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September 16[edit]

What is the correct pronunciation of George Pólya's last name?[edit]

What is the correct pronunciation of George Pólya's last name? (I consider the way he pronounced his name to be the correct pronunciation.) —Preceding unsigned comment added by 98.114.146.57 (talk) 04:20, 16 September 2009 (UTC)[reply]

here is it [1]--84.220.118.109 (talk) 06:08, 16 September 2009 (UTC)[reply]

Hungarian pronunciation: [ˈpoːjɒ ˈɟ͡ʝørɟ͡ʝ]. — Emil J. 10:35, 16 September 2009 (UTC)[reply]

How many combinations are there?[edit]

We have a lock box on our house similar to this: http://www.buyasafe.net/S6-Supra-Pushbutton-Lockbox-Surface-Mount./M/B000M7OXPY.htm?traffic_src=froogle

My question is, how many possible combinations are there?

Here are a few things to consider: 1) There are 10 possible digits (numbers 0 through 9). 2) The order the numbers are pressed doesn't matter. 3) We chose a 4-digit combination; however, we could have selected anywhere from a 0 - 10 digit combination.

To break the 4 digit combination, the odds would be 10 x 9 x 8 x 7 = 5,040, right? However, when you consider we could have selected any number of digits, how does this affect the total number of possible combinations?

Thanks in advance for your help! —Preceding unsigned comment added by 209.206.158.57 (talk) 17:02, 16 September 2009 (UTC)[reply]

Your 5,040 isn't right, because that distinguishes different orders. There are 5,040 4-number permutations using 1-10 w/out repetition. You want to divide that by 4! = 24, meaning there are only 210 4-number combinations.

Now, if you add up the number of 0-number combinations (1) + the number of 1-number combinations (10) + the number of 2-number combinations (45) + . . . + the number of 10-number combinations (1), you'll get 2^10 = 1,024.

One way to see why it's 2^10 is that each possible code either uses or doesn't use each of the 10 numbers. Thus, you either use '1' in your code, or you don't - 2 options. You either use '2' or you don't - 2 options. . . You either use '10' or you don't - 2 options. Multiply all those 2's together, boom.

Does that answer your question? -GTBacchus^(talk) 17:19, 16 September 2009 (UTC)[reply]

Sorry if I'm being a bit slow, but I don't see how that's right. Surely a code using a '5' say, could use it more than once? As a disproof (I think!), imagine a similar lock using two digits, which can either be 1,0 or not used. That leaves 6 combinations, which isn't 2^2.--Leon (talk) 17:33, 16 September 2009 (UTC)[reply]

No, I've used locks like this. When you press a button, it stays pressed, and having all n buttons down at once, with no others down, is how it opens. Thus, with two buttons, labelled '1' and '2', the 4 possible codes are {}, {1}, {2}, and {1,2}. We're really just looking at the power set of {1,...,n}. -GTBacchus^(talk) 18:45, 16 September 2009 (UTC)[reply]

I'm sure GTBacchus is right. But for an edit conflict, I would have posted a similar answer. The person who asked the question stipulated that the order the numbers are pressed in doesn't matter. This implies a simple mechanism which doesn't care if a button is pressed once or many times. If you have two digits, you have two 1-digit combinations, not six. RupertMillard (Talk) 17:53, 16 September 2009 (UTC)[reply]

Apologies: I neglected to check the link, and I presumed that a number could be selected multiple times!--Leon (talk) 18:53, 16 September 2009 (UTC)[reply]

Here's a puzzle. How many distinct, valid combinations are years since 1900? Harder - how many are dates? (eg. 0916) How many combinations does this total for you to try to crack the safe if you knew the combination was a date or a year? I think the answer is the days of our age, but I'm not certain. RupertMillard (Talk) 18:55, 16 September 2009 (UTC)[reply]

To your second question... combinations representing dates can involve 2, 3, or 4 buttons (1/10, 1/23 and 12/03 being the first occurrence of each). I count 82 combinations that can represent 149 different dates. If a thief can try a combination in 5 seconds, that means 410 seconds, which is just under 7 minutes to run through all date combinations. I would note that 0916 is not actually an option, because there's no zero button. If there were a 0 instead of a 10, then only 49 combinations represent dates, which is just over 4 minutes of burgling. -GTBacchus^(talk) 20:10, 16 September 2009 (UTC)[reply]

Interpretation help[edit]

Hi. I have a small problem with some homework but it's only in understanding what the question means.

