Math Notes Fall 25

August 20th 2025

Proof by multiplicative inverse

We have functions $f(x), g(x), h(x)$ .

Given $h(x) f(x) = 1$ . (they are multiplicative inverses)

We would like to determine whether $f(x) = g(x)$ .

Naturally, $h(x)g(x) = 1\ \implies\ f(x) = g(x)$

Application

I discovered this trick when searching for a proof to Euler’s Formula:

e^{it} = \cos{t} + i\sin{t}

We define

\begin{aligned} f(x) &= e^{it}\\ g(x) &= \left(\cos{t} + i\sin{t}\right) \end{aligned}

Naturally, $e^{it}h(x) = 1$ , so

h(x) = e^{-it}

So, if we can prove that $h(x)g(x) = 1$ , then it is clear that $f(x)=g(x)$ . Introduce $j(x)$ for brevity.

h(x)g(x) = j(x) = e^{-it}\left(\cos{t} + i\sin{t}\right)

Differentiate $j(x)$

\begin {aligned} j'(x) &= e^{-it}\left(-\sin{t} + i\cos{t}\right) - i e^{-it}\left(\cos{t} + i\sin{t}\right)\\ &= e^{-it}\left(-\sin{t} + i\cos{t}\right) + e^{-it}\left(-i\cos{t} + \sin{t}\right)\\ &= e^{-it}\left(i\cos{t} -\sin{t} + \sin{t} -i\cos{t} \right)\\ &= 0 \end{aligned}

That is, $j'(x) =0$ for all $x$ , which means that $j(x)$ is constant for all $x$ .

Find value of $j(x)$

j(0) = e^{0}\left(\cos{0} + i\sin{0}\right) = 1

So, $j(x) = 1$ for all $x$ , which shows that $g(x)$ is the multiplicative inverse of $h(x)$ . Given that $g(x)$ is also the multiplicative inverse of $h(x)$ :

f(x) = g(x)

Source: https://math.stackexchange.com/a/8612

Complex Exponentials and Trigonometric functions

The value of the exponential function with an imaginary input is of interest.

e^{ix} = ?

Using the well known series expansion of the exponential function:

e^{x} = \sum_{n=0}^{\infty} \frac{x^{n}}{n!}

Substitute in imaginary constant

e^{ix} = \sum_{n=0}^{\infty} \frac{(i)^{n}x^n}{n!}

There is no direct way to simplify this expression as $(i)^n$ doesn’t directly work out to anything nice. This expression is equivalent to a few other, more-useful ones.

\begin{aligned} e^{ix} &= \sum_{n=0}^{\infty} \frac{(i)^{n}x^n}{n!}\\ &= \sum_{n=0}^{\infty} \frac{(i)^{2n}x^{2n}}{(2n)!} + \frac{(i)^{2n+1}x^{2n+1}}{(2n+1)!}\\ &= \sum_{n=0}^{\infty} \frac{(i)^{4n}x^{4n}}{(4n)!} + \frac{(i)^{4n+1}x^{4n+1}}{(4n+1)!} + \frac{(i)^{4n+2}x^{4n+2}}{(4n+2)!} + \frac{(i)^{4n+3}x^{4n+3}}{(4n+3)!} \end{aligned}

Now the powers of $i$ can be simplified

\begin{aligned} i^{4n} &= i^{4^n} = 1\\ i^{4n+1} &= i^{4n}i = i\\ i^{4n+2} &= i^{4n}i^2 = -1\\ i^{4n+3} &= i^{4n}i^3 = -i \end{aligned}

Back-substituting these into the above sum

e^{ix} = \sum_{n=0}^{\infty}\frac{x^{4n}}{(4n)!}+i\frac{x^{4n+1}}{(4n+1)!}-\frac{x^{4n+2}}{(4n+2)!}-i\frac{x^{4n+3}}{(4n+3)!}

Remember the power series for sine and cosine

\begin{aligned} \sin(x) &=\sum_{n=0}^{\infty} \frac{x^{2n+1}(-1)^{n}}{(2n+1)!}\\ \cos(x) &=\sum_{n=0}^{\infty} \frac{x^{2n}(-1)^{n}}{(2n)!} \end{aligned}

