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help finding certain generating functions using a general formula?

so basically I found this formula :

Attachment 22631

(sorry I tried using latex but I kept getting errors)

and I'm not quite sure if I understand how to use it, so if any of you could give me some examles of how it's used I would be much grateful.

I did use a different way to find the generating function for the sum starting at n=1.

but when I got the exact same thing for n=2, I wasn't sure anymore.

(I think it's 1 + x/(1-x)^2 ? -damn I'm having problems with this latex)

so if you could please show me how to use this formula for n=2 or for any n, it would be great.

p.s - it is possible that I'm also confused by the Leibniz notations since I'm used to Lagrange

thanks.

Re: help finding certain generating functions using a general formula?

Is this what you want to know?

$\displaystyle \begin{gathered} \sum\limits_{n = 0}^\infty {x^n } = \frac{1}{{1 - x}} \hfill \\ \sum\limits_{n = 0}^\infty {nx^{n - 1} } = \frac{1}{{\left( {1 - x} \right)^2 }} \hfill \\ \sum\limits_{n = 0}^\infty {nx^n } = \frac{x}{{\left( {1 - x} \right)^2 }} \hfill \\ \end{gathered}$

BTW: the code is

[TEX]\begin{gathered} \sum\limits_{n = 0}^\infty {x^n } = \frac{1}{{1 - x}} \hfill \\ \sum\limits_{n = 0}^\infty {nx^{n - 1} } = \frac{1}{{\left( {1 - x} \right)^2 }} \hfill \\ \sum\limits_{n = 0}^\infty {nx^n } = \frac{x}{{\left( {1 - x} \right)^2 }} \hfill \\ \end{gathered}[/TEX]

Re: help finding certain generating functions using a general formula?

Well, in class we were shown that $\displaystyle \sum\limits_{n = 1}^\infty {nx^n } = \frac{x}{{\left( {1 - x} \right)^2 }}$.

and you say that $\displaystyle \sum\limits_{n = 0}^\infty {nx^n } = \frac{x}{{\left( {1 - x} \right)^2 }}$.

so is true to say that $\displaystyle \sum\limits_{n \geq 0}^\infty {nx^n } = \frac{x}{{\left( {1 - x} \right)^2 }}$ ?

If so, just out of curiosity, how is it same for all $\displaystyle n \geq 0$?

and if it's not true, what would be the case for n = 2 ?

thanks a lot.

p.s - I was using MATH before (Doh)

Re: help finding certain generating functions using a general formula?

Quote:

Originally Posted by

**Laban** Well, in class we were shown that $\displaystyle \sum\limits_{n = 1}^\infty {nx^n } = \frac{x}{{\left( {1 - x} \right)^2 }}$.

and you say that $\displaystyle \sum\limits_{n = 0}^\infty {nx^n } = \frac{x}{{\left( {1 - x} \right)^2 }}$.

so is true to say that $\displaystyle \sum\limits_{n \geq 0}^\infty {nx^n } = \frac{x}{{\left( {1 - x} \right)^2 }}$ ?

If so, just out of curiosity, how is it same for all $\displaystyle n \geq 0$?

and if it's not true, what would be the case for n = 2 ?

Well $\displaystyle n\cdot x=0$ so $\displaystyle \sum\limits_{n = 0}^\infty {nx^n } = \sum\limits_{n = 1}^\infty {nx^n } \ne \sum\limits_{n \geqslant 2}^\infty {nx^n } $

Re: help finding certain generating functions using a general formula?

not sure I got you there...basically my specific problem is this :

$\displaystyle \sum\limits_{n = 2}^\infty {nx^n } = ?$

I'm probably missing something in your explanations, so if you could help me out with the above problem maybe I'll get the general case, or at least figure out where I'm stuck.

thanks

Re: help finding certain generating functions using a general formula?

Answer your own question:

$\displaystyle \sum\limits_{n = 1}^\infty {nx^n } = x + \sum\limits_{n = 2}^\infty {nx^n } = \frac{x}{{\left( {1 - x} \right)^2 }}$

Re: help finding certain generating functions using a general formula?

well...(Smirk) I guess it's $\displaystyle \sum\limits_{n = 2}^\infty {nx^n } = \frac{x}{{\left( {1 - x} \right)^2 }} - x$ then

Ok but I think I got it, it was the summation that confused me. just to make sure, I'll try it for n=3.

$\displaystyle \sum\limits_{n = 3}^\infty {nx^n } = \frac{x}{{\left( {1 - x} \right)^2 }} - x - 2x^2$

thanks a lot man, you're the best :)

Re: help finding certain generating functions using a general formula?

oh, I did forget to ask : how is the formula, I gave at first, includes the subtraction part?