Fourier series

bobbym · 2014-02-20 02:21:17

Hi;

At the request of Agnishom, this thread will now deal with Fourier series and how to do them using your package.

We will start with the simple y = x between -π and π.

We have the equations:

f[x_] := x;
a[n_] := 1/\[Pi] Integrate[f[x] Cos[n*x], {x, -\[Pi], \[Pi]}]
b[n_] := 1/\[Pi] Integrate[f[x] Sin[n*x], {x, -\[Pi], \[Pi]}]

a[0],a[1],...a[4] will all be 0

b[0] = 0
b[1] = 2
b[2] = -1
b[3] = 2 / 3
b[4] = - 1 / 2

So the Fourier series is

b[0] + b[1] Sin[x] + b[2] Sin[2 x] + b[3] Sin[3 x]+b[4] Sin[4x]

Plot[{2 Sin[x] - Sin[2 x] + 2/3 Sin[3 x] - 1/2 Sin[4 x], 
  x}, {x, -\[Pi], \[Pi]}]

Agnishom · 2014-02-20 02:54:00

Hmm

(%i1) load(fourie)$

(%i2) f(x) = y$

(%i3) fourier(f,x,1);
(%t3)                               a  = f
                                     0

                                   2 f sin(%pi n)
(%t4)                         a  = --------------
                               n       %pi n

(%t5)                               b  = 0
                                     n

(%o5)                           [%t3, %t4, %t5]

EDIT: Sorry, I did not see the edited portion of your last post, the last time

Last edited by Agnishom (2014-02-20 02:55:48)

bobbym · 2014-02-20 02:59:09

Please use Mathematica else we will get confused over differences in packages. Also follow what I did in post 1 so that you can see the workings.

Agnishom · 2014-02-20 03:03:03

bobbym wrote:

We have the equations:

f[x_] := x;
a[n_] := 1/\[Pi] Integrate[f[x] Cos[n*x], {x, -\[Pi], \[Pi]}]
b[n_] := 1/\[Pi] Integrate[f[x] Sin[n*x], {x, -\[Pi], \[Pi]}]

(%i1) f(x) := x$

(%i2) a(n) := (1/%pi)*(integrate(f(x)*cos(n*x),x,-%pi,%pi))$

(%i3) b(n) := (1/%pi)*(integrate(f(x)*sin(n*x),x,-%pi,%pi))$

Agnishom · 2014-02-20 03:04:06

bobbym wrote:

Please use Mathematica else we will get confused over differences in packages. Also follow what I did in post 1 so that you can see the workings.

OK, sorry for the confusion. I will do that when I get a chance to reboot.

bobbym · 2014-02-20 03:06:22

Hi;

This is what it looks like.

Agnishom · 2014-02-20 03:20:43

I will do the Mathematica part once I reboot. For now, Is this ok?

(%i1)  f(x) := x$

(%i2) a(n) := (1/%pi)*(integrate(f(x)*cos(n*x),x,-%pi,%pi))$

(%i3) b(n) := (1/%pi)*(integrate(f(x)*sin(n*x),x,-%pi,%pi))$

(%i4) b(0) + b(1)*sin(x) + b(2)*sin(2*x) + b(3)*sin(3*x)+b(4)*sin(4*x);
                   sin(4 x)   2 sin(3 x)
(%o4)            - -------- + ---------- - sin(2 x) + 2 sin(x)
                      2           3
(%i5) print("Hmmm");
Hmmm 
(%o5)                                Hmmm
(%i6) plot2d (%o4, [x, -%pi, %pi])$

The image is attached

bobbym · 2014-02-20 03:26:02

That looks okay, very good.

Agnishom · 2014-02-20 03:35:02

a[0],a[1],...a[4] will all be 0

So, why calculate any of them? Why declare the a[n_] function at all?

bobbym · 2014-02-20 03:38:43

Because most of the time they will not be 0. Usually there are cos and sin terms.

Agnishom · 2014-02-20 03:44:13

What does a fourier series look like without the sigma notaiton

bobbym · 2014-02-20 03:54:28

Sigma notation was not used. That is a fine point. It looks like a big polynomial except powers of x are replaced with trig functions.

