How to prove a numerical identity?












3












$begingroup$


Let be a function power function



f[s_, r_] := Piecewise[{{s^r, s >= 0}, {0, True}}];


And its convolution



Conv[s_, r_] := Sum[f[k, r]*f[s - k, r], {k, -Infinity, +Infinity}];


Proposition 1. Let be a real coefficients $A_{m,j}$ defined as follows



A[n_, k_] := 0

A[n_, k_] := (2 k + 1)*Binomial[2 k, k]*

   Sum[A[n, j]*Binomial[j, 2 k + 1]*(-1)^(j - 1)/(j - k)*

     BernoulliB[2 j - 2 k], {j, 2 k + 1, n}] /; 2 k + 1 <= n

A[n_, k_] := (2 n + 1)*Binomial[2 n, n] /; k == n;


Then for every $n,m in mathbb{N}$ there is an identity,



$$n^{2m+1}+1=sum_{r=0}^{m}A_{m,r}mathrm{Conv}_r[n], n>0 tag1$$
where $mathrm{Conv}_r[n]=(f_{r}*f_{r})[n]=sum_{t}f_r(t)f_r(n-t)$.



Expression $(1)$ is implemented in Mathematica as folows



MainIdentity[m_, n_] := Sum[A[m, r]*Conv[n, r], {r, 0, m}];


And numerically it is equal to $n^{2m+1}$ for any naturals $m,n$, for example the tabular arrangement 



n = 3; Table[MainIdentity[m, n], {m, 0, 11}]


gives



{4, 28, 244, 2188, 19684, 177148, 1594324, 14348908, 129140164}


Which is set of $3^{2m+1}+1, m=0,1,2,3.. $. But when the condition is checked by mathematica with == operator, 



FullSimplify[MainIdentity[m, n], Assumptions -> n > 0] == 

 First@FullSimplify[n^(2 m + 1) + 1, Assumptions -> n > 0]


it gives False.




Question 1: Is there any other methods of comparison, so we can verify the formula $(1)$ ?



Question 2: Execution time of the Mathematica implementation of $(1)$ is very slow, can we optimaze the solution in order to decrese exec. time ?






expression-test convolution testing-and-verification







share|improve this question











$endgroup$












  • $begingroup$
    An important question: What is 0^0 for you?
    $endgroup$
    – Henrik Schumacher
    16 hours ago






  • 1




    $begingroup$
    For Q1 you can use PossibleZeroQ to check if the difference between the expressions is zero: reference.wolfram.com/language/ref/PossibleZeroQ.html
    $endgroup$
    – Roman
    16 hours ago












  • $begingroup$
    What's the purpose of First in [...] == First@FullSimplify[n^(2 m + 1) + 1, Assumptions -> n > 0]. Not that it returns 1!
    $endgroup$
    – Henrik Schumacher
    15 hours ago










  • $begingroup$
    @HenrikSchumacher I think now it is common agreement that $0^0 = 1$, I prefer to reffer it to Knuth's Concrete mathematics. Concerning First I dont really know Mathematica well, so I just entered my function into the pattern I found in one of my previous questions here
    $endgroup$
    – Petro Kolosov
    15 hours ago








  • 1




    $begingroup$
    When n or k is not numerical, the conditions 2 k + 1 <= n and k == n are not evaluated and A[n,k] evaluates to the first, unconditioned definition of A which is 0. So in total, your definition of A is not appropriate for symbolic evaluation.
    $endgroup$
    – Henrik Schumacher
    15 hours ago
















3












$begingroup$


Let be a function power function



f[s_, r_] := Piecewise[{{s^r, s >= 0}, {0, True}}];


And its convolution



Conv[s_, r_] := Sum[f[k, r]*f[s - k, r], {k, -Infinity, +Infinity}];


