Posts for OmnipotentEntity

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For the record. Nitrocellulose is definitely diamagnetic. I have an estimate of -212.66 ≤ Xd (μemu/mol) ≤ -57.5 based on Pascal Constants. Can you give a reference to that paper Bobo?
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I haven't been able to reproduce the game on my flash cart. I get mostly e5 for the first turn, but for the second turn I get a lot of Nc6, some Qf6, very few d5, and one f6. This is after a couple of hundred attempts over an hour or so. So this path, if possible on a cart, is very unlikely. f5 is a rather bad move, so I guess it's not surprising that the engine almost never chooses it.
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No. It's super easy. To find the norm of any n-dimensional vector just collapse n-dimensional space down to two dimensions. One of which is in the direction of one of the axes, and the other dimension in the direction of the vector perpendicular to the chosen axis. Let's take a specific example: (4, 5, 1, 2), and a specific axis (the third). We now know that the length of this vector is sqrt(1^2 + x^2) where x^2 is the distance along the vector pointing in the (4, 5, 2) direction. This vector is in fact (4, 5, 2) itself. Now we can just simply repeat, let's select the 2nd direction so we have (4, 5, 2) being split into 5 and (4, 2) with length sqrt(5^2 + y^2), and again (4, 2) can be split into 4 and 2 with length sqrt(4^2 + 2^2). Now we know y = sqrt(4^2 + 2^2) so x is sqrt(5^2 + sqrt(4^2 + 2^2)^2) or just sqrt(5^2 + 4^2 + 2^2). And finally the total lengthis sqrt(1^2 + sqrt(5^2 + 4^2 + 2^2)^2) or sqrt(1^2 + 5^2 + 4^2 + 2^2). You can generalize this easily using more formal language. But it's easy to "see" that this problem can be broken down into steps like this, and because the choice of axes is arbitrary we can simplify any vector into a 2D space of our choosing.
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BrunoVisnadi wrote:
It's interesting to think about it, as I'm pretty sure if you remove any sequence of numbers, it will also converge. I wonder what the limit will be if, for instance, we remove only the terms that have a ''1234567890''.
Because this is an interesting problem, you might be dismayed to find that someone has already found a fast method for calculating these values to high precision. For the string "1234567890" for instance, the answer is approximately: 23025850922.0270354982950673268734899111435414556239773730714362058360\ 911501346900226904036846956235736570665999 If you're interested in the technical details, here is the paper: http://eprints.maths.ox.ac.uk/1106/1/NA-06-17.pdf
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Then you can replace zed with any of the following: b, c, g, p, t, v d can't be used because it's the differential, e can't be used because we're using it as a constant. t might work better because it's a more often used variable, and z generally denotes a complex number, which isn't strictly required in this context.
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If you took the time to talk to the person wearing the shirt they'd have probably told you that it's a limerick. The Integral z squared dz From one to the cube root of three times the cosine of three pi over nine is the log of the cube root of e.
