Advanced Calculus Study Guide 1

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The Real and Complex Number Systems Definitions

(a) Let S be a set. An order on S is a relation denoted by <, with the properties:

(i) If x, y ∈ S then one and only on of the statements

x < y, x = y, y < x (Law of Trichotomy) is true.

(ii) If x, y, z ∈ S such that x < y and y < z then x < z, i.e. the transitivity holds.

(b) An ordered set is a set S in which an order is defined.

Examples N, Z and Q are ordered sets if x < y is defined to mean y − x is a positive.

(c) Suppose S is an ordered set and E ⊂ S.

(i) If there exists a β ∈ S such that x ≤ β for every x ∈ E, we say that E is bounded above, and call β an upper bound of E.

(ii) If there exists a α ∈ S such that α ≤ x for every x ∈ E, we say that E is bounded below, and call α a lower bound of E.

(d) Suppose S is an ordered set, E ⊂ S and E is bounded above. Suppose there exists an α ∈ S with the following properties:

(i) α is an upper bound of E.

(ii) If γ < α then γ is not an upper bound of E, i.e. there exists an x ∈ E such that γ < x ≤ α.

Then α is calledthe least upper bound (or the supremum) of E, and we write α = lub E (or sup E).

(e) Suppose S is an ordered set, E ⊂ S and E is bounded above. Suppose there exists an α ∈ S with the following properties:

(i) α is a lower bound of E.

(ii) If γ > α then γ is not a lower bound of E, i.e. there exists an x ∈ E such that γ > x ≥ α.

Then α is calledthe greatest lower bound (or the infimum) of E, and we write α = glb E (or inf E).

Examples (a) Let

S = Q

A = {p ∈ Q | p > 0 and p² < 2}

B = {p ∈ Q | p > 0 and p² > 2}.

Then

(i) A is bounded above by every member of B,

(ii) Since B has no smallest member in Q, A has no least upper bound in Q.

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(iv) Since A has no largest member in Q, B has no greatest lower bound in Q.

(b) Let

S = Q

E₁ = {r ∈ Q | r < 0}

E₂ = {r ∈ Q | r ≤ 0}.

Then

(i) E₁ ⊂ E₂,

(ii) sup E₁ = sup E₂ = 0, (iii) 0 6∈ E₁ and 0 ∈ E₂. (c) Let

S = Q

E = {1

n | n ∈ N}.

Then

(i) sup E = 1 ∈ E, (ii) inf E = 0 6∈ E.

Definitions

(a) An ordered set S is said to have the least-upper-bound (lub) property if the following is true:

If E ⊂ S, E is not empty and E is bounded above, then sup E exists in S.

(b) An ordered set S is said to have the greatest-lower-bound (glb) property if the following is true:

If E ⊂ S, E is not empty and E is bounded below, then inf E exists in S.

Remarks

(a) If E = {p ∈ Q | p² < 2}, since sup E = √

2 /∈ Q, Q does not have the least-upper-bound property.

(b) An ordered set S has the lub property if and on if S has the glb property.

(c) Theorem Suppose S is an ordered set with the lub property, B ⊂ S, B 6= ∅ and B is bounded below. Let L be the set of all lower bounds of B. Then

α = sup L exists in S and α = inf B.

In particular, inf B exists in S.

(d) Properties of sup E

(i) If sup E exists then it is unique.

(ii) If α = sup E exists then either α ∈ E or α /∈ E.

(iii) If γ < sup E, then ∃ x ∈ E such that γ < x ≤ sup E.

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(iv) If E ⊆ F , then sup E ≤ sup F .

Proof Let x ∈ E. Then x ∈ F . Thus x ≤ sup F . Thus sup F is an upper bound of E.

Thus we say sup E ≤ sup F.

Definitions

(a) A field is a set F with two operations, called addition + and multiplication ·, which satisfy the Axioms for addition, the Axioms for multiplication and the distributive law, i.e. a field F is an abelian group under the addition +, with 0 as additive identity; the nonzero elements are an abelian group under the multiplication ·, with 1 as multiplicative identity;

the multiplication is distributive over the addition.

(b) An ordered fieldis a field F which is also an ordered set such that (i) if x, y, z ∈ F and y < z then x + y < x + z,

(ii) if x, y ∈ F, x > 0 and y > 0 then xy > 0.

Example Q is an ordered field.

Definitions

(a) A sequence {x_n} ⊂ R is said to converge to x ∈ R if for every > 0 there is K ∈ N such that if n ≥ K then |x_n− x| < .

