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I do have somewhat of a reasoning for this.

$$S = 1 + x + x^{2} + x^{3} +.. + x^{n} $$

Denoting the derivative of $S$ as $T$

$$T = 1 + 2x + 3x^{2} + 4x^{3} +... + nx^{n-1}$$

$$xT = x + 2x^{2} + 3x^{3} +... + nx^{n}$$

Writing the derivative of this series as $V$

$$V = 1 + 2^{2}x + 3^{2}x^{2} +... + n^{2}x^{n-1}$$

$$xV = x + 2^{2}x^{2} + 3^{2}x^{2} +... + n^{2}x^{n}$$

Which is ${\sum{n^{2}x^{n}}}$

However, I am unable to start from $\sum{n^{2}}$and finish where I started. Can someone give me systematic steps on how to manipulate ${\sum{n^{2}x^{n}}}$into a generating function?

The answers are a little different from what I expected.

Let me explain with $\sum nx^{n}$. This function is not the series of any known function but with one slight manipulation, we can change it.

$$\sum x\frac {d x^{n}}{dx}$$ $$=x \frac{d}{dx} {\sum x^{n}} = x\frac {d}{dx} \frac{1}{1-x}$$ $$=\frac {x}{ (1-x)^2 }$$

I wanted a similar trick for evaluating $\sum n^{2}x^{n}$ $$=\sum nx \frac{d}{dx} x^{n}$$ And, then what ? How can I finish this ? I know how to start at $S$ and finish here, but I'm having trouble doing the algebra of starting with this sum and finishing where I started.

Saikat
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  • Start with $\sum_1^nk^2x^k$. Multiply by $1-x$. Multiply by $1-x$ again. – Gerry Myerson Feb 19 '16 at 11:46
  • I wanted to write sigma of $ (n^2x^n) $ but something else came up. – Saikat Feb 19 '16 at 11:48
  • Divide by $x$, then integrate (twice) instead of differentiating and multiplying by $x$ (twice)? – Steve Kass Feb 19 '16 at 17:35
  • You should work with the function $$f(t) = \sum_{k=0}^{n}\exp(kt) = \frac{1-\exp((n+1)t)}{1-\exp(t)}$$ instead – Count Iblis Feb 19 '16 at 17:37
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    Why not use the direct way? You are after $U=\sum\limits_nn^2x^n$, you know that $U=x(xS')'$ and you (should) know that $S=(1-x)^{-1}$. Hence $S'=(1-x)^{-2}$, $xS'=x(1-x)^{-2}$, $(xS')'=(1+x)(1-x)^{-3}$, and finally $U=x(1+x)(1-x)^{-3}$. – Did Feb 19 '16 at 18:22
  • Any thoughts on the answer and the comments, 230452? – Gerry Myerson Feb 20 '16 at 23:49
  • Are you still here? – Gerry Myerson Feb 22 '16 at 04:27
  • @GerryMyerson I updated my question to clarify what I wanted. – Saikat Feb 22 '16 at 12:53
  • I think what you're after is what's in the comment from @Did, but I'll post it as an answer. – Gerry Myerson Feb 22 '16 at 21:45
  • I am curious to know: what was unclear in my comment (since, apparently, @Gerry's answer made you see the light while my comment did not)? – Did Feb 23 '16 at 06:27
  • @Did xD. I come on this site through iPad app and the text in comments displays as it is written, and is not rendered MathJax. So, it didn't make sense to me. Anyway, don't be sad. I up voted your comment too :) – Saikat Feb 23 '16 at 10:06
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    Not sad, thanks, but if the situation is the one you describe, it seems mandatory to append a sign to each question you ask, explicitely recommending to not post comments. You know, by courtesy, so that people do not lose their time... – Did Feb 23 '16 at 11:01

3 Answers3

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Following the comment from Did, we go $$\sum_nn^2x^n=\sum_nx{d\over dx}x{d\over dx}x^n=x{d\over dx}x{d\over dx}\sum_nx^n$$ and you know what to do from there.

Gerry Myerson
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If you have:

$\begin{align} A(z) = \sum_{n \ge 0} a_n z^n \end{align}$

then it is easy to see that:

$\begin{align} z \frac{\mathrm{d}}{\mathrm{d} z} A(z) = \sum_{n \ge 0} n a_n z^n \end{align}$

To multiply each coefficient by $n^r$, apply the above $r$ times.

Furthermore, it is easy to check that:

$\begin{align} \frac{A(z)}{1 - z} \sum_{n \ge 0} \left(\sum_{0 \le k \le n} a_n \right) z^n \end{align}$

In your case:

$\begin{align} \frac{1}{1 - z} &= \sum_{n \ge 0} z^n \\ z \frac{\mathrm{d}}{\mathrm{d} z} \left( z \frac{\mathrm{d}}{\mathrm{d} z} \frac{1}{1 - z} \right) &= z \frac{\mathrm{d}}{\mathrm{d} z} \frac{z}{(1 - z)^2} \\ &= \frac{z (1 + z)}{(1 - z)^3} \end{align}$

It is easy to extract the coefficient of $z^n$ of this divided by $1 - z$:

$\begin{align} [z^n] \frac{z + z^2}{(1 - z)^4} &= [z^n] (z + z^2) \sum_{k \ge 0} (-1)^k \binom{-4}{k} z^k \\ &= ([z^{n - 1}] + [z^{n - 2}]) \sum_{k \ge 0} \binom{k + 4 - 1}{4 - 1} z^k \\ &= \binom{n - 1 + 3}{3} + \binom{n - 2 + 3}{3} \\ &= \binom{n + 2}{3} + \binom{n + 1}{3} \\ &= \frac{(n + 2) (n + 1) n}{3!} + \frac{(n + 1) n (n - 1)}{3!} \\ &= \frac{(n + 1) n (2 n + 1)}{6} \end{align}$

Another way, as you state, is to start with:

$\begin{align} 1 + z + \dotsb + z^n = \frac{1 - z^{n + 1}}{1 - z} \end{align}$

mangle it twice as above, and evaluate for $z = 1$. That requires using l'Hôpital to compute a limit, the resulting expression isn't defined at $z = 1$.

vonbrand
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Let $a_n=n^2$. You want to find the generating function of $a_n$.

You have $$a_{n+1}+a_{n-1}=2a_n+2$$

Therefore $$a_{n+1}x^{n+1}+x^2 (a_{n-1}x^{n-1})=2x (a_n x^n)+2x (x^n)$$

Now add these relations by $n$ to get an equation in your generating function.

N. S.
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