Fourier Series using fourie.mac

My package pw.mac can do the integration necessary to solve the Fourier
series directly without using "fourie.mac" as long as you remember how to do
it.
Here is a solution using pw.mac.  All the needed functions are documented on
my web site.

This result is based on the fact that for x in the interval (-%pi/2, %pi/2)

abs(sin(x)) = pw([-%pi/2,-sin(x),0,sin(x),%pi/2],x) = signum(x) * sin(x)

(%i1) (load(pw), ratprint:false);
(%o1) false
(%i2) now:elapsed_run_time();
(%o2) 2.95
(%i3) declare(n,integer);
(%o3) done
(%i4) assume(n>0);
(%o4) [n>0]
(%i5) f:[-%pi/2, -sin(x), 0, sin(x), %pi/2];
(%o5) [-%pi/2,-sin(x),0,sin(x),%pi/2]
(%i6) a0 : pwdefint(pw(f,x), x, -%pi/2, %pi/2)/%pi;
(%o6) 2/%pi
(%i7) an : (s:1/(%pi/2)*pwdefint(pw(f,x)*cos(n*%pi*x/(%pi/2)), x, -%pi/2,
%pi/2),radcan(s));
(%o7)
-((4*n-2)*cos((2*%pi*n+%pi)/2)+(-4*n-2)*cos((2*%pi*n-%pi)/2)+4)/(4*%pi*n^2-%pi)
(%i8) bn : (s:(1/(%pi/2))*pwdefint(pw(f,x)*sin(n*%pi*x/(%pi/2)), x, -%pi/2,
%pi/2), radcan(s));
(%o8) 0
(%i9) (define(a(n), an),define(b(n), bn))$
(%i10) g(x, terms)
:=a0+sum(b(m)*sin(m*%pi*x/(%pi/2))+a(m)*cos(m*%pi*x/(%pi/2)), m, 1, terms)$
(%i11) define(h(x), periodic(pw(f,x),x,-%pi/2,%pi/2))$
(%i12) n:5;
(%o12) 5
(%i13) load(draw)$
(%i14) draw2d(
    color=brown,
    yrange=[-0,1],
    label_alignment=left,
    explicit(h(x),x,-%pi,%pi),
    color=blue,
    label(["f(x)=abs(sin(x))",-2,.3]),
    label([concat("with fourier series for n = ",n), -2,.25]),
    explicit(g(x,5),x,-%pi,%pi))$
(%i15) elapsed_run_time()-now;
(%o15) 2.44


To prove the identity

(%i1) display2d:false;

(o1) false
(%i2) abs(sin(x));
(o2) abs(sin(x))
(%i3) abs2signum(%);
(o3) sin(x)*signum(sin(x))
(%i4) periodic(sin(x),x,-%pi/2,%pi/2);
(o4) -cos(%pi*((x+%pi/2)/%pi-floor((x+%pi/2)/%pi)))
(%i5) integrate(%,x);
(o5) -'integrate(cos(%pi*((x+%pi/2)/%pi-floor((x+%pi/2)/%pi))),x)
(%i6) changevar(%,y-(x+%pi/2)/%pi,y,x);
(%o6) sin(%pi*floor(y)-%pi*y)
(%i7) subst((x+%pi/2)/%pi,y,%);
(o7) sin(%pi*floor((x+%pi/2)/%pi)-x-%pi/2)
(%i8) abs(sin(x))=pw([-%pi/2,-sin(x),0,sin(x),%pi/2],x);
(o8) abs(sin(x)) =
sin(x)*(signum(x)-signum(x-%pi/2))/2-sin(x)*(signum(x+%pi/2)-signum(x))/2
(%i9) simp_assuming(%,x<%pi/2,x>-%pi/2);
(o9) abs(sin(x)) = (signum(x)+1)*sin(x)/2-(1-signum(x))*sin(x)/2
(%i10) radcan(%);
(o10) abs(sin(x)) = signum(x)*sin(x)
(%i11)

You can get pw.mac from my site.
http://mysite.verizon.net/res11w2yb/id2.html

Hope this helps.

Richard Hennessy



2009/7/28 Leo Butler <l.butler at ed.ac.uk>

>
>
> On Tue, 28 Jul 2009, Rafa Topolnicki wrote:
>
> < Leo Butler pisze:
> < >
> < > On Mon, 27 Jul 2009, Rafa? Topolnicki wrote:
> < >
> < > < Hi,
> < > <
> < > < During testing Fourier Series Expansion in maxima I found a problem I
> < > < can't cope with.
> < > <
> < > < (%i3)                            load(fourie)
> < > < (%o3)         /usr/share/maxima/5.18.1/share/calculus/fourie.mac
> < > < (%i4)                         f(x) := abs(sin(x))
> < > < (%o4)                         f(x) := abs(sin(x))
> < > < (%i5)                    flist : fourier(f(x), x, %pi)
> < > <                                           2
> < > < (%t5)                              a  = ---
> < > <                                      0   %pi
> < > <
> < > <                       cos(%pi n)   cos(%pi n)      1         1
> < > <                    2 (---------- - ---------- + ------- - -------)
> < > <                        2 n + 2      2 n - 2     2 n + 2   2 n - 2
> < > < (%t6)        a  = -----------------------------------------------
> < > <                n                         %pi
> < > <
> < >
> < > The easiest thing to do would be to add a line
> < >
> < > %t6 : a[n] = if n=1 then 'limit( subst(n=_n,rhs(%t6)),_n,n ) else
> < > %rhs(%t6);
>
> You need to overcome the evaluation rules in sum. Here is one way to do
> this using a lambda-function:
>
> (%i2) load("fourie");
> (%o2) "/home/work/maxima/maxima-5.18.1-clisp/share/calculus/fourie.mac"
> (%i3) f(x):=abs(sin(x));
> (%o3) f(x):=abs(sin(x))
> (%i4) flist : fourier(f(x),x,%pi);
> (%t4) a[0] = 2/%pi
>
> (%t5) a[n] =
> 2*(cos(%pi*n)/(2*n+2)-cos(%pi*n)/(2*n-2)+1/(2*n+2)-1/(2*n-2))/%pi
>
> (%t6) b[n] = 0
>
> (%o6) [%t4,%t5,%t6]
> (%i7) %t5 : a[n] = 'apply(lambda([s], if s=1 then 0 else
> rhs(''%t5)),[n]);
> (%o7) a[n] = 'apply(lambda([s],
>                           if s = 1 then 0
>                               else rhs(
>                               a[n] = 2*(cos(%pi*n)/(2*n+2)
>                                        -cos(%pi*n)/(2*n-2)+1/(2*n+2)
>                                        -1/(2*n-2))
>                                    /%pi)),[n])
> (%i8) fourexpand(flist,y,%pi,3)$
> (%i9) ev(%,nouns);
> (%o9) 2/%pi-4*cos(2*y)/(3*%pi)
>
>
> Leo
>
> --
> The University of Edinburgh is a charitable body, registered in
> Scotland, with registration number SC005336.
>
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