Aaaah, sorry. Yeah, the function of z is the 1/4(exp(i*z)....thing)

That has z-bar's in it. This should concern you when you're applying theorems.

Well, I was in my math class (12th grade) and we are learning probabilities and statistics.

We were talking about expected value.

Here is the problem:

We have 20 balls in a bag, 4 greens, 6 orange and 10 red ones.

If we grab a green ball we get 2€ of profit, 1€ if we grab an orange and 0€ if we grab a red one.(...)
The rest of the problem doesn't really matters.

Imagine we weren't giving any money and we could grab a ball, what is the expected value?

We calculated it this way

0*(1/2)+1*(3/10)+2*(1/5)=0.7

And if we bet 1€, and we lose all we grab a red one but we get the profit from the other balls plus the amount we bet we are getting an +EV. My math teacher told me we would lose 0.3€ every time we would bet, I insinted telling we would win 0.2€ every time we bet.

Is the math correct:

-1*(1/2)+1*(3/10)+2*(1/5)=0.2 - Is it the correct way of calculating it?

Posted here too:

http://forumserver.twoplustwo.com/47.../#post22794524

You pay $1 to play. Do you get $2 back for a green ball OR do you get $2 PLUS the $1 you paid?

Your calculation should (as I'm reading this) be:
-1*(1/2)+0*(3/10)+1*(1/5)=-0.3

Plus.

If that were the case, you'd be correct.

hi first time posting here i have a couple of questions if you guys could help me thanks!
(1)The probability of being dealt a royal flush (ace, king, queen, jack and ten in the same suit)dn poker is about 1.3 x 10^-8. Suppose that an avid poker player sees 100 hands a week, 52 weeks a year, for 20 years. What is the probability that she sees a royal straight flush dealt at least once?
(2) Suppose that there are 15 races a meeting, with 10 horses competing in each. Among the punters at the meeting, 5% are professional gamblers and the rest are amateurs. Suppose further that professionals have a 20% chance of correctly picking the winner in any given race, while amateurs do no better than random. If I correctly pick the winner in 5 of the 15 races, what is the probability that I am a professional gambler?
(3) A manufacturer cookies claims that every bite will contain at least one chocolate drop. Suppose that all chocolate chip cookies have a radius of 3cm and that there are an average of 8 chocolate drops in each cookie. If you take a bite of 7.07cm2, what is the probability that the manufacturers claim will be correct?

For 2) let A be the event that you are a professional gambler, and B be the event that you pick 5 winners. Find P(A|B) = P(A \int B) / P(B), noting that P(B) = P(A) x P(B|A) + P(notA) x P(B|notA), and P(A \int B) = P(B|A) x P(A). P(A), P(B|A) and P(B|notA) are directly obtainable from the information.

thanks mate appreciate it

(3) A manufacturer cookies claims that every bite will contain at least one chocolate drop. Suppose that all chocolate chip cookies have a radius of 3cm and that there are an average of 8 chocolate drops in each cookie. If you take a bite of 7.07cm2, what is the probability that the manufacturers claim will be correct?

This question made me hungry for cookies. Then I realized I didn't have any cookies so I did the question instead.

Computing the area of a cookie, we get that each bite consumes a quarter of the cookie. Therefore, the probability that a given chocolate chip fails to be in the quadrant of the cookie that we bit is 3/4. Hence the probability that all 8 escape our bite is (3/4)^8 = 0.1. Since 1-0.1=0.9, the probability that the manufacturer's claim was correct is 90%.

The likelihood of the claim being correct is not at all effected by any amount
of bites eaten or the amount of chocolate in
said bites - your teacher is sitting at his house right
now going "oh my gosh it will be soooo fun owning them tomorrow "

Let f(x)=(x+1) / (x-1). Show that there's no value of c such that f(2)-f(0)=f '(c)(2-0). Why this does not contradict Mean Value Theorem?

Attempt:
I've found the derivative of f(x) and put in the the equation f(2)-f(0)=f '(c)(2-0) in place of f'(c) to get:

4 = 2[((x+1)((x+1)^3)-(x+1)) / (x-1)^2]

I'm pretty sure its a dead end trying to solve the equation from here. Anyone have any ideas?

Is your function continuous on the interval in question?

Question did not specify an interval... so I guess we take it that f(x) is just not continuous at x=1

oh wait, I guess that's why it does not contradict the mean value theorem, f(x) does is not continuous throughout. But how do we prove that there's no value of c such that f(2)-f(0)=f '(c)(2-0)?

Use

f(x) = (x+1)/(x-1) = (x+1-2+2)/(x-1) = (x-1+2)/(x-1) = 1 + 2/(x-1)

to get a simpler form of f'(x), then try to solve f'(c) = 2.

