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Unformatted text preview: nverges boundedly to the constant
function 0.
Finally we observe that
fn 359 Þ 0 f n x dx Þ 0 nxe n x dx
1 as n 1 2 1en 1
n
2 22 0 Ý. d. f n x nx exp nx 2 for each x 0, 1 and each positive integer n.
3
2.5
2
1.5
1
0.5
0 0.2 0.4 x 0.6 0.8 1 For each n and x the equation f n x 0 says that
2n 2 x 2 exp nx 2 0
n exp nx 2
which gives us x 1
2n and we observe that 1
n exp 1
2
2
2n
from which we deduce that the sequence f n fails to be boundedly convergent on the interval
0, 1 . As in the earlier examples the sequence f n converges pointwise on 0, 1 to the
constant function 0.
Finally we observe that
fn Þ 0 f n x dx Þ 0 nx exp
1 as n 1 nx 2 dx 1e
2 n 1
2 1
2 Ý. e. f n x nx exp n 2 x for each x 0, 1 and each positive integer n.x exp x
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0 0.2 0.4 x 0.6 0.8 1 For each n and x, the equation f n x 0 says that
n exp n 2 x n 3 x exp n 2 x 0
which gives us x 1
n2 and we observe that
1
1
ne .
n2
Ý and therefore the sequence f n converges uniformly to the
fn We deduce that sup f n 0 as n
constant function 0 on 0, 1 .
Finally we observe that Þ 0 f n x dx Þ 0 nx exp
1 1
as n 1
n 3 exp n 2 Ý. 360 n 2 x dx
1
n exp n 2 1
n3 0 2. Given that f n x x n for all x and n, prove that the series f n converges pointwise, but not uniformly, on
the interval 0, 1 and that f n converges uniformly on the interval 0, whenever 0 1.
The fact that
f n converges pointwise on the interval 0, 1 follows from the fact that
Ý xn
n0 whenever 0 1
1 x x 1. Now given any n we have
n 1
1 x xn x1 0 1 1 sup 1 sup sup x n 1
1x j0 x x n 1
1x 0 x1 0 x1 Ý and so, certainly, the sequence of numbers
n 1 sup
does not approach 0 as n
3. 1 xn x 0 x1 j0 Ý. Given that f n x sin nx /n 2 for all n and x, prove that the series f n converges uniformly on R. Use Scientific Notebook to sketch some of the graphs of these functions to motivate your conclusions.
1
0.8
0.6
0.4
0.2
4 2 0
0.2 2x 4 0.4
0.6
0.8
1 Since
1
n2
f n follows at once from the comparison test. f n x 
for all n and x the uniform convergence of 4. Prove that the series x n /n! converges uniformly in x on every bounded interval but does not converge
uniformly in x on the entire line R. Hint: You do not need to know that the sum of this series is e x for each x in order to answer this
question. Note that whenever x 0 and n is a positive integer we have
Ý
j0 xj
j! n x n 1
n1 ! xj
j! j0 and therefore if n is a positive integer we have
Ý sup
j0 xj
j! n xj
j! j0 x0 5. Prove that the series
2n ! n
x
n! 2 4n 361 Ý. does not converge uniformly in x on the interval
, whenever 0 1. Hint: Whenever 0 1, 1 but that it does converge uniformly on the interval x 1 and n is a positive integer we have
Ý
j1 n 2j ! j
x
4 j j! 2 j1 2j ! j
x
4 j j! 2 Ý
jn1 2j ! j
x.
4 j j! 2 Now suppose that n is any given positive integer. Using the fact that the series
choose an integer k n 1 such that
k 2j !
4 j j! 2 diverges, we 2j !
1.
4 j...
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 Fall '08
 STAFF
 Math, Calculus

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