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Geek Culture / Math Question 2

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mamaji4
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Posted: 24th Oct 2007 12:53 Edited at: 24th Oct 2007 16:20
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This is for an assignment I am doing:
I am looking for two functions F and G such that
FoG = GoF
where 'o' denotes 'composite'
where F is not the inverse function of G and vice versa

e.g. If F = x^y , G cannot be the inverse of F
I require the function values to be in the base 10 and not in Naperian logarithm form.

Please help. Chris K. Peter H. Anybody?
And I can do without the jokes about FoGGY math.
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Peter H
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Posted: 24th Oct 2007 14:51 Edited at: 24th Oct 2007 15:29
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G(x) = x
F(x) = x

that's probably cheating isn't it... sorry too early for me right now

[edit] ... maybe i've gone crackers but this
"I am looking for two functions F and G such that
FoG = GoF"
looks alot like the definition of inverse functions to me

[edit2] the answer might be that this is impossible... but i hate choosing that because it makes me look stupid if it is possible

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demons breath
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Posted: 24th Oct 2007 15:39
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it does seem impossible but, I don't know it probably is.

and no the definition of inverse would be FoG=1 wouldn't it?

Oh wait, isn't that reciprocal.

Are they the same thing?

Your method works though lol. If they're the same then it would work...

oh **** that smells like my lunch burning

I'll ponder more upon this later

*runs off to the kitchen*

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mamaji4
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Posted: 24th Oct 2007 15:54 Edited at: 24th Oct 2007 15:57
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@Peter H
My mistake. I forgot to mention that the F and G are distinct. i.e. F <> G
I'm up the creek because I need F and G such that neither is the inverse of the other.

@DB
Yup. You're right. If G is the inverse function of F then the problem is trivial.
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demons breath
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Posted: 24th Oct 2007 15:57
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you have no boundaries on what the functions can be?

you don't have like a limit on indices or anything?

this seems pretty impossible...

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mamaji4
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Posted: 24th Oct 2007 15:59
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Oh. And one more thing. F,G need not be a functions of a single variable. Multivariable functions are allowed.
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mamaji4
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Posted: 24th Oct 2007 16:01 Edited at: 24th Oct 2007 16:20
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@DB

How shall I put it. The problem has been driving me nuts for the past few weeks and I need it in all urgency.
I posted on some Math forums also but no one came up with a feasible solution.
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demons breath
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Posted: 24th Oct 2007 16:28
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so F could be like 17x^y+bc^32-SQRT(a*(b/c)) or something like that? In that case, I'm not going to have a clue to be honest. And I wanted to do maths at uni, this is kind of disheartening

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mamaji4
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Posted: 24th Oct 2007 16:37
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@DB
Geeze man, go easy on yourself DB. There are some problems that are always perplexing in math. Doesn't mean you don't have to get a strong foundation in one of the most essential subjects, especially if you want to do a CS major. I would say take Math. It will take your analytical skills to the next level.
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demons breath
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Posted: 24th Oct 2007 17:33
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Quote: "especially if you want to do a CS major"

haha mamaji worse than that I want to do a maths major, so...

but yeah sorry man I can't figure out this one. Is it just specific numbers that will work or are you searching for a general rule?

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mamaji4
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Posted: 24th Oct 2007 19:35
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I need two single or multi variable functions which satisfy the above condition of equality.
I'm personally beginning to think that maybe it is an imposssiblility, because the equality FoG = GoF is how you state the inverse theorem.
If G is the inverse function of F then
FoG = GoF - equn.(1)

So if that is a fundamental theorem, I am essentially trying to disprove it, by providing an exception to the rule.
I wonder if I would have to theoretically prove that G need NOT be an inverse of F, and yet equn.(1) would hold true.
It would be easier than groping in the dark with a trial and error approach trying to find F and G when it is theoretically an impossiblity. It might just save me a lot of effort.
But then providing a theroetical proof has two caveats
1) I would actually have to arrive at a contradiction to disprove the above theorem
2) Even after disproving it, I wouldn't be sure that my proof is accurate unless I had confirmation from others.
So until then I might arrive at a solution by random trial and error.
I'm stuck
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Diggsey
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Posted: 24th Oct 2007 19:40
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F = x+2
G = x+1

I'm probably misunderstanding the question though! If you could put it in psuedo code or something I might understand

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mamaji4
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Posted: 24th Oct 2007 23:14 Edited at: 24th Oct 2007 23:21
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@Diggsey

