2:00 AM
I'm a bit confused
@PeterTamaroff sorry, I was slow in answering. I'm not used to chat rooms (you all "talk" so rapidly) and I write too much!
$$t_n = 2 t_{n-1} + (-1)^n$$
let's solve this
@Argon I mean $$t_{2n}=2t_{2n-1}+1 Ok yes I like that idea 2:00 AM @cassandra0 Another easier way could be telescopy, maybe, that will removed the (-1)^n user19161 @PeterTamaroff Yes, I know, but what you said is meaningless. @WillHunting It's getting close to Dec 4th user19161 @amWhy Well, yeah, but there won't be an earthquake or anything, just a mild tremor perhaps. =) @WillHunting Well... 2:02 AM 0 I saw the proof by Chebychev that there are constants c_1,c_2:$$c_1 \frac{\log(x)}{x} < \pi(x) < c_2 \frac{\log(x)}{x}$$and the Prime Number Theorem states$$\lim_{x \to \infty} \frac{\pi(x)}{\log(x)/x} = 1$$I don't understand what exactly PNT is telling us, and why is it s... I wanted to you if this is good? do you want me to add any remarks @cassandra0 How does one attack the recursion? to the question user19161 @PeterTamaroff See, we are talking about whether or not E is dense in X. I don't see how isolated points come into the picture. Yes, I do know what isolated points are. @Argon, I'm just trying to do Peters idea fist of all @Argon I have an idea. 2:02 AM @PeterTamaroff Ok so$$t_{n+2} = 2 (2 t_n + (-1)^n) + (-1)^{n+1}$$@Argon Generating functions Yeah! @cassandra0 Ok, yes Hence$$t_{2(m+1)} = 4 t_{2m} + 2 - 1$$Ok :) I see where this is going now... I like it! user19161 2:06 AM So @amwhy which is the month, 5 or 9? The GF of t_n is$$T(x)=\frac {t_0}{1-2x}+\frac{1}{(1-x)(1-2x)}$$GF?$$t_{2(m+1)+1} = 4 t_{2m+1} - 1$$that's the odds and evens separated out Ok @Argon generating function 2:09 AM I think it goes 1, 4+1, 4^2+4+1, 4^3+4^2+4+1, ... so would that be 4^{n+1}-1 for the even part @PeterTamaroff I think that is too fancy for my purposes @cassandra0 Ok @Argon It is uuusome @PeterTamaroff Baoss I like generating functions too the next bit is like 4^3 - 4^2 + 4 - 1 I don't know the closed form for that @cassandra0 I think I do It's a geometric series 2:10 AM oh yeah so it is Yay! @PeterTamaroff How did you think I should do it w/ GFs? you know what strikes me as odd here @cassandra0 2n+1? lol Hahaha! 2:12 AM well yes sort o it's just funny there's two cases I guess you might 3 cases if there was something like (\sqrt[3]{-1})^n in it Bleh user19161 @Argon What is GF, girlfriend? Aaron is a ladies man! 4 mins ago, by Peter Tamaroff @Argon generating function @Argon Just consider$$t_n=t_{n-1}+(-1)^{n}$$and sum over n=1 to \infty @WillHunting 2:13 AM Who cares about convergence anyways? @PeterTamaroff Ok :) By sum do you mean t_0+t_1+\cdots user19161 @peter I hereby declare you as Potato.$$G = \sum_{i} t_i x^i$$@Argon No, I mean \sum t_n x^n @PeterTamaroff Classic power series. Ok :) 2:15 AM @WillHunting [ERROR 435 - TRY AGAIN ] @WillHunting If you try again, you will probably get error 436 no one is answeing my question on the website @cassandra0 What was it? I linked here it a amoment ago user19161 @peter is very naughty today. He has said a number of naughty things today. I don't know what he is thinking, but I think he is probably wrong. 2:16 AM @WillHunting How could @Peter ever be wrong? user19161 @Argon Well, I have not told you what I think he is thinking, so yes, he could be partially wrong. @WillHunting Spit your blabbery user19161 But since I don't even know what Peter wants to say, I won't say anything. QED. @amWhy, did you see my question? @Argon,$$
