### Direct Product of Groups, Example, Inverse

Labels: Cayley, Direct Product, Example, Groups, Table

This is a short piece to summarize whats going in my life.

Labels: Chebyshev, Differential Equation, Polynomials, Proof, Properties, Recurrence

Yesterday I went to Avenida Chile after getting my new cedular. It was 4:50 pm and the very next movie was La Doncella. The movie had Spanish Subtittles. So i decided to go ahead and watch it. The cinematography is spectacular. The story is about a conmen who lists the help of a family of con artists so that he can marry a rich heiress. The heiress is lesbian and falls in love with the handmaid who he had connived with. Well there is another side of the story so i will not spoil the fun for you. A great movie.

Note to prove this we need to use Cauchy Schwarz Inequality. Also notice that triangle inequality is addition.
In short:
Triangle inequality says the sum of two sides is bigger than 3rd side.
Cauchy inequality says the product of two sides is bigger than their dot product

Labels: Cauchy Schwarz, inequality, Proof, Real Analysis, Triangle

Take two different rational numbers $r_1,r_2$, w.l.o.g we can assume that $r_2 > r_1 \Rightarrow r_2-r_1 > 0$. Now we choose an irrational number between $0$ and $1$. Before we do that we must recall that reciprocal of an irrational number is an irrational number ie. $\frac{1}{\text{irrational}}= \text{irrational}$ and rational + irrational = irrational.
There are many choices of irrational numbers to start with lets start with $\frac{1}{\sqrt{5}}$.
$\Rightarrow 0\frac{1}{\sqrt{5}}$ Multiplyting all the sides by a positive number $r_2-r_1$ we get
$0 < (r_2-r_1)\frac{1}{\sqrt{5}} < r_2-_1$
Adding $r_1$ to all the sides we get
$r_1 < r_1+\frac{r_2-r_1}{\sqrt{5}}< r_2$.
As $r_1+\frac{r_2-r_1}{\sqrt{5}}$ is irrational numbers between $r_1$ and $r_2$.

Labels: Density, Irrational Numbers, Proof, Rational Numbers, Real Analysis

While there are many ways to derive the recursive formula of Fibonacci Sequence. The formula to find the nth term of a Fibonacci sequence is a beautiful formula. We are not going to derive here but prove the formula using mathematical Induction

We use Strong Induction to prove the Binet Formula because we will be invoking the recursive formula of fibonacci sequence to prove the $n+1$ case.

The Binet formula is $\frac{\phi^n-\alpha^n}{\sqrt{5}}$ $\phi = \frac{1+\sqrt{5}}{2}$ and $\alpha = \frac{1-\sqrt{5}}{2}$.

Verify the base cases: for $n = 1$ we have $\frac{\phi-\alpha}{\sqrt{5}}= 1$

Similarly for $n = 2$ we have $\frac{\phi^2-\alpha^2}{\sqrt{5}}$

$\Rightarrow \frac{(\phi-1)-(\alpha-1)}{\sqrt{5}}$

$\Rightarrow \frac{\phi-\alpha}{\sqrt{5}}$

which we know from first statement is true To prove the last step we realize that $F_{n+1}=F_n+F_{n-1}$

Thus we have $\Rightarrow F_{n+1}= \frac{\phi^n-\alpha^n}{\sqrt{5}}+\frac{\phi^{n-1}-\alpha^{n-1}}{\sqrt{5}}$

$\Rightarrow F_{n+1}= \frac{\phi^n+\phi^{n-1}-\alpha^n-\alpha^{n-1}}{\sqrt{5}}$

$\Rightarrow F_{n+1}= \frac{\phi^{n-1}(\phi+1)-\alpha^{n-1}(\alpha+1)}{\sqrt{5}}$

$\Rightarrow F_{n+1}= \frac{\phi^{n-1}(\phi^2)-\alpha^{n-1}(\alpha^2)}{\sqrt{5}}$

$\Rightarrow F_{n+1}= \frac{\phi^{n+1}-\alpha^{n+1}}{\sqrt{5}}$

Hence Proved

Labels: Binet Formula, Fibonacci Numbers, Proof, Strong Induction

Here is one way to think about Fibonacci Numbers.

