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When do there exist two functions that satisfy the equation?
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How can we pick two non-zero functions $f$ and $g$ so that for all integer $kge 0$
$$sum_n=1^infty f(n)g(n)^k = left(sum_n=1^infty f(n)g(n) right)^k$$
Assume $f$, $g$, and $k$ allow for convergence. Let neither $f$ or $g$ involve $k$. The lower index could start at zero (or anywhere else) if that helps.
Note that $k=1$ implies
$$sum_n=1^infty f(n) = 1$$
As @Kavi pointed out, solutions for $f$ are trivial when $g=1$. I would ideally wish for a general set of solutions, but I am definitely interested in any interesting not-obvious solutions.
Note: I made quite a few edits to simplify my question.
sequences-and-series functional-equations
asked Jul 15 at 22:22
tyobrien
1,098412
add a comment |Â
up vote
0
down vote
favorite
How can we pick two non-zero functions $f$ and $g$ so that for all integer $kge 0$
$$sum_n=1^infty f(n)g(n)^k = left(sum_n=1^infty f(n)g(n) right)^k$$
Assume $f$, $g$, and $k$ allow for convergence. Let neither $f$ or $g$ involve $k$. The lower index could start at zero (or anywhere else) if that helps.
Note that $k=1$ implies
$$sum_n=1^infty f(n) = 1$$
As @Kavi pointed out, solutions for $f$ are trivial when $g=1$. I would ideally wish for a general set of solutions, but I am definitely interested in any interesting not-obvious solutions.
Note: I made quite a few edits to simplify my question.
sequences-and-series functional-equations
asked Jul 15 at 22:22
tyobrien
1,098412
Can you specify any further about $k$? It seems like $f=h=0$ will do the job, depending on your restrictions for $k.$
– Cameron Buie
Jul 15 at 22:29
How about $g(n,s)equiv1 , f(n,s)=frac 1 2^n$?
– Kavi Rama Murthy
Jul 17 at 0:25
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How can we pick two non-zero functions $f$ and $g$ so that for all integer $kge 0$
$$sum_n=1^infty f(n)g(n)^k = left(sum_n=1^infty f(n)g(n) right)^k$$
Assume $f$, $g$, and $k$ allow for convergence. Let neither $f$ or $g$ involve $k$. The lower index could start at zero (or anywhere else) if that helps.
Note that $k=1$ implies
$$sum_n=1^infty f(n) = 1$$
As @Kavi pointed out, solutions for $f$ are trivial when $g=1$. I would ideally wish for a general set of solutions, but I am definitely interested in any interesting not-obvious solutions.
Note: I made quite a few edits to simplify my question.
sequences-and-series functional-equations
asked Jul 15 at 22:22
tyobrien
1,098412
How can we pick two non-zero functions $f$ and $g$ so that for all integer $kge 0$
$$sum_n=1^infty f(n)g(n)^k = left(sum_n=1^infty f(n)g(n) right)^k$$
Assume $f$, $g$, and $k$ allow for convergence. Let neither $f$ or $g$ involve $k$. The lower index could start at zero (or anywhere else) if that helps.
Note that $k=1$ implies
$$sum_n=1^infty f(n) = 1$$
As @Kavi pointed out, solutions for $f$ are trivial when $g=1$. I would ideally wish for a general set of solutions, but I am definitely interested in any interesting not-obvious solutions.
Note: I made quite a few edits to simplify my question.
sequences-and-series functional-equations
asked Jul 15 at 22:22
tyobrien
1,098412
edited Jul 17 at 4:11
asked Jul 15 at 22:22
tyobrien
1,098412
asked Jul 15 at 22:22
tyobrien
1,098412
asked Jul 15 at 22:22
tyobrien
1,098412
1,098412
Can you specify any further about $k$? It seems like $f=h=0$ will do the job, depending on your restrictions for $k.$
– Cameron Buie
Jul 15 at 22:29
How about $g(n,s)equiv1 , f(n,s)=frac 1 2^n$?
– Kavi Rama Murthy
Jul 17 at 0:25
add a comment |Â
Can you specify any further about $k$? It seems like $f=h=0$ will do the job, depending on your restrictions for $k.$
– Cameron Buie
Jul 15 at 22:29
How about $g(n,s)equiv1 , f(n,s)=frac 1 2^n$?
– Kavi Rama Murthy
Jul 17 at 0:25
Can you specify any further about $k$? It seems like $f=h=0$ will do the job, depending on your restrictions for $k.$
– Cameron Buie
Jul 15 at 22:29
Can you specify any further about $k$? It seems like $f=h=0$ will do the job, depending on your restrictions for $k.$
– Cameron Buie
Jul 15 at 22:29
How about $g(n,s)equiv1 , f(n,s)=frac 1 2^n$?
– Kavi Rama Murthy
Jul 17 at 0:25
How about $g(n,s)equiv1 , f(n,s)=frac 1 2^n$?
– Kavi Rama Murthy
Jul 17 at 0:25
add a comment |Â
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Can you specify any further about $k$? It seems like $f=h=0$ will do the job, depending on your restrictions for $k.$
– Cameron Buie
Jul 15 at 22:29
How about $g(n,s)equiv1 , f(n,s)=frac 1 2^n$?
– Kavi Rama Murthy
Jul 17 at 0:25