{\rm li} (x) = \gamma + \ln \ln x + \sqrt{x} \sum_{n=1}^\infty \frac{ (-1)^{n-1} (\ln x)^n} {n! \, 2^{n-1}} \sum_{k=0}^{\lfloor (n-1)/2 \rfloor} \frac{1}{2k+1} . $$spotted that from wikipedia @cassandra0 SO COOL! 2:19 AM I thought you might like it :D I love the \ln \ln x :) I'm looking up li because it's another version of the PNT @PeterTamaroff Do you think I should do something like this: en.wikipedia.org/wiki/… @Argon Yes. @cassandra0 Yup, and a better approximation then \frac{x}{\log x} 2:20 AM why is it better? by the way I think it was Gauss who originally conjectured pi ~ li @WillHunting I used my real (first) name when I posted a "congrats" answer to Brian M Scott. I'm getting bold! @PeterTamaroff I'm a bit confused as to why f=xf+x^2 f+x @cassandra0 No, I didn't. How far to scroll up? @amWhy, it's here @Argon Fibonacci! 2:23 AM @Argon, I want to explain that @cassandra0 Shoot! user19161 @amWhy You wrote it wrongly. As far as I know, the first letter should be in majuscule and not minuscule form. =) @cassandra0 Ahhh! yes, I see it. @WillHunting Hahahaa... Let$$G = \sum_{i=0}^\infty c_i x^i$$any generating function at all, and we should set all the negative c_{-k} = 0 (for later). user19161 @amWhy I was trying to sound pretentious. I could have just said capital letters and small letters. =) 2:24 AM @Argon Man @Argon nice answer re: modular arithmetic 9 upvotes!$$x^k G = \sum_{i=0}^\infty c_{i-k} x^i$$@WillHunting You sounded pretentious, so it worked! @amWhy ? @PeterTamaroff ? user19161 @amWhy Yay! 2:24 AM @Argon amWhy what? @Argon, notice that I changed the index instead of changing the exponent - of course these are equivalent but this is the key. @amWhy " nice answer re: modular arithmetic 9 upvotes!"? @cassandra0 Ok @Argon Ohhh, I'm sorry, I clicked on the wrong post. I meant to post to cassandra0 @amWhy Oh, hahaha Therefore if we have a linear relation between coefficients e.g. c_{i} - c_{i-1} - c_{i-2} = 0 for the fibonacci numbers we get a polynomial relation for G: (1 - x - x^2)G = 1 + x 1 and 1 \cdot x are the initial conditions. user19161 2:27 AM @amWhy That's called victim of the moving target. @WillHunting That was it, exactly! user19161 @PeterTamaroff Pedro's handwriting looks as beautiful as Pedro! @amWhy, oh that's funny I don't know why they like that calculation so much @PeterTamaroff You have neat writing 2:28 AM @Argon, so is anything I said unclear @cassandra0 A bit We got this for fibonacci numbers$$G = \frac{1 + x}{1 - x - x^2}G = 1 + 2x + 3x^2 + 5x^3 + 8x^4 + 13x^5 + 21x^6 + 34x^7 + 55x^8 + 89x^9 + ...$$How is that polynomial relation obtained? @cassandra0 Well, it seemed to be a hit! It's based on$$x^k G = \sum_{i=0}^\infty c_{i-k} x^i$$2:30 AM @PeterTamaroff Hey there! @Argon One learns how to survive as time goes by @robjohn Man. @PeterTamaroff You should "read" mine, haha @robjohn I aced my linear algebra test. Hehehehe @PeterTamaroff Good job! Congratulations! If c_{i} - c_{i+1} - c_{i+2} = 0 then$$0 = \sum_{i=0}^\infty (c_{i} - c_{i+1} - c_{i+2}) x^i = \sum_{i=0}^\infty c_{i} x^i - \sum_{i=0}^\infty c_{i+1} x^i - \sum_{i=0}^\infty c_{i+2} x^i= (1 - x - x^2)\sum_{i=0}^\infty c_{i} x^i$$2:31 AM @PeterTamaroff YAY! user19161 @PeterTamaroff Everyone is man to you. @cassandra0 Which is 0? @PeterTamaroff I broke 40K today :-) @robjohn Yeah. user19161 @robjohn Thanks to someone. =) 2:33 AM @WillHunting Well, I'm uncomfortable talking to unicorns. @Argon, basically this^ although ttechnically it's c_{i} - c_{i+1} - c_{i+2} = 0 only for i \ge 2 there are two initial conditions c_0 and c_1 we have to define @WillHunting Couldn't have done it otherwise, I'm sure ;-) that's where the 1 + x came from I am very confused sorry :( user19161 2:34 AM @Argon So am I. I am always confused.