Math Notes Spring 25

May 26th 2025

Fibonacci Moduli

For the Fibonacci sequence, defined by $F_{1} = 1, F_{2} = 1, F_{n} = F_{n - 1} + F_{n - 2}$

I believe the following is true (Haven’t found any counterexamples)

(F_{n} \mod n) \neq 0 when (n \mod 6) \in {2,3,4}

Calculus of inverses

Let $f (x), g (x)$ be functions that are defined and differentiable over some interval. Further, let

\begin{aligned} f (g (x)) & = x \\ g (f (x)) & = x \end{aligned}

Such that $g, f$ are inverses.

Differentiate by chain rule.

f^{'} (g (x)) g^{'} (x) = 1

g^{'} (x) = \frac{1}{f^{'} (g (x))}

This already has some neat results

\begin{aligned} f (x) & = \sin (x) \\ g (x) & = \sin^{- 1} (x) \\ \frac{d}{d x} \sin^{- 1} (x) & = \frac{1}{\cos (\sin^{- 1} (x))} \end{aligned}

\begin{aligned} f (x) & = \cos (x) \\ g (x) & = \cos^{- 1} (x) \\ \frac{d}{d x} \cos^{- 1} (x) & = - \frac{1}{\sin (\cos^{- 1} (x))} \end{aligned}

\begin{aligned} f (x) & = \tan (x) \\ g (x) & = \tan^{- 1} (x) \\ \frac{d}{d x} \tan^{- 1} (x) & = \frac{1}{\sec^{2} (\tan^{- 1} (x))} \end{aligned}

These aren’t the standard forms you’re familiar with, but they are equivalent.

By this we can create a very annoying looking integral (for fun of course).

\int \frac{1}{\sec^{2} (\tan^{- 1} (x)) \tan^{- 1} (x)^{2}} d x

Textbook derivatives

I know, it’s in the textbook, but I never really learned the derivative of $f (x) = \log_{n} (x)$

It’s not too complicated though using the logarithm change-of-base identity.

\frac{d}{d x} \log_{n} (x) = \frac{d}{d x} \frac{\ln (x)}{\ln (n)} = \frac{1}{x \ln (n)}

The exponential form isn’t that bad either.

\begin{aligned} g (x) & = n^{x} \\ \ln (g (x)) & = x \ln (n) \\ \frac{d}{d x} \ln (g (x)) & = \frac{d}{d x} x \ln (n) \\ \frac{1}{g (x)} g^{'} (x) & = \ln (n) \\ \frac{d}{d x} n^{x} & = \ln (n) n^{x} \end{aligned}

Of course, given that $n^{x}$ and $\log_{n} (x)$ are inverses, we could find also find the exponential through the identity found in the previous section.

\begin{aligned} f (x) & = \log_{n} (x) \\ g (x) & = n^{x} \\ \frac{d}{d x} n^{x} & = \frac{1}{\frac{1}{\ln (n) n^{x}}} = \ln (n) n^{x} \end{aligned}

Inverse Inverse Trigonometry

In the section on derivatives of inverses, these came up

\begin{array}{r} \cos (\arcsin (x)) \\ \sin (\arccos (x)) \\ \sec^{2} (\arctan (x)) \end{array}

These forms can be simplified to remove any notion of trigonometry.

Draw a right triangle with angle $θ$ , adjacent side length $\sqrt{1 - x^{2}}$ , opposite side length $x$ , and hypotenuse side length $1$ .

Of course, $\sin (θ) = x$ , so $\arcsin (x) = θ$ within bounds of course.

And by definition $\cos (θ) = \sqrt{1 - x^{2}}$ , so

\cos (\arcsin (x)) = \sqrt{1 - x^{2}}, where - 1 \leq x \leq 1

Some more identities found by similar method

\begin{aligned} \cos (\arcsin (x)) & = \sqrt{1 - x^{2}}, & where & - 1 \leq x \leq 1 \\ \sin (\arccos (x)) & = \sqrt{1 - x^{2}}, & where & - 1 \leq x \leq 1 \\ \tan (\arcsin (x)) & = \frac{x}{\sqrt{1 - x^{2}}}, & where & - 1 \leq x \leq 1 \\ \tan (\arccos (x)) & = \frac{\sqrt{1 - x^{2}}}{x}, & where & - 1 \leq x \leq 1 \end{aligned}

Arctangent is a bit more complex to deal with. Let’s do $\sec^{2} (\arctan (x))$

Draw a right triangle with angle $θ$ , adjacent side length $A$ , opposite side length of $O$ , and hypotenuse side length $1$ .

