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Yes, I'm here again, with another misunderstanding about the solution.

Question: Two positive integers $a$ and $b$ are chosen randomly, uniformly, and independently from the set of positive integers less than or equal to $1000$. What is the expected value of the number of quadrants through which the graph $x^a+y^b=1$ will pass?

Solution: Without loss of generality suppose that $a$ (or $b$, respectively) is even. Then $(x, y)$ is a solution if and only if $(−x, y)$ is a solution. Now we show that $y$ can be positive or negative. Note that $$y=\sqrt[b]{1-x^a}.$$ If $b$ is odd, then as long as $x$ is large enough, the expression will be negative. If $b$ is even, then symmetry means that $-\sqrt[b]{1-x^a}$ is a solution whenever $\sqrt[b]{1-x^a}$ is a solution.

Answer is $3.75$.

The solution shows that $y$ can be positive or negative, but there is nothing else. It also says $(x, y)$ is a solution if and only if $(-x, y)$ is, which doesn't have proof. It also doesn't explain how you get to $3.75$. Can someone please provide a better solution?

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    The argument assumes, first, that $a$ is even. Given that, then of course $(x,y)$ is a solution iff $(-x,y)$ is since $x^a=(-x)^a$. – lulu Aug 30 '19 at 01:22
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    Note: I don't understand why they say you can assume "without loss of generality" that at least one of the two is even. The curve makes sense with both $a,b$ odd so I think you need to consider that case as well. – lulu Aug 30 '19 at 01:25
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    Well, I guess it's clear that if both $a,b$ are odd then the thing passes through three quadrants. You can have both $(x,y)$ positive or one positive and one negative, but you can't have both negative. So that case is not difficult. But you still need to consider it. – lulu Aug 30 '19 at 01:27

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First let's establish the cases:

  • If $a$ and $b$ are both even, the curve $x^a + y^b = 1$ is closed and passes through all four quadrants.
  • If exactly one of $a$ and $b$ is even, the curve is not closed, but still passes through all four quadrants.
  • If neither of $a$ and $b$ is even (that is, they are both odd), the curve passes through three quadrants—the first, third, and fourth.

There are a number of ways to show this. Your algebraic argument can be strengthened and analyzed on a case-by-case basis. You can also use calculus. For any $a$ and $b$, $(1,0)$ and $(0,1)$ are on the curve $x^a+y^b = 1$. If we differentiate implicitly treating $y$ as a function of $x$: $$ ax^{a-1} + b y^{b-1} \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = - \frac{ax^{a-1}}{by^{b-1}} $$ Apparently $\frac{dy}{dx} = 0$ at $(0,1)$, which means the tangent line to the curve is horizontal there. So the curve passes into both the first (on the right of $(0,1)$) and third (on the left) quadrants.

Similarly, $\frac{dx}{dy} = 0$ at $(1,0)$, which means the tangent line to the curve is vertical there, and the curve passes into the fourth quadrant as well.

If $b$ is even, then $(0,-1)$ is on the curve, and $\frac{dy}{dx} = 0$ there. So the curve has a horizontal tangent there, and because of this passes into the fourth quadrant.

If $a$ is even, then $(-1,0)$ is on the curve, and $\frac{dx}{dy} = 0$ there. So the curve has a vertical tangent there, and because of this also passes into the fourth quadrant.

As lulu points out, if $a$ and $b$ are both odd, then $x^a + y^b < 0$ for all $(x,y)$ in the third quadrant. So the curve will only pass through three quadrants.

Now let $X(a,b)$ be the random variable which assigns to $(a,b)$ the number of quadrants $x^a+y^b = 1$ passes through. We already know there are only two values of $X$: $3$ and $4$. So $$ E(X) = 3\cdot P(X=3) + 4\cdot P(X=4) $$ We also know that $X=3$ if $a$ and $b$ are both odd and $X=4$ otherwise. Let $\mathrm{OO}$ be the event that $a$ and $b$ are both odd. Then $$ E(X) = 3 \cdot P(\mathrm{OO}) + 4 \cdot (1-P(\mathrm{OO})) = 4 - P(\mathrm{OO}) $$ The last thing to determine is $P(\mathrm{OO})$. But this is pretty clear. The probability that $a$ is odd is $\frac{1}{2}$ since there are 1000 outcomes for $a$ and half of them are odd. The probability that $b$ is odd is also $\frac{1}{2}$. Since the two events are independent, the probability that they are both odd is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$. Thus $$ E(X) = 4 - \frac{1}{4} = 3.75 $$

Matthew Leingang
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