# Power series

• May 15th 2012, 12:35 PM
icelated
Power series
I have a problem i cant seem to figure out. I am trying to find the power series expansion for the function

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

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

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

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

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

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

$f(x) = \frac {9} {0 - (-x -5 ) }$
But, somehow this isnt correct..
• May 15th 2012, 12:45 PM
Plato
Re: Power series
Quote:

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

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

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

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

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

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

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

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

What is the point?
What is the question?
What are you trying do?
• May 15th 2012, 12:50 PM
icelated
Re: Power series
Sorry, i am trying to find the power series expansion for the function
• May 15th 2012, 01:44 PM
skeeter
Re: Power series
$f(x) = \frac{9}{10 - (-x+5)} = \frac{9}{10} \cdot \frac{1}{1 - \frac{5-x}{10}}$

$\frac{9}{10} \left(1 - \frac{(x-5)}{10} + \frac{(x-5)^2}{10^2} - \frac{(x-5)^3}{10^3} + ... \right)$
• May 15th 2012, 02:18 PM
icelated
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
• May 15th 2012, 02:43 PM
skeeter
Re: Power series
Quote:

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 = ?
• May 15th 2012, 03:04 PM
icelated
Re: Power series
Sorry, still confused. I thought we are to follow this form (x-c) -c
• May 15th 2012, 03:11 PM
skeeter
Re: Power series
Quote:

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?
• May 15th 2012, 04:29 PM
HallsofIvy
Re: Power series
Another way to do that is to write
$\frac{9}{x+5}= \frac{9}{x-(-5)}$
and now divide both numerator and denominator by -5:
$\frac{-\frac{9}{5}}{1- \frac{x}{5}}$

If you let $y= x/5$ that becomes
$-\frac{9}{5}\frac{1}{1- y}$
which is, of course, the sum of the geometric series
$-\frac{9}{5}\sum y^n= -\frac{9}{5}\sum \frac{x^n}{5^n}$
• May 15th 2012, 05:17 PM
skeeter
Re: Power series
Quote:

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

...
• May 15th 2012, 07:31 PM
icelated
Re: Power series
Quote:

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 ??
• May 16th 2012, 03:28 AM
skeeter
Re: Power series
Quote:

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.