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Secant series

  1. Jul 21, 2010 #1
    I understand how this works:

    [tex]\cos x = \frac{1}{0!} - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \frac{x^{10}}{10!} + \ldots[/tex]

    But what about this?

    [tex]\frac{1}{\cos x} = \frac{1}{0!} + \frac{x^2}{2!} + \frac{5x^4}{4!} + \frac{61x^6}{6!} + \frac{1385x^8}{8!} + \frac{50521x^{10}}{10!} + \ldots[/tex]

    Is there a way to take the reciprocal of an infinite series or is it necessary to take subsequent derivatives of secant and write the Taylor expansion that way?

    Thanks,
    Unit
     
  2. jcsd
  3. Jul 21, 2010 #2
    Suppose [itex]f(x)[/itex] is analytic in a domain containing [itex]x_{0}[/itex] and [itex]f(x_{0}) \neq 0[/itex]. Then, [itex]1/f(x)[/itex] is also analytic. Both of these functions have Taylor series expansions around [itex]x_{0}[/itex]:

    [tex]
    f(x) = \sum_{n = 0}^{\infty}{a_{n} \, (x - x_{0})^{n}}, \; a_{0} \neq 0
    [/tex]


    [tex]
    \frac{1}{f(x)} = \sum_{n = 0}^{\infty}{b_{n} \, (x - x_{0})^{n}}
    [/tex]

    Multiplying the two series term by term, we get:

    [tex]
    \sum_{n = 0}^{\infty}{\left(\sum_{m = 0}^{n}{a_{n - m} \, b_{m}}\right) \, (x - x_{0})^{n}} = 1
    [/tex]

    from where, we get the following conditions for the coefficients [itex[\{b_{n}\}[/itex]:

    [tex]
    a_{0} \, b_{0} = 1 \; \Rightarrow \; b_{0} = \frac{1}{a_{0}}
    [/tex]

    [tex]
    \sum_{m = 0}^{n}{a_{n - m} \, b_{m}} = 0, \; n \ge 1 \; \Rightarrow \; \sum_{m = 1}^{n}{a_{n - m} \, b_{m}} = -\frac{a_{n}}{a_{0}}
    [/tex]

    The matrix of this system is:

    [tex]
    A = \left(\begin{array}{ccccc}
    a_{0} & 0 & \ldots & 0 & \ldots \\

    a_{1} & a_{0} & \ldots & 0 & \ldots \\

    \ldots & & & & \\

    a_{n - 1} & a_{n - 2} & \ldots & a_{0} & \ldots \\

    \ldots & & & &
    \end{array}\right)
    [/tex]

    It is an infinitely dimensional lower triangular matrix. If we restrict it to the first n rows, then the determinant is simply [itex]\det A_{n} = a^{n}_{0} \neq 0[/itex], so the matrix is never singluar and the inverse always exists. However, it is pretty difficult to find an analytical form for it in the general case.
     
  4. Aug 17, 2010 #3
    Dickfore is correct about the difficulties in finding inverse of infinity series. I have written a paper about finding power series of tan x + sec x. And I think Unit may take a look in it. For the secant series, it just corresponds to the even power of the series, as tan x is odd, and sec x is even.

    http://www.voofie.com/content/117/an-explicit-formula-for-the-euler-zigzag-numbers-updown-numbers-from-power-series/" [Broken]
     
    Last edited by a moderator: May 4, 2017
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