Agnishom · 2014-02-20 04:16:36

Show me one such poly please.

By the way, the attached image looks interesting.

(%i12) b(0);(%o12)                                 0
(%i13) b(0) + b(1)*sin(x)
;
(%o13)                             2 sin(x)
(%i14) b(0) + b(1)*sin(x) + b(2)*sin(2*x);
(%o14)                        2 sin(x) - sin(2 x)
(%i15) b(0) + b(1)*sin(x) + b(2)*sin(2*x) + b(3)*sin(3*x);
                       2 sin(3 x)
(%o15)                 ---------- - sin(2 x) + 2 sin(x)
                           3
(%i16) b(0) + b(1)*sin(x) + b(2)*sin(2*x) + b(3)*sin(3*x)+b(4)*sin(4*x);                   sin(4 x)   2 sin(3 x)
(%o16)           - -------- + ---------- - sin(2 x) + 2 sin(x)
                      2           3
(%i17) b(0) + b(1)*sin(x) + b(2)*sin(2*x) + b(3)*sin(3*x)+b(4)*sin(4*x)+b(5)*sin(5*x);
           2 sin(5 x)   sin(4 x)   2 sin(3 x)
(%o17)     ---------- - -------- + ---------- - sin(2 x) + 2 sin(x)

(%i22) plot2d ([%o12,%o13,%o14,%o15,%o16,%o17,f], [x, -%pi, %pi],[box, false])$

I remember that M has an Animate command that could produce a cooler animation

bobbym · 2014-02-20 04:25:08

Hi;

I did not say that they were polynomials, they just look like them.

Agnishom · 2014-02-20 04:36:49

Why is this not a polynomial?

bobbym · 2014-02-20 04:42:06

A polynomial is an expression made up of integer powers of x.

Agnishom · 2014-02-20 05:26:03

Non-negative inteher

bobbym · 2014-02-20 05:32:23

Yes, that is correct.

Tomorrow we will try another.

Agnishom · 2014-02-20 13:15:07

It should have been integer rather than inteher.

Tomorrow with respect to whom?

bobbym · 2014-02-20 13:50:34

For you of course.

Agnishom · 2014-02-20 14:39:49

OK, it is now tomorrow with respect to me

bobbym · 2014-02-20 14:59:00

Yes, I need some time to prepare another example. Also, I am making dinner.

Agnishom · 2014-02-20 15:09:37

Will you feed me tonight?

bobbym · 2014-02-20 15:11:10

Of course, ever had pizza?

Agnishom · 2014-02-20 15:17:13

Of course, do you know what a gel candle is?

Math Is Fun Forum

#1 2014-02-20 02:21:17

Fourier series

#2 2014-02-20 02:54:00

Re: Fourier series

#3 2014-02-20 02:59:09

Re: Fourier series

#4 2014-02-20 03:03:03

Re: Fourier series

#5 2014-02-20 03:04:06

Re: Fourier series

#6 2014-02-20 03:06:22

Re: Fourier series

#7 2014-02-20 03:20:43

Re: Fourier series

#8 2014-02-20 03:26:02

Re: Fourier series

#9 2014-02-20 03:35:02

Re: Fourier series

#10 2014-02-20 03:38:43

Re: Fourier series

#11 2014-02-20 03:44:13

Re: Fourier series

#12 2014-02-20 03:54:28

Re: Fourier series

#13 2014-02-20 04:16:36

Re: Fourier series

#14 2014-02-20 04:25:08

Re: Fourier series

#15 2014-02-20 04:36:49

Re: Fourier series

#16 2014-02-20 04:42:06

Re: Fourier series

#17 2014-02-20 05:26:03

Re: Fourier series

#18 2014-02-20 05:32:23

Re: Fourier series

#19 2014-02-20 13:15:07

Re: Fourier series

#20 2014-02-20 13:50:34

Re: Fourier series

#21 2014-02-20 14:39:49

Re: Fourier series

#22 2014-02-20 14:59:00

Re: Fourier series

#23 2014-02-20 15:09:37

Re: Fourier series

#24 2014-02-20 15:11:10

Re: Fourier series

#25 2014-02-20 15:17:13

Re: Fourier series

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