Skip to main content
MathOverflow

Return to Answer

revived stale link to the Mathematica package
Source Link
მამუკა ჯიბლაძე
მამუკა ჯიბლაძე
  • 18.4k
  • 3
  • 85
  • 175

Since you are using Mathematica, you definitely want to take a look at the extremely useful package HolonomicFunctionsHolonomicFunctions by Christoph Koutschan.

In your particular example,

Annihilator[f1[d], {S[d]}]

shows that this function satisfies the recurrence \begin{equation} (4+4d)f_1(d+4)+(4-7d)f_1(d+2)-(6-3d)f_1(d)=0. \end{equation}

Once known, Mathematica itself can check symbolically that both of your functions satisfy this recurrence:

(4+4d)f1[d+4] + (4-7d)f1[d+2] - (6-3d)f1[d] // FullSimplify
(4+4d)f2[d+4] + (4-7d)f2[d+2] - (6-3d)f2[d] // FullSimplify

After checking initial conditions (which Mathematica can do) it follows that $f_1(d)=f_2(d)$ for all even integers $d$

But as I'm typing I see that the OP just figured all of this out by himself... ;) So let me just mention that one strategy now could be to look at $f_1-f_2$, show that it satisfies the necessary exponential growth conditions (should be alright after combining the poles; apart from these each function seems to be good by itself), and apply Carlson's Theorem. I hope that helps...

Since you are using Mathematica, you definitely want to take a look at the extremely useful package HolonomicFunctions by Christoph Koutschan.

In your particular example,

Annihilator[f1[d], {S[d]}]

shows that this function satisfies the recurrence \begin{equation} (4+4d)f_1(d+4)+(4-7d)f_1(d+2)-(6-3d)f_1(d)=0. \end{equation}

Once known, Mathematica itself can check symbolically that both of your functions satisfy this recurrence:

(4+4d)f1[d+4] + (4-7d)f1[d+2] - (6-3d)f1[d] // FullSimplify
(4+4d)f2[d+4] + (4-7d)f2[d+2] - (6-3d)f2[d] // FullSimplify

After checking initial conditions (which Mathematica can do) it follows that $f_1(d)=f_2(d)$ for all even integers $d$

But as I'm typing I see that the OP just figured all of this out by himself... ;) So let me just mention that one strategy now could be to look at $f_1-f_2$, show that it satisfies the necessary exponential growth conditions (should be alright after combining the poles; apart from these each function seems to be good by itself), and apply Carlson's Theorem. I hope that helps...

Since you are using Mathematica, you definitely want to take a look at the extremely useful package HolonomicFunctions by Christoph Koutschan.

In your particular example,

Annihilator[f1[d], {S[d]}]

shows that this function satisfies the recurrence \begin{equation} (4+4d)f_1(d+4)+(4-7d)f_1(d+2)-(6-3d)f_1(d)=0. \end{equation}

Once known, Mathematica itself can check symbolically that both of your functions satisfy this recurrence:

(4+4d)f1[d+4] + (4-7d)f1[d+2] - (6-3d)f1[d] // FullSimplify
(4+4d)f2[d+4] + (4-7d)f2[d+2] - (6-3d)f2[d] // FullSimplify

After checking initial conditions (which Mathematica can do) it follows that $f_1(d)=f_2(d)$ for all even integers $d$

But as I'm typing I see that the OP just figured all of this out by himself... ;) So let me just mention that one strategy now could be to look at $f_1-f_2$, show that it satisfies the necessary exponential growth conditions (should be alright after combining the poles; apart from these each function seems to be good by itself), and apply Carlson's Theorem. I hope that helps...

Source Link
Armin Straub
Armin Straub
  • 1.4k
  • 2
  • 18
  • 24

Since you are using Mathematica, you definitely want to take a look at the extremely useful package HolonomicFunctions by Christoph Koutschan.

In your particular example,

Annihilator[f1[d], {S[d]}]

shows that this function satisfies the recurrence \begin{equation} (4+4d)f_1(d+4)+(4-7d)f_1(d+2)-(6-3d)f_1(d)=0. \end{equation}

Once known, Mathematica itself can check symbolically that both of your functions satisfy this recurrence:

(4+4d)f1[d+4] + (4-7d)f1[d+2] - (6-3d)f1[d] // FullSimplify
(4+4d)f2[d+4] + (4-7d)f2[d+2] - (6-3d)f2[d] // FullSimplify

After checking initial conditions (which Mathematica can do) it follows that $f_1(d)=f_2(d)$ for all even integers $d$

But as I'm typing I see that the OP just figured all of this out by himself... ;) So let me just mention that one strategy now could be to look at $f_1-f_2$, show that it satisfies the necessary exponential growth conditions (should be alright after combining the poles; apart from these each function seems to be good by itself), and apply Carlson's Theorem. I hope that helps...