Generating function for Fibonacci-like sequence
$begingroup$
I have been reading the book Irresistible Integrals lately, and it inspired this problem.
For some constants $a,bin Bbb R$ we define the sequence $S(a,b)={e_n(a,b):ninBbb Z_{geq0}}$ by the recurrence
$$e_{n+2}(a,b)=r_{n+2}=ar_{n+1}+br_{n}qquad ninBbb N; ngeq2$$
With initial conditions $$r_0=r_1=1$$
We then immediately see that the Fibonacci numbers are given by the case
$${F_n}=S(1,1)$$
I have been struggling to find a closed form for
$$E(x;a,b)=R(x)=sum_{ngeq0}r_nx^n$$
Since $$F(x)=E(x;1,1)=frac{x}{1-x-x^2}$$
I have very little doubt that a general closed form exists.
My attempts:
$$r_{n+2}=ar_{n+1}+br_n$$
$$r_{n+2}x^n=ar_{n+1}x^n+br_nx^n$$
$$sum_{ngeq0}r_{n+2}x^n=asum_{ngeq0}r_{n+1}x^n+bR(x)$$
If I'm not mistaken,
$$sum_{ngeq0}r_{n+2}x^n=frac1{x^2}(R(x)-1-x)$$
and similarly
$$sum_{ngeq0}r_{n+1}x^n=frac1x(R(x)-1)$$
Since I fear I have made a really simple algebra mistake that I have failed to catch, I will show all of my steps.
$$frac1{x^2}(R(x)-1-x)=frac{a}x(R(x)-1)+bR(x)$$
$$R(x)-1-x=axR(x)-ax+bx^2R(x)$$
$$(1-ax-bx^2)R(x)=(1-a)x+1$$
$$R(x)=E(x;a,b)=frac{(1-a)x+1}{1-ax-bx^2}$$
But plugging in $a=b=1$ shows that I messed up, because
$$frac{1}{1-x-x^2}neqfrac{x}{1-x-x^2}$$
Did I do something wrong? What is the correct generating function $E(x;a,b)$ and how do you find it?
Thanks.
sequences-and-series elementary-number-theory power-series generating-functions fibonacci-numbers
asked Jan 7 at 17:53
clathratusclathratus
4,563337
$endgroup$
|
show 1 more comment
$begingroup$
I have been reading the book Irresistible Integrals lately, and it inspired this problem.
For some constants $a,bin Bbb R$ we define the sequence $S(a,b)={e_n(a,b):ninBbb Z_{geq0}}$ by the recurrence
$$e_{n+2}(a,b)=r_{n+2}=ar_{n+1}+br_{n}qquad ninBbb N; ngeq2$$
With initial conditions $$r_0=r_1=1$$
We then immediately see that the Fibonacci numbers are given by the case
$${F_n}=S(1,1)$$
I have been struggling to find a closed form for
$$E(x;a,b)=R(x)=sum_{ngeq0}r_nx^n$$
Since $$F(x)=E(x;1,1)=frac{x}{1-x-x^2}$$
I have very little doubt that a general closed form exists.
My attempts:
$$r_{n+2}=ar_{n+1}+br_n$$
$$r_{n+2}x^n=ar_{n+1}x^n+br_nx^n$$
$$sum_{ngeq0}r_{n+2}x^n=asum_{ngeq0}r_{n+1}x^n+bR(x)$$
If I'm not mistaken,
$$sum_{ngeq0}r_{n+2}x^n=frac1{x^2}(R(x)-1-x)$$
and similarly
$$sum_{ngeq0}r_{n+1}x^n=frac1x(R(x)-1)$$
Since I fear I have made a really simple algebra mistake that I have failed to catch, I will show all of my steps.
$$frac1{x^2}(R(x)-1-x)=frac{a}x(R(x)-1)+bR(x)$$
$$R(x)-1-x=axR(x)-ax+bx^2R(x)$$
$$(1-ax-bx^2)R(x)=(1-a)x+1$$
$$R(x)=E(x;a,b)=frac{(1-a)x+1}{1-ax-bx^2}$$
But plugging in $a=b=1$ shows that I messed up, because
$$frac{1}{1-x-x^2}neqfrac{x}{1-x-x^2}$$
Did I do something wrong? What is the correct generating function $E(x;a,b)$ and how do you find it?
