Timeline for Expressing power sum symmetric polynomials in terms of lower degree power sums
Current License: CC BY-SA 2.5
17 events
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| Oct 9, 2013 at 22:31 | answer | added | Igor Gornyi | timeline score: 0 | |
| Aug 22, 2013 at 21:29 | comment | added | Ian Agol | @ darij grinberg: it seems that there might be some $1/i!$'s (or something) missing from your formula? Compare to Akkerman's answer for computing $p_{N+1}$. | |
| Sep 13, 2010 at 13:07 | vote | accept | Peter Erskin | ||
| Aug 31, 2010 at 8:59 | answer | added | Akkerman | timeline score: 9 | |
| Aug 27, 2010 at 15:41 | history | edited | Peter Erskin | CC BY-SA 2.5 |
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| Aug 27, 2010 at 15:29 | history | edited | Peter Erskin | CC BY-SA 2.5 |
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| Aug 27, 2010 at 12:02 | history | edited | JBorger |
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| Aug 27, 2010 at 11:16 | answer | added | Gjergji Zaimi | timeline score: 7 | |
| Aug 27, 2010 at 10:59 | history | edited | Peter Erskin | CC BY-SA 2.5 |
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| Aug 27, 2010 at 10:34 | history | edited | Peter Erskin | CC BY-SA 2.5 |
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| Aug 27, 2010 at 9:42 | comment | added | Peter Erskin | I am not sure that the formula for the Schur functions which I quoted from Wikipedia (see my question) includes only $p_j$ with $j\le N$ when the degree of the Schur function is higher than $N$. Apparently, in this case one has $p_k$ with $k > N$ on the r.h.s. However, the corresponding coefficients might be identically zero --- I did not check. If this is not the case, it is interesting to learn what is the expansion of the Schur function $s_\lambda$ of degree higher than $N$ in terms of only the first $N$ power sums $p_1,...,p_N$. | |
| Aug 27, 2010 at 9:33 | history | edited | Peter Erskin | CC BY-SA 2.5 |
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| Aug 27, 2010 at 9:00 | comment | added | Peter Erskin | Thanks! If I understand correctly, this yields a nice relation between the power sum polynomials $p_1,\ldots,p_k$ for $k > N$. However, as you mentioned, this is not an explicit formula. Indeed, in order to expand $$p_k(x_1,\ldots,x_N)$$ in terms of only $N$ power sums $$p_1(x_1,\ldots,x_N),\ldots,p_N(x_1,\ldots,x_N)$$ that form a complete basis for symmetric polynomials in $N$ variables, one should further apply your formula iteratively for each of the lower degree power sums $p_{k-1},\ p_{k-2},\ldots, p_{N+1}$ that appear in your formula for $p_k$. | |
| Aug 26, 2010 at 16:41 | answer | added | Robin Chapman | timeline score: 3 | |
| Aug 26, 2010 at 15:46 | comment | added | darij grinberg | Actually, you can write it as $\sum\limits_{i=0}^k\left(-1\right)^i\sum\limits_{j_1,\ j_2,\ ...,\ j_i\geq 1;\ j_1+j_2+...+j_i=n}\dfrac{1}{j_1j_2...j_i}p_{j_1}p_{j_2}...p_{j_i}=0$ for $k>N$ unless I have made a mistake. | |
| Aug 26, 2010 at 15:40 | comment | added | darij grinberg | Not an explicit formula, but the Newton identity $\sum\limits_{n=1}^{\infty}p_nT^n=T\frac{d}{dT}\log\left(\prod\limits_{i=1}^N\dfrac{1}{1-x_iT}\right)$ (this is an identity of formal power series, where $\log$ is the natural logarithm, and $x_1$, $x_2$, ..., $x_N$ are our variables) yields $\exp\left(-\sum\limits_{n=1}^{\infty}\frac{1}{n}p_nT^n\right)=\prod\limits_{i=1}^N\left(1-x_iT\right)$, which is a polynomial of degree $N$, so that all coefficients before $T^{N+1}$, $T^{N+2}$, ... are zero - and this yields exactly the formulas you want. | |
| Aug 26, 2010 at 15:11 | history | asked | Peter Erskin | CC BY-SA 2.5 |