Real Analysis by Terence Tao! Oh yeah!

April 5, 2021April 5, 2021 | Cosmoshivani

Never thought, the day would come when I’ll read this masterpiece – Real Analysis. Today I’m going to review a few things that got stuck in my mind.

I should be sleeping now, my phone’s battery is just 6%. But I’ll just write the things I remember. Also I dont have the book in my phone. So, lets goooo!

Okay so it starts with “Why bother” section. A perfect start for people like me. In it, he said, Analysis is the Why of calculus. And most of us get satisfying feeling in knowing the Why even if knowing How does the job.

He then asks a lot of questions, like what is the smallest positive number after 0? And similar ones..

Then there are a lot of examples showing how the known methods of computation in calculus can lead to wrong conclusion.

These examples include convergence of series, for one.

Like, if you add 1 – 1 + 1 – 1 + …

like this, 1-(1-1)-(1-1)-.. then its 1.

but if you sum it like this, (1-1)+(1-1)+(1-1)+…=0

so which one of these is correct?

Another example was, swapping the rows and columns and finding that either way, the summation is the same. Though, it doesn’t work if the summation involved infinite number of elements.

There was also an example of swapping the variables in a second order partial derivative; the answer was different after swapping.

Similar example was shown for interchanging the limit and integration.

We have used these methods and techniques but have no idea how and why they work and most importantly as shown in all those examples, where they don’t.

He promised that we will get all these answers as we read more.

But I’m not very sure if I will.

I’d like to try atleast a few more chapters.

Seeya!

Disclaimer: I am neither a mathematician nor I strive to be one. Period.

Inner and Outer Product in Quantum Mechanics

March 24, 2021March 24, 2021 | Cosmoshivani

Alright, This is NOT going to be a very revealing kind of post like was planning but here you go…

In quantum mechanics, the particles are represented as state vectors. These state vectors are part of a state space for a given system or a Hilbert Space.

Let’s take two state vectors, $\psi$ and $\phi$ from a given state space. Now, we can perform some operations on these, just like we do with coordinate vectors.

But to define the inner product of state vectors we’ll have to define a dual space too. Given a basis, we can write these state vectors as matrices. Kets are written as column vectors and they are elements of state space, whereas the Bra vectors are written as row vectors and are elements of the dual space.

So, a ket vector is written as $| \psi>$ and a bra vector as $< \psi|$ . Their inner product is defined as,

$< \psi|\psi>$

The resulting vector is either a scalar or a complex number.

If the inner product is one then, the vector is said to be normalized and if it’s zero, then they are orthogonal.

Whereas, the outer product results in a matrix or an operator and it’s written like this,

$|\psi> < \psi|$

I’m not really sure if all the outer products result in actual operators in quantum mechanics. Because in QM, they are mostly Hermitian Operators. I’d really like to think more about it in future. 🤔 Hopefully 😄😄

see ya!

Inner and Outer product in Linear Algebra

March 1, 2021March 1, 2021 | Cosmoshivani

Let’s take two coordinate vectors, $\textbf v$ and $\textbf u$ . Both of them can be represented as column matrices (why not as row matrices?) of order $m \times 1$ and $n \times 1$ .

The inner product of these two coordinate vectors is, $\textbf v^T \cdot \textbf u$ .

$\textbf v = \begin{pmatrix} a_1 \\ a_2 \\ a_3\\...\\a_m \end{pmatrix}$

$\textbf v^T = \begin{pmatrix} a_1 & a_2 & a_3 & ... & a_m \end{pmatrix}$

$\textbf u = \begin{pmatrix} b_1 \\ b_2 \\ b_3\\...\\ b_n \end{pmatrix}$

$\textbf v^T \cdot \textbf u = \begin{pmatrix} a_1 & a_2 & a_3& ... & a_m \end{pmatrix} \cdot \begin{pmatrix} b_1 \\ b_2 \\b_3 \\ ... \\b_n \end{pmatrix}$

$\textbf v^T \cdot \textbf u = ( a_1 b_1 + a_2 b_2 + a_3 b_3 + ... + a_m b_n)$

Notice that it only works if $m = n$ and it always results in a matrix of order $1 \times 1$ .

The outer product of these two vectors is $\textbf v \cdot \textbf u^T$ .

$\textbf v \cdot \textbf u^T = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \\ ... \\ a_m \end{pmatrix} \cdot \begin{pmatrix} b_1 & b_2 & b_3 & ... & b_n \end{pmatrix}$

$\textbf v \cdot \textbf u^T = \begin{pmatrix} a_1 b_1 & a_1 b_2 & a_2 b_3 &...&a_1 b_n\\a_2 b_1 & a_2 b_2 & a_2 b_3 &...& a_2 b_n\\a_3 b_1 & a_3 b_2 & a_3 b_3 &...&a_3 b_n\\...& ... & ... & ... & ... \\a_m b_1 & a_m b_2 & a_m b_3 & ... & a_m b_n \end{pmatrix}$

If you have a column vector of order $m \times 1$ and a row vector of order $1 \times n$ respectively, their outer product will be a matrix of order $m \times n$ . Here $m$ need not be equal to $n$ .

