Numerical Solution of $\frac{x}{1-e^{-x}} -5 = 0$

Question

I am working on a problem at the moment which cuts down to the following question:

How do I get a numerical solution for:

$$\frac{x}{1-e^{-x}} -5 = 0?$$

I've been thinking about using Newton's method here but I am not quite sure how to get a reasonable starting point. Is there any other approach or maybe even just a method that provides me a suitable starting point?

Thank you very much!
FunkyPeanut

Answer 1

If you want to approximate it, you can show that it is an increasing function and only 1 real solution exists. Then, use the fact that $e^{-5}<<1$ (Well, its $\approx0.006$). Then you can approximately neglect the denominator to $5-5=0$. Hence, the solution is slightly less than $5$. If calculators are allowed, you can easily, check near which value the sign changes.

Answer 2

It's easy to write it in another way

$$x-5=-5\cdot e^{-x}$$

and draw two graphs. Then you see, that you have two solutions, first near $x\in[-1,0]$, second near $5$.

Answer 3

Without getting into the theory, you can often setup a fixed point iteration in lieu of the more powerful Newton's method. What I mean is consider the sequence $$ x_{n+1} = 5(1-\mathrm{e}^{-x_n}). $$

If it converges, it will converge to your answer. And we know from another comment that $x\approx 5$, so that makes an excellent starting guess -- $x_0=5$. Here are the first few terms in the sequence: $$ \{5, 4.96631, 4.96516, 4.96512, 4.96511\} $$

After only 4 iterations (easily done in a calculator, btw), we have to 5 decimal places the correct answer. Not bad.