tag:blogger.com,1999:blog-20067416.post3497665668396004621..comments2023-12-21T06:35:36.624-05:00Comments on Recursivity: A Functional EquationUnknownnoreply@blogger.comBlogger16125tag:blogger.com,1999:blog-20067416.post-33324035572880226002012-02-28T03:30:21.783-05:002012-02-28T03:30:21.783-05:00I had occasion to solve a somewhat similar equatio...I had occasion to solve a somewhat similar equation in January. The following technique failed for mine, but seems to have worked for yours!<br /><br />But maybe the Maple won't render well...<br /><br />> g(t) = f(2^t);<br /> / t\<br /> g(t) = f\2 /<br />> g(t+1) = 2^t*g(t);<br /> t <br /> g(t + 1) = 2 g(t)<br />> rsolve(%, g(t));<br /> (1/2 t (t - 1)) <br /> 2 g(0)<br />> eval(%, t = log[2](z));<br /> / /ln(z) \\ <br /> |ln(z) |----- - 1|| <br /> | \ln(2) /| <br /> |-----------------| <br /> \ 2 ln(2) / <br /> 2 g(0)<br />> simplify(%);<br /> / -ln(z) + ln(2)\ <br /> |- --------------| <br /> \ 2 ln(2) / <br /> z g(0)<br />> f(1)*z^(ln(z)/(2*ln(2))-1/2);<br /> / ln(z) \<br /> |------- - 1/2|<br /> \2 ln(2) /<br /> f(1) z <br />> (eval(%, z = 2*z))/%;<br /> /ln(2 z) \<br /> |------- - 1/2|<br /> \2 ln(2) /<br /> (2 z) <br /> --------------------<br /> / ln(z) \ <br /> |------- - 1/2| <br /> \2 ln(2) / <br /> z <br /><br />simplify((2*z)^((1/2)*ln(2*z)/ln(2)-1/2)/z^((1/2)*ln(z)/ln(2)-1/2));<br /> z<br />>Rob Chttps://www.blogger.com/profile/17415636282470554460noreply@blogger.comtag:blogger.com,1999:blog-20067416.post-67920942142409662522012-02-20T11:02:38.520-05:002012-02-20T11:02:38.520-05:00JimV
Your latest function satisfies
f(2z) = 2 zf(z...JimV<br />Your latest function satisfies<br />f(2z) = 2 zf(z)<br />(there is an extra 2)<br />so it's not correct.<br /><br />By the way, the question really means that f should be an analytic extension of the function which assigns the number 2^{n(n-1)/2} to the integer 2^n for all nonnegative integers n.Takis Konstantopouloshttps://www.blogger.com/profile/14675216467783238403noreply@blogger.comtag:blogger.com,1999:blog-20067416.post-75437479350143770762012-02-18T12:09:02.882-05:002012-02-18T12:09:02.882-05:00You're wrong, onebrow. For example, f(2) = 1,...You're wrong, onebrow. For example, f(2) = 1, but your formula gives 2.Jeffrey Shallithttps://www.blogger.com/profile/12763971505497961430noreply@blogger.comtag:blogger.com,1999:blog-20067416.post-85630222831489600322012-02-18T10:45:00.056-05:002012-02-18T10:45:00.056-05:00Some of these seem very complex.
