Math Notes Fall 24

January 6th 2025

Often times when trying to teach myself math I come across little tidbits that, while likely not original, are neat and obscure. Here are some.

Integrals of Step Functions

The integral of the function $⌊ x ⌋$ with respect to $x$ is equal to

\begin{array}{lrlr} \int ⌊ x ⌋ d x = \\ = & \sum_{n = 0}^{⌊ x ⌋ - 1} n + (x - ⌊ x ⌋) (⌊ x ⌋) \\ = & \frac{1}{2} (⌊ x ⌋ - 1) (⌊ x ⌋ - 1 + 1) + (x - ⌊ x ⌋) ⌊ x ⌋ \\ = & (x - \frac{1}{2}) ⌊ x ⌋ - \frac{1}{2} ⌊ x ⌋^{2} \end{array}

A similar process can be done for the ceiling function $⌈ x ⌉$ .

\begin{array}{lrlr} \int ⌈ x ⌉ d x = \\ = & \sum_{n = 0}^{⌈ x ⌉} n + (x - ⌈ (x) ⌉) ⌈ x ⌉ \\ = & \frac{1}{2} ⌈ x ⌉ (⌈ x ⌉ + 1) + (x - ⌈ x ⌉) ⌈ x ⌉ \\ = & (x + \frac{1}{2}) ⌈ x ⌉ - \frac{1}{2} ⌈ x ⌉^{2} \end{array}

I’ll spare the fine details on the final one. The ‘round to the nearest integer’ function $nint (x)$ can also be integrated.

\int nint (x) d x = x (nint (x)) - \frac{1}{2} nint (x)^{2}

Fibonacci Sequence

The Fibonacci sequence is defined as

\begin{array}{lrlr} F_{1} = & F_{2} = 1 \\ F_{n + 2} = & F_{n + 1} + F_{n} \end{array}

Are there any Fibonacci numbers $F_{n}$ such that $F_{n + 1} > 2 F_{n}$ ?

F_{n + 1} = F_{n} + F_{n - 1}

Substitute into inequality

F_{n} + F_{n - 1} > 2 F_{n}

F_{n - 1} > F_{n}

Which is never true for the sequence.

Another question: which Fibonacci numbers $F_{n}$ are such that $F_{n + 2} > 2 F_{n}$ ?

Again substitute

F_{n + 1} + F_{n} > 2 F_{n}

F_{n + 1} > F_{n}

Which is true for all Fibonacci numbers except for when $n = 1$ .

In a similar spirit, it is obvious that

F_{n} = F_{n - 1} + F_{n - 2}

But this form can be reduced further to be a sum of $F_{n - 2}$ and $F_{n - 3}$

\begin{array}{lrlr} F_{n} = & F_{n - 1} + F_{n - 2} \\ = & F_{n - 2} + F_{n_{3}} + F_{n - 2} \\ = & 2 F_{n - 2} + F_{n - 3} \\ and again \\ = & 2 F_{n - 3} + 2 F_{n - 4} + F_{n - 3} \\ = & 3 F_{n - 3} + 2 F_{n - 4} \end{array}

The coefficients on the reduced forms are interesting to me. Given coefficients $a_{k}, b_{k}$ in a specific level of reduced Fibonacci, find $a_{k + 1}, b_{k + 1}$

a_{k} F_{n - 1} + b_{k} F_{n - 2} = a_{k} (F_{n - 2} + F_{n - 3}) + b_{k} F_{n - 2} = (a_{k} + b_{k}) F_{n - 2} + a_{k} F_{n - 3}

We define $a_{1} = b_{1} = 1$ . and,

\begin{array}{lrlr} a_{k + 1} = & a_{k} + b_{k} \\ b_{k + 1} = & a_{k} \end{array}

a_{k + 2} = a_{k + 1} + b_{k + 1} = a_{k + 1} + a_{k}

Notice something peculiar about the above identity: it’s also the Fibonacci sequence!

This allows us to define a specific level of reduced Fibonacci symbolically. Let $k > 1$

F_{n} = a_{k} F_{n - k + 1} + b_{k} F_{n - k}

F_{n} = F_{k} F_{n - k + 1} + F_{k - 1} F_{n - k}

As an example $n = 10$ and $k = 4$

F_{10} = F_{4} F_{7} + F_{3} F_{6}

55 = 3 \cdot 13 + 2 \cdot 8

Generalized Geometric Series

The geometric series is defined as

\sum_{n = 1}^{\infty} r^{n}

Where $- 1 < r < 1$ . The algebraic way to solve for the closed form $x$ of the series is as follows

\begin{array}{lrlr} x & = r + r^{2} + r^{3} + \dots + r^{n} \\ x & = r (1 + r + r^{2} + \dots + r^{n - 1}) \\ x & = r (1 + x) \\ x & = \frac{r}{1 - r} \end{array}

