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anonymous
 5 years ago
how do you get the power series of 2/((x1)^3)?
anonymous
 5 years ago
how do you get the power series of 2/((x1)^3)?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i know it involves calc.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can use the formula from geometric series, knowing that [1/(1x)] is equal to the sum from 0 to infinity of x^n, provided that the absolute value of your x is less than 1. Just reformulate your original function into the form of (1/1x) and substitute your values in to the summation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[2/(x1)^3=2/(1x)^3\] we know thatf(x)= \[1/(1x)=\sum_{0}^{\infty}x^n\] f'(x)=\[1/(1x)^2=\sum_{1}^{\infty}n(x)^{n1}\] f''(x)= \[2(1x)/(1x)^4=2/(1x)^3=\sum_{2}^{\infty}n(n1)x^{n2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[1/1x = \sum_{n=0}^{\infty} x ^{n}\] For example you can just take your 2/((x1)^3) and rearrange it into 2* 1/(x1)^3. Then you can pull out the 2 and work your power series expansion as 2 times the sum of the series for the function 1/(x1)^3. You can throw the 2 back in there when you're done just to make it pretty.
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