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

UNDER DEVELOPMENT

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

UNDER CONSTRUCTION

The sinc function

UNDER CONSTRUCTION

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.