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Published December 21, 2020

I have been reading a Calculus book recently and came across formulas for min and max:

\begin{align} min(x, y) = \frac{x + y - |y - x|}{2} \\ max(x, y) = \frac{x + y + |y - x|}{2} \end{align}

Here $$\frac{x + y}{2}$$ is the average of the two numbers or the midpoint between the two on the number line.

Also $$\frac{|y - x|}{2}$$ is half the distance between y and x. Adding it to the midpoint gives the maximum (rightmost) value and subtracting it from the midpoint gives the minimum (leftmost) value.

This is particularly interesting coming from a programming background where:

min(x, y) := x <= y ? x : y
max(x, y) := x >= y ? x : y


I never thought of coming up with a non-procedural formula for either of these.

We can prove that $$min(x, y) = \frac{x + y - |y - x|}{2}$$ and $$max(x, y) = \frac{x + y + |y - x|}{2}$$ as follows:

If $$x \leq y$$, then $$|y - x| = y - x$$. So $$x + y - |y - x| = x + y - y + x = 2x = 2 \cdot min(x, y)$$, and $$x + y + |y - x| = x + y + y - x = 2y = 2 \cdot max(x, y)$$.

In both cases, swap $$x$$ and $$y$$ to prove the formulas for $$x \geq y$$.

Published December 21, 2020

Euler's formula states that for any real number $$x$$:

$e^{ix} = \cos x + i \sin x$

When $$x = \pi$$, Euler's formula evaluates to:

$e^{i\pi} + 1 = 0$

This is known as Euler's identity and it links five fundamental mathematical constants ($$\pi$$, $$e$$, $$i$$, $$0$$, and $$1$$) in a beautifully simple equation. Richard Feynman called the equation "the most remarkable formula in mathematics."

The formula can be interpreted as saying that the function $$e^{i\phi}$$ is a unit complex number, i.e. it traces out the unit circle in the complex plane as $$\phi$$ ranges through the real numbers. We can derive Euler's formula from the Maclaurin series expansions of $$\sin x$$, $$\cos x$$, and $$e^x$$:

\begin{align} \sin x &= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{2n+1} \\ &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \\ \cos x &= \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{2n} \\ &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \\ e^x &= \sum_{n = 0}^{\infty} \frac{x^n}{n!} \\ &= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \end{align}

Thus:

\begin{align} e^{ix} &= \sum_{n = 0}^{\infty} \frac{(ix)^n}{n!} \\ e^{ix} &= 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \frac{(ix)^6}{6!} + \frac{(ix)^7}{7!} + \cdots \\ &= 1 + ix - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \frac{ix^5}{5!} - \frac{x^6}{6!} - \frac{ix^7}{7!} + \cdots \\ &= \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} - \cdots \right) + \left( ix - \frac{ix^3}{3!} + \frac{ix^5}{5!} - \frac{ix^7}{7!} + \cdots \right) \\ &= \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} - \cdots \right) + i\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \right) \\ &= \cos x + i\sin x \end{align}

This yields Euler's identity. Substitute $$\pi$$ for $$x$$. Since $$\cos \pi = -1$$ and $$\sin \pi = 0$$, it follows that:

$e^{i\pi} = -1 + 0i$

or

$e^{i\pi} + 1 = 0$