Proving an infinite product for generating function












1












$begingroup$


The following is from chapter 1, problem 10 from Wilf's Generatingfunctionology. The first part of the problem states that you are given a function defined for $ngeq 1$ with the following relations, $f(1)=1, f(2n)=f(n), f(2n+1)=f(n)+f(n+1)$, with $$F(x)=sumlimits_{ngeq 1}f(n)x^{n-1}$$



and show that this satisfies the relation, $F(x)=(1+x+x^2)F(x^2)$. I have done this, it is the second part of the question I'm concerned about as it's slightly outside of the content covered in the chapter. It asks to prove that, $$F(x)=prodlimits_{jgeq 0}^{infty}left{1+x^{2^j}+x^{2^{j+1}}right}$$ Obviously, this is true for finite $n$ as we can inductively just keep applying the relation, ie $F(x^2)=(1+x^2+x^4)F(x^4)$, so plugging into the original gives, $F(x)=(1+x+x^2)(1+x^2+x^4)F(x^4)$ and so on, but I'm sort of at a loss how to prove this infinite product in general, or just a technique to do so. This book has solutions, but the explanation for this part is not given and just stated as "obvious", which it is not to me. Any help appreciated.






combinatorics generating-functions







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  • $begingroup$
    You have a "$+f(n+1)$" floating in the air in your question...
    $endgroup$
    – Jean Marie
    Dec 28 '18 at 9:15
















1












$begingroup$


The following is from chapter 1, problem 10 from Wilf's Generatingfunctionology. The first part of the problem states that you are given a function defined for $ngeq 1$ with the following relations, $f(1)=1, f(2n)=f(n), f(2n+1)=f(n)+f(n+1)$, with $$F(x)=sumlimits_{ngeq 1}f(n)x^{n-1}$$



and show that this satisfies the relation, $F(x)=(1+x+x^2)F(x^2)$. I have done this, it is the second part of the question I'm concerned about as it's slightly outside of the content covered in the chapter. It asks to prove that, $$F(x)=prodlimits_{jgeq 0}^{infty}left{1+x^{2^j}+x^{2^{j+1}}right}$$ Obviously, this is true for finite $n$ as we can inductively just keep applying the relation, ie $F(x^2)=(1+x^2+x^4)F(x^4)$, so plugging into the original gives, $F(x)=(1+x+x^2)(1+x^2+x^4)F(x^4)$ and so on, but I'm sort of at a loss how to prove this infinite product in general, or just a technique to do so. This book has solutions, but the explanation for this part is not given and just stated as "obvious", which it is not to me. Any help appreciated.






combinatorics generating-functions







share|cite|improve this question









$endgroup$












  • $begingroup$
    You have a "$+f(n+1)$" floating in the air in your question...
    $endgroup$
    – Jean Marie
    Dec 28 '18 at 9:15














1












1








1





$begingroup$


The following is from chapter 1, problem 10 from Wilf's Generatingfunctionology. The first part of the problem states that you are given a function defined for $ngeq 1$ with the following relations, $f(1)=1, f(2n)=f(n), f(2n+1)=f(n)+f(n+1)$, with $$F(x)=sumlimits_{ngeq 1}f(n)x^{n-1}$$



and show that this satisfies the relation, $F(x)=(1+x+x^2)F(x^2)$. I have done this, it is the second part of the question I'm concerned about as it's slightly outside of the content covered in the chapter. It asks to prove that, $$F(x)=prodlimits_{jgeq 0}^{infty}left{1+x^{2^j}+x^{2^{j+1}}right}$$ Obviously, this is true for finite $n$ as we can inductively just keep applying the relation, ie $F(x^2)=(1+x^2+x^4)F(x^4)$, so plugging into the original gives, $F(x)=(1+x+x^2)(1+x^2+x^4)F(x^4)$ and so on, but I'm sort of at a loss how to prove this infinite product in general, or just a technique to do so. This book has solutions, but the explanation for this part is not given and just stated as "obvious", which it is not to me. Any help appreciated.






