generating function for $F_5n/(5F_n)$On the generating function of the Fibonacci numbersFibonacci Generating Function of a Complex VariableGenerating function for kind of sum of Fibonacci numbersQuestion about generating function of kind of fibonacci partial sumHow is the Binet's formula for Fibonacci reversed in order to find the index for a given Fibonacci number?Proof about specific sum of Fibonacci numbersExponential generating function and FibonacciLet $ninmathbbN$. Show that $F_n ne 2F_m$ for any $mge3.$Solve using generating functions the following recurrenceHelp with Partial Differential Equation Arising from Generating Function
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generating function for $F_5n/(5F_n)$
On the generating function of the Fibonacci numbersFibonacci Generating Function of a Complex VariableGenerating function for kind of sum of Fibonacci numbersQuestion about generating function of kind of fibonacci partial sumHow is the Binet's formula for Fibonacci reversed in order to find the index for a given Fibonacci number?Proof about specific sum of Fibonacci numbersExponential generating function and FibonacciLet $ninmathbbN$. Show that $F_n ne 2F_m$ for any $mge3.$Solve using generating functions the following recurrenceHelp with Partial Differential Equation Arising from Generating Function
$begingroup$
Let $F_n$ be the $n$-th Fibonacci number with $F_0=0$, $F_1=1$, and $F_k=F_k-1+F_k-2$ for $kge2$. Prove
$$sum^infty_n=0a_nx^n=frac1-4x-9x^2+6x^3+x^41-5x-15x^2+15x^3+5x^4-x^5$$ where $a_n=F_5n/(5F_n).$
generating-functions fibonacci-numbers
asked Mar 10 at 19:51
Qiang LiQiang Li
1,80722638
$endgroup$
add a comment |
$begingroup$
Let $F_n$ be the $n$-th Fibonacci number with $F_0=0$, $F_1=1$, and $F_k=F_k-1+F_k-2$ for $kge2$. Prove
$$sum^infty_n=0a_nx^n=frac1-4x-9x^2+6x^3+x^41-5x-15x^2+15x^3+5x^4-x^5$$ where $a_n=F_5n/(5F_n).$
generating-functions fibonacci-numbers
asked Mar 10 at 19:51
Qiang LiQiang Li
1,80722638
$endgroup$
$begingroup$
You can do it fairly mechanically using Binet's formula $F_n = fracphi^n - varphi^nphi - varphi$, although you might consider that cheating. More interesting might be to find a combinatorial proof.
$endgroup$
– Qiaochu Yuan
Mar 10 at 21:58
$begingroup$
This sequence satisfies the recurrence relation $a_n+4 - 4a_n+3 - 19a_n+2 - 4a_n+1 + a_n + 5 = 0$.
$endgroup$
– Chip Hurst
Mar 11 at 0:01
$begingroup$
@ChipHurst how have you derived that?
$endgroup$
– Qiang Li
Mar 11 at 2:11
add a comment |
$begingroup$
Let $F_n$ be the $n$-th Fibonacci number with $F_0=0$, $F_1=1$, and $F_k=F_k-1+F_k-2$ for $kge2$. Prove
$$sum^infty_n=0a_nx^n=frac1-4x-9x^2+6x^3+x^41-5x-15x^2+15x^3+5x^4-x^5$$ where $a_n=F_5n/(5F_n).$
generating-functions fibonacci-numbers
asked Mar 10 at 19:51
Qiang LiQiang Li
1,80722638
$endgroup$
Let $F_n$ be the $n$-th Fibonacci number with $F_0=0$, $F_1=1$, and $F_k=F_k-1+F_k-2$ for $kge2$. Prove
$$sum^infty_n=0a_nx^n=frac1-4x-9x^2+6x^3+x^41-5x-15x^2+15x^3+5x^4-x^5$$ where $a_n=F_5n/(5F_n).$
generating-functions fibonacci-numbers
generating-functions fibonacci-numbers
asked Mar 10 at 19:51
Qiang LiQiang Li
1,80722638
asked Mar 10 at 19:51
Qiang LiQiang Li
1,80722638
asked Mar 10 at 19:51
Qiang LiQiang Li
1,80722638
asked Mar 10 at 19:51
Qiang LiQiang Li
1,80722638
asked Mar 10 at 19:51
Qiang LiQiang Li
1,80722638
1,80722638
$begingroup$
You can do it fairly mechanically using Binet's formula $F_n = fracphi^n - varphi^nphi - varphi$, although you might consider that cheating. More interesting might be to find a combinatorial proof.