Thanks.
sequences-and-series elementary-number-theory power-series generating-functions fibonacci-numbers
asked Jan 7 at 17:53
clathratusclathratus
4,563337
$endgroup$
$begingroup$
If $F_0=F_1=1$ then $sum_0^infty F_nx^n=frac1{1-x-x^2}$.
$endgroup$
– Lord Shark the Unknown
Jan 7 at 17:55
$begingroup$
@LordSharktheUnknown So I did it right?
$endgroup$
– clathratus
Jan 7 at 17:58
$begingroup$
But you got $frac x{1-x-x^2}$.
$endgroup$
– Lord Shark the Unknown
Jan 7 at 18:00
$begingroup$
@LordSharktheUnknown that's what the book gave me
$endgroup$
– clathratus
Jan 7 at 18:02
1
$begingroup$
@JeanMarie Less I think...
$endgroup$
– clathratus
Jan 7 at 18:06
|
show 1 more comment
$begingroup$
I have been reading the book Irresistible Integrals lately, and it inspired this problem.
For some constants $a,bin Bbb R$ we define the sequence $S(a,b)={e_n(a,b):ninBbb Z_{geq0}}$ by the recurrence
$$e_{n+2}(a,b)=r_{n+2}=ar_{n+1}+br_{n}qquad ninBbb N; ngeq2$$
With initial conditions $$r_0=r_1=1$$
We then immediately see that the Fibonacci numbers are given by the case
$${F_n}=S(1,1)$$
I have been struggling to find a closed form for
$$E(x;a,b)=R(x)=sum_{ngeq0}r_nx^n$$
Since $$F(x)=E(x;1,1)=frac{x}{1-x-x^2}$$
I have very little doubt that a general closed form exists.
My attempts:
$$r_{n+2}=ar_{n+1}+br_n$$
$$r_{n+2}x^n=ar_{n+1}x^n+br_nx^n$$
$$sum_{ngeq0}r_{n+2}x^n=asum_{ngeq0}r_{n+1}x^n+bR(x)$$
If I'm not mistaken,
$$sum_{ngeq0}r_{n+2}x^n=frac1{x^2}(R(x)-1-x)$$
and similarly
$$sum_{ngeq0}r_{n+1}x^n=frac1x(R(x)-1)$$
Since I fear I have made a really simple algebra mistake that I have failed to catch, I will show all of my steps.
$$frac1{x^2}(R(x)-1-x)=frac{a}x(R(x)-1)+bR(x)$$
$$R(x)-1-x=axR(x)-ax+bx^2R(x)$$
$$(1-ax-bx^2)R(x)=(1-a)x+1$$
$$R(x)=E(x;a,b)=frac{(1-a)x+1}{1-ax-bx^2}$$
But plugging in $a=b=1$ shows that I messed up, because
$$frac{1}{1-x-x^2}neqfrac{x}{1-x-x^2}$$
Did I do something wrong? What is the correct generating function $E(x;a,b)$ and how do you find it?
Thanks.
sequences-and-series elementary-number-theory power-series generating-functions fibonacci-numbers
asked Jan 7 at 17:53
clathratusclathratus
4,563337
$endgroup$
I have been reading the book Irresistible Integrals lately, and it inspired this problem.
For some constants $a,bin Bbb R$ we define the sequence $S(a,b)={e_n(a,b):ninBbb Z_{geq0}}$ by the recurrence
$$e_{n+2}(a,b)=r_{n+2}=ar_{n+1}+br_{n}qquad ninBbb N; ngeq2$$
With initial conditions $$r_0=r_1=1$$
We then immediately see that the Fibonacci numbers are given by the case
$${F_n}=S(1,1)$$
I have been struggling to find a closed form for
$$E(x;a,b)=R(x)=sum_{ngeq0}r_nx^n$$
Since $$F(x)=E(x;1,1)=frac{x}{1-x-x^2}$$
I have very little doubt that a general closed form exists.
