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Transcript
```Math 312 S11 Day 6
Note: We started by reviewing what we did with section 9.2, so I am including that
along with the new stuff we did.
Section 9.2 Pointwise Convergence of Sequences of Functions
1. Definition: Given a function f: D→R and a sequence of functions {fn: D→R},
we say that the sequence {fn} converges pointwise on D to f, provided that for
each point x in D, lim fn (x) = f (x).
n→∞
2. Example. Consider the sequence of functions fn: [0,1]→R defined by fn(x) = xn.
Does this sequence of functions converge pointwise to a function? If so, what
⎧0 for x < 1
since if x is one, we have the
1
for
x
=
1
⎩
Answer: Yes. It converges to f (x) = ⎨
constant sequence fn(1) = 1. And we know that if 0≤x<1 then the sequence xn
converges to zero.
3. Consider the sequence of functions fn: [0,1]→R defined by fn(x) =
⎧n 2 x
if 0 ≤ x < 1n
⎪ 2
1
2
⎨−n x + 2n if n ≤ x < n
⎪0
if 2n ≤ x ≤ 1
⎩
a. Draw a sketch of f2, f3, f4 and fn.
Answer: The first function is a “tent” that is symmetrical and peaks at
½ with a height of 2. The second peaks at 1/3 at a height of 3, but
comes back down to zero at 2/3 and is zero from there. As you go, the
peaks move to the left and get higher and higher, but the point at
which they become zero is moving closer and closer to zero.
b. Compute
∫
1
0
fn (x)dx
for each n. (What is the limit of this sequence
of integrals?)
Answer: Each integral is the area of a triangle. The height is n and the
width is 2/n. so you get 1 for each integral. This of course converges
to 1!
c. What function, f, does the sequence of functions converge to
pointwise?
Any positive number is less than some value of 2/n (by A.P.) So for
any point in (0,1) the sequence fn(x) converges to zero. The sequence
may get pretty big while the point is mapping to places close to the
peaks of really tall tents, but eventually it will slide quickly down and
then be zero forever more.
At 0 and 1, the function values are clearly always zero. Thus the
sequence of functions converges pointwise to the function defined by
f(x) = 0
d. What is
∫
1
0
f (x)dx ?
4. So the first example tells us that a sequence of continuous functions can
converge to a function that is not continuous!
5. The second example tells us that the integral of the limit of a sequence of
functions is not necessarily the same as the limit of the sequence of integrals.
6. These results are of course extremely disappointing! Whatever shall we do?
9.3 Uniform Convergence of Sequences of Functions
1. Definition: Given a function f: D→R and a sequence of functions {fn: D→R},
we say that the sequence {fn} converges uniformly on D to f, provided that for
every ε > 0, there is an index N such that
fn (x) − f (x) < ε for every n ≥ N
and every point x in D.
2. How does uniform convergence compare to pointwise convergence?
a. What is different about the definitions? (Try writing out the definition
of pointwise convergence using ε and N before comparing.)
If you write out pointwise convergence with epsilon and N, you will see
that the difference is that in uniform convergence the “for every x” only
comes at the very end after the “there is an N” part. That means that
given an epsilon you must pick an N that works for all values of x. With
pointwise the “for every x” comes before the “there is an N”, so you can
fix your x before looking for an N.
b. Is every uniformly convergent sequence of functions also pointwise
convergent?
Answer: Yes, if there is one awesome N that works for every x, then
given any x, you can just use that N.
c. Is every pointwise convergent sequence also uniformly convergent?
(Prove or disprove that the two examples above are uniformly
convergent).
The first example fn(x) = xn is not uniformly convergent. Although, if you picked
any number smaller than 1 to “end” the interval, it would be. Informally, the reason
it is not uniformly convergent is because the closer you get to 1, the bigger N you
will need to get within epsilon of zero. The way you prove it is you show that there
exists an epsilon such that for any N, you can find an x that doesn’t work.
Proof: First note that if it converged uniformly it would have to converge to
⎧0 for x < 1
since that is what it converges to pointwise. Let epsilon = ½.
f (x) = ⎨
1
for
x
=
1
⎩
N
And let N be a natural number. Now let x0 =
N
⎛ 2⎞
2
then fN(x0) = x0N= ⎜ N ⎟ = 2/3
3
⎝ 3⎠
which is farther than epsilon away from zero even though N≥N .
If we defined the sequence of functions on [0,1/2] instead, it would converge
uniformly. What you could do is, given an epsilon, find the N that works for x = ½.
This is the worst one (slowest converging point) so you should be able to prove it
works for all x’s in [0,1/2].
You use a similar kind of approach to show the other example,
⎧n 2 x
⎪ 2
⎨−n x + 2n
⎪0
⎩
if 0 ≤ x <
1
n
2
n
if
1
n
≤x<
if
2
n
≤ x ≤1
, doesn’t work. Choose epsilon to be ½ again.