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That's a good one Warp. Let's talk about matrix eigenvalue decomposition! So it is possible to represent both the Fibonnaci and Lucas recurrence relation as the same matrix with different initial conditions. This is to say, if you have a state vector (Fn-1, Fn) you can multiply it by a matrix to get the next state vector (Fn, Fn + 1) Our initial vector for Fibonacci numbers is (0, 1). For Lucas numbers it's (2, 1). Note: It seems that the limit you're computing is Ln / Fn+1 according to the canonical numbering. So the formula for the vector containing the (n-1)th and nth Fibonacci number is (0, 1) * ( ( 0, 1 ), ( 1, 1) )n But computing the nth power of a matrix is hard. So we need to find a better way. To begin, the matrix is simply:
A = ( ( 0, 1 ),
      ( 1, 1) )
Next, we calculate the eigenvalues of this matrix: ( ( 0-y, 1 ), ( 1, 1-y) ) (1-y)(-y) - 1 = 0 y2 - y - 1 = 0 y1 = (1 + sqrt(5))/2 y2 = (1 - sqrt(5))/2 (Note: this is the underlying mathematical reason that both the Fibonacci and Lucas sequence ratios approach the golden ratio as n goes to infinity.) So now that we have the eigenvalues we want to change the representation of the matrix from A into Q Y Q-1, where Y is just the matrix with the eigenvalues down the diagonal. We want to do this because then we can raise this to the nth power easily. (An is hard to calculate because it means lots of matrix multiplication. On the other hand, (Q Y Q-1)n is easy because the Q-1 and Q cancel and we're left with Q Yn Q-1. Because Y is a diagonal matrix, it is simple to calculate the nth power of, you just compute the nth power of each diagonal member.) It can be shown with a proof that Q must be the eigenvectors of A, but I'll omit that for brevity. Maybe this can be my challenge to you. So to calculate the eigenvectors of A we just solve the system (A - yI) v = 0 for v. I'll spare the calculations because they're pretty much trivial we get: v1 = (-y2, 1) v2 = (-y1, 1) So Q becomes: ( ( -y2, -y1 ), ( 1, 1) ) And Q-1 is computed just by taking the inverse of Q: ( ( 1/sqrt(5), (1 + sqrt(5))/(2 sqrt(5)) ), ( -(1/sqrt(5)), (-1 + sqrt(5))/(2 sqrt(5)) ) ) Now we can use this to find an explicit formula for the nth Fibonacci (and Lucas) number. It's just: (0, 1) Q Yn Q-1 And because we just care about a single value we can simplify it to a single formula: (-(1/2 (1 - sqrt(5)))^n + (1/2 (1 + sqrt(5)))^n)/sqrt(5) Which is the famous explicit formula for the Fibonacci sequence. Similarly you can use the initial vector for the Lucas sequence to find the explicit formula for that. It is: -2 + (1/2 (1 - sqrt(5)))^n + (1/2 (1 + sqrt(5)))^n Because the limit you're looking for is Ln / Fn+1 we have: (-2 + (1/2 (1 - sqrt(5)))^n + (1/2 (1 + sqrt(5)))^n) / ((-(1/2 (1 - sqrt(5)))^(n+1) + (1/2 (1 + sqrt(5)))^(n+1))/sqrt(5)) Which is messy looking but as n approaches infinity is: 1/2 ( 5 - sqrt(5) ) For Ln / Fn the formula is: (-2 + (1/2 (1 - sqrt(5)))^n + (1/2 (1 + sqrt(5)))^n) / ((-(1/2 (1 - sqrt(5)))^n + (1/2 (1 + sqrt(5)))^n)/sqrt(5)) And the limit is simply: sqrt(5) Which is a lot nicer. :)
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This was a problem from my Calc 2 course in 2002: If you take the harmonic series and remove only the terms that contain a 9 will the series converge or diverge? Prove it. If you want to cheat, this series is known as the Kempner series.
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Here's the method I used to solve the integral: f(a) = integral e^(-a^2 x^2)/(x^2+1) dx f'(a) = integral -2a x^2 e^(-a^2 x^2)/(x^2+1) dx = integral -2a (x^2 + 1 - 1) e^(-a^2 x^2)/(x^2+1) dx = integral -2a e^(-a^2 x^2) dx + integral 2a e^(-a^2 x^2)/(x^2+1) dx f'(a) = -2sqrt(pi) + 2a f(a) This is just a linear non-homogenous first-order ODE. With solution: f(a) = pi e^(a^2) (-erf(a) + C) Because f(0) = pi we get C = 1 and 1 - erf(a) = erfc(a) so f(a) = pi e^(a^2) erfc(a)
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There are at least two other ways to solve this integral other than contour integration. Though I didn't realize that the answer was... just sitting on wikipedia for me to lift it. That would have been nice to know.
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Here's a fun integral that I had to solve recently. Integral -inf to inf exp(-a^2 x^2)/(1 + x^2) dx Interested to see what approaches you take. (You may assume a is positive.)
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Warp wrote:
I wish somebody would integrate the math plugin into the forum... http://tasvideos.org/forum/viewtopic.php?t=19519
I'm pretty sure it won't be that easy. I remember back in the day Bisqwit telling me that he couldn't even update to phpbb 3.0 because there was so much backend integration between the site and the forum. All work that would have to be redone. (In fact, I believe we're actually still on phpBB 2.0). The mod I linked is for phpbb 3.0.x so it would need to be reworked for 2.0. It's not as simple as just installing it.