My attempts:
$$r_{n+2}=ar_{n+1}+br_n$$
$$r_{n+2}x^n=ar_{n+1}x^n+br_nx^n$$
$$sum_{ngeq0}r_{n+2}x^n=asum_{ngeq0}r_{n+1}x^n+bR(x)$$
If I'm not mistaken,
$$sum_{ngeq0}r_{n+2}x^n=frac1{x^2}(R(x)-1-x)$$
and similarly
$$sum_{ngeq0}r_{n+1}x^n=frac1x(R(x)-1)$$
Since I fear I have made a really simple algebra mistake that I have failed to catch, I will show all of my steps.
$$frac1{x^2}(R(x)-1-x)=frac{a}x(R(x)-1)+bR(x)$$
$$R(x)-1-x=axR(x)-ax+bx^2R(x)$$
$$(1-ax-bx^2)R(x)=(1-a)x+1$$
$$R(x)=E(x;a,b)=frac{(1-a)x+1}{1-ax-bx^2}$$
But plugging in $a=b=1$ shows that I messed up, because
$$frac{1}{1-x-x^2}neqfrac{x}{1-x-x^2}$$
Did I do something wrong? What is the correct generating function $E(x;a,b)$ and how do you find it?
Thanks.
sequences-and-series elementary-number-theory power-series generating-functions fibonacci-numbers
sequences-and-series elementary-number-theory power-series generating-functions fibonacci-numbers
asked Jan 7 at 17:53
clathratusclathratus
4,563337
asked Jan 7 at 17:53
clathratusclathratus
4,563337
edited Jan 11 at 1:09
clathratus
asked Jan 7 at 17:53
clathratusclathratus
4,563337
asked Jan 7 at 17:53
clathratusclathratus
4,563337
asked Jan 7 at 17:53
clathratusclathratus
4,563337
4,563337
$begingroup$
If $F_0=F_1=1$ then $sum_0^infty F_nx^n=frac1{1-x-x^2}$.
$endgroup$
– Lord Shark the Unknown
Jan 7 at 17:55
$begingroup$
@LordSharktheUnknown So I did it right?
$endgroup$
– clathratus
Jan 7 at 17:58
$begingroup$
But you got $frac x{1-x-x^2}$.
$endgroup$
– Lord Shark the Unknown
Jan 7 at 18:00
$begingroup$
@LordSharktheUnknown that's what the book gave me
$endgroup$
– clathratus
Jan 7 at 18:02
1
$begingroup$
@JeanMarie Less I think...
$endgroup$
– clathratus
Jan 7 at 18:06
|
show 1 more comment
$begingroup$
If $F_0=F_1=1$ then $sum_0^infty F_nx^n=frac1{1-x-x^2}$.
$endgroup$
– Lord Shark the Unknown
Jan 7 at 17:55
$begingroup$
@LordSharktheUnknown So I did it right?
$endgroup$
– clathratus
Jan 7 at 17:58
$begingroup$
But you got $frac x{1-x-x^2}$.
$endgroup$
– Lord Shark the Unknown
Jan 7 at 18:00
$begingroup$
@LordSharktheUnknown that's what the book gave me
$endgroup$
– clathratus
Jan 7 at 18:02
1
$begingroup$
@JeanMarie Less I think...
$endgroup$
– clathratus
Jan 7 at 18:06
$begingroup$
If $F_0=F_1=1$ then $sum_0^infty F_nx^n=frac1{1-x-x^2}$.
$endgroup$
– Lord Shark the Unknown
Jan 7 at 17:55
$begingroup$
If $F_0=F_1=1$ then $sum_0^infty F_nx^n=frac1{1-x-x^2}$.
$endgroup$
– Lord Shark the Unknown
Jan 7 at 17:55
$begingroup$
@LordSharktheUnknown So I did it right?
$endgroup$
– clathratus
Jan 7 at 17:58
$begingroup$
@LordSharktheUnknown So I did it right?
$endgroup$
– clathratus
Jan 7 at 17:58
$begingroup$
But you got $frac x{1-x-x^2}$.
$endgroup$
– Lord Shark the Unknown
Jan 7 at 18:00
$begingroup$
But you got $frac x{1-x-x^2}$.
$endgroup$
– Lord Shark the Unknown
Jan 7 at 18:00
$begingroup$
@LordSharktheUnknown that's what the book gave me
$endgroup$
– clathratus
Jan 7 at 18:02
$begingroup$
@LordSharktheUnknown that's what the book gave me
$endgroup$
– clathratus
Jan 7 at 18:02
1
1
$begingroup$
@JeanMarie Less I think...
$endgroup$
– clathratus
Jan 7 at 18:06
$begingroup$
@JeanMarie Less I think...
