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 elementary superfunctions Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 04/25/2010, 11:35 AM (This post was last modified: 04/25/2010, 12:03 PM by Kouznetsov.) (04/25/2010, 10:48 AM)bo198214 Wrote: Challenge: Is there an elementary superfunction of a polynomial that has no real fixed point?I thnk so. How about transformation of line 5, $b=2$? Let $H_0(z)=z^2$ ; $P(z)=z+1$; $Q(z)=z-1$; Let $H_1(z)=P(H_0(Q(z)))=1+(z-1)^2$; Equation $z=1+z^2-2z+1$; (BAD)is equivalent of equation $z^2-z+2=0$ (BAD)Gives the fixed points $z=1/2 \pm \sqrt{1/4-2}=1/2\pm i \sqrt{7/4}$ Sorry I lost the signum! should be is equivalent of equation $z^2-3z+2=0$ Gives the fixed points $z=1/2 \pm \sqrt{9/4-2}=1/2\pm \sqrt{1/4}$ I try again: $P(z)=z+a$; $Q(z)=z-a$; Let $H_1(z)=P(H_0(Q(z)))=a+(z-a)^2$; Equation $z=a+z^2-2az+a^2$; is equivalent of equation $z^2-(2a+1)z+a^2+a=0$ Gives the fixed points $z=(2a+1)/2 \pm 1/2\sqrt{(4a^2+4a+1-4(a+a^2)}$ $z=(2a+1)/2 \pm 1/2\sqrt{1}$ Henryk, it is not so easy... Sorry... but.. Wait... May I use complex $a$ or the basefunction is supposed to be real? Do you mean "real polynomial"? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/25/2010, 12:12 PM (This post was last modified: 04/25/2010, 12:13 PM by bo198214.) (04/25/2010, 11:35 AM)Kouznetsov Wrote: How about transformation of line 5, $b=2$? I think this approach is going to fail. Because then the superfunction is of the form $\eta(e^{b^x})$ which means that it is a regular superfunction at some fixed point with $f'(z_0)=b$. The only exception would be that we have a complex fixed point (of a real polynomial) with a real derivative $b$, is that possible? Quote:Henryk, it is not so easy... Sorry... but.. Wait... May I use complex $a$ or the basefunction is supposed to be real? Do you mean "real polynomial"? Yes real base and super function. Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 04/25/2010, 12:42 PM (04/25/2010, 12:12 PM)bo198214 Wrote: .. $\eta(e^{b^x})$ ..is a regular superfunction at some fixed point with $f'(z_0)=b$. The only exception would be that we have a complex fixed point (of a real polynomial) with a real derivative $b$, is that possible?I do not understand the question. Is $\eta$ allowed to be signular at the fixed point? Quote:Yes real base and super function. We could find numerically the superfunction for basefunction H(z)=1+z^2, describe the properties, and nominate it to the Mathematical Community as new element of the set of Special Functions. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/25/2010, 01:10 PM (04/25/2010, 12:42 PM)Kouznetsov Wrote: (04/25/2010, 12:12 PM)bo198214 Wrote: .. $\eta(e^{b^x})$ ..is a regular superfunction at some fixed point with $f'(z_0)=b$. The only exception would be that we have a complex fixed point (of a real polynomial) with a real derivative $b$, is that possible?I do not understand the question. Is $\eta$ allowed to be signular at the fixed point? $\eta=P$. Quote:We could find numerically the superfunction for basefunction H(z)=1+z^2, describe the properties, and nominate it to the Mathematical Community as new element of the set of Special Functions. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/25/2010, 01:51 PM (04/25/2010, 01:42 PM)Ansus Wrote: Quote:Is there an elementary superfunction of a polynomial that has no real fixed point? f(x)=x+1 F(x)=C+x Haha, good hint to exclude the trivial case. Btw. answers are not posted directly in the open problems thread. I explained that thoroughly in the beginning of that thread. Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 04/25/2010, 01:52 PM (This post was last modified: 04/25/2010, 01:55 PM by bo198214.) (04/25/2010, 01:10 PM)bo198214 Wrote: .. $\eta=P$.Ah, I understand. It can be singular, pulling out any property you like, from the exponential tail. $e^{b^x}$ may approach its limiting value, and $P(e^{b^x})$ may do that you like, at least until the closest cutline. While within the range of the holomorphism, you may deform and squeese the superfunction as you want with the transform indicated; it may be good for the application, or for the representation, or for the evaluation; no other deep meaning, that would allow or prohibit some particular properties. Roughly, you can convert ANY real holomorphic monotonic function to ANY other real holomorphic monotonic function, at least until to reach a singularity or a cutline. tommy1729 Ultimate Fellow Posts: 1,359 Threads: 331 Joined: Feb 2009 04/25/2010, 02:34 PM (04/25/2010, 10:53 AM)bo198214 Wrote: Is there an elementary real function $F$, such that $F(1+F^{-1}(x))$ is a real polynomial of degree at least 2 without real fixed points. this question or similar has occured before. some papers have been written about it , i considered similar questions and i believe it appeared on the math forum ... ill have to dig ... i believe hypergeometric solutions were found ... but sometimes a hypergeo can be expressed by elementary functions ... personally i considered inverse hypergeometric functions , but those also can be simplified sometimes. so im optimistic ... have you tried simple cases such as a*exp(b^x)*exp(c*x) + d ? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/25/2010, 05:15 PM (04/25/2010, 02:34 PM)tommy1729 Wrote: some papers have been written about it , i considered similar questions and i believe it appeared on the math forum ... ill have to dig ... i believe hypergeometric solutions were found ... but sometimes a hypergeo can be expressed by elementary functions ... personally i considered inverse hypergeometric functions , but those also can be simplified sometimes. so im optimistic ... Oh that would be great if you could recover some references! Quote:have you tried simple cases such as a*exp(b^x)*exp(c*x) + d ? I am still in the process of trying, however like always not much dedicatable time. « Next Oldest | Next Newest »

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