# Expansion of y=sinh-1(x) in terms of x using inversion of power series method

1. Nov 1, 2006

### mind2work

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 doesn't make sense it's because I'm confused, as you can tell. I expanded the sinhy and expressed it as a power series in y. I need to invert the x=y+(1/3)y^3+(1/5)y^5+... to then express it in terms of power terms of x. Does this sound right??

2. Nov 1, 2006

### arildno

Now, assume there exists a power series of y in terms of x, i.e:
$$y=\sum_{i=0}^{\infty}a_{i}x^{i}$$
Insert this in the y's places, compute&enjoy the first a's.

3. Nov 1, 2006

### mind2work

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??

4. Nov 1, 2006

### arildno

You're right.
Compare the coefficients of each power of x.
When you insert the power series for y, the right-hand side must be IDENTICALLY EQUAL to x, that gives you the equations to determine the a's.

5. Nov 1, 2006

### mind2work

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??

6. Nov 1, 2006

### mind2work

Ok, do you mean compare the coefficients of the y terms from the expansion of the sinhy to the coefficients of the expansion you just talked about??

7. Nov 1, 2006

### mind2work

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.

8. Nov 2, 2006

### arildno

First of all, $a_{0}=0$, since there isn't any constant term on your left-hand side.
Thus, the only y-power containing a $x^{1}$-term on the right-hand side is the $y^{1}$-term.
Thus, we get $a_{1}=1$, by comparing coefficients.

Note now that the only place where there is a $x^{2}$ term is in the $y^{1}$-term, thus $a_{2}=0$
Inductively, it follows that all even coefficients must be zero, that is we may write:
$$y=\sum_{i=0}^{\infty}a_{2i+1}x^{2i+1}[/itex] Now, an $x^{3}$-term appears in both the y-and $y^{3}$-term, and the coefficients must satisfy the equation: [tex]0=a_{3}+\frac{a_{0}^{3}}{3}=a_{3}+\frac{1}{3}\to{a}_{3}=-\frac{1}{3}[/itex] To calculate $a_{5}$, there are three terms appearing, from the $y,y^{3},y^{5}$-terms, respectively. This yields the equation: [tex]a_{5}+\frac{1}{3}*3a_{1}^{2}*a_{3}+\frac{a_{1}^{5}}{5}=0[/itex] or: [tex]a_{5}-\frac{1}{3}+\frac{1}{5}=0\to{a}_{5}=\frac{2}{15}$$
Thus, to fifth order, we have:
$$y=x-\frac{x}{3}+\frac{2}{15}x^{5}$$

Now, since I haven't anything better to do, here's the equation by which we may determine $a_{7}$:
$$a_{7}+\frac{1}{3}(3*a_{1}^{2}*a_{5}+3*a_{3}^{2}*a_{1})+\frac{1}{5}(5*a_{1}^{4}*a_{3})+\frac{1}{7}a_{1}^{7}=0$$
Do you get the pattern now?

Last edited: Nov 2, 2006
9. Nov 2, 2006

### mind2work

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...

10. Nov 3, 2006

### arildno

You're probably right.