(b) Let E be a subset of R. We define the closure of E,denoted ¯E, by {y | ∃ {x_n} ⊂ E s.t. lim

n→∞x_n = x}.

(c) Let E be a subset of S. Then E is called adense subset of S if ¯E = S.

Definitions

(a) Monotone Sequence Property Let F be an ordered field. We say F has the monotone sequence property if every bounded above monotone increasing sequence converges.

(b) Completeness Property An ordered field is said to be completeif it obeys the monotone sequence property.

Remarks

(a) R = {r | ∃ {pn} ⊂ Q s.t. lim

n→∞p_n = r}, i.e. Q is a dense subset of R.

(b) Proposition If the ordered field F has the least-upper-bound property then F has the monotone sequence property.

Theorem There exists an ordered field R which has the least-upper-bound property. Moreover, R contains Q as a subfield.

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(a) (Archimedian Property of R) If x, y ∈ R and x > 0 then there is a positive integer n such that nx > y.

(b) (Q is dense in R) If x, y ∈ R and x < y then there is a p ∈ Q such that x < p < y.

Proof y − x > 0. Thus ∃ n ∈ N such that n(y − x) > 1 by Archimedean property. Also by the Arch. property, {nx} is not bounded (x 6= 0). Choose m such that m − 1 ≤ nx < m, where m ∈ N. Then x < m

n < y because ny > nx + 1 ≥ m.

(c) For every real x > 0 and every integer n > 0, there is one and only one real y such that yⁿ = x.

(d) If a and b are positive real numbers and n is a positive integer, then abn

= a^1/nb^1/n.

Definition Let n ∈ N. Real Euclidean Space of Dimension n si the set of all n-tuples Rⁿ = {(x₁, . . . , x_n) | x_i ∈ R}. Addition is component-wise, with ~0 as the additive identity. However, there is no multiplication that satisfies the axioms of a field.

Remarks

(a) Rⁿ is a vector space, not a field.

(b) Rⁿ has an inner product:

~ x · ~y =

n

X

i=1

x_iy_i and norm

k~xk =√

~ x · ~x.

Complex Numbers Remarks

(a) Rⁿ can be given structure of a field. Consider

(+) (a, b) + (c, d) = (a + c, b + d) (·) (a, b) · (c, d) = (ac − bd, ad + bc).

With this structure (R², +, ·) is called C.

(b) C is a field.

(c) (0, 0) is the additive identity. (1, 0) is the multiplicative identity.

(d) (a, 0) ↔ a ∈ R. (0, 1) is called i. Note

(0, 1) · (0, 1) = (−1, 0).

We can write (a, b) ↔ a + bi.

Definition The length or modulus of z is |z| =√ z ¯z.

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Note Complex numbers have the following properties:

(a) z + w = ¯z + ¯w.

(b) zw = ¯z · ¯w (c) |z| = |¯z|

(d) |zw| = |z||w|

Triangle Inequality |z + w| ≤ |z| + |w|.

Proof

|z + w|² = (z + w)(¯z + ¯w)

= z ¯z + w ¯w + zRe(z ¯w)

≤ |z|² + |w|²+ 2|z ¯w|, because modulus includes imaginary part

= (|z| + |w|)²

Notation. Henceforth, ~a, ~b ∈ Cⁿ will be simply denoted a, b.

Definition Cⁿ = {(z₁, · · · , z_n) | z_i ∈ C}. The inner product of a, b is given by

ha, bi =

n

X

j=1

aj¯bj ≥ 0.

Cauchy Schwarz Theorem Let ¯a, ¯b ∈ Cⁿ. Then

a_jb_j|² ≤X

|a_j|²X

|b_j|²

Proof Let λ ∈ R. Consider

ha, λbi = X

a_jλ¯b_j = λX

a¯b_j = λha, bi.

Also note that kλbk = λkbk. Thus

ha, bi ≤ kak · kbk ⇐⇒ ha, bi ≤ kakkλbk, λ ≤ 0.

This allows us to say without the loss of generality, kbk = 1. Now we will decompose a into its components parallel and perpendicular to b.

proj_ba = ha, bi

kbk b = ha, bib.

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Let c = proj_b⊥a = a − proj_ba. (c is a vector but here we’ll leave off the arrow.) We then compute kak² = ha, ai

= hha, bib + c, ha, bib + ci

= |ha, bi|²kbk²+ kck²

= |ha, bi|²+ kck²

≥ |ha, bi|²,

since kck² is nonnegative. Furthermore, since kbk = 1, we can conclude kakkbk ≥ |ha, bi|, as desired.