# using sinh(x) to find series representation of arcsinh(x)

From "Complex Variables Demystified", 2008, page 102:

Given:

$$sinh(z)=\frac{e^z-e^{-z}}{2}$$

find the series representation for arcsinh(x).

Solution:

(1) The Maclaurin theorem can be used to write a series representation of sinh(x). This is given by:

$$sinh(x)=x+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+\cdots=\sum_{n=0}^{\infty}{\frac{1}{(2n+1)!}x^{2n+1}}$$

(2) The inverse will have some series expansion which we write as:

$$\sinh^{-1}x=b_0+b_1x+b_2x+b_3x+\cdots$$

(3) We label the coefficents in the series expansion of sinh by:

$$a_j$$

We find that:

$$b_0=a_0=0$$

$$b_1=\frac{1}{a_1}=1$$

$$b_2=-\frac{a_2}{a_3}=0$$

$$b_3=\frac{1}{\left(a_1\right)^5}\left(2\left(a_2\right)^2-a_1a_3\right)=-\frac{1}{6}$$

(4) Therefore it follows that:

$$arcsinh(x)=x-\frac{1}{6}x^3+\cdots$$

My Question is as follows:

What are they doing in (3) to equate an and bn coefficients from sinh and arcsinh?

• This is time consuming and does not give a patter to the coefficients $b_n$. Better argue that derivative of $\sinh x$ is $\cos x=\sqrt{1+\sinh^2x}$ so that derivative of $\sinh^{-1}x$ is $(1+x^2)^{-1/2}$. Use binomial theorem and then integrate term by term. – Paramanand Singh Nov 11 '18 at 14:53

Composition. They know that $$\sinh^{-1}(\sinh x)=x$$. But this means that$$b_0+b_1\left(a_0+a_1x+a_2x^2+\cdots\right)+b_2\left(a_0+a_1x+a_2x^2+\cdots\right)^2+\cdots=x.$$Therefore, expanding this you get:
• $$b_0=0$$ (this uses the fact that $$a_0=0$$);
• $$b_1a_1=1$$ and therefore $$b_1=\dfrac1{a_1}$$;
• $$b_1a_2+b_2{a_0}^2+2b_2a_0a_1=0$$ and, since $$a_0=a_2=0$$, this is just the equality $$0=0$$.