• 3 Vote(s) - 4.33 Average
• 1
• 2
• 3
• 4
• 5
 Searching for an asymptotic to exp[0.5] tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 09/10/2014, 11:02 PM (This post was last modified: 09/10/2014, 11:15 PM by tommy1729.) There is alot of work to do for " fake function theory ". It seems to be growing superexponentially So far the main idea was to approximate a nonentire function by an entire one with all derivatives positive at 0. And then considering alternative solutions and zero's of those functions. Recently we also started considering fake function theory to approximate a nonentire function with an entire one , WITHOUT the restriction of the signs of the derivatives. ( mick's fake ln(x^2+1) ). Least squares methods can be used to measure the " quality " of the fake function. Connections to many other fields of math seem to occur. BUT there is ALSO a second possible deviation from the main idea : Approximate an entire function with not all derivatives nonnegative at 0 with an entire function with all derivatives positive at 0. I call this " hyperfake function theory ". It is a subset of " fake function theory ". A fake function can be hyper or not , and all hyperfake are fakes. I assume series multisections to be important. Natural questions : A) Is exp(x)/2 the best hyperfake function for sinh(x) ? Is it unique in some sense ? In this example there are an infinite amount of negative derivatives (for the given function) but perhaps even more intresting is the case when there are only a finite number of negative derivatives. Maybe call that " Ultrafake function theory ". This leads to imho the following natural question : Ultrafake exercise 1 B) find the fake function for exp(x) - (x^3 / 5). Note : ultrafake function theory seems to make more sense when the given function f(z) satisfies : x > 0 => f(x) > 0 , f ' (x) > 0 and f " (x) > 0. ( the selfreference could not be more subtle, right !? ) ... Not completely sure about the consistancy of ultrafake ... Mainly because exp(x) - ultrafake(exp(x) - (x^3 / 5)) should be a Taylor series with all + derivatives but grow about O(x^3) ?? Do fake function theory algorithms give an error for exp(x) - (x^3 / 5) ? Anyways hyperfake seems a good concept. regards tommy1729 " purpose is the biggest gift after money , love and health " tommy1729 tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 09/11/2014, 08:02 AM (This post was last modified: 09/11/2014, 08:17 AM by tommy1729.) My guess is that fake( exp(x) - (x^3 / 5) ) is simply : exp(x) - Constant. But I have not tried anything yet. I assume the best possible approximation ( by using fake - for exp(x) - ( x^3 / 5 ) - ) is approximately exp(x) - a x^b - constant. Where b is between 0 and 1. Also a possibility is with q_i very small : 1 + q_1 x + q_2 x^2 + q_3 x^3 + x^4/4! + x^5/5! + ... or 0 + q_1 x + q_2 x^2 + q_3 x^3 + x^4/4! + x^5/5! + ... I wonder. It seems 0 + q_1 x + q_2 x^2 + q_3 x^3 + x^4/4! + x^5/5! + ... cannot be improved. ( because O(x^3) cannot be given by O(x^(3+eps)) ) Therefore it would be a good (benchmark) test for our fake algorithms I guess. regards tommy1729 sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 09/11/2014, 02:13 PM (This post was last modified: 09/11/2014, 04:02 PM by sheldonison.) Yeah, long way to go, and I'm not sure I'm up for the task of making it all rigorous, and figuring out what's new and what the relevant prior mathematical work is. So far, I've only done the calculations for three function, $\exp^{0.5}\;\; \exp(x)\ln(x)\;\;\exp(0.5\ln(x)^2)$, for the last case, I have a closed form solution, that I can rigorously prove, post#85. They all behave rather differently, so claiming there is some sort of unified interpolation theory seems a bit of a stretch... But in all cases, we are trying to