Yes. This is because addition is associative for real numbers.
i.e. (a + x) + b = a + (x + b)
Similar to exponentiation being commutative
i.e. x^mn = x^nm
which leads us to
F = x^m
G = x^n
as another possible solution also satisfying the equality FoG = GoF

The fault is mine for not explaining the reason I need the above equality.
I need to find G such that FoGoF = G
This is only possible if I can find the function G that satisfies
FoG = GoF and G must not be the inverse of F
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mamaji4
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Posted: 25th Oct 2007 00:57
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Ok. Thanks all for your input. I think I'm pretty close to the solution, though not too close. But I think I have a handle on the answer. Ooh my head hurts.
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Sopo the tocho
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Posted: 25th Oct 2007 01:00 Edited at: 25th Oct 2007 01:00
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beware, threads can get locked for ask questions...


Intel Pentium core 2 duo T6600 2,6 mhz 4mb, 4 gb ram 600 mhz ddr2
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mamaji4
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Posted: 25th Oct 2007 01:06
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huh
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demons breath
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Posted: 25th Oct 2007 01:14
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sopo just had a thread locked. I don't think this one will be though, because sopo's could have been asked to much better effect elsewhere and was to be honest a bit rubbish.

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Sopo the tocho
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Posted: 25th Oct 2007 01:48 Edited at: 25th Oct 2007 04:17
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erased 33

erased
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Seppuku Arts
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Posted: 25th Oct 2007 02:39
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Sopo, the mods aren't associated with the company, they keep the place tidy. The mods decide a clean up is necessary to get things back into shape and threads are getting locked that they feel don't need to be here as there a places where certain threads are much more relevant.

Also, please don't be vulgar or rude - heck don't direct those comments at people, for one thing there are minors on the forum and somebody can easily be offended.

And please don't take the lock personally, it's not a personal thing.



Chances are you're going to be told off, slapped on the wrists or something.

I love Nancy DrewG, but not insert brain here
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Jimpo
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Posted: 25th Oct 2007 03:00
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Quote: "FoGoF = G"

Why wouldn't something simple like Diggsey said work?
F=x
G=x+1

Shooter - Game Finished
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mamaji4
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Posted: 25th Oct 2007 14:08 Edited at: 25th Oct 2007 15:28
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Quote: "Yes. This is because addition is associative for real numbers.
i.e. (a + x) + b = a + (x + b)"


Yes, it would. As I mentioned earlier.
For F = x
G = x + 1
a = 0 and b = 1 in the above equation on associativity
Also, F = x is an identity function over the composite operation(and Peter did suggest it in an earlier post)
i.e. GoF = G which is not suitable for the problem at hand.

I am looking for a an exponential function to the base 10 (although not a strict exponentiation) as I mentioned in the first post and although I suggested a solution
F = x^m
G = x^n
they are not identical, for the purpose of the problem I have. They are similar functions of the type x^y
And though these don't work for FoGoF = G - equn. 2
I have an exponential function which does work for equn. 2 but then fails to satisfy the condition that G cannot be a similar exponentiation function as F
In effect I would literally have to post all my working proofs and the requirements and specifications thereof which seems a bit cumbersome. I do seem to have found a function G that is not similar to F(i.e. not an exponentiation) and yet satisfies the requirement of equn. 2
Anyways, thanks for all the inputs and the help.
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demons breath
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Posted: 25th Oct 2007 16:01
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how's it an exponential function with a base? isn't that a logarithmic function?

x^m has no base 10 in

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mamaji4
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Posted: 25th Oct 2007 16:09 Edited at: 25th Oct 2007 16:10
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Quote: "I am looking for a an exponential function to the base 10 (although not a strict exponentiation)"


Yup. It is not strictly exponential, and has another part which is to be taken to the base 10. Can't afford to have the Naperian e creeping in anywhere in the original function.
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Chris K
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Posted: 25th Oct 2007 16:20
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What was wrong with f( x ) = x + 1 and g( x ) = x + 2?

fg( x ) = gf( x ) = x + 3...

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demons breath
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Posted: 25th Oct 2007 16:51 Edited at: 25th Oct 2007 16:52
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He added that FoGoF = G

x+4 =/= x+2


EDIT: and is it just your natural aversion to ln and e^x or is it not supposed to be those?

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mamaji4
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Posted: 25th Oct 2007 17:05
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Geeze DB, get off my back.
And no I don't have an aversion to e. I love it.
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demons breath
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Posted: 25th Oct 2007 17:06
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hahaha sorry man just trying to help. I'll be quiet now.

