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beero1000
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You need to think more carefully about analyticity at singularities...
Kharakov
Quantum Hot Dog
hmmm weird double post... I'll leave the "updated" thing...
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As n approaches zero, the function gets closer and closer to log(x/y). In fact at n= 10^-40, it gives 40 digits of log(2). At n=10^-30, 30 digits... At n=0, it gives them all. There isn't a singularity for the function.
Although there is for the analytic continuation of it....
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As n approaches zero, the function gets closer and closer to log(x/y). In fact at n= 10^-40, it gives 40 digits of log(2). At n=10^-30, 30 digits... At n=0, it gives them all. There isn't a singularity for the function.
Although there is for the analytic continuation of it....
beero1000
Veteran Member
I don't think that's true. Specifically, an analytic function has a convergent Taylor series in a neighborhood of every point of its domain. You'll see theorems like "f(z) extends analytically onto the whole plane, except for a pole at ...."
Kharakov
Quantum Hot Dog
The series is analytic, and the function f(n,x,y)= [1- (y/x)^n] /n maps to it everywhere except at n=0. What do I call f(n,x,y) in relation to the series g(n,x,y)?
As n--> 0, f(n,x,y) --> log(x/y). For the series, there is no singularity at n=0, it reduces to the Taylor series for log(x/y).
As n--> 0, f(n,x,y) --> log(x/y). For the series, there is no singularity at n=0, it reduces to the Taylor series for log(x/y).
beero1000
Veteran Member
It's a
removable singularity (indeterminate form), so it can be fixed by defining f(0,x,y) to be the limit of f(n,x,y) as n goes to 0, which makes the extended function analytic. However, if you do that, I think that all you end up saying is that the limit of log(x/y) as y goes to 0 is the same as the limit of 1/n as n goes to 0. I have no problem with that. 
Jayjay
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I just saw an episode of Brooklyn Nine-Nine that had the following "brain-teaser":
There are 12 people on an Island. 11 of them weigh the same, and one weighs either more or less. There is a see-saw on the Island, but you can only use it three times. How do you figure out the person who weighs a different amount?
Now, this is fairly easy to solve either by trial and error... but it got me thinking a generalized form where there are N people on the island.
A) What is the smallest number of see-saw weightings needed (assuming that the see-saw is big enough) to solve the puzzle?
B) What is the algorithm?
I cannot prove it, but my first intuitive guess would be that since there are three results for each weighting (either one side weighs more than the other, or they are equal), and since we don't know whether the different weight is more or less and thereby theoretically need one additional bit of information, the number of weightings needed is K where N < 3^K / 2.
As for the algorithm... your thoughts?
There are 12 people on an Island. 11 of them weigh the same, and one weighs either more or less. There is a see-saw on the Island, but you can only use it three times. How do you figure out the person who weighs a different amount?
Now, this is fairly easy to solve either by trial and error... but it got me thinking a generalized form where there are N people on the island.
A) What is the smallest number of see-saw weightings needed (assuming that the see-saw is big enough) to solve the puzzle?
B) What is the algorithm?
I cannot prove it, but my first intuitive guess would be that since there are three results for each weighting (either one side weighs more than the other, or they are equal), and since we don't know whether the different weight is more or less and thereby theoretically need one additional bit of information, the number of weightings needed is K where N < 3^K / 2.
As for the algorithm... your thoughts?
fromderinside
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6-6
3-3
1-1 determined.
Actually in the first two tests one group will weigh less/more than the other. The final test is either the one that weighs more/less, or, if equal the one left out.
You can write the equation. I'm sure of it.
beero1000
Veteran Member
The exact bound is N = (3K - 3)/2. Your information theoretic bound gives N ≤ (3K - 1)/2, but a more sophisticated analysis disallows equality in that bound (i.e. since you know exactly one coin is counterfeit, at least one weighing should be uneven, etc, which lowers the number of distinguished possibilities). The maximal number of coins pattern is then 3, 12, 39, 120, ...
I've seen an algorithm to achieve the optimal number of weighings attributed to Freeman Dyson using trinary labels, but I'd have to look it up or think for a while to re-derive it.
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I don't think that works. You don't know whether the counterfeit coin is heavy or light.
Bomb#20
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Everybody's wrong. Okay, as far as I can tell, Jayjay's answer is right; but his information theoretic argument for it is wrong. You don't need to distinguish 2N possibilities because you don't have to find out whether the odd man out is heavy or light. So in principle N doesn't need to be less than 3^K/2. (Of course you usually find out anyway, so there's a limit to how much benefit you get from solving the problem without finding out. I haven't found any solutions where N > 3^K/2.)
fromderinside's algorithm won't work -- if you start with 6-6 then it might take you three more weighings.
