[math-fun] Identity for Pi
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) 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 ???
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 limit for 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 ???
``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<mailto: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 limit for 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 to mathfuneavesdroppers+unsubscribe@googlegroups.com<mailto:mathfuneavesdroppers+unsubscribe@googlegroups.com>. For more options, visit https://groups.google.com/d/optout.
"parabola" perhaps! WFL On 8/29/18, Mike Hirschhorn <m.hirschhorn@unsw.edu.au> wrote:
``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<mailto: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 limit for 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 to mathfuneavesdroppers+unsubscribe@googlegroups.com<mailto:mathfuneavesdroppers+unsubscribe@googlegroups.com>. For more options, visit https://groups.google.com/d/optout. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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.
Bill, The Good Samaritan :D On 29 Aug 2018, at 03:29, Bill Gosper wrote: <snip>
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 <snip> The values of x, verifying the identity seem to form a parable whose vertex
<snip>
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π == 2t + 2Sum[(Cosh[n*t]*(Cos[t]*Sech[t])^n*Sin[n*t])/n, {n,∞}], (1≤t<2 ?) Is anybody playing with this? I find it incredible. Is it one of those Borwein or Zagier type high precision frauds? Bibasic telescopy? Am I missing something completely obvious? Inconclusive plotting suggests it holds in a region of the complex plane shaped like a football silhouette centered at t = π/2 + 0i of width > 1 and height < i. (DIY or stay tuned.) —rwg On Tue, Aug 28, 2018 at 7:29 PM Bill Gosper <billgosper@gmail.com> wrote:
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 limit for 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 ???
I suspect it’s true, not fraudulent. Try: p = 42; Simplify[D[ 2 t + 2 Sum[(Cosh[n t] (Cos[t] Sech[t])^n Sin[n t])/n, {n, p}], {t, p}] /. t -> Pi/2] -Veit
On Aug 29, 2018, at 1:14 PM, Bill Gosper <billgosper@gmail.com> wrote:
π == 2t + 2Sum[(Cosh[n*t]*(Cos[t]*Sech[t])^n*Sin[n*t])/n, {n,∞}], (1≤t<2 ?)
Is anybody playing with this? I find it incredible. Is it one of those Borwein or Zagier type high precision frauds? Bibasic telescopy? Am I missing something completely obvious? Inconclusive plotting suggests it holds in a region of the complex plane shaped like a football silhouette centered at t = π/2 + 0i of width > 1 and height < i. (DIY or stay tuned.) —rwg
participants (6)
-
Bill Gosper -
David Makin -
françois mendzina essomba2 -
Fred Lunnon -
Mike Hirschhorn -
Veit Elser