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United States10328 Posts
On August 17 2011 23:15 EsX_Raptor wrote:
However, according to de Moivre's formula, we four out e^(i * x) = cos(x) + i * sin(x) for all x. This illustrates how closely related the exponential function is to the trigonometric functions.
rawrgh that's not De Moivre 
De Moivre can be proven elementarily, and states that (cos n*x + i sin n*x) = (cos x + i sin x)^n for positive integer n. The fact that cos x + i sin x = e^(i*x) is deeper and requires that you do the Taylor series thing.
Since I haven't had enough quantum yet to think about entanglement, here are a few cute combinatorics problems:
1. For n a positive integer, let A_1, A_2, ... A_{n+1} be distinct subsets of {1, 2, ..., n}, each containing exactly 3 elements. Show that some two of these subsets have exactly one common element.
2. Show that any n points, not all collinear, determine at least n distinct lines.
+ Show Spoiler [hint] +
edit:
http://www.teamliquid.net/forum/viewmessage.php?topic_id=246103¤tpage=4#61 on entanglement 
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Well, Nawyria does answer the question in that thread you linked lol.
The key is that if you already knew the state, it is impossible to determine whether you did the first measurement or they did. You don't actually gain any information, classically speaking.
edit - How do you do the n points problem by linear algebra? I can think of a proof by induction, which tbh was the immediately obvious method for me. Not a clue atm how to proceed with linear algebra though, maybe cause I haven't used it in a while lol.
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Foolishness
United States3044 Posts
On August 18 2011 11:48 EtherealDeath wrote:
Well, Nawyria does answer the question in that thread you linked lol.
The key is that if you already knew the state, it is impossible to determine whether you did the first measurement or they did. You don't actually gain any information, classically speaking.
edit - How do you do the n points problem by linear algebra? I can think of a proof by induction, which tbh was the immediately obvious method for me. Not a clue atm how to proceed with linear algebra though, maybe cause I haven't used it in a while lol.
Likewise I think I can do the proof using graph theory and k-partite sets (proof by induction seems much simpler though ), but I can't think of a linear algebra and now I'm probably going to lose sleep over it.
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On August 18 2011 10:36 ]343[ wrote:Show nested quote +On August 17 2011 23:15 EsX_Raptor wrote:
However, according to de Moivre's formula, we four out e^(i * x) = cos(x) + i * sin(x) for all x. This illustrates how closely related the exponential function is to the trigonometric functions.
rawrgh that's not De Moivre De Moivre can be proven elementarily, and states that (cos n*x + i sin n*x) = (cos x + i sin x)^n for positive integer n. The fact that cos x + i sin x = e^(i*x) is deeper and requires that you do the Taylor series thing. Since I haven't had enough quantum yet to think about entanglement, here are a few cute combinatorics problems: 1. For n a positive integer, let A_1, A_2, ... A_{n+1} be distinct subsets of {1, 2, ..., n}, each containing exactly 3 elements. Show that some two of these subsets have exactly one common element. 2. Show that any n points, not all collinear, determine at least n distinct lines. + Show Spoiler [hint] +edit: http://www.teamliquid.net/forum/viewmessage.php?topic_id=246103¤tpage=4#61 on entanglement
Just to comment on Euler's famous equality, one doesn't need to do the Taylor series, just basic derivation works (and the proof is a little sexier). I could be wrong though. Anyway, I've always thought of it like this.
f(x) = e^(i*x)
Break up the function into its real and imaginary components.
f(x) = g(x) + i*h(x)
Differentiate
f ' (x) = i*e^(ix) = i*f(x)
f ' (x) = g ' (x) + i*h ' (x)
Equate things
f ' (x) = i*f(x) = i*g(x) - h(x) = g ' (x) + i*h ' (x)
Equate the real and imaginary parts of f ' (x)
g ' (x) = - h(x)
h ' (x) = g(x)
We end up with two equations about the components functions. h(x)'s derivative is g(x) and g(x)'s derivative is the negative of h(x). So h(x) is the sine function and g(x) is the cosine function. So e^(ix) = cos(x) + isin(x). If x is pi, e^(ix) is negative one (cos(pi) is -1 and sin(pi) is 0).
Hopefully this is settled without nasty sigmas!
As for your first problem, it's kind of late, so I could be completely wrong, but I'm not getting it. If you want to have multiple subsets, n has to be greater than 3. So let's say n = 4. The subsets are {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, and {2, 3, 4}. There are no two sets here that share only one common element. If n were 5 however, it works because there are 5 unique elements that two sets of 3 can share with one (and only one) in both. So this seems to work for n > 4, or well, any n such that there are 5 unique elements in them? This seems a little simple and I'll probably have to edit this out later in shame.
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On August 17 2011 23:26 Iranon wrote:Show nested quote +On August 17 2011 23:15 EsX_Raptor wrote:
To understand the equation
e^(pi * i) = -1,
it is necessary to understand what raising a number to an imaginary power means.
However, according to de Moivre's formula, we four out e^(i * x) = cos(x) + i * sin(x) for all x. This illustrates how closely related the exponential function is to the trigonometric functions.
It thus follows that
e^(pi * i) = cos(pi) + i * sin(pi) = -1 + i * 0 = -1.
Do write out the Euler theorem! For a neat bit of insight, look at the Taylor series for e^x. Replace x with iz in the e^z series. Now look at the Taylor series for cos x and sin x...
I was looking for the Taylor series actually. Writing it out for e^x, filling Pi*i, re-ordening the terms and getting cos(x) + i * sin(x). I was amazed the first time I did that.
EDIT: http://www.math.toronto.edu/mathnet/questionCorner/epii.html
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On August 18 2011 11:48 EtherealDeath wrote:
Well, Nawyria does answer the question in that thread you linked lol.
The key is that if you already knew the state, it is impossible to determine whether you did the first measurement or they did. You don't actually gain any information, classically speaking.
The problem is with the way you stated your scenario. It's quite correct to say that entanglement cannot be used to instantaneously transmit classical information, but that wasn't what you described, and you didn't even mention spatial separation or a second experimenter in your original post.
On August 18 2011 02:14 EtherealDeath wrote:
Hmm...
Explain why the following scenario is true.
Suppose we have two entangled particles, let's say their state is a|1>|0>+b|0>|1> to be simple, where the values of a and b are not particularly important save that their squares sum to 1, but let's use one of the Bell States, that is a=b=1/sqrt(2). Now we measure one of the particles to determine it's actual state. We know that instantaneously, the state of the other particle is set. However, it is impossible to determine any classical information from this instantaneous effect - i.e., cannot gain any physical information. Why? Or rather, what information do you think you could gain that you didn't have before?
It is fair to say that no net information is gained in this operation, but only because the particles are in a known, pure state both before and after the measurement.
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+ Show Spoiler +Show that any n points, not all collinear, determine at least n distinct lines.
People have said that it was possible by induction but have not done so.
I interpret the question to mean that a distinct line must only cross 2 points.
This is logical becuase the n points are not collinear.
At least n lines are possible becuase there is a line from one of the points to all other points.
In the case that there are collinear points that lie on one of these lines preventing you from finding n distinct lines immediately, one must simply select these "wasted/overlapped" points as the new "starting point" and revisit the other points to make new lines.
I couldn't figure out how to prove a general case with linear algebra aha :S
All the reading about twin paradox etc. was fun.
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EPT
