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I'm following a paper trying to find a way to repeat the done computations using Mathematica. I'm beginner using Mathematica and I already read that:

  • I could define functions by recursion;
  • I could ask the software to find a generating function for a given family;
  • The Wolfram application can be used to change the initial conditions in a given family and ask again for the generating function;

Let the following $P_k(c)$ be a polynomial in $c$ satisfying the following recursion relation $$(6+2k)P_k(c)=4kcP_{k-2}(c)-2(k-3)P_{k-4}(c).$$

If we take initial conditions $P_{-3}(c)=P_{-2}(c)=P_{-1}(c)=0$ and $P_{-4}(c)=1$ then we arrive at a generating function $$P_{-4}(c,z):=\sum_{k\geq-4} P_{-4,k}(c)z^{k+4}=\sum_{k\geq0}P_{-4,k-4}(c)z^k.$$

How could I repeat this using Mathematica? Is there any easy way to change this initial conditions using the software (and then find the generating function)?

Thank you so much!

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  • $\begingroup$ Show us the code you’ve tried so far $\endgroup$ – MarcoB Oct 12 at 3:25
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First a correction. The coefficient of z^n in the generating function should be p[n,c] and not p[n-4,c]. With this in mind, we may define (for simplicity I only sum up to 6. This is good enough here to make the point. In a real application you would sum up to Infinity and search for a closed form):

p[-3, c] = p[-2, c] = p[-1, c] = 0;
p[-4, c] = 1;
p[n_, c_] := 1/(6 + 2 n) (4 n c p[n - 2, c] - 2 (n - 3) p[n - 4, c])
genf[c_, z_] := Sum[p[i, c] z^i, {i, 0, 6}]

When we now take the m'th derivative of genf and set z->0, we should get p[m,c].

Table[Print@ StringForm["p[,c] == , D[genf[c,z],{z,1}]/.z->0 == ``", i, p[i, c], D[genf[c, z], {z, i}] /. z -> 0], {i, 0, 4}]

p[0,c] == 1 , D[genf[c,z],{z,0}]/.z->0 == 1

p[1,c] == 0 , D[genf[c,z],{z,1}]/.z->0 == 0

p[2,c] == (4 c)/5 , D[genf[c,z],{z,2}]/.z->0 == (4 c)/5

p[3,c] == 0 , D[genf[c,z],{z,3}]/.z->0 == 0

p[4,c] == 1/14 (-2+(64 c^2)/5) , D[genf[c,z],{z,4}]/.z->0 == 1/14 (-2+(64 c^2)/5)

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  • $\begingroup$ I am not sure why you say that the coefficient of $z^n$ should be $P_{-4,k}$. Could you explain? Second, I understand that using this algorithm you describe here, we'll find each $P_{k}(c)$, but, for me, it is note clear how I could use it to find the generating function. For this example, I know that the generating function, in terms of an elliptic integral is $$ z\sqrt{1-2cz^2+z^4}\int \frac{4cz^2-1}{z^2(z^4-2cz^2+1)^{3/2}}dz. $$ Is there any way to make Mathematica help me on this computation? Last observation is that, maybe you need to edit part of given answer to the code format. $\endgroup$ – Felipe Oct 12 at 14:56
  • $\begingroup$ Well, to get the full blown generating function, you would have to evaluate: genf[c_, z_] = Sum[p[i, c] z^i, {i, 0, Infinity}]. But this makes only sense if there exists a closed form for the sum. $\endgroup$ – Daniel Huber Oct 12 at 15:14

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