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#201 Re: Dark Discussions at Cafe Infinity » obsessed with proofs » 2005-12-30 18:48:50
Proofs use the rules and definitions of math. I don't think you can prove the rules; you must define them before you can prove things (chicken and egg, sort of thing). Right?
2+2 = 4 is an identity. It is also a proof in and of itself, because it is evident from the rules of addition that 2+2 = 4; IOW, no further steps are required to show that it is true.
OK, Ok, ok, I'll do it:
Assert: 2+2 = 4.
By the rules of arithmetic, 2+2 = 4.
Substitute: 4 = 4
Therefore, 2+2 = 4.
Woohoo!
#202 Re: Dark Discussions at Cafe Infinity » obsessed with proofs » 2005-12-30 16:37:03
I'm not a stickler for proofs, but I believe it's to avoid basing any work on unfounded assertions, obvious though they may seem.
Perhaps the proof is not difficult, but, because it's a concept that's simply and intuitively grasped, it is not necessary to the text, and therefore beyond its scope.
#203 Re: Maths Is Fun - Suggestions and Comments » Emoticons - Smilie Expressions » 2005-12-29 17:31:52
I disagree. I personally would rather type two characters to get a smily than mess with the mouse.
#204 Re: This is Cool » Interesting suggestions about floor[x] » 2005-12-28 08:13:49
Impressive! How about this:
Floor(x) = x - (x % 1)
Where % is the modulus operator.
#205 Re: Jokes » identities » 2005-12-27 09:15:50
#206 Re: This is Cool » Merry Christmas everyone!!! » 2005-12-25 19:36:26
To all you aussies out there: your country makes the best sheepskin slippers in the world. I am in your debt.
#207 Re: Help Me ! » Venn Diagrams » 2005-12-21 06:27:31
I made one for #16 and started on the other one, but it turned out more complicated than I thought so I'll have to have a go at it later.
#208 Re: Dark Discussions at Cafe Infinity » Can you work in group or who'll be the survivor? » 2005-12-19 18:02:34
IV: White.
#209 Re: Dark Discussions at Cafe Infinity » School's out! » 2005-12-16 10:58:10
No...NOW school's out! 
#210 Re: Help Me ! » Help me » 2005-12-15 05:28:45
Agreed. I was talking more about this:
But you need this for 3 dice. So that's:
546/1296 * 546/1296 * 546/1296 = 162,771,336 / 2,176,782,336
Or a 7.478% chance.
It's really counterintuitive that the probability goes down when you play with three dice. But what I missed is that it's the probability of finishing with all of them having turned up sixes, which makes sense. 8O)
#211 Re: Dark Discussions at Cafe Infinity » Monkey Island » 2005-12-15 05:18:49
I found versions 1 and 2 on a Mac abandonware site a few months back, and played through them for the first time. The first was the better of the two, but they were both quite hilarious.
#212 Re: Help Me ! » Help with Sigma Notation » 2005-12-15 05:06:03
n is the number of iterations.
#213 Re: Help Me ! » Help me » 2005-12-14 19:45:30
Something about that just doesn't feel quite right, but I'm not good enough with probabilities to figure it out. So, I'll toss out my thinking for review.
We can lump the three dice into one theoretical 18-sided die (with 3 sixes). The probability of hitting a six on the first roll is 3/18 = 1/6. Should we wind up removing one die for the next roll, our chances are 2/12 = 1/6. For one die, of course, it's 1/6. So, as we roll along, our chances of hitting a six on one of the dice doesn't decrease.
Making final answer Ricky's first bit, or 546/1296.
#214 Re: Help Me ! » Help with Sigma Notation » 2005-12-14 19:30:48
I think this one is cleaner:
#215 Re: Help Me ! » Log Question » 2005-12-14 19:17:56
First rewrite them in exponential form:
9^x = a
12^x = b
16^x = a + b → 16^x = 9^x + 12^x
Solve for x.
You know, I'm not sure that's even possible without the aid of technology. My calculator says that:
x = ln[ (√(5) - 1)/2) ] / ln(3/4).
That's log base 3/4 of (√(5) - 1)/2), which is 1.67272.
Guess what? 12^1.67272 / 9^1.67272 = 63851199804262/39462211701495. More sanely, it = 1.61803.
Anyone who can solve 16^x = 9^x + 12^x for x by hand deserves a trophy.
#216 Re: Help Me ! » Trigonometry » 2005-12-14 06:19:28
Those angles be not on the unit circle, matey. Aaargh!
Ahem. I don't think they have coterminal angles on the unit circle, either.
Which means...use a calculator! Right? Don't tell me there's a goofy trig identity that will come to the rescue. I hates trig identities.
#217 Re: Help Me ! » Differentiation. » 2005-12-14 06:15:03
Arrrgh, mathsy be a quick one. 