$endgroup$
– Qiaochu Yuan
Mar 10 at 21:58
$begingroup$
This sequence satisfies the recurrence relation $a_n+4 - 4a_n+3 - 19a_n+2 - 4a_n+1 + a_n + 5 = 0$.
$endgroup$
– Chip Hurst
Mar 11 at 0:01
$begingroup$
@ChipHurst how have you derived that?
$endgroup$
– Qiang Li
Mar 11 at 2:11
add a comment |
$begingroup$
You can do it fairly mechanically using Binet's formula $F_n = fracphi^n - varphi^nphi - varphi$, although you might consider that cheating. More interesting might be to find a combinatorial proof.
$endgroup$
– Qiaochu Yuan
Mar 10 at 21:58
$begingroup$
This sequence satisfies the recurrence relation $a_n+4 - 4a_n+3 - 19a_n+2 - 4a_n+1 + a_n + 5 = 0$.
$endgroup$
– Chip Hurst
Mar 11 at 0:01
$begingroup$
@ChipHurst how have you derived that?
$endgroup$
– Qiang Li
Mar 11 at 2:11
$begingroup$
You can do it fairly mechanically using Binet's formula $F_n = fracphi^n - varphi^nphi - varphi$, although you might consider that cheating. More interesting might be to find a combinatorial proof.
$endgroup$
– Qiaochu Yuan
Mar 10 at 21:58
$begingroup$
You can do it fairly mechanically using Binet's formula $F_n = fracphi^n - varphi^nphi - varphi$, although you might consider that cheating. More interesting might be to find a combinatorial proof.
$endgroup$
– Qiaochu Yuan
Mar 10 at 21:58
$begingroup$
This sequence satisfies the recurrence relation $a_n+4 - 4a_n+3 - 19a_n+2 - 4a_n+1 + a_n + 5 = 0$.
$endgroup$
– Chip Hurst
Mar 11 at 0:01
$begingroup$
This sequence satisfies the recurrence relation $a_n+4 - 4a_n+3 - 19a_n+2 - 4a_n+1 + a_n + 5 = 0$.
$endgroup$
– Chip Hurst
Mar 11 at 0:01
$begingroup$
@ChipHurst how have you derived that?
$endgroup$
– Qiang Li
Mar 11 at 2:11
$begingroup$
@ChipHurst how have you derived that?
$endgroup$
– Qiang Li
Mar 11 at 2:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This seems quite tough so I'm thinking you must have some results from a course text book that are relevant. Anyway, tackling it from the back end, the given generating function's denominator can be factorised;
$$frac1-4x-9x^2+6x^3+x^41-5x-15x^2+15x^3+5x^4-x^5$$
$$=frac1-4x-9x^2+6x^3+x^4(x^2-7x+1)(x^2+3x+1)(1-x)$$
and then partial fractions yield,
$$frac2-7x5(x^2-7x+1)+frac3x+25(x^2+3x+1)+frac15(1-x)$$
These quadratic denominators in a generating function scream Fibonacci;
$$frac1x^2+3x+1$$
$$=1-3x+8x^2-21x^3+55x^4-144x^5+dots$$
$$=F_2-F_4x+F_6x^2-F_8x^3+F_10x^4-F_12x^5+dots$$
and
$$frac1x^2-7x+1$$
$$=1+7x+48x^2+329x^3+2255x^4+15456x^5+105937x^6+726103x^7+dots$$
$$=fracF_43+fracF_83x+fracF_123x^2+fracF_163x^3+fracF_203x^3+dots$$
and these and their ilk in themselves would be interesting to explore.
