Let's discuss Mathematics

[Atticus]

I do like that solution you gave, Mise!

It seems this kind of problems have always one twist that takes you some time to realize: that you can turn the gloves inside-out, or that you can weight more balls at the same time. In this case it was that not all have to follow the same rule in the room.

 

[Garonius]

I do like that solution you gave, Mise!

It seems this kind of problems have always one twist that takes you some time to realize: that you can turn the gloves inside-out, or that you can weight more balls at the same time. In this case it was that not all have to follow the same rule in the room.

Most people figure it out given enough time. It's actually more tricky than you think at first glance though. That said the only real way to do it is wait five years and then ask as there's a one in a million chance you will be wrong. But of course not everyone has some advanced calculus tricks at their disposal so it's pleasing to see people scratching their heads over this.

 

[Bill3000]

 

[dutchfire]

332.6 m^3?

 

[Mise]

Yeah, that's what I get.
http://www.wolframalpha.com/input/?i=%28%2812^2%29%2F3%29^%283%2F2%29

 

[SS-18 ICBM]

A problem.

A casino offers you a coin flipping game with the following rules:

1. Flip a fair coin. (We may assume it's perfectly fair.)
2. If heads, the game ends and you win $N, where initially N = 1.
3. If tails, double N and repeat step 2.

Eventually you'll come up tails and win some amount of money. If you flip tails many times it may in fact be a pretty huge amount, since the prize has doubled so many times before you come up heads.

Using the expected value concept, how much money can you pay for this game and still come out ahead? How much would you actually be willing to pay, if it were personally offered you?

Savvy readers may ask how much money the casino has. Start off assuming infinite wealth, and then, if you'd like, repeat for a casino of large but finite wealth. Go!

Thoughts, comments?

 

[Mise]

So, P(N=1)=0.5, P(2)=0.25, P(3)=0.125, etc
E(X)=1/2 + 2/2^2 + 3/2^3 + ...

Wolfram Alpha sez it sums to 2. So you should pay $2 if you are risk neutral. I dunno what I'd pay IRL, maybe $0.50 I guess.

EDIT: Oh, wait, I didn't double N on the numerator... doh.

In that case the expected outcome is infinite. I would still only pay a few dollars. Expectation value ignores the fact that I can't do it an infinite number of times with an infinite amount of money. And, if I had an infinite amount of money, the marginal value of any gains is negligible...

 

[uppi]

I'm not sure if I've asked this here before, but if I have two groups of uneven sample size and I want the standard deviation of the difference between them, how do I calculate it?

I think I need to weight the variances by dividing by the sample size, add them and subtract 2*(deviation1*deviation2/total sample size) but I'm not sure how to weight the deviations/variances.

I am not sure what exactly you mean by the difference between them and I am assuming you mean the difference of the mean of both groups.

If so, I'd do the following:
1. Calculate the standard deviation of both groups.
2. Divide both by the square root of the respective sample size to get the standard deviation of the mean.
3. Multiply both with the Student's t-distribution (with a chosen confidence interval) to guess the standard deviation for an infinite sample size.
4. If you believe both groups to be independent from each other, add the resulting deviations quadratically (sqrt(x^2 + y^2))

That's how I would calculate the error in physics, but I am not sure it is done the same way in your field.

 

[dutchfire]

@SS-18 ICBM
That question is as old as Bernouilli

 

[SS-18 ICBM]

Never said it was new. And is there anything Wolfram Alpha can't do, math-wise?

 

[Bill3000]

 

[SS-18 ICBM]

450 .

 

[dutchfire]

In xx0 - xx9 he's got (0+9)*10/2=45 ho's. xx can be anywhere in 0 to 99, so 100 possibilities. So I'd say 4500.

 

[SS-18 ICBM]

Damn, off by an order of magnitude.

 

[Garonius]

Since they are easy not much beyod GCSE maths mostly why not try a more difficult integral:


integral of sin(x) 1/x dx from -infinity to infinity=pi


Show all working.

 

[dutchfire]

Since they are easy not much beyod GCSE maths mostly why not try a more difficult integral:

Show all working.

Your image isn't showing properly on my pc.
For other people with this problem, the integral is: sin(x) 1/x dx from -infinity to infinity=pi

edit: now your image is working for me. That's strange.

 

[Garonius]

Your image isn't showing properly on my pc (physicsforum doesn't allow hotlinking of images).
For other people with this problem, the integral is: sin(x) 1/x dx from -infinity to infinity=pi

edit: now your image is working for me. That's strange.

Indeed I'd appreciate it if people worked it out though rather than going to that site and just fishing it out.

If you could edit that quote to get rid of the image, I've written it out now.

 

[Atticus]

Spoiler :

I won't bother to pick up a pen, but instead guess that it involves integration in parts twice and then manipulating the equation you get.

That's probably one of the most classic basics of analysis problems.

 

[ParadigmShifter]

I like the mathroom problems

Hopefully we will have one about a conical spliff with a density of pot equal to rho soon.

EDIT: If Abe smokes a conical spliff, whilst Ben smokes a cylindrical spliff, both with the same width (radius) at the midpoint, who
gets more stoned, and if anyone how much more stoned do they get?

Assume no roach and no-one soggies the end.

EDIT2: Ben could be smoking a pencil (wider at the near end than the lit end), does that make a difference?

 

[Mise]

Cylinder would be bigger, right? Cos the base is a square of width x rather than a circle of radius x? Although, burning the end of the conical one would result in a higher THC intake per second at the start, which would probably make it more potent than the cylinder...