combinatorics generating-functions







share|cite|improve this question









$endgroup$




The following is from chapter 1, problem 10 from Wilf's Generatingfunctionology. The first part of the problem states that you are given a function defined for $ngeq 1$ with the following relations, $f(1)=1, f(2n)=f(n), f(2n+1)=f(n)+f(n+1)$, with $$F(x)=sumlimits_{ngeq 1}f(n)x^{n-1}$$



and show that this satisfies the relation, $F(x)=(1+x+x^2)F(x^2)$. I have done this, it is the second part of the question I'm concerned about as it's slightly outside of the content covered in the chapter. It asks to prove that, $$F(x)=prodlimits_{jgeq 0}^{infty}left{1+x^{2^j}+x^{2^{j+1}}right}$$ Obviously, this is true for finite $n$ as we can inductively just keep applying the relation, ie $F(x^2)=(1+x^2+x^4)F(x^4)$, so plugging into the original gives, $F(x)=(1+x+x^2)(1+x^2+x^4)F(x^4)$ and so on, but I'm sort of at a loss how to prove this infinite product in general, or just a technique to do so. This book has solutions, but the explanation for this part is not given and just stated as "obvious", which it is not to me. Any help appreciated.






combinatorics generating-functions




combinatorics generating-functions






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asked Dec 28 '18 at 6:12









Jimmy2GoonsJimmy2Goons

16015




16015












  • $begingroup$
    You have a "$+f(n+1)$" floating in the air in your question...
    $endgroup$
    – Jean Marie
    Dec 28 '18 at 9:15


















  • $begingroup$
    You have a "$+f(n+1)$" floating in the air in your question...
    $endgroup$
    – Jean Marie
    Dec 28 '18 at 9:15
















$begingroup$
You have a "$+f(n+1)$" floating in the air in your question...
$endgroup$
– Jean Marie
Dec 28 '18 at 9:15




$begingroup$
You have a "$+f(n+1)$" floating in the air in your question...
$endgroup$
– Jean Marie
Dec 28 '18 at 9:15










1 Answer
1






active

oldest

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2












$begingroup$

Two parts:




  1. Does the given product satisfy the relationship?


  2. Is it a unique solution? I.e. can there be more than one function which satisfies the relationship and also evaluates to $f(1)$ at $x=0$? Suppose $F(x) = (1 + x+ x^2)F(x^2)$, $G(x) = (1 + x + x^2)G(x^2)$, $F(0) = G(0) = 1$. Then $F(x) - G(x) = (1 + x + x^2)left(F(x^2) - G(x^2)right)$ and you can do the same expansion. Then suppose that the first coefficient of $x^a$ which differs is finite, and derive a contradiction by expanding $k + lg a$ times.







share|cite|improve this answer









$endgroup$













  • $begingroup$
    For uniqueness, you might also consider $H(x) = F(x) / G(x)$, so $H(x) = H(x^2)$ for all $x$. If $H$ is continuous this implies $H(x)$ is a constant, and since $H(1) = 1$, $H(x) = 1$ for all $x$.
    $endgroup$
    – awkward
    Dec 28 '18 at 13:34










  • $begingroup$
    I played around with a few ideas like that, but I'm not sure how to justify the assumption that $H$ is continuous.
    $endgroup$
    – Peter Taylor
    Dec 28 '18 at 14:13











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1 Answer
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2












$begingroup$

Two parts:




  1. Does the given product satisfy the relationship?


  2. Is it a unique solution? I.e. can there be more than one function which satisfies the relationship and also evaluates to $f(1)$ at $x=0$? Suppose $F(x) = (1 + x+ x^2)F(x^2)$, $G(x) = (1 + x + x^2)G(x^2)$, $F(0) = G(0) = 1$. Then $F(x) - G(x) = (1 + x + x^2)left(F(x^2) - G(x^2)right)$ and you can do the same expansion. Then suppose that the first coefficient of $x^a$ which differs is finite, and derive a contradiction by expanding $k + lg a$ times.







share|cite|improve this answer









$endgroup$













  • $begingroup$
    For uniqueness, you might also consider $H(x) = F(x) / G(x)$, so $H(x) = H(x^2)$ for all $x$. If $H$ is continuous this implies $H(x)$ is a constant, and since $H(1) = 1$, $H(x) = 1$ for all $x$.
    $endgroup$
    – awkward
    Dec 28 '18 at 13:34










  • $begingroup$
    I played around with a few ideas like that, but I'm not sure how to justify the assumption that $H$ is continuous.
    $endgroup$
    – Peter Taylor
    Dec 28 '18 at 14:13
















2












$begingroup$

Two parts:




  1. Does the given product satisfy the relationship?


  2. Is it a unique solution? I.e. can there be more than one function which satisfies the relationship and also evaluates to $f(1)$ at $x=0$? Suppose $F(x) = (1 + x+ x^2)F(x^2)$, $G(x) = (1 + x + x^2)G(x^2)$, $F(0) = G(0) = 1$. Then $F(x) - G(x) = (1 + x + x^2)left(F(x^2) - G(x^2)right)$ and you can do the same expansion. Then suppose that the first coefficient of $x^a$ which differs is finite, and derive a contradiction by expanding $k + lg a$ times.