I just have 2^[n...Some of these seem very complex.<br /><br />I just have 2^[n(n+1)/2)]One Browhttps://www.blogger.com/profile/11938816242512563561noreply@blogger.comtag:blogger.com,1999:blog-20067416.post-90978365351711312792012-02-13T11:22:57.480-05:002012-02-13T11:22:57.480-05:00After double checking I take back my earlier comme...After double checking I take back my earlier comment too.<br /><br />In the end I got what VKS got. <br /><br />(giving credit to a search for 1,1,2,8,64,1024 in OEIS)David Swartnoreply@blogger.comtag:blogger.com,1999:blog-20067416.post-44920438379091458822012-02-13T01:37:41.051-05:002012-02-13T01:37:41.051-05:00No it isn't. I shouldn't do this late at ...No it isn't. I shouldn't do this late at night without checking my work before posting - exponents add, not multiply;<br /><br />f(2^n) = 2^(n-1)*2^(n-2)* ,,, 1 <br /><br />= 2^[(n-1)*(n-2)/2]<br /><br />= 2^{Gamma(n)/[2*Gamma(n-2)]}<br /><br />so I think what you want is<br /><br />f(z) = 2^{Gamma(x)/[2*Gamma(x-2)]}<br /><br />where x = ln(z)/ln(2)JimVhttps://www.blogger.com/profile/10198704789965278981noreply@blogger.comtag:blogger.com,1999:blog-20067416.post-654143683960925752012-02-13T01:00:34.739-05:002012-02-13T01:00:34.739-05:00Dang, what I actually meant to say is f(2^n) = 2^G...Dang, what I actually meant to say is f(2^n) = 2^Gamma(n), where Gamma(n) = (n-1)!, but you want f(z), where z=2^x,<br />or x=ln(z)/ln(2), so:<br /><br />f(z) = 2^Gamma(ln(z)/ln(2)) - that's my final answer.JimVhttps://www.blogger.com/profile/10198704789965278981noreply@blogger.comtag:blogger.com,1999:blog-20067416.post-69706556902290959582012-02-12T22:31:54.377-05:002012-02-12T22:31:54.377-05:00Whoops, I meant 2 raised to the power Gamma(z), th...Whoops, I meant 2 raised to the power Gamma(z), that is.JimVhttps://www.blogger.com/profile/10198704789965278981noreply@blogger.comtag:blogger.com,1999:blog-20067416.post-13535989495949431292012-02-12T22:00:01.756-05:002012-02-12T22:00:01.756-05:002 raised to the power Gamma(n).2 raised to the power Gamma(n).JimVhttps://www.blogger.com/profile/10198704789965278981noreply@blogger.comtag:blogger.com,1999:blog-20067416.post-73243518976567967012012-02-11T20:56:22.154-05:002012-02-11T20:56:22.154-05:00simplified a little:
z^(ln(z)/ln(4) + 1/2)simplified a little:<br /><br />z^(ln(z)/ln(4) + 1/2)Johnhttps://www.blogger.com/profile/10876775111703252840noreply@blogger.comtag:blogger.com,1999:blog-20067416.post-65006814184424737012012-02-11T20:32:12.417-05:002012-02-11T20:32:12.417-05:00z^((1+ln(z)/ln(2))/2)z^((1+ln(z)/ln(2))/2)Johnhttps://www.blogger.com/profile/10876775111703252840noreply@blogger.comtag:blogger.com,1999:blog-20067416.post-6147861821792829552012-02-11T17:51:13.398-05:002012-02-11T17:51:13.398-05:00f(x) = Sqrt(x) ^ [ log_2(x) - 1 ]
which is the ...f(x) = Sqrt(x) ^ [ log_2(x) - 1 ] <br /><br />which is the same as VKS's function, slightly rewritten.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-20067416.post-12617999906935459392012-02-11T15:38:56.841-05:002012-02-11T15:38:56.841-05:00Let g(z) = f(2^z), so that g(z) = 2^(z-1) * g(z-1)...Let g(z) = f(2^z), so that g(z) = 2^(z-1) * g(z-1) and g(0) = 1. Then g(z) = 2^(z*(z-1)/2), and f(z) = g(log_2 z).Pavel Bakhilauhttps://www.blogger.com/profile/07744199542325433927noreply@blogger.comtag:blogger.com,1999:blog-20067416.post-18807684866497203902012-02-11T09:57:48.808-05:002012-02-11T09:57:48.808-05:00Two exponentiated by itself log(n) times seems to ...Two exponentiated by itself log(n) times seems to fit.David Swartnoreply@blogger.comtag:blogger.com,1999:blog-20067416.post-72247093965206107652012-02-11T08:40:10.115-05:002012-02-11T08:40:10.115-05:00Let g(x) = x(x-1)/2 , so that g(x+1) = x + g(x).
...Let g(x) = x(x-1)/2 , so that g(x+1) = x + g(x).<br /><br />Then if we define f(x) = 2^g(log_2(x)), then f(2x) = x*f(x) and f(1) = 1 as required.VKShttps://www.blogger.com/profile/18165497929233440473noreply@blogger.comtag:blogger.com,1999:blog-20067416.post-87917450064809263642012-02-11T08:06:33.817-05:002012-02-11T08:06:33.817-05:00is exp( log(z)*log(z/2) / 2log(2)) reasonable enou...is exp( log(z)*log(z/2) / 2log(2)) reasonable enough?fudohttps://www.blogger.com/profile/09004773966721427129noreply@blogger.com