Is this a new Fibonacci Identity? [closed] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)fibonacci identity using generating functionPrimality criterion for generalized Fermat numbers similar to the LLT ?Solvable parametric $7$th and $13$th degree equations using $eta(q)/eta(q^p)$?To what extent is it possible to generalise a natural bijection between trees and $7$-tuples of trees, suggested by divergent series?Two spaces attached to mod 2 level 9 modular forms--a conjectural Hecke isomorphismCan someone explain this appearance of the Fibonacci series in the formula of the Fibonacci series?On the automorphism group of binary quadratic formsFor what (other) families of graphs does the clique-coclique bound hold?Coefficients $U_m(n,k)$ in the identity $n^2m+1=sumlimits_0leq k leq m(-1)^m-kU_m(n,k)cdot n^k$A question on the Faulhaber's formula
Is this a new Fibonacci Identity? [closed]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)fibonacci identity using generating functionPrimality criterion for generalized Fermat numbers similar to the LLT ?Solvable parametric $7$th and $13$th degree equations using $eta(q)/eta(q^p)$?To what extent is it possible to generalise a natural bijection between trees and $7$-tuples of trees, suggested by divergent series?Two spaces attached to mod 2 level 9 modular forms--a conjectural Hecke isomorphismCan someone explain this appearance of the Fibonacci series in the formula of the Fibonacci series?On the automorphism group of binary quadratic formsFor what (other) families of graphs does the clique-coclique bound hold?Coefficients $U_m(n,k)$ in the identity $n^2m+1=sumlimits_0leq k leq m(-1)^m-kU_m(n,k)cdot n^k$A question on the Faulhaber's formula
$begingroup$
I have found the following Fibonacci Identity (and proved it).
If $F_n$ denotes the nth Fibonacci Number, we have the following identity
beginequation
F_n-r+hF_n+k+g+1 - F_n-r+gF_n+k+h+1 = (-1)^n+r+h+1 F_g-hF_k+r+1
endequation
where $F_1 = F_2 = 1$, $r leq n$, $h leq g$, and $n, g, k in mathbbN$.
It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few.
I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see what the connections are. Thanks for your patience and input!
nt.number-theory co.combinatorics
asked Apr 1 at 22:34
GrassiGrassi
13927
$endgroup$
closed as off-topic by user44191, Stopple, Max Alekseyev, Zhi-Wei Sun, Jan-Christoph Schlage-Puchta Apr 2 at 12:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question does not appear to be about research level mathematics within the scope defined in the help center." – user44191, Stopple, Max Alekseyev, Zhi-Wei Sun, Jan-Christoph Schlage-Puchta
|
show 1 more comment
$begingroup$
I have found the following Fibonacci Identity (and proved it).
If $F_n$ denotes the nth Fibonacci Number, we have the following identity
beginequation
F_n-r+hF_n+k+g+1 - F_n-r+gF_n+k+h+1 = (-1)^n+r+h+1 F_g-hF_k+r+1
endequation
where $F_1 = F_2 = 1$, $r leq n$, $h leq g$, and $n, g, k in mathbbN$.
It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few.
I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see what the connections are. Thanks for your patience and input!
nt.number-theory co.combinatorics
asked Apr 1 at 22:34
GrassiGrassi
13927
$endgroup$
closed as off-topic by user44191, Stopple, Max Alekseyev, Zhi-Wei Sun, Jan-Christoph Schlage-Puchta Apr 2 at 12:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question does not appear to be about research level mathematics within the scope defined in the help center." – user44191, Stopple, Max Alekseyev, Zhi-Wei Sun, Jan-Christoph Schlage-Puchta
1
$begingroup$
This can be simplified to $F_a - rF_b + k + 1 - F_b - r F_a + k + 1 = (-1)^a + r + 1 F_b - a F_k + r + 1$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
$endgroup$
– user44191
Apr 1 at 22:47
1
$begingroup$
Sorry, I meant 4 variables instead of 5. I'm pretty sure this can be reduced by changes of variables (with no Fibonacci arithmetic) to Vajda's identity.
$endgroup$
– user44191
Apr 1 at 22:52
2
$begingroup$
This is a disguised version of Vajda's identity (with minor amounts of arithmetic for powers of $-1$); try doing variable substitutions to see for yourself.
