Simple Algebra, or Not?

Well, I won't be doing much explaining in this post, so if you don't understand maybe I'll post another explanation for the second part; Luoning's tag has directed me to a very nice solution for determining fractional expressions for recurring decimals - it's a method taught to me by Mrs Pauline Kan way back in JC1, and I always turn to it for some mathematical fun when dealing with young Secondary School kids during tuition.

The question goes like this, given a recurring decimal like the one below, can you determine a fraction that equals it:

Well, the trick is to recognise that you can make use of the commutative properties of algebraic manipulation as follows:

And then by equating the two equations we must concur that:

And therefore:

But of course, for those budding Mathematicians out there, you may have seen this as a simple Geometric Progression sum to infinity, and therefore the general formula follows directly for such a sum:

I'm taking for granted that all of you know what Geometric Progressions are, haha.

For The Record

Well, here's just something Dr. Phil Chan, a really really really interesting Physics Professor, mentioned during my course in my first Semester at NUS:

Do you understand why? Haha. I probably won't be revealing this even if you don't get it, because Physicists truly are number one! :p

How Many Ways?

So, given this picture of four rooms interlinked by pink gates or doors, do you think you can find a route through the doors such that you pass through every door only once? Or do you think there's no such route?

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Well, Luoning nailed down the essence of my explanation; basically, it's got to do with the number of entrances to each room you see in the picture above. To facilitate the explanation, let me redraw the network of rooms above as such:

Notice I've represented each room by a black dot, which I shall now refer to as a node; so there's nodes A, B, C and D, each representing the rooms. Now, notice each node has lines connecting it to other nodes - these lines represent the doors or entrances leading from one room to another room.

Looking at this diagram, you'll find that it's much more easier to figure out possible routes.

So what did Luoning say again? Ah yes, the even or odd number of entrances! Indeed! Ask yourself what it means to be able to trace out a path where you don't re-use an entrance: this means that you must be able to arrive at the room the same number of times as you leave the room.

For you to arrive at the room the same number of times as you leave the room, one obvious condition is needed: you need each room to have an even number of entrances. Try it!

Of course, there is another condition: you could also have two rooms having an odd number of entrances but of course you need these rooms to be directly linked to one another as well. The direct linkage of these two rooms then means that the two rooms can be visualised as one big room and therefore reduces the number of effective entrances by one, turning an odd number of entrances into an even number of entrances!

Well I didn't think of this myself, but I did think of how to put forward this explanation myself! So give me some credit, won't ya? Haha.

So for this problem, of course no single route can be found! :)

[Edit: Hmm, this post isn't that well explained - I'll come back to this once I've figured out Graph Theory for myself!]

Double Pendulum

The double pendulum is always an interesting question because it's a special physical system composed of well, surprise surprise, two pendulums! Haha. Anyway, here's a question Funman asked me, so I might as well share it:

Well, just click on it to enlarge it!

What say I master Lagrangian Mechanics before I attempt this? :p

Complex Power

Well, here's something I asked one of my students, Cassandra, and let's see how many people actually manage to figure it out by, say, the end of the week:

Have a go at it!

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Some of you might be thinking this problem might be adequately solved if one considers Euler’s exponential form of complex numbers:

But here’s another problem; we know that:

In fact, the number i is obtained each time we make a full revolution in the imaginary phase space, such that:

Does that mean now that:

Go figure!

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Well, the answer actually is very simple: you just have to define the allowed values of the argument that define the imaginary number i, such that it can be placed to the power of something. :)

Oh yeah, this answer was contributed by Kenneth Tay Jingyi!

Expansion Woes

My my, have you ever worried about how things expand when heated? I once explained this to one of my friends in Physics in my first year of University, and looking back now, it’s rather interesting. The question he asked me was:

“Eh, if you got this ring right, when you heat it and it expands… what will it look like?”

Oh yes, he asked me with reference to this diagram:

And I wonder... Mmhmm...

A Light Burden

Did you know that radiation carries with it a kind of pressure known as radiation pressure? That means that when I shine a torch on you, it's supposed to be pushing you as it gets absorbed or scattered off of you. Let's try and determine the force exerted on the Earth by the light rays from the Sun, and you'll see what a staggering figure this is!

Assuming that the Sun is a perfect sphere, of radius R = 695 500 km and that it is a perfect blackbody with emissivity e = 1.00, and a roughly constant surface temperature of T = 5 778 K, we can use the Stefan-Boltzmann Law to determine the power of radiation emitted by the sun:


Where σ is the Stefan Boltzmann constant and A is the surface area of the Sun. And yes I've conveniently assumed the Earth is of so a low temperature with respect to the Sun such that it hardly radiates back any radiation back to the Sun.

Let us denote the distance of the Earth from the sun to be d (which is an enormous d = 149 600 000 km) and its radius as r = 6371 km. With these figures, we can then determine the intensity of solar light received at the position of our Earth, which is the initial power emitted by the Sun divided by the surface area of a sphere of radius d since the light from the Sun is isotropic in all directions:


Since all light rays coming from such a far off source will be parallel to one another, we can take the cross sectional area of the Earth to be the area that intercepts the light rays, which can be taken to be:


The power intercepted by the Earth is therefore:

Now, we know that for electromagnetic radiation, pc = E, where p is the equivalent momentum of a quantum of energy, c is the speed of light in a vacuum, and E is the energy of radiation considered. This simply means that the power is related to the momentum as such:


Don't forget, we know that the force exerted by the radiation on Earth is equivalent to the rate of change of momentum of the radiation photons (of course, we're assuming all radiation is absorbed, which is an approximation!), and thus the force exerted on Earth by the Sun's light is:



And this force is about 10^8 N! Which turns out to be an astronomical number! :) So why do you think the Earth is not being pushed away by the Sun? Easy, there's still the force of gravity, which when worked out by Newton's Law of Gravitation:


And this amounts to about 10^22 N! Do you see the difference in the two forces? The gravitational force far outweighs the force exerted by light, and thus continues to aid us in our journey through intergalactic space! :)