2
$\begingroup$

enter image description hereI want to know if there is a package in Mathematica that can give back some recursion given a list of polynomial in $f_{n}(h)$. For example I know there exists package RISC guess that where I give some list of rational numbers it can give me the recursion it satisfies. My list is as follows [1,0,-2,$-h^2+3$,$5h^2+1$,$-5h^2-11$,$-35h^2+15$,$-21h^4+140h^2+13$,$189h^4-84h^2-77$] I want a recursion of the from $$ \sum_{k=0}^{N}a_{k}(h)[n]f(n+k) =0 $$

where $a_{k}(h)[n]$ is a polynomial in $n$ over the rational field $Q(h)$.

An example done in Guess package is as follows enter image description here

[{1,1},{2,0},{3,-2}, {4,-h^2+3},{5,5*h^2+1},{5,5*h^2+1},{6,-5*h^2-11},{7,-35*h^2+15},{8,-21*h^4+140*h^2+13},{9,189*h^4-84*h^2-77},{10,-462*h^4-840*h^2+86},{11,-1848*h^4+2640*h^2+144},{12,-1485*h^6+15015*h^4-495*h^2-595},{13,19305*h^6-27027*h^4-16445*h^2+495},{14,-78507*h^6-102102*h^4+41041*h^2+1520},{15,-173745*h^6+636636*h^4+11375*h^2-4810},{16,-225225*h^8+2922920*h^6-816816*h^4-282100*h^2+2485},{17,3828825*h^8-9432280*h^6-4084080*h^4+559300*h^2+15675},{18,-22141548*h^8-15801500*h^6+19399380*h^4+474300*h^2-39560},{19,-15796638*h^8+218196836*h^6-15872220*h^4-4399260*h^2+6290},{20,-59520825*h^10+897498459*h^8-554268000*h^6-128741340*h^4+6807225*h^2+159105},{21,1249937325*h^10-4414328919*h^8-1127011600*h^6+476030940*h^4+11062275*h^2-324805},{22,-9445293585*h^10-903687785*h^8+10988363100*h^6-171609900*h^4-63677075*h^2-87075},{23,6418990305*h^10+101767530865*h^8-21565644100*h^6-3397876020*h^4+73363675*h^2+1592843}]

Improve this question
$\endgroup$
16
  • $\begingroup$ It would be useful to have an example that can be cut and pasted. $\endgroup$
    – Daniel Lichtblau
    Commented Apr 17, 2018 at 14:44
  • $\begingroup$ I copy paste an example if it helps. It from RISC GUESS pacakge where they can do it for rational numbers but for me it's polynomials. $\endgroup$
    – GGT
    Commented Apr 17, 2018 at 23:27
  • 1
    $\begingroup$ I cannot cut and paste an image... $\endgroup$
    – Daniel Lichtblau
    Commented Apr 18, 2018 at 13:47
  • $\begingroup$ I see so which format is good for you ? I can send you a pdf file ? $\endgroup$
    – GGT
    Commented Apr 18, 2018 at 23:39
  • $\begingroup$ Just put plain ascii text, in Mathematica input form, into the actual question. So one "row" could be the Mathematica list {10,-462*h^4-840*h^2+86} for example. $\endgroup$
    – Daniel Lichtblau
    Commented Apr 19, 2018 at 14:15

1 Answer 1

Reset to default
1
$\begingroup$ 
pollist = {{1, 1}, {2, 0}, {3, -2}, {4, -h^2 + 3}, {5, 5*h^2 + 1}, {5,
     5*h^2 + 1}, {6, -5*h^2 - 11}, {7, -35*h^2 + 15}, {8, -21*h^4 + 
     140*h^2 + 13}, {9, 
    189*h^4 - 84*h^2 - 77}, {10, -462*h^4 - 840*h^2 + 
     86}, {11, -1848*h^4 + 2640*h^2 + 144}, {12, -1485*h^6 + 
     15015*h^4 - 495*h^2 - 595}, {13, 
    19305*h^6 - 27027*h^4 - 16445*h^2 + 495}, {14, -78507*h^6 - 
     102102*h^4 + 41041*h^2 + 1520}, {15, -173745*h^6 + 636636*h^4 + 
     11375*h^2 - 4810}, {16, -225225*h^8 + 2922920*h^6 - 816816*h^4 - 
     282100*h^2 + 2485}, {17, 
    3828825*h^8 - 9432280*h^6 - 4084080*h^4 + 559300*h^2 + 
     15675}, {18, -22141548*h^8 - 15801500*h^6 + 19399380*h^4 + 
     474300*h^2 - 39560}, {19, -15796638*h^8 + 218196836*h^6 - 
     15872220*h^4 - 4399260*h^2 + 6290}, {20, -59520825*h^10 + 
     897498459*h^8 - 554268000*h^6 - 128741340*h^4 + 6807225*h^2 + 
     159105}, {21, 
    1249937325*h^10 - 4414328919*h^8 - 1127011600*h^6 + 
     476030940*h^4 + 11062275*h^2 - 324805}, {22, -9445293585*h^10 - 
     903687785*h^8 + 10988363100*h^6 - 171609900*h^4 - 63677075*h^2 - 
     87075}, {23, 
    6418990305*h^10 + 101767530865*h^8 - 21565644100*h^6 - 
     3397876020*h^4 + 73363675*h^2 + 1592843}}[[All, 2]];