We know that $\tan (θ) = \frac{O}{A}$ so set $x = \frac{O}{A}$ . Also known is $O^{2} + A^{2} = 1$

\begin{aligned} A^{2} + x^{2} A^{2} & = 1 \\ A^{2} (1 + x^{2}) & = 1 \\ A^{2} & = \frac{1}{1 + x^{2}} \\ A & = \frac{1}{\sqrt{1 + x^{2}}} \\ A x & = O \\ O & = \frac{x}{\sqrt{1 + x^{2}}} \end{aligned}

And so, the triangle has been completed.

Finally, $\csc^{2} (θ) = \frac{1}{A^{2}}$

This leads to another set of identities

\begin{aligned} \sec^{2} (\arctan (x)) & = 1 + x^{2} \\ \cos (\arctan (x)) & = \frac{1}{\sqrt{1 + x^{2}}} \\ \sin (\arctan (x)) & = \frac{x}{\sqrt{1 + x^{2}}} \end{aligned}

As a sidenote, it seems weird to me that

\begin{aligned} f (x) & = \sin (\arctan (x)) \\ f^{- 1} (x) & = \tan (\arcsin (x)) \end{aligned}

Complete Factorization

Quick! what are all the factors of $720$ ?

Instead of giving you an answer, here’s an algorithm to find it.

First, by repeated division, find the prime factors.

\begin{array}{lrlr} 720 / 2 & = 360 \\ 360 / 2 & = 180 \\ 180 / 2 & = 90 \\ 90 / 2 & = 45 \\ 45 / 3 & = 15 \\ 15 / 3 & = 5 \\ 5 / 5 & = 1 \end{array}

F_{0} = [2, 2, 2, 2, 3, 3, 5]

Find the all unique factors and their cartesian products in $F_{0}$ to create $P_{0}$ .

P_{0} = {2, 3, 5, 6, 10, 15, 30}

S_{0} = P_{0}

Remove one of each unique factor from $F_{0}$ to make $F_{1}$ .

F_{1} = [2, 2, 2, 2, 3, 3, 5]

Find the all unique factors and their cartesian products in $F_{1}$ to create $P_{1}$ .

P_{1} = {2, 3, 6}

Multiply each number in $P_{1}$ by each number in $S_{0}$ to find $S_{1}$ Remove any duplicate values and remove any values which had been found in previous $S_{n}$ .

\begin{array}{lrlr} S_{1} = { & 4, 6, 10, 12, 20, 30, 60 \\ 6, 9, 15, 18, 30, 45, 90 \\ 12, 18, 30, 36, 60, 90, 180} \\ S_{1} = { & 4, 10, 12, 20, 60, 9, 18, 45, 90, 36, 180} \end{array}

Remove one of each unique factor from $F_{1}$ to make $F_{2}$ .

F_{2} = [2, 2, 2, 2, 3, 3, 5]

Find the all unique factors and their cartesian products in $F_{2}$ to create $P_{2}$ .

P_{2} = {2}

Multiply each number in $P_{2}$ by each number in $S_{1}$ to find $S_{2}$ Remove any duplicate values and remove any values which had been found in previous $S_{n}$ .

\begin{array}{lrlr} S_{2} & = {8, 20, 24, 40, 120, 18, 36, 90, 180, 72, 360} \\ S_{2} & = {8, 24, 40, 120, 72, 360} \end{array}

Remove one of each unique factor from $F_{2}$ to make $F_{3}$ .

F_{3} = [2, 2, 2, 2, 3, 3, 5]

Find the all unique factors and their cartesian products in $F_{3}$ to create $P_{3}$ .

P_{3} = {2}

Multiply each number in $P_{3}$ by each number in $S_{2}$ to find $S_{3}$ Remove any duplicate values and remove any values which had been found in previous $S_{n}$ .

S_{3} = {16, 48, 80, 240, 144, 720}

Remove one of each unique factor from $F_{3}$ to make $F_{4}$ .

