how can I work out a taylor series expansion of arcsinh( x) using taylor series expansion of sinh( x)?

Answer 1

Not sure how to answer your specific question (finding one series from a knowledge of the other), but this is the path to finding the Taylor expansion of arcSinh from knowing the differential properties of Sinh...

General inverse theory:
Let the function y = F(x); let it's inverse be G such that

G(y) = x

Differentiate...

(dG/dy )(dy/dx) = 1

and again...

(d^2G/dy^2)(dy/dx)^2 + (dG/dy)(d^2y/dx^2) = 0

Let's get specific:

let y = Sinh(x)
dy/dx = Cosh(x) = Sqrt(y^2+1)
d^2y/dx^2 = Sinh(x) = y

Substitute in...
(d^2G/dy^2)(y^2+1) + (dG/dy)(y) = 0

Assume that
G = sum(a_n * y^n) over n = 0 to infinity (I won't say this again...)

Then dG/dy = sum({a_[n+1] * y^n}/n!)
d^2G/dy^2 = sum({a_[n+2] * y^n}/n!)

so
(y^2+1)*sum({a_[n+2] * y^n}/n!) + y*sum({a_[n+1] * y^n}/n!) = 0

multiply out...
sum({a_[n+2] * y^[n+2]}/n!) + sum({a_[n+2] * y^n}/n!) + sum({a_[n+1] * y^[n+1]}/n!) = 0

taking a couple of terms out of the sums in order to tidy up...
a_2 + a_3 * y + a_1 * y + sum({a_[n+2]/n! + a_[n+4]/(n+2)! + a_[n+2]/(n+1)!}y^[n+2]) = 0

equate coefficients of y^n...
a_2 = 0
a_1+a_3 = 0
a_[n+2]/n! + a_[n+4]/(n+2)! + a_[n+2]/(n+1)! = 0

we now have a recurrence relationship between a_[n+2] and a_[n+4]
since a_2 = 0, all coefficients of even index must be zero

From way back at the start, we have
(dG/dy )(dy/dx) = 1
i.e
(dG/dy ) = 1/(dy/dx) = 1/Cosh(x) = 1/Sqrt(y^2+1) in our case

dG/dy = sum({a_[n+1] * y^n}/n!) = 1/Sqrt(y^2+1)

and then let y = 0, resulting in

a_1 = 1

So we have the value of a_1, an equation relating a_1 and a_3, a recurrence relationship between a_[n+2] and a_[n+4], and the knowledge that all terms a_n for even n are zero. The rest is (profoundly tedious) algebra...

Answer 2

Not possible.

Try this:

Integral [1 + x^2](^-1/2).dx = sinh(^-1){x}

Find the binomial series expansion for [1 + x^2](^ -1/2)

and proceed by term by term integration.

Answer 3

Sounds messy, seek a arcsinh(x) series somewhere.