1
$\begingroup$

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 $d_i\ge1,n_i>0,x<0$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.