Well, all it's asking for is you to differentiate

f(x) = 1 + 2/(x-1)

so
f'(x) = d/dx [1 + 2/(x-1)]
f'(x) = 2*d/dx[(x-1)^-1]
f'(x) = -2/(x-1)^2

f'(c)=2 is asking for you to find the value of x (ie, when it's 'c') where f'(x)=2

2 = -2/(c-1)^2

And solve for 'c' from there.....well, actually, no, you don't solve it from there, because that doesn't have a rational solution. It has a solution, but if you're being asked a question of this form, then it'd be a given that your solution isn't a complex number, so, well, I'm not sure there.

However, if you've just mistyped that and it's actually f(c)=2, then yes, that does have a solution

2 = 1 + 2/(c-1)

Edit: Bah, my mistake, I thought you were asking for a solution, rather than working on Dumbos question from above. My apologies.

does anyone have a good resource for understanding how to calculate the Jordan Canonical Form of a matrix? my professor explained it really poorly and the book only has it in an appendix. I can't seem to find a good resource online that gives an intuitive explanation of how to generally calculate these

This isn't a homework question but I figured I'd ask here.

Yesterday morning at around 6am during twilight, I was outside and noticed Venus was low on the horizon. It caught my eye because it was brighter than I'd ever seen it and seemed very large, probably due to its proximity to the horizon. It also had many thin rays shooting off of it in several directions, and I know this sounds hokey but almost had an angelic appearance. Short rays on top, longer rays on the sides like wings and the longest rays were at the bottom like a long gown that flowed down and away from the sun.

What really seemed strange was that a couple of the rays had rays jutting off of them at slight angles instead of emanating straight out from Venus like the others. I observed this for several minutes, and to test if my eyes were playing tricks on me, I would look away and look back and still see the exact same phenomenon, the same two rays originating from other rays instead of Venus.

Has anyone else seen anything like this or read about it? I went out this morning to see if it happened again but it was too cloudy, there were no rays and Venus wasn't nearly as bright or large looking. I've Googled around a bit but can find nothing.

Horn and Johnson, Matrix Analysis. Sec 3.1 gives the relevant derivations and the theory, Sec 3.2 gives the practical details about implementation.

Basically, the multiplicity of $\lambda$ as a root of the minimal polynomial tells you the size of the largest Jordan block associated to $\lambda$, and the dimension of the eigenspace of each eigenvector tells you the number of blocks associated to it.

For small matrices, this is typically sufficient. This is insufficient, though, when you get to things like a 7x7 matrix M whose minimal polynomial is (x-2)^3, and the eigenspace associated with \lambda = 2 is 3 dimensional. Could be a (3,3,1) Jordan structure or a (3,2,2).

Assume M is actually just a matrix of Jordan blocks (and convince yourself that everything I write from here on out holds if not, since we can write M = PJP^(-1), where J is a matrix of Jordan blocks). Notice that (from the minimal poly), (M-2I)^3 = 0, and (M-2I)^2 is nonzero. This says that the max Jordan block is of size 3 (why? Think about what cubing does to the 3x3 Jordan block with 0 on the diagonal. To the 4x4? To the 2x2?). Ah, so how can we count the number of Jordan blocks of size 3? Let's kill all the smaller ones! Convince yourself that rk((M-2I)^2) = rk ((J-2I)^2) = # of size 3 Jordan blocks. And rk(J-2I) = # Jordan blocks of size > 1.

Of course, if M has more than one eigenvalue, you use the same logic. Suppose the eigenvalues of M are 2 and 5, that the characteristic polynomial is (x-2)^4 (x-5)^10, and that the minimal polynomial is (x-2)^2 (x-5)^4. What info [which ranks of which matrices] would you need to compute the Jordan block structure of M?

HTH,
-BW

PS - Thanks for the review.
PPS - I'm sure Dummit and Foote does this in painful detail also. I used that book in grad school. But H&J is on my desk atm and is both friendly and practical.

Is it true that, given f holomorphic on the disk and continuous on the closure of the disk, that f has only finitely many zeros?

This is driving me crazy, any help appreciated.

I think yes. Something about infinitely many points in a compact set having a limit point and then something about a holomorphic function's zeros being a discrete set.

But I haven't thought about complex analysis in 5 years, and I don't have a reference with me at the moment.

You should verify *everything* I ever say regarding analysis, because there is a huge chance it's wrong.

True, if a holomorphic function has an accumulation point of its zeros in its domain, then it is constant. And if f vanished along an arc on the boundary of the disk, then an application of the Schwarz reflection principle would tell us f vanished identically.

But why can't f's zeros accumulate to isolated points on the boundary?!

they probably can

i think i followed up until the bolded, i have never seen the notation rk() before. I'm going to try and go back a rewatch (recorded lectures ftw) how the TA did it and hopefully with what you said i can make some sense out of it. its sort of frustrating because the professor is pretty bad at lecturing and this is the first thing i haven't been able to figure out on my own.