In my Analysis class today I defined the trigonometric functions {\sin} and {\cos} by means of the complex exponential. As usual I noted that for real {x} we have {|\mathrm{e}^{\mathrm{i} x}| = 1}, i.e. {\mathrm{e}^{\mathrm{i} x}} lies on the complex unit circle. Then I drew the following picture:

 

075_exp

This was meant to show that the real part and the imaginary part of {\mathrm{e}^{\mathrm{i} x}} are what is known as {\cos(x)} and {\sin(x)}, respectively.

After the lecture a student came to me and noted that we could have started with {a>1} and note that {|a^{\mathrm{i} x}|=1} and could do the same thing. The question is: Does this work out? My initial reaction was: Yeah, that works, but you’ll get a different {\pi}

But then I wondered, if this would lead to something useful. At least for the logarithm one does a similar thing. We define {a^x} for {a>0} and real {x} as {a^x = \exp_a(x) = \exp(x\ln(a))}, notes that this gives a bijection between {{\mathbb R}} and {]0,\infty[} and defines the inverse function as

\displaystyle  \log_a = \exp_a^{-1}.

So, nothing stops us from defining

\displaystyle  \cos_a(x) = \Re(a^{\mathrm{i} x}),\qquad \sin_a(x) = \Im(a^{\mathrm{i} x}).

Many identities are still valid, e.g.

\displaystyle  \sin_a(x)^2 + \cos_a(x)^2 = 1

or

\displaystyle  \cos_a(x)^2 = \tfrac12(1 + \cos_a(2x)).

For the derivative one has to be a bit more careful as it holds

\displaystyle  \sin_a'(x) = \ln(a)\cos_a(x),\qquad \cos_a'(x) = -\ln(a)\sin_a(x).

Coming back to “you’ll get a different {\pi}”: In the next lecture I am going to define {\pi} by saying that {\pi/2} is the smallest positive root of the functions {\cos}. Naturally this leads to a definition of “{\pi} in base {a}” as follows:

Definition 1 {\pi_a/2} is the smallest positive root of {\cos_a}.

How is this related to the area of the unit circle (which is another definition for {\pi})?

The usual analysis proof goes by calculating the area of a quarter the unit circle by integral {\int_0^1 \sqrt{1-x^2} dx}.

Doing this in base {a} goes by substituting {x = \sin_a(\theta)}:

\displaystyle  \begin{array}{rcl}  \int\limits_0^1\sqrt{1-x^2}\, dx & = & \int\limits_0^{\pi_a/2}\sqrt{1-\sin_a(\theta)^2}\, \ln(a)\cos_a(\theta)\, d\theta\\ & = & \ln(a) \int\limits_0^{\pi_a/2} \cos_a(\theta)^2\, d\theta\\ & = & \ln(a) \frac12 \int\limits_0^{\pi_a/2}(1 + \cos_a(2\theta))\, d\theta\\ & = & \frac{\ln(a)}{2} \Big( \frac{\pi_a}{2} + \int\limits_0^{\pi_a/2}\cos_a(2\theta)\, d\theta\\ & = & \frac{\ln(a)\pi_a}{4} + 0. \end{array}

Thus, the area of the unit circle is now {\ln(a)\pi_a}

Oh, and by the way, you’ll get the nice identity

\displaystyle  \pi_{\mathrm{e}^\pi} = 1

(and hence, the area of the unit circle is indeed {\ln(\mathrm{e}^\pi)\pi_{\mathrm{e}^\pi} = \pi})…

Advertisements