Hello I need help with this question,
Prove that $g(x)=e^{-x^2}$ has a unique fixed point on the interval [0,1]
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Are you familiar with derivatives? – Jonathan Y. Sep 17 '13 at 00:11
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The continuous function $f(x) = e^{-x^2} - x$ is positive at 0 and negative at 1, so by the Intermediate Value Theorem there is a point $a \in [0,1]$ with $e^{-a^2} - a = 0$ or $e^{-a^2} = a$. This fixed point is unique because $e^{-x^2}$ is monotonically decreasing on $[0,1]$.
AGM
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how can I show that f(x) is monotonically decreasing on the interval? (without graph) – PooperScooper Sep 17 '13 at 00:21
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$-x$ is decreasing, so it would suffice to show that so is $e^{-x^2}$. Indeed, $e^{-x^2}-e^{-y^2} = e^{-x^2}(1-e^{x^2-y^2}) = e^{-x^2}(1-e^{(x-y)(x+y)})$, which is positive iff $(x-y)(x+y)<0$, i.e. iff $x<y$. – Jonathan Y. Sep 17 '13 at 00:25
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Hint: This is tantamount to showing that the function $h(x):=e^{-x^2}-x$ has a unique root on $[0,1]$.
Note that $h(0)=1$ and $h(1)=\frac{1}{e}-1<0$. The function is continuous, so it must cross the axis at least once. Does it happen to be the case that $h$ is either strictly monotone increasing or strictly monotone decreasing? That would do it. (Why?)
Nick Peterson
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