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I have a midterm tomorrow and while I was looking through old exams from my professor I stumbled on a problem for which I'm not able to see the solution.

We want to find the rots of $f(x) = e^x - xe^a$ with $a>1$.

Consider the fixed point functions $g_1(x) = e^x/e^a$ and $g_2(x) = a + \ln(x)$.

First, I had to show that $f(x)$ has two root $P$ and $Q$ such as $0<P<1<a<Q$ which I did using the Intermediate value theorem and the fact that $g_1(x)$ and $g_2(x)$ are strictly increasing.

My problem is this:

(a) Show that $g_1(x)$ and $g_2(x)$ have exactly two fixed points each and they coincide with the roots of $f(x)$.

(b) Then show that $g_1(x)$ doesn't converge to $Q$ and $g_2(x)$ doesn't converge to $P$.

I tried to show (a) by setting $g_1(x) = e^x/e^a = x$ and $g_2(x) = a + \ln(x) = x$ but I got stuck.

I also tried arguing that if:

  1. $g_1(x) \in C[0,1]$ and $g_1(x) \in [0,1]$ $\forall x \in [0,1]$
  2. $g_1'(x) \in C[0,1]$ and $\exists K$ $0<K<1$ s.a $|g_1'(x)| \leq K$ in $[0,1]$

Then there id a unique fixed point in $[0,1]$ and $x_{n+1} = g_1(x_n)$ converges to $P$

And the same argument for an interval $[a,a+1]$ so that there would be two unique fixed point, but the conditions don't hold for that interval.

Some help would be greatly appreciated, I really am stuck on this problem and the midterm I'm preparing for is tomorrow afternoon.

Zak Soliman
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2 Answers2

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Note that if:

  • $g_1(x)=x\Longrightarrow e^x/e^a=x\Longrightarrow e^x=e^ax\Longrightarrow e^x-e^ax=0\equiv f(x)=0$

how $f$ have only two roots then $g_1$ have exactly two fixed points.

  • $g_2(x)=x\Longrightarrow a + \ln{x}=x\Longrightarrow \ln{e^a}+\ln{x}=\ln{e^x}\Longrightarrow \ln{e^ax}=\ln{e^x}\Longrightarrow e^ax=x \Longrightarrow x-e^ax=0\equiv f(x)=0$

how $f$ have only two roots then $g_2$ have exactly two fixed points.

AsdrubalBeltran
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Notice that $$f(P)=0\Rightarrow e^P-e^aP=0\Rightarrow e^P=e^aP\Rightarrow P=\frac{e^P}{e^a}=g_1(P)$$ In the same way for $Q$. So $g_1(x)$ has two fixed points. Now $$f(P)=0\Rightarrow e^P-e^aP=0\Rightarrow e^P=e^aP\Rightarrow P=a+\ln(P)=g_2(P)$$ The same technique for $Q$.

Arian
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