Math Q&A Game


by Gokul43201
Tags: game, math
mansi
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#55
May4-05, 09:21 AM
P: 58
i know it's a bad idea but looks like this is gonna be the end of this sticky...


anyways...i want to work on the problem i posted....chingkui,could you elaborate the circle part...didn't quite get that...
chingkui
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#56
May4-05, 11:20 PM
P: 195
the reason to just look at [0,1] is that if it is dense there, then, it is dense in [n,n+1] (by translation) and you know that the set Z+[tex]\sqrt{2}[/tex]Z is dense in R.

As for n[tex]\sqrt{2}[/tex] mod(1), let's clarify what it means:
[tex]\sqrt{2}[/tex] mod(1)=0.4142...=[tex]\sqrt{2}[/tex]-1
2[tex]\sqrt{2}[/tex] mod(1)=0.8284...=[tex]\sqrt{2}[/tex]-2
3[tex]\sqrt{2}[/tex] mod(1)=0.2426...=[tex]\sqrt{2}[/tex]-4
So, it is just eliminating the integer part so that the number fall between 0 and 1.
This is what I mean by "circle", it is like you have a circle with circumference 1, you start at point 0, go clockwise for[tex]\sqrt{2}[/tex] unit, you will pass the point 0 and reach 0.4142..., and go clockwise for another [tex]\sqrt{2}[/tex], you will get to 0.8284...

Now, if [tex]\sqrt{2}[/tex]Z (mod 1) is not dense in [0,1], then there are a<b, where a=m[tex]\sqrt{2}[/tex] (mod 1) and b=n[tex]\sqrt{2}[/tex] (mod 1) such that nothing between a and b is a multiple of [tex]\sqrt{2}[/tex] (mod 1). Note that [tex]\mu[/tex]=b-a>0 is irrational.

The set S={c|a<c<b} is a "forbidden region" (meaning that S is free of multiple of [tex]\sqrt{2}[/tex] (mod 1). Then, for any integer k, k[tex]\sqrt{2}[/tex]+S (mod 1) is also forbidden. Pick any integer M>1/[tex]\mu[/tex], the total length of the M+1 sets S, [tex]\sqrt{2}[/tex]+S,..., M[tex]\sqrt{2}[/tex]+S exceed 1, so, at least two of them, say [tex]S_{p}[/tex]=p[tex]\sqrt{2}[/tex]+S and [tex]S_{q}[/tex]=q[tex]\sqrt{2}[/tex]+S must intersect. If [tex]S_{p}[/tex]!=[tex]S_{q}[/tex], then a boundary point is inside a forbidden region, wlog, say upper boundary of [tex]S_{p}[/tex] is inside [tex]S_{q}[/tex], then, since that boundary point is just b+p[tex]\sqrt{2}[/tex] (mod 1), which should not be in any forbidden region, we are left with [tex]S_{p}[/tex]=[tex]S_{q}[/tex]. But this is impossible, since [tex]\sqrt{2}[/tex] is irrational. So, we have a contradiction.
chingkui
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#57
May5-05, 04:48 PM
P: 195
Where is Steven's proof???
snoble
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#58
May6-05, 02:15 AM
P: 127
I have posted the proof to the paired set problem in a seperate thread to avoid clutter. Please make any comments about it there.
http://www.physicsforums.com/showthread.php?t=74511

thanks,
Steven
mansi
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#59
May14-05, 12:56 PM
P: 58
here's a proof of the question i had posted....thanks to matt grime(he sent me the proof) and i guess chingkui's done the same thing...so he gets to ask the next question...

Let b be the square root of two, and suppose that the numbers
If nb mod(1) are dense in the interval [0,1), then m+nb is dense in R. Proof, let r be in R since there is an n such that nb and r have as many places after the decimal point in common as you care, we just subtract or add an integer onto nb so that m+nb have the same bit before the decimal point too. Thus m+nb and r agree to as much precision as you care.

So it suffices to consider nb mod 1, the bits just after the decimal point.

now, there is a nice map from [0,1) to the unit circle in the complex plane, which we call S^1


t --> exp(2pisqrt(-1)t)


the map induced on the circle by t -->t+b is a rotation by angle 2(pi)b radians.

it is a well known result in dynamical systems that such rigid rotations have dense orbits if and only if b is irrational, and then all orbits are dense.

the orbit of t is just the images of t got by applying the rotation repeatedly. Thus the orbit of 0 is just the set of all points nb mod(1), whcih is dense.

The proof of density isn't too hard, though you need to know about compactness and sequential compactness, at least in the proof I use.


Let r be a rotation by angle 2pib for some irrational b, then the images of t, namely

t+b,t+2b,t+3b... must all be distinct, otherwise

t+mb=t+nb mod 1 for some m=/=n

that is there is an integer k such that

k+mb=nb, implying b is rational.

thus the set of images of t must all be distinct. S^1 is compact, thus there is a convergent subsequence.

Hence given e>0, there are points t+mb and t+nb such that

|nb-mb|<e mod 1.

Let N=n-m, and let p and q be the points

mb and nb mod 1.


Consider the interval between p and q, and its image under rotating by 2pibN. These intervals are no more than e long, and the cover the circle/interval, hence the forward orbit under rotating by 2pibN is dense, thus so is rotating by 2pib, and hence all forward orbits are dense as required.
chingkui
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#60
May23-05, 02:36 PM
P: 195
Sorry for not posting so long. I have a counting question that is not very difficult (at least after I saw the solution), some of you might know the answer already. If you have already seen it somewhere, please wait 1 or 2 days before posting, let people think about it first. Thanks.