Moving to the front end, $F_5n$ is always a multiple of 5, so I can see why the 'divide by 5' is in there, and it's in the partial fraction decomposition too. This can be proved by induction, and is a well known result; e.g. http://www.math.utep.edu/Faculty/duval/class/2325/104/fib.pdf
The $F_5n$ can be broken down into an expression involving $F_n$ and $F_n+1$ with the neat algorithm on page four on the following document;
http://www.maths.qmul.ac.uk/~pjc/comb/ch4s.pdf
Wikipedia has this decomposition for $F_4n$
$$F_4n=4F_nF_n+1(F_n+1^2+2F_n^2)-3F_n^2(F_n^2+2F_n+1^2)$$
I found an $F_5n$ decomposition (Falcon, Plaza 2007) as;
$$F_5n=F_n+1^5+4F_n^5-F_n-1^5+10F_n+1F_n^3F_n-1$$
I don't know if that's going to help.
There is an interesting result;
$$fracF_ktF_t=sum_i=0^(k-3)/2(-1)^itL_(k-2i-1)t+(-1)^frac(k-1)t2$$
k is odd and > 2 and L is the Lucas number. (Vajda 85)
This and 300 more Fibonacci relationships are at http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormulae.html
It may not be relevant but I note that the convolution of
$$frac1x^2+3x+1 times frac1x^2-7x+1$$
is
$$(1.1)+(1.7+(-3).1)x+(1.48+(-3).7+8.1)x^2+(1.329+(-3).48+8.7+(-21).1)x^2$$
$$+(1.2255+(-3).329+8.48+(-21).7+55.1)x^3$$
That is
$$1+4x+35x^2+220x^3+10556x^4+dots$$
and the $frac1(1-x)$ will be taking the partial sums of this.
Update : $F_n^4$ has the generating function
$$frac1-4x-4x^2+x^31-5x-15x^2+15x^3+5x^4-x^5$$
answered Mar 10 at 20:11
Martin HansenMartin Hansen
39213
$endgroup$
add a comment |
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$begingroup$
This seems quite tough so I'm thinking you must have some results from a course text book that are relevant. Anyway, tackling it from the back end, the given generating function's denominator can be factorised;
$$frac1-4x-9x^2+6x^3+x^41-5x-15x^2+15x^3+5x^4-x^5$$
$$=frac1-4x-9x^2+6x^3+x^4(x^2-7x+1)(x^2+3x+1)(1-x)$$
and then partial fractions yield,
$$frac2-7x5(x^2-7x+1)+frac3x+25(x^2+3x+1)+frac15(1-x)$$
These quadratic denominators in a generating function scream Fibonacci;
$$frac1x^2+3x+1$$
$$=1-3x+8x^2-21x^3+55x^4-144x^5+dots$$
$$=F_2-F_4x+F_6x^2-F_8x^3+F_10x^4-F_12x^5+dots$$
and
$$frac1x^2-7x+1$$
$$=1+7x+48x^2+329x^3+2255x^4+15456x^5+105937x^6+726103x^7+dots$$
$$=fracF_43+fracF_83x+fracF_123x^2+fracF_163x^3+fracF_203x^3+dots$$
and these and their ilk in themselves would be interesting to explore.
Moving to the front end, $F_5n$ is always a multiple of 5, so I can see why the 'divide by 5' is in there, and it's in the partial fraction decomposition too. This can be proved by induction, and is a well known result; e.g. http://www.math.utep.edu/Faculty/duval/class/2325/104/fib.pdf
The $F_5n$ can be broken down into an expression involving $F_n$ and $F_n+1$ with the neat algorithm on page four on the following document;
http://www.maths.qmul.ac.uk/~pjc/comb/ch4s.pdf
Wikipedia has this decomposition for $F_4n$
$$F_4n=4F_nF_n+1(F_n+1^2+2F_n^2)-3F_n^2(F_n^2+2F_n+1^2)$$
I found an $F_5n$ decomposition (Falcon, Plaza 2007) as;
$$F_5n=F_n+1^5+4F_n^5-F_n-1^5+10F_n+1F_n^3F_n-1$$
I don't know if that's going to help.
