# 
# 
> restart;
> #Here we define several functions with support in the interval [-1,1].
> f0 := x->piecewise(x>-1 and x<1, 1, 0);
> f1 := x-> piecewise(x>-1 and x<1, (1-x^2), 0);
> f2 := x-> (f1(x))^2;
> #We plot the functions defined above. f0 has a jump discontinuity, f1
> is continuous and f2 is once differentiable.
> plot([f0(x), f1(x),f2(x)],x=-2..2);
> #We can use maple to symbolically integrate and find the Fourier
> transforms of the functions defined above.
> int((1-x^2)*exp(-I*x*y),x=-1..1);
> int((1-x^2)^2*exp(-I*x*y),x=-1..1);
> #Using these formulae we define the functions g and h. They are the
> Fourier transforms of f1 and f2 respectively. This is a much more
> efficient way to work than to define the function in terms of the
> integral.
> g := y-> piecewise(y=0,4/3, 4*(sin(y)-y*cos(y))/y^3);
> h := y-> piecewise(y=0, 16/15,
> 16*((3*sin(y)-3*y*cos(y)-y^2*sin(y))/y^5));
> # We plot the Fourier transforms. The slowest decaying graph is the
> F.T. of f0 the middle one is the F.T. of f1 and the fastest decaying
> graph is the F.T. of f2. Note how smoothness of a function translates
> into decay of its Fourier transform.
> plot([2*sin(y)/y,(3/2)*g(y), 15/8*h(y)],y=-40..40,discont=true);
> #In the next formula we define the partial Fourier inverse of g:
> instead of integrating from -infinity to infinity, we integrate from
> -R to R.
> F := (R,x) -> int((4/Pi)*(sin(y)-y*cos(y))/y^3*cos(x*y),y=0..R);
> #We plot the partial Fourier inverse for several values of R along
> with the graph of f1. The partial inverse quickly does a good job away
> from x=1. 
> plot([F(5,x),F(10,x),F(15,x),F(20,x),f1(x)],x=0..2);
> #Here we check things outside the interval [-1,1]. Note that the
> partial Fourier inverses oscillate around the true value of zero.
> plot([F(5,x),F(10,x),F(15,x),F(20,x)],x=10..15);
> with(plots):
> #In the next line we look at the Fourier transform of the function f1
> translated by different amounts. The animation shows the F.T. of
> f1(x-t) for $t=0,1,...,20.
> animate({cos((t/4)*y)*g(y),-sin((t/4)*y)*g(y)},y=-10..10,t=0..20,numpo
> ints=200);# 
> #Here we examine the effect of scaling a function on its Fourier
> transform. The next animation shows the Fourier transforms of the
> functions f1(t*x) for t=1/4, 2/4,...5. Note that the peak goes down as
> t increases. 
> animate(4*g(t*y/4)/t,y=.01..10,t=1..20);
> #To see better what is happenning in the previous example, we rescale
> so that all the functions take the same value at zero. These are the
> Fourier transforms of the functions t*f1(t*x) for t in the same range
> as before.
> animate(4*g(t*y/4),y=.01..10,t=1..20);
> #Try these routines out with functions of your own design!
> 
>