Math 499, Problem Set #1 Solutions. 1. Solve the following two ...

Math 499, Problem Set #1 Solutions.
1. Solve the following two systems as efficiently as possible.
(a) xy = 12 √ 6, yz = 54 √ 2, xz = 48 √ 3.
Solution: Multiplying the three equations yields (xyz) 2 = 186624 and thus
xyz = ±432. Now, substituting yz = 54 √ 2 into xyz = ±432 and solving for x, we
obtain x = ±4 √ 2. Similarly, we can solve for y and z so that
(x, y, z) = (±4 √ 2, ±3 √ 3, ±6 √ 6).
(b) (x + y)(x + y + z) = 66, (y + z)(x + y + z) = 99, (x + z)(x + y + z) = 77.
Solution: Adding the three equations yields
(2x + 2y + 2z)(x + y + z) = 2(x + y + z) 2 = 242. So, we find that x + y + z = ±11.
Substituting this into the original system reduces it to x + y = ±6, y + z = ±9, and
x + z = ±7. Substituting each of these equations back into x + y + z = ±11, we
conclude that (x, y, z) = (±2, ±4, ±5).
2. (a) Suppose that a, b, c are the roots of a cubic polynomial with leading coefficient 1.
What is the equation of the cubic equation in terms of a, b, c
Solution: Let p(t) be the cubic polynomial in question. By the Factor Theorem,
t − a, t − b, and t − c are factors of p(t). Hence, p(t) = k(t − a)(t − b)(t − c) for some
constant k, because the degrees of both sides are equal. Since p(t) has leading
coefficient 1, we see that k = 1. Therefore,
p(t) = (t − a)(t − b)(t − c) = t 3 − (a + b + c)t 2 + (ab + ac + bc)t − abc.
(b) Solve the system x + y − z = 0, xz − xy + yz = 27, xyz = 54.
Solution: Using part (a), let a = x, b = y, and c = −z. So, a, b, c are the roots of the
cubic equation t 3 − 27t + 54 = 0. To solve this, note that the only possible rational
roots are factors of the constant term 54 (by the Rational Root Theorem). By trial
and error, we see that t = 3 is a root and thus t − 3 is a factor. By division, we can
factor this as (t − 3)(t 2 + 3t − 18) = 0, which is readily solved to find that
t = 3, 3, −6. Hence (a, b, c) = (3, 3, −6), (3, −6, 3), (−6, 3, 3).
Finally, (x, y, z) = (3, 3, 6), (3, −6, −3), (−6, 3, −3).
3. Factor x 7 + 1 as the product of linear and quadratic factors with real coefficients.
Solution: First, we find the zeros of x 7 + 1 using De Moivre’s Theorem. Since
x 7 = −1 = cos π + i sin π, we have x k = cos( π+2πk ) + i sin( π+2πk ) for k = 0, 1, ..., 6.
7 7
I’ll write these out, and show how the complex roots pair up via conjugacy:
x 0 = cos( π) + i sin( π), x 7 7 6 = cos( 13π
13π
−π
−π
) + i sin( ) = cos( ) + i sin( ) = x 7 7 7 7 0,
x 1 = cos( 3π) + i sin( 3π), x 7 7 5 = cos( 11π
11π
−3π
−3π
) + i sin( ) = cos( ) + i sin( ) = x 7 7 7 7 1,
x 2 = cos( 5π) + i sin( 5π), x 7 7 4 = cos( 9π) + i sin( 9π −5π
−5π
) = cos( ) + i sin( ) = x 7 7 7 7 2,
and x 3 = cos π + i sin π = −1.
Now, we are almost done. By the Factor Theorem, x − x k is a factor of x 7 + 1.