There is an interesting result;
$$fracF_ktF_t=sum_i=0^(k-3)/2(-1)^itL_(k-2i-1)t+(-1)^frac(k-1)t2$$
k is odd and > 2 and L is the Lucas number. (Vajda 85)
This and 300 more Fibonacci relationships are at http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormulae.html
It may not be relevant but I note that the convolution of
$$frac1x^2+3x+1 times frac1x^2-7x+1$$
is
$$(1.1)+(1.7+(-3).1)x+(1.48+(-3).7+8.1)x^2+(1.329+(-3).48+8.7+(-21).1)x^2$$
$$+(1.2255+(-3).329+8.48+(-21).7+55.1)x^3$$
That is
$$1+4x+35x^2+220x^3+10556x^4+dots$$
and the $frac1(1-x)$ will be taking the partial sums of this.
Update : $F_n^4$ has the generating function
$$frac1-4x-4x^2+x^31-5x-15x^2+15x^3+5x^4-x^5$$
answered Mar 10 at 20:11
Martin HansenMartin Hansen
39213
$endgroup$
add a comment |
$begingroup$
This seems quite tough so I'm thinking you must have some results from a course text book that are relevant. Anyway, tackling it from the back end, the given generating function's denominator can be factorised;
$$frac1-4x-9x^2+6x^3+x^41-5x-15x^2+15x^3+5x^4-x^5$$
$$=frac1-4x-9x^2+6x^3+x^4(x^2-7x+1)(x^2+3x+1)(1-x)$$
and then partial fractions yield,
$$frac2-7x5(x^2-7x+1)+frac3x+25(x^2+3x+1)+frac15(1-x)$$
These quadratic denominators in a generating function scream Fibonacci;
$$frac1x^2+3x+1$$
$$=1-3x+8x^2-21x^3+55x^4-144x^5+dots$$
$$=F_2-F_4x+F_6x^2-F_8x^3+F_10x^4-F_12x^5+dots$$
and
$$frac1x^2-7x+1$$
$$=1+7x+48x^2+329x^3+2255x^4+15456x^5+105937x^6+726103x^7+dots$$
$$=fracF_43+fracF_83x+fracF_123x^2+fracF_163x^3+fracF_203x^3+dots$$
and these and their ilk in themselves would be interesting to explore.
Moving to the front end, $F_5n$ is always a multiple of 5, so I can see why the 'divide by 5' is in there, and it's in the partial fraction decomposition too. This can be proved by induction, and is a well known result; e.g. http://www.math.utep.edu/Faculty/duval/class/2325/104/fib.pdf
The $F_5n$ can be broken down into an expression involving $F_n$ and $F_n+1$ with the neat algorithm on page four on the following document;
http://www.maths.qmul.ac.uk/~pjc/comb/ch4s.pdf
Wikipedia has this decomposition for $F_4n$
$$F_4n=4F_nF_n+1(F_n+1^2+2F_n^2)-3F_n^2(F_n^2+2F_n+1^2)$$
I found an $F_5n$ decomposition (Falcon, Plaza 2007) as;
$$F_5n=F_n+1^5+4F_n^5-F_n-1^5+10F_n+1F_n^3F_n-1$$
I don't know if that's going to help.
There is an interesting result;
$$fracF_ktF_t=sum_i=0^(k-3)/2(-1)^itL_(k-2i-1)t+(-1)^frac(k-1)t2$$
k is odd and > 2 and L is the Lucas number. (Vajda 85)
This and 300 more Fibonacci relationships are at http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormulae.html
It may not be relevant but I note that the convolution of
$$frac1x^2+3x+1 times frac1x^2-7x+1$$
is
$$(1.1)+(1.7+(-3).1)x+(1.48+(-3).7+8.1)x^2+(1.329+(-3).48+8.7+(-21).1)x^2$$
$$+(1.2255+(-3).329+8.48+(-21).7+55.1)x^3$$
That is
$$1+4x+35x^2+220x^3+10556x^4+dots$$
and the $frac1(1-x)$ will be taking the partial sums of this.