The significance of inner product in quantum mechanics becomes clear right away, but what about the outer product?

To be continued in the next article: Inner and Outer Product in Quantum Mechanics

A complex-looking simple problem

February 10, 2021February 10, 2021 | Cosmoshivani

I would like to share the solution of a very simple problem. I tried solving it during an exam. It looked really complex (to me) at first but then I finally did it!

Problem: There is a square of side $r$ . There are two circles drawn with radius equal to the side of the square. What is the probability of a particle to be found in the shaded (yellow) region?

Usually when we solve the probability problems (not that I am an expert), the formula that we use is, (number of desired items e.g. red balls) divided by (total number of items e.g. red and yellow balls). Similarly, here we can use, (area of shaded region) divided by (total area). To calculate it, we need to figure out the area of the yellow region.

It can be seen, that there are two circles here. And the area of circle is, $\pi r^2$ , but the circle inside the square is just $\frac{1}{4}^{th}$ , so area of first circle inside the square is, $\frac{1}{4} \pi r^2$ . The area of square itself will be $r^2$ . Now if we subtract the area of the circle from the square, we get the area of one purple region. $r^2 - \frac{1}{4} \pi r^2 = r^2(1 - \frac{\pi}{4} )$ . The area of second purple region will also be the same, and hence the total area is, $2 [r^2(1 - \frac{\pi}{4})]$ . In the end what we need to do is, subtract this from the area of square to get the area of yellow region. Which is going to be, $r^2 - 2 [r^2(1 - \frac{\pi}{4})] = -r^2 + \frac{\pi}{2}r^2$ .

Finally, we can get the probability of the particle to be found in the yellow region, which is, $\frac{[-r^2 + \frac{\pi}{2}r^2]}{r^2} = \frac{\pi}{2} - 1$ .

Math reading challenge 2020

January 17, 2021 | Cosmoshivani

So, I found this article by Evelyn lamb in her blog Roots of Unity at the end of 2019. It had some prompts to help you find math related books. I wasn’t able to complete all of them but I did try some.

I read three books for the prompt, “A work of fiction in which a main character is a mathematician“. The very first book I read was, “The devotion of suspect X“. It was beautiful. Set in Japan, it’s about a mathematician Ishigami, who is a High school teacher. One day, when he was trying to take his life, someone rings the doorbell. His new neighbors, Yasuko and her daughter Misato. Without knowing, they save his life. And from that point, he does everything to protect this family next door. It’s mysterious, and very emotional at times. Next was, “The Housekeeper and the Professor“. Due to a terrible accident in 1970, a mathematician is now only able to remember last 80 minutes of his life. Like a hard disk which has only 80 minutes of memory limit; anything new is recorded on top of that. So whenever he wakes up in morning, he doesn’t remember a single thing that happened yesterday. Many housekeepers are assigned to take care of him but no one lasts. But it’s about to change. The tenth housekeeper, who is the youngest in the company takes the job and then starts a story of friendship between the housekeeper, her son and the professor. They go on many trips and he tells them interesting things like the importance of square root, amicable numbers, perfect numbers, triangle numbers etc. But the good days come to an end. His 80 minute memory tape is also broken and he cannot retain even a minute of new memories. Sadly, in the end, he is transferred to a new care facility. Then I read “Uncle Petros and Goldbach’s Conjecture“. Its story revolves around a mathematician who is obsessed with the Goldbach’s conjecture (Every even number greater than $2$ can be written as a sum of prime numbers). And how he puts in so much effort to discourage his nephew from pursuing higher mathematics, because he doesn’t want him to get mad and obsessed like himself. It also has a very tragic ending.

For the next prompt, “A graphic novel about math or mathematicians“, I read “Logicomix: An Epic Search for Truth“. It had really awesome script and beautiful drawings. I read about Russel’s paradox for the first time in it, which is introduced by Bertie Russell himself. It goes something like this, if a set consists of all the sets that do not contain themselves, then does that set contain itself as an element? If it does then it does not follow the property that it should not contain itself; If it doesn’t contain itself then it does follow the property and thus it belongs to the set but then again if it does it shouldn’t and if it doesn’t then it should.

For the most part, the story follows the life of Bertrand Russell. There are many events of his life from childhood to a university student; how he started questioning everything that others took for granted. It also contains many dialogues and fictional meetings between famous mathematicians, which makes it even more interesting.

So basically, all I read was mathematical fiction. But I did learn some concepts in number theory and logic here and there.

