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Everything posted by Shadow

  1. If by z you mean "output", then yes.
  2. Any way that works and gives you some meaningful information, for example the absolute value ([math]|a+bi| = \sqrt{a^2 + b^2} \in \mathbb{R}[/math])
  3. After a bit of playing around with WA I discovered that the k*(k+1) should actually be k^2 (or at least the sum works out this way). So who knows, maybe it was a sign error somewhere, maybe something else. But it's irrelevant, I was only after the method of solving this limit. Many thanks for your help, it was much appreciated
  4. Okay, I think I could probably work that out on my own. But I'm still confused as to why I got 3/8*Sqrt[e] instead of 1/8*Sqrt[e].
  5. I'm no expert on this, frankly I'd say we're at about the same level, but I'm next to certain you get the landscape just by plotting the zeta function. The Z function is a complex function, taking as input and returning as output a complex number. I would imagine taking the absolute value (which always returns a positive real number regardless of the input) of the outputs would give you the landscape you're talking about, ie. the landscape is made out of all the points with coordinates (Re(z), Im(z), Abs[Zeta(z)]), where z is any complex number. If you go through all the complex numbers, you get the whole landscape. But again, this is only a conclusion drawn from common sense and what little I know about the function. I would prefer someone more familiar with the Zeta function to confirm my reasoning. Also, I'm not sure what level of mathematics you're familiar with, so I hope the above is understandable enough.
  6. Gotcha. Okay, so when I expand (1-1/2n)^n I get -(1/2) +(-1 + n)/(8 n) -(((-2 + n) (-1 + n))/(48 n^2)) +((-3 + n) (-2 + n) (-1 + n))/(384 n^3) -(((-4 + n) (-3 + n) (-2 + n) (-1 + n))/(3840 n^4) . . . and when I expand the series I get -(1/2) +1/8 -(1/48) +1/384 -(1/3840) . . . The denominators agree up to a power of n, so when we find the common denominator and do the difference, we get rid of the highest power of n in the numerator (which is the same power as the highest power in the denominator), but when we multiply by the n in the limit we get a fraction that has the same highest power of n in both the numerator and denominator, with the coefficient of the highest power in the denominator is always one, and the coefficient of the highest power in the numerator of the kth fraction should be [math]\frac{(-1)^{k+1}*k*(k+1)}{2^{k+1}*k!}[/math], which is what the fraction will converge to. Unfortunately, a sum of these terms converges to 3/8*Sqrt[e], instead of 1/8*Sqrt[e], so I must've made a mistake somewhere, but I can't figure it out. Also, how would one go about showing that the sum of [math]\frac{(-1)^{k+1}*k*(k+1)}{2^{k+1}*k!}[/math] converges to 3/8*Sqrt[e]?
  7. That's exactly what I tried to do, but as I said, if I replace the inverse square root with something that converges to it, it almost always changes the result of the limit. For example, if I used what you suggest the result would be trivially zero and if I used say [math]\frac{1}{\sqrt{e}} = \lim_{n -> \infty} (1 + \frac{1}{n})^{-\frac{n}{2}}[/math], the limit would converge to [math]\frac{3}{8\sqrt{e}}[/math] instead of [math]\frac{1}{8\sqrt{e}}[/math]. Thanks. I'm not that well versed in power series yet, so I'm not too sure how to go from here: [math]\frac{1}{\sqrt{e}} = e^{-\frac{1}{2}} =\sum_{k=0}^\infty \frac{(-1)^k}{2^k(k!)}[/math] I can see that the first terms cancel out, but how do I get anything containing n (or 1/n) out of the power series? The only thing that occurs to me is to rewrite the power series as [math]\lim_{n->\infty} \sum_{k=0}^n ...[/math], but the closed form for [math]\sum_{k=0}^n ...[/math] contains gamma functions, which are beyond my level of comprehension and more importantly the inverse square root of e, which just gets us back were we started. Also, neither L'Hospital's rule nor power series should be used. Frankly, given the fact that this is just a random exercise a friend sent me I'll settle for being able to solve it with the two, but if anyone sees an alternate, non-L'Hospital/power series method, it would be much apreciated.
