scubature-1.0.0.0: README.md

# scubature

Pure Haskell implementation of simplicial cubature (integration over a simplex).

```haskell
integrateOnSimplex
    :: (VectorD -> VectorD)   -- integrand
    -> Simplices              -- domain of integration (union of the simplices)
    -> Int                    -- number of components of the integrand
    -> Int                    -- maximum number of evaluations
    -> Double                 -- desired absolute error
    -> Double                 -- desired relative error
    -> Int                    -- integration rule: 1, 2, 3 or 4
    -> IO Results             -- values, error estimates, evaluations, success
```

## Example

![equation](http://latex.codecogs.com/gif.latex?%5Cint_0%5E1%5Cint_0%5Ex%5Cint_0%5Ey%5Cexp%28x+y+z%29%5C,%5Cmathrm%7Bd%7Dz%5C,%5Cmathrm%7Bd%7Dy%5C,%5Cmathrm%7Bd%7Dx=%5Cfrac%7B1%7D%7B6%7D%28e-1%29%5E3%5Capprox%20.8455356853)

Define the integrand:

```haskell
import Data.Vector.Unboxed as V
:{
f :: Vector Double -> Vector Double
f v = singleton $ exp (V.sum v)
:}
```

Define the simplex:

```haskell
simplex = [[0, 0, 0], [1, 1, 1], [0, 1, 1], [0, 0, 1]]
```

Integrate:

```haskell
import Numeric.Integration.SimplexCubature
integrateOnSimplex f [simplex] 1 100000 0 1e-10 3
-- Results { values = [0.8455356852954488]
--         , errorEstimates = [8.082378899762402e-11]
--         , evaluations = 8700
--         , success = True }
```

For a scalar-valued integrand, it's more convenient to define... a scalar-valued
integrand! That is:

```haskell
:{
f :: Vector Double -> Double
f v = exp (V.sum v)
:}
```

and then to use `integrateOnSimplex'`:

```haskell
integrateOnSimplex' f [simplex] 100000 0 1e-10 3
-- Result { value         = 0.8455356852954488
--        , errorEstimate = 8.082378899762402e-11
--        , evaluations   = 8700
--        , success       = True }
```

## Integration on a spherical triangle

The library also allows to evaluate an integral on a spherical simplex on the
unit sphere (in dimension 3, a spherical triangle).

For example take the first orthant in dimension 3:

```haskell
import Numeric.Integration.SphericalSimplexCubature
o1 = orthants 3 !! 0
o1
-- [ [1.0, 0.0, 0.0]
-- , [0.0, 1.0, 0.0]
-- , [0.0, 0.0, 1.0] ]
```

And this integrand:

```haskell
:{
integrand :: [Double] -> Double
integrand x = (x!!0 * x!!0 * x!!2) + (x!!1 * x!!1 * x!!2) + (x!!2 * x!!2 * x!!2)
:}
```

Compute the integral (the exact result is `pi/4 ≈ 0.7853981634`):

```haskell
integrateOnSphericalSimplex integrand o1 20000 0 1e-7 3
-- Result { value         = 0.7853981641913279
--        , errorEstimate = 7.71579524444753e-8
--        , evaluations   = 17065
--        , success       = True }
```