Proposition 1. Let be a real coefficients $A_{m,j}$ defined as follows



A[n_, k_] := 0

A[n_, k_] := (2 k + 1)*Binomial[2 k, k]*

   Sum[A[n, j]*Binomial[j, 2 k + 1]*(-1)^(j - 1)/(j - k)*

     BernoulliB[2 j - 2 k], {j, 2 k + 1, n}] /; 2 k + 1 <= n

A[n_, k_] := (2 n + 1)*Binomial[2 n, n] /; k == n;


Then for every $n,m in mathbb{N}$ there is an identity,



$$n^{2m+1}+1=sum_{r=0}^{m}A_{m,r}mathrm{Conv}_r[n], n>0 tag1$$
where $mathrm{Conv}_r[n]=(f_{r}*f_{r})[n]=sum_{t}f_r(t)f_r(n-t)$.



Expression $(1)$ is implemented in Mathematica as folows



MainIdentity[m_, n_] := Sum[A[m, r]*Conv[n, r], {r, 0, m}];


And numerically it is equal to $n^{2m+1}$ for any naturals $m,n$, for example the tabular arrangement 



n = 3; Table[MainIdentity[m, n], {m, 0, 11}]


gives



{4, 28, 244, 2188, 19684, 177148, 1594324, 14348908, 129140164}


Which is set of $3^{2m+1}+1, m=0,1,2,3.. $. But when the condition is checked by mathematica with == operator, 



FullSimplify[MainIdentity[m, n], Assumptions -> n > 0] == 

 First@FullSimplify[n^(2 m + 1) + 1, Assumptions -> n > 0]


it gives False.




Question 1: Is there any other methods of comparison, so we can verify the formula $(1)$ ?



Question 2: Execution time of the Mathematica implementation of $(1)$ is very slow, can we optimaze the solution in order to decrese exec. time ?






expression-test convolution testing-and-verification







share|improve this question











$endgroup$












  • $begingroup$
    An important question: What is 0^0 for you?
    $endgroup$
    – Henrik Schumacher
    16 hours ago






  • 1




    $begingroup$
    For Q1 you can use PossibleZeroQ to check if the difference between the expressions is zero: reference.wolfram.com/language/ref/PossibleZeroQ.html
    $endgroup$
    – Roman
    16 hours ago












  • $begingroup$
    What's the purpose of First in [...] == First@FullSimplify[n^(2 m + 1) + 1, Assumptions -> n > 0]. Not that it returns 1!
    $endgroup$
    – Henrik Schumacher
    15 hours ago










  • $begingroup$
    @HenrikSchumacher I think now it is common agreement that $0^0 = 1$, I prefer to reffer it to Knuth's Concrete mathematics. Concerning First I dont really know Mathematica well, so I just entered my function into the pattern I found in one of my previous questions here
    $endgroup$
    – Petro Kolosov
    15 hours ago








  • 1




    $begingroup$
    When n or k is not numerical, the conditions 2 k + 1 <= n and k == n are not evaluated and A[n,k] evaluates to the first, unconditioned definition of A which is 0. So in total, your definition of A is not appropriate for symbolic evaluation.
    $endgroup$
    – Henrik Schumacher
    15 hours ago














3












3








3





$begingroup$


Let be a function power function



f[s_, r_] := Piecewise[{{s^r, s >= 0}, {0, True}}];


And its convolution



Conv[s_, r_] := Sum[f[k, r]*f[s - k, r], {k, -Infinity, +Infinity}];


Proposition 1. Let be a real coefficients $A_{m,j}$ defined as follows



A[n_, k_] := 0

A[n_, k_] := (2 k + 1)*Binomial[2 k, k]*

   Sum[A[n, j]*Binomial[j, 2 k + 1]*(-1)^(j - 1)/(j - k)*

     BernoulliB[2 j - 2 k], {j, 2 k + 1, n}] /; 2 k + 1 <= n

A[n_, k_] := (2 n + 1)*Binomial[2 n, n] /; k == n;


Then for every $n,m in mathbb{N}$ there is an identity,



$$n^{2m+1}+1=sum_{r=0}^{m}A_{m,r}mathrm{Conv}_r[n], n>0 tag1$$
where $mathrm{Conv}_r[n]=(f_{r}*f_{r})[n]=sum_{t}f_r(t)f_r(n-t)$.



