1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Nov 1, 2006 #1
    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. jcsd
  3. Nov 1, 2006 #2

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Now, assume there exists a power series of y in terms of x, i.e:
    [tex]y=\sum_{i=0}^{\infty}a_{i}x^{i}[/tex]
    Insert this in the y's places, compute&enjoy the first a's.
     
  4. Nov 1, 2006 #3
    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??
     
  5. Nov 1, 2006 #4

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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.
     
  6. Nov 1, 2006 #5
    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??
     
  7. Nov 1, 2006 #6
    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??
     
  8. Nov 1, 2006 #7
    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.
     
  9. Nov 2, 2006 #8

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

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

    Note now that the only place where there is a [itex]x^{2}[/itex] term is in the [itex]y^{1}[/itex]-term, thus [itex]a_{2}=0[/itex]
    Inductively, it follows that all even coefficients must be zero, that is we may write:
    [tex]y=\sum_{i=0}^{\infty}a_{2i+1}x^{2i+1}[/itex]

    Now, an [itex]x^{3}[/itex]-term appears in both the y-and [itex]y^{3}[/itex]-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 [itex]a_{5}[/itex], there are three terms appearing, from the [itex]y,y^{3},y^{5}[/itex]-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}[/tex]
    Thus, to fifth order, we have:
    [tex]y=x-\frac{x}{3}+\frac{2}{15}x^{5}[/tex]

    Now, since I haven't anything better to do, here's the equation by which we may determine [itex]a_{7}[/itex]:
    [tex]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[/tex]
    Do you get the pattern now?
     
    Last edited: Nov 2, 2006
  10. Nov 2, 2006 #9
    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...
     
  11. Nov 3, 2006 #10

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Expansion of y=sinh-1(x) in terms of x using inversion of power series method
Loading...