Parabola... My friend Bill, if I can afford it, is a nice man ... High precision fraud ??? Better bet to the pocker. It's possible, it's possible ... I even think that I do not recognize myself in this bullshit, I really have to be silly ... Fortunately there is dramaturgy to console me Le mercredi 29 août 2018 à 03:36:03 UTC+1, Mike Hirschhorn <m.hirschhorn@unsw.edu.au> a écrit : #yiv6855577298 #yiv6855577298 -- P {margin-top:0;margin-bottom:0;}#yiv6855577298 ``parable''? From: mathfuneavesdroppers@googlegroups.com <mathfuneavesdroppers@googlegroups.com> on behalf of Bill Gosper <billgosper@gmail.com> Sent: 29 August 2018 12:29:35 To: françois mendzina essomba2 Cc: math-fun@mailman.xmission.com Subject: Re: Identity for Pi On Tue, Aug 28, 2018 at 2:23 PM françois mendzina essomba2 <m_essob@yahoo.fr> wrote: Hello, Here is a identity that j found, with a ratio between cosine and hyperbolic cosine. Pi=2*mu*x+2*sum((1/n)*(cos(mu*x)/cosh(alpha*x))^n*cosh(mu*n*x)*sin(alpha*n*x),n=1..infinity); This identity seems true for the following conditions: mu = alpha and x in [1/mu , 2/mu] The value of x, for which this identity is the most convergent is close to : x = (1.5)*(1/mu) Umm, would you believe π/(2𝜇 ?-) All smirking aside, for general t, e.g.,Pi == 2*t + 2*Sum[(Cosh[n*t]*(Cos[t]*Sech[t])^n*Sin[n*t])/n, {n, Infinity}] (t≠π/2)I don't recall anything like it. OtOH, at my age, I don't recall a lot of things.--rwg Note, as usual, the n=0 term would explain the prepended term if we took a limitfor 0/0. Maybe we should redefine Sum to always take limits. The values of x, verifying the identity seem to form a parable whose vertex corresponds to (1/2) * (1 / mu), which I found curious0 Why ??? -- You received this message because you are subscribed to the Google Groups "mathfuneavesdroppers" group. To unsubscribe from this group and stop receiving emails from it, send an email tomathfuneavesdroppers+unsubscribe@googlegroups.com. For more options, visit https://groups.google.com/d/optout.