$$x^k \sum_{i=0}^\infty c_{i} x^i = \sum_{i=0}^\infty c_{i-k} x^i$$do you understand this? this is where it all comes from @WillHunting You are a Jambul @cassandra0 Yes I understand that bit @cassandra0 Shouldn't you be changing some indices in the summation too?= @PeterTamaroff, I actually reindexed the sum which is why we have c_{i-k} @Argon OH NO! That is wrong. It should be -x/(1+x) not 1/(1+x) 2:36 AM @Argon, so for example$$(5 - 3x^2) \sum_{i=0}^\infty c_{i} x^i = \sum_{i=0}^\infty (5 c_i - 3 c_{i-2}) x^i$$@robjohn I have a silly question, user19161 Hmm, if I can get 100 every day in December, I should reach 10k this year. But that would be very taxing... @cassandra0 Ok Suppose \{f_n\} are bounded over [a,b] and f_n\to f uniformly. THen f is bounded. How do I specify ymin and ymax in a (2d) plot in mathematica? 2:38 AM I getting |f(x)|<M_N+\epsilon @robjonh For any given \epsilon and large N @Argon, that's really the heart of it. For fibonacci numbers F_{n} - F_{n-1} - F_{n-2} = 0 we get$$(1 - x - x^2) G = (F_{0} - F_{-1} - F_{-2}) x^0 + (F_{1} - F_{0} - F_{-1}) x^1 + (F_{2} - F_{1} - F_{0}) x^2 + \ldots = 1$$therefore$$G = \frac{1}{1 - x - x^2}$$Cools I'm wrong about this F_1 - F_0 - F_{-1} = 0 I corrected it now F_{-n} = 0? yes @Argon, so that's how to turn recurrences into generating functions. Then performing long division computes fibonacci numbers (or powers of two or whatever, any linear recurrence) 2:41 AM @robjohn ? So shouldn't it equal \frac{1}{1-x}? we can use G to find closed form formulas too \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots @PeterTamaroff Sorry, I was afk. What's up? this is the recurrence relation a_n - a_{n-1} = 0 @cassandra0 I mean, why does the sum equal 1? 2:42 AM which sum? love joe satriani$$ (F_{0} - F_{-1} - F_{-2}) x^0 + (F_{1} - F_{0} - F_{-1}) x^1 + (F_{2} - F_{1} - F_{0}) x^2 + \ldots = 1$$Isn't this just 1+x+x^2+\cdots? @Argon, F_i - F_{i-1} - F_{i-2} = 0 for all i > 2 because we're doing fibonicci numbers @cassandra0 There was a show here! G3 2:43 AM @PeterTamaroff I haven't gotten into his music, but I haven't heard enough yet that means there's just couple terms at the very beginning (the initial conditions of the recurrence) that aren't zero @robjohn I really meant to tell you to "Just look up" and see what I had asked Hehehehehe @cassandra0 So why does it equal 1? @Argon Each term is zero @Argon, well the fibonacci sequence F_{-2},F_{-1},F_0,F_1,F_2,... is 0,0,1,1,2,... 2:45 AM Except the first @PeterTamaroff Ok, now I see so (F_{0} - F_{-1} - F_{-2}) x^0 + (F_{1} - F_{0} - F_{-1}) x^1 + (F_{2} - F_{1} - F_{0}) x^2 + \ldots = (1 - 0 - 0) 1 + (1 - 1 - 0) x + (2 - 1 - 1)x^2 + (3 - 2 - 1)x^3 + (5 - 3 - 2)x^4 + ...) @robjohn 8 mins ago, by Peter Tamaroff Suppose \{f_n\} are bounded over [a,b] and f_n\to f uniformly. THen f is bounded. 