work with interpolating a function that eventually has all positive derivatives, and as you rotate around the unit circle (perhaps multiple times), the function eventually gets arbitrarily small in magnitude. And I think maybe that's the unifying thread. Also, the error term is proportional to 1/x, because we generate an entire Taylor series, but the function we are generating an asymptotic for has a Laurent series, since it is not entire. For the ln(x) or case, we have conjectured equivalent integral formation, f(z) =$\int\frac{1-\exp(-x)}{x}$ Since f(z) is entire, if we plug exp(z)f(z) into our interpolation function, we get out exactly exp(z)f(z). Also, if we plug $\ln(x)\exp(x)$ into our interpolation engine, we also get an entire function. The difference of two entire functions has to be an entire function; in this case, conjectured equal to zero. So, if I can prove exp(z)f(z) eventually has all positive derivatives, then I think I could rigorously prove they are equivalent, since I'm just taking the "fake" Cauchy integral... Speaking of all positive derivatives, another question is the Gaussian approximation, post#16, which can approximate all of these functions reasonably well, accurate for $a_n$ to one part per thousand or better for large n. In fact, for my third example here, the Gaussian approximation is coincidentally, exact! However, like the exp^{0.5} case, for this function we wrap the function we are interpolating around the unit circle multiple times, or an infinite number of times in this case, where we choose n to get optimal results. So claiming there is some sort of unified interpolation theory again seems like a huge bit of a stretch... $g(x)=\sum_{-n}^{+n} f(\exp(\ln(x)+2n\pi i))\;\;$ summing up multiple loops around the unit circle sometimes helps convergence. Integrate $\int_{-n\pi}^{n\pi}$ instead of $\int_{-\pi}^{\pi}$ - Sheldon tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 09/12/2014, 07:49 AM Im probably crazy but Is there an entire asymptotic to exp , G(z) , that has a pseudoperiod of pi i instead of period 2 pi i ? The inverse would then by an asymptotic to ln. I was thinking about sqrt( (e^2)^x ) but it either reduces to exp(x) or it has sqrt branches. Let g(x) be asymptotic to 1/2 for real x < 0 and asymptotic to 1 for x > 1. Then g(x) ln(x) seems to be the inverse of G(z). I ask this for better insight to the connections of complex analysis and fake function theory. This could help to formalize things. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 09/12/2014, 06:35 PM A few simpler questions for a change. How does the difference between fake_exp^[0.5](z) and Kneser_exp^[0.5](z) look like ? And how about the difference of fake_exp^[0.5](z) with 2sinh^[0.5](z) ? In particular I consider there differences for Re(z) > 0. It is tempting to conjecture where the difference is small , such as in the region where the fake f(z) satisfies f(f(z)) ~ exp(z) pretty well. But a simple plot might disprove that. Of course this relates to the difference of 2 sinh^[0.5](z) with Kneser_exp^[0.5](z) as well. For positive reals this is all pretty Obvious ofcourse , but away from the real line the question becomes more serious. regards tommy1729 sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 09/13/2014, 07:15 PM (This post was last modified: 09/13/2014, 07:40 PM by sheldonison.) (09/12/2014, 06:35 PM)tommy1729 Wrote: How does the difference between fake_exp^[0.5](z) and Kneser_exp^[0.5](z) look like ? .... And how about the difference of fake_exp^[0.5](z) with 2sinh^[0.5](z) ? .... In particular I consider there differences for Re(z) > 0.Below, is a log_10/log_10 plot of both ratios, from 0.1 to 10^6. The fakeexp asymptotic is accurate to >30 decimal digits at 10^6, using only 55-60 terms. btw, there is