share|cite|improve this answer









$endgroup$













  • $begingroup$
    For uniqueness, you might also consider $H(x) = F(x) / G(x)$, so $H(x) = H(x^2)$ for all $x$. If $H$ is continuous this implies $H(x)$ is a constant, and since $H(1) = 1$, $H(x) = 1$ for all $x$.
    $endgroup$
    – awkward
    Dec 28 '18 at 13:34










  • $begingroup$
    I played around with a few ideas like that, but I'm not sure how to justify the assumption that $H$ is continuous.
    $endgroup$
    – Peter Taylor
    Dec 28 '18 at 14:13














2












2








2





$begingroup$

Two parts:




  1. Does the given product satisfy the relationship?


  2. Is it a unique solution? I.e. can there be more than one function which satisfies the relationship and also evaluates to $f(1)$ at $x=0$? Suppose $F(x) = (1 + x+ x^2)F(x^2)$, $G(x) = (1 + x + x^2)G(x^2)$, $F(0) = G(0) = 1$. Then $F(x) - G(x) = (1 + x + x^2)left(F(x^2) - G(x^2)right)$ and you can do the same expansion. Then suppose that the first coefficient of $x^a$ which differs is finite, and derive a contradiction by expanding $k + lg a$ times.







share|cite|improve this answer









$endgroup$



Two parts:




  1. Does the given product satisfy the relationship?


  2. Is it a unique solution? I.e. can there be more than one function which satisfies the relationship and also evaluates to $f(1)$ at $x=0$? Suppose $F(x) = (1 + x+ x^2)F(x^2)$, $G(x) = (1 + x + x^2)G(x^2)$, $F(0) = G(0) = 1$. Then $F(x) - G(x) = (1 + x + x^2)left(F(x^2) - G(x^2)right)$ and you can do the same expansion. Then suppose that the first coefficient of $x^a$ which differs is finite, and derive a contradiction by expanding $k + lg a$ times.








share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 28 '18 at 12:22









Peter TaylorPeter Taylor

9,12712343




9,12712343












  • $begingroup$
    For uniqueness, you might also consider $H(x) = F(x) / G(x)$, so $H(x) = H(x^2)$ for all $x$. If $H$ is continuous this implies $H(x)$ is a constant, and since $H(1) = 1$, $H(x) = 1$ for all $x$.
    $endgroup$
    – awkward
    Dec 28 '18 at 13:34










  • $begingroup$
    I played around with a few ideas like that, but I'm not sure how to justify the assumption that $H$ is continuous.
    $endgroup$
    – Peter Taylor
    Dec 28 '18 at 14:13


















  • $begingroup$
    For uniqueness, you might also consider $H(x) = F(x) / G(x)$, so $H(x) = H(x^2)$ for all $x$. If $H$ is continuous this implies $H(x)$ is a constant, and since $H(1) = 1$, $H(x) = 1$ for all $x$.
    $endgroup$
    – awkward
    Dec 28 '18 at 13:34










  • $begingroup$
    I played around with a few ideas like that, but I'm not sure how to justify the assumption that $H$ is continuous.
    $endgroup$
    – Peter Taylor
    Dec 28 '18 at 14:13
















$begingroup$
For uniqueness, you might also consider $H(x) = F(x) / G(x)$, so $H(x) = H(x^2)$ for all $x$. If $H$ is continuous this implies $H(x)$ is a constant, and since $H(1) = 1$, $H(x) = 1$ for all $x$.
$endgroup$
– awkward
Dec 28 '18 at 13:34




$begingroup$
For uniqueness, you might also consider $H(x) = F(x) / G(x)$, so $H(x) = H(x^2)$ for all $x$. If $H$ is continuous this implies $H(x)$ is a constant, and since $H(1) = 1$, $H(x) = 1$ for all $x$.
$endgroup$
– awkward
Dec 28 '18 at 13:34












$begingroup$
I played around with a few ideas like that, but I'm not sure how to justify the assumption that $H$ is continuous.
$endgroup$
– Peter Taylor
Dec 28 '18 at 14:13




$begingroup$
I played around with a few ideas like that, but I'm not sure how to justify the assumption that $H$ is continuous.
$endgroup$
– Peter Taylor
Dec 28 '18 at 14:13


















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