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I am faced with this function (warning, I am not good at math)

$x(t+1)=0.5 \ln x(t)+1$ initial condition = 1 . I know the fixed point is 1 because $0.5 \ln (1)+1=1$ but I wanted to know the general method to obtain this for other logarithmic equations. For instance, if I were to find the fixed point of $$ x(t+1)=0.5 \ln x(t)+3 $$ How should I solve for $x$ when I have both $x$ and $\ln x$?

Another issue:

With linear equations I usually either: set $x(t+1)=x(t)$ and solve for $x$, or subtract $x$.

Are these two methods equal? They seem to be in linear and quadratic equations. What about for logarithmic equations like the one above? Is this the approach to use?

izœc
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user127623
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1 Answers1

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If I properly understand your post, you have difficulties solving for $x$ the equation $$f(x)=x-\frac{\log (x)}{2}-3=0$$ This kind of equation cannot be solved on the basis of elementary functions, but, as Mhenni Benghorbal mentioned, the solution can be obtained using Lambert function.

For this specific case, the solutions are given by $$x_1=-\frac{1}{2} W\left(-\frac{2}{e^6}\right)$$ and $$x_2=-\frac{1}{2} W_{-1}\left(-\frac{2}{e^6}\right)$$ in which appear the two branches of Lambert function.

Let us assume that you do not use Lambert function. In such a case, only numerical methods could be used and, from far away, the simplest will be be Newton. If you plot your function, you will notice taht there is one $x$ intercept close to $0$ (in fact $0.002491$) and another one between $3$ and $4$ (in fact $3.64694$); what I gave inside parenthese are the numerical values obtained using Lambert function.

Let us go back to Newton iteration for which we shall say that the estimate (we shall name it $x_{old}$) is equal to $3$. Newton iteration will update the guess according to the following scheme
$$x=x-\frac{f(x)}{f'(x)}$$ In the left hand side, will be used the guess $x_{old}$ and the result will be assigned to the new estimate of the solution (say $x_{new}$) and the process will be repeated replacing the value of $x_{old}$ by the so computed value of $x_{new}$.

Doing so, you will have the following successive iterates : 3.00000, 3.65917, 3.64695, 3.64694 this last value corresponding to six exact significant digits.

Starting the process at $x=4$ , the successive iterates would have been : 4.00000, 3.64931, 3.64695, 3.64694.

For sure, you could continue this iterative process unitl you reached the desired level of accuracy.

I suggest you look at the link provided by Mhenni Benghorbal since this Newton method allows to solve almost any equation.

If there is anything you want me to clarify, just post.

Claude Leibovici
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