I tried to do this and got coefficients for the x terms, if my memory serves me right, sinh-1(x)=x-1/6(x)+3/40(x^3). I turned the homework in this morning. I am going to take your notes here and rework the problem when I get it back. I know this will be on our next test. Thanks a lot for the help...
arildno,
Let me try it and see what happens. I'm trying to mentally visualize too far ahead. It will probably be more obvious when I expand everything. Check back later this evening if you can and don't mind.
Ok, another brain cramp. Once I put the power series in for each y, expand, and sum I get and expression like x=(a1)*x-term+(a3)*x-term^3+...How am I going to compare coefficients with one x on the left side and many x-terms on the right side??
arildno,
Thanks for the help. That seems too easy. So I would insert the power series you mentioned in the place of each y. For the y^3 term, for example, I would cube the power series. I would assume you must expand the power series to get a collection of x terms and sum them up??
I need to express the sinh-1(x) as a power series in terms of powers of x. I have written the expression as x=sinhy and expanded the sinhy using the exponential series to give x = y+(1/3)y^3+(1/5)y^5+... I guess I need to expand the y=sinh-1(x) and compare or equate the coefficients. If this...