- #1

member 428835

- Homework Statement
- Given two random numbers ##x##, ##y##, what is the probability ##x/y## is closer to an odd integer than an even integer?

- Relevant Equations
- Nothing comes to mind.

Closer to odd number implies ##|y/x - (2n+1)| < 1/2## for ##n = 0,1,2...##. Then

$$-\frac 1 2 < \frac y x - (2n+1) < \frac 1 2 \implies\\

y < (2n + 1.5)x,\\

y > (2n + 0.5)x$$

for each ##n##. We note ##x \in (0,1)## implies ##y## can be larger than 1 since the slope is greater than 1 (but we know it can't be). So the integration limits would not be uniform, and thus more work is needed. Fortunately, the problem is identical if we take ##x/y##, which implies the slopes are inverse of what I wrote, and therefore always less than 1, which implies the integration limits can are ##0,1##. This implies probability ##P## is equal to

$$P = \sum_n \int_0^1 \left( \frac{1}{2n + 1.5}x - \frac{1}{2n + 0.5}x \right) \, dx \implies\\

P = \frac 1 2 \sum_n \frac{1}{0.5 + 2n} - \frac{1}{1.5 + 2n}$$. This sum is ##P = 1/5-1/7+1/9-1/11+...## which is similar to expansion of ##\log(2)## but that's all I know. This sum is not telescoping (at least not immediately so). Intuitively we know this must converge (just plot the integrals for various ##n## and it's clear the area is finite). But I don't know how to proceed. Or is leaving this in open form the best we can do?

EDIT: linebreak doesn't seem to work. Have I missed something in latest syntax?

$$-\frac 1 2 < \frac y x - (2n+1) < \frac 1 2 \implies\\

y < (2n + 1.5)x,\\

y > (2n + 0.5)x$$

for each ##n##. We note ##x \in (0,1)## implies ##y## can be larger than 1 since the slope is greater than 1 (but we know it can't be). So the integration limits would not be uniform, and thus more work is needed. Fortunately, the problem is identical if we take ##x/y##, which implies the slopes are inverse of what I wrote, and therefore always less than 1, which implies the integration limits can are ##0,1##. This implies probability ##P## is equal to

$$P = \sum_n \int_0^1 \left( \frac{1}{2n + 1.5}x - \frac{1}{2n + 0.5}x \right) \, dx \implies\\

P = \frac 1 2 \sum_n \frac{1}{0.5 + 2n} - \frac{1}{1.5 + 2n}$$. This sum is ##P = 1/5-1/7+1/9-1/11+...## which is similar to expansion of ##\log(2)## but that's all I know. This sum is not telescoping (at least not immediately so). Intuitively we know this must converge (just plot the integrals for various ##n## and it's clear the area is finite). But I don't know how to proceed. Or is leaving this in open form the best we can do?

EDIT: linebreak doesn't seem to work. Have I missed something in latest syntax?