Problem Solving Answers to Problem Set 3

06 July 2012

1. Let f : ℝ^{2} → ℝ^{2} be given by f(x, y) = (x^{2} − y^{2})e^{−x^{2}−y^{2}}

(a) Prove that f attains its minimum and maximum.

(b) Determine all points (x, y) such that

\frac{∂f}{∂x} = \frac{∂f}{∂y} = 0

and determine for which of them f has global or local minimum or maximum.

Answer:

(a) Let

R = x^{2} + y^{2}.

Then

|f(x, y)| ≤ Re^{−R}.

Since the function on the right → 0 as R → ∞, while

f(1, 0) = e^{−1}, f(0, 1) = −e^{−1},

it follows that

|f(x, y)| < e^{−1}

if R > R_{0}, for some R_{0} > 0. Hence the minimum and maximum of f must be the minimum and maximum inside the bounded region R ≤ R_{0}. But we know that a continuous function will attain its minimum and maximum within a bounded region.

(b) Since

f(y, x) = −f(x, y),

it is sufficient to find stationary points when x ≥ y. If x = y then f(x, y) = 0. Differentiating at (x, x) with respect to x,

\frac{∂f}{∂x} = 2xe^{−2x^{2}}.

Similarly

\frac{∂f}{∂y} = −2xe^{−2x^{2}}.

Thus the only stationary point on the line x = y occurs at (0, 0). Since

f(ϵ, 0) > 0, f(0, ϵ) < 0

for small ϵ, it follows that (0, 0) is not a local minimum or maximum. Now let us assume x > y. Taking logarithms,

log f = log(x^{2} − y^{2}) − x^{2} − y^{2}.

Thus

\frac{∂f/∂x}{f} = \frac{2x}{x^{2} − y^{2}} − 2x,

while

\frac{∂f/∂y}{f} = −\frac{2y}{x^{2} − y^{2}} − 2y,

We are assuming that (x, y) ≠ (0, 0). If x = 0 but y ≠ 0 then

y^{2} = 1.

Hence y = −1 since we are assuming that x > y. Similarly, if y = 0 but x ≠ 0 then

x^{2} = 1,

ie x = 1. If x ≠ 0 and y ≠ 0 then

x^{2} − y^{2} = 1 and x^{2} − y^{2} = −1,

which is impossible.

Thus the stationary points are:

(0, 0), (0, ±1), (±1, 0).

Since

f(0, ±1) = −e^{−1},

and the global minimum is attained, these must be global minima. Similarly, since

f(±1, 0) = e^{−1},

these must be global maxima.

2. A sequence u(n) satisfies the relation

u(n + 2) = [u(n + 1)]^{2} − u(n),

with u(1) = 13 and u(2) = 45. Prove that 2012 divides infinitely many terms of the sequence. Answer: It is sufficient to consider the sequence mod 2012. Note that we can continue the sequence backwards as well as forwards. In particular we can go back to u(0), with

u(0) = 45^{2} − 13 = 2025 − 13 = 2012.

Now consider consecutive pairs

(u(n), u(n + 1)) mod 2012.

Since there are only a finite number of pairs modulo 2012, there must be a repeat. Let the first repeat be

u(n + r) ≡ u(n), u(n + r + 1) ≡ u(n + 1) (mod 2012)

From the recurrence relation, if n > 0.

u(n + r − 1) ≡ u(n − 1) (mod 2012)

taking the repetition one step earlier.

Hence the first repetition comes at the beginning,

u(r) ≡ u(0), u(r + 1) ≡ u(1) (mod 2012)

But then it follows that the sequence must recur every r steps, and in particular

0 ≡ u(0) ≡ u(r) ≡ u(2r) ≡ u(3r) ≡ ⋯ (mod 2012).