Thursday, 6 June 2024
A Twist in Classical Proof of Infinitude of Primes
Monday, 17 July 2023
Informal Introduction to Continuous functions
- the graph of such a function can be drawn without lifting the pen
- there is no sudden jump in values
- if we can be guarantee that change in output can be made as small as we please by making the change in input sufficiently small.
- Calculus by M. Spivak
Wednesday, 12 July 2023
Naïve Set Theory and Paradoxes
- $(W_1, \leq_1)$ isomorphic to $(W_2, \leq_2)$
- $(W_1, \leq_1)$ isomorphic to a proper initial segment of $(W_2, \leq_2)$
- $(W_2, \leq_2)$ isomorphic to a proper initial segment of $(W_1, \leq_1)$
- Foundations of Mathematics by Kenneth Kunen
- Naive Set Theory by Paul Halmos
Thursday, 29 June 2023
The Nine Point Circle
The nine point circle (also known as Feuerbach's circle) is one of the most interesting topics in elementary geometry. It doesn't require much prerequisites and so this post can be read by a high school student who has some idea about geometry of circle and triangle.
Just a reminder: All three altitudes of a triangle meet at a single point, which we call orthocenter and a set of points is said to be concyclic if they all lie in a circle.
Consider $\triangle ABC$ and let $O$ be the orthocenter (shown in green). Let $A_a$ be the feet of altitude from vertex $A$ on side $BC$, $A_m$ be the midpoint of side $BC$ and $A_o$ be the midpoint of line segment $AO$ (we have used the colors red, blue and orange respectively for these points). Similar notations corresponding to vertex $B$ and $C$. Then the nine points (which sometimes may not be distinct) $A_o, B_o, C_o$ (midpoints between orthocenter and vertices) $A_m, B_m, C_m$ (midpoints of sides) and $A_a, B_a, C_a$ (feet of altitudes) lie in a same circle which is commonly called the nine point circle (shown with dotted circle).
Try interacting with the GeoGebra applet given below ($\Omega$ is the center of nine point circle):
To prove this we need some elementary results:
- A convex quadrilateral is cyclic (ie. its vertices are concyclic) if and only if its opposite angles are supplementary (ie. $180^{\circ}$). One half of this result is Euclid's proposition $22$ in Book III.
- Line segment formed by connecting the midpoints of two sides of a triangle will be parallel to the third side and have half of its length. This called midpoint theorem.
- The center of the circumcircle of a right triangle lies on its hypotenuse. This is converse of a famous result called Thales' theorem.
- A Sequel to the First Six Books of Euclid by John Casey
- David Joyce's web version of Euclid's Elements
Sunday, 25 June 2023
Riemann Sphere and Stereographic Projection
Sometimes while studying complex valued functions, it is sometimes useful to consider the extended complex plane $\mathbb{C} \cup \{ \infty \}$, where we append an additional point $\infty$ to the set of usual complex numbers. This is especially helpful in study of Mobius maps $z \mapsto \frac{az+b}{cz+d}$ where $ad-bc \not = 0$.
Riemann sphere provides us with a model of extended complex plane. This makes it in useful in complex analysis because it allows us to make sense of division by zero in some circumstance ($\frac{0}{0}$ is still undefined).
We start with the unit sphere $S^2$ in $\mathbb{R}^3$ ie. the spherical surface with center at origin and unit radius (set of points $(x,y,z)$ given by $x^2 +y^2 +z^2=1$). First, we shall show there is a bijection between the $R^2$ and $S^2$ minus a single point. (We identity $xy$-plane with $\mathbb{R}^2$)
Start with a point $P$ on $S^2$ other than the north pole $N = (0,0,1)$. Draw a ray starting from $N$ passing through $P$. This ray will intersect the $xy$-plane at a unique point $Q$. The map that takes $P$ to $Q$ is called as stereographic projection and is denoted as $\Pi$.