7 mins ago, by Peter Tamaroff I getting |f(x)|<M_N+\epsilon @robjonh 7 mins ago, by Peter Tamaroff For any given \epsilon 7 mins ago, by Peter Tamaroff and large N @PeterTamaroff for some n, \|f_n-f\|_\infty<1 put that together with \|f_n\|_\infty=M and what do you get? @robjohn Oh, man, Well, I like incognito. hhahaha 2:48 AM still no one answered my question :( @robjohn Why do you write ||_{\infty}?? \infty norm is |x|? @PeterTamaroff because that is what you use for uniform convergence, is it not? @robjohn Well, UC is |f-f_n|<\epsilon for all x, given any \epsilon @PeterTamaroff which is saying the same thing? @robjohn I thought the \infty norm was \max =P 2:50 AM @Argon, anything unclear left or am I just making it more confusing each time I say something more? @robjohn But I now see it is the uniform norm too. I mean, it has another meaning. Just the same word. In mathematical analysis, the uniform norm (or sup norm) assigns to real- or complex-valued bounded functions f defined on a set S the non-negative number :\|f\|_\infty=\|f\|_{\infty,S}=\sup\left\{\,\left|f(x)\right|:x\in S\,\right\}. This norm is also called the supremum norm, the Chebyshev norm, or the infinity norm. The name "uniform norm" derives from the fact that a sequence of functions \{f_n\} converges to f under the metric derived from the uniform norm if and only if f_n converges to f uniformly. If we allow unbounded functions, this formula does not yield a norm or metric i... @PeterTamaroff I am not talkiing about a.e. @cassandra0 Can we just review what was covered? @Argon, I'd love to! @robjohn Almost everywhere? 2:52 AM$$F_n-F_{n-1}-F_{n-2}=0$$\forall n \ge 2 @robjohn What you mean is that one can write$$\sup_{n \geq N}|f-f_n|<\epsilon$$correct? @Argon, that's the definition of fibonacci numbers, right? @cassandra0 Yes along with F_0 = 1, F_1 = 1 2:53 AM @Argon \forall n\in \Bbb Z! @Argon, so this is just a nice example but this theory works for any linear recurrence @PeterTamaroff Not by our defintion - F_{-1} = 0 my use of negatives is not standard and maybe confusing.. it's just how I do it : user19161 I am going to take a shower and nap now. See you guys in my dreams. @cassandra0 It is really useful. 2:54 AM I find it good @WillHunting Bye user19161 @Argon I will start work on the book cover! =) Fundamental Theorem:$$x^k \sum_{i=0}^\infty c_{i} x^i = \sum_{i=0}^\infty c_{i-k} x^i$$@WillHunting Hahaha! @cassandra0 Ok, got that part (There is also a nice \frac{d}{dx} \sum_{i=0}^\infty c_{i} x^i = \sum_{i=0}^\infty i c_{i-1} x^i) 2:56 AM @cassandra0 Hahaha cool! user19161 @amWhy That would be my secret for now... @WillHunting What would be what secret for now? ;-) user19161 @amWhy You need to click on the left arrow to see which I am replying to. That's another tip for you. =) @cassandra0$$t_n - 2 t_{n-1} - (-1)^n = 0$$@Argon, so this is not linear we can't apply the technique 2:59 AM DARRRRRRRRNNN wait!! @WillHunting Thanks, got it. Well, I hope all is Good Will Hunting, soon! I'm not so good myself, but that would be my secret... there isa very beautiful theorem Good @Will Hunting Let a_n and b_n be linear recurrences 2:59 AM Done :) user19161 @amWhy OK, perhaps we might share some secrets in future... then c_n defined by c_{2n} = a_n, c_{2n+1} = b_n is a linear recurrence @WillHunting Do take a nice "nap"...I think I'm heading there myself. @cassandra0 ok @WillHunting Yes, let's do. When you're ready ;-) 3:00 AM This is what you did before? so let's just think about it a bit..... if A(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots  generates (a_n) and B(x) = b_0 + b_1 x + b_2 x^2 + b_3 x^3 + \ldots  generates (b_n) user19161 @Argon =) then C(x) = A(x^2) + x B(x^2) generates (c_n) proved! 