Expression $(1)$ is implemented in Mathematica as folows



MainIdentity[m_, n_] := Sum[A[m, r]*Conv[n, r], {r, 0, m}];


And numerically it is equal to $n^{2m+1}$ for any naturals $m,n$, for example the tabular arrangement 



n = 3; Table[MainIdentity[m, n], {m, 0, 11}]


gives



{4, 28, 244, 2188, 19684, 177148, 1594324, 14348908, 129140164}


Which is set of $3^{2m+1}+1, m=0,1,2,3.. $. But when the condition is checked by mathematica with == operator, 



FullSimplify[MainIdentity[m, n], Assumptions -> n > 0] == 

 First@FullSimplify[n^(2 m + 1) + 1, Assumptions -> n > 0]


it gives False.




Question 1: Is there any other methods of comparison, so we can verify the formula $(1)$ ?



Question 2: Execution time of the Mathematica implementation of $(1)$ is very slow, can we optimaze the solution in order to decrese exec. time ?






expression-test convolution testing-and-verification







share|improve this question











$endgroup$




Let be a function power function



f[s_, r_] := Piecewise[{{s^r, s >= 0}, {0, True}}];


And its convolution



Conv[s_, r_] := Sum[f[k, r]*f[s - k, r], {k, -Infinity, +Infinity}];


Proposition 1. Let be a real coefficients $A_{m,j}$ defined as follows



A[n_, k_] := 0

A[n_, k_] := (2 k + 1)*Binomial[2 k, k]*

   Sum[A[n, j]*Binomial[j, 2 k + 1]*(-1)^(j - 1)/(j - k)*

     BernoulliB[2 j - 2 k], {j, 2 k + 1, n}] /; 2 k + 1 <= n

A[n_, k_] := (2 n + 1)*Binomial[2 n, n] /; k == n;


Then for every $n,m in mathbb{N}$ there is an identity,



$$n^{2m+1}+1=sum_{r=0}^{m}A_{m,r}mathrm{Conv}_r[n], n>0 tag1$$
where $mathrm{Conv}_r[n]=(f_{r}*f_{r})[n]=sum_{t}f_r(t)f_r(n-t)$.



Expression $(1)$ is implemented in Mathematica as folows



MainIdentity[m_, n_] := Sum[A[m, r]*Conv[n, r], {r, 0, m}];


And numerically it is equal to $n^{2m+1}$ for any naturals $m,n$, for example the tabular arrangement 



n = 3; Table[MainIdentity[m, n], {m, 0, 11}]


gives



{4, 28, 244, 2188, 19684, 177148, 1594324, 14348908, 129140164}


Which is set of $3^{2m+1}+1, m=0,1,2,3.. $. But when the condition is checked by mathematica with == operator, 



FullSimplify[MainIdentity[m, n], Assumptions -> n > 0] == 

 First@FullSimplify[n^(2 m + 1) + 1, Assumptions -> n > 0]


it gives False.




Question 1: Is there any other methods of comparison, so we can verify the formula $(1)$ ?



Question 2: Execution time of the Mathematica implementation of $(1)$ is very slow, can we optimaze the solution in order to decrese exec. time ?






expression-test convolution testing-and-verification




expression-test convolution testing-and-verification






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 15 hours ago







Petro Kolosov

















asked 16 hours ago









Petro KolosovPetro Kolosov

367




367












  • $begingroup$
    An important question: What is 0^0 for you?
    $endgroup$
    – Henrik Schumacher
    16 hours ago






  • 1




    $begingroup$
    For Q1 you can use PossibleZeroQ to check if the difference between the expressions is zero: reference.wolfram.com/language/ref/PossibleZeroQ.html
    $endgroup$
    – Roman
    16 hours ago












  • $begingroup$
    What's the purpose of First in [...] == First@FullSimplify[n^(2 m + 1) + 1, Assumptions -> n > 0]. Not that it returns 1!
    $endgroup$
    – Henrik Schumacher
    15 hours ago










