Dave

Darn, almost got there. Maybe tomorrow

Bloody hard stuff to watch; I can't stand it most of the time and have to leave, keeping up with the score on BBC News.

I will tell you that the correct answer for question 2 is -2/125.

Okay, I'm going to get you started on the second question, since it involves some tricky algebra. In this case, (for y=1/x2), we have P being (5, 1/25) and Q being (5+h, 1/(5+h)2). So our gradient function will be [math]\frac{\frac{1}{(5+h)^2} - \frac{1}{25}}{h}.[/math] Simplifying this down is by no means a simple task. The best way to approach this kind of thing is to combine the numerator into a single fraction. To do this, we simply do the same thing that you'd do when you're doing, say, 1/2 - 1/3: you convert them both into sixths by multiplying the first fraction by 3 and the second fraction by 2. So we get: [math]\frac{\frac{1}{25+10h+h^2} - \frac{1}{25}}{h} = \frac{25 - (25+10h+h^2)}{25h(25+10h+h^2)}[/math]. I'll let you take it from here, but it should be fairly obvious now.

Right, let me try and sort out this problem. What we're trying to do is find the exact value of the gradient of a curve at one particular value of x - for example, x = 2. Again, let's consider a point P. We're going to want P to be the point at which we want to measure the gradient, so the co-ordinates of P are (2, 4) - 4 because we're using the equation y = x2. Now we have our point Q, which is a little bit up the line from P. So we're going to let P have the x-value x = 2+h, where h is very small. Now for y = x2, we're going to have the point Q at (2+h, (2+h)2). So, as before, we're going to work out the gradient of PQ. This is exactly the same process as you'd use to work out the gradient of a line lying between two points: we simply take the difference of the y-coordinates and divide this by the difference of the x-coordinates. We're going to get the gradient to be equal to: [math]\frac{(2+h)^2 - 4}{(2+h) - 2} = \frac{(4+4h+h^2) - 4}{h} = \frac{4h+h^2}{h} = 4+h[/math]. So this is the gradient of the line PQ. As h goes towards zero, we're going to get the gradient going towards 4, so this is our gradient at the point x=2. I really don't know how to explain this any more clearly without someone telling me the specific problem that they have with my explanation. I'm going into my second year of maths at uni, and (without trying to sound arrogant) I find this relatively easy now which makes it fairly difficult to explain. Replies are welcome

Fair enough. However, it still stands that there will only ever be a h on the bottom of the fraction when you consider this particular limit. There is only so much I can explain over an internet forum; as I've said, it's only a supplement to proper teaching. Trying to convey an abstract idea like this over the net is not an easy thing to accomplish, and I'm sorry for the confusion. If it would help, I will post the application of this method for differentiating x2 for any value of x, rather than just a specific value.

They're just bad. Especially the ones about the molecular biologist.

Hmm. I guess he must be talking about the chemical thing like Sayo said. Damn, this is gonna bug me now.

I think before space tourism was going to take off, we'd need some kind of space elevator a la 3001. That way it makes it easy to get into space and fly to various places, although I don't know the technicalities involved in this; I daresay they're rather large.

Highly doubt it's any of those. Although the one involving albatrosses made me laugh

I shouldn't try it, it's pretty hard to get it working like that. Maybe I'll knock something up at some point, probably involve making two images of the same thing and using a rollover or something.

And please: * If you know anyone that's interested in this stuff, tell them about the thread. * Rate the thread according to how good you think it is

Yes, that's really the worst bit; whilst I was writing it I thought it might confuse people a bit. That's the most abstract part of the entire thing. When you do differentiation from day to day, you'll find that you more or less never use a limit. It's only in very rare circumstances (or if you're doing a maths degree) that you will. But I do think it's important to realise and understand what's going on when you differentiate a function. Next lesson will probably be published either on Wednesday or Thursday; doesn't take me too long to write, it's just thinking of examples. I'll be extending the method and starting to talk about some real applications to different functions.

rofl Really makes me want to click it

C/C++ may be more technical, but you can make practically anything in them. I don't think a lot of enterprise systems are made using Java either. But yes, Visual Basic is an ideal language to start on, and it'll ease you into programming nicely.

Although to be honest I don't know why you wouldn't want to run xfree. There's no point in not having it.

To be fair, when the water starts boiling off, there will be a decrease in GPE because the beaker is losing mass; but that's besides the point.

Hey, I'm Catholic (although not a very good one ). Welcome to the club.

I was going to say the union of professional tennis players, but then I realised the T and P are the other way around.

That was an absolutely cracking game. Although I did almost die when Henman lost the service game that would've won him the match. Puts me through agony every time.

Gah, lazy admins

Cheers Congrats on your high result as well

rofl In retrospect, I shouldn't've posted the second question because it does require quite a lot of tricky algebra to get the limit out. Nevermind. Okay, question number 1. We want to find the gradient of y = x3 at x = 5. Let's do exactly the same as what we did before, and pick a point P. When we pick this point, we want it to be the point at which we want the gradient. So in our case, P is going to be the point (5, 53) = (5, 125). Now we want to pick Q, the point which we're going to move towards P. We basically want the x-coordinate to be slightly bigger than the one for P, so let's assign some value h to be the distance between P and Q. Then the co-ordinate representing Q will be the same as P's, plus some value - we'll call it h. So we have our two points: P at (5, 125) and Q at (5+h, (5+h)3). Now we're going to find the gradient of the line PQ, because we know that as we decrease the distance between the points, we're going to get a better and better approximation for the gradient of x3 at x = 5. So our gradient is: [math]\frac{(5+h)^3 - 125}{h} = \frac{h^3 + 15h^2 + 75h}{h} = h^2 + 15h + 75[/math]. Now the bit to where people are getting confused. Basically, we want to work out the limit of this function as h gets very small. The easiest way to do this is just to put h=0 into the equation and we're going to see that the first two terms disappear, leaving us with the correct answer of 75. Don't worry too much if you're quite confused by this; it's hard to get your head around at first. Once we get into the differentiation a bit more, it'll become easier to understand.

Group theory is very interesting because you can use it to define basic addition and multiplication in a rigourous sense; indeed, if you start with set theory you can basically build up the natural numbers, the rationals, reals, etc and all the operations that can be done with them.

Cheers