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

• mind2work
In summary, to express sinh-1(x) as a power series in terms of powers of x, we can use the exponential series to expand sinhy, equate the coefficients, and solve for the power series of y in terms of x. This involves comparing the coefficients of each power of x and setting them equal to the corresponding coefficients in the y expansion. We can then use this power series to express sinh-1(x) as a series in powers of x.
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??

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.

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??

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.

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??

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??

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.

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:
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...

You're probably right.
Note that you forgot the factorial sign in your expansion of sinh(x).
When I posted my reply, I forgot that those !'s should be included.

## 1. What is the formula for expanding y=sinh-1(x) in terms of x using inversion of power series method?

The formula for expanding y=sinh-1(x) in terms of x using inversion of power series method is:
y = x + (1/6)x^3 + (3/40)x^5 + (5/112)x^7 + (35/1152)x^9 + ...

## 2. How does the inversion of power series method work for expanding y=sinh-1(x) in terms of x?

The inversion of power series method involves using the Taylor series expansion of sinh-1(x) to find its inverse function, x = sinh(y). This inverse function can then be expressed as a power series in terms of y, which is then substituted back into the original equation to obtain the expansion of y=sinh-1(x) in terms of x.

## 3. What is the significance of using the inversion of power series method for expanding y=sinh-1(x) in terms of x?

The inversion of power series method allows for a more accurate and efficient way of expanding y=sinh-1(x) in terms of x, as compared to other methods such as using the inverse hyperbolic function identities. It also allows for a wider range of values to be used for x, as the convergence of the series is not limited by the range of x values.

## 4. Can the expansion of y=sinh-1(x) in terms of x using inversion of power series method be used to approximate other functions?

Yes, the inversion of power series method can be applied to other functions as well, as long as they have an inverse that can be expressed as a power series. This method is commonly used in fields such as physics, engineering, and finance to approximate complex functions and equations.

## 5. Are there any limitations to using the inversion of power series method for expanding y=sinh-1(x) in terms of x?

One limitation of this method is that it may not converge for all values of x. It is important to check the convergence of the series before using this method. Additionally, the accuracy of the expansion may decrease as the number of terms in the series increases, so it is important to choose an appropriate number of terms for the desired level of accuracy.

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