LECTURE 1M
SHENGYANG ZHANG
Theorem. (last quarter) Rn is complete
Quick Properties. :
lim an + bn = lim an + lim bn
lim an bn = lim an lim bn
1
lim a1 = lim an if lim an = 0 (and terms not 0).
n
follow from analysis facts for continuous functions. PSET 1.
LECTURE 1W
SHENGYANG ZHANG
1. Limits and continuous functions.
Theorem. If f is continuous and (an ) converges, then limn f (an ) =
n=1
f (limn an ).
Example 1.
lim (1 +
n
1
lim f ( )
n
1
= f ( lim )
n n
continuous, f (0) = (1 + 0) = 1.
=
for f (x) = (1 +
LECTURE 2M
SHENGYANG ZHANG
Claim: for |x| < 1, limn nk xn = 0.
Proof. write |x| =
nk
1
1+c , c
> 0,
|terms|, (1+c)n 0 claim as n
lemma: if lim |an | = 0, lim an = 0.
Two methods:
1. use limn xn = 0.
1
write y k+1 = x y = x k+1 , 0 < y < 1.
nk xn = (ny n
LECTURE 2W
SHENGYANG ZHANG
1
1
Theorem. A) e = lim(1 + n )n = n! .
n=0
n
x
x
B) lim(1 + n )n =
n! (for x 0 for now)
Proof. of Theorem A.
1st step:
n=0
1
n!
1
1
1
+
+
+
2 23 234
1
1
1
+
+
= leq1 + 1 + +
2 22 222
= 3
= 1+1+
1 n
)
n
n
n
1k
= =
(1)nk
n
k
(1
LECTURE 2F
SHENGYANG ZHANG
Root Test. Given power series cn xn let R = lim sup 1 |c |1/n Radios
n=0
n n
of convergence.
Notes: if lim sup |cn |1/n = 0. Then the sequence R = +
If lim sup |cn |1/n = +
then we say R = 0.
converge |x| < R
Then cn xn =
n=0
di
LECTURE 3W
SHENGYANG ZHANG
Theorem. Root test for power series
cn xn converges for |x| < R, diverges for |x| > R, where R =
1
1
lim sup |cn | n
.
Prop. Root test
1
an converges if L < 1, diverges if L > 1 for L = lim sup |an | n .
Just comparison with geo
LECTURE 3F
SHENGYANG ZHANG
Power Series are continuous. (inside radius of convergence)
1
f (x) = cn xn , R =
1
n=0
lim sup |cn | n
Claim: f : (R, R) R is continuous.
Proof. Need to show > 0, p (R, R), r > 0 s.t. in(R, R), |x p| <
r |f (x) f (p)| < .
Fix p
LECTURE 4M
SHENGYANG ZHANG
Derivatives, little o and Big O. Recall: f is o(g) as x p means
limxp f (x) = 0
g(x)
in terms of s: > 0, r > 0 s.t. |x p| < r then
i.e. |f (x)| < |g(x)|.
f (x)
g(x)
< .
f smaller than any multiple of g in the limit
Big O: c > 0
LECTURE 4W
SHENGYANG ZHANG
Claim: If f : S R R di at p, g : T R R di at f (p) then
g f S R R is di at p with (g f ) (p) = g (f (p)f (p)
Derivative of composition = product of derivatives
dg
dg df
=
dx
df dx
Proof. f di at p f (x) = f (p) + f (p)(x p) + o(
LECTURE 4F
SHENGYANG ZHANG
Rolles Theorem. f is continuous on [a, b] and dierentiable on (a, b), f (a) =
f (b) then c (a, b) s.t. f (s) = 0.
Key Point: If f has a max or min at c (a, b) and f is di at c, then
f (c) = 0.
Proof. WLOG suppose max
limxc f (x)
LECTURE 5M
SHENGYANG ZHANG
Taylors Theorem. Given f : [a, b] R continuous on [a, b] di n + 1
times on [a, b).
Then c (a, b) s.t. f (b) = f (a) + f (a)(b a) + f
(a)
2
(b a)2 + f
(n+1)
a)n + f n+1!(c) (b a)n+1 .
f (k) (a)
n
k
k=0
k! (b a) is Pn (b) deg n T
LECTURE 5W
SHENGYANG ZHANG
Taylors Theorem V.1. Given f continuous on [a, b], (n + 1) times di on
[a, b).
c (a, b) s.t.
n
f (b) =
k=0
f (k) (a)
f (n+1) (c)
(b a)k +
(b a)n+1
k!
