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Taylor Series using Geometric Series and Power Series |
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| Sep29-10, 09:00 PM | #1 |
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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! |
| 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 |
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[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? |
| 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.
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| Sep30-10, 09:48 PM | #5 |
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How does this look? (See figure attached) |
| Sep30-10, 10:48 PM | #6 |
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Looks good!
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