8/(2+x)$ in a series of powers of x-2?

2. Well in your case it is not hard, since you will have only two terms. Let $f(x)=8/2+x$. Since all higher derivatives of $f$ vanish (and since you can see that it’s linear), $f(x)=a_0+a_1(x-2)$. Solve for $a_i$.

3. Originally Posted by Tikoloshe
Well in your case it is not hard, since you will have only two terms. Let $f(x)=8/2+x$. Since all higher derivatives of $f$ vanish (and since you can see that it’s linear), $f(x)=a_0+a_1(x-2)$. Solve for $a_i$.
I should have written the function as 8/(2+x). I assume this will make a difference.

4. Oops. I guess I was too quick to answer. You can show that the nth derivative is given by $f^{(n)}(x)=(-1)^n\frac{8n!}{(x+2)^{n+1}}$. Then, we know that $f^{(n)}(2)=(-1)^n\frac{8n!}{(2+2)^{n+1}}=2n!\left(\frac{-1}{4}\right)^n$. Since f is analytic at 2, we can write its power series as
$f(x)=\sum_{n=0}^\infty f^{(n)}(2)\frac{(x-2)^n}{n!}=2\sum_{n=0}^\infty \left(\frac{-1}{4}\right)^n(x-2)^n$.

5. Thanks for the reply, but I am afraid I don't fully understand. Could you provide the basic steps you would take, not necesarrily for this problem, but expanding functions into powers series in general?

6. Originally Posted by Mattpd
How would I expand the function $8/(2+x)$ in a series of powers of x-2?
What Tikoloshe did was use the formula for the "Taylor's series" at x= 2.

Simpler, I think, is: Write the function at $\frac{8}{2+ x- 2+ 2}= \frac{8}{4+ (x- 2)}= \frac{2}{1+ \frac{1}{4}(x-2)}= \frac{4}{1- (-(x-2))}$.

Now, use the fact that sum of a geometric series is given by $\sum_{n=0}^\infty ar^n= \frac{a}{1- r}$
with a= 4 and r= -(x-2).

7. @HallsofIvy: I don’t think your answer is correct. When you have $a=4$ and $r=-(x-2)$, then your series converges to $\frac{4}{x-1}$, which is not the original function.

If you want to use a geometric series method, you could use $a=8$ and $r=-(x+1)$, but then your power series is no longer in factors of (x-2).

Or you could stop at $\frac{8}{x+2}=\frac{2}{1+ \frac{1}{4}(x-2)}$, and simply let $a=2$ and $r=-\frac{1}{4}(x-2)$, which gives my original answer.

@Mattpd: My method is quite a brute force method, which one often uses when learning about power series. Basically, Taylor’s theorem states that under certain conditions, $f(x)=\sum_{n=0}^\infty f^{(n)}(c)\frac{(x-c)^n}{n!}$, where $f^{(n)}(c)$ is shorthand for the $n$th derivative of $f$ evaluated at $c$. Depending on what function you have, it may be possible to find (by induction) an exact expression for $f^{(n)}(c)$.

The method HallsofIvy gave (which is correct despite a slight haste in the actual solution), reduces the problem to one for which the answer is known (namely it tries to make your function look like the expression for the sum of a geometric series).

8. You are right, I changed what I had written on the very first line, but left the final (incorrect) part of that line and then used it!

$
\frac{8}{2+ x- 2+ 2}= \frac{8}{4+ (x- 2)}= \frac{2}{1+ \frac{1}{4}(x-2)}= \frac{4}{1- (-(x-2))}
$
should be $
\frac{8}{2+ x- 2+ 2}= \frac{8}{4+ (x- 2)}= \frac{2}{1+ \frac{1}{4}(x-2)}
$

so that we have a geometric series with a= 2 and $r= -\frac{1}{4}(x- 2)$. Thanks for catching that.