  • $begingroup$
    @HenrikSchumacher I think now it is common agreement that $0^0 = 1$, I prefer to reffer it to Knuth's Concrete mathematics. Concerning First I dont really know Mathematica well, so I just entered my function into the pattern I found in one of my previous questions here
    $endgroup$
    – Petro Kolosov
    15 hours ago








  • 1




    $begingroup$
    When n or k is not numerical, the conditions 2 k + 1 <= n and k == n are not evaluated and A[n,k] evaluates to the first, unconditioned definition of A which is 0. So in total, your definition of A is not appropriate for symbolic evaluation.
    $endgroup$
    – Henrik Schumacher
    15 hours ago


















  • $begingroup$
    An important question: What is 0^0 for you?
    $endgroup$
    – Henrik Schumacher
    16 hours ago






  • 1




    $begingroup$
    For Q1 you can use PossibleZeroQ to check if the difference between the expressions is zero: reference.wolfram.com/language/ref/PossibleZeroQ.html
    $endgroup$
    – Roman
    16 hours ago












  • $begingroup$
    What's the purpose of First in [...] == First@FullSimplify[n^(2 m + 1) + 1, Assumptions -> n > 0]. Not that it returns 1!
    $endgroup$
    – Henrik Schumacher
    15 hours ago










  • $begingroup$
    @HenrikSchumacher I think now it is common agreement that $0^0 = 1$, I prefer to reffer it to Knuth's Concrete mathematics. Concerning First I dont really know Mathematica well, so I just entered my function into the pattern I found in one of my previous questions here
    $endgroup$
    – Petro Kolosov
    15 hours ago








  • 1




    $begingroup$
    When n or k is not numerical, the conditions 2 k + 1 <= n and k == n are not evaluated and A[n,k] evaluates to the first, unconditioned definition of A which is 0. So in total, your definition of A is not appropriate for symbolic evaluation.
    $endgroup$
    – Henrik Schumacher
    15 hours ago
















$begingroup$
An important question: What is 0^0 for you?
$endgroup$
– Henrik Schumacher
16 hours ago




$begingroup$
An important question: What is 0^0 for you?
$endgroup$
– Henrik Schumacher
16 hours ago




1




1




$begingroup$
For Q1 you can use PossibleZeroQ to check if the difference between the expressions is zero: reference.wolfram.com/language/ref/PossibleZeroQ.html
$endgroup$
– Roman
16 hours ago






$begingroup$
For Q1 you can use PossibleZeroQ to check if the difference between the expressions is zero: reference.wolfram.com/language/ref/PossibleZeroQ.html
$endgroup$
– Roman
16 hours ago














$begingroup$
What's the purpose of First in [...] == First@FullSimplify[n^(2 m + 1) + 1, Assumptions -> n > 0]. Not that it returns 1!
$endgroup$
– Henrik Schumacher
15 hours ago




$begingroup$
What's the purpose of First in [...] == First@FullSimplify[n^(2 m + 1) + 1, Assumptions -> n > 0]. Not that it returns 1!
$endgroup$
– Henrik Schumacher
15 hours ago












$begingroup$
@HenrikSchumacher I think now it is common agreement that $0^0 = 1$, I prefer to reffer it to Knuth's Concrete mathematics. Concerning First I dont really know Mathematica well, so I just entered my function into the pattern I found in one of my previous questions here
$endgroup$
– Petro Kolosov
15 hours ago






$begingroup$
@HenrikSchumacher I think now it is common agreement that $0^0 = 1$, I prefer to reffer it to Knuth's Concrete mathematics. Concerning First I dont really know Mathematica well, so I just entered my function into the pattern I found in one of my previous questions here
$endgroup$
– Petro Kolosov
15 hours ago






1




1




$begingroup$
When n or k is not numerical, the conditions 2 k + 1 <= n and k == n are not evaluated and A[n,k] evaluates to the first, unconditioned definition of A which is 0. So in total, your definition of A is not appropriate for symbolic evaluation.
$endgroup$
– Henrik Schumacher
15 hours ago




