# Let's discuss Mathematics

Discussion in 'Science & Technology' started by ParadigmShifter, Mar 16, 2009.

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2. ### Miseisle of lucy

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

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Partial fraction decomposition is the devil!

4. ### QuackersThe Frog

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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

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

6. ### QuackersThe Frog

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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....

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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...

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Mathematically speaking, why is there rioting in London?

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

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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.

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

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13. ### Miseisle of lucy

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A-level maths textbook?

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Have a gold star then!

15. ### AtticusDeityRetired Moderator

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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

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...any chance I can move to Finland?

17. ### QuackersThe Frog

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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.

18. ### sanabasPsycho BunnyHall of Fame Staff

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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.

19. ### cytherLord of the Dance

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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:

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).

20. ### SpoonwoodGrand Philosopher

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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?