22 lines
781 B
TeX
22 lines
781 B
TeX
\beginsection 18.6
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Prove that $x=\cos x$ for some $x$ in $(0,\pi/2)$.
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\medskip
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Solution: We have $\cos(0)=1$ and $\cos(\pi/2)=0$.
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The problem is that the interval is open, not closed as required by IVT.
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However, since $\cos(0)\ne0$ and $\cos(\pi/2)\ne\pi/2$ we know that
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$\cos x=x$ does not occur at the endpoints.
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Therefore it is safe to use the closed interval $[0,\pi/2]$.
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%(Is this o.k. or b.s.?)
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Use the trick from Example 1: $f(x)=x-\cos x$.
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We have $f(0)=0-1=-1$ and $f(\pi/2)=\pi/2-0=\pi/2$ and
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therefore $f(0)<0<f(\pi/2)$.
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By IVT there exits an $x_0$ such that $f(x_0)=0$.
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Because $f(x_0)=x_0-\cos x_0=0$, we have a point $x_0$
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at which $x=\cos x$.
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We have already shown that $x\ne\cos x$ at either $0$ or $\pi/2$ so
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$x=\cos x$ must occur on the interval $(0,\pi/2)$.
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