#218 Re: Help Me ! » Differentiation. » 2005-12-14 06:13:47
sin(ax)
The chain rule is most convenient for this one. (If you haven't learned it, the chain rule is a rule for differentiating composed functions. If ƒ(x) = sinx and g(x) = ax, then sin(ax) = ƒ(g(x)). The chain rule goes like this: (ƒ(g(x)))′ = ƒ(g(x))′ * g′(x).)
So. (sinx)′ = cosx, and (ax)′ = a. Plug these into the chain rule, and out pops a*cos(ax).
The same principle applies to the rest of the functions in that list. Another example is (ln(ax))′ = a/ax = 1/x.
sin²(x)
Rewrite this one as (sinx)(sinx). You can then use the product rule ((ƒ(x)g(x))′ = ƒ(x)g′(x) + ƒ′(x)g(x)).
(sin²(x))′ = sinxcosx + cosxsinx = 2sinxcosx (there's probably and identity to simplify that further, but I can never remember those darn things).
Again, the same idea applies to the rest of the list.
ax*sin(x)
Product rule again. ax*cosx + a*sinx.
Or did you mean to say a*sin(x)? That's just a*cos(x). Remember that (aƒ(x))′ = a(ƒ′(x))
a x sin(bx + c)
This one's an application of chain and product rules both.
sin(bx + c)′ = b*cos(bx + c)
(ax*sin(bx+c))′ = abx*cos(bx+c) + a*sin(bx+c)
I hope that helps you. If you were looking more for explanations of why the various rules works, there are some good proofs in my calculus book...
#219 Re: Help Me ! » logarithims, remind me again? » 2005-12-13 16:36:55
(e^x - e^-x)/2 = 7
My calculator gives the same answer as your book, but I'm darned to heck if I know how it got there.
#220 Re: Help Me ! » Equation » 2005-12-13 15:57:15
a^x*a^y = a^(x+y), but it only works for a common base. Therefore,
ƒ(AxBy) = ƒ(Ax) * ƒ(By) = [ƒ(Ax) * ƒ(By)]^[(x+y)(1/x+y)] = [ƒ(Ax)^x * ƒ(By)^y]^[1/(x+y)]
if and only if ƒ(Ax) = ƒ(By).
#221 Re: Dark Discussions at Cafe Infinity » Can you work in group or who'll be the survivor? » 2005-12-12 05:08:39
So we have time to plan. Let's get the silly questions out of the way first.
1) Can't we just tell each other our hat's colors, since we have time to plan?
2) What is stopping the people in front from looking down the line in the other direction?
3) The person in back can see all of the hats. Can't he just be spokesperson and tell the colors?
4) I assume the natives are watching and wouldn't be happy if we just took our own hat off to look at the color.
5) Can we use nonverbal communication? For example, starting at the back of the line, we tap shoulders. A tap on the left indicates a black hat, and a tap on the right a white. Each person would pass on the sequence of taps until the person in front receives nine taps. Everyone then knows their hat (except he in the back).
6) Look up into the sun, only hold your hand up so it blocks the sun's sphere from your view. If you notice light reflected onto your arm/hand, you have a white hat; if not, it's black.
7) jU appears to have become an orangutan, like the Librarian in Terry Pratchett books.
#222 Re: Dark Discussions at Cafe Infinity » Can you work in group or who'll be the survivor? » 2005-12-11 17:53:24
I want IV!
#223 Re: Dark Discussions at Cafe Infinity » Speed Reading » 2005-12-09 05:50:02
PE, or Physical Education, is to teach you about Physical activities. Not to tell you the correct posture for life, your brain and body develop their own correct posture as they go on. It does what it says on the tin and it does it well.
Yes, your body adapts to whatever posture you habitually assume. The trouble is, there's gravity and whatnot, and we end up with lots of chronic back and neck pain (not serious, but irritating, no?).
Your perception of what PE does is wrong.
No, I just want to extend it.
Speed Reading shouldn't be taught. It's a bad habit that often leaves things missed out. I can do it and I often miss bits of posts.
Like all tools, you have to know when and how to use it. I'm told retention does not suffer if done properly (read: with lots of practice), and my experience is that that is true, though I've not pushed up to true "speed reading" speeds.
#224 Re: Help Me ! » Linear Equations » 2005-12-08 14:37:50
It's because, when solving things algebraically, whatever you do to the equation, you must do to both sides--every part of both sides. For example:
x/3 + 6 = 12
x/3 + 6 - 6 = 12 - 6
x/3 = 6
3(x/3) = 3(6)
x = 18
OR
x/3 + 6 = 12
3(x/3 + 6) = 3(12)
x + 18 - 18 = 36 - 18
x = 18
It really doesn't matter in what order you do things, but personally, I like the first method better. 
#225 Re: Maths Is Fun - Suggestions and Comments » Minesweeper MathsIsFun Style » 2005-12-08 12:56:31
2) I say, can win if all clear spaces are correctly marked.
3) I don't think it makes it too easy, since that info is readily available to the player that is willing to sum them up. Who wants to do that?
4) Or stops when "new game is clicked"; animations get old when we are forced to wait for them.
7) I'll draw up a suggestion later tonight.