[math-fun] I think, I'm Siamese
hello... tanh(x) =-csch(2*x)+1/(-csch(4*x)+1/(-csch(8*x)+1/(-csch(16*x)+1/(-csch(32*x)+1/(-csch(64*x)+1/(-csch(128*x)+1/(…))))))); approximation. tanh(x)=-csch(2*x)+1/(-csch(4*x)+1/(-csch(8*x)+1/(-csch(16*x)+1/(-csch(32*x)+1/(-csch(64*x)+1/(-csch(128*x)+1/(1))))))); FME...
Cool! Any chance for a nice approximation of sinh(x) and/or asinh(x) ? At 07:38 AM 1/2/2018, françois mendzina essomba2 via math-fun wrote:
hello... tanh(x) =-csch(2*x)+1/(-csch(4*x)+1/(-csch(8*x)+1/(-csch(16*x)+1/(-csch(32*x)+1/(-csch(64*x)+1/(-csch(128*x)+1/(Â ))))))); approximation. tanh(x)=-csch(2*x)+11/(-csch(4*x)+1/(-csch(8*x)+1/(-csch(16*x)+1/(-csch(32*x)+1/(-csch(64*x)+1/(-csch(128*x)+1/(1))))))); FME...
The formula presented by pr. Rich comes to enlighten my understanding of this continuous fraction. tanh (x) + csch (2x) = coth (2x) = 1 / tanh (2x) And it brings me to give a general writing for this type of continuous fraction: f (x) = 1 / f (kx) + h (x) where f and h are functions and k any real. It would be enough then to find a real function which is written on this form to obtain a continuous fraction. And more generally, we will have: f (x) = a / f (kx) + b, a and b being constants or variables. I derive from this result an approximate value of sinh (x) carried by the function a / f (kx) = cosh (3 * x) / sinh (x): sinh(x)=-1/(2*tanh(x/2))+cosh(3*x/2)/(-1/tanh(x/4)+2*cosh(3*x/4)/(-1/tanh(x/8)+2*cosh(3*x/8)/(-1/tanh(x/16)+cosh(3*x/16)/sinh(x/16)))); for the tangent function here an approximation of the fraction with k = 7 ... tan(x)=csc(2*x)-1/(csc(4*x)-1/(csc(8*x)-1/(csc(16*x)-1/(csc(32*x)-1/(csc(64*x)-1/(csc(128*x)-1/(tan(128*x)))))))); and more... tan(x) =csc(2*x)-1/(csc(4*x)-1/(csc(8*x)-1/(csc(16*x)-1/(csc(32*x)-1/(csc(64*x)-1/(csc(128*x)-1/(…1/(csc(2^(k)*x)-1/tan(2^(k)*x))))))))); Le Mercredi 3 janvier 2018 1h53, Henry Baker <hbaker1@pipeline.com> a écrit : Cool! Any chance for a nice approximation of sinh(x) and/or asinh(x) ? At 07:38 AM 1/2/2018, françois mendzina essomba2 via math-fun wrote:
hello... tanh(x) =-csch(2*x)+1/(-csch(4*x)+1/(-csch(8*x)+1/(-csch(16*x)+1/(-csch(32*x)+1/(-csch(64*x)+1/(-csch(128*x)+1/(…))))))); approximation. tanh(x)=-csch(2*x)+11/(-csch(4*x)+1/(-csch(8*x)+1/(-csch(16*x)+1/(-csch(32*x)+1/(-csch(64*x)+1/(-csch(128*x)+1/(1))))))); FME...
On Tue, Jan 2, 2018 at 2:58 PM, françois mendzina essomba2 <m_essob@yahoo.fr
wrote:
hello...
tanh(x) =-csch(2*x)+1/(-csch(4*x)+1/(-csch(8*x)+1/(-csch(16*x)+1/(- csch(32*x)+1/(-csch(64*x)+1/(-csch(128*x)+1/(…)))))));
approximation.
tanh(x)=-csch(2*x)+1/(-csch(4*x)+1/(-csch(8*x)+1/(-csch(16* x)+1/(-csch(32*x)+1/(-csch(64*x)+1/(-csch(128*x)+1/(1)))))));
FME...
Dr. Essomba, perhaps because I am old and rusty, I don't recall ever seeing such a result. The convergence rate of a continued fraction is an increasing function of the (magnitudes of the) denominators, which vanishes along with them. The only way yours makes sense numerically is with that 1 you have tacked onto the tail. The value is chaotically sensitive to replacing that "1" with a variable. Mathematica: Coth@x == ContinuedFractionK[1, -Csch[2^k x], {k, \[Infinity]}] ContinuedFractionK::div: The continued fraction \!\(\*UnderoverscriptBox[\(\[ContinuedFractionK]\), \(k = 1\), \(\[Infinity]\)]\*FractionBox[\(1\), \(-Csch[\*SuperscriptBox[\(2\), \(k\)]\ x]\)]\) does not converge. And yet, with that "1", the "convergence", if you could call it that, is fast and furious. Have you ever seen such continued fractions in the literature? --Bill Gosper
Dear Pr. Gosper, following our last exchange. You did well to recall the very unusual nature of this continuous fraction that manages to be efficient in terms of calculations despite a theoretical convergence rate of zero. I admit that the formula that I transmitted to you in relation to tan (x) always intrigues me particularly. Le Mardi 2 janvier 2018 23h58, françois mendzina essomba2 <m_essob@yahoo.fr> a écrit : hello... tanh(x) =-csch(2*x)+1/(-csch(4*x)+1/(-csch(8*x)+1/(-csch(16*x)+1/(-csch(32*x)+1/(-csch(64*x)+1/(-csch(128*x)+1/(…))))))); approximation. tanh(x)=-csch(2*x)+1/(-csch(4*x)+1/(-csch(8*x)+1/(-csch(16*x)+1/(-csch(32*x)+1/(-csch(64*x)+1/(-csch(128*x)+1/(1))))))); FME...
recalling the formula tanh(x) + csch(2x) = coth(2x) = 1/tanh(2x) might help here. ----- Quoting françois mendzina essomba2 via math-fun <math-fun@mailman.xmission.com>:
Dear Pr. Gosper, following our last exchange. You did well to recall the very unusual nature of this continuous fraction that manages to be efficient in terms of calculations despite a theoretical convergence rate of zero. I admit that the formula that I transmitted to you in relation to tan (x) always intrigues me particularly.
Le Mardi 2 janvier 2018 23h58, françois mendzina essomba2 <m_essob@yahoo.fr> a écrit :
hello...
tanh(x) =-csch(2*x)+1/(-csch(4*x)+1/(-csch(8*x)+1/(-csch(16*x)+1/(-csch(32*x)+1/(-csch(64*x)+1/(-csch(128*x)+1/(?)))))));
approximation.
tanh(x)=-csch(2*x)+1/(-csch(4*x)+1/(-csch(8*x)+1/(-csch(16*x)+1/(-csch(32*x)+1/(-csch(64*x)+1/(-csch(128*x)+1/(1))))))); FME...
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Bill Gosper -
françois mendzina essomba2 -
Henry Baker -
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