$endgroup$
– user44191
Apr 1 at 23:25
1
$begingroup$
The identity is with arguments renamed the same as $F_a F_b - F_c F_a+b-c = (-1)^a+1F_c-aF_b-c$. Read my essay "In the elliptic realm" to see connection.
$endgroup$
– Somos
Apr 2 at 2:55
1
$begingroup$
I don't really see why this question should be put on hold. The answer is "no, this identity is not new" but the question itself seems totally legitimate and not obvious.
$endgroup$
– Sam Hopkins
Apr 2 at 20:36
|
show 1 more comment
$begingroup$
I have found the following Fibonacci Identity (and proved it).
If $F_n$ denotes the nth Fibonacci Number, we have the following identity
beginequation
F_n-r+hF_n+k+g+1 - F_n-r+gF_n+k+h+1 = (-1)^n+r+h+1 F_g-hF_k+r+1
endequation
where $F_1 = F_2 = 1$, $r leq n$, $h leq g$, and $n, g, k in mathbbN$.
It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few.
I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see what the connections are. Thanks for your patience and input!
nt.number-theory co.combinatorics
asked Apr 1 at 22:34
GrassiGrassi
13927
$endgroup$
I have found the following Fibonacci Identity (and proved it).
If $F_n$ denotes the nth Fibonacci Number, we have the following identity
beginequation
F_n-r+hF_n+k+g+1 - F_n-r+gF_n+k+h+1 = (-1)^n+r+h+1 F_g-hF_k+r+1
endequation
where $F_1 = F_2 = 1$, $r leq n$, $h leq g$, and $n, g, k in mathbbN$.
It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few.
I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see what the connections are. Thanks for your patience and input!
nt.number-theory co.combinatorics
nt.number-theory co.combinatorics
asked Apr 1 at 22:34
GrassiGrassi
13927
asked Apr 1 at 22:34
GrassiGrassi
13927
asked Apr 1 at 22:34
GrassiGrassi
13927
asked Apr 1 at 22:34
GrassiGrassi
13927
asked Apr 1 at 22:34
GrassiGrassi
13927
13927
closed as off-topic by user44191, Stopple, Max Alekseyev, Zhi-Wei Sun, Jan-Christoph Schlage-Puchta Apr 2 at 12:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question does not appear to be about research level mathematics within the scope defined in the help center." – user44191, Stopple, Max Alekseyev, Zhi-Wei Sun, Jan-Christoph Schlage-Puchta
closed as off-topic by user44191, Stopple, Max Alekseyev, Zhi-Wei Sun, Jan-Christoph Schlage-Puchta Apr 2 at 12:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question does not appear to be about research level mathematics within the scope defined in the help center." – user44191, Stopple, Max Alekseyev, Zhi-Wei Sun, Jan-Christoph Schlage-Puchta
1
$begingroup$
This can be simplified to $F_a - rF_b + k + 1 - F_b - r F_a + k + 1 = (-1)^a + r + 1 F_b - a F_k + r + 1$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
$endgroup$
– user44191
Apr 1 at 22:47
1
$begingroup$
Sorry, I meant 4 variables instead of 5. I'm pretty sure this can be reduced by changes of variables (with no Fibonacci arithmetic) to Vajda's identity.
$endgroup$
– user44191
Apr 1 at 22:52
2
$begingroup$
This is a disguised version of Vajda's identity (with minor amounts of arithmetic for powers of $-1$); try doing variable substitutions to see for yourself.
$endgroup$
– user44191
Apr 1 at 23:25
1
$begingroup$
The identity is with arguments renamed the same as $F_a F_b - F_c F_a+b-c = (-1)^a+1F_c-aF_b-c$. Read my essay "In the elliptic realm" to see connection.
$endgroup$
– Somos
Apr 2 at 2:55
1
$begingroup$
I don't really see why this question should be put on hold. The answer is "no, this identity is not new" but the question itself seems totally legitimate and not obvious.
$endgroup$
– Sam Hopkins
Apr 2 at 20:36
|
show 1 more comment
1
$begingroup$
This can be simplified to $F_a - rF_b + k + 1 - F_b - r F_a + k + 1 = (-1)^a + r + 1 F_b - a F_k + r + 1$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
$endgroup$
– user44191
Apr 1 at 22:47
1
$begingroup$
Sorry, I meant 4 variables instead of 5. I'm pretty sure this can be reduced by changes of variables (with no Fibonacci arithmetic) to Vajda's identity.