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FractalFusion wrote:
OmnipotentEntity wrote:
But if $$a$$ is not a real number then bringing $$\mathrm{e}^{a t}$$ into the $$\Re$$ isn't allowed. But we get the same answer. So if we try to compute the integral in this way it's an abuse of notation.
Instead of using Re(eiωx) for cos(ωx), use (1/2)(eiωx+e-iωx) instead. (This follows from Euler's formula.) Then it will work out to be the same answer. This way, we do not need to worry about abuse of notation; it works when a is complex. In fact, it now works even if ω is complex. (Euler's formula holds for complex values as well.) By the way, the image you uploaded isn't visible in imgur without going into expand mode. It may be because the background color is regarded as transparent, even though it should be white. Edit: The image is directly linked now, which solves the problem.
I did mention that it works with the cosine formula. I was hoping to use the abuse of notation, provided I knew exactly where it works and where it breaks down.
OmnipotentEntity wrote:
(Yes, it is possible to also just use the identity $$\cos(\omega x) = \frac1{2} \left( \mathrm{e}^{i \omega x} + \mathrm{e}^{-i \omega x} \right)$$ and while that's easier than the integration by parts, it's more involved than the abuse.)
Sorry about the transparency issue, the images were always direct links, but I guess imgur wasn't behaving as I expected.
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Using latex notation. Sorry about how messy it is. This post rendered as a document: Page 1 Page 2 Let's say that we have an integral: $$\int \cos(\omega x) \mathrm{e}^{a x} dx $$ One method of working this integral is to use integration by parts twice to obtain a copy of the original integral, but that's long and kind of annoying: $$\begin{split} \int \cos(\omega x) \mathrm{e}^{a x} dx &= \frac1{a} \cos(\omega x) \mathrm{e}^{a x} - \int -\frac{\omega}{a} \sin(\omega x) \mathrm{e}^{a x} dx \\ &= \frac1{a} \cos(\omega x) \mathrm{e}^{a x} + \frac{\omega}{a} \int \sin(\omega x) \mathrm{e}^{a x} dx \\ &= \frac1{a} \cos(\omega x) \mathrm{e}^{a x} + \frac{\omega}{a} \left(\frac1{a} \sin(\omega x) \mathrm{e}^{a x} - \int \frac{\omega}{a} \cos(\omega x) \mathrm{e}^{a x} dx\right) \\ &= \frac1{a} \cos(\omega x) \mathrm{e}^{a x} + \frac{\omega}{a^2} \sin(\omega x) \mathrm{e}^{a x} - \frac{\omega^2}{a^2} \int\ \cos(\omega x) \mathrm{e}^{a x} dx \\ \left(a^2 + \omega^2\right)\int \cos(\omega x) \mathrm{e}^{a x} dx &= a \cos(\omega x) \mathrm{e}^{a x} + \omega \sin(\omega x) \mathrm{e}^{a x} \\ \int \cos(\omega x) \mathrm{e}^{a x} dx &= \frac{a}{a^2 + \omega^2} \cos(\omega x) \mathrm{e}^{a x} + \frac{\omega}{a^2 + \omega^2} \sin(\omega x) \mathrm{e}^{a x} \\ \end{split}$$ Instead, we can use the complex domain and compute (assuming $$x$$ is real): $$\int \Re\left\{\mathrm{e}^{i \omega x}\right\} \mathrm{e}^{a x} dx $$ And assuming that $$a$$ is also a real number we can bring it into the Real part and compute the integral simply: $$\begin{split} &= \int \Re\left\{\mathrm{e}^{(a + i \omega) x}\right\} dx \\ &= \Re\left\{\frac1{a + i \omega}\mathrm{e}^{(a + i \omega) x}\right\} \\ &= \Re\left\{\frac{a - i \omega}{a^2 + \omega^2} \mathrm{e}^{ i \omega x} \mathrm{e}^{a x}\right\} \\ &= \frac{a}{a^2 + \omega^2} \cos(\omega x) \mathrm{e}^{ax} + \frac{\omega}{a^2 + \omega^2} \sin(\omega x) \mathrm{e}^{a x} \\ \end{split}$$ But if $$a$$ is not a real number then bringing $$\mathrm{e}^{a t}$$ into the $$\Re$$ isn't allowed. But we get the same answer. So if we try to compute the integral in this way it's an abuse of notation. It does work in this instance, and other simple cases like $$x \cos(\omega x) \mathrm{e}^{a x}$$. Because this sort of integrand appears in the Laplace Transform reasonably commonly, it might be useful to know when this sort of abuse will work, and when it breaks down. (Yes, it is possible to also just use the identity $$\cos(\omega x) = \frac1{2} \left( \mathrm{e}^{i \omega x} + \mathrm{e}^{-i \omega x} \right)$$ and while that's easier than the integration by parts, it's more involved than the abuse.) I asked around and most answers I got had to do with complex analysis and the idea of analytic continuation. If we have: $$f(x) cos(\omega x) \mathrm{e}^{a x}$$ Then if $$f(x)$$ is an real-analytic function on the domain of integration then this should be a valid abuse of notation. Is this correct? Or does it need to be complex analytic? Or am I completely off track?