F_{4} = [2, 2, 2, 2, 3, 3, 5]

Seeing as there are no more prime factors left in $F$ , Halt. The factors of $720$ are found in the union $S_{0}, S_{1}, S_{2}, S_{3}$ and of course $1$ .

\begin{array}{lrlr} S & = {1} \cup S_{0} \cup S_{1} \cup S_{2} \cup S_{3} \\ S & = { \\ 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 30, 36, \\ 40, 45, 48, 60, 72, 80, 90, 120, 144, 180, 240, 360, 720 \\ } \end{array}

I don’t know the name of this algorithm. I figured it out myself, but the simplicity of the algorithm leads me to believe it has been known for quite some time. There are likely more efficient algorithms, but compared to the naive algorithm of dividing by all numbers less than, it is appreciably efficient. An advantage of this algorithm is that it can be done on pen-and-paper and also can be done using a computer.

Example Python Program

Squares mod 100

Two neat little identities for squares mod 100.

n^{2} \equiv (n + 50)^{2} \mod 100

And

n^{2} \equiv (50 - n)^{2} \mod 100

These can be algebraically shown to be true

n^{2} \equiv (n^{2} + 2500 + 100 n) \mod 100

What about the generic case of a modulus $k$ ? What is the specific $m$ to be added to $n$ to make the squares congruent?

n^{2} \equiv (n + m)^{2} \mod k

Expand

n^{2} \equiv (n^{2} + m^{2} + 2 m n) \mod k

The trivial case is for $m^{2} \equiv 0 \mod k$ and $2 m n \equiv 0 \mod k$ . Let’s just ignore the possibility that both of those aren’t true and yet still $(m^{2} + 2 m n) \equiv 0 \mod k$

The smallest natural number $m$ such that $2 m n \equiv 0 \mod k$ for any integer $n$ is clearly $k / 2$ if $k$ is even, and $k$ if $k$ is odd.

The smallest natural number $m$ such that $m^{2} \equiv 0 \mod k$ is a bit trickier.

$m^{2} = k^{2}$ Obviously works, but the case of $m^{2} = (k / 2)^{2}$ only works if $k$ is divisible by $4$

m = \frac{k}{2}

m^{2} = \frac{k^{2}}{2^{2}}

\frac{m^{2}}{k} = \frac{k}{4}

So for

n^{2} \equiv (n + m)^{2} \mod k

If $k$ is divisible by $4$

m = k / 2

Otherwise:

m = k

Partial Geometric sums

So, the geometric series: if you’ve taken some Calculus, you know of it.

r = 1 + x + x^{2} + x^{3} + \dots

Solve for $r$

r = 1 + r x

r - r x = 1

r (1 - x) = 1

r = \frac{1}{1 - x}

And there’s the formula the textbook gave ya'

\sum_{j = 0}^{\infty} x^{j} = \frac{1}{1 - x}

But what about the partial sums?

\sum_{j = 0}^{n} x^{j} = ?

Decompose into difference of sums

\sum_{j = 0}^{n} x^{j} = \sum_{j = 0}^{\infty} x^{j} - \sum_{j = n + 1}^{\infty} x^{j}

We know the value of the left sum, and the right sum can be found similarly.

h = x^{n + 1} + x^{n + 2} + x^{n + 3} + \dots

h = x^{n + 1} + x (x^{n + 1} + x^{n + 2} + \dots)

h = x^{n + 1} + x (h)

h = \frac{x^{n + 1}}{1 - x}

So the partial sum can be found

\sum_{j = 0}^{n} x^{j} = \frac{1}{1 - x} - \frac{x^{n + 1}}{1 - x} = \frac{1 - x^{n + 1}}{1 - x}

Some alternate forms are common

\sum_{j = 0}^{n - 1} x^{j} = \frac{1 - x^{n}}{1 - x} = \frac{x^{n} - 1}{x - 1}

Also note that, now that the upper bound of the sum is not infinity, the sum is defined for values $| x | \geq 1$

