# Thread: Power series

1. ## Power series

I have a problem i cant seem to figure out. I am trying to find the power series expansion for the function

$\displaystyle f(x) = \frac {9} {5 + x}$
c = 5

Now, would i arrange the 5 + x to force a negative?

$\displaystyle f(x) = \frac {9} {5 - (-x)}$

then, subtract C? I think this is where i get confused on how to handle C correctly!

$\displaystyle f(x) = \frac {9} {-5 + 5 - (-x -5 ) }$

$\displaystyle f(x) = \frac {9} {-5 + 5 - (-x -5 ) }$

$\displaystyle f(x) = \frac {9} {0 - (-x -5 ) }$
But, somehow this isnt correct..

2. ## Re: Power series

Originally Posted by icelated
I have a problem i cant seem to figure out.

$\displaystyle f(x) = \frac {9} {5 + x}$
c = 5

Now, would i arrange the 5 + x to force a negative?

$\displaystyle f(x) = \frac {9} {5 - (-x)}$

then, subtract C? I think this is where i get confused on how to handle C correctly!

$\displaystyle f(x) = \frac {9} {-5 + 5 - (-x -5 ) }$

$\displaystyle f(x) = \frac {9} {-5 + 5 - (-x -5 ) }$

$\displaystyle f(x) = \frac {9} {0 - (-x -5 ) }$
But, somehow this int correct..
What is the point?
What is the question?
What are you trying do?

3. ## Re: Power series

Sorry, i am trying to find the power series expansion for the function

4. ## Re: Power series

$\displaystyle f(x) = \frac{9}{10 - (-x+5)} = \frac{9}{10} \cdot \frac{1}{1 - \frac{5-x}{10}}$

$\displaystyle \frac{9}{10} \left(1 - \frac{(x-5)}{10} + \frac{(x-5)^2}{10^2} - \frac{(x-5)^3}{10^3} + ... \right)$

5. ## Re: Power series

Why 10 - (-x+5) can you show me how to properly add the C? i just dont see how you got 10

6. ## Re: Power series

Originally Posted by icelated
Why 10 - (-x+5) can you show me how to properly add the C? i just dont see how you got 10
uhh ... 10 - 5 = ?

7. ## Re: Power series

Sorry, still confused. I thought we are to follow this form (x-c) -c

8. ## Re: Power series

Originally Posted by icelated
Sorry, still confused. I thought we are to follow this form (x-c) -c
no, you want the form k - (x-c)

in this case, k - [-(x - c)] ... see why k = 10?

9. ## Re: Power series

Another way to do that is to write
$\displaystyle \frac{9}{x+5}= \frac{9}{x-(-5)}$
and now divide both numerator and denominator by -5:
$\displaystyle \frac{-\frac{9}{5}}{1- \frac{x}{5}}$

If you let $\displaystyle y= x/5$ that becomes
$\displaystyle -\frac{9}{5}\frac{1}{1- y}$
which is, of course, the sum of the geometric series
$\displaystyle -\frac{9}{5}\sum y^n= -\frac{9}{5}\sum \frac{x^n}{5^n}$

10. ## Re: Power series

Originally Posted by HallsofIvy
Another way to do that is to write
$\displaystyle \frac{9}{x+5}= \frac{9}{x-(-5)}$
and now divide both numerator and denominator by -5: <--- check this step ...
$\displaystyle \frac{-\frac{9}{5}}{1- \frac{x}{5}}$
...

11. ## Re: Power series

Originally Posted by skeeter
no, you want the form k - (x-c)

in this case, k - [-(x - c)] ... see why k = 10?
I understand the form but we are to subtract c from x and then add c to k ? You seem to be forcing 10. I have not done this in the homework at all.
So, would this be correct 9 / -5 + (x - 5) - 5 ??

12. ## Re: Power series

Originally Posted by icelated
I understand the form but we are to subtract c from x and then add c to k ? You seem to be forcing 10. I have not done this in the homework at all.
So, would this be correct 9 / -5 + (x - 5) - 5 ??
exactly ... to get the denominator in the desired form to write a power series centered at 5

no, 9/[-5 + (x-5) - 5] = 9/(x - 15)

... probably because you have not come across a problem (like this one) that does not follow a "cook-book" procedure. I recommend you speak w/ your instructor about it.