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, $h (x) g (x) = 1 ⟹ f (x) = g (x)$

Application

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

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

We define

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

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

h (x) = e^{- i t}

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^{- i t} (\cos t + i \sin t)

Differentiate $j (x)$

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

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

Now the powers of $i$ can be simplified

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

Back-substituting these into the above sum

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

Remember the power series for sine and cosine

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

Reorder series

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

Split sums based on evens, and odds

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

Notice

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

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

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

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

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

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

Clever proof of solutions to linear diophantine equation

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

a x + b y = c

a, x, b, y, c \in 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 = a x + b y = a x^{'} + b y^{'}

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{b n}{d} = x - x^{'}

Likewise of $y$

\frac{a n}{d} = y - y^{'}

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

\frac{b n}{d} + x = x^{'}

\frac{a n}{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 $y$ , $y = x q$ .