Tag Archives: English

The Maths Behind Logistic Regression

I got it wrong šŸ™

What are Your Odds?

Let’s have a look at this god-tier math puzzle. Only one out of seven gets it right, and the other six don’t. So, what are the odds for solving it correctly? 1 to 6. Generally, if \(p\) is the probability that someone will get it right, then his/her odds are \(p/(1-p)\).

However, it isn’t necessarily true that \(p=1/7\) for every person because some people are smarter, some have better education and so on. Hence, \(p\) also depends on the person attempting the puzzle. In a Bayesian framework, we capture this dependence with conditional probability.

Continue reading The Maths Behind Logistic Regression

Want to Fight Climate Change? Don’t Waste Food

source: http://res.publicdomainfiles.com/pdf_view/152/13985772022293.jpg

A few days ago, my roommate and I were getting dinner at an Japanese restaurant. While we waited for food, we were having a brief discussion about the recent heat waves in Europe. Both of us felt very sad about these changes. Continue reading Want to Fight Climate Change? Don’t Waste Food

A Week of Poetry: Day 3

Sonnet 130

By William Shakespeare

My mistressā€™ eyes are nothing like the sun;
Coral is far more red than her lipsā€™ red;
If snow be white, why then her breasts are dun;
If hairs be wires, black wires grow on her head.
I have seen roses damaskā€™d, red and white,
But no such roses see I in her cheeks; Continue reading A Week of Poetry: Day 3

How Game of Thrones Should Have Ended

This is a very unusual post for this blog. I hope my regular readers will bear with me.

I love Game of Thrones. I have read the books and the companion novels. I have been following the series for years. I am familiar with most of the fan theories too. Tonight the series finale aired. I didn’t like how the show ended. It felt rushed and very baffling. Continue reading How Game of Thrones Should Have Ended

Prime Counting Function and Chebyshev Bounds

The distribution of primes plays a central role in number theory. The famous mathematician Gauss had conjectured that the number of primes between \(1\) and \(n\) is roughly \(n/\log n\). This estimation gets more and more accurate as \(n\to \infty\). We use \(\pi(n)\) to denote the number of primes between \(1\) and \(n\). So, mathematically, Gauss’s conjecture is equivalent to the claim

\[\lim_{n\to\infty}\frac{\pi(n)}{n/\log n}=1\]

Continue reading Prime Counting Function and Chebyshev Bounds

Lie Detectors and the Story of Halting Turing Machines

 

Is it possible to design a machine that detects lies? Sure, people have already built devices called polygraphs that monitor blood pressure, respiration, pulse etc. to determine if a person is giving false information. However, that is not same as “detecting lies” because polygraphs merely measure physical effects of telling lies. On the top of that, they are hella inaccurate. I am talking about REAL lie detectors, ie. machines that can instantly validate the statements themselves. Can we actually make such machines? (*insert new startup idea*) Continue reading Lie Detectors and the Story of Halting Turing Machines

A Special Case of Zsigmondy’s Theorem

Theorem (Zsigsmondy):Ā 

  1. For two coprime positive integersĀ \(a\) and \(b\) and for any positive integer \(n\), \(a^n-b^n\) has a prime divisor that does not divide \(a^k-b^k\) for all positive integers \(k< n\) with the following exceptions:
    • \(n = 2\) and \(a+b\) is a power of two.
    • \(n=6, a=2, b=1\).
  2. For two coprime positive integersĀ \(a\) and \(b\) and for any positive integer \(n\), \(a^n+b^n\) has a prime divisor that does not divide \(a^k+b^k\) for all positive integers \(k< n\) with the following exception:
    • \(n=3, a=2, b=1\).

This theorem is very helpful in solving many olympiad number theory problems. However, its use is often frowned upon, as this theorem is quite hard to prove using only elementary mathematics. (I am aware of the proof using Cyclotomic Polynomials, but that’s not “elementary enough” in my opinion.)

In olympiad mathematics, one can get away in most of the cases by just using this theorem for a fixed pair of \(n\) and \(k\). This modification allows us to provide a very simple proof using LTE. Here I’ll restate this particular case, and then prove the first part. The second part can be proven analogously.

Theorem (Zsigsmondy Special Case):Ā 

  1. For two coprime positive integersĀ \(a\) and \(b\) and any two positive integers \(n\) and \(k\) with \(k<n\), \(a^n-b^n\) has a prime divisor that does not divide \(a^k-b^k\) with the following exception:
    • \(n = 2, k=1\) and \(a+b\) is a power of two.
  2. For two coprime positive integersĀ \(a\) and \(b\) and any two positive integers \(n\) and \(k\) with \(k<n\), \(a^n+b^n\) has a prime divisor that does not divide \(a^k+b^k\) with the following exception:
    • \(n=3, k=1,a=2, b=1\).

Proof of 1:

Suppose \(a^n-b^n\) and \(a^k-b^k\) share same set of prime divisors. This implies \(a^n-b^n\) and \(a^{\gcd(n,k)}-b^{\gcd(n,k)}\) also share same set of prime divisors. Assuming \(A=a^{\gcd(n,k)}\) and \(B=b^{\gcd(n,k)}\), we could follow the rest of this argument and get to a contradiction. So, without loss of generality, we may assume \(k=1\).

Now we consider two cases:

Case 1: \(n\) is a power of \(2\).
If \(n=2\) and \(a+b\) is a power of two, we arrive at one of the listed exceptions. Now note that,
\(\begin{align*}&\gcd(a+b, a^2+b^2)\\
=&\gcd(a+b, (a+b)^2-a^2-b^2)\\
=&\gcd(a+b,2ab)\\
=&\gcd(a+b,2)\end{align*}\)
This implies both \(a+b\) and \(a^2+b^2\) can’t be powers of two unless \(a+b=2\implies a=b=1\). Hence at least one of them must have an odd prime divisor. Furthermore, this prime divisor does not divide \(a-b\) since \(\gcd(a-b, a^2+b^2)=\gcd(a-b,2)\) (following the previous steps). However, if \(n>2\), then \(a^2+b^2|a^n-b^n\), implying that odd prime will divide \(a^n-b^n\). This is a contradiction.

Case 2: \(n=2^md\) with \(d>1\) being odd.
Without loss of generality, we may assume \(n>1\) is odd, since \(a^d-b^d|a^n-b^n\) and it is sufficient to show that \(a^d-b^d\) has a prime divisor that does not divide \(a-b\).Ā Ā From LTE, \(v_p(a-b)+v_p(n)=v_p(a^n-b^n)\) for each odd prime \(p|a-b\). Furthermore, \[\frac{a^n-b^n}{a-b}\equiv na^{n-1}\equiv 1\pmod 2\] implying \(v_2(a-b)=v_2(a^n-b^n)\). Therefore, we can conclude \[\frac{a^n-b^n}{a-b}=\prod_{p|\gcd(n,a-b)}p^{v_p(n)}\leq n\] which is impossible for \(n>1\). This raises another contradiction and we are done.