Let's discuss Mathematics

[Quackers]

Well does anybody know a place where I can do some exercises? Online?

I think I just need some practice to get it into ma brain before I start doing moddeling using straight lines....

 

[ParadigmShifter]

Shouldn't you be in London rioting anyway Quacko?

Did you not do A-level maths before starting a degree in economics?

EDIT: And you should use brackets in your example, do you mean

y = (x-3) / (x+2)

or

y = x - (3/x) + 2

or something else? The latter is the usual precedence rules for division...

 

[IdiotsOpposite]

Mathematically speaking, why is there rioting in London?

Remember to use the words "set", "subset", and "population" in your answer.

 

[ParadigmShifter]

Students aren't happy about plans to raise fees to up to £9000. A subset of them are rioting. Some of them set off fire extinguishers. They probably haven't had sets in a long while.

 

[IdiotsOpposite]

Nice job. I wish I could pity the students, but I'm paying the equivalent of £14730 (according to Wolfram-Alpha), so...

 

[ParadigmShifter]

Here's Wolfram Alpha on what Quackers probably meant. Click "show steps" to show the working oots:

http://www.wolframalpha.com/input/?i=solve+y+=+(x-3)/(x+2)+for+x

 

[IdiotsOpposite]

Here's Wolfram Alpha on what Quackers probably meant. Click "show steps" to show the working oots:

http://www.wolframalpha.com/input/?i=solve+y+=+(x-3)/(x+2)+for+x

Yeah, that's just what I did. And what I said...

 

[Mise]

Well does anybody know a place where I can do some exercises? Online?

I think I just need some practice to get it into ma brain before I start doing moddeling using straight lines....

A-level maths textbook?

 

[ParadigmShifter]

Yeah, that's just what I did. And what I said...

Have a gold star then!

 

[Atticus]

We had to pay equivalent of 83 euros when I begun my studies in 1996, and something like 90 euros when I ended them in 2003. That's how we roll in Librulistan

 

[IdiotsOpposite]

...any chance I can move to Finland?

 

[Quackers]

Shouldn't you be in London rioting anyway Quacko?

Did you not do A-level maths before starting a degree in economics?

EDIT: And you should use brackets in your example, do you mean

y = (x-3) / (x+2)

or

y = x - (3/x) + 2

or something else? The latter is the usual precedence rules for division...

I was accepted with just a B in GCSE mathematics (A in exam, c in coursework ). I haven't done maths in two years so I need to brush up on just everything tbh.

I mean the former.

 

[sanabas]

2 ways to do it:

First, multiply both sides by (x+2)

xy + 2y = x-3

Not obvious where to go from there, but remember you're trying to get x by itself. So let's get all the terms with x on one side, all the terms without on the other, by subtracting xy from both sides and adding 3 to both sides.

2y + 3 = x - xy

factorise the RHS

2y + 3 = x (1-y)

Divide both sides by (1-y)

x = (2y+3)/(1-y)

Second way to do it is to do the division (x-3)/(x+2) at the start, or just write x-3 as x+2-5, either of those gives you

y = 1 - 5/(x+2), which is more straightforward to rearrange:

y-1 = -5/(x+2)

x+2 = -5/(y-1)

x+2 = 5/(1-y)

x = 5/(1-y) - 2

x = [5 - 2(1-y)]/(1-y)

x = (2y+3)/(1-y)


And yeah, best way to practice is just head for the library and find a textbook of the right level.

 

[cyther]

So yesterday I was bored after a few hours of not knowing what to do, I started playing with the Riemann Zeta function on Mathematica.

I really couldn't find anything interesting about that right away (yeah I know about the Riemann hypothesis, I meant more of myself not the function). So I started pressing the ' key to take derivatives. I was bothered by the notation and wanted to show the guy sitting next to me how silly the notation was. That's when I noticed this:

Zeta''''''(2) was about 720.

I removed one of the 's and saw that the result I got was about -120.

So I kept going down to Zeta'(2) and got about 1. I also went up to around 60 and it was still about n! (There was an error of around .001).

Has anyone seen / can explain the connection between the zeta and n! (or gamma shifted over if you prefer)?

I figured that what I have is:

The nth derivative of the Riemann Zeta function evaluated at 2 is about equal to n! * (-1)^n.

Edit:

Did some more testing and found out that:

The nth derivative of the Riemann Zeta function evaluated at 0 is about equal to n! * (-1).

