Monday, July 31, 2017

Direct Product of Groups, Example, Inverse

Groups of order 1 to 4

Symmetric Group of order 3

Symmetric Group Order 3 and Composition

Sunday, July 30, 2017

Quick review of linear algebra and abstract algebra

Friday, July 28, 2017

Chebyshev Polynomials and their properties

The 4 subspaces

Thursday, July 27, 2017

Elementary linear algebra for high school

Wednesday, July 26, 2017

La Doncella (The handmaiden) Review

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.

Monday, July 24, 2017

Pythagorean Triples General Formula

Pythagorean Triples (Consecutive odd integers as one side)

Linear Algebra discussion

Sunday, July 23, 2017

Matrix Vector Product (Two ways to think about it)

Saturday, July 22, 2017

Complex numbers and Trig Identities

real analysis basics (rough draft)

Thursday, July 20, 2017

Triangle Inequality vs Cauchy Schwarz

Triangle Inequality n dimensional Proof

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

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Cauchy Schwarz n dimensional proof

Tuesday, July 18, 2017

Basic Facts on Primes for middle school

Friday, July 14, 2017

Harmonic Series Inequality by Mathematical Induction

Thursday, July 13, 2017

Triangle and Reverse Triangle Inequality Proof

Wednesday, July 12, 2017

Minimal Set theory for high school students

Tuesday, July 11, 2017

Bounded Sets

Archimedean Property: 5 ways to say the same thing

Limit Theorems 2

Limit Theorems 1

Proof: Convergent Sequence are bounded

Monday, July 10, 2017

Proof that between two rational numbers we can include an irrational number or Irrational numbers are dense in rational

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 $r_2$.

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Saturday, July 08, 2017

Binet Formula for Fibonacci Sequence

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

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Fibonacci Number thorugh Combinatorics

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.

Case n=2 balls -- 3 ways.
BB (2 Blue)

BR (1 Blue 1 Red)

Case n = 3 balls -- 5ways
BBB (3 blue)

BBR (2 Blue 1 Red)

RBR (1 Blue 2 Red)

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.

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Wednesday, July 05, 2017

Peano's Axioms

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 $S=\mathbb{N}$

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Bernoulli Inequality AM-GM proof

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

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Bernoulli Inequality by Mathematical Induction

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

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Primality test sigm(n)+phi(n) = n*d(n)

Monday, July 03, 2017

Various Means and their Order

e^pi versus pi^e (Which is bigger)

Regularised Sum of Infinite Natural Number = -1/12

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Fundamental theorem of Arithmetic Proof


Proof of Archimedean Property


Set of Rational Numbers is Dense in Real Numbers

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