It's probably because you said
∏=3.1415926535(etc)406286
instead of
∏=3.1415926535(etc)406286...

Anyways...

Do you have any tricks to calculate the probability of an event?

The ACT really isn't a knowledge test. It's more logic. I got high scores on it without a calculator.

Yes, by all means lurkstick around until someone is wrong so you can jump up and point it out.

...?

I feel like everybody is just out to make me look stupid. I mean I was identified as a genius! I sat 6th chair at state orchestra as a sophomore! Cut me some slack!

I hope you're joking.

I think he is. Or she. I really don't know or care.

So you're used to dealing with mundanes and can lord it over them with your mighty intellect?

It's not going to happen here. Get used to it. Here any slight defect in your posts can and usually will be pointed out. There is no malice involved. You can use that response as an opportunity to improve your precision, or you can go and sulk in a corner. Your call.

I was never really that good at math anyway. I like the arts better. Puzzles are fun too I guess.

f(x) = x³ + xlog(x) - x, g(x) = x⁴+x

How would I go about proving that f(x) = O(g(x)) using the definition of Big-O? f(x) may not be O(g(x)) but from looking at it I think it is. I just don't know how to prove it.

For x ≥ 1, x³ + xlog(x) - x ≤ x⁴ + x(x) + x⁴ ≤ x⁴ + x⁴ + x⁴ ≤ 3x⁴ ≤ 3(x⁴ + x).

Thus f(x) is O(x) with witnesses N ≥ 1, k = 3.

How do you come up with the (x)+x⁴ in the second part of the first inequality and why do you do that?

There are three terms in the expression for f(x): "x³", "xlog(x)", and "-x". We want to show that the sum of these terms is less than or equal to another expression which we can trivially show is O(g(x)). The easiest way to do that is to show that each of the terms is less than or equal to x⁴, for x ≥ 1, and combine them. Note that since both x and log(x) are ≤ x for x ≥ 1, we have that xlog(x) ≤ x(x) ≤ x⁴. As for the right term, it should be apparent that -x ≤ x⁴ for x ≥ 1. And the same holds for the first term.

Edit: Fixed a "≤".

Do you have to do anything special when the bases are different? Like f(x) = log₃n and g(x) = log²n

log_bx is lower order than x, regardless of the value of b.

More specifically, it's because to change a base of a logarithm, you simply divide by the logarithm of the new base, which is a constant term.

log_bx = log_ax / log_ab

log_ab is a constant and so it doesn't matter to big O.

Gotcha. Okay last one and I have to use limits for this one. f(x) = n² and g(x) = n^logn. Should I use L'hopital's Rule or is there a better way?

How would I go about solving the equation x^5-5x^3+4x=0?

You can factor out an x to get x(x⁴ - 5x² + 4) = 0, which gives you one root (x = 0). Then you can factor the remainder of the expression into (x² - 1)(x² - 4) = (x - 1)(x + 1)(x - 2)(x + 2), giving you the remaining roots.

This equation is solvable this way only because, once the first x is factored out, we have a quadratic equation in terms of x².

This has confused me for a while now:
Someone shoots an arrow at a target. Before it reaches the target, it has to travel the first half of the distance. Then it must travel half of the remaining distance. And so on. How does it reach the target?

It must obviously reach the target (assuming that it is a straight shot, gravity is not taken into account, the force is great enough, etc.) so how DOES it do so?

Okay... So where's the issue?

The issue is that the arrow must move through an infinite number of intervals to reach the target. The solution is a proper understanding of infinitesimals and infinite series. The sum of the distances traveled by the arrow is ∑d/2ⁿ for n = 1 to ∞, where d is the total distance to the target. And value of that infinite series is d. No paradox.

If only someone had taught Zeno calculus, we could have avoided this whole mess.

Indeed!

Well thank you! Much clearer! (mud is clearer than total darkness, right?)

I sort of understand where you're coming from, but I haven't taken a course that taught those kinds of things yet.