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Changing initial conditions in a recursive polynomial family definition. Then, finding the generation function for this family

Mathematica Asked on January 3, 2021

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
$$P_{-4}(c,z):=sum_{kgeq-4} P_{-4,k}(c)z^{k+4}=sum_{kgeq0}P_{-4,k-4}(c)z^k.$$

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!

One Answer

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)

Answered by Daniel Huber on January 3, 2021

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