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I would like to determine a sequence f[n]. I have obtained the first few terms:

fun[0]=(2^(2 + 2/a) π^(3/2) Gamma[1/2 + 1/a])/Gamma[1 + 1/a];
fun[1]=(2^(3 + 2/a) π^(3/2) Gamma[1/2 + 1/a])/Gamma[2 + 1/a];
fun[2]=(2^(4 + 2/a) (1 + a^2 (-1 + 3 a)) π^(3/2)Gamma[1/2 + 1/a])/(a^2 (1 + a) Gamma[3 + 1/a]);
fun[3]=(3 2^(6 + 2/a) (3 + a (3 + a (-4 + a (2 + 5 a)))) π^(3/2)Gamma[3/2 + 1/a])/(a (2 + a)^3 Gamma[4 + 1/a]);
fun[4]=(3 2^(7 + 2/a) (3 + a (6 + a (57 + a (132 + a (24 + a (-66 + a (59 + 5 a (22 + 7 a)))))))) π^(3/2)Gamma[3/2 + 1/a])/(a^3 (2 + a)^3 (3 + a)^2 Gamma[5 + 1/a]);
fun[5]=(15 2^(7 + 2/a) (15 + a (30 + a (105 + a (240 + a (52 + a (-146 + a (59 + 7 a (26 + 9 a)))))))) π^(3/2) Gamma[1/2 + 1/a])/(a^4 (3 + a)^2 (4 + a)^2 Gamma[6 + 1/a]);

Now I wonder if it is possible to extrapolate the above and obtain a closed form expression for f[n] using Mathematica? I can generate more terms for even higher n if needed. Thanks for any suggestion.

EDIT:

I was asked in the comments to mention where the sequence of interest arises. The following is the background:

In eq. (10.35) of his book "Symmetric functions and Hall polynomials" I.G.Macdonald gives the following scalar product, under which Jack polynomials with different partitions $\mu\neq\lambda$ are orthogonal

$$\langle J^a_\lambda(z_1,z_2),J^a_\mu(z_1,z_2)\rangle'_2=\frac{1}{2}\int_T J^a_\lambda(z_1,z_2)\overline{J^a_\mu(z_1,z_2)}\prod_{i\neq j}\left(1-\frac{z_i}{z_j}\right)^{1/a}dz^2$$

where the integration contour is $T=\{(z_1,z_2)\in\mathbb{C}^2:|z_1|=1,|z_2|=1\}$. Therefore, the integral equals $c_{\lambda,a}\delta_{\mu,\lambda}$, where $\delta_{\mu,\lambda}=\left\{{1~\text{if}~\lambda=\mu}\atop{0~\text{if}~\lambda\neq\mu}\right.$ is the Kronecker delta. However, Macdonald does not give the normalization $c_{\lambda,a}$ for the scalar product. I would like to figure out what the normalization $c_{\mu,a}$ is in the case of Jack polynomials with two variables. Using eq. (10.15) of this paper we have an explicit expression for the polynomials. Then one can observe that the $\mu$ dependance in $c_{\mu,a}$ is just the difference of the two entries in the partition $n=\mu_1-\mu_2$. So basically, what we see above are explicit results for fun[n]=$2c_{n,a}$.

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    $\begingroup$ You might have better luck if you factor out the gamma functions and then incorporate them later. $\endgroup$ – J. M. is away Aug 10 '16 at 16:19
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    $\begingroup$ You could try FindSequenceFunction $\endgroup$ – JungHwan Min Aug 10 '16 at 16:28
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    $\begingroup$ obviously start by dividing by Gamma[1/2+1/a]/Gamma[n+1+1/a] ( use FullSimplify on that result. ). Is the procedure used to obtain the series useful? $\endgroup$ – george2079 Aug 10 '16 at 22:05
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    $\begingroup$ Occasionally for me searching the Online Encyclopedia of Integer Sequences has yielded useful insight. e.g. oeis.org/search?q=15+30+105+240+52+146+59 $\endgroup$ – Mr.Wizard Aug 11 '16 at 15:13
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    $\begingroup$ Could you explain where these terms came from? $\endgroup$ – Chip Hurst Aug 11 '16 at 15:36
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I dont have a solution, but maybe it is useful to show a way to approach this problem:

first turn your expressions into a list so we can operate on it as a list:

s0 = Table[fun[i], {i, 0, 5}];

divide by some obvious common factors (should have included 4 here..):

 s1 = s0/(\[Pi]^(3/2) Gamma[1/2 + 1/a] 2^(2/a))

enter image description here

now use MapIndexed to factor some indexed quantites:

 s2 = MapIndexed[ # Gamma[First@#2 + 1/a] & , s1]

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FullSimplify clears the Gamma 's , nice..

 s3 = FullSimplify@s2

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the first two terms are now independent of a , so maybe group terms..

s4 = Total /@ Partition[s3, 2] // FullSimplify

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I'm kind of stuck here.. ( a few more terms would help ), but basically whittle the list down by inspection to something that FindSequenceFunction can handle.

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  • $\begingroup$ Thank you for your help! I don't quite understand what happens in the last step. You add the expressions somehow? How does this help to find the sequence? $\endgroup$ – Kagaratsch Aug 11 '16 at 14:24
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    $\begingroup$ the last step is a bit of a shot in the dark, maybe if you find a simple expression for f[2 n ]+f[2 n+1], n=0,1.. it would be helpful. Or maybe its useless, so go back and look at s3 $\endgroup$ – george2079 Aug 11 '16 at 14:52
  • $\begingroup$ Oh, ok, now I see. $\endgroup$ – Kagaratsch Aug 11 '16 at 21:37

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