In Lecture 9 we showed that if ∑n=1∞ |an|+|bn| is convergent, then
a0 + ∑n=1∞ ancos(nx)+ bnsin(nx)
converges to a 2π periodic, continuous function f(x) whose Fourier series is the trigonometric series. The error from truncating the series at the Nth term is
|SN(x) - f(x)| < ∑n=N+1∞ |an|+|bn|.
We also showed that if a function is continuous and has piecewise continuous first and second derivatives, then the same estimate on the truncation error holds. Let's look at a simple example. Consider the 2π periodic function f(x), where f(x) = |x| on [-π,π]. This is continuous and has piecewise continuous derivatives. Its Fourier series is
f(x) = &pi/2 - (4/π)∑n=1∞ cos((2n-1)x)/(2n-1)2 = π/2 - (4/π)(cos(x) + cos(3x)/32 + ...),
and it converges to it for all x. All the bn's are 0, and so are all bn's with n even, except for a0. Thus, the two partial sums S2N-1 and S2N are equal, since they differ by 0·cos(2Nx). The truncation error satisfies
|f(x) -S2N| < |a2N+1| + |a2N+3| + |a2N+5| + ... = (4/π)((2N+1)-2 + (2N+3)-2 + (2N+3)-2 + ...) = (4/π)∑n=N∞ (2n+1)-2
We can estimate the series on the right using a variant of the integral test: Let g(x) be a continuous, positive, decreasing function defined on an interval including [M,∞). If cn = g(n), then
∑n=M+1∞ cn < ∫M∞ g(x)dx
Applying this to ∑n=N∞ (2n+1)-2, we get
∑n=N∞ (2n+1)-2 < ∫N-1∞ (2x+1)-2dx = ½ (2N-1)-1.
Thus, we see that the truncation error satisfies |f(x) -S2N| < (2/π)(2N-1)-1. We will test this numerically shortly. Again, the important thing to notice about the truncation error estimate is that it does not depend on x.
Our example exhibits an important type of behavior that is frequently found in series approximation of all types, not just Fourier series. Here we will state it only for Fourier series.
Definition. Let SN(x) be the Nth partial sum of the Fourier series for a function f(x). We say that the Fourier series for f is uniformly convergent to f(x) on an interval [a,b] if and only ifThe corollary in the previous lecture actually provides conditions on the function f for its Fourier series to be uniformly convergent. These amount to requiring that f be continuous, and that both f′ and f" be piecewise continuous. Our example satisfies these conditions.limN -> ∞ maxa<x<b|f(x)-SN(x)| = 0.
Numerical example The Matlab code below produces a plot of f(x), the 2π periodic extension of |x| on [-π,π], the Fourier series for f truncated at 2N, and plots of f(x)+(2/π)/(2*N-1), f(x) - (2/π)/(2*N-1). Notice that the S2N falls between the upper and lower errors, as it should.
x=linspace(-2*pi,2*pi,1001); y=abs(mod(x+pi,2*pi)-pi); N=5; a=[pi/2,zeros(1,2*N)]; a(2:2:end)=-(4/pi)*(1:2:(2*N)).^(-2); y_2N=part_sum_c(a,0,x); err=(2/pi)/(2*N-1); plot(x,y,'-.g',x,y+err,'-.r',x,y-err,'-.r',x,y_2N,'b')
Least squares We closed with a brief discussion of curve fitting via the method of least squares. There is an important connection between this and our next topic, convergence in the mean.