hyp2exp

[robmio]

(09-05-2022 11:50 AM)Albert Chan Wrote:  

(09-04-2022 08:00 PM)parisse Wrote:  No, but it's fairly easy: subst(expression,exp,cosh+sinh)

Above exp2hyp expression might be "simplified" to sin/cos
Just keep in mind, e^(i*x) = cos(x) + i*sin(x) = cosh(i*x) + sinh(i*x)

      cos(x) = cosh(i*x)      i*sin(x) = sinh(i*x)
      cosh(x) = cos(i*x)      i*sinh(x) = sin(i*x)

CAS> subst(e^(i*x), exp, x -> cosh(x) + sinh(x))

cos(x) + i*sin(x)

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Same trick can be used for a "better" exp2trig

CAS> f := e^x - e^-x                                   // = 2*sinh(x)
CAS> exp2trig(f)      → e^x - e^-x
CAS> f(exp = (x -> cos(i*x) + sin(i*x)/i))     → -2*i*sin(i*x)

Or, in steps, since sinh/cosh tends to "simplify" to sin/cos

CAS> f(exp=cosh+sinh)(x=x/i)(x=x*i)          → -2*i*sin(i*x)

Hello, is there an algorithm that translates the logarithmic formulas of the inverse hyperbolic functions into the inverse hyperbolic functions represented with "acosh", "asinh", "atanh", etc.?
For instance:

e^x*ln(x+√(x^2+1))-ln(ln(√(x+1)*√(x-1)+x)) --instruction--> exp(x)*asinh(x)-ln(acosh(x))

Best regards, robmio.