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
asked 16 hours ago
Petro KolosovPetro Kolosov
367
$endgroup$
|
show 2 more comments
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
asked 16 hours ago
Petro KolosovPetro Kolosov
367
$endgroup$
$begingroup$
An important question: What is0^0for you?
$endgroup$
– Henrik Schumacher
16 hours ago
1
$begingroup$
For Q1 you can usePossibleZeroQto 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 ofFirstin[...] == First@FullSimplify[n^(2 m + 1) + 1, Assumptions -> n > 0]. Not that it returns1!
$endgroup$
– 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. ConcerningFirstI 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
$begingroup$
Whennorkis not numerical, the conditions2 k + 1 <= nandk == nare not evaluated andA[n,k]evaluates to the first, unconditioned definition ofAwhich is0. So in total, your definition ofAis not appropriate for symbolic evaluation.
$endgroup$
– Henrik Schumacher
15 hours ago
|
show 2 more comments
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
asked 16 hours ago
Petro KolosovPetro Kolosov
367
$endgroup$
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
asked 16 hours ago
Petro KolosovPetro Kolosov
367
asked 16 hours ago
Petro KolosovPetro Kolosov
367
edited 15 hours ago
Petro Kolosov
asked 16 hours ago
Petro KolosovPetro Kolosov
367
asked 16 hours ago
Petro KolosovPetro Kolosov
367
asked 16 hours ago
Petro KolosovPetro Kolosov
367
367
$begingroup$
An important question: What is0^0for you?
$endgroup$
– Henrik Schumacher
16 hours ago
1
$begingroup$
For Q1 you can usePossibleZeroQto 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 ofFirstin[...] == First@FullSimplify[n^(2 m + 1) + 1, Assumptions -> n > 0]. Not that it returns1!
$endgroup$
– 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. ConcerningFirstI 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
$begingroup$
Whennorkis not numerical, the conditions2 k + 1 <= nandk == nare not evaluated andA[n,k]evaluates to the first, unconditioned definition ofAwhich is0. So in total, your definition ofAis not appropriate for symbolic evaluation.
$endgroup$
– Henrik Schumacher
15 hours ago
|
show 2 more comments
$begingroup$
An important question: What is0^0for you?
$endgroup$
– Henrik Schumacher
16 hours ago
1
$begingroup$
For Q1 you can usePossibleZeroQto 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 ofFirstin[...] == First@FullSimplify[n^(2 m + 1) + 1, Assumptions -> n > 0]. Not that it returns1!
$endgroup$
– 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. ConcerningFirstI 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
$begingroup$
Whennorkis not numerical, the conditions2 k + 1 <= nandk == nare not evaluated andA[n,k]evaluates to the first, unconditioned definition ofAwhich is0. So in total, your definition ofAis 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
|
show 2 more comments
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
answered 16 hours ago
Henrik SchumacherHenrik Schumacher
59.4k582165
$endgroup$
$begingroup$
As I see you use==on the values of Timings or ? Still, the main aim is to show thatMainIdentity[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 ofConvandConv2in order to check that the produce the same results. The timings are supposed to show you that the new definition ofConvandflead to 10000-times faster evaluation for numeric input.
$endgroup$
– Henrik Schumacher
15 hours ago
add a comment |
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1 Answer
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1 Answer
1
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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
answered 16 hours ago
Henrik SchumacherHenrik Schumacher
59.4k582165
$endgroup$
$begingroup$
As I see you use==on the values of Timings or ? Still, the main aim is to show thatMainIdentity[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 ofConvandConv2in order to check that the produce the same results. The timings are supposed to show you that the new definition ofConvandflead to 10000-times faster evaluation for numeric input.
$endgroup$
– Henrik Schumacher
15 hours ago
add a comment |
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
answered 16 hours ago
Henrik SchumacherHenrik Schumacher
59.4k582165
$endgroup$
$begingroup$
As I see you use==on the values of Timings or ? Still, the main aim is to show thatMainIdentity[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 ofConvandConv2in order to check that the produce the same results. The timings are supposed to show you that the new definition ofConvandflead to 10000-times faster evaluation for numeric input.
$endgroup$
– Henrik Schumacher
15 hours ago
add a comment |
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
answered 16 hours ago
Henrik SchumacherHenrik Schumacher
59.4k582165
$endgroup$
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
answered 16 hours ago
Henrik SchumacherHenrik Schumacher
59.4k582165
answered 16 hours ago
Henrik SchumacherHenrik Schumacher
59.4k582165
answered 16 hours ago
Henrik SchumacherHenrik Schumacher
59.4k582165
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 thatMainIdentity[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 ofConvandConv2in order to check that the produce the same results. The timings are supposed to show you that the new definition ofConvandflead to 10000-times faster evaluation for numeric input.
$endgroup$
– Henrik Schumacher
15 hours ago
add a comment |
$begingroup$
As I see you use==on the values of Timings or ? Still, the main aim is to show thatMainIdentity[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 ofConvandConv2in order to check that the produce the same results. The timings are supposed to show you that the new definition ofConvandflead 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
add a comment |
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$begingroup$
An important question: What is
0^0for you?$endgroup$
– Henrik Schumacher
16 hours ago
1
$begingroup$
For Q1 you can use
PossibleZeroQto 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
Firstin[...] == First@FullSimplify[n^(2 m + 1) + 1, Assumptions -> n > 0]. Not that it returns1!$endgroup$
– 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
FirstI 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
norkis not numerical, the conditions2 k + 1 <= nandk == nare not evaluated andA[n,k]evaluates to the first, unconditioned definition ofAwhich is0. So in total, your definition ofAis not appropriate for symbolic evaluation.$endgroup$
– Henrik Schumacher
15 hours ago