The Pi Manifesto is complete bonkers.

An enlightening discussion about pi and tau.

The Pi Manifesto is complete bonkers.

by rdococ » Fri Oct 18, 2013 1:54 am

This is an effort to counteract all of the arguments stated in The Pi Manifesto.

Area definition of [tex]\pi[/tex]
[tex]\pi[/tex]ists can't get it in their heads that integrals bring a factor of a half, described by the power rule! That's why
[tex]{A \over r^2} = \pi[/tex]
and not [tex]\tau[/tex].
Conclusion: The area of a circle is based on a natural half factor, and so are all of the other integrals that evaluate to [tex]\pi[/tex].

Squaring the circle
This amounts to the same integral as
[tex]A = \int \tau r \ dr = \frac{1}{2} \tau r^2[/tex]
However, this makes an interesting remark of its own.
Conclusion: Squaring the circle is a draw at the best for [tex]\pi[/tex].

Gaussian Integral
The [tex]2[/tex] DOES belong to the [tex]\pi[/tex], by, as The Tau Manifesto said, squaring the un-normalized distribution then switching to polar co-ordinates.
Conclusion: As with area, the un-normalized distribution had a half factor in it, and switching to polar co-ordinates is the only addition to the argument, and it is a winner for [tex]\tau[/tex].

Cauchy Distribution

Triangles add up to [tex]\pi[/tex]
The vertices of any triangle add up to [tex]\pi[/tex], granted, but what do the vertices of a quadrilateral add up to? It'd be nice if it were [tex]\pi[/tex], but no, it is the almighty [tex]\tau[/tex]!
[tex]\alpha + \beta + \gamma + \delta = \tau[/tex]
Conclusion: Lots of polygons add up to lots of multiples of [tex]\pi[/tex]. For example, there's a nice case for [tex]\alpha + \beta + \gamma + \delta + a = 3 \pi[/tex], but that doesn't mean we should isolate [tex]1 \pi[/tex] for its use in triangles.

Area of a regular n-gon inscribed into a unit circle
Credit to Joseph Lindenburg for this formula.
You can simplify the formula given for this argument into a [tex]\tau[/tex]tology.
[tex]A = n\sin\left(\frac{\pi}{n}\right)\cos\left(\frac{\pi}{n}\right) = \left(\frac{n}{2}\right)\sin\left(\frac{2\pi}{n}\right) = \left(\frac{n}{2}\right)\sin\left(\frac{\tau}{n}\right)[/tex]
Conclusion: This belongs in The [tex]\tau[/tex] Manifesto.

Trigonometric functions
Let's explore the meaning of the functions a little further.
[tex]\sin{\alpha}[/tex] Gives you the y position of the polar point at (r = 1, \theta = \alpha).
[tex]\cos{\alpha}[/tex] Gives you the x position of the polar point at (r = 1, \theta = \alpha).
[tex]\tan{\alpha}[/tex] Gives you the slope of the cartesian vector (0, 0)->(cos \alpha, sin \alpha).
The third of these functions has a period of [tex]\pi[/tex] because [tex]{-y \over -x} = {y \over x}[/tex]. Thus, again, there is a factor of a half in the tangent function.
Conclusion: Because [tex]\tan{\alpha} = \tan(angleInverse(\alpha))[/tex] (where angleInverse is the theta such that [tex]\sin{\theta} = -\sin{\alpha}[/tex] and [tex]\cos{\theta} = -\cos{\alpha}[/tex]), you will realize that there is a factor of 1/2 and that's why it looks like [tex]\pi[/tex] wins.

The erf function

The sinc function

The gamma function
Not only does [tex]\Gamma(1/2) = \sqrt{\pi}[/tex], and that is 1/2, not 1, but this is also defined as an integral!
Conclusion: Again, the integrals bring a factor in. [tex]\tau[/tex] wins.

Euler's Reflection Formula
This involves the gamma function, thus integrals.
Conclusion: The [tex]\pi[/tex] Manifesto gets away too often with integrals.

Volume of unit n-ball
This minimizes to the same integral as the area of a circle.
Conclusion: This is related to integrals again too.

Area of an ellipse
This is similar to that of a circle, but the radius measured in the y axis is different to that in the x axis.
Conclusion: This is not exactly an integral, but [tex]\tau[/tex] still wins.

All the other integrals
Integrals bring out a factor of a half, so [tex]\tau[/tex] wins automatically in such cases for [tex]\pi[/tex].
Conclusion: Yet again, many of the cases taken in The [tex]\pi[/tex] Manifesto are just integrals.

Euler's identity
This is an interesting case. But there is a special form of Euler's identity using [tex]\tau[/tex]:
[tex]0 = 1 + e^{-i\frac{\tau}{2}}[/tex]
which contains addition, subtraction, multiplication and division in order,
0, 1 and 2 in numerical order,
AND [tex]e[/tex], [tex]i[/tex], and [tex]\tau[/tex] in alphabetical order, all in one formula!
Conclusion: The original argument for Euler's identity with [tex]\tau[/tex] isn't that great. But this one shows that Euler's identity works BETTER with [tex]\tau[/tex], not [tex]\pi[/tex].

Conclusion: [tex]\tau[/tex] wins the argument.
Elementary School
Elementary School
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Re: The Pi Manifesto is complete bonkers.

by 1=0 » Sun Mar 19, 2017 2:14 pm

Integrals don't always bring in a factor of 1/2, but I have looked into the integrals in the Pi Manifesto, and sure enough, they did bring out factors of 1/2. It seems very suspicious to me that the Pi Manifesto would use so many integrals, and I think the reason is that integrals can bring in fractional factors, so there are examples of integrals with values of pi. Every example in the list of formulas involved an integral, and to me that shows that integrals are the only place where it was even possible to find cases where pi is simpler. That section would be much stronger if it included examples from all over mathematics.
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