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Loose Ends and Fun Stuff
Adrian Down 16779577
August 9, 2005
1 Continuous functions on Rn
Last time, we saw some continuous function from Rk^ to R. For example, any polynomial function. A projection onto the ith^ variable, called πi, is a function of the form
f (x 1 ,... , xk) = xi
Continuity of these functions relied on: xn^ → x in Rk^ ⇔ xni → xi in R.
Theorem. f : R → Rk^ is continuous if and only if each fi is continuous, where fi : R → R and f (x) = (f 1 (x),... , fk(x)).
Proof.
xn → in R ⇒ fi(xn) → fi(x), ∀i ⇒ (f 1 (xn),... , fk(xn)) → (f 1 (x),... , fk(x))
More examples:
f : Rn^ → Rm, f (x) = (f 1 (x),... , fm(x)), ∀x ∈ Rn
f is continuous if and only if fi : Rn^ → R is continuous ∀i
2 Convex sets
Definition: E ⊆ Rk^ is convex if ∀x, y ∈ E, ∀t ∈ [0, 1], tx + (1 − t)y ∈ E.
Examples:
- Br(x) ⊆ Rk^ is convex ∀r, x (by the triangle inequality)
- any convex polygon
Convex sets have many nice properties
Theorem. E convex ⇒ E connected
Proof. E convex ⇒ E path-connected ⇒ E connected
3 Fun Stuff: e is irrational
Recall the definition of e that we built from first principles. first defined the logarithm,
L(x) =
∫ (^) x
1
dt t
E(x) = L−^1 (x) ⇒ E(x) =
∑^ ∞
n=
xn n!
E(x) has radius of convergence = +∞ ⇒ E(x) exists ∀x ∈ R. We defined the notion E(x) = ex. Definition: e = E(1) =
n= 1 n!.
Theorem. e is irrational
Theorem. f is not differentiable at x, ∀x ∈ R
Proof. Relies on the intermediate value theorem. Let x ∈ R. Write x as a decimal,
x = x 0 .x 1 x 2 x 3... Fix n ≥ 1, Let y 0 = x 0 .x 1 x 2... xn y 1 = y 0 + 10−n
(y 1 is obtained by adding 1 to the last digit of the decimal expansion). So y 0 ≤ x since we truncated the decimal, and x ≤ y 1 , since we added 1 to the last place of the truncation. Also,
10 nπy 0 , 10 nπy 1 integer multiples of π ⇒ fn(y 0 ) = 2−n^ · cos(10nxy 0 ) = 2−^1 · (−1)xn fn(y 1 ) = 2−n^ · (−1)xn+ ⇒ |fn(y 0 ) − fn(y 1 )| = 2 · 2 −n^ = 2^1 −n That was all warm up. Now things get interesting. k ≥ n ⇒ 10 k^ · yi is even for i = 0, 1 ⇒ fk(y 0 ) = fk(y 1 ) = 2−k^ · cos(2N π) = 2−k
For 1 ≤ k < n, |fk(y 0 ) − fk(y 1 )| = f (^) k′(c)|y 0 − y 1 | for some c ∈ (y 0 , y 1 ) by the Mean Value Theorem.
|f (^) k′(c)| · |y 0 − y 1 | ≤ (2−k 10 kπ) · (10−n) = 2−n^ · π · 5 k−n
⇒ |f (y 0 ) − f (y 1 )| ≥ |fn(y 0 ) − fn(y 1 )| −
k 6 =n
|fn(y 0 ) − fk(y 1 )|
by the triangle inequality. Use the estimates from before and get rid of the series terms that are 0 to get
≥ 21 −n^ −
∑^ n
k=
2 −nπ · 5 k−n
= 2^1 −n^ − 2 −n^ · 5 −n^ · π · 5 n^ − 5 4
2 −n^ · (2 − π 4
π 4
1 ⇒ 2 −n^ · (2 − π 4 ) > 2 −n ⇒ |f (y 0 ) − f (x)| + |f (x) − f (y 1 )| ≥ |f (y 0 ) − f (y 1 )| > 2 −n
One of f (y 0 ), f (y 1 ) has |f (yi) − f (x)| > 2 −n−^1. Let i = 0 or 1 such that |f (yi) − f (x)| > 2 −n−^1 , and let zn = yni. Then zn → x as n → ∞, but
|f (zn) − f (x)| |zn − x|
2 −n−^1 10 −n^
5 n 2
Thus f is not differentiable at x.
Say that I wanted to make some math, like f +g ⊆ F¯ kn^ and then put words ina+ bgn∃∀ * αβγ