Update : $F_n^4$ has the generating function
$$frac1-4x-4x^2+x^31-5x-15x^2+15x^3+5x^4-x^5$$
answered Mar 10 at 20:11
Martin HansenMartin Hansen
39213
$endgroup$
add a comment |
$begingroup$
This seems quite tough so I'm thinking you must have some results from a course text book that are relevant. Anyway, tackling it from the back end, the given generating function's denominator can be factorised;
$$frac1-4x-9x^2+6x^3+x^41-5x-15x^2+15x^3+5x^4-x^5$$
$$=frac1-4x-9x^2+6x^3+x^4(x^2-7x+1)(x^2+3x+1)(1-x)$$
and then partial fractions yield,
$$frac2-7x5(x^2-7x+1)+frac3x+25(x^2+3x+1)+frac15(1-x)$$
These quadratic denominators in a generating function scream Fibonacci;
$$frac1x^2+3x+1$$
$$=1-3x+8x^2-21x^3+55x^4-144x^5+dots$$
$$=F_2-F_4x+F_6x^2-F_8x^3+F_10x^4-F_12x^5+dots$$
and
$$frac1x^2-7x+1$$
$$=1+7x+48x^2+329x^3+2255x^4+15456x^5+105937x^6+726103x^7+dots$$
$$=fracF_43+fracF_83x+fracF_123x^2+fracF_163x^3+fracF_203x^3+dots$$
and these and their ilk in themselves would be interesting to explore.
Moving to the front end, $F_5n$ is always a multiple of 5, so I can see why the 'divide by 5' is in there, and it's in the partial fraction decomposition too. This can be proved by induction, and is a well known result; e.g. http://www.math.utep.edu/Faculty/duval/class/2325/104/fib.pdf
The $F_5n$ can be broken down into an expression involving $F_n$ and $F_n+1$ with the neat algorithm on page four on the following document;
http://www.maths.qmul.ac.uk/~pjc/comb/ch4s.pdf
Wikipedia has this decomposition for $F_4n$
$$F_4n=4F_nF_n+1(F_n+1^2+2F_n^2)-3F_n^2(F_n^2+2F_n+1^2)$$
I found an $F_5n$ decomposition (Falcon, Plaza 2007) as;
$$F_5n=F_n+1^5+4F_n^5-F_n-1^5+10F_n+1F_n^3F_n-1$$
I don't know if that's going to help.
There is an interesting result;
$$fracF_ktF_t=sum_i=0^(k-3)/2(-1)^itL_(k-2i-1)t+(-1)^frac(k-1)t2$$
k is odd and > 2 and L is the Lucas number. (Vajda 85)
This and 300 more Fibonacci relationships are at http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormulae.html
It may not be relevant but I note that the convolution of
$$frac1x^2+3x+1 times frac1x^2-7x+1$$
is
$$(1.1)+(1.7+(-3).1)x+(1.48+(-3).7+8.1)x^2+(1.329+(-3).48+8.7+(-21).1)x^2$$
$$+(1.2255+(-3).329+8.48+(-21).7+55.1)x^3$$
That is
$$1+4x+35x^2+220x^3+10556x^4+dots$$
and the $frac1(1-x)$ will be taking the partial sums of this.