Here is the question:

I was thinking about this problem with a number of friends when we were cutting a birthday cake: How many pieces can be produced at most when we are allowed to cut n times (no need to have equal area/volume).

More precisely, in 2 dimensions, we want to know with n straight lines (extend to infinite at both ends), how many pieces at most can we cut the plane (R^2) into? In this case, the answer is 1+1+2+3+...+n=1+n(n+1)/2.

In 3D, it looks a lot more complicated. Again we have n infinite planes, and we wish to cut R^3 into as many pieces as possible. Does anyone know the answer?

How about in m dimensions? n hyperplanes to cut R^m.
chingkui
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#61
Jun23-05, 02:32 PM
P: 195
Dear all,

It has been exactly one month since I post the question, is the question too difficult or just not interesting at all?
If anyone is still interested, here is one hint: the formula in R^3 is a recursive formula that actually depends on the formula in R^2. That is as much hint as I can give, and I am almost writing down the solution.
If this still doesn't generate any interest and response in say 2 weeks, I will probably post another question if everyone agree.
mustafa
mustafa is offline
#62
Jul2-05, 03:27 PM
P: 19
I think this recursion might be correct, not sure though:
1+1+2+4+7+11+...+(1+n(n-1)/2) = (n+1)(n^2-n+6)/6
chingkui
chingkui is offline
#63
Aug2-05, 11:20 AM
P: 195
Sorry for replying so late. mustafa is right, and he can ask the next question.
The recursive relation I mentioned is P3(n)=P3(n-1)+P2(n-1)
where P2(n) is the number of pieces with n cut in R^2
and P3(n) is the number of pieces with n cut in R^3
Solving the relations, we can get mustafa's formula.
DeadWolfe
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#64
Oct17-05, 06:58 PM
P: 461
Well mustafa?
siddharth
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#65
Dec12-05, 04:38 AM
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Since mustafa doesn't seem to be online anymore, here is a question to revive this thread

SECOND EDIT:

Let [tex] f(\frac{xy}{2}) = \frac{f(x)f(y)}{2} [/tex] for all real [itex] x [/itex] and [itex] y [/itex]. If [tex] f(1)=f'(1) [/itex], Prove that [tex] f(x)=x [/itex] or [tex] f(x)=0 [/tex] for all non zero real [itex] x [/itex].
matt grime
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#66
Dec12-05, 05:02 AM
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Is that the correct question? f(x)=x seems to do the trick a little too obviously.
siddharth
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#67
Dec12-05, 05:07 AM
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Quote Quote by matt grime
Is that the correct question? f(x)=x seems to do the trick a little too obviously.
Oops, I should have seen that coming.
I was expecting someone to prove that the function has to be of this form from the given conditions, not guess the answer.
matt grime
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#68
Dec12-05, 05:10 AM
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Then add the rejoinder that they must prove that this is the only possible answer (if indeed it is; since i didn't prove it but merely guessed by inspection i can't claim that 'prize'; of course it is explicit that f is differentiable, hence continuous)

EDIT: obviosuly it isn't the only solution: f(x)=0 for all x will do.
D H
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#69
Dec12-05, 06:39 AM
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It doesn't work for arbitrary [itex]c[/itex].

If [itex]f(x)=cx[/itex] then [itex]f(\frac{xy}{2}) = c \frac{xy}{2}[/itex] and [itex]\frac{f(x)f(y)}{2} = c^2 \frac{ xy}2[/itex]. Equating these two yields [itex]c=c^2[/itex] so the only solutions for [itex]c[/itex] are the two Matt has already found.
siddharth
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#70
Dec12-05, 08:12 AM
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Yes, I see that.
Can you prove that the only possible solutions are
[tex] f(x)=x [/tex]
and
[tex] f(x)=0[/tex]
from the given conditions?
Sorry for the poorly worded question.
matt grime
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#71
Dec12-05, 01:49 PM
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Quote Quote by D H
It doesn't work for arbitrary [itex]c[/itex].

If [itex]f(x)=cx[/itex] then [itex]f(\frac{xy}{2}) = c \frac{xy}{2}[/itex] and [itex]\frac{f(x)f(y)}{2} = c^2 \frac{ xy}2[/itex]. Equating these two yields [itex]c=c^2[/itex] so the only solutions for [itex]c[/itex] are the two Matt has already found.
What doesn't work for arbitrary c?
AKG
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#72
Dec12-05, 07:36 PM
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From f(2x/2) = f(x), we get that either:

a) f = 0, OR
b) f(2) = 2

From case b), using that f(0) = f(0)f(x)/2 for all x, we get either:

b1) f(0) = 0, OR
b2) f(x) = 2 for all x

b2) is impossible given the condition that f'(1) = f(1), so we have two cases overall:

a) f = 0
b) f(0) = 0 and f(2) = 2

In general, it holds that f(x) = +/- f(-x) since f(xx/2) = f((-x)(-x)/2). In fact, by looking at f((-x)y/2) = f(x(-y)/2), we can make an even stronger claim that either for all x, f(x) = f(-x) or for all x, f(x) = -f(-x).

Note that this gives a solution f(x) = |x| which satisfies the criteria (it is not required that f be differentiable everywhere, only at 1 is necessary) but is neither 0 nor identity.

Suppose f(x) = 0 for some non-zero x. Then for all y, f(y) = f(x(2y/x)/2) = 0, so f(0) = 0, and either f(x) is non-zero for all other x, or f(x) is zero for all other x:

a) f = 0
b) f(0) = 0, f(2) = 2, f is either odd or even, and f(x) is non-zero for x non-zero.


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