$endgroup$
– user44191
Apr 1 at 22:52
2
$begingroup$
This is a disguised version of Vajda's identity (with minor amounts of arithmetic for powers of $-1$); try doing variable substitutions to see for yourself.
$endgroup$
– user44191
Apr 1 at 23:25
1
$begingroup$
The identity is with arguments renamed the same as $F_a F_b - F_c F_a+b-c = (-1)^a+1F_c-aF_b-c$. Read my essay "In the elliptic realm" to see connection.
$endgroup$
– Somos
Apr 2 at 2:55
1
$begingroup$
I don't really see why this question should be put on hold. The answer is "no, this identity is not new" but the question itself seems totally legitimate and not obvious.
$endgroup$
– Sam Hopkins
Apr 2 at 20:36
1
1
$begingroup$
This can be simplified to $F_a - rF_b + k + 1 - F_b - r F_a + k + 1 = (-1)^a + r + 1 F_b - a F_k + r + 1$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
$endgroup$
– user44191
Apr 1 at 22:47
$begingroup$
This can be simplified to $F_a - rF_b + k + 1 - F_b - r F_a + k + 1 = (-1)^a + r + 1 F_b - a F_k + r + 1$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
$endgroup$
– user44191
Apr 1 at 22:47
1
1
$begingroup$
Sorry, I meant 4 variables instead of 5. I'm pretty sure this can be reduced by changes of variables (with no Fibonacci arithmetic) to Vajda's identity.
$endgroup$
– user44191
Apr 1 at 22:52
$begingroup$
Sorry, I meant 4 variables instead of 5. I'm pretty sure this can be reduced by changes of variables (with no Fibonacci arithmetic) to Vajda's identity.
$endgroup$
– user44191
Apr 1 at 22:52
2
2
$begingroup$
This is a disguised version of Vajda's identity (with minor amounts of arithmetic for powers of $-1$); try doing variable substitutions to see for yourself.
$endgroup$
– user44191
Apr 1 at 23:25
$begingroup$
This is a disguised version of Vajda's identity (with minor amounts of arithmetic for powers of $-1$); try doing variable substitutions to see for yourself.
$endgroup$
– user44191
Apr 1 at 23:25
1
1
$begingroup$
The identity is with arguments renamed the same as $F_a F_b - F_c F_a+b-c = (-1)^a+1F_c-aF_b-c$. Read my essay "In the elliptic realm" to see connection.
$endgroup$
– Somos
Apr 2 at 2:55
$begingroup$
The identity is with arguments renamed the same as $F_a F_b - F_c F_a+b-c = (-1)^a+1F_c-aF_b-c$. Read my essay "In the elliptic realm" to see connection.
$endgroup$
– Somos
Apr 2 at 2:55
1
1
$begingroup$
I don't really see why this question should be put on hold. The answer is "no, this identity is not new" but the question itself seems totally legitimate and not obvious.
$endgroup$
– Sam Hopkins
Apr 2 at 20:36
$begingroup$
I don't really see why this question should be put on hold. The answer is "no, this identity is not new" but the question itself seems totally legitimate and not obvious.
$endgroup$
– Sam Hopkins
Apr 2 at 20:36
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
Here is an expanded comment of user44191. The basic observation is that one can extend $F_n$ to all $nin mathbb Z$ by requiring $F_-n=(-1)^n+1F_n$. Then by Vajda's formula, one has $$F_n'+a'F_n'+b'-F_n'F_n'+a'+b'=(-1)^n'F_a'F_b'=(-1)^n'+a'+1F_-a'F_b',$$ where one uses the extension above in the last equality. Now by the following substitutions, the above identity leads to the one given by user44191: $$n'=b-r,a'=a-b,b'=k+r+1,$$ using the fact that $a-r+1$ is congruent to $a+r+1$ mod $2$.
answered Apr 2 at 0:08
Cherng-tiao PerngCherng-tiao Perng
895159
$endgroup$
add a comment |
$begingroup$
This identity is a special case of Euler's Identity for Continuants. It is a Pfaffian of degenerate $4times 4$ matrix.