  8. I came across the limit [math]\lim_{n -> \infty} n(\frac{1}{\sqrt{e}} - {(1 - \frac{1}{2n})}^{n})[/math], which converges to [math]\frac{1}{8\sqrt{e}}[/math] and I have no idea how to evaluate it. The only thing I could think of is to replace the inverse square root of e with something that converges to the same value, except that if I do that it changes the limit (unless I use [math]{(1 - \frac{1}{2n})}^{n - \frac{1}{4}}[/math], but I've only determined this experimentally with the help of WolframAlpha). Any ideas? Also, not that I think it would make any difference in this particular case, but you're not supposed to use either L'Hospital or Taylor.
  9. [math]\frac{(x+2)^{x+2}}{(x+1)^{x+1}} - \frac{(x+1)^{x+1}}{x^x} = \frac{(x+2)^{x+1} \cdot (x+2)}{(x+1)^{x+1}} - \frac{(x+1)^x \cdot (x+1)}{x^x} = (1 + \frac{1}{x+1})^{x+1}\cdot(x+2) - (1 + \frac{1}{x})^x\cdot (x+1)[/math] = [math](1 + \frac{1}{x+1})^{x+1}\cdot(x+1) - (1 + \frac{1}{x})^x\cdot (x+1) + (1 + \frac{1}{x+1})^{x+1}[/math] At this point, I would argue that [math](1 + \frac{1}{x+1})^{x+1} \sim (1 + \frac{1}{x})^x[/math] as [math]x \to \infty \wedge (x+1) = (x+1) \Rightarrow (1 + \frac{1}{x+1})^{x+1} \cdot (x+1) \sim (1 + \frac{1}{x})^x \cdot (x+1)[/math] as [math] x \to \infty \Rightarrow \lim_{x\to\infty} (1 + \frac{1}{x+1})^{x+1}\cdot(x+1) - (1 + \frac{1}{x})^x\cdot (x+1) + (1 + \frac{1}{x+1})^{x+1} = \lim_{x\to\infty} (1 + \frac{1}{x+1})^{x+1} = e[/math] But I'd be interested to know if this limit can be solved without the use of asymptotic equalities; it seems L'Hôpital wouldn't lead anywhere.
  10. You might be interested in reading Where Mathematics Comes From by Lakoff and Núñez; it deals with exactly what you've asked, namely "What does [math]e^{i\pi}[/math] mean". Note that I haven't finished the book yet and reactions to it are mixed at best. But it still might be an interesting read.
  11. http://www.bbc.co.uk/news/technology-16163931 http://web.media.mit.edu/~raskar/trillionfps/ I would be curious as to how they deal with storing the frames. If we assume 1kb/frame, which is probably way less than the actual size, we get 30 PB of data per second, which according to WA is about 1/30 of the estimated data content of the internet.
  12. I did not know that, but nevertheless this one does; the solution is [math] T = \left( \begin{array}{cccc} -1 & 0 & 0 & 0\\ -1 & 0 & -1 & 0\\ 1 & 0 & 1 & 1\\ -1 & 1 & -1 & 0 \end{array}\right) [/math]
  13. Given the matrix [math] A = \left( \begin{array}{cccc} 2 & 1 & 0 & -1\\ 0 & 2 & -1 & -1\\ 0 & 0 & 3 & 1\\ 0 & 0 & -1 & 1 \end{array}\right) [/math], find its Jordan matrix and the matrix T, for which [math]A=TJT^{-1}[/math]. All four of the eigenvalues are 2, and the Jordan matrix will consist of two Jordan blocks; [math] J = \left( \begin{array}{cccc} 2 & 1 & 0 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & 2 & 1\\ 0 & 0 & 0 & 2 \end{array}\right) [/math]. Thus, if we label the eigenvectors [math]h_1[/math], [math]h_2[/math], [math]h_3[/math], [math]h_4[/math], we have [A].[[math]h_1[/math] | [math]h_2[/math] | [math]h_3[/math] | [math]h_4[/math] ] = [[math]h_1[/math] | [math]h_2[/math] | [math]h_3[/math] | [math]h_4[/math] ].[J], and from this A[math]h_1[/math] = 2[math]h_1[/math] A[math]h_2[/math] = [math]h_1[/math] + 2[math]h_2[/math] A[math]h_3[/math] = 2[math]h_3[/math] A[math]h_4[/math] = [math]h_3[/math] + 2[math]h_4[/math] But this would mean that [math]h_1[/math] = [math]h_3[/math] and thus [math]h_2[/math] = [math]h_4[/math], which is nonsense. And I'm stumped as to what to do about it. Any ideas?