"Let $g(z)={\frac {z-i}{z+i}}$ where z is a complex number. Writing $g^{2}(z)$ for $g(g(z))$ and $g^{-1}(z)$ for the inverse function, what is the image of the real line under the successive maps $g$ , $g^{2}$ , $g^{3}$ ? What is the image of this set of images under these maps?"

Letting z=x+iy, does the first part just mean determine $g(x)$ , $g^{2}(x)$ , $g^{3}(x)$ ? And I don't have a clue for the second part. Any ideas? Thanks. 92.2.21.125 (talk) 19:56, 16 September 2009 (UTC)[reply]

They want you to determine the images of R under these linear fractional transformations, that is, the sets A_n:=gⁿ(R)={gⁿ(x): x in R}, for n=1,2, and 3. The other question is not clearly stated indeed; probably they just want gⁿ(A_k), which is trivially A_n+k; or they mean the image of the union of the sets A_k under the maps gⁿ. --84.221.68.30 (talk) 20:19, 16 September 2009 (UTC)[reply]

(edit conflict) Finding the "image of the real line" means that, if you plug in every point on the real line to g, and then look at where they all end up, what shape on the complex plane does that give you? For example, the point 0 gets sent to -1, the point 1 gets sent to g(1) = (1-i)/(1+i) = -i, etc. Try plugging in some more real numbers, and see if you can make a sketch of where the real line is going, point by point. Once you have an idea, try and prove your conjecture.

The second part of the question seems to be saying, do the same thing, but instead of just the real line as your starting set, use the set $\mathbb {R} \cup g(\mathbb {R} )\cup g^{2}(\mathbb {R} )\cup \cdots$ . Does that help at all? -GTBacchus^(talk) 20:29, 16 September 2009 (UTC)[reply]

OK I'm with the first part of the question now - just determine the sets (/regions of the complex plane) produced by the results of

g(x)

,

g^{2}(x)

,

g^{3}(x)

. That was mainly what I was thinking before, it's just that the wording '... the successive maps...' seemed to imply you might do

g^{3}(g^{2}(g(x)))

but clearly not. You both seem to be going for the

g^{n}(\mathbb {R} \cup g(\mathbb {R} )\cup g^{2}(\mathbb {R} )\cup \cdots )

idea for the second part so I'll try that. Thank you both! 92.2.21.125 (talk) 20:43, 16 September 2009 (UTC)[reply]

But notice that what is the image of the real line means: what set is g(R) --you should find that it's a circle. So you do not have to find what is the image of every single real point g(x). Also, recall that the composition of linear fractional transformations is easily computed multiplying the associated 2x2 matrices of coefficients. Here g³ sounds like a sort of well known map ;-) --84.221.68.30 (talk) 21:00, 16 September 2009 (UTC)[reply]

Of course nobody's going to try every single real point. We're mortal. I think he's supposed to figure out whether it's a circle; that's why it's homework. Let's not give away so much, eh? -GTBacchus^(talk) 21:18, 16 September 2009 (UTC)[reply]

For extra credit (and the pleasure of a Eureka moment) visualise and describe the action of g(z) on the Riemann sphere. Gandalf61 (talk) 09:42, 17 September 2009 (UTC)[reply]

Ooh, nice! Thank you. :) -GTBacchus^(talk) 14:44, 17 September 2009 (UTC)[reply]

I've just worked through g(R) and, once simplified, have $g(x)={\frac {x^{2}-1}{x^{2}+1}}-{\frac {2ix}{x^{2}+1}}$ . Now I can show that the modulus of this is 1 but when I graph this, I most certainly do not get a circle. Where am I going wrong? 92.2.18.79 (talk) 14:24, 17 September 2009 (UTC)[reply]

Right, for

g^{2}(x)