Concrete mathematics gives the following reference:
As Michael Somos mentioned in his comment it is a part of "elliptic realm" where different identities arise as determinants of degenerate matrices. These matrices are degenerate because they are submatrices of infinite matrices of finite rank. For examle the matrix with entries $a_m,n=s_m+ns_m-n$ $(m,nin mathbbZ)$ where $s_n$ is the Somos-$4$ seqence has rank $2$. For Somos-$6$ corresponding matrix has rank $4$ etc.
answered Apr 2 at 4:07
Alexey UstinovAlexey Ustinov
7,00945980
$endgroup$
add a comment |
$begingroup$
"Vajda's identity" is really Tagiuro's identity: A. Tagiuri, Di alcune successioni ricorrenti a termini interi e positivi, Periodico di Matematica 16 (1900–1901), 1–12.
See also https://math.stackexchange.com/questions/1356391/is-there-a-name-for-this-fibonacci-identity.
answered Apr 2 at 1:16
Ira GesselIra Gessel
8,4322642
$endgroup$
$begingroup$
While probably historically interesting, this doesn't seem relevant to the question.
$endgroup$
– LSpice
Apr 2 at 15:19
2
$begingroup$
The question is "I am still wondering if anyone has seen this identity before?" I am saying, yes, this identity (or an equivalent identity) was seen by Tagiuri in 1900.
$endgroup$
– Ira Gessel
Apr 2 at 16:26
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is an expanded comment of user44191. The basic observation is that one can extend $F_n$ to all $nin mathbb Z$ by requiring $F_-n=(-1)^n+1F_n$. Then by Vajda's formula, one has $$F_n'+a'F_n'+b'-F_n'F_n'+a'+b'=(-1)^n'F_a'F_b'=(-1)^n'+a'+1F_-a'F_b',$$ where one uses the extension above in the last equality. Now by the following substitutions, the above identity leads to the one given by user44191: $$n'=b-r,a'=a-b,b'=k+r+1,$$ using the fact that $a-r+1$ is congruent to $a+r+1$ mod $2$.
answered Apr 2 at 0:08
Cherng-tiao PerngCherng-tiao Perng
895159
$endgroup$
add a comment |
$begingroup$
Here is an expanded comment of user44191. The basic observation is that one can extend $F_n$ to all $nin mathbb Z$ by requiring $F_-n=(-1)^n+1F_n$. Then by Vajda's formula, one has $$F_n'+a'F_n'+b'-F_n'F_n'+a'+b'=(-1)^n'F_a'F_b'=(-1)^n'+a'+1F_-a'F_b',$$ where one uses the extension above in the last equality. Now by the following substitutions, the above identity leads to the one given by user44191: $$n'=b-r,a'=a-b,b'=k+r+1,$$ using the fact that $a-r+1$ is congruent to $a+r+1$ mod $2$.
answered Apr 2 at 0:08
Cherng-tiao PerngCherng-tiao Perng
895159
$endgroup$
add a comment |
$begingroup$
Here is an expanded comment of user44191. The basic observation is that one can extend $F_n$ to all $nin mathbb Z$ by requiring $F_-n=(-1)^n+1F_n$. Then by Vajda's formula, one has $$F_n'+a'F_n'+b'-F_n'F_n'+a'+b'=(-1)^n'F_a'F_b'=(-1)^n'+a'+1F_-a'F_b',$$ where one uses the extension above in the last equality. Now by the following substitutions, the above identity leads to the one given by user44191: $$n'=b-r,a'=a-b,b'=k+r+1,$$ using the fact that $a-r+1$ is congruent to $a+r+1$ mod $2$.
answered Apr 2 at 0:08
Cherng-tiao PerngCherng-tiao Perng
895159
$endgroup$
Here is an expanded comment of user44191. The basic observation is that one can extend $F_n$ to all $nin mathbb Z$ by requiring $F_-n=(-1)^n+1F_n$. Then by Vajda's formula, one has $$F_n'+a'F_n'+b'-F_n'F_n'+a'+b'=(-1)^n'F_a'F_b'=(-1)^n'+a'+1F_-a'F_b',$$ where one uses the extension above in the last equality. Now by the following substitutions, the above identity leads to the one given by user44191: $$n'=b-r,a'=a-b,b'=k+r+1,$$ using the fact that $a-r+1$ is congruent to $a+r+1$ mod $2$.
answered Apr 2 at 0:08
Cherng-tiao PerngCherng-tiao Perng
895159
answered Apr 2 at 0:08
Cherng-tiao PerngCherng-tiao Perng
895159
answered Apr 2 at 0:08
Cherng-tiao PerngCherng-tiao Perng
895159
answered Apr 2 at 0:08
Cherng-tiao PerngCherng-tiao Perng
895159
895159
add a comment |
add a comment |
$begingroup$
This identity is a special case of Euler's Identity for Continuants. It is a Pfaffian of degenerate $4times 4$ matrix.
