Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Are there power series for X^2

  1. May 1, 2010 #1
    title is pretty much the jist of it.
  2. jcsd
  3. May 1, 2010 #2


    User Avatar
    Homework Helper

    x^2 is pretty much the power series for x^2. A power series (about zero) is defined by

    [tex]f(x) = \sum_{n=0}^\infty a_n x^n[/tex]

    x^2 is just a power series where all of the a_n except a_2 are zero.

    Of course, that's not the whole story. That's for a power series about the point x = 0. About some arbitrary point x_0 a power series is defined by

    [tex]f(x) = \sum_{n=0}^\infty a_n (x-x_0)^n[/tex]

    In this case then we can write a slightly non-trivial power series for x^2. Noting that x = x - x_0 + x_0, and expanding (x-x_0+x_0)^2 in terms of (x-x_0) will get you the power series.
  4. May 2, 2010 #3


    User Avatar
    Science Advisor

    The power series for [itex]x^2[/itex] about x= 0 (its MacLaurin series) is just [itex]0+ 0x+ 1x^2+ 0x^2+ \cdot\cdot\cdot= x^2[/itex] itself. To find its power series about x= a (the general Taylor's series), let u= x- a. Then the power series for x, about x= a, is the power series for u about u= a- a= 0, [itex]u^2[/itex]. And since u= x- a, the power series for [itex]x^2[/itex] about x= a is [itex](x- a)^2= x^2- 2ax+ a^2[/itex]
    Last edited by a moderator: May 2, 2010
  5. May 2, 2010 #4
    Any polynomial in the form:

    [tex]a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x + a_0[/tex]

    Its own unique power series (around 0, in this case).
  6. May 2, 2010 #5


    User Avatar
    Homework Helper

    Careful, technically what you've given is the power series of [itex](x-a)^2[/itex] about the point x = 0. It is not the power series of x^2 about x = a. That should look like const + A*(x-a) + B*(x-a)^2.

    To the OP, to get the power series of x^2 about the point x = a, take the form HallsofIvy gives and rearrange it to give

    [tex]x^2 = -a^2 + 2ax + (x-a)^2.[/tex]

    Now, fiddle with the first to terms of the right hand side of the equation to make them look a constant plus another constant*(x-a).
  7. May 2, 2010 #6
    You could also do e^x and then e^(2lnx) as a series, which would converge to x^2.
  8. May 2, 2010 #7
    So that would be a power series in log x, and not a power series in x.
    I guess the OP was not specific, so this, too, answers the question.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook