1 |gn(x) – g(x)| = - |n(1+x²) Thus, given e > 0, we can choose N > 1/e (which does not depend on x), and it follows that n 2N implies Ign(x) – 9(x)| < E for all r e R. By Definition 6.2.3, In +0 uniformly on R. (ii) Look back at Example 6.2.2 (i), where we saw that fn(x) = (x2 + nx)/n converges pointwise on R to f(x) uniform. To see this write = x. On R, the convergence is not x' + nx |fn(x) – f(x)| = and notice that in order to force |fn (x) – f(x)| < e, we are going to have to choose N> Although this is possible to do for each r e R, there is no way to choose a single value of N that will work for all values of x at the same time. On the other hand, we can show that fn + f uniformly on the set [-b, b]. By restricting our attention to a bounded interval, we may now assert that Given e > 0, then, we can choose N > independently of x E [-b, b]. Graphically speaking, the uniform convergence of fn to a limit f on a set A can be visualized by constructing a band of radius ±e around the limit func- tion f. If fn + f uniformly, then there exists a point in the sequence after which each fn is completely contained in this e-strip (Fig. 6.4). This image should be compared with the graphs in Figures 6.1-6.2 from Example 6.2.2 and the one in Figure 6.5. VI
1 |gn(x) – g(x)| = - |n(1+x²) Thus, given e > 0, we can choose N > 1/e (which does not depend on x), and it follows that n 2N implies Ign(x) – 9(x)| < E for all r e R. By Definition 6.2.3, In +0 uniformly on R. (ii) Look back at Example 6.2.2 (i), where we saw that fn(x) = (x2 + nx)/n converges pointwise on R to f(x) uniform. To see this write = x. On R, the convergence is not x' + nx |fn(x) – f(x)| = and notice that in order to force |fn (x) – f(x)| < e, we are going to have to choose N> Although this is possible to do for each r e R, there is no way to choose a single value of N that will work for all values of x at the same time. On the other hand, we can show that fn + f uniformly on the set [-b, b]. By restricting our attention to a bounded interval, we may now assert that Given e > 0, then, we can choose N > independently of x E [-b, b]. Graphically speaking, the uniform convergence of fn to a limit f on a set A can be visualized by constructing a band of radius ±e around the limit func- tion f. If fn + f uniformly, then there exists a point in the sequence after which each fn is completely contained in this e-strip (Fig. 6.4). This image should be compared with the graphs in Figures 6.1-6.2 from Example 6.2.2 and the one in Figure 6.5. VI
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
(i) Let gn(x) =1/n(1 + x2) For any fixed x ∈ R, we can see that limgn(x) = 0 so that g(x) = 0 is the pointwise limit of the sequence (gn) on R. Is this convergence uniform? The observation that 1/(1 + x2) ≤ 1 for all x ∈ R implies that
.
Transcribed Image Text:1
|gn(x) – g(x)| =
-
|n(1+x²)
Thus, given e > 0, we can choose N >
1/e (which does not depend on x),
and it follows that
n 2N implies Ign(x) – 9(x)| < E
for all r e R. By Definition 6.2.3, In +0 uniformly on R.
(ii) Look back at Example 6.2.2 (i), where we saw that fn(x) = (x2 + nx)/n
converges pointwise on R to f(x)
uniform. To see this write
= x. On R, the convergence is not
x' + nx
|fn(x) – f(x)| =
and notice that in order to force |fn (x) – f(x)| < e, we are going to have
to choose
N>
![Although this is possible to do for each r e R, there is no way to choose
a single value of N that will work for all values of x at the same time.
On the other hand, we can show that fn + f uniformly on the set [-b, b].
By restricting our attention to a bounded interval, we may now assert that
Given e > 0, then, we can choose
N >
independently of x E [-b, b].
Graphically speaking, the uniform convergence of fn to a limit f on a set
A can be visualized by constructing a band of radius ±e around the limit func-
tion f. If fn + f uniformly, then there exists a point in the sequence after which
each fn is completely contained in this e-strip (Fig. 6.4). This image should be
compared with the graphs in Figures 6.1-6.2 from Example 6.2.2 and the one
in Figure 6.5.
VI](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff90ff6dc-7620-4c93-8d48-784c6819ef76%2F30a653cb-641f-4d9b-b569-938c5bbfed6f%2F4ju13w8.png&w=3840&q=75)
Transcribed Image Text:Although this is possible to do for each r e R, there is no way to choose
a single value of N that will work for all values of x at the same time.
On the other hand, we can show that fn + f uniformly on the set [-b, b].
By restricting our attention to a bounded interval, we may now assert that
Given e > 0, then, we can choose
N >
independently of x E [-b, b].
Graphically speaking, the uniform convergence of fn to a limit f on a set
A can be visualized by constructing a band of radius ±e around the limit func-
tion f. If fn + f uniformly, then there exists a point in the sequence after which
each fn is completely contained in this e-strip (Fig. 6.4). This image should be
compared with the graphs in Figures 6.1-6.2 from Example 6.2.2 and the one
in Figure 6.5.
VI
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