Concrete mathematics gives the following reference:
As Michael Somos mentioned in his comment it is a part of "elliptic realm" where different identities arise as determinants of degenerate matrices. These matrices are degenerate because they are submatrices of infinite matrices of finite rank. For examle the matrix with entries $a_m,n=s_m+ns_m-n$ $(m,nin mathbbZ)$ where $s_n$ is the Somos-$4$ seqence has rank $2$. For Somos-$6$ corresponding matrix has rank $4$ etc.
answered Apr 2 at 4:07
Alexey UstinovAlexey Ustinov
7,00945980
$endgroup$
add a comment |
$begingroup$
This identity is a special case of Euler's Identity for Continuants. It is a Pfaffian of degenerate $4times 4$ matrix.
Concrete mathematics gives the following reference:
As Michael Somos mentioned in his comment it is a part of "elliptic realm" where different identities arise as determinants of degenerate matrices. These matrices are degenerate because they are submatrices of infinite matrices of finite rank. For examle the matrix with entries $a_m,n=s_m+ns_m-n$ $(m,nin mathbbZ)$ where $s_n$ is the Somos-$4$ seqence has rank $2$. For Somos-$6$ corresponding matrix has rank $4$ etc.
answered Apr 2 at 4:07
Alexey UstinovAlexey Ustinov
7,00945980
$endgroup$
add a comment |
$begingroup$
This identity is a special case of Euler's Identity for Continuants. It is a Pfaffian of degenerate $4times 4$ matrix.
Concrete mathematics gives the following reference:
As Michael Somos mentioned in his comment it is a part of "elliptic realm" where different identities arise as determinants of degenerate matrices. These matrices are degenerate because they are submatrices of infinite matrices of finite rank. For examle the matrix with entries $a_m,n=s_m+ns_m-n$ $(m,nin mathbbZ)$ where $s_n$ is the Somos-$4$ seqence has rank $2$. For Somos-$6$ corresponding matrix has rank $4$ etc.
answered Apr 2 at 4:07
Alexey UstinovAlexey Ustinov
7,00945980
$endgroup$
This identity is a special case of Euler's Identity for Continuants. It is a Pfaffian of degenerate $4times 4$ matrix.
Concrete mathematics gives the following reference:
As Michael Somos mentioned in his comment it is a part of "elliptic realm" where different identities arise as determinants of degenerate matrices. These matrices are degenerate because they are submatrices of infinite matrices of finite rank. For examle the matrix with entries $a_m,n=s_m+ns_m-n$ $(m,nin mathbbZ)$ where $s_n$ is the Somos-$4$ seqence has rank $2$. For Somos-$6$ corresponding matrix has rank $4$ etc.
answered Apr 2 at 4:07
Alexey UstinovAlexey Ustinov
7,00945980
edited Apr 2 at 5:36
answered Apr 2 at 4:07
Alexey UstinovAlexey Ustinov
7,00945980
answered Apr 2 at 4:07
Alexey UstinovAlexey Ustinov
7,00945980
answered Apr 2 at 4:07
Alexey UstinovAlexey Ustinov
7,00945980
7,00945980
add a comment |
add a comment |
$begingroup$
"Vajda's identity" is really Tagiuro's identity: A. Tagiuri, Di alcune successioni ricorrenti a termini interi e positivi, Periodico di Matematica 16 (1900–1901), 1–12.
See also https://math.stackexchange.com/questions/1356391/is-there-a-name-for-this-fibonacci-identity.
answered Apr 2 at 1:16
Ira GesselIra Gessel
8,4322642
$endgroup$
$begingroup$
While probably historically interesting, this doesn't seem relevant to the question.
$endgroup$
– LSpice
Apr 2 at 15:19
2
$begingroup$
The question is "I am still wondering if anyone has seen this identity before?" I am saying, yes, this identity (or an equivalent identity) was seen by Tagiuri in 1900.