I have

g^{2}(x)=(1-i){\frac {x+1}{x-1}}

but don't know where to go from here. Taking the modulus still leaves it in terms of x, which doesn't seem helpful. Any suggestions? Thanks 92.2.18.79 (talk) 14:48, 17 September 2009 (UTC)[reply]

Yeah, double-check your work. That's not g^2(x). -GTBacchus^(talk) 14:55, 17 September 2009 (UTC)[reply]

My mistake.

g^{2}(x)=-i{\frac {x+1}{x-1}}

. So does this just mean that every point on the real line, excluding x=1, is mapped onto a point on the imaginary line? 92.2.18.79 (talk) 15:06, 17 September 2009 (UTC)[reply]

Yes, you've just shown that

g^{2}(\mathbb {R} )\subset i\mathbb {R}

. If you can show inclusion in the other direction, then you'll have

g^{2}(\mathbb {R} )

! (That's an exclamation point at the end, and not a factorial, of course.) -GTBacchus^(talk) 15:24, 17 September 2009 (UTC)[reply]

OK,

g^{3}(x)=x

and so every point on the real line is mapped onto itself. Now for the final part of the question, I just need to check the circle. I managed to get

g(x)={\frac {x^{2}-1}{x^{2}+1}}-{\frac {2ix}{x^{2}+1}}

and that its modulus is one. So is the equation of this circle given by

(x-1)^{2}+y^{2}=1

? 92.2.18.79 (talk) 18:31, 17 September 2009 (UTC)[reply]

If the modulus of every point on the circle is 1, then where should the center of the circle be? -GTBacchus^(talk) 19:04, 17 September 2009 (UTC)[reply]

Hmm, I sense you mean the origin. This is the first time I've ever considered the modulus of the points on the circle, rather than the distance of the points o the circle from the centre, to determine the centre. Thanks GTBacchus. 92.2.18.79 (talk) 19:11, 17 September 2009 (UTC)[reply]

You're quite welcome. Complex analysis is fun; enjoy! -GTBacchus^(talk) 19:15, 17 September 2009 (UTC)[reply]

Expectation[edit]

Resolved

So, I have emailed my TA a question once and his response indicated he did not understand my question at all. I just got back from asking him other questions in person and he again did not understand what I was even asking, though I explained it several times. The professor does not have any office hours. So, I come here.

I'm doing a simple problem and I need help in understanding. The problem says to show if you have random variables X and Y then

E(max(X, Y)) = EX + EY - E(min(X, y))

They even give a hint to show that X + Y = max(X, Y) + min(X, Y). Well, this is a problem that is intuitively simple but I don't know how to explain it. I know if I am dealing with 2 real numbers x and y, then x + y = max(x, y) + min(x, y) and the proof is easy. So, it seems obvious to me from this that the same equation with random variables must be true. But, just saying it is obvious does not prove it. This is my only difficulty with this problem. Clearly, if I prove the equation with random variables, I can just take the Expectaion of both sides and use linearity and then subtract the min from both sides.

One thought is to let A = X + Y and B = max + min, new random variables. Then, P(A = a) = P(X + Y = a) = P(X = a - Y)... I don't know. We have only talked about functions of one random variable, but not functions of multiple random variables.StatisticsMan (talk) 20:48, 16 September 2009 (UTC)[reply]

According to Random variable#Formal definition X and Y are both functions. Taemyr (talk) 21:15, 16 September 2009 (UTC)[reply]

I know they are functions. What does that have to do with my question? StatisticsMan (talk) 21:22, 16 September 2009 (UTC)[reply]

ecThis: you already have all you need. Recall that X and Y are in particular functions Ω →R, so at any ω in Ω, X(ω) and Y(ω) are certain real numbers, and for them you do know the equality:

X(ω) + Y(ω) = max( X(ω), Y(ω)) + min( X(ω), Y(ω) );

this equality for all ω in Ω just means the equality between functions

X + Y = max(X, Y) + min(X, Y);

so you conclude as you said by linearity of the expectation. --pma (talk) 21:24, 16 September 2009 (UTC)[reply]

Well, I guess it has everything to do with my question! And, I believe I understand this. Thanks a lot! StatisticsMan (talk) 02:41, 17 September 2009 (UTC)[reply]