Alexander Bass

Math Notes Fall 24

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⌋\lfloor x \rfloor with respect to xx is equal to

∫⌊x⌋ dx==∑n=0⌊x⌋−1n+(x−⌊x⌋)(⌊x⌋)=12(⌊x⌋−1)(⌊x⌋−1+1)+(x−⌊x⌋)⌊x⌋=(x−12)⌊x⌋−12⌊x⌋2 \begin{align} \int \lfloor x \rfloor \, dx =& \\ =& \sum_{n=0}^{\lfloor x \rfloor -1} n + (x-\lfloor x \rfloor )(\lfloor x \rfloor )\\ =& \frac{1 }{2} (\lfloor x \rfloor -1)(\lfloor x \rfloor -1 +1 ) + (x-\lfloor x \rfloor )\lfloor x \rfloor \\ =& \left( x-\frac{1}{2} \right) \lfloor x \rfloor - \frac{1}{2} \lfloor x \rfloor^2 \end{align}

(Constant of integration omitted)

A similar process can be done for the ceiling function ⌈x⌉\lceil x \rceil.

∫⌈x⌉ dx==∑n=0⌈x⌉n+(x−⌈(x)⌉)⌈x⌉=12⌈x⌉(⌈x⌉+1)+(x−⌈x⌉)⌈x⌉=(x+12)⌈x⌉−12⌈x⌉2 \begin{align} \int \lceil x \rceil \, dx =& \\ =&\sum_{n=0}^{\lceil x \rceil } n +(x-\lceil (x) \rceil )\lceil x \rceil \\ =& \frac{1}{2} \lceil x \rceil(\lceil x \rceil +1 ) + (x-\lceil x \rceil )\lceil x \rceil \\ =& \left( x+\frac{1}{2} \right)\lceil x \rceil -\frac{1}{2} \lceil x \rceil^2 \end{align}

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

∫nint⁡(x) dx=x(nint⁡(x))−12nint⁡(x)2 \int \operatorname{nint}(x) \, dx = x(\operatorname{nint} (x)) - \frac{1}{2} \operatorname{nint}(x)^2

Fibonacci Sequence

The Fibonacci sequence is defined as

F1=F2=1Fn+2=Fn+1+Fn \begin{align} F_{1} =& F_{2} = 1\\ F_{n+2} =& F_{n+1} + F_{n} \end{align}

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

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

Substitute into inequality

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

Which is never true for the sequence.

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

Again substitute

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

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

In a similar spirit, it is obvious that

Fn=Fn−1+Fn−2 F_n = F_{n-1} + F_{n-2}

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

Fn=Fn−1+Fn−2=Fn−2+Fn3+Fn−2=2Fn−2+Fn−3and again=2Fn−3+2Fn−4+Fn−3=3Fn−3+2Fn−4 \begin{align} F_n =& F_{n-1} + F_{n-2}\\ =& F_{n-2} + F_{n_3} + F_{n-2}\\ =& 2F_{n-2} + F_{n-3}\\ & \text{and again}\\ =& 2F_{n-3} + 2F_{n-4} + F_{n-3}\\ =& 3F_{n-3} + 2F_{n-4} \end{align}

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

akFn−1+bkFn−2=ak(Fn−2+Fn−3)+bkFn−2=(ak+bk)Fn−2+akFn−3 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 a1=b1=1a_1 =b_1 = 1. and,

ak+1=ak+bkbk+1=ak \begin{align} a_{k+1} =& a_k + b_k \\ b_{k+1} =& a_k \end{align} ak+2=ak+1+bk+1=ak+1+ak 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>1k > 1

Fn=akFn−k+1+bkFn−k F_n = a_k F_{n-k+1} + b_k F_{n-k} Fn=FkFn−k+1+Fk−1Fn−k F_n = F_k F_{n-k+1} + F_{k-1} F_{n-k}

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

F10=F4F7+F3F6 F_{10} = F_{4}F_{7} + F_{3} F_{6} 55=3⋅13+2⋅8 55 = 3\cdot 13 + 2 \cdot 8

Generalized Geometric Series

The geometric series is defined as

∑n=1∞rn \sum_{n=1}^{\infty} r^n

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

x=r+r2+r3+…+rnx=r(1+r+r2+…+rn−1)x=r(1+x)x=r1−r \begin{align} x &= r + r^2+r^3 + \ldots + r^n\\ x &= r(1 +r + r^2 + \ldots + r^{n-1})\\ x &= r(1+x)\\ x &= \frac{r}{1-r} \end{align}

How about some other values for the exponent of rr? Let’s try the odd numbers 2n−12n-1

x=r+r3+r5+…+r2n−1x=r(1+r2+r4+…+r2n−2)x=r(1+r(r+r3+…+r2n−3))x=r(1+rx)x=r1−r2 \begin{align} x &= r+r^3+r^5+\ldots + r^{2n-1}\\ x &= r(1+r^2+r^4+\ldots + r^{2n-2})\\ x &= r(1+r(r+r^3+\ldots + r^{2n-3}))\\ x &= r(1+rx)\\ x&= \frac{r}{1-r^2} \end{align} ∑n=1∞r2n−1=r1−r2 \sum_{n=1}^{\infty} r^{2n-1} = \frac{r}{1-r^2}