$endgroup$
– Ira Gessel
Apr 2 at 16:26
add a comment |
$begingroup$
"Vajda's identity" is really Tagiuro's identity: A. Tagiuri, Di alcune successioni ricorrenti a termini interi e positivi, Periodico di Matematica 16 (1900–1901), 1–12.
See also https://math.stackexchange.com/questions/1356391/is-there-a-name-for-this-fibonacci-identity.
answered Apr 2 at 1:16
Ira GesselIra Gessel
8,4322642
$endgroup$
$begingroup$
While probably historically interesting, this doesn't seem relevant to the question.
$endgroup$
– LSpice
Apr 2 at 15:19
2
$begingroup$
The question is "I am still wondering if anyone has seen this identity before?" I am saying, yes, this identity (or an equivalent identity) was seen by Tagiuri in 1900.
$endgroup$
– Ira Gessel
Apr 2 at 16:26
add a comment |
$begingroup$
"Vajda's identity" is really Tagiuro's identity: A. Tagiuri, Di alcune successioni ricorrenti a termini interi e positivi, Periodico di Matematica 16 (1900–1901), 1–12.
See also https://math.stackexchange.com/questions/1356391/is-there-a-name-for-this-fibonacci-identity.
answered Apr 2 at 1:16
Ira GesselIra Gessel
8,4322642
$endgroup$
"Vajda's identity" is really Tagiuro's identity: A. Tagiuri, Di alcune successioni ricorrenti a termini interi e positivi, Periodico di Matematica 16 (1900–1901), 1–12.
See also https://math.stackexchange.com/questions/1356391/is-there-a-name-for-this-fibonacci-identity.
answered Apr 2 at 1:16
Ira GesselIra Gessel
8,4322642
answered Apr 2 at 1:16
Ira GesselIra Gessel
8,4322642
answered Apr 2 at 1:16
Ira GesselIra Gessel
8,4322642
answered Apr 2 at 1:16
Ira GesselIra Gessel
8,4322642
8,4322642
$begingroup$
While probably historically interesting, this doesn't seem relevant to the question.
$endgroup$
– LSpice
Apr 2 at 15:19
2
$begingroup$
The question is "I am still wondering if anyone has seen this identity before?" I am saying, yes, this identity (or an equivalent identity) was seen by Tagiuri in 1900.
$endgroup$
– Ira Gessel
Apr 2 at 16:26
add a comment |
$begingroup$
While probably historically interesting, this doesn't seem relevant to the question.
$endgroup$
– LSpice
Apr 2 at 15:19
2
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The question is "I am still wondering if anyone has seen this identity before?" I am saying, yes, this identity (or an equivalent identity) was seen by Tagiuri in 1900.
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– Ira Gessel
Apr 2 at 16:26
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While probably historically interesting, this doesn't seem relevant to the question.
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– LSpice
Apr 2 at 15:19
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While probably historically interesting, this doesn't seem relevant to the question.
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– LSpice
Apr 2 at 15:19
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The question is "I am still wondering if anyone has seen this identity before?" I am saying, yes, this identity (or an equivalent identity) was seen by Tagiuri in 1900.
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– Ira Gessel
Apr 2 at 16:26
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The question is "I am still wondering if anyone has seen this identity before?" I am saying, yes, this identity (or an equivalent identity) was seen by Tagiuri in 1900.
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– Ira Gessel
Apr 2 at 16:26
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This can be simplified to $F_a - rF_b + k + 1 - F_b - r F_a + k + 1 = (-1)^a + r + 1 F_b - a F_k + r + 1$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6.
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– user44191
Apr 1 at 22:47
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Sorry, I meant 4 variables instead of 5. I'm pretty sure this can be reduced by changes of variables (with no Fibonacci arithmetic) to Vajda's identity.
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– user44191
Apr 1 at 22:52
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This is a disguised version of Vajda's identity (with minor amounts of arithmetic for powers of $-1$); try doing variable substitutions to see for yourself.
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– user44191
Apr 1 at 23:25
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The identity is with arguments renamed the same as $F_a F_b - F_c F_a+b-c = (-1)^a+1F_c-aF_b-c$. Read my essay "In the elliptic realm" to see connection.
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– Somos
Apr 2 at 2:55
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I don't really see why this question should be put on hold. The answer is "no, this identity is not new" but the question itself seems totally legitimate and not obvious.
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– Sam Hopkins
Apr 2 at 20:36