Math Is Fun Forum

I'm not sure.  Intuitively I would think that you could integrate in any order, but I haven't actually tried.  I set up the integrations from most to least tightly bound because that was the easiest way for me to visualize and set up the problem.

It gets really complicated really fast for something like bell curves (I don't think they even have closed form solutions), but a uniform probability distribution is pretty straightforward.  You take the integral of 1 from the lowest value to the highest value that you're interested in.  For example, to find the probability that a randomly drawn number is between 1/2 and 3/4:

which is what you'd expect.  This makes it really nice for calculating problems involving multiple probabilities.  For example, to find the probability that the first number drawn is less than 1/2 and the second number is between 1/3 and 2/3:

Again, this matches our intuition.  For this problem we are drawing 3 numbers so we need a triple integral.  I skipped showing the innermost integral in my first post since it was late and I didn't want to do that much LaTeX, and doing an integral of 1 is straightforward enough.

The important thing to keep in mind for this problem is that you have 2 breakpoints: 1/3 and 2/3.  For example, consider a simplified problem where we want to find the probability that 2 randomly drawn numbers are separated by 1/3 or less.  If our first number x is between 1/3 and 2/3 then there our second number y can be anywhere between x - 1/3 and x + 1/3, which has a 2/3 chance of happening.  But if x is less than 1/3 then y must be between 0 and x + 1/3, which has a x + 1/3 chance of happening.  So we need to handle these situations differently.

For the main problem we identified 4 distinct cases.  Here's my logic behind each of them:

Case 1 - x is less than 1/3, y is less than x.  Since x and y are both less than 1/3 this means that it is impossible for z to be too small, or in other words z is bounded below by 0.  It is bounded from above by the smaller of x and y, which in this case is y.  So overall we have 0 <= x < 1/3, 0 <= y <= x, and 0 <= z <= y + 1/3.  The integral for this is

Case 2 - x is less than 1/3, y is greater than x but less than 1/3.  Again, since x and y are both less than 1/3 then z is bounded below by 0.  As with case 1 it is bounded above by the smaller of x and y, which in this case is x.  Overall we have 0 <= x < 1/3, x <= y < 1/3, 0 <= z <= x + 1/3.  Our integral for this is

Case 3 - x is less than 1/3, y is greater than 1/3 but less than x + 1/3.  Now z is bounded below by y and above by x, giving us 0 <= x < 1/3, 1/3 <= y <= x + 1/3, y - 1/3 <= z <= x + 1/3.  This gives us

Case 4 - x is between 1/3 and 2/3, y is greater than x - 1/3 and less than x.  In this case z is bounded below by x and above by y, giving us 1/3 <= x <= 2/3, x - 1/3 <= y <= x, x - 1/3 <= z <= y + 1/3.

There is exactly 1 equivalent mirror case for each of these 4 cases, so adding these probabilities together and multiplying by 2 will give you the final answer.

Another way of stating this problem is "find the probability that at least one of the two rolls is greater than 4".  The easiest way to solve this problem is to find the opposite, i.e. find the probability that neither roll is greater than 4.  With one roll there is a 4/6 = 2/3 chance that the roll is not greater than 4.  With two rolls you need both to be 4 or less, so the probability is 2/3 * 2/3 = 4/9.  This means there is a 4/9 chance that neither roll is greater than 4, so there is a 5/9 chance that at least one roll is greater than 4.

I think he's asking about the cardinality of R versus the cardinality of C.

Is there a typo in this problem?  First of all I assume the last term in the sum is supposed to be 1/√n, not 1/n, correct?  The left inequality, n <= 1 + 1/√2 + 1/√3 + ... + 1/√n is false, in fact the exact opposite is true for all positive integers.  And the right inequality is trivial, considering that the sum grows logarithmically while n*√5 grows linearly.

It's a typo, he meant k^4 + 2k^3 + 2k^2 + k.

Hah, I can do a proof concerning quadratic equations but can't solve a simple one

As far as I can tell this isn't true.  For example, try a = -1, b = 1, c = -1.  -x^2 + x - 1 does not have 2 distinct real roots, a + b > c, but f(x) < 0 for all real values of x.

By inspection this is true for n = 0 and n = 1.  Prove it is not true for n >= 2:

Well, umm, you see, the answer is that you do have to, and my solution is wrong.  I have what should be the correct expression now, but I'm not sure how to solve it.  Here's what I have so far.  Instead of integrating from -1 to 1, you have to integrate between where the circle intersects y = c.

Again we set this to pi/3:

I'm not sure where to go with it from here.

Thinking about it some more, we can actually narrow our search even further.  Remembering that we asserted a <= b <= c, this means that

must be true.  We showed before that the RHS of this inequality is smallest when b is smallest, which in our case would be b = a.  And the minimum value allowed for c is c = a, so if the inequality

is not true then there is no solution for that value of 'a'.  Solving the inequality we get

This inequality does not hold for a >= 4, so we only need to search on a = 2 and a = 3 (note that this inequality does not hold for a = 2, but because of the quirk we noted earlier the RHS of the original inequality actually increases for a short while as b increases so it does allow for possible answers).  This leaves us with something like 10 possible triplets to check, easily done with trial and error.

There's a handful I found mathematically (i.e. not trial-and-error).  WLOG choose one variable to fix and solve for the other two, we'll go with

.  To keep from tripping over ourselves on notation let's use x and y for b and c, as they will be the variables we're solving for.  So we have

We want to find a y such that x is an integer.  The easiest way is to set the denominator to 1, so let's do that.

As long as y is an integer x will be too, so you can grab some triplets this way.  That doesn't rule out other solutions, I'll let you know if I find any more.