  14. No idea, but I seem to recall he said something about that in the video.
  15. Okay, after digging around a little bit I discovered the original solution in its original language, which makes understanding it a lot easier and also cuts away a lot of the ugliness I was perceiving. But it still seems like overkill to me. I have since discovered that the solution I had in mind when I started this topic would be invalid. DrRocket, yours was the approach I was trying to use since I first saw the problem, but a professor of mine shot it down with showing that for a = 1, b = 1, 8 = a*2^2 + b*2^2 and 9 = a*0^2 + b*3^2. Stupidly, it never occurred to me to try and show this applied for almost all integers. Thanks for the idea.
  16. I originally got the link from here, which says: There is also a brochure you can download, which may or may not go into further details; I'm not a chemist so I'm not entirely sure. This seems to be the company's website; experimental data in particular may be interesting, but again, I have no way of judging the quality of the information they present. Does any of this change anything?
  17. Why isn't there one of these in every household? Why haven't I heard anything about this up until now?
  18. During preparations for a mathematical contest I might take part in, I've come across an interesting problem: Let [math] n \in \mathbb{N}[/math] and [math]a, b \in \mathbb{Z}[/math] be integers such that the set [math]\mathbb{Z} \setminus \{ ax^n+by^n, x, y \in \mathbb{Z} \}[/math] is finite. Prove that n = 1. The "official" solution is the ugliest thing I've ever seen, and so incomprehensible that even my professors didn't understand it. My gut was/is telling me that there was a simpler solution, so I've been trying to find one. I've gotten to the point where all I have to do is prove that the set [math]\mathbb{Z} \setminus \{ x^{2k+1}+y^{2k+1}, x, y \in \mathbb{Z}, k \in \mathbb{N} \}[/math] is finite, ie. that there are infinitely many numbers [math]c \in \mathbb{Z}[/math] that cannot be expressed as [math] x^{2k+1}+y^{2k+1}[/math] for a given k and some integers x, y. And I'm stuck. The only hunch I have is that c will be dependent on k, but other than that, I have nothing. I've been clueless for a week or so now, so I'd really appreciate any ideas any of you might have.
  19. Just to avoid possible confusion, this is supposed to be [math]du = 2x \: dx[/math].
  20. What you posted is Sacks spiral, the same as Ulams spiral except using an Archimedean spiral instead of a square one, and has nothing whatsoever to do with sums of primes. In any case, what I meant by plot was [math]f(x) = \sum_{i=1}^{x} P_{i}[/math].
  21. Think. If that fails, plot.
  22. True, though up until now I believed [math]0 \cdot \infty [/math] was indeterminate regardless of the context. Are you saying it's zero?
  23. Like I said before, I agree with 2, I disagree with 1. While [math] \lim_{x \to 0} \frac{1}{\frac{1}{x}} = 0[/math], this does not mean that [math]\frac{1}{\frac{1}{0}} = 0[/math], in the same way that [math]\lim_{x \to \infty} \frac{1}{x} = 0[/math] does not mean that [math]\frac{1}{\infty} = 0[/math].
  24. I'm not a math teacher, but I'm as positive as I can be about this. [math]\frac{1}{\frac{1}{x}} = x \Leftrightarrow x \neq 0[/math]. And you can really ignore all this and just think of it this way. If you evaluate, for example [math]\frac{3}{\frac{3}{2}}[/math], by flipping the denominator, you get the same result as you would by first evaluating [math]\frac{3}{2}[/math] and then dividing 3 with it. However, if you evaluate [math]\frac{1}{\frac{1}{0}}[/math] by flipping the denominator, you would end up with zero, and if you do it the other way, you end up with undefined. This alone should set off alarm bells. I'd be more comfortable if someone with a higher education than mine would confirm this, but I'm pretty confident I'm right.
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