Circle Series Reciprocal: Taking the Reciprocal of an Infinite Series

[Unit]

I understand how this works:

$$\cos x = \frac{1}{0!} - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \frac{x^{10}}{10!} + \ldots$$

But what about this?

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

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

 

[Dickfore]

Suppose $f(x)$ is analytic in a domain containing $x_{0}$ and $f(x_{0}) \neq 0$. Then, $1/f(x)$ is also analytic. Both of these functions have Taylor series expansions around $x_{0}$:

$$f(x) = \sum_{n = 0}^{\infty}{a_{n} \, (x - x_{0})^{n}}, \; a_{0} \neq 0$$$$\frac{1}{f(x)} = \sum_{n = 0}^{\infty}{b_{n} \, (x - x_{0})^{n}}$$

Multiplying the two series term by term, we get:

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

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

$$a_{0} \, b_{0} = 1 \; \Rightarrow \; b_{0} = \frac{1}{a_{0}}$$

$$\sum_{m = 0}^{n}{a_{n - m} \, b_{m}} = 0, \; n \ge 1 \; \Rightarrow \; \sum_{m = 1}^{n}{a_{n - m} \, b_{m}} = -\frac{a_{n}}{a_{0}}$$

The matrix of this system is:

$$A = \left(\begin{array}{ccccc}<br /> a_{0} & 0 & \ldots & 0 & \ldots \\<br /> <br /> a_{1} & a_{0} & \ldots & 0 & \ldots \\<br /> <br /> \ldots & & & & \\<br /> <br /> a_{n - 1} & a_{n - 2} & \ldots & a_{0} & \ldots \\<br /> <br /> \ldots & & & & <br /> \end{array}\right)$$

It is an infinitely dimensional lower triangular matrix. If we restrict it to the first n rows, then the determinant is simply $\det A_{n} = a^{n}_{0} \neq 0$, 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.

 

[ross_tang]

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/"