56 lines
1.5 KiB
TeX
56 lines
1.5 KiB
TeX
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\section{More examples}
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\subsection{Fran\c cois Vi\`ete}
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Fran\c cois Vi\`ete was the first to discover an exact formula for $\pi$.
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Here is his formula.
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\begin{displaymath}
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{2\over\pi}={\sqrt2\over2}\times{\sqrt{2+\sqrt2}\over2}\times
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{\sqrt{2+\sqrt{2+\sqrt2}}\over2}\times\cdots
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\end{displaymath}
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%We can flip it around and write the formula like this.
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%\begin{displaymath}
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%\pi=2\times{2\over\sqrt2}\times{2\over\sqrt{2+\sqrt2}}\times
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%{2\over\sqrt{2+\sqrt{2+\sqrt2}}}\times\cdots
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%\end{displaymath}
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Let $a_0=0$ and $a_{n}=\sqrt{2+a_{n-1}}$.
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Then we can write
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\begin{displaymath}
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{2\over\pi}={a_1\over2}\times{a_2\over2}\times
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{a_3\over2}\times\cdots
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\end{displaymath}
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%
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Solving for $\pi$ we have
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\begin{displaymath}
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\pi=2\times{2\over a_1}\times{2\over a_2}\times{2\over a_3}\times\cdots=2\prod_{k=1}^\infty
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{2\over a_k}
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\end{displaymath}
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%
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Let us now use Eigenmath to compute $\pi$ according to Vi\`ete's formula.
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Of course, we cannot calculate all the way out to infinity, we have to stop somewhere.
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It turns out that nine factors are just enough to get six digits of accuracy.
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\medskip
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\verb$a(n)=test(n=0,0,sqrt(2+a(n-1)))$
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\verb$float(2*product(k,1,9,2/a(k)))$
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$$3.14159$$
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\medskip
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\noindent
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The function $a(n)$ calls itself $n$ times so overall there are
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54 calls to $a(n)$.
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By using a different algorithm with temporary variables, we can get the
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answer in just nine steps.
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\medskip
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\verb$a=0$
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\verb$b=2$
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\verb$for(k,1,9,a=sqrt(2+a),b=b*2/a)$
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\verb$float(b)$
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$$3.14159$$
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