Mathematical Methods for Physicists: A concise introduction - Site Map

SERIES OF FUNCTIONS AND UNIFORM CONVERGENCE
(c) jsin…nx†=n=nj 1=n ˆ M n . However, P M n does not converge. The M test
cannot be used in this case and we cannot conclude anything about the
uniform convergence by this test.
A uniformly convergent in®nite series of functions has many of the properties
possessed by the sum of ®nite series of functions. The following three are particularly
useful. We state them without proofs.
(1) If the individual terms u n …x† are continuous in [a, b] and if P u n …x† converges
uniformly to the sum S…x† in [a, b], then S…x† is continuous in [a, b].
Brie¯y, this states that a uniformly convergent series of continuous functions
is a continuous function.
(2) If the individual terms u n …x† are continuous in [a, b] and if P u n …x† converges
uniformly to the sum S…x† in [a, b], then
or
Z b X 1
a
Z b
nˆ1
a
S…x†dx ˆ X1
u n …x†dx ˆ X1
nˆ1
nˆ1
Z b
a
Z b
a
u n …x†dx
u n …x†dx:
Brie¯y, a uniform convergent series of continuous functions can be integrated
term by term.
(3) If the individual terms u n …x† are continuous and have continuous derivatives
in [a, b] and if P u n …x† converges uniformly to the sum S…x† while
P dun …x†=dx is uniformly convergent in [a, b], then the derivative of the
series sum S…x† equals the sum of the individual term derivatives,
d
S…x† ˆX1
dx
nˆ1
d
dx u n…x†
or
( )
d X 1
u
dx n …x†
nˆ1
ˆ X1
nˆ1
d
dx u n…x†:
Term-by-term integration of a uniformly convergent series requires only continuity
of the individual terms. This condition is almost always met in physical
applications. Term-by-term integration may also be valid in the absence of uniform
convergence. On the other hand term-by-term di€erentiation of a series is
often not valid because more restrictive conditions must be satis®ed.
Problem A1.14
Show that the series
sin x sin 2x
1 3 ‡
2 3 ‡‡
is uniformly convergent for x .
523
sin nx
n 3 ‡