1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

82 Chapter 1 Fourier Series and Integrals

|f(x)−SN(x)|isnearlyequalto1,soconvergenceisnotuniform. (Inciden-
tally, the graphs in Fig. 9 also show the partial sums off(x)overshooting their
mark nearx=0. This feature of Fourier series is calledGibbs’ phenomenon
and always occurs near a jump.) On the other hand, Fig. 10 shows graphs of a
“sawtooth” function and the partial sums of its Fourier series. The maximum
deviation always occurs atx=0, and the convergence is uniform.
One of the ways of proving uniform convergence is by examining the coef-
ficients.

Theorem 1.If the series

∑∞

n= 1 (|an|+|bn|)converges, then the Fourier series

a 0 +

∑∞

n= 1

ancos

(nπx

a

)

+bnsin

(nπx

a

)

converges uniformly in the interval−a≤x≤a and, in fact, on the whole interval
−∞<x<∞.

Example.
For the function

f(x)=|x|, −π<x<π,

the Fourier coefficients are

a 0 =π

2

, an=^2

π

cos(nπ)− 1

n^2

, bn= 0.

Since the series

∑∞

n= 11 /n^2 converges, the series of absolute values of the coeffi-
cients converges, and so the Fourier series converges uniformly on the interval
−π≤x≤πto|x|. The Fourier series converges uniformly to the periodic
extension off(x)on the whole real line (see Fig. 10).

Another way of proving uniform convergence of a Fourier series is by exam-
ining the functionfthat generates it.

Theorem 2. If f is periodic and continuous and has a sectionally continuous
derivative, then the Fourier series corresponding to f converges uniformly to f(x)
on the entire real axis.

While this theorem is stated for a periodic function, it may be adapted to a
functionf(x)given on the interval−a<x<a.Iftheperiodic extensionoff
satisfies the conditions of the theorem, then the Fourier series offconverges
uniformly on the interval−a≤x≤a.

Example.
Consider the function

f(x)=x, − 1 <x< 1.