Dave

what kind of maths did it encompass? i have absolutely no idea about US grading systems nice job though.

in terms of learning, if you're going to start learning to program from C++ first with no prior knowledge, then you're going to have a steep learning curve. i started off with BBC BASIC, moved onto qbasic, vb, C, C++, and a couple of auxiliary languages like PHP, tcl, etc. if you want to start learning to program, vb is ideal, as it's an extremely easy language to learn. it'll also get you into the idea of OOP, which is a key concept of C++. in general, you can do anything in C++ that you care to put your mind to, but it is a harder language to learn from scratch. also, experience of C before going to C++ is rather essential unless you're really good

Quoted from BBC News: "Britain is planning its first solo space science mission in 20 years, BBC News Online has learned. Scientists hope to send a satellite into deep space to study solar influences on climate change. The Earthshine mission would showcase British expertise and provide vital data on climate change. Principal investigator Mike Lockwood believes going it alone will deliver answers more quickly than joining forces with other nations." I think this is quite an interesting article in respect to the fact that the UK has more or less always not believed in a committed space programme, and has merely only contributed to the ESA. I hope there are more projects like this to come, as Britain and the world in general can only benefit mankind from this sort of scientific activity. Read more.

my own personal method when it comes to mental math additions is to break it down into tens and units. so if i had 42 + 12 i'd say it's 42 + 10 + 2 (obviously not for something that simple). this way generally works best for me. i can kindof do the same thing for subtraction and multiplication. i'm not very good at division, however. apparently (or so i've been told), kids who are trained using abaci (i mean really trained like hell, hours per day) have amazing mental math capabilities, doing stuff like 1531 * 1084 in their heads. it comes from a basic understanding of the way numbers work.

i suggest large amounts of rocket fuel and a sturdy cardboard box from your local supermarket

clearly strange quarks rule

quite. i doubt the forums were set up so that we can all argue over which one is better.

well, i finished, so since you offered to host it here (cheers), i've included it as an attachment (it's only 3k or so). this "version" is a lot different. it has a config file (mandelbrot.conf) which details all of the things that you can do with it. it's set up to give a fairly nice initial image there's also the ability to save it as a jpeg or a png. this means youll have to have the jpeg library installed as well. have fun mandelbrot.tar.gz

bloody tripod. oh well, i'm working on a better version. it's nearly finished, i just have to fix up the conf file a bit, it's having a bit of a paddy with the colours atm.

i have to also get it working better with small numbers, cos right now it really does suck at that

it looks like this particular forum hasn't had many posts in general, which i think is a shame considering that the topic of it (fractals and chaos theory) is such an interesting and diverse topic. so, i decided to write my own mandelbrot set generator, even though there are hundreds around i suppose, and i've included the source code which is available here. you'll need a linux machine to compile and run the program, but a Makefile is included. You'll also the need the png and gd libraries and header files for gd. program usage: ./mandelbrot x1 x2 y1 y2 width iterations/pixel it's a fairly simple program to use. the mandelbrot set is defined in the complex plane, so the first four values you can input are the x and y co-ordinates on the plane for which you want to generate the image. the point (x1, y1) is measured from the top left hand corner. so the co-ordinates -2, 1, 1, -1 would generate the familiar picture of the mandelbrot set. the other two parameters speak for themselves; the width of the image you want (the height will be automatically calculated) and the iterations/pixel. it takes ~30 seconds to generate a 600x400 png with 1000 iterations/pixel on my Pentium II 400MHz. output file is mandelbrot.png. bugs/other stuff the colour is a bit crappy atm, i haven't worked out a decent way of doing a palette. if the mandelbrot.png file exists already, then it can cause problems when trying to generate another image, and it might segfault. just delete the old image file. happy mandelbrot set generating.

you're okay until you get to this line: and then you meet the problem, how do you integrate something with respect to x and y, which you can't in this particular instance. i'd get another question personally

hmm, i guess it was when i was 3 or 4. i had a book called "the planets" and remember reading it from front to cover about 40 times. after that, i studied a lot of science and maths. then i kinda got more interested in the mathematical side of it when i went into secondary education, and got hooked on the maths side of it when my teacher showed me some basic calculus when i was 14. having said that, i still do a lot of physics and mechanics and like the modules we do on stuff like the doppler effect, wave theories, etc.