Apparent size

October 26, 2019May 5, 2020 | Cosmoshivani

Sitting at the dining table, once I was calculating how planets appear to the human eye from our planet. I was trying to calculate the angular diameter or apparent size of planets using trigonometry. I have no idea where that notebook is, but today, I found an awesome blog which gave me enough background to start doing some calculations again.

This time I am going to calculate the apparent size of ISS. I have seen it floating across the sky a few times.

Angular diameter, $\delta = 2 sin^{-1} \frac{D}{2d}$
$d$ : Distance of ISS from Earth = $400 km$
$D$ : Length of ISS = $109 m$
$\delta$ : angular diameter of ISS = ?
$\delta = 2 sin^{-1} \frac{109}{2 \times 400 \times 10^3}$
$= 2 sin^{-1} (0.000136)$
$= 2 \times 0.0079 = 0.0156 \deg$
$0.0156 \deg = 0.9 arcminute$
Its almost equal to the apparent diameter given on Wikipedia, which is 1 arcminute.

Series Expansion

February 12, 2018February 1, 2021 | Cosmoshivani

A function can be expanded using Taylor series which is given by,

$f(x) = f(a) + \frac{x-a}{1!} f^{'}(a) + \frac{(x-a)^{2}}{2!} f^{''}(a) + \frac{(x-a)^{3}}{3!} f^{'''}(a) + ...$

If $a=0$ , it’s called Maclaurin series.

$f(x) = f(0) + \frac{x}{1!} f^{'}(0) + \frac{(x)^{2}}{2!} f^{''}(0) + \frac{(x)^{3}}{3!} f^{'''}(0) + ...$

Let’s figure out the Maclaurin series for all trigonometric functions,

$f(x) = sinx$ $f(0) = 0$
$f^{'}(x) = cosx$ $f^{'}(0) = 1$
$f^{''}(x) = -sinx$ $f^{''}(0) = 0$
$f^{'''}(x) = -cosx$ $f^{'''}(0) = -1$

$sinx = \frac{x}{1!} - \frac{(x)^{3}}{3!} + ...$

$f(x) = cosx$ $f(0) = 1$
$f^{'}(x) = -sinx$ $f^{'}(0) = 0$
$f^{''}(x) = -cosx$ $f^{''}(0) = -1$
$f^{'''}(x) = sinx$ $f^{'''}(0) = 0$

$cosx = 1 - \frac{(x)^{2}}{2!} + ...$

$f(x) = tanx$
$f(0) = 0$
$f^{'}(x) = sec^{2}x = 1+tan^{2}x = 1+f(x)^{2}$
$f^{'}(0)=1$
$f^{''}(x) = 2f(x) f^{'}(x)$
$f^{''}(0) = 0$
$f^{'''}(x) = 2f^{'}(x)^{2}+2f(x)f^{''}(x)$
$f^{'''}(0) = 2$

$tanx = \frac{x}{1!} + \frac{2(x)^{3}}{3!} + ...$

$f(x) = secx$
$f(0) = 1$
$f^{'}(x) = f(x) tanx$
$f^{'}(0) = 0$
$f^{''}(x) = f^{'}(x) tanx + f(x) f(x)^{2}$
$f^{''}(0) = 1$
$f^{'''}(x) = f{''}(x) tanx + f^{'}(x) f(x)^{2} + f^{'}(x) f(x)^{2} +2 f(x)^{2} f^{'}(x)$

$secx = 1 + \frac{(x)^{2}}{2!} + ...$

$f(x) = cotx$ $f(0) = \infty$
$f(x) = cosecx$ $f(0) = \infty$
How can you write series expansion if the value of function is infinite at $x=0$ ? I am still trying to figure out a satisfactory answer to this question.

Update:

If a function is not analytic at a point then it cannot be expanded using Taylor series. We have another series expansion for such points and its called Laurent’s expansion.

So, lets evaluate Laurent series expansion of $f(z) = cotz$ at $z=0$
$f(z)=\frac{1}{tanz}$
$cotz = \frac{1}{z + \frac{z^{3}}{3!} + \frac{2(z)^{5}}{15} + ...}$
$cotz = \frac{1}{z[1 + \frac{z^{2}}{3!} + \frac{2(z)^{4}}{15} + ...]}$

Now we can use the long division method to divide $\frac{1}{z}$ by $\frac{1}{1 + \frac{z^{2}}{3!} + \frac{2(z)^{4}}{15} + ...}$ .

So, we get

$cotz = \frac{1}{z} - \frac{z}{3} -\frac{z^{3}}{45} - ...$

Similarly we can evaluate Laurent series expansion of $cosecz$ .

$f(z) = cosecz$ at $z=0$
$f(z) = \frac{1}{sinz}$
$cosecz = \frac{1}{z- \frac {z^{3}}{3!} + \frac{z^{5}}{5!} - ...}$
$cosecz = \frac{1}{z[1- \frac {z^{2}}{3!} + \frac{z^{4}}{5!} - ...]}$