This is just a thread for Math boffins, to try their hand at solving each other's equations. I think this has been done before in another thread, but that was a while back.

I don't think will be very difficult to do, but I'll still type it out.

1. Using the binomial theorem, expand (3x + 2)^3, simplifying each term of the expansion.

It's sure to be easy. I've got the answers if anybody needs them.

Binomial theorem (general case)

(x + y)^n = SUM (i = 0 to n)((nCi)(x^n-1)(y^i))

Binomial theorem for case where n = 3

(x + y)³ = x³ + 3x²y + 3xy² + y³

(3x + 2)³ = (3x)³ + (3)(3x)²(2) + (3)(3x)(2)² + 2³

= 27x³ + 54x² + 36x + 8


Fair enough.

2. What are the three cube roots of 1? (a.k.a. the cube roots of unity)

Random guess alart!

1, -1 and i?

Random guess alart!

1, -1 and i?

Sorry, I'm afraid not. -1 cubed is -1, i cubed is -i. You got 1 right, though :p

There are three? Gosh. Hmm.

1, obviously.

So... -(i ^(2/3)) must work. And... er...... (-1)^(2/3)

Which is to say 1 and -0.5 +/- 0.866025...i. I think.

Good old Google Calculator.

There are three? Gosh. Hmm.

1, obviously.

So... -(i ^(2/3)) must work. And... er...... (-1)^(2/3)

Which is to say 1 and -0.5 +/- 0.866025...i. I think.

Good old Google Calculator.

Yep. 1, (-1/2 + i√3/2) and (-1/2 - i√3/2)

Alternatively, define (-1/2 + i√3/2) as ω and we can have them as 1, ω and ω²

On an Argand diagram, the three points form an equilateral triangle, centre 0.

Gosh. That was surprisingly interesting.

Okay, how about this. A series of events (such as radioactive decay) is a Poisson process, which is to say that the odds of an event happening at any given moment don't vary with time or depend in any way on the other events. In this case, you expect, on average, one event every 100 seconds.

A clock starts immediately after one instance of the event and stops immediately upon the next event. What is the most likely number of seconds to have registered on the clock?

100 seconds? :/ *is idiot*

It's not, no. That's the expectation value (well, 99 is really, as clocks tend to start at zero) but that's not what I asked for...

I know this seems silly. But would there be a delay between the clock realising the event has ended/begun?

I don't follow.

But it's maths -- not engineering. Let's assume instantaneous events and no dead-times on instruments or reaction delays, and so forth.

In other words, given a mean of 100 and the fact that it's a Poisson, find the mode.

Well, IIRC it'll be less than 100. But it's been at least three years since I last covered Poissons. So I may be entirely wrong.

It's quite a lot less than that, yes, although I don't know that it's strictly the mode of the distribution.

Do it in discrete seconds and work out the odds of the clock stopping on each one. A pattern forms.

It's quite a lot less than that, yes, although I don't know that it's strictly the mode of the distribution.

Do it in discrete seconds and work out the odds of the clock stopping on each one. A pattern forms.

Although, hang on: with a lambda of 100, the distribution's practically normal anyway.

I expect so. But the probability of the clock stopping on t seconds is a function of t, and one number is more likely than any other.

http://upload.wikimedia.org/math/5/5/9/55978f02e2b22e9a93943595030ecf64.png
That's the equation to work it out. *attempts* It's probably above my station.

Holy crap, that's an ugly equation.

Alright, every second there's a 1% chance of anythign happening, right?
So the odds of the stopwatch stopping on zero seconds are 1%, right?
What are the odds of it stopping on one second? Two seconds?

This thread confuses the hell out of me. But I did watch and enjoy the movie PI today, which is about as close to math I am willing to get.

Holy crap, that's an ugly equation.

Alright, every second there's a 1% chance of anythign happening, right?
So the odds of the stopwatch stopping on zero seconds are 1%, right?
What are the odds of it stopping on one second? Two seconds?I don't think that's how a Poisson distribution works.

I'm pretty sure that it is.

A Poisson event can happen with equal probability at any moment. And if you're using some other definition of "Poisson distribution" then let me know what that is, but more importantly, please mentally just delete the word "poisson" from the question and answer it as it was clearly meant instead of derailing the thread into an irrelevant argument.

I'm using both the definition I remembered at school and this one (http://en.wikipedia.org/wiki/Poisson_distribution). Now to be fair, what you're saying may still be a property of it, but it doesn't help explain the question. What you're actually asking reminds me more of the geometric distribution, but continuous.

But oh well. I'll use the discrete geometric distribution to get a cricket ground figure.

Geometric distribution: X = number of trials required to obtain first success (Number of seconds required to obtain first click, probability of click given as 0.01).

f(x) = P(X = x) = p*(p-1)^(x-1) = (0.01)*(0.99)^(x-1)

x = 1, f(x) = 0.01
x = 2, f(x) = 0.0099
x = 3, f(x) = 0.009801
x = 4, f(x) = 0.00970299

... and it's not going to get any bigger, because we're multiplying by a number less than 1 each time (even if it is 0.99). So the mode must be 1. For reference, the standard deviation is 99.5 to three significant figures.