We shall use some properties of vectors to derive stereographic projection explicitly. $Q - N$ is parallel to $P - N$, so $P - N = k(Q - N)$ for some scalar $k$. Denote $P=(x,y,z)$ and $Q=(u,v,0)$. Then, we get $(x,y,z) = (ku,kv,1-k)$ so that $k=1-z$. Note $z \not = 1$ since $P \not = N$.
$$\therefore \Pi (x,y,z) = (u,v,0) = \left( \frac{x}{1-z}, \frac{y}{1-z}, 0 \right)$$
Since $x^2 + y^2 + z^2 = 1$, we have $k^2u^2 + k^2v^2 + (1-k)^2 = 1$. Simplifying (while noting $k \not = 0$ since $z \not = 1$),
$$k = \frac{2}{u^2+v^2 +1}$$
$$\therefore P = (x,y,z) = \left( \frac{2u}{u^2+v^2 +1}, \frac{2v}{u^2+v^2 +1}, \frac{u^2 +v^2 -1}{u^2+v^2 +1} \right)$$
It takes some simple calculations to verify that $\Pi : S^2 \setminus \{ N \} \to \mathbb{R}^2$ is a bijection and the map $ (u,v,0) \mapsto \left( \frac{2u}{u^2+v^2 +1}, \frac{2v}{u^2+v^2 +1}, \frac{u^2 +v^2 -1}{u^2+v^2 +1} \right)$ is its inverse.
Interact with GeoGebra applet to see how stereographic projection actually works.
In fact our map $\Pi$ is conformal (ie. it preserves angles) but proof of this assertion requires some tools from differential geometry.
We can extend $\Pi$ to the whole sphere $S^2$ by simply defining $\Pi (N) = \infty$. Here, $\infty$ is just a symbol. And similarly extend $\Pi^{-1} : \mathbb{R}^2 \cup \{ \infty \} \mapsto S^2$. Since, we can interpret $ \mathbb{C}$ as $\mathbb{R}^2$ (from a topological point of view), we can also define $\Pi^{-1} : \mathbb{C} \cup \{ \infty \} \to S^2$ as $$z \mapsto \left( \frac{2 \Re(z)}{|z|^2 +1}, \frac{2 \Im(z)}{|z|^2 +1}, \frac{|z|^2 -1}{|z|^2 +1} \right)$$
where $\Re(z)$ and $\Im(z)$ are real and imaginary part of $z$ respectively, and
$$\infty \mapsto (0,0,1)$$
PS: From a topological point of view, we can define a topology $\tau_1$ on $\mathbb{C} \cup \{\infty\}$ such that our usual topology $\tau_2$ on $\mathbb{C}$ is a subspace of $\tau_1$ and $\tau_1$ is one point compactification of $\tau_2$.
References:
- A Pathway to Complex Analysis by S. Kumaresan
- Topology by J. Munkres
- A Comprehensive introduction to Differential Geometry Vol. 2 by M. Spivak
Wednesday, 21 June 2023
What is a matrix?
- $X+Y = Y+X$
- $X+(Y+Z) = (X+Y)+Z$
- There exist $\textbf{0} \in \mathbb{R}^n$ such that for all $X \in \mathbb{R}^n$, $X +\textbf{0} = X$. Just choose $\textbf{0} = \left( 0, 0,...,0 \right)$.
- For all $X \in \mathbb{R}^n$ there exist $X' \in \mathbb{R}^n$ such that $X+X'=\textbf{0}$. Take $X' = (-1)X$ and it is not hard to see that this is the only possible choice for $X'$.
- $(k_1k_2) X = k_1 (k_2 X)$
- $(k_1 + k_2) X = k_1 X + k_2 X$
- $k (X +Y) = k X + k Y$
- $0X = \textbf{0}$ and $1X = X$
- Linear Algebra $4^{th}$ ed. by Friedberg, Insel and Spence
- Linear Algebra Done Right $3^{rd}$ ed. by Axler
- What Is Mathematics? by Courant and Robbins
Mathematical Biscuit I
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A naive approach to set theory creates a lot of foundational problems, famous examples being paradoxes of Russell, Cantor and Burali-Forti. ...
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Sometimes while studying complex valued functions, it is sometimes useful to consider the extended complex plane $\mathbb{C} \cup \{ \infty...