Using Complex numbers in Classical Mechanics

April 18, 2018April 18, 2018 / 153armstrong / Leave a comment

When one is solving problems on the two dimensional plane and you are using polar coordinates, it is always a challenge to remember what the velocity/acceleration in the radial and angular directions ( $v_r , v_{\theta}, a_r, a_{\theta}$ ) are. Here’s one failsafe way using complex numbers that made things really easy :

$z = re^{i \theta}$

$\dot{z} = \dot{r}e^{i \theta} + ir\dot{\theta}e^{i \theta} = (\dot{r} + ir\dot{\theta} ) e^{i \theta}$

From the above expression, we can obtain $v_r = \dot{r}$ and $v_{\theta} = r\dot{\theta}$

$\ddot{z} = (\ddot{r} + ir\ddot{\theta} + i\dot{r}\dot{\theta} ) e^{i \theta} + (\dot{r} + ir\dot{\theta} )i \dot{\theta} e^{i \theta}$

$\ddot{z} = (\ddot{r} + ir\ddot{\theta} + i\dot{r}\dot{\theta} + i \dot{r} \dot{\theta} - r\dot{\theta}\dot{\theta} )e^{i \theta}$

$\ddot{z} = (\ddot{r} - r(\dot{\theta})^2+ i(r\ddot{\theta} + 2\dot{r}\dot{\theta} ) )e^{i \theta}$

From this we can obtain $a_r = \ddot{r} - r(\dot{\theta})^2$ and $a_{\theta} = (r\ddot{\theta} + 2\dot{r}\dot{\theta})$ with absolute ease.

Something that I realized only after a mechanics course in college was done and dusted but nevertheless a really cool and interesting place where complex numbers come in handy!

Fibonacci sequence in the hiding

July 30, 2017 / 153armstrong / Leave a comment

$Daily-Life-Math-Facts-2$

What ?!! There exists such an elegant decimal representation of the Fibonacci sequence? Well yes! and the only thing that you need to know to prove this is that if the Fibonacci numbers were the coefficients to a power series expansion, then the Fibonacci generating function is given as follows:

$1x + 1x^{2} + 2x^{3} + 3x^{4} + 5x^{5} + \hdots = \frac{x}{1-x-x^{2}}$

Subsituting the value of $x = \frac{1}{10}$ , we get :

$\frac{1}{10} + \frac{1}{10}^{2} + 2(\frac{1}{10})^{3} + 3(\frac{1}{10})^{4} + 5(\frac{1}{10})^{5} + \hdots = \frac{\frac{1}{10}}{1-\frac{1}{10}-\frac{1}{10}^{2}}$

$0.1 + 0.01 + 0.002 + 0.0003 + 0.00005 + \hdots = \frac{10}{89}$

$0.01 + 0.001 + 0.0002 + 0.00003 + 0.000005 + \hdots = \frac{1}{89}$

Proved. 😀

Prof.Ghrist at his best!

July 28, 2017July 28, 2017 / 153armstrong / Leave a comment

Screenshot from 2017-07-28 10:55:27

To understand why this is true, we must start with the Fundamental Theorem of Vector calculus. If $F$ is a conservative field ( i.e $F = \nabla \phi$ ), then

$\int\limits_{A}^{B} F.dr = \int\limits_{A}^{B} \nabla\phi .dr = \phi_{A} - \phi_{B}$

What this means is that the value is dependent only on the initial and final positions. The path that you take to get from A to B is not important.

Screenshot from 2017-07-28 11:29:46

Now if the path of integration is a closed loop, then points A and B are the same, and therefore:

$\int\limits_{A}^{A} F.dr = \int\limits_{A}^{A} \nabla\phi .dr = \phi_{1} - \phi_{1} = 0$

Now that we are clear about this, according to Stokes theorem the same integral for a closed region can be represented in another form:

$\int_{C} F.dr = \int\int_{A} (\nabla X F) .\vec{n} dA = 0$

From this we get that Curl = $\nabla X F = 0$ for a conservative field (i.e $F = \nabla \phi$ ). Therefore when a conservative field is operated on by a curl operator ( $\nabla X$ ), it yields 0.

Bravo Prof.Ghrist! Beautifully said 😀

Beautiful Proofs(#3): Area under a sine curve !

June 7, 2017June 9, 2017 / 153armstrong / Leave a comment

So, I read this post on the the area of the sine curve some time ago and in the bottom was this equally amazing comment :

Screenshot from 2017-06-07 00:19:11

$\sum sin(\theta)d\theta =$ Diameter of the circle/ The distance covered along the x axis starting from $0$ and ending up at $\pi$ .

And therefore by the same logic, it is extremely intuitive to see why:

$\int\limits_{0}^{2\pi} sin/cos(x) dx = 0$

Because if a dude starts at $0$ and ends at $0/ 2\pi/ 4\pi \hdots$ , the effective distance that he covers is 0.

If you still have trouble understanding, follow the blue point in the above gif and hopefully things become clearer.

Tricks that I wish I knew in High School : Trigonometry (#1)

June 4, 2017June 6, 2017 / 153armstrong / Leave a comment

I really wish that in High School the math curriculum would dig a little deeper into Complex Numbers because frankly Algebra in the Real Domain is not that elegant as it is in the Complex Domain.

To illustrate this let’s consider this dreaded formula that is often asked to be proved/ used in some other problems:

$cos(nx)cos(mx) =$ ?

