Let's discuss Mathematics

I think quackers means just algebra, rearranging the equation so it says x=<stuff> rather than y=<stuff> or similar.

An example would be handy to show the general principle of it.

Wolfram alpha will do it though

e.g.

http://www.wolframalpha.com/input/?i=solve+y=12x^2+2x+for+x

 

"Whatever you do to one side you must do to the other"?

 

Partial fraction decomposition is the devil!

 

yeah thats what I mean.

will have a look at the link

some easy ones I can work out. but stuff like

express x in terms of y when

y=x-3/x+2

 

Multiply both sides by (x+2) to get a linear equation in x. Then solve for x.

 

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....

 

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...

 

Mathematically speaking, why is there rioting in London?

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

 

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.

 

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

 

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...

 

A-level maths textbook?

 

Have a gold star then!

 

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

 

...any chance I can move to Finland?

 

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.

 

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.

 

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).

 

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?