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Taylor Series using Geometric Series and Power Series

 
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Sep29-10, 09:00 PM   #1
 

Taylor Series using Geometric Series and Power Series


1. The problem statement, all variables and given/known data
See figure attached.


2. Relevant equations



3. The attempt at a solution

Okay I think I handled the lnx portion of the function okay(see other figure attached), but I'm having from troubles with the,

[tex]\frac{1}{x^{2}}[/tex]

[tex]\int x^{-2} = \frac{-1}{x} + C[/tex]

How do I deal with the C?

If I can get,

[tex]\frac{-1}{x}[/tex]

I can work with it to get something like the following,

[tex]\frac{\text{first term of geometric series}}{1 - \text{common ratio}}[/tex]

So what do I do about the C? Once I figure this out I can make more of an attempt into shaping,

[tex] \frac{-1}{x}[/tex]

into the form mentioned above.

Any ideas?

Thanks again!
Attached Thumbnails
TSQ.jpg   TSQA1.jpg  
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Sep30-10, 06:27 AM   #2
 
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You're going about it backwards. Use

[tex]\frac{1}{x^2} = -\frac{d}{dx}\left(\frac{1}{x}\right)[/tex]
Sep30-10, 09:07 PM   #3
 
Quote by vela View Post
You're going about it backwards. Use

[tex]\frac{1}{x^2} = -\frac{d}{dx}\left(\frac{1}{x}\right)[/tex]
Alrighty I think I've got a series for,

[tex]\frac{1}{x^{2}}[/tex]

See figure attached. Is this correct?

I can't seem to figure out how to express it sigma notation however.

Any ideas?
Attached Thumbnails
TSQA2.jpg  
Sep30-10, 09:09 PM   #4
 
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Taylor Series using Geometric Series and Power Series


No, it looks like you integrated the series, but you want to differentiate -1/x to get 1/x2.
Sep30-10, 09:48 PM   #5
 
Quote by vela View Post
No, it looks like you integrated the series, but you want to differentiate -1/x to get 1/x2.
Whoops!

How does this look? (See figure attached)
Attached Thumbnails
TSQA3.jpg  
Sep30-10, 10:48 PM   #6
 
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Looks good!
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