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Find an expression for $S_2$ in terms of $S_0$ and $S_1$. Do not need to simplify. If someone could explain this, that would be a real life saver.

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$S_2=x\,\frac{\partial}{\partial x}S_1$ –  alex.jordan Dec 5 '13 at 7:15
    
Vandermonde says these three things are linearly independent for $n\geq2$, so you aren't going to find a linear relation between them. If you allow a derivative operator, you have my comment above. Otherwise you need to allow nonlinear relations as in Andres' answer below. –  alex.jordan Dec 5 '13 at 7:19

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I am not sure what kind of expression you are after. Here is a suggestion:

Note that $xS_2=\sum_{k=1}^nk^2x^{k+1}=\sum_{k=2}^{n+1}(k-1)^2x^k$, so $$(1-x)S_2=S_2-xS_2=x-n^2x^{n+1}+\sum_{k=2}^n(k^2-(k-1)^2)x^k, $$ or $$ (1-x)S_2=x-n^2x^{n+1}+\sum_{k=2}^n(2k-1)x^k=x-n^2x^{n+1}+2(S_1-x)-(S_0-1-x), $$ that is,
$$ (1-x)S_2=x-n^2x^{n+1}+2S_1-2x-S_0+1+x $$ or, finally, $$ (1-x)S_2=1-n^2x^{n+1}+2S_1-S_0. $$ Is this the kind of relation you are seeking?

Another typical approach, but more advanced, uses calculus: $S_1'=\sum_{k=1}^n k^2x^{k-1}$, so $$ S_2=xS_1'. $$ A similar relation between $S_1$ and $S_0'$, and the realization that $$S_0=\frac{1-x^{n+1}}{1-x} $$ allows us to find explicit formulas for $S_1,S_2$ in terms of $x$, when $x\ne 1$. (When $x=1$, something different is needed, but it can be proved that the first expression is $n+1$, the second is $\displaystyle \frac {n(n+1)}2$, and the last is $\displaystyle \frac{n(n+1)(2n+1)}6$.) But I suspect this is not the approach you seek.

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There is an example in the book before that gives $$S_0=\frac{1-x^{n}}{1-x} $$. Then In the exorcise that follows we are supposed to do $S_1=\sum_{k=1}^n kx^{k}$ using the same simplifying process. I get stuck at $$ (1-x)S_1=x-(n+1)x^{n+1} $$ which im not sure how to continue or is correct. Then we are supposed to find for $$S_2 $$ then finally make it in terms as per the orignal question. That may or may not make it clearer –  Neal Dec 5 '13 at 7:32
    
@Neal You have a small typo, it should be $S_0-xS_0=1-x^{n+1}$. –  Andres Caicedo Dec 5 '13 at 7:41
    
Ignore that comment the example in the book for $$S_0 = \sum_{k=0}^{n-1} x^{k}$$ –  Neal Dec 5 '13 at 7:42

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