Why does fixed point iteration only produce the solution greater than $1$ to the equation $Mx = e^x$ for $x \in \Bbb R$?

Question

The equation $Mx = e^x$, when $M > 0$. I know that the first solution must be at the tangent where the line $Mx$ crosses $e^x$, so $M$x has gradient $e^x$. This leads to $x(e^x) = e^x$, $x = 1$

But all of the $M$ values greater than $e$ must yield exactly $2$ solutions. One that is larger than one and the other occurring between $0$ and $1$, (Rolle's theorem)

An approximation to a solution for any given $M$ is $\log(M)$, since $\log(M) + \log(x) = x$. This can be improved to give us $\log(M) + \log\log(M) = x$. The error to this approximation is $\log(1 - (\log(x))/(x))$, which rapidly approaches $0$ as $x$ gets larger. The error is this because $\log\log M = \log(x-\log(x)) = \log(x) + \log(1 - \frac{\log(x)}{x})$

Here's what I can't work out, $y = \log M$ has precisely one solution, whereas $y = x - \log(x)$ has two. Why does the approximation $\log M + \log\log M$ always gives the solution above $1$, and somehow ignores the other solution?

Answer 1

It is true that $\frac{\log x}{x}$ is small when $x$ is large, and therefore the relative error of $\log M$ is small then. But when $0 < x < 1$, $\frac{\log x}{x}$ is not necessarily small. It can be quite large, in fact.

The error term $\log\left(1 - \frac{\log x}{x}\right)$ also can be large when $0 < x < 1$.

In short, the assumptions you made in taking your approximations are not true for $x < 1$, so you cannot expect those assumptions to find an approximate solution for $x$ there.

Answer 2

Transform your equation to

$$ln(M) + ln(x) = x$$

Then, look at the plot of $ln(x)$ and $x$:

As described in your question: If the curve $ln(x)$ is shifted upwards or downwards by $ln(M)$, it can intersect the main diagonal never (for $M \lt e$), once (for $M = e$) or two times (for $M \gt e$).

Answer 3

You said in the post somewhere that $M > e$. Then $\log(M)>1$ and $\log(\log(M))>0$, thus $\log(M)+\log(\log(M)) > 1$. Therefore this approximation gives a solution above 1.

If $M < e$, then $\log(M)+\log(\log(M)) < 1$, but then the equation doesn't have a solution bigger than 1, as the derivative of $e^x-cx$ is $e^x-c$, which is positive when $x>1>\log(c)$ when $c<1$, and $e-c>0$ as well.

Answer 4

Cobweb plots, whether you've heard of them or seen them, are an excellent way to visualize the process of iteration.

In your case you're iterating $$f(x) = \log M + \log x $$ starting at $$x_0 = 1 $$ with the intent of solving $$x = f(x)$$

You should be able to see from the plot that the iteration converges.

Using linear stability analysis, you can find which of the two fixed points is stable. To do this you only need to computer $f'(x) = {1 \over x}$. It's a general theorem that when $|f'(x)| < 1$ the fixed point is stable and when $|f'(x)| > 1$ the fixed point is unstable, which means that in this case the only stable fixed point is the one for which $x > 1$.

This is a good reference for the topic.

Answer 5

set $\mu=\log M$ so that the equation becomes: $$ x = e^{x-\mu} $$ suppose, for an example, that $M=e^2$ so $\mu=2$. take $x=0.3$ as a first guess, $x_0$. then $$ x_1 = e^{ 0.3 - 2} = 0.1826... \\ x_2 = e^{0.1826 -2} = 0.1624... $$ whilst the convergence is not particularly impressive, this iteration does home in on the root in $[0,1]$

in effect there are two 'directions' in which we can perform this iteration - the log route or the exp route - corresponding to a transformation and its inverse. of the two fixed points, one is reached by each route.

Answer 6

Can I just add something? With the method used to obtain the better approximation for the first root '$log(M) +log(log(M)) ≈ x$'. We can change $x ≈ \frac{1}{M}$ into '$x ≈ e^{1/M - log(M)}$'. Since, $logx = x - log(M)$ and $x$ is small. Though I'm not bothered to work out the error to that approximation