For free coefficients

coef0[n_] = FindSequenceFunction[Coefficient[#, h, 0] & /@ pollist, n]

gives

DifferenceRoot[{[FormalY],[FormalN]}[Function]{([FormalN]+4) (2936 [FormalN]^2-13255 [FormalN]+13103) [FormalY]([FormalN]+3)+(11818 [FormalN]^3-39180 [FormalN]^2+12312 [FormalN]+47530) [FormalY]([FormalN]+1)+(4441 [FormalN]^3-7769 [FormalN]^2-21002 [FormalN]+47530) [FormalY]([FormalN]+2)-5 ([FormalN]-5) [FormalN] (1431 [FormalN]-1199) [FormalY]([FormalN])==0,[FormalY](1)==1,[FormalY](2)==0,[FormalY](3)==-2}][n]

Check:

Table[coef0[k], {k, 1, 24}]

{1,0,-2,3,1,1,-11,15,13,-77,86,144,-595,495,1520,-4810,2485,15675,-39560,6290,159105,-324805,-87075,1592843}

For the coefficients at $h^2$

coef2[n_] = FindSequenceFunction[Coefficient[#, h, 2] & /@ pollist, n]

DifferenceRoot[{[FormalY],[FormalN]}[Function]{(550 [FormalN]^3-2640 [FormalN]^2+1220 [FormalN]+1110) [FormalY]([FormalN]+1)+(237 [FormalN]^3-1527 [FormalN]^2+2442 [FormalN]-672) [FormalY]([FormalN]+2)+25 ([FormalN]-5) [FormalN] (17 [FormalN]-5) [FormalY]([FormalN])+([FormalN]-2) ([FormalN]-1) (76 [FormalN]-297) [FormalY]([FormalN]+3)==0,[FormalY](1)==0,[FormalY](2)==0,[FormalY](3)==0,[FormalY](4)==-1,[FormalY](5)==5}][n]

etc.

BTW recurrences becomes simpler if to begin from the third line:

coef0[n_] = FindSequenceFunction[Coefficient[#, h, 0] & /@ pollist[[3 ;;]], n]

DifferenceRoot[{[FormalY],[FormalN]}[Function]{(5 [FormalN]+10) [FormalY]([FormalN])+(2 [FormalN]+7) [FormalY]([FormalN]+1)+([FormalN]+5) [FormalY]([FormalN]+2)==0,[FormalY](1)==-2,[FormalY](2)==3,[FormalY](3)==1,[FormalY](4)==1,[FormalY](5)==-11}][n]

Improve this answer
$\endgroup$
5
  • $\begingroup$ Thanks for that. As I understand you gave the recursion for the constant term of the polynomials. What is Formal[Y], Formal[N] denotes? Is it possible to give a recursion for a bivariate table? The entry of the table would be $(d,coeff[f_d,h,2g])$. $f_d$ denote the $d-th$ polynomial in the list. $\endgroup$
    – GGT
    Commented Apr 21, 2018 at 0:01
  • $\begingroup$ Yes, for the coefficients of $h^2$ too, changed the answer. As for Formal[Y] , Formal[N] it's mathematica's symbols introduced to not use y, n because those variables can be defined already. $\endgroup$
    – Andrew
    Commented Apr 21, 2018 at 5:29
  • $\begingroup$ you mean you have a different recursion for coefficients of $h^2$ ? $\endgroup$
    – GGT
    Commented Apr 21, 2018 at 7:18
  • $\begingroup$ To make sure the above recursion for the constant coefficient read as follows $ (5n+10)*y(n)+(2n+7)*y(n+1)+(2n+5)*y(n+2)=0$. $\endgroup$
    – GGT
    Commented Apr 21, 2018 at 7:34
  • $\begingroup$ @GGT yes, for different coefficients recursions are different. $\endgroup$
    – Andrew
    Commented Apr 21, 2018 at 7:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.