# Category Archives: My Research

Something interesting which I believe to have discovered

# Prime Polynomial Theorem

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I just wanted to point towards a nice theorem, analogous to the Prime Number Theorem, which is not talked about much:

# irreducible monic polynomials with coefficients in $\mathbb{F}_q$ and of degree $n \sim \frac{q^n}{n}$, for a prime power $q$.

The proof of this theorem follows from Gauss’ formula:

# monic irreducible polynomialswith coefficients in $\mathbb{F}_q$ and of degree $n$ = $\displaystyle{\frac{1}{n}\sum_{d|n}\mu\left(\frac{n}{d}\right)q^d}$, by taking $d=n$.

For details, see first section of this: http://alpha.math.uga.edu/~pollack/thesis/thesis-final.pdf

# Recursion and Periodicity

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One of the simplest recursive formula that I can think of is the one which generates the Fibonacci sequence, $F_{n+1} = F_n +F_{n-1}, n\geq 1$ with $F_0 = F_1=1$. So, I will illustrate a following general concept about recursions using Fibonacci sequence.

A sequence generated by a recursive formula is periodic modulo k, for any positive integer k greater than 1.

I find this fact very interesting because it means that a sequence which is strictly increasing when made bounded using the modulo operation (since it will allow only limited numbers as the output of recursion relation), will lead to a periodic cycle.

Following are the first 25 terms of the Fibonacci sequence:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025.

And here are few examples modulo k, for k=2,3,4,5,6,7,8

As you can see, the sequence repeats as soon as 1,0 appears. And from here actually one can see why there should be a periodicity.

For the sequence to repeat, what we need is a repetition of two consecutive values (i.e. the number of terms involved in the recursive formula) in the sequence of successive pairs. And for mod k, the choices are limited, namely k^2.  Now, all we have to ensure is that “1,0” should repeat. But since consecutive pairs can’ repeat (as per recursive formula) the repetition of “1,0” must occur within the first k^2.

For rigorous proofs and its relation to number theory, see: http://math.stanford.edu/~brianrl/notes/fibonacci.pdf

# Repelling Numbers

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An important fact in the theory of prime numbers is the Deuring-Heilbronn phenomenon, which roughly says that:

The zeros of L-functions repel each other.

Interestingly, Andrew Granville in his article for The Princeton Companion to Mathematics remarks that:

This phenomenon is akin to the fact that different algebraic numbers repel one another, part of the basis of the subject of Diophantine approximation.

I am amazed by this repelling relation between two different aspects of arithmetic (a.k.a. number theory). Since I have already discussed the post Colourful Complex Functions, wanted to share this picture of the algebraic numbers in the complex plane, made by David Moore based on earlier work by Stephen J. Brooks:

In this picture, the colour of a point indicates the degree of the polynomial of which it’s a root, where red represents the roots of linear polynomials, i.e. rational numbers,  green represents the roots of quadratic polynomials, blue represents the roots of cubic polynomials, yellow represents the roots of quartic polynomials, and so on.  Also, the size of a point decreases exponentially with the complexity of the simplest polynomial with integer coefficient of which it’s a root, where the complexity is the sum of the absolute values of the coefficients of that polynomial.

Moreover,  John Baez comments in his blog post that:

There are many patterns in this picture that call for an explanation! For example, look near the point $i$. Can you describe some of these patterns, formulate some conjectures about them, and prove some theorems? Maybe you can dream up a stronger version of Roth’s theorem, which says roughly that algebraic numbers tend to ‘repel’ rational numbers of low complexity.

To read more about complex plane plots of families of polynomials, see this write-up by John Baez. I will end this post with the following GIF from Reddit (click on it for details):

# Prime Consequences

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Most of us are aware of the following consequence of Fundamental Theorem of Arithmetic:

There are infinitely many prime numbers.

The classic proof by Euclid is easy to follow. But I wanted to share the following two analytic equivalents (infinite series and infinite products) of the above purely arithmetical statement:

• $\displaystyle{\sum_{p}\frac{1}{p}}$   diverges.

