Processes which propagate faster than exponentially
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From: Robert on 15 Jan 2010 14:16 On 15 Jan, 19:09, "I.N. Galidakis" <morph...(a)olympus.mons> wrote: > Robert wrote: > > Well, for anyone that hasn't kill filed me, the "worthless caveat" is > > essential to the OP's question. > > > Would > > > as x->oo, for all t, f(x)/e^xt -> oo > > > be an adequate definition for a function with faster than exponential > > growth? > > Yes. Sorry for the (created) confusion. > > As far as I am concerned, I mean anything like: f(x) ~ a^x, with a > e. > > Thanks again to all the responders. > -- > Ioannis in that case androcles is right and there are millions of examples. so the question is pretty trivial. a^x, even with a > e, *is* exponential growth. a^x = e^(x ln(a)). in other words all the a not being e does is change the growth constant, but it's still exponential.
From: Rod on 15 Jan 2010 14:19 "I.N. Galidakis" <morpheus(a)olympus.mons> wrote in message news:1263582569.638801(a)athprx04... > Robert wrote: > >> Well, for anyone that hasn't kill filed me, the "worthless caveat" is >> essential to the OP's question. >> >> Would >> >> as x->oo, for all t, f(x)/e^xt -> oo >> >> be an adequate definition for a function with faster than exponential >> growth? > > Yes. Sorry for the (created) confusion. > > As far as I am concerned, I mean anything like: f(x) ~ a^x, with a > e. > > Thanks again to all the responders. > -- > Ioannis > a^x = (exp(ln(a)))^x = exp(ln(a) x) = exp(b.x) so this is not faster than exp things like exp(a.x^2) are faster and exp(exp(x)) is really fast. Look up tetration for something super fast but I don't know any physical process related to it.
From: I.N. Galidakis on 15 Jan 2010 14:27 Robert wrote: > On 15 Jan, 19:09, "I.N. Galidakis" <morph...(a)olympus.mons> wrote: >> Robert wrote: >>> Well, for anyone that hasn't kill filed me, the "worthless caveat" is >>> essential to the OP's question. >> >>> Would >> >>> as x->oo, for all t, f(x)/e^xt -> oo >> >>> be an adequate definition for a function with faster than exponential >>> growth? >> >> Yes. Sorry for the (created) confusion. >> >> As far as I am concerned, I mean anything like: f(x) ~ a^x, with a > e. >> >> Thanks again to all the responders. >> -- >> Ioannis > > in that case androcles is right and there are millions of examples. so > the question is pretty trivial. > > a^x, even with a > e, *is* exponential growth. > > a^x = e^(x ln(a)). in other words all the a not being e does is change > the growth constant, but it's still exponential. Sorry, typo on my part. Sentence should read: > As far as I am concerned, I mean anything greater than: f(x) ~ a^x, with a > e. But I am interested in these cases as well. Any examples with a > e? -- Ioannis
From: Robert on 15 Jan 2010 14:36 On 15 Jan, 19:27, "I.N. Galidakis" <morph...(a)olympus.mons> wrote: > Robert wrote: > > On 15 Jan, 19:09, "I.N. Galidakis" <morph...(a)olympus.mons> wrote: > >> Robert wrote: > >>> Well, for anyone that hasn't kill filed me, the "worthless caveat" is > >>> essential to the OP's question. > > >>> Would > > >>> as x->oo, for all t, f(x)/e^xt -> oo > > >>> be an adequate definition for a function with faster than exponential > >>> growth? > > >> Yes. Sorry for the (created) confusion. > > >> As far as I am concerned, I mean anything like: f(x) ~ a^x, with a > e. > > >> Thanks again to all the responders. > >> -- > >> Ioannis > > > in that case androcles is right and there are millions of examples. so > > the question is pretty trivial. > > > a^x, even with a > e, *is* exponential growth. > > > a^x = e^(x ln(a)). in other words all the a not being e does is change > > the growth constant, but it's still exponential. > > Sorry, typo on my part. Sentence should read: > > > As far as I am concerned, I mean anything greater than: f(x) ~ a^x, with a > > > e. > > But I am interested in these cases as well. Any examples with a > e? > -- > Ioannis well. your question then doesn't have much physical meaning. as any real world exponential growth has a growth factor k (and an initial value A) A e^(k t) this scales the process to whatever units you are using. because of the scaling anything with a > e could, with a different scale, be written as a = e or even a < e. they are essentially the same.
From: I.N. Galidakis on 15 Jan 2010 14:48
Robert wrote: > On 15 Jan, 19:27, "I.N. Galidakis" <morph...(a)olympus.mons> wrote: >> Robert wrote: >>> On 15 Jan, 19:09, "I.N. Galidakis" <morph...(a)olympus.mons> wrote: >>>> Robert wrote: >>>>> Well, for anyone that hasn't kill filed me, the "worthless caveat" is >>>>> essential to the OP's question. >> >>>>> Would >> >>>>> as x->oo, for all t, f(x)/e^xt -> oo >> >>>>> be an adequate definition for a function with faster than exponential >>>>> growth? >> >>>> Yes. Sorry for the (created) confusion. >> >>>> As far as I am concerned, I mean anything like: f(x) ~ a^x, with a > e. >> >>>> Thanks again to all the responders. >>>> -- >>>> Ioannis >> >>> in that case androcles is right and there are millions of examples. so >>> the question is pretty trivial. >> >>> a^x, even with a > e, *is* exponential growth. >> >>> a^x = e^(x ln(a)). in other words all the a not being e does is change >>> the growth constant, but it's still exponential. >> >> Sorry, typo on my part. Sentence should read: >> >>> As far as I am concerned, I mean anything greater than: f(x) ~ a^x, with a > >> >> e. >> >> But I am interested in these cases as well. Any examples with a > e? >> -- >> Ioannis > > well. your question then doesn't have much physical meaning. as any > real world exponential growth has a growth factor k (and an initial > value A) > > A e^(k t) > > this scales the process to whatever units you are using. because of > the scaling anything with a > e could, with a different scale, be > written as a = e or even a < e. > > they are essentially the same. Excellent. So is there then general agreement that there are no *naturally occuring* (excluding Rod's examples on Wiki which I am not sure they qualify as *natural*) processes in nature which propagate faster than exponential? -- Ioannis |