How to show a zero of a function is a fixpoint of another function?

Question

$g(x)=x \log(x+1)+x-1$ where the log has the base $e$. Set $G_1(x)=1/(\log(x+1)+1)$ and $G_2(x)=1-x\log(x+1)$. Show that the zero $x^*$ of $g(x)$ is a fixpoint of $G_1(x)$ as well as $G_2(x)$.

My thought about the problem:

I know how to show there exists an $x^*\in[0,1]$ using the Bolzano's Theorem. But I cannot get the explicit form of $x^*$, how can I then use it to find the fixpoint of $G_1(x)$ and $G_2(x)$?

I know a fixpoint of $G_1(x)$ means $G_1(X^*)=x^*$, but we don't know $x^*$?

Another thing that confuses me is if I want to find an iterated function, which one is the better candidate $G_1$ or $G_2$? Because as far I know about fixpoint is it always has something to do with iteration.

Can anybody please give some light on the problem?

Thanks

Answer 1

Let $y=x^*$. This will simplify my answer.

Thus $y \log(y+1)+y-1=0 \iff \log(y+1)=\frac{1-y}y$.

Substitute this to calculate $G_1(y)$ and $G_2(y)$ in a simpler form $\Rightarrow Q.E.D.$

Edit: When substitute you get: $$G_1(y)=\frac1{\log(y+1)+1}=\frac1{\frac{1-y}y+1}=\frac1{\frac1y}=y$$ $$G_2(y)=1-y\log(y+1)=1-y\frac{1-y}y=1-(1-y)=y$$

Answer 2

Hint: If $x^*$ is a root/zero of $g(x)$ then $\log(x^*+1)=\frac1{x^*}-1$.

What comes out if you work out $G_1(x^*)$ and $G_2(x^*)$ under this condition?

Answer 3

Since $g(x^*)=0$ we have $x^*\log(x^*+1)+x^*-1=0$. Notice that this can be rearranged in the following two ways:

1. $x^*\ =\ \dfrac1{\log(x^*+1)+1}$
2. $x^*\ =\ 1-x^*\log(x^*+1)$

Answer 4

No need to have the value of the zero.

If x* is a zero of g(x) then g(x*)=0, that means xlog(x+1)+x*-1=0. Bring terms on the right side of =. You then have x* = 1 - xlog(x+1) = G2(x*). So x* is a fixed point for G2(x).

The same way, if x* is a zero of g(x), xlog(x+1)+x*-1=0, so xlog(x+1)+x*=1. From the definition of g(x), we see that 0 cannot be its zero. We can then divide both sides of the equation by x* as x*≠0.

We have log(x*+1)+1=1/x*. Reversing both sides of the equations gives you 1/(log(x*+1)+1) = x*, meaning G1(x*)=x*. Then x* is a fixed point for G1(x).

Good luck.