3:02 AM Good night everybody! I need to sleep now :) @Argon g'night. Me too. @cassandra0 Bye but C(x) = a_0 + b_1 x + a_2 x^2 + b_2 x^3 + \ldots @amWhy Good night! 3:03 AM just so you see it What about coffee people!? What about coffee!? @Argon Sleep well! @amWhy You too! user19161 Let's all meet in our dreams. Over and out! @robjohn, i was hoping ,if you saw it > 15 years old, Gauss proposed that pi(n)âˆ¼n/(ln n). 15 years old.. > Gauss later refined his estimate to pi(n)âˆ¼Li(n) 3:08 AM I wrote up a quick spiel @cassandra0 He had tables and tables and calculated and calculated. Yes. One hell of a mathematician. @cassandra0 Let it be recorded that I upvoted you first. First I think I'll take a look into this In conclusion, cass, if you want to understand the meaning behind elements of analytic number theory, it is necessary to understand the meaning and motivation behind asymptotic thinking. @anon Well said. @PeterTamaroff Eh? I'm the only one that's upvoted the question... @anon Your answer 3:13 AM But you linked to cass. :-) @cassandra0 I just got back from fixing dinner for my dog. I answered your question, but anon has a much larger post. I would have answered it sooner, too, if it weren't for meddling plot options in mathematica. #nostalgia This is interesting: Let x_n be such that |x_{n+1}-x_n|<c^n for some 0<c<1 then x_n is Cauchy. (I proved it, but yet...) sure, geometric series bounding, yadda yadda I was thinking about finite calc finite differences 3:18 AM is x^2 \sim 2x^2? oh they don't have the same leading coeffcient @cassandra0 No, they are not \sim to each other. thanks so much why would \pi(n) actually be \sim to anything, let alone such a simple function as \frac{x}{\log(x)} Because primes are musical and grow swimmingly on large scales. Also I like the words musical and swimmingly. you know what's really weird$$\pi(x) \sim \frac{x}{\log(x)}$$now let \pi_{a,q} count the number of primes \equiv a \pmod q for (a,q) = 1. We have$$\pi_{a,q}(x) \sim \frac{1}{\varphi(q)} \frac{x}{\log(x)}$$so the prime number theorem isn't just a simple function like that it's ONE multiplied by a simple function if there was some irrational number like \gamma it would seem less weird well, in a sense there is a weird constant in the expression - the natural base e. 3:24 AM oh that's a really good point, it must be the natural log, musn't it? Any other wouldn't be \sim. I was wondering earlier if I could bound like$$c_1 2^n < 3.7^n < c_2 2^n $$and you can't changing the base of your logarithms results in a scalar multiple change in the function. that is,$$\pi(x)\sim (\log_e b)\frac{x}{\log_b x}.$$wow I'm really amazed now this is due to the change-of-base formula$$\log_bc=\frac{\log_ac}{\log_ab}$$this is truly a result how the primes behave in the large scale, not just a measure of their growth behavior, growth, poh-tay-to, poh-tah-to. 3:28 AM well I mean Chebychev is just a growth result asymptotic growth thanks very much I am really glad to understand this now I learned about big O in computing so I've a had a different view on these asymptotic notations this \sim is much more fine viewing it as an equiv. class is very helpful too. (\sin(x)+5)x^2 is bounded between 4x^2 and 6x^2 but isn't \sim a x^2 for any a right 3:46 AM wikipedia says "  \pi(x)\sim\frac{x}{\ln x}.