Reorder series

e^{ix} = \sum_{n=0}^{\infty}\frac{x^{4n}}{(4n)!} - \frac{x^{4n+2}}{(4n+2)!} + i \frac{x^{4n+1}}{(4n+1)!} - i\frac{x^{4n+3}}{(4n+3)!}

Split sums based on evens, and odds

\begin{aligned} e^{ix} &= \left[{\sum_{n=0}^{\infty}\frac{x^{4n}}{(4n)!} - \frac{x^{4n+2}}{(4n+2)!}}\right]\\ &+ \left[i\sum_{n=0}^{\infty} \frac{x^{4n+1}}{(4n+1)!} - \frac{x^{4n+3}}{(4n+3)!}\right] \end{aligned}

Notice the first sum is simply iterating over the even numbers, and the second is iterating over the odds. Simplify

\begin{aligned} e^{ix} &= \left[{\sum_{n=0}^{\infty}\frac{x^{2n}(-1)^n}{(2n)!}}\right]\\ &+ \left[i\sum_{n=0}^{\infty} \frac{x^{2n+1}(-1)^{n}}{(2n+1)!} \right] \end{aligned}

Notice

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

With Negatives

Recall these two properties of sine and cosine

\begin{aligned} \sin(-x) &= -\sin(x)\\ \cos(-x) &= \cos(x)\\ \end{aligned}

With that, use the exponential function with a negative imaginary number

e^{-ix} = \cos{x} - i\sin{x}

We can now solve for cosine and sine using the exponential function.

e^{-ix} + e^{ix} = 2\cos{x}

\cos{x} = \frac{e^{ix} + e^{-ix}}{2}

e^{ix} - e^{-ix} = 2\sin{x}

\sin{x} = \frac{e^{ix} - e^{-ix}}{2}

Clever proof of solutions to linear diophantine equation

A simple result in elementary number theory relates to the linear diophantine equation

ax + by = c

a,x,b,y,c \in \mathbb{Z}

A property of this equation (which is not proven here) is that it has infinitely many $(x,y)$ solutions if $c$ is a multiple of $\gcd(a,b)$ , otherwise it has none.

The clever part is, after finding the first solution we can find the rest. Suppose there exists a solution $(x,y)$ , and that $(x',y')$ is another solution:

c = ax+by = ax'+by'

Rearrange equation

a(x-x') = -b(y-y')

Let $d = \gcd(a,b)$ . Naturally, $a$ and $b$ are both divisible by $d$ . Divide by $d$

\frac{a}{d}(x-x') = -\frac{b}{d}(y-y')

Now, $a/d$ and $b/d$ share no common factors, therefor $a/d$ does not divide $b/d$ and $b/d$ does not divide $a/b$ . Note however that $b/d$ does divide the right side of the equation, and therefor must divide the left side of the equation. Because we already determined that $b/d$ does not divide $a/d$ , $b/d$ must divide $(x-x')$ , thus $(x-x')$ must be a multiple of $b/d$ (likewise $(y-y')$ is a multiple of $a/d$ ).

State that $(x-x')$ is a multiple of $b/d$ in a new equation:

\frac{bn}{d} = x - x'

Likewise of $y$

\frac{an}{d} = y - y'

Thus, given the initial solutions of $x,y$ , the other solutions are found as:

\frac{bn}{d}+x = x'

\frac{an}{d}+y = y'

This proof is not complete but highlights the clever trick of extracting an equation based on divisibility. It comes down to the (often subtle) fact that when $x$ divides $y$ , $y=xq$ .

Complex Difference of Squares

Sometimes the solution to math puzzles lies in the difference off squares identity, $x^2-y^2 = (x+y)(x-y)$ . I’ve often thought that it’s a shame how there is no similar identity for $x^2+y^2$ , but… there is!

x^2+y^2 = (x+iy)(x-iy)

Where $i$ is the imaginary constant.