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FractalFusion wrote:
Warp wrote:
OmnipotentEntity wrote:
And the answer is yes. Because we know that lim(x->0) -x / x3 = -infinity. So we just need to set g(x) to -x and we set 1/ln(f(x)) = x3 which gives f(x) = exp(x-3)
That seems to indeed work.
The function f(x) given above does not have a (two-sided) limit as x->0. The one I came up with was f(x)=exp(-x-4), g(x)=x2, and a=0. In this case, f(x), g(x) and f(x)g(x) all have the (two-sided) limit 0 as x->0.
I had mentioned this, but because it was below the picture I guess it was overlooked. I had already known it wasn't a two-sided limit. But I had lobbed the question back to Warp to try to find the solution you found.
OmnipotentEntity wrote:
However, f(x)'s limit at 0 is not two sided. But I bet you can find an example where all limits are two sided armed with this method. (There's a simple example very close in form to the example I gave.)
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Warp wrote:
Is it possible to have two functions f(x) and g(x) such that lim(x->a) f(x) = 0, lim(x->a) g(x) = 0 lim(x->a) f(x)g(x) = 0 and f(x) is not just the constant 0 (nor reduces to the constant zero, ie. eg. "f(x)=x-x" would be too boring of an answer)?
We can transform lim(x->a) f(x)g(x) into lim(x->a) exp(g(x) / (1 / ln(f(x)))) which is a 0/0 indeterminate form. So we can ask is it possible to find a pair of functions such that lim(x->a) exp(g(x) / (1 / ln(f(x)))) = -infinity ? And the answer is yes. Because we know that lim(x->0) -x / x3 = -infinity. So we just need to set g(x) to -x and we set 1/ln(f(x)) = x3 which gives f(x) = exp(x-3) f(x)^g(x) However, f(x)'s limit at 0 is not two sided. But I bet you can find an example where all limits are two sided armed with this method. (There's a simple example very close in form to the example I gave.)
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Updating the Linux and/or MacOSX support is something I would like very much. It would make my life much easier.
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Huh, that's too bad. I really enjoyed this movie.
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Playing around with it, for x negative, it seems like any C will do, if you allow complex values in intermediate steps. In fact, if you allow complex values, any C will do for any value of x and you'll get back your (1 + sqrt(4x + 1))/2. Except of course for the special case where x = 0 and C = 0.
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FractalFusion wrote:
What does sqrt(x+sqrt(x+sqrt(x+sqrt(...)))) mean exactly? If you mean a0=C, ai=sqrt(x+ai-1), limit as i goes to infinity, then there are values of C for which ai converges to (1 + sqrt(4x + 1)) / 2 for any x>=-1/4.
That's fair. And I suppose in the case of x > 0 then the particular value of C = 0 will converge. I wonder what is the convergence criteria for any x. If x > 0 then even if C is very large and negative, the repeated square rooting will drag the arg back toward 0. I suppose this means that for x > 0 for the expression to converge places no requirements on C. If x = 0 then even if C is very large and negative, you'll get a pull to one. However, if C = 0 then you won't converge. So for x = 0, C can be any value except 0. For x negative, it's non-trivial and I need to finish up some work. So I'll either revisit it or someone else will.