Now in the complex domain:

$cos(x) = \frac{e^{ix} + e^{-ix}}{2}$

And therefore:

$cos(mx) = \frac{e^{imx} + e^{-imx}}{2}$

$cos(nx) = \frac{e^{inx} + e^{-inx}}{2}$

$cos(mx)cos(nx) = \left( \frac{e^{imx} + e^{-imx}}{2} \right) \left( \frac{e^{inx} + e^{-inx}}{2} \right)$

$cos(mx)cos(nx) = \frac{1}{4} \left( e^{i(m+n)x} + e^{-i(m+n)x} + e^{i(m-n)x} + e^{-i(m-n)x} \right)$

$cos(mx)cos(nx) = \frac{1}{2} \left( \left( \frac{e^{i(m+n)x} + e^{-i(m+n)x}}{2} \right) + \left( \frac{e^{i(m-n)x} + e^{-i(m-n)x}}{2} \right) \right)$

$cos(mx)cos(nx) = \frac{1}{2} \left( cos(m+n)x + cos(m-n)x \right)$
And similarly for its variants like $cos(mx)sin(nx)$ and $sin(mx)sin(nx)$ as well.

****

Now if you are in High School, that’s probably all that you will see. But if you have college friends and you took a peak what they rambled about in their notebooks, then you might this expression (for $m \neq n$ ):

$I = \int\limits_{-\pi}^{\pi} cos(mx)cos(nx) dx \\$

But you as a high schooler already know a formula for this expression:

$I = \int\limits_{-\pi}^{\pi} \left( cos(m+n)x + cos(m-n)x \right)dx \\$

$I = \int\limits_{-\pi}^{\pi} cos(\lambda_1 x) dx + \int\limits_{-\pi}^{\pi} cos(\lambda_2 x) dx \\$

where $\lambda_1$ , $\lambda_2$ are merely some numbers. Now you plot some of these values for lambda i.e ( $\lambda = 1,2, \hdots$ ) and notice that since integration is the area under the curve, the areas cancel out for any real number.

and so on….. Therefore:

$I = \int\limits_{-\pi}^{\pi} cos(mx)cos(nx)dx = 0$

This is an important result from the view point of Fourier Series!

Why is the area under one hump of a sine curve exactly 2?

May 7, 2017June 6, 2017 / 153armstrong / 1 Comment

Girls' Angle

I was talking with a student recently who told me that he always found the fact that $latex int_0^{pi} sin x , dx = 2$ amazing. “How is it that the area under one hump of the sine curve comes out exactly 2?” He asked me if there is an easy way to see that, or is it something you just have to discover by doing the computation.

If you’ve wondered about this too, perhaps you’ll find the following of interest.

View original post 162 more words

Beautiful proofs(#2): Euler’s Sum

March 5, 2017October 27, 2017 / 153armstrong / Leave a comment

$1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \hdots = \frac{\pi^2}{6}$

Say what? This one blew my mind when I first encountered it. But it turns out Euler was the one who came up with it and it’s proof is just beautiful!

Prerequisite
Say you have a quadratic equation $f(x)$ whose roots are $r_1,r_2$ , then you can write $f(x)$ as follows:

$f(x) = (x-r_1)(x-r_2) = 0$ (or)

$f(x) = (r_1-x)(r_2-x) = 0$ (or)

$f(x) = (1- \frac{x}{r_1})(1- \frac{x}{r_2}) = 0$

$f(x) = 1 - (\frac{1}{r_1} + \frac{1}{r_2}) + \frac{x^2}{r_1 r_2} = 0$

As for as this proof is concerned we are only worried about the coefficient of $x$ , which you can prove that for a n-degree polynomial is:

$a_1 = - (\frac{1}{r_1} + \frac{1}{r_2} + \hdots + + \frac{1}{r_n})$

where $r_1,r_2 \hdots r_n$ are the n-roots of the polynomial.

Now begins the proof

It was known to Euler that

$f(y) = \frac{sin(\sqrt{y})}{\sqrt{y}} = 1 - \frac{1}{3!}y + \hdots$

But this could also be written in terms of the roots of the equation as:

$f(y) = \frac{sin(\sqrt{y})}{\sqrt{y}} = 1 - (\frac{1}{r_1} + \frac{1}{r_2} + \hdots + + \frac{1}{r_n})y + \hdots$

Now what are the roots of $f(y)$ ?. Well, $f(y) = 0$ when $\sqrt{y} = n \pi$ i.e $y = n^2 \pi^2$ *

The roots of the equation are $y = \pi^2, 4 \pi^2, 9 \pi^2, \hdots$

Therefore,

$f(y) = \frac{sin(\sqrt{y})}{\sqrt{y}} = 1 - \frac{1}{3!}y + \hdots = 1 -( \frac{1}{\pi^2} + \frac{1}{4 \pi^2} + \hdots )y + \hdots$

Comparing the coefficient of y on both sides of the equation we get that:

$\frac{1}{6} = \frac{1}{\pi^2} + \frac{1}{4 \pi^2} + \frac{1}{ 9 \pi^2} + \hdots$

$\zeta(2) = \frac{\pi^2}{6} = 1 + \frac{1}{4} + \frac{1}{9} + \hdots$

Q.E.D

* n=0 is not a root since
$\frac{sin(\sqrt{y})}{\sqrt{y}} = 1$ at y = 0

** Now if all that made sense but you are still thinking : Why on earth did Euler use this particular form of the polynomial for this problem, read the first three pages of this article. (It has to do with convergence)

Ecstasy Shots

A poetry of logical ideas

beautiful proofs