$endgroup$
– clathratus
Jan 7 at 18:06
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
As Lord Shark the Unknown noted, your mistake is not in any of this algebraic manipulation, but rather in mixing up two common definitions of the Fibonacci sequence, only one of which gives a factor of $x$ in the numerator. Either way, sanity checks are always your friend. Expanding $frac{x}{1-x-x^2}$ as a power series clearly gives $x+cdots$; it can't start with a non-zero constant term, since it vanishes at $x=0$. But suppose we stick with that definition of $F_n$: then in the question's notation, ${F_{n+1}}=S(1,,1)$. In other words, your general result is consistent with the Fibonacci sanity check you attempted after all.
answered Jan 7 at 18:00
J.G.J.G.
27.1k22843
$endgroup$
$begingroup$
Thank you! I can see why they are called sanity checks
$endgroup$
– clathratus
Jan 7 at 18:03
1
$begingroup$
@clathratus There's also a name for the error we caught this time.
$endgroup$
– J.G.
Jan 7 at 18:04
$begingroup$
Do you know if my general generating function is correct?
$endgroup$
– clathratus
Jan 7 at 18:07
1
$begingroup$
@clathratus It looks right, yes. Apart from me following your proof, we can verify its series begins $1+x$ as expected, which is enough to determine the rest of the series.
$endgroup$
– J.G.
Jan 7 at 18:12
add a comment |
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$begingroup$
As Lord Shark the Unknown noted, your mistake is not in any of this algebraic manipulation, but rather in mixing up two common definitions of the Fibonacci sequence, only one of which gives a factor of $x$ in the numerator. Either way, sanity checks are always your friend. Expanding $frac{x}{1-x-x^2}$ as a power series clearly gives $x+cdots$; it can't start with a non-zero constant term, since it vanishes at $x=0$. But suppose we stick with that definition of $F_n$: then in the question's notation, ${F_{n+1}}=S(1,,1)$. In other words, your general result is consistent with the Fibonacci sanity check you attempted after all.
answered Jan 7 at 18:00
J.G.J.G.
27.1k22843
$endgroup$
$begingroup$
Thank you! I can see why they are called sanity checks
$endgroup$
– clathratus
Jan 7 at 18:03
1
$begingroup$
@clathratus There's also a name for the error we caught this time.
$endgroup$
– J.G.
Jan 7 at 18:04
$begingroup$
Do you know if my general generating function is correct?
$endgroup$
– clathratus
Jan 7 at 18:07
1
$begingroup$
@clathratus It looks right, yes. Apart from me following your proof, we can verify its series begins $1+x$ as expected, which is enough to determine the rest of the series.
$endgroup$
– J.G.
Jan 7 at 18:12
add a comment |
$begingroup$
As Lord Shark the Unknown noted, your mistake is not in any of this algebraic manipulation, but rather in mixing up two common definitions of the Fibonacci sequence, only one of which gives a factor of $x$ in the numerator. Either way, sanity checks are always your friend. Expanding $frac{x}{1-x-x^2}$ as a power series clearly gives $x+cdots$; it can't start with a non-zero constant term, since it vanishes at $x=0$. But suppose we stick with that definition of $F_n$: then in the question's notation, ${F_{n+1}}=S(1,,1)$. In other words, your general result is consistent with the Fibonacci sanity check you attempted after all.
answered Jan 7 at 18:00
J.G.J.G.
27.1k22843
$endgroup$
$begingroup$
Thank you! I can see why they are called sanity checks
$endgroup$
– clathratus
Jan 7 at 18:03
1
$begingroup$
@clathratus There's also a name for the error we caught this time.
$endgroup$
– J.G.
Jan 7 at 18:04
$begingroup$
Do you know if my general generating function is correct?
$endgroup$
– clathratus
Jan 7 at 18:07
1
$begingroup$
@clathratus It looks right, yes. Apart from me following your proof, we can verify its series begins $1+x$ as expected, which is enough to determine the rest of the series.
$endgroup$
– J.G.
Jan 7 at 18:12
add a comment |
$begingroup$
As Lord Shark the Unknown noted, your mistake is not in any of this algebraic manipulation, but rather in mixing up two common definitions of the Fibonacci sequence, only one of which gives a factor of $x$ in the numerator. Either way, sanity checks are always your friend. Expanding $frac{x}{1-x-x^2}$ as a power series clearly gives $x+cdots$; it can't start with a non-zero constant term, since it vanishes at $x=0$. But suppose we stick with that definition of $F_n$: then in the question's notation, ${F_{n+1}}=S(1,,1)$. In other words, your general result is consistent with the Fibonacci sanity check you attempted after all.
answered Jan 7 at 18:00
J.G.J.G.