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Masterjun wrote:
FractalFusion wrote:
Note that the strategies above are optimal but it may be possible that they aren't unique, though my guess is that they are unique.
A quick question: Does optimal strategy in this context mean that there are no other strategies against it with an expected win rate of over 50%? Or am I misinterpreting this?
A good way to think about an optimal strategy is, if I'm playing an optimal strategy against you, then my strategy is setup such that every option you choose has an equal expected payoff. It doesn't mean that your win rate is any specific value. Just that you get the same, minimized win probability no matter which choice you pick.
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Warp wrote:
OmnipotentEntity wrote:
You get a similar oxymoron if you substitute x = 0: 1 = sqrt(0 + sqrt(0 + ...)).
Does this mean that you can only say sqrt(x+sqrt(x+sqrt(x+sqrt(...)))) = (1 + sqrt(4x + 1)) / 2 , x > 0 In other words, the equality does not hold for x <= 0? If yes, at which point of the proof does one have to assert that x>0?
I already went over this. y = sqrt(x+sqrt(x+sqrt(x+...))) y = sqrt(x + y) Going from step 1 to 2 you have to assert that the expression converges, and then I go to show (in a loose, nonrigorous way actually, I take that back, my argument makes no fucking sense at all...) that the expression cannot converge for x < 0. This isn't a rigorous argument, because I didn't show that the expression actually does converge anywhere, but assuming it does then its domain is limited to [0, inf), and although I didn't actually show that it doesn't converge for x=0, I'm thinking it doesn't, so its domain of convergence is actually (0, inf). If you want to make it rigorous, I'll leave that to you. EDIT: Here's a second attempt at making the x < 0 cannot converge argument stick: I'm going to approach this from a sign standpoint. If you collect a bunch of is in the sqrts, then they have to all cancel when forming the final value. Which means arg(z) = n pi. If x > 0 then sqrt(-x) = i sqrt(x). sqrt(-x + sqrt(-x)) = i sqrt(x + i sqrt(x)) arg(i sqrt(x)) = pi/2 arg(x + i sqrt(x)) = arctan(sqrt(x)/x) = arctan(1/sqrt(x)) arg(i sqrt(x +i sqrt(x))) = pi/2 + arctan(1/sqrt(x))/2 arg(x + i sqrt(x +i sqrt(x))) = arctan( (pi + arctan(1/sqrt(x))) /2x) From here we can create a general expression. arg(sqrt(-x + sqrt(-x ...)) = w = arctan( (pi + arctan(w)) /2x) Note that we are assuming that the value converges here. If it does not converge, then the original expression does not converge. So this is a safe assumption. tan(w) = (pi + arctan(w))/2x From this we can get x != 0. And because in order for y to be real, w must be n pi, and tan(n pi) = 0 we can replace: 0 = (pi + arctan(n pi))/2x pi + arctan(n pi) = 0 arctan has a range of (-pi/2, pi/2) so pi + arctan(anything) !=0. Therefore, this equation has no solutions that give y with Im{y} = 0. Therefore, over the domain (-inf, 0], sqrt(x + sqrt(x ....)) cannot have a purely real value.
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@Fractal Fusion. Thank for doing this write up. I appreciate it. I've never seen a non-trivial game theory problem attacked before.
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y = sqrt(x+sqrt(x+sqrt(x+...))) y = sqrt(x + y) y^2 = x + y y^2 - y - x = 0 y = (1 + sqrt(4x + 1)) / 2 Going from step 1 to step 2 assumes that the function converges. You have to be careful with this sort of infinite regression. They aren't always well behaved. In this case, because we know that sqrt(a + b) is less than sqrt(a) + sqrt(b) (for a, b > 0), we know that sqrt(y) + sqrt(x) > sqrt(y + x). But if sqrt(x) is out of domain then this equality simply doesn't hold. And because we know that sqrt(y + x) = y it must have a value if the series is convergent. So the series must not be convergent over this domain. (Ignore this shit. Read my next post.) You get a similar oxymoron if you substitute x = 0: 1 = sqrt(0 + sqrt(0 + ...)).
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