Fibonacci summation proof using matrices?
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I have a standard proof for the theorem:
$$sum_^n f_1+f_3+f_5+...+f_2n-1 = f_2n$$
$$f_i$$ refers to the Fibonacci numbers for future reference.
It involves setting p(k) as p(k+1) and proving it through weak induction, it has been graded by my professor and is correct.
However, I have recently came across the fibonacci matrix formulation from here:
How to prove Fibonacci sequence with matrices?
I am curious how I would go about solving this theorem with matrices.
I have tried using the product operator: $$Pi$$
but I am not experienced enough to correctly formulate it so that they equal each other, for example:
$$Pi_i=1^n beginbmatrix1 & 1\1 & 0endbmatrix^2n-1 = beginbmatrix1 & 1\1 & 0endbmatrix^2n$$
Using the product operator may be completely pointless, but I honestly just don't know since I have never really used them before.
Any idea on how the original theorem is shown in matrices?
Thank you for any help in advance.
matrices proof-writing fibonacci-numbers
edited Sep 10 at 20:08
Bernard
112k636104
asked Sep 10 at 19:10
Antas Santas
41
 |Â
show 1 more comment
up vote
0
down vote
favorite
I have a standard proof for the theorem:
$$sum_^n f_1+f_3+f_5+...+f_2n-1 = f_2n$$
$$f_i$$ refers to the Fibonacci numbers for future reference.
It involves setting p(k) as p(k+1) and proving it through weak induction, it has been graded by my professor and is correct.
However, I have recently came across the fibonacci matrix formulation from here:
How to prove Fibonacci sequence with matrices?
I am curious how I would go about solving this theorem with matrices.
I have tried using the product operator: $$Pi$$
but I am not experienced enough to correctly formulate it so that they equal each other, for example:
$$Pi_i=1^n beginbmatrix1 & 1\1 & 0endbmatrix^2n-1 = beginbmatrix1 & 1\1 & 0endbmatrix^2n$$
Using the product operator may be completely pointless, but I honestly just don't know since I have never really used them before.
Any idea on how the original theorem is shown in matrices?
Thank you for any help in advance.
matrices proof-writing fibonacci-numbers
edited Sep 10 at 20:08
Bernard
112k636104
asked Sep 10 at 19:10
Antas Santas
41
Yes, $$f_i$$ refers to the fibonacci numbers, i'll edit that in. Thank you.
– Antas Santas
Sep 10 at 19:25
There is no $i$ in the summands of your sum $sum_i$.
– Dietrich Burde
Sep 10 at 19:27
@DietrichBurde That is a completely different Fibonacci identity.
– Mike Earnest
Sep 10 at 19:31
Ooops, will be corrected soon (this one is the right identity., but not yet with matrices).
– Dietrich Burde
Sep 10 at 19:31
One nice setup for the matrix proof is to begin by noting that $$ pmatrixf_1\f_0 + pmatrixf_3\f_2 + cdots + pmatrixf_2n-1\f_2n-2 = \ pmatrix1\0 + F^2 pmatrix1\0 + cdots + F^2n-2pmatrix1\0 $$ where $F = pmatrix1&1\1&0$
– Omnomnomnom
Sep 10 at 19:32
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a standard proof for the theorem:
$$sum_^n f_1+f_3+f_5+...+f_2n-1 = f_2n$$
$$f_i$$ refers to the Fibonacci numbers for future reference.
It involves setting p(k) as p(k+1) and proving it through weak induction, it has been graded by my professor and is correct.
However, I have recently came across the fibonacci matrix formulation from here:
How to prove Fibonacci sequence with matrices?
I am curious how I would go about solving this theorem with matrices.
I have tried using the product operator: $$Pi$$
but I am not experienced enough to correctly formulate it so that they equal each other, for example:
$$Pi_i=1^n beginbmatrix1 & 1\1 & 0endbmatrix^2n-1 = beginbmatrix1 & 1\1 & 0endbmatrix^2n$$
Using the product operator may be completely pointless, but I honestly just don't know since I have never really used them before.
Any idea on how the original theorem is shown in matrices?
Thank you for any help in advance.
matrices proof-writing fibonacci-numbers
edited Sep 10 at 20:08
Bernard
112k636104
asked Sep 10 at 19:10
Antas Santas
41
I have a standard proof for the theorem:
$$sum_^n f_1+f_3+f_5+...+f_2n-1 = f_2n$$
$$f_i$$ refers to the Fibonacci numbers for future reference.
It involves setting p(k) as p(k+1) and proving it through weak induction, it has been graded by my professor and is correct.
However, I have recently came across the fibonacci matrix formulation from here:
How to prove Fibonacci sequence with matrices?
I am curious how I would go about solving this theorem with matrices.