\! This notation (and the theorem) does not say anything about the limit of the difference of the two functions as x approaches infinity." this is actually wrong, we have |f(x)-g(x)| = o(g(x)) whenever f(x) \sim g(x) that's just proved by multiplying f(x)/g(x) - 1 \to 0 by g(x) and taking absolute value right. the riemann hypothesis puts some bounds on the error term too. in terms of chebychev I think we get some negative -0.94 x/\log(x) < \pi(x) - x/\log(x) < 2^{342} x/ \log(x) the constants are made up but supposed to be realistic I think I'd need absolute value I'm a bit confused user19161 @anon What has potato got to do with what you are saying? I found in apaper c1 = 0.92129, c2 = 1.10555 holds for x > 96097 user19161 @anon I like musical and swimmingly too! 3:55 AM so yeah subtracting x/log(x) really just tells us that pi(x) - x/log(x) is between -1/10th and 1/10th of x/log(x) wait no that's wrong "poh-tay-toe, poh-tah-toe" is a common phrase so |pi(x) - x/log(x)| < 1/10 x/log(x)... but PNT gives |pi(x) - x/log(x)| = o(x/log(x)) i.e. the constant 1/10th gets smaller for any \epsilon>0, there is an X such that \left|\pi(x)-\frac{x}{\log(x)}\right|<\epsilon\frac{x}{\log x} for all x>X. this is a restatement of PNT. omg! You can get Bertrands Postulate from Chebyshevs bound!$$\sum_{p^k\leq x} \log p= x -\sum_{\rho:\ \zeta(\rho)=0} \frac{x^{\rho}}{\rho} +\frac{\zeta'(0)}{\zeta(0)}.$$that's such a cool formula PNT doesn't imply bertrands postulate though or does it? 4:20 AM PNT is asymptotic, so no if you remove that "for all" part on bertrands postulate and replace it with "eventually," you can not only use PNT but obtain stronger results with it. hmm so I get$$\pi(2n) - \pi(n) \ge \frac{2n}{\log(2) + \log(n)} - \frac{n}{\log n} + o(\frac{2n}{\log(2n)})$$if I could show that last term \ge 1 (and I guess that holds if we take n large enough so that the implied constant is < 1/2) that would give bertrand for n large that's pretty simple. pick u in between 1 and 2. then log(2)+log(n)<u*log(n) eventually, so the RHS above is >= (2/u-1)n/log(n)+o(blah), which is eventually >1. ahh very nice I guess the thing about betrand is, as you said, that we really want to have it for all n though 4:45 AM have you guys seen parametrizations \delta, \gamma of the unit circle in \Bbb R^2 (or part of it) so that \delta\circ\gamma^{-1} is not of class C^\infty? I'm looking for two inyective immersions from \Bbb R into \Bbb R^2, F G so that F\circ G^{-1} is not a diffeomorphism by parametrizations do you mean functions \Bbb R/\Bbb Z\to S^1? You may as well view S^1 as a subset of \Bbb C, set \gamma to be the usual parametrization and \delta something like the complex exponential of say the cantor function 5:17 AM @anon sorry for the delay. I mean functions \Bbb R\to S^1. So far I managed to find \delta:(-1,1)\to S^1;\theta\mapsto (\cos\theta,\sin\theta) and \gamma:(-1,1)\to S^1;t\mapsto \left(\dfrac{2t}{1+t^2},\dfrac{1-t^2}{1+t^2}\right), as shown here neither of those is surjective. are you okay with that? Now I have to find \gamma^{-1} @anon yep. Indeed I want something from \Bbb R or an open interval to a some part of S^1 so that the map is injective, as in those cases the domain of \delta is wrong it must be ]0,\pi[ if (\cos\theta,\sin\theta)=\left(\dfrac{2t}{1+t^2},\dfrac{1-t^2}{1+t^2}\right), it is enough to show that the function from (-1,1)\to(0,\pi) is not a diffeomorphism @anon the important part is that they have the same image set \lt 5:48 AM it seems that the function of \theta in terms of t is$$\theta=2\arctan\left( \dfrac{1-t}{1+t} \right)