Mdtryhht

I was solving some semiconductor physics problem and in order to get the temperature I got this nasty equation:

$$ T = dfrac7020dfrac32ln(T)+12.$$

I was thougt that I can solve this kind of equation simply by guessing solution for $T$ and then substituting that answer back into equation and then again substituting answer back into equation and so on until I am satisfied by precision of result. Somehow this method works.

Concretly for my example, my first guess was $T=1$ and I got this sequance of numbers $(585.0, 325.6419704169386, 339.4797907885183, 338.4580701961562, 338.53186591337385,338.52652733834424, ...)$ and they really seem to solve equation better and better.

Questions.

1) What is intuitive way to see why this method works?

2) How can I show rigoursly that this method actualy converges to solution of equation?

3) Obvious generalization for which the method will works seems to be:
$$ x = dfracabln(x)+c. $$ For which $a,b,c$ will this method work? Is this equation special case of some natural generalization of this equation? What are some similar equations which I can solve via this described method?

4) When will sequance of numbers in iteratiton process be finite to exacly solve equatiton? Does that case exist? Is solution to equation:
$$ x = dfracabln(x)+c $$
always (for every $a,b,c$) irational? Is it transcendental? If not, for which $a,b,c$ will that be the case?

Thank you for any help.

This method works because you are looking at a discrete dynamical of the form

$$x_n+1 = f(x_n)$$

where $f$ is a contraction. The rigorous proof is the Banach fixed point theorem.

The equation $$T=fracab log (T)+c$$ has explicit solution(s) in terms of Lambert function.

The result is given by
$$T=fracab, Wleft(fraca be^fraccbright)$$

In the linked page, you will see the different steps.

Applied to your case, this will immeditely give
$$T=frac4680Wleft(4680 e^8right)=338.526887451390053458527935852$$ If you do not access this function, for large values of the argument, use the expansion given in the linked page
$$W(x)=L_1-L_2+fracL_2L_1+frac(L_2-2) L_22 L_1^2+frac(2 L_2^2-9L_2+6) L_26 L_1^3+ ...$$ with $L_1=log (x)$ and $L_2=log (L_1)$

In general, a fixed point $p$ of a function $f(x)$ is an attractor for the iteration $x_n+1 = f(x_n)$ if $|f'(p)| < 1$. Then, if your initial guess is close enough to the fixed point, the iterations will eventually converge to it.
If $|f'(p)| > 1$, the fixed point is a repeller, and the only way to converge to the fixed point is to start exactly there (or happen to land there after a finite
number of iterations).

You have three parameters $a,b,c$, but there are really just two because you can multiply numerator and denominator by the same constant. So let's suppose $b=1$. As Claude remarked, the fixed point is
$$ p = fracaW(a e^c)$$
and this is the only real fixed point if $a,c>0$ (this is easy to see because $f(x)$ is decreasing where it is positive). The curve $f'(p) = -1$ in the $a,c$ plane looks like this:

Above the curve, the fixed point is an attractor. In particular that is always true for $a > e$. However, $a=c=1$ is right on the curve, and it's not clear whether the fixed point would be an attractor in that case (it turns out that it isn't, by taking higher derivatives into account). If $(a,c)$ is below the curve, the fixed point is a repeller.