Is there a closed-form solution to the following equation?

Question

I would like to know if there is a closed-form solution for $x$ in the following equation. If there is no such form, how can you show this?

$$\frac{\sum{d_{i}^{x}\log(d_{i})}}{\sum{d_{i}^{x}}}=\frac{\sum{n_{i}\log(d_{i})}}{\sum{n_{i}}}$$

If it helps, we can assume that $d_i\ge1,n_i>0,x<0$

Answer 1

Note that it is notoriously difficult to prove that an expression can't be written in closed form in terms of elementary functions. But I certainly wouldn't wager on an equation involving the sum of arbitrary exponentials having one.

The left-hand side can be written as $\frac{d}{dt}\log \sum d_i^t\Bigg\vert_{t\to x}$, in case it somehow helps.

Answer 2

Except for very particular cases where a change of variable would reduce the problem to a polynomial of low degree, it is impossible to get a closed form solution for $x$ and numerical methods will be required. However, since some terms can be quite stiff, some preliminary care could be required.

Let us consider the expression$$A=\frac{\sum_{i=1}^N{d_{i}^{x}\log(d_{i})}}{\sum_{i=1}^N{d_{i}^{x}}}$$ Since everything is known beside $x$, let is rewrite it as $$A=\frac{\sum_{i=1}^N{\alpha_i e^{\alpha_i x}}}{\sum_{i=1}^N{e^{\alpha_i x}}}$$ using $\alpha_i=\log(d_i)$ and let us name $B$ the right hand side. This makes the equation to solve $$\sum_{i=1}^N{\alpha_i e^{\alpha_i x}}=B\sum_{i=1}^N{ e^{\alpha_i x}}$$ To make things more linear, take logarithms and define now the equation as $$f(x)=\log\Big(\sum_{i=1}^N{\alpha_i e^{\alpha_i x}} \Big)-\log\Big(\sum_{i=1}^N{ e^{\alpha_i x}} \Big)-\log(B)$$ A plot (as a function of $x$) should very simply reveal the approximate location of the root. At this point, use Newton method as a root finder.

For illustration purposes, I used $N=10$, $d_i=5i+3$, $n_i=3i+5$. The plot shows a root close to $x_0=1$ and Newton method iterates are $0.79793$, $0.810451$, $0.810503$ which is the solution for six significant figures.