Is this a new Fibonacci Identity? [on hold]
$begingroup$
I have found the following Fibonacci Identity (and proved it).
If $F_n$ denotes the nth Fibonacci Number, we have the following identity
begin{equation}
F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1}
end{equation}
where $F_1 = F_2 = 1$, $r leq n$, $h leq g$, and $n, g, k in mathbb{N}$.
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
11927
$endgroup$
put on hold as off-topic by user44191, Stopple, Max Alekseyev, Zhi-Wei Sun, Jan-Christoph Schlage-Puchta 2 days ago
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
If this question can be reworded to fit the rules in the help center, please edit the question.
|
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
begin{equation}
F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1}
end{equation}
where $F_1 = F_2 = 1$, $r leq n$, $h leq g$, and $n, g, k in mathbb{N}$.
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
11927
$endgroup$
put on hold as off-topic by user44191, Stopple, Max Alekseyev, Zhi-Wei Sun, Jan-Christoph Schlage-Puchta 2 days ago
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
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
This can be simplified to $F_{a - r}F_{b + k + 1} - F_{b - r} F_{a + k + 1} = (-1)^{a + r + 1} F_{b - a} F_{k + r + 1}$, 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 $F_a F_b - F_c F_{a+b-c} = (-1)^{a+1}F_{c-a}F_{b-c}$. 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
2 days ago
|
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
begin{equation}
F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1}
end{equation}
where $F_1 = F_2 = 1$, $r leq n$, $h leq g$, and $n, g, k in mathbb{N}$.
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
11927
$endgroup$
I have found the following Fibonacci Identity (and proved it).
If $F_n$ denotes the nth Fibonacci Number, we have the following identity
begin{equation}
F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1}
end{equation}
where $F_1 = F_2 = 1$, $r leq n$, $h leq g$, and $n, g, k in mathbb{N}$.
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
11927
asked Apr 1 at 22:34
GrassiGrassi
11927
asked Apr 1 at 22:34
GrassiGrassi
11927
asked Apr 1 at 22:34
GrassiGrassi
11927
asked Apr 1 at 22:34
GrassiGrassi
11927
11927
put on hold as off-topic by user44191, Stopple, Max Alekseyev, Zhi-Wei Sun, Jan-Christoph Schlage-Puchta 2 days ago
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
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by user44191, Stopple, Max Alekseyev, Zhi-Wei Sun, Jan-Christoph Schlage-Puchta 2 days ago
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
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
This can be simplified to $F_{a - r}F_{b + k + 1} - F_{b - r} F_{a + k + 1} = (-1)^{a + r + 1} F_{b - a} F_{k + r + 1}$, 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 $F_a F_b - F_c F_{a+b-c} = (-1)^{a+1}F_{c-a}F_{b-c}$. 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
2 days ago
|
show 1 more comment
1
$begingroup$
This can be simplified to $F_{a - r}F_{b + k + 1} - F_{b - r} F_{a + k + 1} = (-1)^{a + r + 1} F_{b - a} F_{k + r + 1}$, 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 $F_a F_b - F_c F_{a+b-c} = (-1)^{a+1}F_{c-a}F_{b-c}$. 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
2 days ago
1
1
$begingroup$
This can be simplified to $F_{a - r}F_{b + k + 1} - F_{b - r} F_{a + k + 1} = (-1)^{a + r + 1} F_{b - a} F_{k + r + 1}$, 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 $F_{a - r}F_{b + k + 1} - F_{b - r} F_{a + k + 1} = (-1)^{a + r + 1} F_{b - a} F_{k + r + 1}$, 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 $F_a F_b - F_c F_{a+b-c} = (-1)^{a+1}F_{c-a}F_{b-c}$. 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 $F_a F_b - F_c F_{a+b-c} = (-1)^{a+1}F_{c-a}F_{b-c}$. 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
2 days ago
$begingroup$
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
2 days ago
|
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 $F_{-n}=(-1)^{n+1}F_n$. Then by Vajda's formula, one has $$F_{n'+a'}F_{n'+b'}-F_{n'}F_{n'+a'+b'}=(-1)^{n'}F_{a'}F_{b'}=(-1)^{n'+a'+1}F_{-a'}F_{b'},$$ 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
885159
$endgroup$
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 $a_{m,n}=s_{m+n}s_{m-n}$ $(m,nin mathbb{Z})$ 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
6,98445979
$endgroup$
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
8,4222642
$endgroup$
$begingroup$
While probably historically interesting, this doesn't seem relevant to the question.