How about some other values for the exponent of $r$ ? Let’s try the odd numbers $2 n - 1$

\begin{array}{lrlr} x & = r + r^{3} + r^{5} + \dots + r^{2 n - 1} \\ x & = r (1 + r^{2} + r^{4} + \dots + r^{2 n - 2}) \\ x & = r (1 + r (r + r^{3} + \dots + r^{2 n - 3})) \\ x & = r (1 + r x) \\ x & = \frac{r}{1 - r^{2}} \end{array}

\sum_{n = 1}^{\infty} r^{2 n - 1} = \frac{r}{1 - r^{2}}

How about for the multiples of three $3 n$

\begin{array}{lrlr} x & = r^{3} + r^{6} + r^{9} + \dots + r^{3 n} \\ x & = r^{3} (1 + r^{3} + r^{6} + \dots + r^{3 n - 3}) \\ x & = r^{3} (1 + x) \\ x & = \frac{r^{3}}{1 - r^{3}} \end{array}

\sum_{n = 1}^{\infty} r^{3 n} = \frac{r^{3}}{1 - r^{3}}

Generally, it can be assumed that for any natural number $k > 0$ and for any real $r$ where $- 1 < r < 1$ that

\sum_{n = 1}^{\infty} r^{k n} = \frac{r^{k}}{1 - r^{k}}

To show that this is true, consider the following

\begin{array}{lrlr} x & = r^{k} + r^{2 k} + r^{3 k} + \dots + r^{n k} \\ x & = r^{k} (1 + r^{k} + r^{2 k} + \dots + r^{n k - k}) \\ x & = r^{k} (1 + x) \\ x & = \frac{r^{k}}{1 - r^{k}} \end{array}

How about for numbers which leave a remainder of two when divided by three: $3 n - 1$ ?

\begin{array}{lrlr} x & = r^{2} + r^{5} + r^{8} + \dots + r^{3 n - 1} \\ x & = r^{2} (1 + r^{3} + r^{6} + \dots + r^{3 n - 3}) \end{array}

The earlier found identity allows us to solve this

\begin{array}{lrlr} x & = r^{2} (1 + \sum_{n = 1}^{\infty} r^{3 n}) \\ x & = r^{2} (1 + \frac{r^{3}}{1 - r^{3}}) \\ x & = r^{2} + \frac{r^{5}}{1 - r^{3}} \\ x & = \frac{r^{2}}{1 - r^{3}} \end{array}

\sum_{n = 1}^{\infty} r^{3 n - 1} = \frac{r^{2}}{1 - r^{3}}

That got me thinking, is there a general closed form of

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

where $a, b \in ℕ$ , $0 \leq b < a$ , and $- 1 < r < 1$ ?

\begin{array}{lrlr} x & = r^{a - b} + r^{2 a - b} + r^{3 a - b} + \dots + r^{a n - b} \\ x & = r^{a} (r^{- b} + r^{a - b} + r^{2 a - b} + \dots + r^{a (n - 1) - b}) \\ x & = r^{a} (r^{- b} + x) \\ x & = r^{a - b} + r^{a} x \\ x & = \frac{r^{a - b}}{1 - r^{a}} \end{array}

Which seems to agree with the previous result of $a = 3, b = 1$

\frac{r^{2}}{1 - r^{3}}

And the even earlier result of odd numbers $a = 2, b = 1$

\frac{r}{1 - r^{2}}

Thus I think it’s fair to say

\sum_{n = 1}^{\infty} r^{a n - b} = \frac{r^{a - b}}{1 - r^{a}}

There’s one final case that has not been covered $a n + b$ where $a, b \in ℕ$ and where $a, b > 0$

\sum_{n = 1}^{\infty} r^{a n + b}

\begin{array}{lrlr} x & = r^{a + b} + r^{2 a + b} + r^{3 a + b} + \dots + r^{a n + b} \\ x & = r^{a + b} (1 + r^{a} + r^{2 a} + \dots + r^{a (n - 1)}) \\ x & = r^{a + b} (1 + \frac{r^{a}}{1 - r^{a}}) \\ x & = r^{a + b} + \frac{r^{2 a + b}}{1 - r^{a}} \\ x & = \frac{r^{a + b} - r^{2 a + b} + r^{2 a + b}}{1 - r^{a}} \end{array}

Thus

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

All this would seem to imply that if $f (n)$ is a linear function which maps $ℕ \to ℕ$ then there exists a trivially obtainable closed form for the sum

\sum_{n = 1}^{\infty} r^{f (n)}

I’d like to see this extended to other number systems than natural numbers, but I’m a bit exhausted of series at this point.