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Power series representation |
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| Mar7-08, 08:03 PM | #1 |
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Power series representation
1. The problem statement, all variables and given/known data
Use differentiation to find a power series representation for f(x) = 1/ (1+x)^2 2. Relevant equations geometric series sum = 1/(1+x) 3. The attempt at a solution (1) I see that the function they gave is the derivative of 1/(1+x). (2) Therefore, (-1)*(d/dx)summation(x^n) = -1/(1+x)^2 (3) Differentiating the summation gives: (-1)*[summation (n)x^(n-1)] However, the book is telling me that for my second step (2) I should be getting d/dx [summation (-1)^n (x^n)]. Why is it becoming an alternating series here? |
| Mar7-08, 08:16 PM | #2 |
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Recognitions:
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Remember:
[tex] \frac{1}{1-x} = 1+x+x^2+.... [/tex] so that, after substituting -x for x: [tex] \frac{1}{1+x} = 1 - x + x^2 + ... [/tex] You can remember the denominator in the first equation is 1-x by multiplying both sides by (1-x), giving: [tex] 1 = (1-x)(1+x+x^2+...) = 1 + x + x^2 + ... - x - x^2 -x^3 - ... = 1[/tex] which is consistent, as opposed to what you'd get if you assume 1+x+... was 1/(1+x). (By the way, these manipulations of infinite sums aren't strictly valid, but they can be made more rigorous by restricting to finite sums and taking a limit at the end). |
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