EDIT: Oh wait, the exponential distribution, that's like a continuous version of the geometric distribution. But in that case, the mode is 0.

Well actually I think I said "poisson process" rather than "poisson distribution", but yes, that's the trick I was going for: zero seconds (or one depending on coordinate origins) is more likely than 100 seconds.

My thermal physics lecturer once tried to use this to explain why buses come in threes -- he modelled them as totally random and then said that the most likely time for a bus was straight after another bus.

He was, and is, an idiot, as were and are most of my old lecturers.

Heh.

4. What's the real reason buses come in threes (or, at least, twos)?

Because there's always at least one driver who doesn't know where he's going and has to follow someone else :p

No :p.

Anyone who's played Transport Tycoon or one of its sequels should know the answer to this one.

Because your opponent destroyed the bridge and you only just built another?

Because someone funded road rebuilding?

Because you always purchase buses in threes?

None of those. Pickleworm is closest, in the sense that he said exactly the wrong thing, as opposed to Daniel and Bunsy who were talking about something irrelevant.

Eurpoean double-decker busses completely outclass normal busses, ergo two must be bought to compensate.

Or...

You buy two busses at the same time so that there's one bus going in either direction of the route at all times (ie: halve the wait time). An extra bus or two might be bought for very long routes with high passenger concentration. However, because of the diminishing returns on adding busses, it is never cost or time effective to have more than three or four busses on any one route.

What?

They don't come in packs of two! They come in twos: buses which are usually shceduled to appear individually nevertheless arrive in pairs. Why?

(I may tire of this and answer it myself but I'd rather let you play.)

Although, there are buses in Leeds that go the same rought scheduled to arrive within two minutes of each other, with the next ten minutes later.

The 127 service which runs from Coalville to Leicester (conveniently through Loughborough) is quite possibly the best example of "You wait for ages then two show up at once". If I'm walking through town, I will almost always see two 127's within 20m of each other, going in the same direction, then nothing for 20mins, then another two buses come, again with virtually no separation between them. The service is supposed to be every 10mins.

And, that's for both services - the one going from Leicester and the one going to Leicester.


Trouble is, along the way, there's usually so much traffic coming the other way that the bus behind can't go past the first one when it stops, so it gets stuck behind it. The first one then becomes standing-room-only, while the second one has virtually nobody on it.


Stupid public transport :mad:

Limerick.
http://upload.wikimedia.org/math/b/e/f/bef517b32e950a38140fdf795f326cd6.png

Limerick.
http://upload.wikimedia.org/math/b/e/f/bef517b32e950a38140fdf795f326cd6.png

I've already seen the Wikipedia page, so I won't spoil it.

Thomas is the closest so far.

Sorry to break the thread somewhat, but I need abit of help. :/

Differentiation.
b) Given that y = a/x + 2x^3/2 and dy/dx = 7 when x = 4, find the value of the constant a.

I can handle discovering dy/dx from first principles. But I'm really not sure how to do this one. :/ Any help would be appreciated!

Sorry to break the thread somewhat, but I need abit of help. :/

Differentiation.
b) Given that y = a/x + 2x^3/2 and dy/dx = 7 when x = 4, find the value of the constant a.

I can handle discovering dy/dx from first principles. But I'm really not sure how to do this one. :/ Any help would be appreciated!

y = ax^-1 + 2x^3/2

dy/dx = -ax^-2 + 3x^1/2

7 = -a.4^-2 + 3.4^1/2

7 = -a/16 + 6

1 = -a/16

a = -16

6. 2 is not prime over the Gaussian integers. Discuss

6.2 is not an integer and therefore can't be prime.

That was easy, though I suspect I'll need to know what a Gaussian integer is before I can discuss without being snide.

6.2 is not an integer and therefore can't be prime.

That was easy, though I suspect I'll need to know what a Gaussian integer is before I can discuss without being snide.

The ring of Gaussian integers is all numbers of the form Z + iZ, where Z is the ring of integers.

An ring is any set where at the very least the following axioms hold:

1. There is an operation called addition, which is commutative, associative, has an identity (let's call it 0) and inverse -a such that a + -a = 0
2. There is an operation called multiplication, which is associative and has an identity (let's call it 1).
3. Multiplication is distributive over addition: a * (b + c) = a * b + a * c

There are three additional axioms that may be true

a. Non-triviality: 1 does not equal 0.
b. Multiplication is commutative (these rings are called Abelian rings).
c. Every number in the ring has a multiplicative inverse, 1/a, such that a * 1/a = 1.

Any ring satisfying a, b and c is called a field. The Gaussian integers satisfy a and b, but not c.

Examples of rings that are not fields: The integers, the Gaussian integers, the general set of 2x2 matrices, modular arithmetic with a composite number.
Examples of fields: The rationals, the reals, the complex numbers, modular arithmetic with a prime number.

I might think about this later. Equally, I might not.