Dave

Indeed. There's already god only knows threads on this already.

He means really, really big numbers that are used for encyption purposes.

Looks fine to me. That's certainly we way I derive it. (It works for [math]|x| < 1[/math] btw).

I think it's intended more for personal usage than general public usage. There's already loads of anonymizing services around on the net already.

It's not like it's even expensive anymore; there are companies out there that will provide hosting for stupidly small amounts like $1/mo or even less. It usually involves a decent amount of space and you won't be able to transfer all that much per month, but it's okay for a personal site.

Your web browser will send individual http requests for each of the pages, so this won't help a lot. A better approach is to get the php script to access the page using something like get_file_contents() and then echo the output to screen. This is assuming, of course, that they aren't using content-based filters.

950 = 468 * 2 + 14: so gcd(950, 468) = gcd(468, 14) 468 = 14 * 33 + 6: gcd(468, 14) = gcd(14, 6) 14 = 2*6 + 2: gcd(14, 6) = gcd(6,2) 6 = 3*2 => gcd(6,2) = 2 => gcd(950, 468) = 2

This thread doesn't seem to have a lot to do with Linear Algebra any more, that's for sure.

Just a quick hint: To get [math]\sin(x)[/math] instead of [math]sin(x)[/math], use \sin - looks a lot nicer

Another good one for the UK customers at http://www.adslguide.org.uk/

There's no general rule for integrating [math]f(x)g(x)[/math]. You have to work it out by hand, I'm afraid. The closest you're going to get is integration by parts.

You can ask any questions you like in here, within reason And they obviously have to be math-related.

Not really Plus I'm not quite sure what you're going on about there, for instance - what if x is negative? A nice (and standard) proof goes as follows: [math](|x|+|y|)^2 = |x|^2 + |y|^2 + 2|xy| = x^2 + y^2 + 2|xy|[/math] Also, [math]|x+y|^2 = (x+y)^2 = x^2 + y^2 + 2xy[/math]. So [math](|x|+|y|)^2 - |x+y|^2 = 2|xy| - 2xy[/math] This quantity is certainly greater than or equal to zero, since [math]|xy| \geq xy[/math]. Hence: [math](|x|+|y|)^2 - |x+y|^2 \geq 0[/math], delivering the required result.

Not a problem For the record, here's the proof. Fairly straightforward, just apply seperation of variables: [math]\int \frac{1}{y^2} \ dy = \int x^2 \ dx[/math] So integrating gives: [math]-\frac{1}{y} = \frac{x^3}{3} + c[/math] Now apply initial conditions to get: [math]\frac{1}{y} = \tfrac{1}{3}(1-x^3)[/math] And from there, we get the answer.

A quadratic doesn't have any roots until you set it equal to something For example, [math]x^2 + 5x + 6[/math] has factors [math](x+3)[/math] and [math](x+2)[/math]. If [math]x^2 + 5x + 6 = 0[/math] then the equation has roots -2 and -3, but if I said [math]x^2 + 5x + 6 = 2364[/math] then you'd have different roots.

Whilst my answer is certainly true, it's probably not what you're looking for (the graph up from my post is correct). It's been a while since I've plotted inequalities. If you want real cunning, you should check out the following identity (triangle inequality): [math]|x+y| \leq |x| + |y|[/math] I use this everywhere. It's quite a nice exercise to try and prove it.

Looks good to me (I just solved it myself). I'd integrate the -1 into the bottom to give: [math]y = \frac{3}{1-x^3}[/math]

Sure - I'll use the same notation in the question. Here's the chain rule for differentiation: If F(x) = f(g(x)) then: F'(x) = f'(g(x)).g'(x) In this case, we have F(x) = f(3x), so g(x) = 3x and f is just some function. g'(x) = 3, so by the chain rule, we must have that: [math]\frac{d}{dx} f(3x) = 3f'(3x)[/math] As an example, take f = sin: [math]\frac{d}{dx} \sin(3x) = 3\cos(3x)[/math] Hope this helps.

Having said my above post, I'm not entirely sure I'm right. My brain is still in Analysis III mode; I'll have a think and get back to you.

That's the idea with the three simultaneous equations, yes. However: [math]3x^2 + 6x + 5 \neq (3x + 3)(x+ 2)[/math] Quickly expanding the brackets will show you this. In fact, that quadratic doesn't have any real roots. But for something like [math]x^2 + 5x + 6[/math], you can factor that to [math](x+3)(x+2)[/math]. Then: [math]x^2 + 5x + 6 = 0 \Rightarrow (x+3)(x+2) = 0 \Rightarrow x = -3, -2[/math].

Indeed And thanks.

Erm... no. vBulletin comes ad-free because you pay a large sum of money to own a license. The adverts here are from google adwords. We can't (and don't) ask people to click on them intentionally, as that would violate the T&C of the adwords program. We do make some income from it which goes directly towards paying for servers, bandwidth, etc.

Basically, it wants you to apply the chain rule for differentiation to an arbitrary function f. For example, when you want to work out the derivative of [math]\sin(x^2)[/math], you just use the chain rule to calculate it - i.e: [math]\frac{d}{dx}\sin(x^2) = 2x \cdot \cos(x^2)[/math] Same kind of thing here, only using a general function instead of a specific one.

When they say "show", it's usually equivalent and right to substitute the word "prove" How do we know if something is differentiable? Well, it should obviously be continuous, and moreover, the limit: [math]\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}[/math] should exist. Maybe the fact that [math]|x| = \sqrt{x^2}[/math] would help.

Not that I can see; it seems to be a rather convoluted question. The easiest way to do it that I can see is to just pick three numbers, and then solve the system of 3 simultaneous equations. There are ways of shortening this process, but it involves matrices and row reduction, so I wouldn't suggest it. This is probably one of the quicker ways. FYI: [math]b^2 - 4ac[/math] is the determinant of the quadratic; lets you know how many roots there are and/or if there are any roots of the equation [math]ax^2 + bx + c = 0[/math].