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The following is from Orthogonal Polynomials of Several Variables by Charles F Dunkl and Yuan Xu 2nd edition , Encyclopedia of math..and applications 155 page 320. Here i am assuming (i,j) means the transposition of indices i and j or the 2 cycle of the symmetric group exchanging the 2 indices.


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Is it true that this lemma is not true for a general multi variable function f(x) of the symmetric group. If so can anyone prove it or disprove it. Can anyone show the lemma is true for some restricted class of functions such as polynomials or homogenous polynomials ? I have the guesstimate that it is not true in general for any function. If anyone is familiar or read the book i think he is mistaken where he says chapter 4 and it should be chapter 6 ?

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    $\begingroup$ You're going to need to keep the image but convert its contents into LaTeX for the blind. Also, please put in some more related definitions so we know for instance wth $\Pi^d$ is for example... $\endgroup$ Commented Oct 12 at 12:57
  • $\begingroup$ if i can figure out how to convert it to latex will try. Meanwhile if anyone knows how that could be done feel free to do so. $\endgroup$ Commented Oct 13 at 3:50
  • $\begingroup$ I don't know what anything is defined as :) $\endgroup$ Commented Oct 13 at 6:07
  • $\begingroup$ It seems to me that if it fails then (since only two indices are involved) it fails for $d=2$. So I'd check it on some easy cases like $f(x_1,x_2)=x_1^2+x_2$ to see what happens. But without a definition of $\Pi^d$ and a precise definition of $(i,j)f(x)$ it's not worth trying. $\endgroup$ Commented Oct 13 at 6:58
  • $\begingroup$ I now can't see why you think the lemma 10.2.1 is in doubt. If $f$ is any differentiable function of $x_1,\dots,x_n$ then I think it's a straightforward task to evaluate the LHS and get the RHS. $\endgroup$ Commented Oct 14 at 8:22

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It seems my guesstimate was wrong and it is in general true for f(x) a polynomial. Take simple example in only 2 dimensions. Let f(x)=$x_1^a x_2^b$where a,b are constants and by hand calculation or using computer algebra package such as maxima (as I did) or whatever one can show both (i) and (ii) true in this simple case. Because the scripts 1,2 arbitrary then by linearity we can assume it true for f(x) a polynomial in any number of variables.

Here is the code in used in maxima that i used if anyone is interested:

(kill(a,b,ff,k,x), ff:x[1]^ax[2]^b,tt:x[1](diff(ff,x[1])+k* (ff-subst(y[2]=x[2],subst(x[2]=x[1],subst(x[1]=y[2],ff))))/(x[1]-x[2])) ,disp(["tt"=tt,"ratdenom"=ratdenom(tt),"ratnum"=ratnum(tt)]), f1:x[1]ff ,t1:diff(f1,x[1])+k (f1-subst(y[2]=x[2],subst(x[2]=x[1],subst(x[1]=y[2],f1))))/(x[1]-x[2]) ,disp(["t1"=t1,"ratdenom"=ratdenom(t1),"ratnum"=ratnum(t1)]) ,te:tt-t1+ff+k*subst(y[2]=x[2],subst(x[2]=x[1],subst(x[1]=y[2],ff))) ,ldisp(["From the calculation if lemma 10.2.1 (i) were true in general then", ratsimp(te),"would = 0,and since it is 0 then (i) is in general true "])

,ff:x[1]^ax[2]^b,tt:x[2](diff(ff,x[1])+k* (ff-subst(y[2]=x[2],subst(x[2]=x[1],subst(x[1]=y[2],ff))))/(x[1]-x[2])) ,disp(["tt"=tt,"ratdenom"=ratdenom(tt),"ratnum"=ratnum(tt)]) ,f1:x[2]ff ,t1:diff(f1,x[1])+k (f1-subst(y[2]=x[2],subst(x[2]=x[1],subst(x[1]=y[2],f1))))/(x[1]-x[2]) ,disp(["t1"=t1,"ratdenom"=ratdenom(t1),"ratnum"=ratnum(t1)]),te:tt-t1-k*subst(y[2]=x[2],subst(x[2]=x[1],subst(x[1]=y[2],ff))) ,ldisp(["From the calculation if lemma 10.2.1 (ii) were true in general then", ratsimp(te),"would = 0,and since it is 0 then (ii) is in general true "]))
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  • $\begingroup$ All you need to prove the Lemmas is that $f$ is differentiable function of $x_i$. $\endgroup$ Commented Oct 16 at 6:37
  • $\begingroup$ @Daniel Donnelly The problem i have in converting it to latex is don't have any way of making the scripts he has on D and k and the partial derivative sign he uses. NOne of those have a latex equivalent except maybe the partial derivative sign maybe but not sure. The $\Pi^d$ means any polynomial with d variables say $x_i$ $0$<i<d+1. I think if one just does ctrl + a few times on the keyboard to enlarge it then it should be clear as it is written. $\endgroup$ Commented Oct 17 at 17:32
  • $\begingroup$ \$ \mathscr{D} \$ and \$ \kappa \$ $\endgroup$ Commented Oct 18 at 6:39
  • $\begingroup$ i wonder why $\mathscr{D}$ does not work on my emtex program but it does seem to work here $\endgroup$ Commented Oct 21 at 9:56

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