Mathematics Department Stanford University Math 51H Homework 1

Mathematics Department Stanford University
Math 51H Homework 1
Due at TA section, Friday Sept. 26
Before starting this assignment, please read the section on
“Homework” at http://math51h.stanford.edu/51h-info.pdf
1. Explain why the following “proof” of the Cauchy-Schwarz inequality |x · y| ≤ kxk kyk, where
x, y ∈ Rn , is not valid:
Proof: If either x or y is zero, then the inequality |x · y| ≤ kxk kyk is trivially correct because both
sides are zero. If neither x nor y is zero, then from lecture x · y = kxk kyk cos θ, where, as discussed in
lecture, θ is the angle between x and y. Hence |x · y| = kxk kyk | cos θ| ≤ kxk kyk.
Note: We shall give a detailed discussion of the function cos θ later; for the moment you should of course assume that
cos θ is well-defined and has its usual properties (e.g., it is 2π-periodic, has absolute value ≤ 1 and takes each value in
[−1, 1] exactly once if we restrict θ to the interval [0, π], etc.). Thus | cos θ| ≤ 1 (used in the last step above) is certainly
correct.
2. Use mathematical induction to check that
Pn
j=1 j
3
= n2 (n + 1)2 /4 for n = 1, 2, . . ..
Note: The principle of mathematical induction is one of the basic properties of the natural numbers N = {1, 2, . . .}, and
we shall use it frequently; it says that if a subset P ⊂ N has the properties (a) 1 ∈ P, and (b) n ∈ P ⇒ n + 1 ∈ P for each
n = 1, 2, . . ., then P = N. Slightly more concretely, that is the same as saying that if Pn is a (true or false) proposition
for each n = 1, 2, . . . and if (a) P1 is true, and (b) for each n we are able to check that Pn+1 is true whenever Pn is true,
then Pn is true for all n = 1, 2, . . ..
3. Prove that there exists a cubic polynomial ax3 + bx2 + cx + d whose graph passes through the 4
points (−1, 0), (0, 1), (1, −2), (2, 5). Is this cubic polynomial the unique such?
√
4. Prove that (i) 2 is irrational (i.e. not rational), and (ii) If x, y rational, x 6= 0, and z irrational,
then y + xz, y + x/z are both irrational. (Recall that x rational means that ∃ integers p, q with q 6= 0
such that x = p/q.)
√
√
Hint for (i): If 2 is rational, then we can write 2 = p/q, where p, q are positive integers without common factors.
√
Note: 2 is defined to be the positive real number a such that a2 = 2; we’ll discuss the existence of such square roots in
week 2 as part of our discussion of the real numbers; for the moment you should just assume it exists without discussion.
5. (“In every open interval ∃ both rationals and irrationals.”)
If a, b ∈ R with a < b, prove:
(a) There is a rational r ∈ (a, b).
(b) There is an irrational c ∈ (a, b). Hint: Use part (a) and Q.4.
(c) (a, b) contains infinitely many rationals and infinitely many irrationals.
Hint for (a): Start by selecting an integer q > 0 such that 1q < b − a (i.e. q > 1/(b − a)), so that (qa, qb) is an interval
of length greater than 1; you can of course use the fact that the real numbers can be written as the disjoint union
∪n∈{0,±1,±2,...,} [n, n + 1) (which is a restatement of the important general fact every real x can be uniquely written in
the form x = n + y with n an integer and y ∈ [0, 1)—such n is referred to as the “integer part of x”).
6. By examining the proof of the triangle inequality kx + yk ≤ kxk + kyk given in lecture (recall that
proof began with the identity kx + yk2 = kxk2 + kyk2 + 2x · y), prove that equality holds in the triangle
inequality ⇐⇒ either at least one of x, y is 0 or x, y 6= 0 and y = λx with λ > 0.
7. (Another proof
of the Cauchy-Schwarz inequality.) If a = (a1 , . . . , an ), b = (b1 , . . . , bn ) ∈ Rn , prove
1 Pn
the identity 2 i,j=1 (ai bj − aj bi )2 = kak2 kbk2 − (a · b)2 , and hence prove |a · b| ≤ kakkbk.
Note: For any given aij , i, j = 1, . . . , n,
Pn
i,j=1
aij means
Pn
j=1 (
Pn
i=1
aij )), which is the same as
Pn
i=1 (
Pn
j=1
aij ))
8. Using the dot product, prove, for any vectors x, y ∈ Rn :
(a) The parallelogram law: kx − yk2 + kx + yk2 = 2(kxk2 + kyk2 )
(b) The law of cosines: kx − yk2 = kxk2 + kyk2 − 2kxk kyk cos θ, assuming x, y are non-zero and θ
is the angle between x and y as discussed in lecture.
(c) Give a geometric interpretion of identities (a),(b) (i.e. describe what (a) is saying about the
−→
−→
−→
parallelogram determined by x, y—i.e. OACB where OA = x, OB = y, OC = x + y, and what (b) is
−→
−→
saying about the triangle determined by x and y—i.e. OAB , where OA = x, OB = y).