How do I solve for a variable inside of a summation?

Question

So I have the following equation:

$$Q = 1 - \frac{6}{\pi^2} \displaystyle\sum_{n=0}^{\infty} \frac{1}{n^2} \cdot \exp\left(\frac {-Dn^2\pi^2t} {r^2}\right)$$

How would I solve for $D$ given the values of $Q, r,$ and $t$?

Answer 1

As there is no answer to this question, I share here my approach to express $D$ by itself - this might not be what you want. It feels a little bit wrong and is not complete, but might be useful - if not, than to show that it is a dead end. My comment here, that I upvoted your question because it was thought-provoking, but I could just as well downvote it, as a comment also says, you did not show your approach, so the below could be what you have already tried.

So my approach is to get rid of $exp(x)$ by letting $x = ln(y) \tag{1}$
so that $exp(x) = y$. As everything is a constant in the expression of $x$ except for $n$, let us define $y$ for the sake of simplicity as a function of $n$, and as the expression contains $D$, let us denote it also:
$y = f_D{(n)} \text{(D is in index, it hardly looks like that in my browser)} \tag{2}$
The index of $f$ is just a reminder for later that the definition contains $D$, but $f$ is the same for $\forall D$.
Using $(2)$, we can write $(1)$ as
$x = \frac{-Dn^2\pi^2t} {r^2} = ln(f_D{(n)}) \tag{3}$
and as you have $n^2$ in your formula, let's express it from $(3)$:
$n^2 = \frac{-ln(f_D{(n)})r^2} {D\pi^2t} \tag{4}$
Now let's substitute first $(3)$ to your formula to get
$Q = 1 - \frac{6}{\pi^2} \displaystyle\sum_{n=0}^{\infty} \frac{1}{n^2}f_D{(n)} \tag{5}$
and then substitute $(4)$ into $(5)$ to get
$Q = 1 - \frac{6}{\pi^2} \displaystyle\sum_{n=0}^{\infty} \frac{-D\pi^2t}{ln(f_D{(n)})r^2}f_D{(n)} \tag{6}$
and now factor out and cancel that we can (not dependent of $n$, so that the same in every term of the sum):
$Q = 1 + \frac{6Dt}{r^2} \displaystyle\sum_{n=0}^{\infty} \frac{f_D{(n)}}{ln(f_D{(n)})} \tag{7}$
Now at this point let's get rid of the infinite sum. This is where my approach is incomplete, but I did not have time to investigate this, and as I said above, you might have already tried this. So we have an infinite sum by $n$, that has $D$ as a kind of constant parameter, as the index reminds us so. If $D$ wasn't there, I assume the sum would be expressible by a closed-form expression, in which everything other than functions is a constant. Now as we have $D$, I assume we can define a function $g$, that calculates the infinite sum from the closed-form expression, and $g$ is the same for $\forall D$. Based on this, I define
$g{(D)} = \displaystyle\sum_{n=0}^{\infty} \frac{f_D{(n)}}{ln(f_D{(n)})} \tag{8}$
and substituting $(8)$ into $(7)$ we get
$Q = 1 + \frac{6Dt}{r^2} g{(D)} \tag{9}$
Expressing $D$ we get
$D = \frac{\left(Q - 1\right)r^2}{g{(D)}6t} \tag{10}$
which is expressing $D$ by itself, as we have $g{(D)}$ on the opposite side of the equality. We can do fixed-point iteration to converge to the value of $D$ - if $(10)$ converges, if not, than we might need $g^{-1}$, or variable substitution, as I explain here.

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Update:
Based on the logic of $(8)$ one could also do something trivial, and I document it here:
$h{(D)} = \displaystyle\sum_{n=0}^{\infty} \frac{1}{n^2} \cdot \exp\left(\frac {-Dn^2\pi^2t} {r^2}\right)$
and using it in your original equation we get
$Q = 1 - \frac{6}{\pi^2} h{(D)}$
$h{(D)} = \left(1 - Q\right)\frac {\pi^2} {6}$
$D = h^{-1}{(\left(1 - Q\right)\frac {\pi^2} {6})}$
I don't know if you have verified the existence of $h$ and $h^{-1}$.

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Update 2:
I think my last approach, also ends by using fixed-point iteration: it might be easier to get rid of the infinite sum, if the terms were in a form of $\frac{1}{ze^z}$. It can be achieved by letting $f^{'}_D{(n)} = \frac {Dn^2\pi^2t} {r^2}$,
and doing the following transformations on the original equation:
$Q = 1 - \frac{6}{\pi^2} \displaystyle\sum_{n=0}^{\infty} \frac{1}{n^2} \cdot \exp\left(\frac {-Dn^2\pi^2t} {r^2}\right) =$
$Q = 1 - 6 \displaystyle\sum_{n=0}^{\infty} \frac{1}{n^{2}\pi^2e^{\frac {Dn^2\pi^2t} {r^2}}} =$
$Q = 1 - \frac{6Dt}{r^2} \displaystyle\sum_{n=0}^{\infty} \frac{1}{\frac {Dn^2\pi^2t} {r^2}e^{\frac {Dn^2\pi^2t} {r^2}}}$
which using $f^{'}_D{(n)}$ becomes
$Q = 1 - \frac{6Dt}{r^2} \displaystyle\sum_{n=0}^{\infty} \frac{1}{f^{'}_D{(n)}e^{f^{'}_D{(n)}}}$
and let the infinite sum based on the explanation before $(8)$ above be
$g^{'}{(D)} = \displaystyle\sum_{n=0}^{\infty} \frac{1}{f^{'}_D{(n)}e^{f^{'}_D{(n)}}}$
so we get for $Q$ where we left off
$Q = 1 - \frac{6Dt}{r^2}g^{'}{(D)}$
and from this the final expression is similar to the one at my first approach:
$D = \frac{\left(1 - Q\right)r^2}{g^{'}{(D)}6t}$