Calculus: Analytic Geometry and Calculus, with Vectors

(^618) Series
synthesis. Let (ki, 02, 03, .. - be a set of functions orthonormal over a set E.
Let a,, a2, a3, be given coefficients for which 2jakI2 < -. Then, when
Lebesgue integrals are used, there is a function f for which the formulas
(1) fE l f(x)J2 dx <
Jim
f

I f(x)- I ak4k(x)I2 dx= 0

n'_'°° k=1

are valid. Moreover the numbers a1i a2, as, are the Fourier coefficients

of f, that is,

(2) ak =fEf(x);r(x) dx (k = 1,2,3, ...)

In this formula we have recognized the fact that, in many important applications,

4,k(x) is a complex number and "¢k(x) bar" is the complex conjugate of 4,k(x).

It has not been asserted (and is in fact sometimes untrue) that the series in

(3) f(x) - a,951 (x) + a24,(x) + asq5s(x) +...

converges to f(x). However, as the second formula in (1) shows, the sum of the

first n terms of the series must be a good global approximation to f(x) whenever

n is large. Persons who study Lebesgue integration and Fourier series for a year

or two can learn all about these things.

14 Supposing that Ao, A,, B2, and B,, B2, are bounded sequences

of constants, let

(1) AX) _ !lo + I (Ak cos kx + Bk sin kx)

k-1

for those values of x (if any) for which the series is convergent. The series in

the right member of (1) is called a trigonometric series. Some profound studies

of the series in (1) depend upon use of the function F defined by

(2) F(x) = 'ffUox1 - 4k cos kx + Bk sin kx

k1 k2

the series being convergent for each x because it is dominated by a convergent

series of constants. Show that, when t 0,

(3)

F(x - 2t) - 2F(x) + F(x + 2t)

4t2

_ Jo + I [A cos kx + Bk sin kx]

(sin kt 2

k-1 kt

Remark: Riemann discovered this formula and used it to solve some difficult

problems. In more advanced mathematics, it is proved that the formula

sin kt 2li

(4) l 0 [uo+ uk =

(sin

k=1 )J u0 +k=1uk