Partial Extended-Geometric Sums

This also applies to series like

x^{1} + x^{4} + x^{7} + x^{10} + \dots

or in general terms where $a > b \geq 0$

r = x^{a + b} + x^{2 a + b} + x^{3 a + b} + \dots

Finding a closed form for this is not too much more difficult. First find the closed for to the simpler

h = x^{a} + x^{2 a} + x^{3 a} + \dots

h = x^{a} (1 + x^{a} + x^{2 a} + \dots)

h = x^{a} + x^{a} h

h = \frac{x^{a}}{1 - x^{a}}

Now the more complex form can be found

r = x^{a + b} + x^{2 a + b} + x^{3 a + b} + \dots

r = x^{a + b} + x^{2 a + b} + x^{3 a + b} + \dots

r = x^{b} (x^{a} + x^{2 a} + x^{3 a} + \dots)

r = x^{b} h = \frac{x^{a + b}}{1 - x^{a}}

\sum_{j = 1}^{\infty} x^{a j + b} = \frac{x^{a + b}}{1 - x^{a}}

Following similar reasoning as the regular geometric series, the partial form is found.

m = x^{a n + b} + x^{a (n + 1) + b} + x^{a (n + 2) + b} + \dots

m = x^{b} (x^{a n} + x^{a (n + 1)} + x^{a (n + 2)} + \dots)

m = x^{b} (x^{a n} + x^{a (n + 1)} + x^{a (n + 2)} + \dots)

Pause solving $m$ , introduce new equation, $t$

t = x^{a n} + x^{a (n + 1)} + x^{a (n + 2)} + \dots

t = x^{a n} + x^{a n + 1 a} + x^{a n + 2 a} + \dots

t = x^{a n} (1 + x^{1 a} + x^{2 a} + \dots)

Pause solving $t$ , introduce new equation $o$

o = x^{a} + x^{2 a} + x^{3 a} + \dots

o = x^{a} (1 + o)

o = \frac{x^{a}}{1 - x^{a}}

Resume solving $t$

t = x^{a n} (1 + x^{1 a} + x^{2 a} + \dots)

t = x^{a n} (1 + \frac{x^{a}}{1 - x^{a}})

t = x^{a n} \frac{1}{1 - x^{a}}

Resume solving $m$

m = x^{b} (x^{a n} + x^{a (n + 1)} + x^{a (n + 2)} + \dots)

m = \frac{x^{a n + b}}{1 - x^{a}}

Thus

\sum_{j = n}^{\infty} x^{a j + b} = \frac{x^{a n + b}}{1 - x^{a}}

\sum_{j = 1}^{n - 1} x^{a j + b} = \sum_{j = 1}^{\infty} x^{a j + b} - \sum_{j = n}^{\infty} x^{a j + b}

\sum_{j = 1}^{n - 1} x^{a j + b} = \frac{x^{a + b}}{1 - x^{a}} - \frac{x^{a n + b}}{1 - x^{a}} = \frac{x^{a + b} - x^{a n + b}}{1 - x^{a}} = \frac{x^{a + b} (1 - x^{a (n - 1)})}{1 - x^{a}}

For convenience, increment the value of $n$ by one.

\sum_{j = 1}^{n} x^{a j + b} = \frac{x^{a + b} (1 - x^{a n})}{1 - x^{a}}

When $0 \leq b < a$

I’m not sure if any of that made any sense, but from my bit of verification, I believe it’s true.

As far as verification goes, it wouldn’t hurt to do so through Induction.

First the base case of $n = 1$

\sum_{j = 1}^{1} x^{a j + b} = x^{a + b}

\frac{x^{a + b} (1 - x^{a})}{1 - x^{a}}

The base case is satisfied

The equation is valid if the following holds

\frac{x^{a + b} (1 - x^{a n})}{1 - x^{a}} + x^{a (n + 1) + b} = \frac{x^{a + b} (1 - x^{a (n + 1)})}{1 - x^{a}}

x^{a + b} (1 - x^{a n}) + x^{a (n + 1) + b} (1 - x^{a}) = x^{a + b} (1 - x^{a (n + 1)})

x^{a + b} - x^{a (n + 1) + b} + x^{a (n + 1) + b} (1 - x^{a}) = x^{a + b} - x^{a (n + 2) + b}

x^{a + b} - x^{a (n + 1) + b} + x^{a (n + 1) + b} - x^{a (n + 2) + b} = x^{a + b} - x^{a (n + 2) + b}

x^{a + b} - x^{a (n + 1) + b} + x^{a (n + 1) + b} - x^{a (n + 2) + b} = x^{a + b} - x^{a (n + 2) + b}

And so it has been proven.