How about for the multiples of three 3n3n

x=r3+r6+r9+…+r3nx=r3(1+r3+r6+…+r3n−3)x=r3(1+x)x=r31−r3 \begin{align} x &= r^3+r^6+r^9+\ldots +r^{3n}\\ x &= r^3(1 + r^3+r^6+\ldots +r^{3n-3})\\ x &= r^3(1+x)\\ x &= \frac{r^3}{1-r^3} \end{align} ∑n=1∞r3n=r31−r3 \sum_{n=1}^{\infty} r^{3n} =\frac{r^3}{1-r^3}

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

∑n=1∞rkn=rk1−rk \sum_{n=1}^{\infty} r^{kn} = \frac{r^k}{1-r^k}

To show that this is true, consider the following

x=rk+r2k+r3k+…+rnkx=rk(1+rk+r2k+…+rnk−k)x=rk(1+x)x=rk1−rk \begin{align} x &= r^{k}+ r^{2k} + r^{3k} + \ldots + r^{nk}\\ x &= r^{k}(1+r^{k}+ r^{2k} + \ldots + r^{nk-k})\\ x &= r^{k}(1+x)\\ x &= \frac{r^{k}}{1-r^{k}} \end{align}

How about for numbers which leave a remainder of two when divided by three: 3n−13n-1?

x=r2+r5+r8+…+r3n−1x=r2(1+r3+r6+…+r3n−3) \begin{align} x &= r^{2} + r^{5} + r^{8} + \ldots + r^{3n-1} \\ x &= r^2 (1+ r^3 + r^6 +\ldots + r^{3n-3}) \end{align}

The earlier found identity allows us to solve this

x=r2(1+∑n=1∞r3n)x=r2(1+r31−r3)x=r2+r51−r3x=r21−r3 \begin{align} x &= r^2 \left(1+\sum_{n=1}^{\infty} r^{3n}\right)\\ x &= r^2\left(1+ \frac{r^3}{1-r^3}\right)\\ x &= r^2+ \frac{r^5}{1-r^3}\\ x &= \frac{r^2}{1-r^3} \end{align} ∑n=1∞r3n−1=r21−r3 \sum_{n=1}^{\infty} r^{3n-1} = \frac{r^2}{1-r^3}

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

∑n=1∞ran−b=x \sum_{n=1}^{\infty}r^{an-b} = x

where a,b∈N a,b \in \mathbb{N}, 0≤b<a0 \leq b \lt a, and −1<r<1-1<r<1?

x=ra−b+r2a−b+r3a−b+…+ran−bx=ra(r−b+ra−b+r2a−b+…+ra(n−1)−b)x=ra(r−b+x)x=ra−b+raxx=ra−b1−ra \begin{align} x&= r^{a-b} + r^{2a-b} + r^{3a-b} + \ldots + r^{an-b}\\ x&= r^{a}(r^{-b} + r^{a-b} + r^{2a-b} + \ldots + 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{align}

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

r21−r3 \frac{r^{2}}{1-r^{3}}

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

r1−r2 \frac{r}{1-r^2}

Thus I think it’s fair to say

∑n=1∞ran−b=ra−b1−ra \sum_{n=1}^{\infty}r^{an-b} =\frac{r^{a-b}}{1-r^{a}}

There’s one final case that has not been covered an+ban +b where a,b∈Na,b \in \mathbb{N} and where a,b>0a,b > 0

∑n=1∞ran+b \sum_{n=1}^{\infty} r^{an+b} x=ra+b+r2a+b+r3a+b+…+ran+bx=ra+b(1+ra+r2a+…+ra(n−1))x=ra+b(1+ra1−ra)x=ra+b+r2a+b1−rax=ra+b−r2a+b+r2a+b1−ra \begin{align} x&=r^{a+b} + r^{2a+b} + r^{3a+b} + \ldots +r^{an+b}\\ x&=r^{a+b} (1+ r^{a} + r^{2a} + \ldots +r^{a(n-1)})\\ x&=r^{a+b} \left(1+ \frac{r^{a}}{1-r^{a}}\right)\\ x&= r^{a+b} + \frac{r^{2a+b}}{1-r^{a}}\\ x&= \frac{r^{a+b} - \cancel{r^{2a+b}} + \cancel{r^{2a+b}}}{1-r^a}\\ \end{align}

Thus

∑n=1∞ran+b=ra+b1−ra \sum_{n=1}^{\infty} r^{an+b} = \frac{r^{a+b}}{1-r^a}

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

∑n=1∞rf(n) \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.