ApproxFun is a package for approximating functions. It is in a similar vein to the Matlab
package `Chebfun`

and the Mathematica package `RHPackage`

.

The `ApproxFun Documentation`

contains detailed information, or read on for a brief overview of the package.

The `ApproxFun Examples`

contains many examples of
using this package, in Jupyter notebooks and Julia scripts.

Take your two favourite functions on an interval and create approximations to them as simply as:

```
using LinearAlgebra, SpecialFunctions, Plots, ApproxFun
x = Fun(identity,0..10)
f = sin(x^2)
g = cos(x)
```

Evaluating `f(.1)`

will return a high
accuracy approximation to `sin(0.01)`

. All the algebraic manipulations of functions
are supported and more. For example, we can add `f`

and `g^2`

together and compute
the roots and extrema:

```
h = f + g^2
r = roots(h)
rp = roots(h')
plot(h; label="f + g^2")
scatter!(r, h.(r); label="roots")
scatter!(rp, h.(rp); label="extrema")
```

Notice from above that to find the extrema, we used `'`

overridden for the `differentiate`

function. Several other `Julia`

base functions are overridden for the purposes of calculus. Because the exponential is its own
derivative, the `norm`

is small:

```
f = Fun(x->exp(x), -1..1)
norm(f-f') # 4.4391656415701095e-14
```

Similarly, `cumsum`

defines an indefinite integration operator:

```
g = cumsum(f)
g = g + f(-1)
norm(f-g) # 3.4989733283850415e-15d
```

Algebraic and differential operations are also implemented where possible, and most of Julia's built-in functions are overridden to accept `Fun`

s:

```
x = Fun()
f = erf(x)
g = besselj(3,exp(f))
h = airyai(10asin(f)+2g)
```

We can also solve differential equations. Consider the Airy ODE `u'' - x u = 0`

as a boundary value problem on `[-1000,200]`

with conditions `u(-1000) = 1`

and `u(200) = 2`

. The unique solution is a linear combination of Airy functions. We can calculate it as follows:

```
x = Fun(identity, -1000..200) # the function x on the interval -1000..200
D = Derivative() # The derivative operator
B = Dirichlet() # Dirichlet conditions
L = D^2 - x # the Airy operator
u = [B;L] \ [[1,2],0] # Calculate u such that B*u == [1,2] and L*u == 0
plot(u; label="u")
```

Solve a nonlinear boundary value problem satisfying the ODE `0.001u'' + 6*(1-x^2)*u' + u^2 = 1`

with boundary conditions `u(-1)==1`

, `u(1)==-0.5`

on `[-1,1]`

:

```
x = Fun()
u₀ = 0.0x # initial guess
N = u -> [u(-1)-1, u(1)+0.5, 0.001u'' + 6*(1-x^2)*u' + u^2 - 1]
u = newton(N, u₀) # perform Newton iteration in function space
plot(u)
```

One can also solve a system nonlinear ODEs with potentially nonlinear boundary conditions:

```
x=Fun(identity, 0..1)
N = (u1,u2) -> [u1'(0) - 0.5*u1(0)*u2(0);
u2'(0) + 1;
u1(1) - 1;
u2(1) - 1;
u1'' + u1*u2;
u2'' - u1*u2]
u10 = one(x)
u20 = one(x)
u1,u2 = newton(N, [u10,u20])
plot(u1, label="u1")
plot!(u2, label="u2")
```

There is also support for Fourier representations of functions on periodic intervals.
Specify the space `Fourier`

to ensure that the representation is periodic:

```
f = Fun(cos, Fourier(-π..π))
norm(f' + Fun(sin, Fourier(-π..π))) # 5.923502902288505e-17
```

Due to the periodicity, Fourier representations allow for the asymptotic savings of `2/π`

in the number of coefficients that need to be stored compared with a Chebyshev representation.
ODEs can also be solved when the solution is periodic:

```
s = Chebyshev(-π..π)
a = Fun(t-> 1+sin(cos(2t)), s)
L = Derivative() + a
f = Fun(t->exp(sin(10t)), s)
B = periodic(s,0)
uChebyshev = [B;L] \ [0.;f]
s = Fourier(-π..π)
a = Fun(t-> 1+sin(cos(2t)), s)
L = Derivative() + a
f = Fun(t->exp(sin(10t)), s)
uFourier = L\f
ncoefficients(uFourier)/ncoefficients(uChebyshev),2/π
plot(uFourier)
```

Other operations including random number sampling using [Olver & Townsend 2013]. The
following code samples 10,000 from a PDF given as the absolute value of the sine function on `[-5,5]`

:

```
f = abs(Fun(sin, -5..5))
x = ApproxFun.sample(f,10000)
histogram(x;normed=true)
plot!(f/sum(f))
```

We can solve PDEs, the following solves Helmholtz `Δu + 100u=0`

with `u(±1,y)=u(x,±1)=1`

on a square. This function has weak singularities at the corner,
so we specify a lower tolerance to avoid resolving these singularities
completely.

```
d = ChebyshevInterval()^2 # Defines a rectangle
Δ = Laplacian(d) # Represent the Laplacian
f = ones(∂(d)) # one at the boundary
u = \([Dirichlet(d); Δ+100I], [f;0.]; # Solve the PDE
tolerance=1E-5)
surface(u) # Surface plot
```