MathObj.PowerSeries.Core

Documentation

evaluate :: C a => [a] -> a -> a

evaluateCoeffVector :: C a v => [v] -> a -> v

evaluateArgVector :: (C a v, C v) => [a] -> v -> v

approximate :: C a => [a] -> a -> [a]

approximateCoeffVector :: C a v => [v] -> a -> [v]

approximateArgVector :: (C a v, C v) => [a] -> v -> [v]

Simple series manipulation

alternate :: C a => [a] -> [a]

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

holes2 :: C a => [a] -> [a]

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

holes2alternate :: C a => [a] -> [a]

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

insertHoles :: C a => Int -> [a] -> [a]

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

Series arithmetic

add :: C a => [a] -> [a] -> [a]

sub :: C a => [a] -> [a] -> [a]

negate :: C a => [a] -> [a]

scale :: C a => a -> [a] -> [a]

mul :: C a => [a] -> [a] -> [a]

stripLeadZero :: C a => [a] -> [a] -> ([a], [a])

divMod :: (C a, C a) => [a] -> [a] -> ([a], [a])

divide :: C a => [a] -> [a] -> [a]

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

divideStripZero :: (C a, C a) => [a] -> [a] -> [a]

Divide two series also if the divisor has leading zeros.

progression :: C a => [a]

recipProgression :: C a => [a]

differentiate :: C a => [a] -> [a]

integrate :: C a => a -> [a] -> [a]

sqrt :: C a => (a -> a) -> [a] -> [a]

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.

pow :: C a => (a -> a) -> a -> [a] -> [a]

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

exp :: C a => (a -> a) -> [a] -> [a]

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'

sinCos :: C a => (a -> (a, a)) -> [a] -> ([a], [a])

sinCosScalar :: C a => a -> (a, a)

sin :: C a => (a -> (a, a)) -> [a] -> [a]

cos :: C a => (a -> (a, a)) -> [a] -> [a]

tan :: C a => (a -> (a, a)) -> [a] -> [a]

log :: C a => (a -> a) -> [a] -> [a]

Input series must start with non-zero term.

derivedLog :: C a => [a] -> [a]

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

atan :: C a => (a -> a) -> [a] -> [a]

asin :: C a => (a -> a) -> (a -> a) -> [a] -> [a]

acos :: C a => (a -> a) -> (a -> a) -> [a] -> [a]

compose :: C a => [a] -> [a] -> [a]

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

composeTaylor :: C a => (a -> [a]) -> [a] -> [a]

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.

inv :: (Eq a, C a) => [a] -> (a, [a])

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.

invDiff :: C a => [a] -> (a, [a])