$begingroup$
When n or k is not numerical, the conditions 2 k + 1 <= n and k == n are not evaluated and A[n,k] evaluates to the first, unconditioned definition of A which is 0. So in total, your definition of A is not appropriate for symbolic evaluation.
$endgroup$
– Henrik Schumacher
15 hours ago










1 Answer
1






active

oldest

votes


















3












$begingroup$

Towards Question 2: Make it a finite sum in Conv. Indeed, there are only finitely many nonzero summands. Or better use Dot.



f2[s_, r_] := s^r UnitStep[s];

Conv2[s_, r_] := #.Reverse[#] &[f2[Range[0, s], r]]



m = 10;

aa = Outer[Conv, Range[0, m], Range[1, m]]; // AbsoluteTiming // First

bb = Outer[Conv2, Range[0, m], Range[1, m]]; // AbsoluteTiming // First

aa == bb



11.0072



0.001253



True







share|improve this answer









$endgroup$













  • $begingroup$
    As I see you use == on the values of Timings or ? Still, the main aim is to show that MainIdentity[m_, n_] == n^(2m+1)+1.
    $endgroup$
    – Petro Kolosov
    15 hours ago












  • $begingroup$
    How to check identity $(1)$ in main question ?
    $endgroup$
    – Petro Kolosov
    15 hours ago






  • 1




    $begingroup$
    I applied == to the outputs of Conv and Conv2 in order to check that the produce the same results. The timings are supposed to show you that the new definition of Conv and f lead to 10000-times faster evaluation for numeric input.
    $endgroup$
    – Henrik Schumacher
    15 hours ago












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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Towards Question 2: Make it a finite sum in Conv. Indeed, there are only finitely many nonzero summands. Or better use Dot.



f2[s_, r_] := s^r UnitStep[s];

Conv2[s_, r_] := #.Reverse[#] &[f2[Range[0, s], r]]



m = 10;

aa = Outer[Conv, Range[0, m], Range[1, m]]; // AbsoluteTiming // First

bb = Outer[Conv2, Range[0, m], Range[1, m]]; // AbsoluteTiming // First

aa == bb



11.0072



0.001253



True







share|improve this answer









$endgroup$













  • $begingroup$
    As I see you use == on the values of Timings or ? Still, the main aim is to show that MainIdentity[m_, n_] == n^(2m+1)+1.
    $endgroup$
    – Petro Kolosov
    15 hours ago












  • $begingroup$
    How to check identity $(1)$ in main question ?
    $endgroup$
    – Petro Kolosov
    15 hours ago






  • 1




    $begingroup$
    I applied == to the outputs of Conv and Conv2 in order to check that the produce the same results. The timings are supposed to show you that the new definition of Conv and f lead to 10000-times faster evaluation for numeric input.
    $endgroup$
    – Henrik Schumacher
    15 hours ago
















3












$begingroup$

Towards Question 2: Make it a finite sum in Conv. Indeed, there are only finitely many nonzero summands. Or better use Dot.



f2[s_, r_] := s^r UnitStep[s];

Conv2[s_, r_] := #.Reverse[#] &[f2[Range[0, s], r]]



m = 10;

aa = Outer[Conv, Range[0, m], Range[1, m]]; // AbsoluteTiming // First

bb = Outer[Conv2, Range[0, m], Range[1, m]]; // AbsoluteTiming // First

aa == bb



11.0072



0.001253



True







share|improve this answer









$endgroup$













  • $begingroup$
    As I see you use == on the values of Timings or ? Still, the main aim is to show that MainIdentity[m_, n_] == n^(2m+1)+1.
    $endgroup$
    – Petro Kolosov
    15 hours ago












  • $begingroup$
    How to check identity $(1)$ in main question ?
    $endgroup$
    – Petro Kolosov
    15 hours ago






