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Why is e^(pi i) = -1?

math.toronto.edu

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Re: Why is e^(pi i) = -1?

#51
post #45
post #40

Earlier quoted context omitted.

The area of a circle, for example, would be r^2 pi/2. This is actually my favorite example of why pi is wrong—it's the "exception" that proves the rule. To see why, set τ = C / r = 2 pi, and then consider the following chart of common quadratic forms: integral of u 1/2 u^2 kinetic energy 1/2 m v^2 distance fallen 1/2 g t^2 spring energy 1/2 k x^2 triangular area 1/2 b h circular area 1/2 τ r^2 We see that, far from c…

All of those (except the circle) have the 1/2 because they're integrals of something linear. While it's true that area and integral are closely related (the latter being a special case of the former), a circle is clearly not linear.

johnaspden has it right. To put it more explicitly, we can calculate the area of a circle by integrating the differential element of area dA for an infinitesimal annulus from 0 to r. Now, dA is simply the arclength (circumference C) times the thickness dr; since the circumference scales linearly with radius, this leads to the integral of a linear function as follows:

dA = C dr = τ r dr => A = ½ τ r².

Re: Why is e^(pi i) = -1?

#53
post #45

Earlier quoted context omitted.

All of those (except the circle) have the 1/2 because they're integrals of something linear. While it's true that area and integral are closely related (the latter being a special case of the former), a circle is clearly not linear.

As a circle expands, its area grows proportional to its circumference. The circumference is proportional to the radius. So you're getting the area by integrating a linear thing.

As a circle expands, its area grows proportional to its circumference. The circumference is proportional to the radius.

Exactly. In symbols, this reads dA/dr = Cr. Setting τ = C/r = 6.2831853…, we have dA = C dr = τ r dr => A = ½ τ r². Q.E.D.

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