also a pretty good asymptotic for sinh^{0.5}, almost as good as the one for exp^{0.5}. Tommy might have been interested in sinh(z)/exp(z) which converges exponentially to '1', much quicker than the fakeexp converges. $\exp(x)\sqrt{x} \approx \sum_{n=0}^{\infty} \frac{x^n}{\Gamma(n+0.5)}\;\;$ Have you seen this excellent asymptotic entire series? I can't explain it, other than it matches the "fakefunc" integral     - Sheldon tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 09/13/2014, 11:25 PM (This post was last modified: 09/13/2014, 11:41 PM by tommy1729.) (09/13/2014, 07:15 PM)sheldonison Wrote: $\exp(x)\sqrt{x} \approx \sum_{n=0}^{\infty} \frac{x^n}{\Gamma(n+0.5)}\;\;$ Have you seen this excellent asymptotic entire series? Yes Ive seen this before. Have you heard of hypergeometric functions or generalized hypergeometric functions ? When I was still active at sci.math I mentioned the underrated and insuffiently known properties and powers of the hypergeometric functions and more particular their INVERSES. Inverse hypergeometric functions is one those other " crazy math concepts " apart from tetration , collatz and prime twins. Bill Dubuque and a few others recognized / remembered the power of these functions. But its far from mainstraim. It is amazing how unexpectedly these inverse hypergeometric functions can occur. Literature is very very rare ( like tetration ) and it is not introduced to younger students. From " entire function theory " comes the idea that hidden recursions are often of hypergeometric nature. hypergeometric and their inverse occur in for instance closed form solutions ( with integral or sum ) for half-iterates. They are also very good approximations to other functions. A good understanding of the gamma function is essential. Id say Euler and Gauss were the first to recognise its importance , but that is open for debate. The theory is very uncomplete ( as for tetration I guess ) , and that might be the reason for the avoidance in more classical math. That's enough background. I have to think about this , I have seen it before ... But that is probably over 10 years ago and I have to dig in my memory. Maybe its easy , but I lack time and concentration now. Perhaps this matters or helps : http://www.math.upenn.edu/~wilf/AeqB.pdf edit : this appears as the fake sqrt (after division by exp). I conjecture for x > 2 : $|\exp(x)\sqrt{x} - \sum_{n=0}^{\infty} \frac{x^n}{\Gamma(n+0.5)}\;\;| < sqrt {x+1}$ regards tommy1729 sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 09/13/2014, 11:45 PM (This post was last modified: 09/13/2014, 11:46 PM by sheldonison.) (09/13/2014, 11:25 PM)tommy1729 Wrote: (09/13/2014, 07:15 PM)sheldonison Wrote: $\exp(x)\sqrt{x} \approx \sum_{n=0}^{\infty} \frac{x^n}{\Gamma(n+0.5)}\;\;$ Have you seen this excellent asymptotic entire series? Yes Ive seen this before. Have you heard of hypergeometric functions or generalized hypergeometric functions ? When I was still active at sci.math I mentioned the underrated and insuffiently known properties and powers of the hypergeometric functions and more particular their INVERSES. Inverse hypergeometric functions is one those other " crazy math concepts " apart from tetration , collatz and prime twins. Bill Dubuque and a few others recognized / remembered the power of these functions. But its far from mainstraim. It is amazing how unexpectedly these inverse hypergeometric functions can occur. ..... A good understanding of the gamma function is essential. Hey Tommy, I posted it on mathstack; http://math.stackexchange.com/questions/...237#930237 and got quick responses! I need to get really good at understanding and working with the Gamma function. - Sheldon jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 09/13/2014, 11:49 PM (09/13/2014, 07:15 PM)sheldonison Wrote: $\exp(x)\sqrt{x} \approx \sum_{n=0}^{\infty} \frac{x^n}{\Gamma(n+0.5)}\;\;$ Have you seen this excellent asymptotic entire series? I can't explain it, other than it matches the "fakefunc" integral Hey, this reminds me of the generalization I made for the (unscaled) asymptotic binary partition function: (09/09/2014, 07:43 PM)jaydfox Wrote: $ f(x) = \sum_{k=0}^{\infty}\frac{1}{2^{k(k-1)/2} k!