For the series of a real function f compute the series for x -> f (-x)

For the series of a real function f compute the series for x -> (f x + f (-x)) / 2

For the series of a real function f compute the real series for x -> (f (i*x) + f (-i*x)) / 2

For power series of f x, compute the power series of f(x^n).

Divide two series where the absolute term of the divisor is non-zero. That is, power series with leading non-zero terms are the units in the ring of power series.

Knuth: Seminumerical algorithms

Divide two series also if the divisor has leading zeros.

We need to compute the square root only of the first term. That is, if the first term is rational, then all terms of the series are rational.

Input series must start with a non-zero term, even better with a positive one.

The first term needs a transcendent computation but the others do not. That's why we accept a function which computes the first term.

(exp . x)' =   (exp . x) * x'
(sin . x)' =   (cos . x) * x'
(cos . x)' = - (sin . x) * x'

Input series must start with non-zero term.

Computes (log x)', that is x'/x

Since the inner series must start with a zero, the first term is omitted in y.

Compose two power series where the outer series can be developed for any expansion point. To be more precise: The outer series must be expanded with respect to the leading term of the inner series.

This function returns the series of the inverse function in the form: (point of the expansion, power series).

That is, say we have the equation:

y = a + f(x)

where function f is given by a power series with f(0) = 0. We want to solve for x:

x = f^-1(y-a)

If you pass the power series of a+f(x) to inv, you get (a, f^-1) as answer, where f^-1 is a power series.

The linear term of f (the coefficient of x) must be non-zero.

This needs cubic run-time and thus is exceptionally slow. Computing inverse series for special power series might be faster.