Update : $F_n^4$ has the generating function
$$frac1-4x-4x^2+x^31-5x-15x^2+15x^3+5x^4-x^5$$
answered Mar 10 at 20:11
Martin HansenMartin Hansen
39213
$endgroup$
This seems quite tough so I'm thinking you must have some results from a course text book that are relevant. Anyway, tackling it from the back end, the given generating function's denominator can be factorised;
$$frac1-4x-9x^2+6x^3+x^41-5x-15x^2+15x^3+5x^4-x^5$$
$$=frac1-4x-9x^2+6x^3+x^4(x^2-7x+1)(x^2+3x+1)(1-x)$$
and then partial fractions yield,
$$frac2-7x5(x^2-7x+1)+frac3x+25(x^2+3x+1)+frac15(1-x)$$
These quadratic denominators in a generating function scream Fibonacci;
$$frac1x^2+3x+1$$
$$=1-3x+8x^2-21x^3+55x^4-144x^5+dots$$
$$=F_2-F_4x+F_6x^2-F_8x^3+F_10x^4-F_12x^5+dots$$
and
$$frac1x^2-7x+1$$
$$=1+7x+48x^2+329x^3+2255x^4+15456x^5+105937x^6+726103x^7+dots$$
$$=fracF_43+fracF_83x+fracF_123x^2+fracF_163x^3+fracF_203x^3+dots$$
and these and their ilk in themselves would be interesting to explore.
Moving to the front end, $F_5n$ is always a multiple of 5, so I can see why the 'divide by 5' is in there, and it's in the partial fraction decomposition too. This can be proved by induction, and is a well known result; e.g. http://www.math.utep.edu/Faculty/duval/class/2325/104/fib.pdf
The $F_5n$ can be broken down into an expression involving $F_n$ and $F_n+1$ with the neat algorithm on page four on the following document;
http://www.maths.qmul.ac.uk/~pjc/comb/ch4s.pdf
Wikipedia has this decomposition for $F_4n$
$$F_4n=4F_nF_n+1(F_n+1^2+2F_n^2)-3F_n^2(F_n^2+2F_n+1^2)$$
I found an $F_5n$ decomposition (Falcon, Plaza 2007) as;
$$F_5n=F_n+1^5+4F_n^5-F_n-1^5+10F_n+1F_n^3F_n-1$$
I don't know if that's going to help.
There is an interesting result;
$$fracF_ktF_t=sum_i=0^(k-3)/2(-1)^itL_(k-2i-1)t+(-1)^frac(k-1)t2$$
k is odd and > 2 and L is the Lucas number. (Vajda 85)
This and 300 more Fibonacci relationships are at http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormulae.html
It may not be relevant but I note that the convolution of
$$frac1x^2+3x+1 times frac1x^2-7x+1$$
is
$$(1.1)+(1.7+(-3).1)x+(1.48+(-3).7+8.1)x^2+(1.329+(-3).48+8.7+(-21).1)x^2$$
$$+(1.2255+(-3).329+8.48+(-21).7+55.1)x^3$$
That is
$$1+4x+35x^2+220x^3+10556x^4+dots$$
and the $frac1(1-x)$ will be taking the partial sums of this.
Update : $F_n^4$ has the generating function
$$frac1-4x-4x^2+x^31-5x-15x^2+15x^3+5x^4-x^5$$
answered Mar 10 at 20:11
Martin HansenMartin Hansen
39213
edited 2 days ago
answered Mar 10 at 20:11
Martin HansenMartin Hansen
39213
answered Mar 10 at 20:11
Martin HansenMartin Hansen
39213
answered Mar 10 at 20:11
Martin HansenMartin Hansen
39213
39213
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$begingroup$
You can do it fairly mechanically using Binet's formula $F_n = fracphi^n - varphi^nphi - varphi$, although you might consider that cheating. More interesting might be to find a combinatorial proof.
$endgroup$
– Qiaochu Yuan
Mar 10 at 21:58
$begingroup$
This sequence satisfies the recurrence relation $a_n+4 - 4a_n+3 - 19a_n+2 - 4a_n+1 + a_n + 5 = 0$.
$endgroup$
– Chip Hurst
Mar 11 at 0:01
$begingroup$
@ChipHurst how have you derived that?
$endgroup$
– Qiang Li
Mar 11 at 2:11