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



find the series representation for arcsinh(x).


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


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


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


We find that:





(4) Therefore it follows that:


My Question is as follows:

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

  • $\begingroup$ 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. $\endgroup$ – 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 $0=0$.

And so on…


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