Partial Differential Equations PowerTool

by Dr. Jim Herod

Section 2.3: Convergence of Fourier Series

     In the Section 2.1, we discussed three general types of convergence in C([0, 1]): normed, pointwise, and uniform. Here, we apply these ideas to the special case that we have a function f  and we have the Fourier series

                      .

where   ...   is the usual orthonormal sequence. These results are presented here as Theorems, without proofs. The results may be found in many texts in texts on Fourier Series and Boundary Value Problems.

There are several terms we will use in this module.

A function is sectionally continuous  on an interval [a, b] if it is continuous on that interval except for possibly a finite number of jumps and removable discontinuities.

A function is sectionally smooth  on an interval [a, b] if both f   and   f  ' are sectionally continuous on the interval [a, b].

Examples:  The function f(x) = signum(x) is sectionally continuous on [-1, 1], but the function g(x) = 1/x is not sectionally continuous on that interval. Both are continuous on that interval except at zero. The signum function has a jump at x = 0, but 1/x has a more serious discontinuity.

>	plot([signum(x),1/x+1/10],x=-1..1,y=-3..3,discont=true,    color=[BLACK, RED]);

>

Example: The function  is continuous, but not sectionally smooth on [-1, 1]. It is not sectionally smooth because the derivative goes to  as x approaches zero.

>	Diff(sqrt(abs(x)),x)=diff(sqrt(abs(x)),x); plot([rhs(%),sqrt(abs(x))],x=-1..1,y=-3..3,discont=true,color=[red,black]);

>

Here is the first  result for convergence of series.

THEOREM:  If the function f  is sectionally smooth and periodic with period 2  c , then at each point x the Fourier series for f  converges and

         = [ f(x+) + f(x-)] / 2.

Example:  The function signum(x) is sectionally smooth. Therefore the Fourier series for this function converges to 1 for 0 < x < 1, to  -1 for -1 < x < 0, and to 0 for x = -1, 0, or 1. The Fourier series for the function has period 2. Here is a plot for 5 terms of the series.

>	c:=1; f:=x->signum(x); a[0]:=int(f(x),x=-c..c)/int(1^2,x=-c..c); for n from 1 to 5 do    a[n]:=int(f(x)*cos((n*Pi*x)/c),x=-c..c)/           int(cos(n*Pi*x/c)^2,x=-c..c);    b[n]:=int(f(x)*sin((n*Pi*x)/c),x=-c..c)/           int(sin(n*Pi*x/c)^2,x=-c..c); od; n:='n':

>	s:=x->a[0]+sum(a[n]*cos((n*Pi*x)/c)+b[n]*sin((n*Pi*x)/c),n=1..5);

>	plot([[x,f(x),x=-c..c],[x,s(x),x=-2*c..2*c]]);

>

In this previous example, the function was not continuous. We will see that a discontinuity always leads to the overshoot that can be observed, no matter how many terms of the series are taken. Is it not clear that the function would be getting close to f ( x ) if more terms were used?

In this next Theorem, we see a more powerful result that will insure uniform convergence for all x .

THEOREM:  If the series  converges, then the Fourier series for f  converges uniformly in the interval [-c, c].

Example:  Take the series   , with . Since the number series  converges, the series of absolute values converges, and so a Fourier Series with these coefficients converges. We draw graphs for this series on [- ,  ].

>	c:=Pi; a[0]:=Pi/2; for n from 1 to 5 do     a[n]:=2/Pi*(-1+(-1)^n)/n^2;     b[n]:=0; od; n:='n':

>	s:=x->a[0]+sum(a[n]*cos((n*Pi*x)/c)+b[n]*sin((n*Pi*x)/c),n=1..5);

>	plot(s(x),x=-2*c..2*c);

>

Example:  If { , ... } is an orthogonal sequence and we construct the Fourier Series

then this series will converge in norm if and only if

converges. We established this earlier with the Fourier Inequality. This result and the following example should be contrasted with the previous example. In this example, we have normed convergence, but not uniform convergence.

The example is . I try to convince you and me that the convergence is not uniform by seeing that the series converges in norm, but not to a continuous function.

>	f2:=x->sum(sin(n*x)/n,n=1..15): g2:=x->sum(sin(n*x)/n,n=1..25): h2:=x->sum(sin(n*x)/n,n=1..35):

>	plot({f2(x),g2(x),h2(x)},x=-Pi..Pi,y=-2..2,    color=[red,blue,green]);

>

Here is the last Theorem that we state in this section. It gives a condition for uniform convergence based on properties of the function, not the series  as the last theorem did. After all, it is usually the function we know at the outset, and not properties of the series.

THEOREM:  If f ( x ) is periodic, continuous, and has a sectionally continuous derivative, then the Fourier Series corresponding to f  converges uniformly to f(x)  for the entire real line.

As an example of this Theorem, take the function f ( x ) = | x | on the interval [ ,  ]. It is continuous on that interval and its periodic extension is also continuous. While its derivative is not continuous, it is sectionally continuous. Further, the coefficients are the ones used in the previous example. Check that.

On the other hand, the function f ( x ) = x  on the interval [-1, 1] is continuous there. While the periodic extension is sectionally  continuous, the periodic extension is not continuous . Thus, you cannot expect the Fourier Series to converge uniformly. At these discontinuities, you should expect the series to converge to

          ( f ( c -) + f (- c +) )/2,

the average of the jump from the left and the jump from the right at the points of discontinuities. Recall the first Theorem in this Section

We have presented two type results in this Section. One type result was information about  convergence based on the character of the coefficients. The other result was based on the character of the function to which the series was to converge. The most frequent application is the last result: if the function is periodic, piecewise smooth, and continuous, the Fourier Trigonometric Series converges uniformly. If the function is not continuous, the series will not converge uniformly.

EMAIL: herod@math.gatech.edu   or    jherod@tds.net

URL: http://www.math.gatech.edu/~herod

Copyright ©  2003  by James V. Herod

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