For proof, refer to this discussion: https://math.stackexchange.com/q/361308/214604

• $\displaystyle{\sum_{n=1}^\infty \frac{1}{n^{s}} = \prod_p\left(1-\frac{1}{p^s}\right)^{-1}}$, where $s$ is any complex number with $\text{Re}(s)>1$.

The outline of proof,   when $s$ is a real number, has been discussed here: http://mathworld.wolfram.com/EulerProduct.html

# Solving Logarithmic Equations

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While reading John Derbyshire’s Prime Obsession I came across the following statement (clearly explained on pp. 74):

Any positive power of $\log(x)$ eventually increases more slowly than any positive power of $x$.

It is easy to prove this (existence) analytically, by taking derivative to compare slopes. But algebraically it implies that (for example):

There are either no real solution or two real solutions of the equation
$\log(x) = x^\varepsilon$
for any given $\varepsilon>0$.

Now the question that arises is “How to find this $x$?” I had no idea about how to solve such logarithmic equations, so I took help of Google and discovered this Mathematic.SE post. So, we put $\log(x)=y$ and re-write the equation as:

$y=e^{y\varepsilon}$

Now to be able to use Lambert W function (also called the product logarithm function) we need to re-write the above equation, but I have failed to do so.

But using WolframAlpha I was able to solve $\log(x)=x^2$ to get $x=e^{\frac{-W(-2)}{2}}$ (which is an imaginary number, i.e. no real solution of this equation) but I was not able to figure out the steps involved. So if you have any idea about the general method or the special case of higher exponents, please let me know.

# Combinatorial Puzzle

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This is a continuation of previous post:

How many distinct numbers can be formed by using four 2s and the four arithmetic operations $+,-,\times, \div$.

For example:
$1 = \frac{2+2}{2+2}=\frac{2}{2}\times\frac{2}{2}$
$2 = 2+\frac{2-2}{2}=\frac{2}{2}+\frac{2}{2}$
$3 = 2+2 - \frac{2}{2}$
$4 = 2+2+2-2 = (2\times 2) + (2-2)$
(note that some binary operations do not make sense without parenthesis)

I have no idea about how to approach this problem (since I am not very comfortable with combinatorics). So any help will be appreciated.

Edit[29 May 2017]: This problem has been solved in the comments below.

# Arithmetic Puzzle

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Following is a very common arithmetic puzzle that you may have encountered as a child:

Express any whole number $n$ using the number 2 precisely four times and using only well-known mathematical symbols.

This puzzle has been discussed on pp. 172 of Graham Farmelo’s “The Strangest Man“, and how Paul Dirac solved it by using his knowledge of “well-known mathematical symbols”:

$\displaystyle{n = -\log_{2}\left(\log_{2}\left(2^{2^{-n}}\right)\right) = -\log_{2}\left(\log_{2}\left(\underbrace{\sqrt{\sqrt{\ldots\sqrt{2}}}}_\text{n times}\right)\right)}$

This is an example of thinking out of the box, enabling you to write any number using only three/four 2s. Though, using a transcendental function to solve an elementary problem may appear like an overkill.  But, building upon such ideas we can try to tackle the general problem, like the “four fours puzzle“.

This post on Puzzling.SE describes usage of following formula consisting of  trigonometric operation $\cos(\arctan(x)) = \frac{1}{\sqrt{1+x^2}}$ and $\tan(\arcsin(x))=\frac{x}{\sqrt{1-x^2}}$ to obtain the square root of any rational number from 0:

$\displaystyle{\tan\left(\arcsin\left(\cos\left(\arctan\left(\cos\left(\arctan\left(\sqrt{n}\right)\right)\right)\right)\right)\right)=\sqrt{n+1}}$.

Using this we can write $n$ using two 2s:

$\displaystyle{n = (\underbrace{\tan\arcsin\cos\arctan\cos\arctan}_{n-4\text{ times}}\,2)^2}$

or even with only one 2:

$\displaystyle{n = \underbrace{\tan\arcsin\cos\arctan\cos\arctan}_{n^2-4\text{ times}}\,2}$