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anonymous

  • 5 years ago

how do you get the power series of 2/((x-1)^3)?

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  1. anonymous
    • 5 years ago
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    i know it involves calc.

  2. anonymous
    • 5 years ago
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    You can use the formula from geometric series, knowing that [1/(1-x)] 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 (1/1-x) and substitute your values in to the summation.

  3. anonymous
    • 5 years ago
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    \[2/(x-1)^3=-2/(1-x)^3\] we know thatf(x)= \[1/(1-x)=\sum_{0}^{\infty}x^n\] f'(x)=\[1/(1-x)^2=\sum_{1}^{\infty}n(x)^{n-1}\] f''(x)= \[-2(1-x)/(1-x)^4=-2/(1-x)^3=\sum_{2}^{\infty}n(n-1)x^{n-2}\]

  4. anonymous
    • 5 years ago
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    \[1/1-x = \sum_{n=0}^{\infty} x ^{n}\] For example you can just take your 2/((x-1)^3) and rearrange it into 2* 1/(x-1)^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/(x-1)^3. You can throw the 2 back in there when you're done just to make it pretty.

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