## Rule of 72

Posted: December 27, 2018 in Mathematics

People love money. Specially when the money gets double. So, coming up with an estimate of the time it takes to double the money helps an average person to manage their investment. However, the math to calculate the amount of time to double the money from an investment that yields a fix rate of return could be quite complex for some individuals. Fortunately, the Rule of 72 comes in handy for this purposes.

$\text{Time for Investment to Double} = \dfrac{72}{\text{Rate of Return}}$

For example, if a person invested a certain amount of money at a fixed annual interest rate of 9%, the Rule of 72 will give around 72 / 9 = 8 years for the investment to be doubled. Where the actual calculation should be log(2) / log(1.09) = 8.043, but it’s pretty damn close.

Let’s look at another example, you deposit $100 into a saving account with 4% p.a., the Rule of 72 says it will take about 72 / 4 = 18 years for your money to become$200. The actual calculation should be log(2) / log(1.04) = 17.67, again it’s close enough.

The Rule of 72 is very useful in situation like when calculators aren’t around because there is no way I can calculate log(2) / log(1.04) in my head but I can manage to compute 72 / 4 mentally.

The reason that this rule works is because of the following:

First, we are trying to find the linear approximation of $\log(1+r)$.

Let $f(x) = \log(x)$, then first order taylor series around $a=1$ would be

$f(x) \approx f(1) + f'(1)(x-1) = x-1$

hence,

$\log(x) \approx x-1, \quad x\approx 1$

or

$\log(1+r) \approx r, \quad r\approx0$

So, $2 = \left(1+\frac{r}{100}\right)^t$ becomes

$t = \dfrac{\log 2}{\log(1+r/100)} = \dfrac{\log 2}{r/100}\cdot\dfrac{r/100}{\log(1+r/100)} \approx \dfrac{100 \log 2}{r} \approx \dfrac{69.31}{r} \approx \dfrac{72}{r}$

The number 72 is chosen because in fact it is close to 69.31 and 72 has 12 divisors that is considered a lot. 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72 divide 72.

## Perfect Number & Mersenne Prime

Posted: December 25, 2018 in Mathematics

One of the interesting thing about Number Theory is that quite often a theorem or a conjecture is easy to be stated and yet it is hard to be proven. For example, a conjecture that there are infinitely many perfect numbers. To this date no one has ever proved it true or false. However it is believed to be true by many mathematicians. First of all, what is a perfect number? A perfect number is a positive integer such that the sum of all its proper divisors equal to the number itself. For instance, 6 is a perfect number since 1, 2, and 3 are the proper divisors of 6 and 6 = 1 + 2 + 3. The next perfect number would be 28 since 28 = 1 + 2 + 4 + 7 + 14. This has to be one of the oldest puzzles in the history of mathematics. Euclid (around 300 BC) had shown that when $2^k-1$ is prime, $2^{k-1}(2^k-1)$ is perfect. There is a special name for numbers in the form $2^k-1$. They are called the Mersenne Numbers, which is named after a French priest Marin Mersenne (1588 – 1648) who devoted a lot of his time and effort studying them. And whenever a Mersenne number is prime, it is called a Mersenne Prime. So there is a relationship between Mersenne Prime and Perfect Number. To be more accurate, when you find a Mersenne prime, you will find an even perfect number. It was proven by Euler around the 18th Century that all even perfect numbers must be in the form $2^{k-1}(2^k-1)$. Now, the proof is not really that difficult in today’s standard.

Euclid-Euler Theorem. For $k > 1$, if $2^k-1$ is prime, then $n = 2^{k-1}(2^k-1)$ is perfect. And every even perfect number is of this form.

Proof. Denote the sum of all divisors function as $\sigma$. Let $2^k-1$ be Mersenne Prime and $n = 2^{k-1}(2^k-1)$ be an integer. Since $(2^{k-1}, 2^k-1) = 1$ and $\sigma$ is multiplicative, we then have

$\sigma(n) = \sigma(2^{k-1}(2^k-1)) = \sigma(2^{k-1})\sigma(2^k-1)$

Since the factors of $2^{k-1}$ are $1, 2, 2^2, \ldots, 2^{k-1}$ and $2^k-1$ is Mersenne prime,

$\sigma(2^{k-1}) = 2^k-1 \quad\text{and}\quad \sigma(2^k-1) = 2^k$

$\therefore \sigma(n) = (2^k-1)2^k = 2n$

And this is exactly the definition of a perfect number.

To prove the converse, assume n is an even perfect number such as $n = 2^{k-1}(2a-1)$ where a and k are positive integers and $k\geq2$. Since $(2^{k-1}, 2a-1) = 1$, we have

$\sigma(n) = \sigma(2^{k-1}(2a-1)) = \sigma(2^{k-1})\sigma(2a-1) = (2^k-1)\sigma(2a-1)$

Since n is perfect, then

$\sigma(n) = 2n = 2\cdot 2^{k-1}(2a-1) = 2^k(2a-1)$

$\therefore 2^k(2a-1) = (2^k-1)\sigma(2a-1) \qquad (1)$

Hence, $2^k-1 \;|\; 2^k(2a-1)$, but $(2^k-1, 2^k)=1$, so $2^k-1 \;|\; 2a-1$, which means $2a-1 = (2^k-1)b$ for some integer b. Then equation (1) becomes

$2^k(2^k-1)b = (2^k-1)\sigma(2a-1)$

$2^kb = \sigma(2a-1)$

Since $2a-1$ and $b$ are both divisors of $2a-1$, we have

$2^kb = \sigma(2a-1) \geq (2a-1) + b = (2^k-1)b + b = 2^kb$

This forces the inequality to equality,

$\sigma(2a-1) = (2a-1) + b$

And this implies that $2a-1$ has only two divisors and so $2a-1$ must be prime and $b=1$. Therefore $2a-1 = (2^k-1)b = 2^k-1$ is a prime number, which completes the proof. QED.