27.1k22843
$endgroup$
As Lord Shark the Unknown noted, your mistake is not in any of this algebraic manipulation, but rather in mixing up two common definitions of the Fibonacci sequence, only one of which gives a factor of $x$ in the numerator. Either way, sanity checks are always your friend. Expanding $frac{x}{1-x-x^2}$ as a power series clearly gives $x+cdots$; it can't start with a non-zero constant term, since it vanishes at $x=0$. But suppose we stick with that definition of $F_n$: then in the question's notation, ${F_{n+1}}=S(1,,1)$. In other words, your general result is consistent with the Fibonacci sanity check you attempted after all.
answered Jan 7 at 18:00
J.G.J.G.
27.1k22843
answered Jan 7 at 18:00
J.G.J.G.
27.1k22843
answered Jan 7 at 18:00
J.G.J.G.
27.1k22843
answered Jan 7 at 18:00
J.G.J.G.
27.1k22843
27.1k22843
$begingroup$
Thank you! I can see why they are called sanity checks
$endgroup$
– clathratus
Jan 7 at 18:03
1
$begingroup$
@clathratus There's also a name for the error we caught this time.
$endgroup$
– J.G.
Jan 7 at 18:04
$begingroup$
Do you know if my general generating function is correct?
$endgroup$
– clathratus
Jan 7 at 18:07
1
$begingroup$
@clathratus It looks right, yes. Apart from me following your proof, we can verify its series begins $1+x$ as expected, which is enough to determine the rest of the series.
$endgroup$
– J.G.
Jan 7 at 18:12
add a comment |
$begingroup$
Thank you! I can see why they are called sanity checks
$endgroup$
– clathratus
Jan 7 at 18:03
1
$begingroup$
@clathratus There's also a name for the error we caught this time.
$endgroup$
– J.G.
Jan 7 at 18:04
$begingroup$
Do you know if my general generating function is correct?
$endgroup$
– clathratus
Jan 7 at 18:07
1
$begingroup$
@clathratus It looks right, yes. Apart from me following your proof, we can verify its series begins $1+x$ as expected, which is enough to determine the rest of the series.
$endgroup$
– J.G.
Jan 7 at 18:12
$begingroup$
Thank you! I can see why they are called sanity checks
$endgroup$
– clathratus
Jan 7 at 18:03
$begingroup$
Thank you! I can see why they are called sanity checks
$endgroup$
– clathratus
Jan 7 at 18:03
1
1
$begingroup$
@clathratus There's also a name for the error we caught this time.
$endgroup$
– J.G.
Jan 7 at 18:04
$begingroup$
@clathratus There's also a name for the error we caught this time.
$endgroup$
– J.G.
Jan 7 at 18:04
$begingroup$
Do you know if my general generating function is correct?
$endgroup$
– clathratus
Jan 7 at 18:07
$begingroup$
Do you know if my general generating function is correct?
$endgroup$
– clathratus
Jan 7 at 18:07
1
1
$begingroup$
@clathratus It looks right, yes. Apart from me following your proof, we can verify its series begins $1+x$ as expected, which is enough to determine the rest of the series.
$endgroup$
– J.G.
Jan 7 at 18:12
$begingroup$
@clathratus It looks right, yes. Apart from me following your proof, we can verify its series begins $1+x$ as expected, which is enough to determine the rest of the series.
$endgroup$
– J.G.
Jan 7 at 18:12
add a comment |
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$begingroup$
If $F_0=F_1=1$ then $sum_0^infty F_nx^n=frac1{1-x-x^2}$.
$endgroup$
– Lord Shark the Unknown
Jan 7 at 17:55
$begingroup$
@LordSharktheUnknown So I did it right?
$endgroup$
– clathratus
Jan 7 at 17:58
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But you got $frac x{1-x-x^2}$.
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– Lord Shark the Unknown
Jan 7 at 18:00
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@LordSharktheUnknown that's what the book gave me
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– clathratus
Jan 7 at 18:02
1
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@JeanMarie Less I think...
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– clathratus
Jan 7 at 18:06