1. There are $n$ number of balls of only two colors (Let these be red and blue).

2. The red balls are identical and so are blue balls.

3. No two red balls can be together.

Case n=1 ball -- 2 ways.

Clearly there are two ways to do this. Either you have a red ball or a blue ball.

$\binom{1+1}{0}+\binom{1}{1}=1+1=2$

Case n=2 balls -- 3 ways.

BB (2 Blue)

BR (1 Blue 1 Red)

RB

$\binom{2+1}{0}+\binom{2}{1}=1+2=3$

Case n = 3 balls -- 5ways

BBB (3 blue)

BBR (2 Blue 1 Red)

BRB

RBB

RBR (1 Blue 2 Red)

$\binom{3+1}{0}+\binom{3}{1}+\binom{2}{2}=1+3+1=5$

Case n=k balls $\binom{k+1}{0}+\binom{k}{1}+\binom{k-1}{2}+\cdots$

In Summation notation $\sum_{i=0}^{k}\binom{n+1-i}{i}$

Note the reason for $k+1$ is because we are trying to find the number of space among $k$ blue balls and there are $k+1$ such space and we are trying to find $0$ spot for a red ball.

In second case we have $k-1$ blue balls and we have $k$ spots and we are trying to find one spot for the red balls. So we can generalize this for $k$ balls starting with all blue balls and they have $k+1$ space between them. However we need $0$ red balls to be placed. As total number of balls have to be $k$ In next case we have $k-1$ blue balls and there are $k$ space among these blue balls and we need $1$ space for our red ball and this can be done in $\binom{k-1+1}{1}= \binom{k}{1}$ After this we have $k-2$ blue balls and $2$ red balls. For $k-2$ blue balls we have $k-2+1= k-1$ spaces and two spaces for red balls can be chosen in $\binom{k-1}{2}$ and so on.

Labels: Binomial Coefficients, Combinatorics, Fibonacci Numbers

Following are the 5 Axioms of Peano or also called as Peano's Postulates

N1. $1 \in \mathbb{N}$

N2. If $n \in \mathbb{N} \Rightarrow n+1 \in \mathbb{N}$

N3. $1$ is not the successor of any element in $\mathbb{N}$.

N4. If two numbers $m,n \in \mathbb{N}$ have the same successor then $m=n$.

N5. A subset of $\mathbb{N}$ which contains $1$, and which contains $n+1$ whenever it contains $n$, must equal $\mathbb{N}$

Q. What is the significance of Peano's Axiom ?

Most familiar properties of $\mathbb{N}$ can be proved using Peano's Axioms.

Q. How do you prove N5 ?

Given the set contains $1$. If We will prove by Contradiction Suppose there is a set $S \subseteq \mathbb{N}$ and $S \ne \mathbb{N}$, that means there is a smallest element $n_0 \in \{n\in \mathbb{N}| n \not \in S \}$. Obviously $n_0 \ne 1$ as $1 \in S$. As $n_0$ is the smallest element which is not in $S \Rightarrow n_0-1 \in S$. But if $n_0-1 \in S \Rightarrow n_0-1+1=n_0 \in S$ so we have a $\Rightarrow \Leftarrow$ and our assumption that there exists a number outside set $latex S$ is False and $S=\mathbb{N}$

Labels: Peano's Axiom, Proof, Proof by Contradiction, Real Analysis

This is a proof of Bernoulli Inequality using Arithmetic Geometric Mean for the rational power, where the power is between 0 and 1. Note the inequality is reversed

Labels: AM GM inequality, Bernoulli Inequality, Proof

Proving Bernoulli Inequality for Natural numbers using Mathematical Induction is easy and here is the proof. For rational Number we can use the AM-GM Inequality

Labels: Bernoulli Inequality, Mathematical Induction, Natural Numbers, Proof