 

[Spoonwood]

In civ III each scientific tribe gets a free technology (tech) at an era change. Let's say there are 4 techs available for a specific era change (modern age). The probability of drawing any of the particular techs is .25. Say we have 7 tribes that get a free tech at an age change. The probability of all 7 tribes getting one of the same techs is .25^7=.000061, so it's 4*.25^7=.000244 for all 7 tribes getting any of the same techs. What's the probability of 7 tribes getting exactly 2 of the techs? For 7 tribes getting exactly 3 of the techs? For 7 tribes getting exactly 4 of the techs? How does one compute these?

 

[ParadigmShifter]

Are you talking about the Binomial distribution?

http://en.wikipedia.org/wiki/Binomial_distribution

 

[Spoonwood]

No, at least not directly. That might help indirectly, but I don't know how to use it at the moment for this question. After reading your suggestion, I actually tried to compute with the probability mass function formula, but then recalled I don't have a k=0 case for the binomial distribution. With 7 tribes "n" would equal 7, but then k=number of techs which can't equal any more than 4, when we need cases {0, ..., 7} if n=7. A similar mathematical example is:

Throw a die 7 times. What's the probability that only two of the six numerals on the die appear in the set of 7 throws?

I have a bit better idea now as to how to compute this, but I don't feel sure that I've gotten on the right track here.

Spoiler :

If you have just one scientific tribe, then you have probability of 1 of them getting a tech. If have two scientific tribes, the probability of them together getting two techs is .75, since 1*.75, since the probability of the second tribe getting a tech different than the first tribe is 1-.25=.75. If you have three tribes, and the first two tribes got different techs, the third tribe has a .5 probability of getting a tech the same as the first two. So, we have .375 probability of 3 tribes getting exactly two techs *given that the first two tribes each got different techs* (but the case where the first two tribes got the same first tech isn't taken into account here). So, given that the first two tribes drew distinct techs, the probability of 7 tribes getting only 2 techs is .75*(.5)^(7-2)=.02344. But, this isn't the probability I'm looking for since I didn't consider the possibility of the first two tribes getting the same tech, and there are some of those cases too.

This is a game inspired question. I had a space-race game where my 7 opponents only drew 2 of the 4 modern age techs, which I believe rather unlucky. But, I don't really know the probability of that happening, so I don't know if I just have some bias here.

 

[sanabas]

It just basic probability theory, combinations & permutations.

The chance of a particular permutation is 0.25^7, there are 4^7 = 16384 possible permutations.

Then just count it up.

Number of ways for all 7 to have tech A = 1, and same for them all to have techs B, C, D. So number of ways for all 7 to produce just one tech is 4/16384

Number of ways for only techs A & B to be covered is a bit trickier. There's 1 A & 6 B's (7 ways), 2 A's & 5 B's (7C2 = 21), 3 & 4 (7C3 = 35), ... 6 & 1 (7C6 = 7) Add that up, and there's 126 ways to get a particular 2 techs. There's 6 combinations of 2 techs, so that's a 756/16384 chance.

Number of ways for 3 techs is going to be just like that for 2, except more of a pain to count. So I'm going to be tricky and use what we worked out already. The number of ways to NOT get tech A is 3^7 = 2187. But of those ways, 3 of them had just one tech, and 6 x 126 = 756 had just two techs. That leaves 1428 ways to have all of techs B, C, D represented. Same deal for leaving out the other 3 techs, so there's 4 x 1428 = 5712 ways to have 3 techs total.

That's 6472 ways covered, which means the other 9912 ways will give you all 4 techs.

So, for 7 tribes drawing from 4 techs, the chances are:

1 tech: 4/16384 = 0.024 %

2 techs: 756/16384 = 4.61 %

3 techs: 5712/16384 = 34.86 %

4 techs: 9912/16384 = 60.5 %

Throw a die 7 times. What's the probability that only two of the six numerals on the die appear in the set of 7 throws?

Yep, same working. There's 126 different ways to get a particular two numbers. There's 15 combinations of 2 numbers to get. There's 6^7 = 279936 different ways to roll 7 dice. So your chance is 126 x 15/279936 = 0.675 %

 

[ParadigmShifter]

You can probably work it out with the multinomial distribution too

http://en.wikipedia.org/wiki/Multinomial_distribution

 

[Spoonwood]

Thanks Sanabas. I had to reread your explanation a few times, I think I spotted a typo in the numbers, and using numerals for both the tribes and how many times they drew a tech I found confusing, but that helped. Thanks!