cosine

Yeah, there's another meaning of Hypothesis when you are doing statistics.

Mr. Skeptic brings up a good point... Mary, are your 1000 numbers an ordered sequence?

Oh I see, k in the p equation is the 'really k' from later on! Sorry CalleighMay! Um... are you still working on this problem?

It can be intuitively difficult to imagine such a function because when we draw something it tends to be smooth (a.e. -for technicality's sake). But continuous Brownian Motion is an example of a function that is continuous everywhere but nowhere differentiable. (Note: Brownian Motion is also referred to as a Weiner Process.) Merged post follows: Consecutive posts merged hey insane_alien, I know what you mean by the bean machine, that is the sort of image I have in mind! Actually, what I want to see is that as the balls and spaces get smaller and more plentiful, the end result of the bean machine approximates a bell curve. I just want to see that in math language (aka by passing to the limit).

I see why you might say that, but no it is not like that. You can only take Reimann integrals over continuous domains, but because your theta goes from (-pi, pi), you're not allowed to jump from pi to -pi! If you could do the reimann sum as you were saying, then you would necessarily have: [math] \int^{-\pi+\epsilon}_{\pi-\epsilon} d\theta = -\int^{\pi-\epsilon}_{-\pi+\epsilon} d\theta [/math] However, all is not lost because the region you highlighted is just: [math] \int^{-\pi+\epsilon}_{-\pi} d\theta + \int^{\pi}_{\pi-\epsilon} d\theta [/math]

You can actually do this naïvely with brute force. there are 32 choices for a and 32 choices for c, so you only have 934 possibilities. So take any two numbers next two each other in the sequence and use it as a check for all the possible values of a and c. Store the ones that work for this pair, and then try these out on another pair in the sequence. Do it until needed. This should be O(n).

Hey guys, long time no post! In the past few years I wound up getting a BA in math and then accidently wound up going to grad school. I'm giving an informal presentation in a few days where I will introduce Brownian motion to some colleagues (no laxitive jokes, please!). Anyway, for the presentation I want to pass from discrete Brownian motion to continuous Brownian motion, but I need some help connecting the dots! I'll put what I have so far here and then maybe you can point the way. Thanks! Here we go: Here is a diagram of the simplest Brownian motion: 1 / 0 \ -1 Where each possible outcome (either 1 or -1) has a probability of 1/2. Let's say this happens over the discrete time period /\s. Now let's look at two time steps: 2 / 1 / \ 0 0 \ / -1 | \ | -2 | | [u]/\[/u]s, 2[u]/\[/u]s Now we can define p(t, x), a probability density function, where t is the time, and x is the position on the vertical axis. There is a counting arguement where you define n, the number of time steps that have passed, as t//\s, that allows you to define this function explicitly. I won't go into the arguement here, but instead just show that: p(t, x) = {n}CHOOSE{(n+x)/2} / 2^n Ok, so this part has been straight forward. What I want to do is find the limit of p(t, x) as /\s -> 0. Actually, it will be that p(t, x) is a normal distribution in x, where t is the squareroot of the variance, and 0 is the mean. Any ideas on how to show this though by letting /\s -> 0? Thanks again for the help guys!

your p is very strange as it is the sum of a vector and a scalar value. i j and k are vectors. how they work is like this: you could write the point (1, 2, 3) as 1i + 2j + 3k also this p is weird because density is a scalar value, not a vector value, so there should be no i's j's or k's that don't cancel out in its formula. when you have this figured out the straight forward way to look at this would be to use cylindrical co-ordinates. the cross-section of the object at any given z is a circle of radius 2. thats because cos t i + sin t j is the parametrization of a circle in a single variable. maybe when you come back with more details on p we can help you out more

Philosophy should be a good course for presidential candidates. I seriously think that people who have something to gain by winning the presidency should not be president.

Where's the F?

Step 1. Write a research proposal that requires the calculation of factorials. Step 2. Budget for a program that already calculates factorials. Step 3. Get research grant. Step 4. Purchase and install program. 4 lines of code, easy!

As an aside, I believe this method is part of a larger curriculum called Vedic mathematics, which claims to have roots in the Hindu Vedas, and has some following in Indian education.

I think you're meaning that that doesn't only apply to mathematics, in which case I totally agree. I'm just trying to dispell the illusions that most mathematicians lock themselves up alone in offices and out pops formulae

Well I would argue that it is a social discipline because you the social mathematician is infinitely better than the unsocial one. First off, explaining your thoughts is a very social skill. I appeal to authority now when I agree with the Feynmann/Einstein quote (depending on your source) that "If you can't explain it to a 6 year old [or an interested undergraduate], then you don't understand it enough yourself." And secondly the development of mathematics is much richer and quicker when you're bouncing ideas with a partner instead of only on your own path. I think of it much like this, If you consider a person as a vector, His ideas can span the space with the basis of one vector, aka a 1-dimensional space. Then add another person as a vector into the thought process, and the thoughts spanned by the two are 2-dimensional instead of 1!

I beg to differ! Mathematician A: Let's assume Fermat's Last Theorem to be true... Mathematician B: No. Mathematician A: Please? Mathematician B: No. Mathematician A: Pretty please? It will simplify the proof... Mathematician B: Well, okay then. Mathematician A: Well if we assume Fermat's Last Theorem, then my proof of it follows from our assumption! Haha I win! Mathematician C: B got pwned! Mathematician A: Thanks C.