Beero's upper limit is too low. I found solutions for 4, 13 and 40. (I haven't worked on K=5.) The solution for N=4, K=2 is to weigh 1 against 1, and then weigh one of the ones you weighed in the first round against one of the ones you didn't. This will always identify the person who weighs a different amount but won't always tell you if he's heavy or light.
So the follow-up challenge is to solve the problem when there are 13 people...
beero1000
Veteran Member
Ah, I thought we were looking at the classic problem where you do have to determine whether the counterfeit is heavier or lighter.
If you don't then you do get the N = (3K - 1)/2 bound using trinary labels.
I also found the reference - The Problem of the Pennies. It may be behind a paywall, you can PM me if you can't access it.
If you don't then you do get the N = (3K - 1)/2 bound using trinary labels.
I also found the reference - The Problem of the Pennies. It may be behind a paywall, you can PM me if you can't access it.
fromderinside
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Got it. Need to take averages as well a s direction so that going into the final test compare averages against individual results. If two are equal it is obvious. If the two are unequal the averages of the previous tests will determine the direction of the difference signalling the one nearer to the averages is from the uniform set while the one most different from the averages is the culprit. OK. So a bit more calc, but, still three trials.
beero1000
Veteran Member
Classes start Monday, but until then...
What the balls indeed.
What the balls indeed.
beero1000
Veteran Member
The quality of high school level math texts in the US. Someone at McGraw-Hill made an oopsie...
Kharakov
Quantum Hot Dog
What is the following, as \(a\to\infty\)?
\( \frac {1}{ \frac{\pi^2}{2^{2a+.5}}+\frac{\pi}{2^a} } - \frac{2^a}{\pi}\)
How would you extract \(\pi\) from this assuming the limit \(2=\sqrt{2+\sqrt{2+\sqrt{2+...}}}\)?
\( \frac {1}{ \frac{\pi^2}{2^{2a+.5}}+\frac{\pi}{2^a} } - \frac{2^a}{\pi}\)
How would you extract \(\pi\) from this assuming the limit \(2=\sqrt{2+\sqrt{2+\sqrt{2+...}}}\)?
George S
Veteran Member
Remember \( \frac{sqrt{\pi}}{2}=(1/2)! \), but only if it is relevant.
beero1000
Veteran Member
Change variables \(x = \frac{2^a}{\pi}\) and fiddle...
\(\displaystyle \lim_{a \to \infty} \left(\frac {1}{ \frac{\pi^2}{2^{2a+.5}}+\frac{\pi}{2^a} } - \frac{2^a}{\pi}\right) = \lim_{x \to \infty} \left(\frac {1}{\frac{1}{\sqrt{2}x^2}+\frac{1}{x}} - x\right) = \lim_{x \to \infty} \left(\frac{\sqrt{2}x^2}{1 + \sqrt{2}x} - x\right) = \lim_{x \to \infty} \left(\frac{\sqrt{2}x^2 - x(1+\sqrt{2}x)}{1 + \sqrt{2}x}\right) = \lim_{x \to \infty} \left(\frac{-x}{1 + \sqrt{2}x}\right) = - \frac{1}{\sqrt{2}}\)
Kharakov
Quantum Hot Dog
There is a class of transcendental numbers that I'm revisiting. What I want to do is a bit twisted, but fun. I'll try and explain it, keep in mind that I don't know what I'm talking about, so don't expect strict terminology.
\( f(x,n,a) = \sqrt[n]{x^n-x + \sqrt[n]{x^n-x +\sqrt[n]{x^n-x + \cdots}}}\)
As the number of nestings (nested radicals???), a, approaches infinity, f(x,n,a) approaches x.
\( dx = n x ^{n-1} \approx \frac {x - f(x,n,a)} {x-f(x,n,a+1)} \) or \( \simeq\) ??
\( q= {dx}^{a+1} \, x - {dx}^{a+1} \, f(x,n,a)\)
Should the limit for q be 0 as a --> infinity? What about for x=2, n=2?
Challenge: create a generating function for x>1 and n>1, that creates integer q.
\( f(x,n,a) = \sqrt[n]{x^n-x + \sqrt[n]{x^n-x +\sqrt[n]{x^n-x + \cdots}}}\)
As the number of nestings (nested radicals???), a, approaches infinity, f(x,n,a) approaches x.
\( dx = n x ^{n-1} \approx \frac {x - f(x,n,a)} {x-f(x,n,a+1)} \) or \( \simeq\) ??
\( q= {dx}^{a+1} \, x - {dx}^{a+1} \, f(x,n,a)\)
Should the limit for q be 0 as a --> infinity? What about for x=2, n=2?
Challenge: create a generating function for x>1 and n>1, that creates integer q.