$endgroup$
– LSpice
2 days ago
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
2 days ago
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 $F_{-n}=(-1)^{n+1}F_n$. Then by Vajda's formula, one has $$F_{n'+a'}F_{n'+b'}-F_{n'}F_{n'+a'+b'}=(-1)^{n'}F_{a'}F_{b'}=(-1)^{n'+a'+1}F_{-a'}F_{b'},$$ 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
885159
$endgroup$
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 $F_{-n}=(-1)^{n+1}F_n$. Then by Vajda's formula, one has $$F_{n'+a'}F_{n'+b'}-F_{n'}F_{n'+a'+b'}=(-1)^{n'}F_{a'}F_{b'}=(-1)^{n'+a'+1}F_{-a'}F_{b'},$$ 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
885159
$endgroup$
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 $F_{-n}=(-1)^{n+1}F_n$. Then by Vajda's formula, one has $$F_{n'+a'}F_{n'+b'}-F_{n'}F_{n'+a'+b'}=(-1)^{n'}F_{a'}F_{b'}=(-1)^{n'+a'+1}F_{-a'}F_{b'},$$ 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
885159
$endgroup$
Here is an expanded comment of user44191. The basic observation is that one can extend $F_n$ to all $nin {mathbb Z}$ by requiring $F_{-n}=(-1)^{n+1}F_n$. Then by Vajda's formula, one has $$F_{n'+a'}F_{n'+b'}-F_{n'}F_{n'+a'+b'}=(-1)^{n'}F_{a'}F_{b'}=(-1)^{n'+a'+1}F_{-a'}F_{b'},$$ 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
885159
answered Apr 2 at 0:08
Cherng-tiao PerngCherng-tiao Perng
885159
answered Apr 2 at 0:08
Cherng-tiao PerngCherng-tiao Perng
885159
answered Apr 2 at 0:08
Cherng-tiao PerngCherng-tiao Perng
885159
885159
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 $a_{m,n}=s_{m+n}s_{m-n}$ $(m,nin mathbb{Z})$ 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
6,98445979
$endgroup$
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 $a_{m,n}=s_{m+n}s_{m-n}$ $(m,nin mathbb{Z})$ 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
6,98445979
$endgroup$
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 $a_{m,n}=s_{m+n}s_{m-n}$ $(m,nin mathbb{Z})$ 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
6,98445979
$endgroup$
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 $a_{m,n}=s_{m+n}s_{m-n}$ $(m,nin mathbb{Z})$ 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
6,98445979
edited 2 days ago
answered Apr 2 at 4:07
Alexey UstinovAlexey Ustinov
6,98445979
answered Apr 2 at 4:07
Alexey UstinovAlexey Ustinov
6,98445979
answered Apr 2 at 4:07
Alexey UstinovAlexey Ustinov
6,98445979
6,98445979
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
8,4222642
$endgroup$
$begingroup$
While probably historically interesting, this doesn't seem relevant to the question.
$endgroup$
– LSpice
2 days ago
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
2 days ago
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
8,4222642
$endgroup$
$begingroup$
While probably historically interesting, this doesn't seem relevant to the question.
$endgroup$
– LSpice
2 days ago
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
2 days ago
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
8,4222642
$endgroup$
"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
8,4222642
answered Apr 2 at 1:16
Ira GesselIra Gessel
8,4222642
answered Apr 2 at 1:16
Ira GesselIra Gessel
8,4222642
answered Apr 2 at 1:16
Ira GesselIra Gessel
8,4222642
8,4222642
$begingroup$
While probably historically interesting, this doesn't seem relevant to the question.
$endgroup$
– LSpice
2 days ago
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
2 days ago
add a comment |
$begingroup$
While probably historically interesting, this doesn't seem relevant to the question.
$endgroup$
– LSpice
2 days ago
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
2 days ago
$begingroup$
While probably historically interesting, this doesn't seem relevant to the question.
$endgroup$
– LSpice
2 days ago
$begingroup$
While probably historically interesting, this doesn't seem relevant to the question.
$endgroup$
– LSpice
2 days ago
2
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
2 days ago
$begingroup$
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
2 days ago
add a comment |
1
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This can be simplified to $F_{a - r}F_{b + k + 1} - F_{b - r} F_{a + k + 1} = (-1)^{a + r + 1} F_{b - a} F_{k + r + 1}$, 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 $F_a F_b - F_c F_{a+b-c} = (-1)^{a+1}F_{c-a}F_{b-c}$. 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
2 days ago