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Plot circles using the various historical approximations of Pi?

Posted 13 days ago
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I want to be able to see the over or under approximations as a consequence of using older approximations of Pi such as 355/113 or 25/8 instead of the default value used by Mathematica.

I was not able to see any visible difference by using the ReplaceAll function to replace Pi. What kind of function can I use to to plot a circle to demonstrate these differences to students?

Thank you!

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POSTED BY: R Q
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4 Replies

(1) The Wolfram Language is case-sensitive. So Pi is not the same as pi.

(2) A Circle primitive is going to be a graphic of a circle. No substitution will change that.

(3) It is quite unclear, both from post and notebook, what exactly is the goal. A circle is a circle, independent of how ratio of circumference to diameter is evaluated.

Posted 12 days ago

Thank you for your response, Daniel.

Do I understand correctly that you are saying Pi plays no role in the plotting of a circle? If this is true, I'd like to know how they are plotted.

POSTED BY: R Q
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Posted 12 days ago

I am not privy to how Mathematica does things under the hood, but it is indeed true that one does not need $\pi$ to draw a circle. On the conceptually simpler end of things, there is e.g. Minsky's method, whose (naive) Mathematica implementation might go like

With[{h = N[1/16]}, 
     Graphics[Line[NestList[{{1, -h}, {h, 1 - h^2}}.# &, {1, 0}, Round[8/h]]]]]

and on the fancier end of things, one can use non-uniform rational B-splines (NURBS), which is mathematically equivalent to parametrizing a circle with the stereographic projection:

With[{n = 150}, 
     ParametricPlot[{1 - u^2, 2 u}/(1 + u^2), {u, -n, n}, PlotRange -> All]]

Jim Blinn gives a good survey of various methods here. All of this is independent of computing $\pi$.

POSTED BY: J. M.
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Posted 12 days ago

Thank you for your reply, JM.

If π is not necessary to draw a circle, I will rephrase my question: What can I plot to demonstrate the consequences of using different approximations of π?

POSTED BY: R Q
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