} x^k \\ \Rightarrow f(x) = \sum_{k=0}^{\infty}\frac{1}{2^{k(k-1)/2} \Gamma(k+1)} x^k$ Treating Gamma(k) at negative non-positive integers as infinity, and the reciprocal of such as zero, we can take the limit from negative to positive infinity. And we can replace k with (k+b), where b is zero in the original solution, but can now be treated as any real (well, any complex number, but the complex versions are less interesting). $ f_{\beta}(x) = \sum_{k=-\infty}^{\infty}\frac{2^{-(k+\beta)(k+\beta-1)/2}}{\Gamma(k+\beta+1)} x^{k+\beta}$ We can apply the same generalization to exp(x): $ \exp(x) \approx \sum_{k=-\infty}^{\infty}\frac{x^{k+\beta}}{\Gamma(k+\beta+1)}$ Notice that I put "approximately equal". I haven't checked, but I assume it's exactly equal, but only in the sense that it should satisfy the functional equation exp'(x) = exp(x). Now, set beta = 1/2 and truncate the negative powers of x: $ \exp(x) \approx \sum_{k=0}^{\infty}\frac{x^{k+1/2}}{\Gamma(k+3/2)}$ ...and comparing your power series, it easily follows that your power series is asymptotic to exp(x) x^(1/2). Hmm, oh dear, I think you missed a +1 in the Gamma function in your series? (LOL, it's okay, I usually forget the +1 as well.) If you set beta=-1/2, then all is well! $ \frac{\exp(x)}{\sqrt{x}} \approx \sum_{k=0}^{\infty}\frac{x^{k}}{\Gamma(k+1/2)}$ $ \exp(x)\sqrt{x} \approx \sum_{k=0}^{\infty}\frac{x^{k}}{\Gamma(k+3/2)}$ ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 09/14/2014, 12:00 AM (This post was last modified: 09/14/2014, 12:13 AM by jaydfox.) (09/13/2014, 11:49 PM)jaydfox Wrote: ...and comparing your power series, it easily follows that your power series is asymptotic to exp(x) x^(1/2). Hmm, oh dear, I think you missed a +1 in the Gamma function in your series? (LOL, it's okay, I usually forget the +1 as well.) If you set beta=-1/2, then all is well! Haha, I think my math is wrong somewhere, because I double-checked in Excel, and you were right! I obviously did something wrong in my math, though I haven't figured out what yet... I'll update this post when I get it sorted out, LOL! Update: I think I figured it out. I was doing the Gamma function correctly (beta = -1/2), but I was extracting beta from the exponent incorrectly. I had my equations backwards. So my original equation (with the beta's intact) is still correct... I think... $ \frac{\exp(x)}{\sqrt{x}} \approx \sum_{k=0}^{\infty}\frac{x^{k}}{\Gamma(k+3/2)}$ $ \exp(x)\sqrt{x} \approx \sum_{k=0}^{\infty}\frac{x^{k}}{\Gamma(k+1/2)}$ And generally: $ \exp(x) x^{-\beta} \approx \sum_{k=0}^{\infty}\frac{x^{k}}{\Gamma(k+\beta+1)}$ ~ Jay Daniel Fox « Next Oldest | Next Newest »

 Possibly Related Threads... Thread Author Replies Views Last Post Using a family of asymptotic tetration functions... JmsNxn 15 4,646 08/06/2021, 01:47 AM Last Post: JmsNxn Reducing beta tetration to an asymptotic series, and a pull back JmsNxn 2 896 07/22/2021, 03:37 AM Last Post: JmsNxn A Holomorphic Function Asymptotic to Tetration JmsNxn 2 1,341 03/24/2021, 09:58 PM Last Post: JmsNxn An asymptotic expansion for \phi JmsNxn 1 1,159 02/08/2021, 12:25 AM Last Post: JmsNxn Merged fixpoints of 2 iterates ? Asymptotic ? [2019] tommy1729 1 3,563 09/10/2019, 11:28 AM Last Post: sheldonison Another asymptotic development, similar to 2sinh method JmsNxn 0 4,102 07/05/2011, 06:34 PM Last Post: JmsNxn

Users browsing this thread: 3 Guest(s)