# Infinite series: When sum f(t) = sum f(t)g(t) ?

I was wandering if there are any theorems/ideas that would help me with the following situation (I came up with it myself and have been unable to find anything whatsoever).

Say you have two continuous functions $\;f,g : \mathbb{R}\to \mathbb{R}$ such that

$\displaystyle\sum_{t=1}^\infty f(t) = \sum_{t=1}^\infty f(t)g(t)$

Is there anything that can be said about the relationship between $f$ and $g$?

Trying to figure this out on my own has just led to me going around in circles. The idea of taking the integral of both occurred to me, but that seems to introduce too big of an error term. Also, I tried regarding $\;f$ and $g$ as vectors

i.e. $\;\;\;\vec{f}_2 = f(2)$

and multiplying them by the infinite-dimensional identity matrix $E$...then considering

trace$(E\vec{f}) =$ trace$(E\vec{f}\vec{g})$

But this doesn't seem to introduce anything new. I'm at a loss as to whether this is too general of an idea or if I'm missing something critical here. Any help would be greatly appreciated.

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Are you summing over $t\in\mathbb N$? If so, there is no difference between using continuous functions and using arbitrary sequences. – Alex Becker Sep 9 '12 at 0:52
Yes, I'm summing over $\N$. I guess I threw in continuity since it would allow for integration. But seeing as how it could oscillate wildly between any consecutive $t\in\N$ and integration probably isn't a good idea, continuity is pretty irrelevant. So dropping continuity is fine. – James000 Sep 9 '12 at 0:57
@JamesT you probably want to use \mathbb{N} or \Bbb{N} instead of \N ;) – user2468 Sep 9 '12 at 1:32

Well, $$\displaystyle\sum_{t=1}^\infty f(t) = \sum_{t=1}^\infty f(t)g(t)$$ is the same as $$\displaystyle 0 = \sum_{t=1}^\infty f(t)\big(g(t)-1\big)$$ so you could say that $f$ and $g-1$ are "orthogonal" perhaps.

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You can't really say much. For any fixed $f$, you have infinitely many degrees of freedom in choosing $g$ but only one constraint, leaving you with still infinitely many degrees of freedom.

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Ahh, thanks Alex. I feel kind of silly for not realizing that :). Do you happen to know if there are any additional constraints that would lead to something meaningful? The only obvious thing I can think of is that convergence of one implies convergence of the other...but that's really, really uninteresting. – James000 Sep 9 '12 at 1:06

Your choice of notation is terrible! Since $t \in \mathbb{N}$, I will replace it with $k$. I will also replace $f (t)$ and $g (t)$ with $a_k$ and $b_k$, respectively. Hence, we have that $a, b : \mathbb{N} \to \mathbb{R}$, and

$\displaystyle\sum_{k=1}^{\infty} a_k = \sum_{k=1}^{\infty} a_k \, b_k$.

If you know the sequence $(a_k)_{k \in \mathbb{N}}$, can you say anything about sequence $(b_k)_{k \in \mathbb{N}}$? Not really, because you have one linear equation in an infinite number of unknowns.

Imagine that we restrict $a, b$ to the domain $[n] := \{1,2,\dots,n\}$. We can associate with functions $a,b : [n] \to \mathbb{R}$ vectors $a, b \in \mathbb{R}^n$ and rewrite the sums in the more compressed form

$a^T 1_n = a^T b$

where $1_n$ is the $n$-dimensional vector whose components are all equal to $1$. You thus have one equation in $n$ unknowns, i.e., there are $n-1$ degrees of freedom. To determine $b$ you need a total of $n$ linearly independent equations, and you only have one.

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I don't think the choice of notation is terrible. Unconventional, yes, but it amounts to the same thing, and function calls tend to compose more nicely than subscripts. – Ben Millwood Sep 9 '12 at 11:03