Conference rusure::math

    
    
    Here my idea it looks like the normal f(x) except at the ends then its
    the avg. of the end points. I'm using -1 too 1 as my interval.
    
    					+
    					|
    					|
    					|
    					+       o
    					|       .
    					|      .
    					|      .
    					+     .
    					|    .
    				.	|   .   .
    					| ..
       				      	+.
    				     ..	|
    				    .	|
    				 o..	|
               ----+------+------+------+------+------+------+------+
    					|
    					|
    					|


From -1 to 1

>                     <<< Note 1434.0 by ARCANA::ESTRELLA >>>
>                 -< what does this fourier series look like? >-
>
>
>    							      x
>    What would the fourier sine and cosine series for f(x) = e
>
>    look like graphicly? Would they look the same?

The Fourier series can only be computed for periodic functions.
Do you mean to have e^x sliced up into repeating segments ?

Also, I think there may be some periodic functions for which it isn't possible
to compute a Fourier series, because the coefficient formulae evaluate to
a divergent series (true, Fourier experts?)

if a signal x(t) could be obtained by summing a number of periodic signals
then we say that x(t) has a fourier series expansion.

x(t) could be both periodic, and Aperiodic.

if x(t) is periodic, then to represent it in terms of its building blocks
of signals we say

   x(t)= sum ( a(k) exp(Ikwt) )
          k=-oo..+oo

where w is the frequency of x(t) "fundamental frequency"

example, let x(t) be periodic of frequency 2Pi which can be written as

   x(t)=sum  (  a(k) exp(Ikwt))
      k=-3..+3

   coffeients are a(0)= 1
                  a(1)= a(-1) = 1/4
                  a(2)= a(-2) = 1/2
                  a(3)= a(-3) = 1/3

so x(t)= 1 + 1/4(exp(I2Pit)+exp(-I2Pit)) + 1/2(exp(I4Pit)+ exp(-I4Pit))
         +1/3 (exp(I6Pit)+ exp(-I6Pit))

 i.e. x(t)= 1+ 1/2 cos(2Pit) + cos(4Pit) + 2/3 cos(6Pit)


in general if x(t) is periodic and has a fourier series then you can
find the a(k) terms from

        a(k)= 1  integral( x(t) exp(-Ikwt) dt ) over period T
             ---
              T

where T is any interval

for a periodic x(t) to have a fourier series it must meet what is called
Dirichlet conditions
  1) over any period, x(t) must be absolutely integrable
  2) over any finite interval in time, there is a finite number of 
     maxima and minima.
  3) in any finite interval in time, there is finite number of discontinuities.
  example x(t)= 1/t, period 1 do not have fourier series due to #1
          x(t)= sin(2Pi/t)   0<t<= 1 of period T=1  failed due to #2

an Aperiodic signal can have a fourier series.
the idea is we think of Aperiodic signal xa(t) as the limit of a periodic
signal x(t) as the period becomes large.

i.e. given an Aperiodic signal xa(t) of finite duration , from this we
can construct periodic x(t) for which this xa(t) is ONE period.

to be able to construct the signal xa(t) as sum of signals in this case
we need to construct the fourier transform for xa(t). 

so in summary, we can decompose both periodic (per Dirichlet conditions)
and Aperiodic signals into a linear combination of complex exponentials.

/naser

    

    re .2,
    
    It all depends on what one means by "compute".  If by that we mean
    "approximate" a given function with Fourier series, then we have to ask
    what exactly one means by "approximation".  A standard one is to make
    
     / 1       2
    (   |f - g|  = 0 where g is the Fourier series of f.  Then if 
     )
    / -1
    
     /1   2
    (  |f|   < oo  then we know that f has a Fourier approximation--
     )
    /-1
    
    I think it is called Parsival's theorem.
    
    Eugene

ref .-1 (Eugene)

right, one class of periodic signals that could be represented by
fourier series is that which is square-integrable over a period.
this is the same as saying the energy of the signal over one period is
finite, i.e.
                  /         2
                 (  | x(t) |      < oo
                 /
                 T

Parseal's theory says that the total energy of the signal can be determined
by adding the energy per unit time /+00       2
                                  (     |x(t)| dt
                                  /
                                 -oo

OR
 by computing the energy per unit frequency and integrating over all
frequencies.

i.e.    /+oo        2             /+oo         2
       (      | x(t)|   = 1/2Pi  (       |X(w)|   dw
       /                         /
       -oo                       -oo


I took a course last quarter in signals and system at Northeastern where
we learned this stuff, and things like Laplace transforms, Z transforms etc..
really interesting stuff, place for good algorithms..

This gets more interesting when one starts analyzing signals in 2 or 3
dimensions.

I've looked at papers, where FFT methods are used to solve Partial 
differential equations.. but that for the future .. got'a have solid
foundations before doing the heavy duty stuff.. :-)

/naser
    

    re .4              
    
    I guess I had in mind the Laplace transform, in which you have to
    integrate an integrand of form v(t)*exp( -sigma*t-j*2*pi*f*t).   I was
    concerned that if v(t) is of order greater than exp(sigma*t), then the
    integral may not exist.  Since the function given was an exponential,
    I thought there might be a problem.
                                               
    After looking up the Fourier series coefficient formula, I realised
    it's OK, because the sliced periodic exponential actually
    referred to has series coefficients that are obviously finite.
    This is because the integral is performed over a single period with
    an everywhere-finite integrand.
    
    Sorry about the confusion :-(.