I have tried using the product operator: $$Pi$$
but I am not experienced enough to correctly formulate it so that they equal each other, for example:
$$Pi_i=1^n beginbmatrix1 & 1\1 & 0endbmatrix^2n-1 = beginbmatrix1 & 1\1 & 0endbmatrix^2n$$
Using the product operator may be completely pointless, but I honestly just don't know since I have never really used them before.
Any idea on how the original theorem is shown in matrices?
Thank you for any help in advance.
matrices proof-writing fibonacci-numbers
matrices proof-writing fibonacci-numbers
edited Sep 10 at 20:08
Bernard
112k636104
asked Sep 10 at 19:10
Antas Santas
41
edited Sep 10 at 20:08
Bernard
112k636104
asked Sep 10 at 19:10
Antas Santas
41
edited Sep 10 at 20:08
Bernard
112k636104
edited Sep 10 at 20:08
Bernard
112k636104
edited Sep 10 at 20:08
Bernard
112k636104
112k636104
asked Sep 10 at 19:10
Antas Santas
41
asked Sep 10 at 19:10
Antas Santas
41
asked Sep 10 at 19:10
Antas Santas
41
41
Yes, $$f_i$$ refers to the fibonacci numbers, i'll edit that in. Thank you.
– Antas Santas
Sep 10 at 19:25
There is no $i$ in the summands of your sum $sum_i$.
– Dietrich Burde
Sep 10 at 19:27
@DietrichBurde That is a completely different Fibonacci identity.
– Mike Earnest
Sep 10 at 19:31
Ooops, will be corrected soon (this one is the right identity., but not yet with matrices).
– Dietrich Burde
Sep 10 at 19:31
One nice setup for the matrix proof is to begin by noting that $$ pmatrixf_1\f_0 + pmatrixf_3\f_2 + cdots + pmatrixf_2n-1\f_2n-2 = \ pmatrix1\0 + F^2 pmatrix1\0 + cdots + F^2n-2pmatrix1\0 $$ where $F = pmatrix1&1\1&0$
– Omnomnomnom
Sep 10 at 19:32
 |Â
show 1 more comment
Yes, $$f_i$$ refers to the fibonacci numbers, i'll edit that in. Thank you.
– Antas Santas
Sep 10 at 19:25
There is no $i$ in the summands of your sum $sum_i$.
– Dietrich Burde
Sep 10 at 19:27
@DietrichBurde That is a completely different Fibonacci identity.
– Mike Earnest
Sep 10 at 19:31
Ooops, will be corrected soon (this one is the right identity., but not yet with matrices).
– Dietrich Burde
Sep 10 at 19:31
One nice setup for the matrix proof is to begin by noting that $$ pmatrixf_1\f_0 + pmatrixf_3\f_2 + cdots + pmatrixf_2n-1\f_2n-2 = \ pmatrix1\0 + F^2 pmatrix1\0 + cdots + F^2n-2pmatrix1\0 $$ where $F = pmatrix1&1\1&0$
– Omnomnomnom
Sep 10 at 19:32
Yes, $$f_i$$ refers to the fibonacci numbers, i'll edit that in. Thank you.
– Antas Santas
Sep 10 at 19:25
Yes, $$f_i$$ refers to the fibonacci numbers, i'll edit that in. Thank you.
– Antas Santas
Sep 10 at 19:25
There is no $i$ in the summands of your sum $sum_i$.
– Dietrich Burde
Sep 10 at 19:27
There is no $i$ in the summands of your sum $sum_i$.
– Dietrich Burde
Sep 10 at 19:27
@DietrichBurde That is a completely different Fibonacci identity.
– Mike Earnest
Sep 10 at 19:31
@DietrichBurde That is a completely different Fibonacci identity.
– Mike Earnest
Sep 10 at 19:31
Ooops, will be corrected soon (this one is the right identity., but not yet with matrices).
– Dietrich Burde
Sep 10 at 19:31
Ooops, will be corrected soon (this one is the right identity., but not yet with matrices).
– Dietrich Burde
Sep 10 at 19:31
One nice setup for the matrix proof is to begin by noting that $$ pmatrixf_1\f_0 + pmatrixf_3\f_2 + cdots + pmatrixf_2n-1\f_2n-2 = \ pmatrix1\0 + F^2 pmatrix1\0 + cdots + F^2n-2pmatrix1\0 $$ where $F = pmatrix1&1\1&0$
– Omnomnomnom
Sep 10 at 19:32
One nice setup for the matrix proof is to begin by noting that $$ pmatrixf_1\f_0 + pmatrixf_3\f_2 + cdots + pmatrixf_2n-1\f_2n-2 = \ pmatrix1\0 + F^2 pmatrix1\0 + cdots + F^2n-2pmatrix1\0 $$ where $F = pmatrix1&1\1&0$
– Omnomnomnom
Sep 10 at 19:32
 |Â
show 1 more comment
1 Answer
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up vote
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Letting $F=beginbmatrix1&1\1&0endbmatrix$, so $F^2=F+I$, then the finite geometric series formula gives
$$
I + F^2 + F^4 + dots + F^2n-2 = (F^2n-I)(F^2-I)^-1 = (F^2n-I)F^-1=F^2n-1-F^-1.