  • 1




    $begingroup$
    I applied == to the outputs of Conv and Conv2 in order to check that the produce the same results. The timings are supposed to show you that the new definition of Conv and f lead to 10000-times faster evaluation for numeric input.
    $endgroup$
    – Henrik Schumacher
    15 hours ago














3












3








3





$begingroup$

Towards Question 2: Make it a finite sum in Conv. Indeed, there are only finitely many nonzero summands. Or better use Dot.



f2[s_, r_] := s^r UnitStep[s];

Conv2[s_, r_] := #.Reverse[#] &[f2[Range[0, s], r]]



m = 10;

aa = Outer[Conv, Range[0, m], Range[1, m]]; // AbsoluteTiming // First

bb = Outer[Conv2, Range[0, m], Range[1, m]]; // AbsoluteTiming // First

aa == bb



11.0072



0.001253



True







share|improve this answer









$endgroup$



Towards Question 2: Make it a finite sum in Conv. Indeed, there are only finitely many nonzero summands. Or better use Dot.



f2[s_, r_] := s^r UnitStep[s];

Conv2[s_, r_] := #.Reverse[#] &[f2[Range[0, s], r]]



m = 10;

aa = Outer[Conv, Range[0, m], Range[1, m]]; // AbsoluteTiming // First

bb = Outer[Conv2, Range[0, m], Range[1, m]]; // AbsoluteTiming // First

aa == bb



11.0072



0.001253



True








share|improve this answer












share|improve this answer



share|improve this answer










answered 16 hours ago









Henrik SchumacherHenrik Schumacher

59.4k582165




59.4k582165












  • $begingroup$
    As I see you use == on the values of Timings or ? Still, the main aim is to show that MainIdentity[m_, n_] == n^(2m+1)+1.
    $endgroup$
    – Petro Kolosov
    15 hours ago












  • $begingroup$
    How to check identity $(1)$ in main question ?
    $endgroup$
    – Petro Kolosov
    15 hours ago






  • 1




    $begingroup$
    I applied == to the outputs of Conv and Conv2 in order to check that the produce the same results. The timings are supposed to show you that the new definition of Conv and f lead to 10000-times faster evaluation for numeric input.
    $endgroup$
    – Henrik Schumacher
    15 hours ago


















  • $begingroup$
    As I see you use == on the values of Timings or ? Still, the main aim is to show that MainIdentity[m_, n_] == n^(2m+1)+1.
    $endgroup$
    – Petro Kolosov
    15 hours ago












  • $begingroup$
    How to check identity $(1)$ in main question ?
    $endgroup$
    – Petro Kolosov
    15 hours ago






  • 1




    $begingroup$
    I applied == to the outputs of Conv and Conv2 in order to check that the produce the same results. The timings are supposed to show you that the new definition of Conv and f lead to 10000-times faster evaluation for numeric input.
    $endgroup$
    – Henrik Schumacher
    15 hours ago
















$begingroup$
As I see you use == on the values of Timings or ? Still, the main aim is to show that MainIdentity[m_, n_] == n^(2m+1)+1.
$endgroup$
– Petro Kolosov
15 hours ago






$begingroup$
As I see you use == on the values of Timings or ? Still, the main aim is to show that MainIdentity[m_, n_] == n^(2m+1)+1.
$endgroup$
– Petro Kolosov
15 hours ago














$begingroup$
How to check identity $(1)$ in main question ?
$endgroup$
– Petro Kolosov
15 hours ago




$begingroup$
How to check identity $(1)$ in main question ?
$endgroup$
– Petro Kolosov
15 hours ago




1




1




$begingroup$
I applied == to the outputs of Conv and Conv2 in order to check that the produce the same results. The timings are supposed to show you that the new definition of Conv and f lead to 10000-times faster evaluation for numeric input.
$endgroup$
– Henrik Schumacher
15 hours ago




$begingroup$
I applied == to the outputs of Conv and Conv2 in order to check that the produce the same results. The timings are supposed to show you that the new definition of Conv and f lead to 10000-times faster evaluation for numeric input.
$endgroup$
– Henrik Schumacher
15 hours ago


















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