0
$\begingroup$

I have the following equation

$\lambda_1 ln\left( \epsilon \alpha+1\right) = \lambda_2 ln\left( \epsilon \beta+1\right)$

All being known values but $\epsilon$, which I would like to clear but haven't been able so far.

Any hint is appreciated

[Edit] To add more background about it

This is a physics problem I'm trying to solve, starting with the Plank Law of Radiation

$R(\lambda, T) = \frac{2 \pi h c^2}{\lambda^5}\frac{1}{e^{\frac{hc}{\lambda K T}}-1}\epsilon$

For the same $T$ (unknown) and two $\lambda_1, \lambda_2, \lambda_1 \ne \lambda_2$ (known) I know the result of $R(\lambda, T)$. Being $\pi, h, c, K$ known constants I should be able to obtain $\epsilon$ from it.

Defining $\phi(\lambda) = \frac{2 \pi h c^2}{\lambda^5}$

$R(\lambda, T) = \phi(\lambda) \frac{1}{e^{\frac{hc}{\lambda K T}}-1}\epsilon$

$e^{\frac{hc}{\lambda K T}} = \epsilon \frac{\phi(\lambda)}{R(\lambda, t)}+1$

$\frac{hc}{\lambda K T} = ln \left ( \epsilon \frac{\phi(\lambda)}{R(\lambda, t)}+1 \right )$

Having $\lambda_1$ and $\lambda_2$ I divide both sides of the equation

$\frac{\lambda_2}{\lambda_1} = \frac{ln \left ( \epsilon \frac{\phi(\lambda_1)}{R(\lambda_1, t)}+1 \right )}{ln \left ( \epsilon \frac{\phi(\lambda_2)}{R(\lambda_2, t)}+1 \right )}$

$\endgroup$
  • $\begingroup$ By "clear" do you mean that you'd like to obtain some equation of the form $\epsilon=f(\lambda_1,\lambda_2,\alpha,\beta)$, for some function $f$? That is, are you just trying to solve for $\epsilon$ here? $\endgroup$ – Arbutus Jul 10 '18 at 15:17
  • $\begingroup$ Yes, @Arbutus. I wasn't entirely clear if that was the proper english for it. $\endgroup$ – aleperno Jul 10 '18 at 15:22
  • $\begingroup$ What sorts of things have you tried? Can you solve it if $\lambda_1=\lambda_2$ or $\alpha=\beta$? What about if $\lambda_1$ and $\lambda_2$ are integers? In the latter case (assuming $\lambda_i$ are positive as well), you can exponentiate and use the binomial theorem to obtain a polynomial in $\epsilon$ the roots of which will be your solution. $\endgroup$ – Arbutus Jul 10 '18 at 15:24
  • $\begingroup$ @Arbutus I've added more info on the problem. $\lambda_1 \ne \lambda_2$ and $\alpha \ne \beta$ afaik. $\endgroup$ – aleperno Jul 10 '18 at 15:37
  • $\begingroup$ @JohnGlenn if there's no way to get the exact value, yeah of course. $\endgroup$ – aleperno Jul 10 '18 at 15:45
1
$\begingroup$

As said in coments and answer, in the most general case where the equation cannot be reduced to a polynomial in $\epsilon$, the is no hope for an analytical solution and numerical methods would be required.

Consider that you look for the zero of function$$f(\epsilon)=\lambda_1 \log\left( \epsilon \alpha+1\right) - \lambda_2 \log\left( \epsilon \beta+1\right)$$ $$f'(\epsilon)=\frac{\alpha \lambda_1}{\alpha \epsilon +1}-\frac{\beta \lambda_2}{\beta \epsilon +1}$$ $$f''(\epsilon)=\frac{\beta ^2 \lambda_2}{(\beta \epsilon +1)^2}-\frac{\alpha ^2 \lambda_1}{(\alpha \epsilon +1)^2}$$ The first derivative cancels at $$\epsilon_*=\frac{\beta \lambda_2-\alpha \lambda_1}{\alpha \beta (\lambda_1-\lambda_2)}$$ If the arguments of the logarithms are small compared to $1$, we could use Taylor series and get as an estimate (beside the trivial $\epsilon=0$) $$\epsilon_{est}=2\frac{\alpha \lambda_1-\beta \lambda_2}{\alpha ^2 \lambda_1-\beta ^2 \lambda_2}$$

Admitting that $\epsilon_*>0$, Newton method would need to start above this value (say $\epsilon_0=2 \epsilon_*$).

For illustration purposes, let us use $\lambda_1=e$, $\lambda_2=\pi$, $\alpha=2$, $\beta=3$; these would give $\epsilon_*\approx 0.737$. So, let us start Newton using $\epsilon_0=1.5$ and get the following iterates $$\left( \begin{array}{cc} n & \epsilon_n \\ 0 & 1.500000000 \\ 1 & 4.664842315 \\ 2 & 5.046217899 \\ 3 & 5.056568407 \\ 4 & 5.056575597 \end{array} \right)$$

$\endgroup$
0
$\begingroup$

We can rearrange this equation as $(1+\epsilon\alpha)^{\lambda_1}=(1+\epsilon\beta)^{\lambda_2}$ by taking the multipliers into the exponent and exponentiating both sides. Looking the physics, let's assume all these constants are positive.

This equation is not straightforward to solve. There's one obvious solution of $\epsilon=0$ which holds for all values of other constants. For $\epsilon>0$, I took a glance of graphs for functions of the form $f(x)=(1+ax)^b$ for various values of $a,b,x>0$. These show monotonically increasing functions. If we can find places where the graphs intersect for different values of $a,b$, we find a solution. Let's take a look at some ranges of $\epsilon$ and see what we can determine.

For small $\epsilon$, $f(x)\approx ab \epsilon+1$ and for large $\epsilon$, $f(x)\approx (a \epsilon)^b$. So one function will others in the long run if it has a larger value of $b$, but will outpace another in the short run if it has a larger values of $a$. What this means for our equation is that we will find a non trivial solution if $\alpha>\beta$ but $\lambda_1 <\lambda_2$. A graphical example shows this to hold (in this case at least). non-trivial solution

As for finding a closed form solution of the equation, I doubt there's a way. The possibility of varying numbers of solutions suggests the solution has some complicated multi-valued behavior over some domain, and that is hard to analyze. If anyone knows anything about algebraic geometry, that might be of some use here. A more practical way of finding an answer would be numerically; find roots of the funtion $g(x)=(1+\epsilon\alpha)^{\lambda_1}-(1+\epsilon\beta)^{\lambda_2}$. Hope this helps!

$\endgroup$

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.