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mamaji4
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Posted: 25th Oct 2007 17:19
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Hey James, just saw your photo on tgcapollo and you need a haircut.
Of course I should be the last one to mention that because I'm always in need of one myself.
And it's irritating when you can't post for a week. Gotta wait for a week...
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demons breath
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Posted: 25th Oct 2007 17:31
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SHHH everyone tells me that. One of the main reasons I don't do it.

It's my 18th on sunday though and one of my mates is convinced that to celebrate he's going to shave my head. Someone's going to get really hurt if he tries. Probably me, 'cos he plays rugby, and we'll both be VERY drunk by that point so we won't be holding back if we start scrapping.

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mamaji4
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Posted: 25th Oct 2007 17:35
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lol. I'm lucky I don't have any violent mates.
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Chris K
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Posted: 25th Oct 2007 17:58
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Quote: "FoGoF = G"


Ok then you can have f( x ) = abs( x ) and g( x ) = x^2

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demons breath
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Posted: 25th Oct 2007 18:08 Edited at: 25th Oct 2007 18:08
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and both equations need indices, I think.

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Chris K
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Posted: 25th Oct 2007 18:11
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?!?[/b][/i][b][i][/b][/i][b][i]

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demons breath
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Posted: 25th Oct 2007 18:13
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Quote: "I am looking for a an exponential function to the base 10 (although not a strict exponentiation) as I mentioned in the first post and although I suggested a solution
F = x^m
G = x^n
they are not identical, for the purpose of the problem I have. They are similar functions of the type x^y"



Just out of interest, what is the actual question?

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mamaji4
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Posted: 25th Oct 2007 18:27
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James
I thought you agreed to get off my back. You liar.

Chris
I hate repeating myself but I don't need an identity function over the composite operation.
abs(x) is the same as x , except that x can take positive or negative values.

I hope I can rest in peace now, cause I already have the solution I was looking for.
AAAARRRGHHHH! I can see Chris, Peter and James chasing me with a weapon that seriously resembles a math question. (*I wake up in a cold sweat. Its only a nightmare. Falls asleep again*) Peace shall finally be mine
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demons breath
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Posted: 25th Oct 2007 18:29
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Hahaha I didn't know you actually had the answer.

Care to share it?

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mamaji4
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Posted: 25th Oct 2007 18:38
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Well, ummmm, aahhhh.....

Dang, I knew it was you James. I was wondering who the guy with the masked face was. I quickly grab the math question from James hand and turn it around. He runs , but cannot escape...
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demons breath
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Posted: 25th Oct 2007 19:31
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AGHHHH

impaled by mine own sword

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Chris K
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Posted: 25th Oct 2007 20:14 Edited at: 25th Oct 2007 20:18
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What's your solution then? 1/x and x^m?

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mamaji4
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Posted: 25th Oct 2007 23:43
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Quote: "What's your solution then? 1/x and x^m?"


Nope. I already mentioned I didn't want the functions x^m and x^n.
Your first function can be obtained from x^n where n = -1
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demons breath
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Posted: 25th Oct 2007 23:47
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you DIDN'T want them? I thought that was what you were asking for :s

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mamaji4
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Posted: 25th Oct 2007 23:53
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James, get off my back. Isn't it enough that you appear in my nightmares
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demons breath
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Posted: 25th Oct 2007 23:54
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I'm NEVER GOING TO STOP! seriously... NEVER!

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Chris K
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Posted: 25th Oct 2007 23:55
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So what is your solution? You seem to be changing the question quite a lot... what is your answer? You said you already had one...

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demons breath
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Posted: 25th Oct 2007 23:56
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See - me and chris k are both going to bug the hell out of you until you tell us

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mamaji4
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Posted: 25th Oct 2007 23:58
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That's all you're gonna get outta me. Until you cut your hair.
And I cut mine. Note that the second condition is not under your control James.
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demons breath
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Posted: 26th Oct 2007 00:04
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Harsh. Considering all that time I wasted acting like I could possibly be helpful...

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mamaji4
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Posted: 26th Oct 2007 00:07
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Geeze, I'm not trying to be harsh. It's just that this is like a very secretive thing I'm working on and...
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Peter H
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"and..." if you don't tell us i will haunt your dreams forever.

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mamaji4
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Posted: 26th Oct 2007 09:32
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*shudder.. shudder*
Oh Lord Peter, have mercy on thy slave, I beseech thee...
I shalt stand on yon pulpit and ne'er again henceforth, heretofore and fore'ermore spake in the forked tongue that ist mathematicaah...
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