i've been playing with this one for about 15 minutes now and i can't seem to make any inroads into it myself, so i thought i'd check back here to check the formula is correct. the formula you've given strictly speaking isn't an actual differential equation, but i gathered the formula you want solved is: (y^2 - x^2)dy/dx - 2xy = 0 i've tried this and although i'm no beginner to differential equations, i'm no expert either i can't see any way of seperating the variables as it stands, and if you are a beginner to differential equations, i highly recommend you go and find some more examples, as this one appears to be quite difficult. on the other hand, i may have missed something extremely obvious.

if you want to be pedantic, then |sin(x) + cos(x)| = sqrt(1+sin(2x)) but even so, i thought it was a pretty method for solving a problem that looks fairly nasty.

nice method: you can use the fact that (sin(x))^2 + (cos(x))^2 = 1 to solve this problem. clearly this implies that: :int: sqrt((sin(x))^2 + sin(2x) + (cos(x))^2)dx = :int: sqrt((sin(x))^2 + 2sin(x)cos(x) + (cos(x))^2)dx but (sin(x))^2 + 2sin(x)cos(x) + (cos(x))^2 = (sin(x)+cos(x))^2, so the integral boils down to: :int: (sin(x) + cos(x))dx which i thought was rather nifty in regard to the differential equation problem, i have looked at it, but it is (as you say) a bit of a toughie. i may have a look at it in a bit.

since i'm once again bored, I thought I'd share a problem with you. Solve :int: sqrt(1+sin(2x))dx I've seen a very nice way of doing this that made me kick myself after I'd seen it, so i'll let you all have a look at it before before spoiling the fun have fun.

if you haven't read simon singh's book on fermat's last theorem, i very much suggest you do so, since it really does give some excellent insight to not only the problem itself but the history of mathematics, along with a lot of nice little bits about different branches of maths. he condenses a lot of complicated stuff like galois theory quite nicely without being too technical about it, plus there are some proofs of various things in the back. it's more maths than science (obviously), but i definately recommend it.

i saw something similar to the regular argument used to prove 1=2 the other day, and i thought i'd share it in here since these forums seem a little dead on the maths side sometimes it goes like this: (cos(x))^2 = 1 - (sin(x))^2 then, 1 + cos(x) = 1 + sqrt(1 - (sin(x))^2) squaring, (1+cos(x))^2 = (1+sqrt(1 - (sin(x))^2))^2 now when x = 2*pi/3: (1-1/2)^2 = (1 + sqrt(1-3/4))^2 1/4 = (1+sqrt(1/4))^2 therefore 1=9 i haven't looked at it much, as mainly i find these things tedious, but i thought it was quite neat so here you are. enjoy

in a way i suppose it is cheating, but the answer previously obtained still stands. that is a solution to the differential equation stated, they just wanted to see that it can be written in it's most simplest form if you let the constant of integration be written in a logarithmic form. as for the degree, i'm currently doing my A2 levels at school/college in preparation for a maths degree at warwick or york, whichever i get into. my current plan is to do the 4 year MMath course and then go on to do my PhD in maybe some pure or applied field of maths. (i hope )

i'll do the entire thing (mainly through boredom ) you need to start by seperating the variables, which isn't all that hard: :lint: (1/y)dy = :lint: (1/a)dx therefore ln |y| = x/a + const. now, since this is a constant we're talking about here, we can do more or less anything with it. since we have ln's floating about, it would be sensible to have a constant involving a natural log, since this would still be a constant and it doesn't really do anything wrong. after all, if you had any values of x and y at different points, you'd still get the same answer for the constant. so: ln (y) = x/a + ln c therefore ln (y/c) = x/a y=ce^(x/a) as required. these types of questions are a bit misleading, because you never really know the form that the answer is going to be in. as a general rule, when i integrate something that has natural logs floating around, i tend to use a log constant unless i already know what the answer is going to be. you'll also find that a lot of the time, you'll be getting different values of x and y to find the constant.

i worry that human knowledge will come to such a point that we can't actually teach all (or even a specialized branch (maths, physics etc)) of it.

you could make a load of vbcode tags for MathML i suppose. although you need mozilla or a special plugin for IE, it does make symbols (and indeed the math) look nice

trig functions are pretty much a core part of mathematics: you can't really get away from them. i don't think Newton invented the actual terms sine, cosine and tangent, but i think the names themselves were derived from the unit circle that can be drawn. i'm not exactly sure how they fit in though.

pretty strange paper putting "solve the differential equation dy/dx with initial conditions y = 0" doesn't make any sense whatsoever. mind you, edexcel did put an unsolvable problem on one of the a-level pure maths exam paper, which wasn't really all that clever of them, so nothing ceases to amaze me