I'm trying to solve the following equation for $\delta$:

$$-0.01 = \frac{\ln(1+\delta)}{\ln{\delta}}$$

And I found that while Wolfram Alpha and Mathematica can give me numerical estimates ($\delta = 0.034300977...$), they cannot give me a closed form solution (even including the non-elementary functions that they'll sometimes use.)

Is there a closed-form solution to the problem above, or the more general version below? $$y = \frac{\ln(a+x)}{\ln(b+x)}$$

Or is this a case where there isn't a solution and moreover determining one would require a new function unrelated to those already established (such as the Lambert W function).

I have not taken Complex Analysis, or any other math higher than Differential Equations; however, depending on the complexity I may be able to follow along.

Full context and motivation:

A problem from my undergrad Nuclear Detection course: Given exponentially decaying signals of the form $f(t) = K\mathrm{e}^{-\mu t}$ that are generated periodically with period $x$. We can find the smallest $x$ such that signal pileup is less than 1%, we can see that $1.01 = \sum_n^\infty{\mathrm{e}^{-\mu n x}}$ which yields $x = \frac{\ln{101}}{\mu}$ after some massaging.

However, I noticed that in this instance, neglecting all but the first two terms of the series (ie, 1 and $\mathrm{e}^{-\mu x}$), gave the result $x^{\prime} = \frac{\ln{100}}{\mu}$ for an error percentage between $x^{\prime}$ and $x$ of only 0.216%. So I wanted to figure out what percentage of pileup was required before more than just these terms were required. I selected 1% as my threshold. So I had $$\frac{x}{x^{\prime}} = \frac{\ln(1+\delta) - \ln{\delta}}{-\ln{\delta}}$$ which gives the first formula of the question.

  • $\begingroup$ I think a numerical solution is as good as you can expect to do in this sort of situation. $\endgroup$ – Sean Lake Sep 4 '16 at 2:26
  • $\begingroup$ A numerical solution certainly is useful, and it's all I really need. I was merely curious if there was some sort of closed form solution to this more interesting than let $OMG(x; a, b) = \frac{\ln{a+x}}{\ln{b+x}}$. $\endgroup$ – OmnipotentEntity Sep 4 '16 at 2:32
  • $\begingroup$ Not really an answer, but I suspect that (depending on your choices for $a, b, x$) you'd most likely get a transcendental equation which would only be solvable numerically. $\endgroup$ – jamesh625 Sep 4 '16 at 2:48
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    $\begingroup$ The specific case described can, by exponentiating both sides appropriately, be brought to the form $\delta(1+\delta)^{100}=1$. So this is in fact a polynomial of high degree, and one is looking for its real roots (of which there is but one in this case). More generally, if $y$ is rational in the second equation, then the problem amounts to solving some polynomial equation. My guess is that this isn't going to have a closed-form solution in any reasonable sense. $\endgroup$ – Semiclassical Sep 6 '16 at 4:48

We have $$y=\frac{\ln(a+x)}{\ln(b+x)}\implies (b+x)^y=a+x\implies x=(b+x)^y-a$$ $$\implies x=(b+(b+(b+......+(b+x)^y-a)^y.....-a)^y-a)^y-a)$$ The actual solution may come differently in different values of a ,b and y. Now we come to the specific problem: $$\frac{-1}{100}=\frac{\ln(1+\delta)}{\ln \delta}$$ let $f(\delta)=\ln \delta$ then $$\frac{-1}{100}f(\delta)=f(\delta+1)$$ When you would solve this recurrence you would get something like this:- $$f(\delta)=(-1)^{(n+1)}\frac{f(\delta-n)}{100^{(n+1)}}$$ Now put the value of $f(\delta)$ and put $\lim_{{\delta-n}\to 0}$ such that $n \to 0$

  • $\begingroup$ I apologize, I ought to have been slightly more specific and formal with my terminology. I was seeking a closed form solution. I'm well aware of how to obtain this recurrence relation. $\endgroup$ – OmnipotentEntity Sep 5 '16 at 10:38
  • $\begingroup$ Try to solve the recurrence. No need to apologize $\endgroup$ – Mayank Deora Sep 5 '16 at 12:41
  • $\begingroup$ I have tried so solve the recurrence @OmnipotentEntity. Check the modified answer. $\endgroup$ – Mayank Deora Sep 6 '16 at 2:31
  • $\begingroup$ I have given the correct numerical solution in my original post, and it does not seem to match. $\endgroup$ – OmnipotentEntity Sep 6 '16 at 3:05

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