Is this a new Fibonacci Identity?
$begingroup$
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 7 hours ago
GrassiGrassi
11626
$endgroup$
|
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 7 hours ago
GrassiGrassi
11626
$endgroup$
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
7 hours ago
$begingroup$
It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
$endgroup$
– Grassi
7 hours ago
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
7 hours ago
1
$begingroup$
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
6 hours ago
$begingroup$
Yes, but from what I can tell this is no different than saying Vajda's identity is a disguised version of Catalan's identity. You can't go the other way without adding a variable.
$endgroup$
– Grassi
5 hours 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 7 hours ago
GrassiGrassi
11626
$endgroup$
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 7 hours ago
GrassiGrassi
11626
asked 7 hours ago
GrassiGrassi
11626
asked 7 hours ago
GrassiGrassi
11626
asked 7 hours ago
GrassiGrassi
11626
asked 7 hours ago
GrassiGrassi
11626
11626
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
7 hours ago
$begingroup$
It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
$endgroup$
– Grassi
7 hours ago
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
7 hours ago
1
$begingroup$
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
6 hours ago
$begingroup$
Yes, but from what I can tell this is no different than saying Vajda's identity is a disguised version of Catalan's identity. You can't go the other way without adding a variable.
$endgroup$
– Grassi
5 hours 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
7 hours ago
$begingroup$
It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
$endgroup$
– Grassi
7 hours ago
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
7 hours ago
1
$begingroup$
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
6 hours ago
$begingroup$
Yes, but from what I can tell this is no different than saying Vajda's identity is a disguised version of Catalan's identity. You can't go the other way without adding a variable.
$endgroup$
– Grassi
5 hours 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
7 hours ago
$begingroup$
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
7 hours ago
$begingroup$
It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
$endgroup$
– Grassi
7 hours ago
$begingroup$
It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
$endgroup$
– Grassi
7 hours ago
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
7 hours ago
$begingroup$
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
7 hours ago
1
1
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
$begingroup$
Yes, but from what I can tell this is no different than saying Vajda's identity is a disguised version of Catalan's identity. You can't go the other way without adding a variable.
$endgroup$
– Grassi
5 hours ago
$begingroup$
Yes, but from what I can tell this is no different than saying Vajda's identity is a disguised version of Catalan's identity. You can't go the other way without adding a variable.
$endgroup$
– Grassi
5 hours 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 5 hours ago
Cherng-tiao PerngCherng-tiao Perng
835148
$endgroup$
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 4 hours ago
Ira GesselIra Gessel
8,4122642
$endgroup$
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 1 hour ago
Alexey UstinovAlexey Ustinov
6,91445979
$endgroup$
add a comment |
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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 5 hours ago
Cherng-tiao PerngCherng-tiao Perng
835148
$endgroup$
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 5 hours ago
Cherng-tiao PerngCherng-tiao Perng
835148
$endgroup$
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 5 hours ago
Cherng-tiao PerngCherng-tiao Perng
835148
$endgroup$
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 5 hours ago
Cherng-tiao PerngCherng-tiao Perng
835148
answered 5 hours ago
Cherng-tiao PerngCherng-tiao Perng
835148
answered 5 hours ago
Cherng-tiao PerngCherng-tiao Perng
835148
answered 5 hours ago
Cherng-tiao PerngCherng-tiao Perng
835148
835148
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 4 hours ago
Ira GesselIra Gessel
8,4122642
$endgroup$
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 4 hours ago
Ira GesselIra Gessel
8,4122642
$endgroup$
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 4 hours ago
Ira GesselIra Gessel
8,4122642
$endgroup$
"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 4 hours ago
Ira GesselIra Gessel
8,4122642
answered 4 hours ago
Ira GesselIra Gessel
8,4122642
answered 4 hours ago
Ira GesselIra Gessel
8,4122642
answered 4 hours ago
Ira GesselIra Gessel
8,4122642
8,4122642
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 1 hour ago
Alexey UstinovAlexey Ustinov
6,91445979
$endgroup$
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 1 hour ago
Alexey UstinovAlexey Ustinov
6,91445979
$endgroup$
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 1 hour ago
Alexey UstinovAlexey Ustinov
6,91445979
$endgroup$
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 1 hour ago
Alexey UstinovAlexey Ustinov
6,91445979
edited 15 mins ago
answered 1 hour ago
Alexey UstinovAlexey Ustinov
6,91445979
answered 1 hour ago
Alexey UstinovAlexey Ustinov
6,91445979
answered 1 hour ago
Alexey UstinovAlexey Ustinov
6,91445979
6,91445979
add a comment |
add a comment |
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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
7 hours ago
$begingroup$
It doesn't have 6 variables? It has 5: $n, r, k, h, g$.
$endgroup$
– Grassi
7 hours ago
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
7 hours ago
1
$begingroup$
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
6 hours ago
$begingroup$
Yes, but from what I can tell this is no different than saying Vajda's identity is a disguised version of Catalan's identity. You can't go the other way without adding a variable.
$endgroup$
– Grassi
5 hours ago