- #1

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I know the representation for 1/1-x is x^n so does that mean x/(1-x)^2 is x^n^2? could use some clarification please

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- Thread starter chyeaman
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- #1

- 2

- 0

I know the representation for 1/1-x is x^n so does that mean x/(1-x)^2 is x^n^2? could use some clarification please

- #2

- 135

- 0

[tex]\sum_{n=0}^{n=\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n[/tex]

Looks like for [itex]\frac{x}{(1-x)^2}[/itex] this it will be something like

[tex]x+2x^2+3x^3+4x^4+\ldots+[/tex]

if you center it on 0.

- #3

- 534

- 1

[tex]\frac{1}{1 - x} = \sum_{n = 0}^{\infty} x^n = 1 + x + x^2 + x^3 + x^4 + \dotsb.[/tex]

Take the derivative of both sides to obtain

[tex]\frac{1}{(1 - x)^2} = \sum_{n = 0}^{\infty} (n - 1) x^n = 1 + 2x + 3x^2 + 4x^3 + \dotsb,[/tex]

and multiply by x to get

[tex]\frac{x}{(1 - x)^2} = \sum_{n = 1}^{\infty} n x^n = x + 2x^2 + 3x^3 + 4x^4 + \dotsb.[/tex]

- #4

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Thank you so much for clarifying!

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