(n + 1)!
Remark: ame holds for [b, a], c (b, a].
Cor. Suppose f is dened on sa
LECTURE 6M
SHENGYANG ZHANG
More on ex . f (x) = ex
Proof. Claim: f (0) = 1
1
1
1
e = limn (1 + n )n = n=0 n! = 1 + 1 + 1 + 24
6
xn
x n
x =
If x 0, e
n=0 n! = limn (1 + n )
Remark: also true for x < 0 but we didnt show it
Claim: cn , R0
ex e0
x0 x 0
lim
LECTURE 6W
SHENGYANG ZHANG
puzzle. Name an explicit function f (x) which is
innitely dierentiable
positive for 0 < x < 1
2
e1/x x > 0
zeros for x 0, 1 x g(x) =
0
x0
f (x) = g(x)g(1 x).
Thm. Let cfw_Ui be a collection of open sets in R with Ui = R then ca
LECTURE 1F
SHENGYANG ZHANG
lim sup and lim inf.
Denition-Theorem. (ak ) , E = cfw_L : L is the limit of a subsequence
k=1
(ank ), limk ank .
supcfw_L : L is the limit of a subsequence (ank ) = limk supcfw_an : n k.
k=1
This value is called lim supn an .
Self-Assessment Questions Math 118A, Fall
2009
1. Every rational number can be written in the form x = m/n, where
n > 0, and m and n are integers without any comon divisors. When
x = 0, we take n = 1. Consider the functions f dened on R by
f (x ) =
0 if x
Homework 1 Math 118A, Fall 2009
Due on thursday, October 8, 2009
1. Let A1 , A2 , A3 , . . . be subsets of a metic space.
(a) If Bn = n=1 Ai , prove that Bn = n=1 Ai .
i
i
(b) If B = Ai , prove that B Ai .
i=1
i=1
2. Is every point of every open set E R2
Homework 2 Math 118A, Fall 2009
Due on Thursday, October 15th, 2009
1. Give an example of an open cover of (0, 1) which has no nite subcovering.
2. Regard Q, the set of all rational numbers, as a metric space, with
d(p, q ) = |p q |. Let E be the set of a
Homework 3 Math 118A, Fall 2009
Due on Thursday, October 22nd, 2009
1. Let A and B be separated subsets of Rk , suppose a A, b B , and
dene
p(t) = (1 t)a + tb
for t R. Put A0 = p1 (A), B0 = p1 (B ).
(a) Prove that A0 and B0 are separated subsets of R1 .
(
1. If every subsequence of cfw_pn converges to p then cfw_pn must converge to p, since it is a
subsequence of itself.
Conversely, if cfw_pn converges to p, and we let cfw_nk be an increasing sequence of natural
numbers, then > 0, N > 0 s.t. n N |pn p|
Homework 4 Math 118A, Fall 2009
Due on Thursday, October 29th, 2009
1. Prove that in a metric space (X, d), a sequence cfw_pn nN converges to
p X , if and only if every subsequence of cfw_pn nN converges to p.
2. Find the upper and lower limits of the seq
Homework 5 Math 118A, Fall 2009
Due on Thursday, November 5th, 2009
1. Consider a sequence cfw_sn nN R. Prove the following:
(a)
lim sup sn = inf (sup sk ).
n
n
kn
(b)
lim inf sn = sup(inf sk ).
n
n
kn
2. If sn tn for n N , where N is xed, then
lim inf sn
Homework 6 Math 118A, Fall 2009
Due on Thursday, November 19th, 2009
1. Consider the following space of sequences of real numbers:
lp =
cfw_s n R
|s n | p < ,
n=0
1 p < ,
and
l =
cfw_sn R sup |sn | < .
n0
For 1 p < , dene in lp the function:
s
p
=(
n=0
Homework 7 Math 118A, Fall 2009
Due on Thursday, December 3rd, 2009
1. Investigate the convergence of
an where
(a)
an =
(b)
n+1
n+1
an =
n+1
n+1
n
.
n
.
Solution:
(a) This series converges, because
n+1 n
1
c
=
3/2 ,
an =
n+1
n
( n + 1 + n)(n + 1)
and it
Homework 7 Math 118A, Fall 2009
Due on Thursday, December 3rd, 2009
1. Investigate the convergence of
an where
(a)
an =
(b)
n+1
n+1
an =
n+1
n+1
n
.
n
.
2. Let
an and
bn converge, with bn > 0 for all n. Suppose that
an /bn L. Prove that
n=N
n=N
lim
N
an