$$
Since $F^-1=beginbmatrix0&1\1&-1endbmatrix$, conclude by examining the upper left entry of the equation
$$
I + F^2 + F^4 + dots + F^2n-2 = F^2n-1-F^-1.
$$
and recalling that $F^n=beginbmatrixF_n+1&F_n\F_n&F_n-1endbmatrix$.
answered Sep 10 at 19:26
Mike Earnest
18.6k11950
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
Letting $F=beginbmatrix1&1\1&0endbmatrix$, so $F^2=F+I$, then the finite geometric series formula gives
$$
I + F^2 + F^4 + dots + F^2n-2 = (F^2n-I)(F^2-I)^-1 = (F^2n-I)F^-1=F^2n-1-F^-1.
$$
Since $F^-1=beginbmatrix0&1\1&-1endbmatrix$, conclude by examining the upper left entry of the equation
$$
I + F^2 + F^4 + dots + F^2n-2 = F^2n-1-F^-1.
$$
and recalling that $F^n=beginbmatrixF_n+1&F_n\F_n&F_n-1endbmatrix$.
answered Sep 10 at 19:26
Mike Earnest
18.6k11950
add a comment |Â
up vote
6
down vote
Letting $F=beginbmatrix1&1\1&0endbmatrix$, so $F^2=F+I$, then the finite geometric series formula gives
$$
I + F^2 + F^4 + dots + F^2n-2 = (F^2n-I)(F^2-I)^-1 = (F^2n-I)F^-1=F^2n-1-F^-1.
$$
Since $F^-1=beginbmatrix0&1\1&-1endbmatrix$, conclude by examining the upper left entry of the equation
$$
I + F^2 + F^4 + dots + F^2n-2 = F^2n-1-F^-1.
$$
and recalling that $F^n=beginbmatrixF_n+1&F_n\F_n&F_n-1endbmatrix$.
answered Sep 10 at 19:26
Mike Earnest
18.6k11950
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Letting $F=beginbmatrix1&1\1&0endbmatrix$, so $F^2=F+I$, then the finite geometric series formula gives
$$
I + F^2 + F^4 + dots + F^2n-2 = (F^2n-I)(F^2-I)^-1 = (F^2n-I)F^-1=F^2n-1-F^-1.
$$
Since $F^-1=beginbmatrix0&1\1&-1endbmatrix$, conclude by examining the upper left entry of the equation
$$
I + F^2 + F^4 + dots + F^2n-2 = F^2n-1-F^-1.
$$
and recalling that $F^n=beginbmatrixF_n+1&F_n\F_n&F_n-1endbmatrix$.
answered Sep 10 at 19:26
Mike Earnest
18.6k11950
Letting $F=beginbmatrix1&1\1&0endbmatrix$, so $F^2=F+I$, then the finite geometric series formula gives
$$
I + F^2 + F^4 + dots + F^2n-2 = (F^2n-I)(F^2-I)^-1 = (F^2n-I)F^-1=F^2n-1-F^-1.
$$
Since $F^-1=beginbmatrix0&1\1&-1endbmatrix$, conclude by examining the upper left entry of the equation
$$
I + F^2 + F^4 + dots + F^2n-2 = F^2n-1-F^-1.
$$
and recalling that $F^n=beginbmatrixF_n+1&F_n\F_n&F_n-1endbmatrix$.
answered Sep 10 at 19:26
Mike Earnest
18.6k11950
edited Sep 10 at 21:45
answered Sep 10 at 19:26
Mike Earnest
18.6k11950
answered Sep 10 at 19:26
Mike Earnest
18.6k11950
answered Sep 10 at 19:26
Mike Earnest
18.6k11950
18.6k11950
add a comment |Â
add a comment |Â
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Yes, $$f_i$$ refers to the fibonacci numbers, i'll edit that in. Thank you.
– Antas Santas
Sep 10 at 19:25
There is no $i$ in the summands of your sum $sum_i$.
– Dietrich Burde
Sep 10 at 19:27
@DietrichBurde That is a completely different Fibonacci identity.
– Mike Earnest
Sep 10 at 19:31
Ooops, will be corrected soon (this one is the right identity., but not yet with matrices).
– Dietrich Burde
Sep 10 at 19:31
One nice setup for the matrix proof is to begin by noting that $$ pmatrixf_1\f_0 + pmatrixf_3\f_2 + cdots + pmatrixf_2n-1\f_2n-2 = \ pmatrix1\0 + F^2 pmatrix1\0 + cdots + F^2n-2pmatrix1\0 $$ where $F = pmatrix1&1\1&0$
– Omnomnomnom
Sep 10 at 19:32