Now let N be any natural number (bigger than 1, this sequence doesn’t make sense
for n=1). The idea is simply to choose a x that when plugged into a member of the
sequence of functions (with n bigger than or equal to N) gives you the peak of one
of the triangles.
Well we use the same trick we choose an x that gives us a peak when we plug it
into fN. Well the peak of this function will happen at 1/N. So that is our “bad” x.
fN(1/N) = N which is way farther than epsilon way from zero. (Again we are using
the fact that if it converges uniformly it must converge to what it converges to
pointwise – the zero function).
Note: Don’t just copy the above for your quiz! You need to work out the details
and prove it formally using the definition.
3. Example: The sequence {fn: D →R}defined by fn (x) = 1/n. Converges
uniformly to f(x) = 0.
Proof: There are no x’s in the function rules so you can easily prove this using A.P.
xk
4. Example: The sequence {fn: [0,1]→R} defined by fn (x) = ∑ , converges
k =0 k!
uniformly to the function f: [0,1]→R defined by f(x) = ex.
n
Proof: Recall from Section 8.2 (Theorem 8.10) that fn (x) is the nth Taylor
polynomial for ex at 0, and that for any x in [0,1], 0 < ex - fn (x) <
4
(n + 1)!
Now let ε >0. By the Archimedian Property, we can choose a natural number N
such that
1 ε
< .
N 4
So n ≥ N ⇒
4
4
<
< ε.
(N + 1)! N
4
4
e x − fn (x) <
≤
< ε.
(n + 1)! (N + 1)!
Then
The first funny inequality just comes from the Taylor polynomial error estimate
5. Definition: The sequence of functions {fn: D →R} is said to be uniformly
Cauchy if for every ε > 0, there is an index N such that
fn+k (x) − fn (x) < ε
for every index n ≥ N, every natural number k, and every point x in D.
6. Theorem 9.29 (Weierstrass Uniform Convergence Criterion).
The sequence of functions {fn: D →R} converges uniformly to a function f: D →R
if and only if it is uniformly Cauchy.
Proof.
“⇒” Suppose that the sequence converges uniformly to f.
Let ε > 0.
We can choose an index N s.t. n ≥ N ⇒
fn (x) − f (x) <
D.
ε
for every point x in
2
Then if k is a natural number and n ≥ N, we have that for every point x in D,
fn+k (x) − fn (x) =
fn+k (x) − f (x) + f (x) − fn (x) ≤ fn+k (x) − f (x) + f (x) − fn (x) < ε.
Thus the sequence of functions is uniformly Cauchy.
“⇐” Suppose that the sequence of functions is uniformly Cauchy. Let x be a point
in D. Then fn(x) is a Cauchy sequence of real numbers. This means that fn(x)
converges to some real number, call it f(x). So for each x in D we have a number
f(x) with the property that fn(x) converges to f(x). [This means that the sequence of
functions converges pointwise to the function f(x).]
Now we will show that it converges uniformly to this function.
Let ε > 0. We can choose an index N s.t.
fn+k (x) − fn (x) < ε/2 for every index
n ≥ N, every natural number k, and every point x in D.
Will show that
fn (x) − f (x) < ε for every n ≥ N and every point x in D.
Suppose n ≥ N and x is in D. (Note that little n is now fixed for the rest of the
proof.) Choose N1 s.t. fm (x) −
pointwise convergence).
f (x) < ε/2 for every m ≥ N1 (here we are using
Now let k = N1 – n if N1 is bigger than n (if not, just let k = 1). In either case, we
know that n+k is at least as big as N1 and it is already bigger than N because n is.
fn+k (x) − f (x) < ε/2 (using the pointwise convergence) and
fn+k (x) − fn (x) < ε/2 (using uniform Cauchy).
Then
fn (x) − f (x) =
fn (x) − fn+k (x) + fn+k (x) − f (x) ≤ fn (x) − fn+k (x) + fn+k (x) − f (x)
Thus
< ε.
So the sequence of functions converges uniformly to f.
Section 9.5 Power Series
1. Definition: Given a sequence of real numbers {cn} indexed
by the non-negative integers, we define the domain of
∞
convergence of the power series
∑c x
k
k
to be the set of all
k=0
∞
numbers x such that the series
∑c x
k
k
converges. Denote
k=0
the domain of convergence by D. We then define a function
∞
⎡ n
k⎤
f: D→R by f(x)= lim ⎢ ∑ ck x ⎥ = ∑ ck x k for all x in D.
n→∞
⎣ k=0
⎦ k=0
∞
1 k
∑ k + 1x . By the
k=0
Ratio Test, this series converges absolutely if x is in (-1, 1).
2. Example: Consider the power series
[The limit of the sequence
x k+1
k+2
xk
k +1
=
k +1
x as k goes to
k+2
infinity is x .]
By the alternating series test, the series converges if x = -1.
And by the limit comparison test (using the harmonic series),
it does not converge if x = 1. So the domain of convergence is
[-1, 1).
```
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