$f(x) = (x-2)^2 - ln(x)$ on the interval $[1,2]$ has a unique root? I thought it in this way: $f(x) = 0$ is equivalent to say $(x-2)^2 = ln(x)$, which is equivalent to $g(x) = exp((x-2)^2)$ has a unique fixed point, but the $L$ turns out to be that it can be greater than 1.
how to prove contractivity
0
$\begingroup$
numerical-methods
fixed-point-theorems
-
0sorry, unique root – 2017-02-05
-
0The fixed-point equation formed the other way around, $g(x)=2-\sqrt{\ln x}$, would probably work better for contractivity. – 2017-02-06
1 Answers
1
I suggest an alternative approach. We want to show that $$f(x) = (x-2)^2 - \ln(x)$$ has a unique root on $[1,2]$. We have
$$f'(x) = 2(x-2) - \frac{1}{x}$$ so clearly $f'(x) < 0$ for $x \in [1,2]$ showing that $f$ is strictly monotonically decreasing on $[1,2]$, and thus can have at most one root on $[1,2]$. Because $f(1) > 0$ and $f(2) < 0$, by the intermediate value theorem we see that $f$ in fact has a unique root in $[1,2]$.
-
0why the fixed point method doesn't work? – 2017-02-05
-
0the banach fixed point theorem is not an equivalence statement. If your map is a contraction with constant < 1, you have a unique fixed point. The converse, however isn't true in general. – 2017-02-05