The Official Math/Physics/Whatever Homework questions thread

[Subfallen]

Also,

which is in the unit disk, so the numerator must be smaller than the denominator, making calculus unnecessary even for that.

Or just...

|1-zr|^2 - |z-r|^2 = (1-|z|^2)(1-r^2) > 0

...yes? I think the (B - A)/(C - A) < B/C step is most opaque to me, since I completely failed to notice it after five hours.

[jason1990]

Originally Posted by Subfallen

Or just...

|1-zr|^2 - |z-r|^2 = (1-|z|^2)(1-r^2) > 0

...yes?

This looks right, and it shows that ϕ_r(z) is in the unit disk. A similar inequality holds when r is complex. But this should really be proven prior to this last step we are discussing. If nothing else, z in this last step is really wϕ_z2(z₁), and we would like to know that |z| < 1.

I had a friend who worked on some very complicated research, but his papers were always much shorter than mine. I asked him why he thought that was. His answer was that much of his work involved long strings of equalities. According to him, most of these could be omitted and left to the reader, because there are only a limited number of ways to get from one end of a string of equalities to the other. My work, on the other hand, involves long strings of inequalities. And, again according to him, these need to be explicitly described to the reader, because the reader would not be able to easily reproduce them.

I am not sure how much truth is in my friend's comments. I do, however, feel there is a lot of art in doing inequalities. Whatever the case may be, there is no denying that inequalities are at the heart of analysis.

[smcdonn2]

3.4.15
Let Y be the random variable described in question 3.4.3. Define W = 3Y+2. Find f_W(w). For which values of w is f_W(w) does not equal 0.

3.4.3
let f_Y(y)=(3/2)y^2, -1<=y<=1

I guess my problem here is he part define W=3Y+2 am I substituting the equation in 3.4.3 for Y in 3.4.15?

[smcdonn2]

IM confused on multiplying cycles

2 examples

(1,4,5,6)(2,1,5)

A={1,2,3,4,5,6}

multiplying we get

(1 2 3 4 5 6)
(6 4 3 5 2 1)

2ND EXAMPLE

(2,1,5)(1,4,5,6)

multiplying we get

(1 2 3 4 5 6)
(4 1 3 2 6 5)

same set A, If you could explain this to me it would be great, my book is a little vague.

[closer]

Originally Posted by smcdonn2

IM confused on multiplying cycles

(1,4,5,6)(2,1,5)

Always go right to left in terms of cycles. In each individual cycle go from left to right.

Start on the right hand side. 2 goes to 1, where does 1 go in the next cycle? It goes to 4.

1 goes to 5, then 5 goes to 6, and 6 doesn't go anywhere. So 1 goes to 6

5 goes to 2, and 2 doesn't go anywhere else. So 5 goes to 2.

3 goes to itself since it does not appear in any of the cycles.
4 goes to 5.
6 goes to 1.

(1 2 3 4 5 6)
(6 4 3 5 2 1)

You can always check by re-writing the cycles like
(1,4,5,6) =
1,2,3,4,5,6
4,2,3,5,6,1

(2,1,5)=
1,2,3,4,5,6
5,1,3,4,2,6

Then multiplying, but you will find with some practice (1,4,5,6)(2,1,5) is much faster.

[Englanese]

I am in a beginner game theory course and I am having some trouble when it comes to games that involve more than 3 players. I don't understand how to set up the games in strategic form and I was wondering if someone could help me out.
One of my home work questions is:

Suppose several friends go out to dinner with the understanding that the bill will be divided equally. The problem is that someone might order something expensive, knowing that part of the cost will be paid by others. To analyze the situation, suppose there are n diners and for simplicity they have the same food preferences. The accompanying table states the price of each of three dishes on the menu and how much each person values it. Value is measured by the maximum amount the person would be willing to pay for the meal.

dish value price surplus
pasta $21 $14 $7
salmon $26 $21 $5
steak $29 $30 -$1

a) suppose there are two diners. What will they order (at a Nash equilibrium)?
b)suppose there are four diners. What will they order (at a Nash equilibrium)?

I solved question a) no problem and got (salmon, salmon) but I have absolutely no idea how to do question b). I know this forum has a rule to try it yourself but I honestly don't know where to start.

[furyshade]

Let A;B;C;D be sets in some universal set u. Assume that A U B = u = C U D and
A - C = null set. Prove that u = B U D.

i can get that A is a sub set of C so C U B = u but can't seem to get how you get that to C U D = u. i feel it has something to do with subtracting the absolute compliment of C from D but not really sure.

[lastcardcharlie]

Originally Posted by furyshade

Let A;B;C;D be sets in some universal set u. Assume that A U B = u = C U D and
A - C = null set. Prove that u = B U D.

Counter-example: A = 1, 2; B = 3, 4; C = 1, 2, 3; D = 4.

[furyshade]

Originally Posted by lastcardcharlie

Counter-example: A = 1, 2; B = 3, 4; C = 1, 2, 3; D = 4.

it isn't just saying A U B = C U D, it is saying that A U B = u = C U D, where u is the set of all possible elements. from the book it seems this is a naive set theory problem so basically we are allowed a set of all sets. so basically A U u = u and A intersection U = A. basically A is some set of elements and B is at least all the elements not in B, same with C and D.

[furyshade]

disregard my question, the professor emailed out a correction on the problem and it is really easy. for anyone interested it is A intersection C = null not A - C = null

[Reefypoopoo]

drinking game question:

I pull one card, opponent pulls 2 cards from a deck. Highest card of the 3 wins, loser drinks once. Either of opponents cards can win.

What % of the time am I drinking?

[Pyromantha]

2/3

[Reefypoopoo]

lmao that was so incred easy once I thought about it for 3 secs. Thx

[goblinhat]

Quite a broad question.

If you were to pick one scientific breakthrough in physics which had a large impact on other topics in physics research and other engineering areas. Which would be easy to discuss what would you pick?

[Reefypoopoo]

F = ma

maybe http://en.wikipedia.org/wiki/Maxwell%27s_equation