So the problem of finding even perfect numbers boils down to the hunt for Mersenne primes. And I remember the last time I wrote the article on the largest known prime, the last greatest Mersenne prime was $2^{77232917}-1$. But now the record has been broken again. On December 7, 2018 the largest known prime (or Mersenne prime) is found to be

$2^{82589933}-1$

which makes the largest known perfect number to be

$2^{82589932}(2^{82589933}-1)$

which is 49,724,095 digits long!

## Proof of the Square of Standard Normal is a Chi-Square(1) Distribution

Posted: December 3, 2018 in Mathematics

Theorem. If $Z \sim N(0, 1)$, then $W = Z^2 \sim \chi^2(1)$.

Proof

The idea of the proof is to show that the pdf of random variable W is the same as the pdf of chi-square random variable with 1 degree of freedom, $\chi^2_1$. That is

$f_W(w) = \dfrac{1}{2^{1/2}\Gamma(1/2)}w^{1/2-1}e^{-w/2}, w>0$

So, our strategy is to find F(w), the cdf of W, and then differentiate it with respect to w to get f(w), the pdf of W.

Here we start,

$F_W(w) = P(W

$=P(|Z|<\sqrt{w}) = P(-\sqrt{w}

$= \displaystyle \int_{-\sqrt{w}}^{\sqrt{w}} \frac{1}{\sqrt{2\pi}} e^{-z^2/2} dz$

Now, since standard normal distribution is symmetric about zero. hence,

$F_W(w) = \displaystyle 2\int_0^{\sqrt{w}} \frac{1}{\sqrt{2\pi}} e^{-z^2/2} dz$

Now, differentiating F(w) with respect to w to get f(w).

$\displaystyle f_W(w) = F_W'(w) = \frac{d}{dw} 2\int_0^{\sqrt{w}} \frac{1}{\sqrt{2\pi}} e^{-z^2/2} dz$

Now, by the Fundamental Theorem of Calculus and chain rule, we get

$\displaystyle f_W(w) = \frac{2}{\sqrt{2\pi}} e^{-w/2}\cdot\frac{1}{2\sqrt{w}}$

$\displaystyle = \frac{1}{\sqrt{2\pi}} w^{-1/2} e^{-w/2}, \quad 0 < w < \infty$

Now since $\Gamma(\frac12) = \sqrt\pi$, we get

$f_W(w) = \dfrac{1}{2^{1/2}\Gamma(1/2)} w^{1/2-1}e^{-w/2}, w>0$

$\therefore W=Z^2 \sim \chi^2(1). \quad \square$

## The Number 153 in the Bible

Posted: December 3, 2018 in Mathematics

The number 153 appears in the Gospel of John 21:1-11 in the following.

After this Jesus revealed himself again to the disciples by the Sea of Tiberias, and he revealed himself in this way. Simon Peter, Thomas (called the Twin), Nathanael of Cana in Galilee, the sons of Zebedee, and two others of his disciples were together. Simon Peter said to them, “I am going fishing.” They said to him, “We will go with you.” They went out and got into the boat, but that night they caught nothing.

Just as day was breaking, Jesus stood on the shore; yet the disciples did not know that it was Jesus. Jesus said to them, “Children, do you have any fish?” They answered him, “No.” He said to them, “Cast the net on the right side of the boat, and you will find some.” So they cast it, and now they were not able to haul it in, because of the quantity of fish. That disciple whom Jesus loved therefore said to Peter, “It is the Lord!” When Simon Peter heard that it was the Lord, he put on his outer garment, for he was stripped for work, and threw himself into the sea. The other disciples came in the boat, dragging the net full of fish, for they were not far from the land, but about a hundred yards off.

When they got out on land, they saw a charcoal fire in place, with fish laid out on it, and bread. Jesus said to them, “Bring some of the fish that you have just caught.” So Simon Peter went aboard and hauled the net ashore, full of large fish, 153 of them. And although there were so many, the net was not torn.

Why the number 153 is mentioned specifically in the gospel? There is no clear answer to this question. 153 may looks like a random number that doesn’t have any meaning in the story. So, let’s try to examine the number 153.

153 is obviously divisible by 9 because 1 + 5 + 3 = 9. So let’s divide that by 9 and we’ll get 17. Hence,

$153 = 9 \cdot 17 = 3^2 \cdot 17$

Now, since two times 9 is 18 and this implies that 153 is the 17th triangular number.

$T_{17} = \dfrac{17(17+1)}{2} = 17 \cdot 9 = 153$

What so special about the 17th triangular number? 17 is a prime number. 17 is 10 plus 7. There is a 10 commandments and God creates the earth in 7 days. Could this be the hidden message? Or maybe this

$17 = 10 + 7 = (2\times5) + (2+5)$

Anyhow, I don’t know neither. But mathematically, the number 153 indeed has some nice properties. Such as:

1. 153 is a 17th triangular number. $153 = \dfrac{(17+1)(17)}{2}$
2. 153 is equal to the sum of the first 5 factorials. $153 = 1! + 2! + 3! + 4! + 5!$
3. 153 is an Armstrong number. $153 = 1^3 + 5^3 + 3^3$

An Armstrong number is a positive integer that is equal to the sum of its digits each raised to the power of the number of digits.