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...