Since (x − x k )(x − x k ) = [(x − cos( kπ)) − i sin( kπ)][(x − cos( kπ)) + i sin( kπ)] =
7 7 7 7
x 2 − (2 cos( kπ ))x + 1, we conclude that the desired factorization is
7
x 7 + 1 = (x + 1)(x 2 − (2 cos( π))x + 7 1)(x2 − (2 cos( 2π))x + 7 1)(x2 − (2 cos( 3π ))x + 1).
7
4. If A and B are square matrices of the same dimension and AB 2 − A is invertible,
show that BA − A is also invertible.
Solution: Since AB 2 − A is invertible, we have |AB 2 − A| ̸= 0. Since
|AB 2 − A| = |A(B 2 − I)| = |A(B − I)(B + I)| = |A||B − I||B + I| ̸= 0. So, |A| ̸= 0,
|B − I| ̸= 0, and |B + I| ̸= 0. Therefore, |BA − A| = |(B − I)A| = |B − I||A| ≠ 0,
and so BA − A is invertible.
5. Let a, b > 0. Show that lim
n→∞
(a n + b n ) 1/n = max{a, b}.
Solution: Use the fact that lim n→∞ c n = 0 when |c| < 1. If a = b, then this reduces
to lim n→∞ (2a n ) 1/n = lim n→∞ 2 1/n a = 1 · a = a = max{a, b}. Without loss of
generality, now assume that a > b. Then, lim n→∞ (a n + b n ) 1/n =
lim n→∞ [a n (1 + ( b a )n )] 1/n = lim n→∞ a(1 + ( b a )n ) 1/n = a(1 + 0) 0 = a = max{a, b}.
6. Show that there does not exist a differentiable function f : R → R such that
f(f(x)) = e −x for all x ∈ R. (It may be useful to first establish that is f ◦ f has a
unique fixed point, then so does f.)
Solution: First, we show that if x = a is the only fixed point of f ◦ f, then x = a is
also the only fixed point of f. Let b = f(a), so that f(b) = f(f(a)) = a. Hence,
f(b) = a and f(a) = b, and so we have f(f(b)) = f(a) = b. Since f ◦ f has only one
fixed point, we must have b = a, which implies that f(b) = f(a) = a. That is, a is a
fixed point of f. Moreover, it is the only one, because every fixed point of f is fixed
point of f ◦ f, and f ◦ f has only one of them.
Now, suppose that there exists f such that f(f(x)) = e −x . Check that there is only
one root to the equation e −x = x; call it a. So, x = a is the only fixed point of f ◦ f.
On one hand, since a is a fixed point, the Chain Rule gives
(f ◦ f) ′ (a) = f ′ (f(a)) · f ′ (a) = [f ′ (a)] 2 ≥ 0. On the other hand,
(f ◦ f) ′ (a) = −e −x < 0. So, we have a contradiction. .
7. Show that
∫ ∞
1
(
1
3√
t3 + 1 − 1 )
dt converges.
t + 1
Solution: We first of all rewrite the integrand before applying the Comparison Test.
First of all, we combine the fractions together.
1
3√
t3 + 1 − 1
t + 1 = (t + 1) − 3√ t 3 + 1
(t + 1) · 3√ t 3 + 1 .

Next, we rationalise the numerator via difference of cubes formula:
Simplifying yields
(t + 1) − 3√ t 3 + 1
(t + 1)( 3√ t 3 + 1) · (t + 1)2 + 2(t + 1) 3√ t 3 + 1 + 3√ (t 3 + 1) 2 )
(t + 1) 2 + 2(t + 1) 3√ t 3 + 1 + 3√ (t 3 + 1) 2 .
3t 2 + 3t
(t + 1) 3√ t 3 + 1 · ((t + 1) 2 + 2(t + 1) 3√ t 3 + 1 + 3√ (t 3 + 1) 2 ) .
Next, cancel a common factor:
3t
3√
t3 + 1 · ((t + 1) 2 + 2(t + 1) 3√ t 3 + 1 + 3√ (t 3 + 1) 2 ) .
We can increase its value by eliminating factors from the denominator. So, we obtain
1
3√
t3 + 1 − 1
t + 1 <
3t
3√
t3 (t 2 + 2t 3√ t 3 + 3√ (t 3 ) 2 = 1 t 2 for all t > 0.
Since ∫ ∞ dt
converges (to 1), we conclude that the integral in question also converges
1 t 2
by the Comparison Test.