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 f(x) == f(10x)
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octopi




PostPosted: Sun Dec 07, 2003 10:13 pm   Post subject: f(x) == f(10x)

100 bits for someone who solves the following problem:

Find a function f(x), where f(x) is equal to f(10x) for all values of x,
The varible x must be used in the function.

example of how not to do it:
f(x) = 4
f(x) can't be 4 (because, the result would be 4 for all values of x), and x is not used in the function.

f(x) = 4+x-x
NO, because when simplified you get 4, cheap bastards.


Good luck.

Any cheapory in your anwsers will result in me not giving you bits...aka stupid anwser thats not right. but I didn't cover above in my examples of how not to do it.
(like, f(x)= 0x, or f(x)= x^0)
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Mazer




PostPosted: Mon Dec 08, 2003 9:16 am   Post subject: Re: f(x) == f(10x)

octopi wrote:
f(x)= 0x


crap!!! now i have to come up with something else! Evil or Very Mad
McKenzie




PostPosted: Mon Dec 08, 2003 10:05 am   Post subject: (No subject)

code:

int f(int x)
{
     while(x % 10 == 0)
          x = x / 10;
     return x;
}
Tony




PostPosted: Mon Dec 08, 2003 10:13 am   Post subject: (No subject)

Laughing

wait Confused Is the question about a math function or a programming function Thinking

btw, McKenzie - since when do you have those billion bits?
and for that matter... where did half of my bits go? Thinking
Latest from compsci.ca/blog: Tony's programming blog. DWITE - a programming contest.
McKenzie




PostPosted: Mon Dec 08, 2003 10:36 am   Post subject: (No subject)

I think it is supposed to be a math question. I thought a compsci answer was better Wink bits...what bits?
Tony




PostPosted: Mon Dec 08, 2003 10:41 am   Post subject: (No subject)

McKenzie wrote:
bits...what bits?


Thinking umm... can Andy create bits now that he's a mod Thinking

*Tony takes out his sniper rifle and goes to investigate* Sniper
Latest from compsci.ca/blog: Tony's programming blog. DWITE - a programming contest.
Blade




PostPosted: Mon Dec 08, 2003 1:20 pm   Post subject: (No subject)

wow, how quickly topics get changed around from the initial equation..
Andy




PostPosted: Mon Dec 08, 2003 7:29 pm   Post subject: (No subject)

f(x)=[1/x]

take off the top bar of the square bracket. that is actually a mathematical operator, it means the largest integer smaller than the number contained.... heh that works... but only for integral values of x
AsianSensation




PostPosted: Mon Dec 08, 2003 7:33 pm   Post subject: (No subject)

actually that one works for every single number except for 1 itself.

[1/1] = 1, [1/10] = 0
Andy




PostPosted: Mon Dec 08, 2003 7:40 pm   Post subject: (No subject)

dope!

fine [1/x+1]
AsianSensation




PostPosted: Mon Dec 08, 2003 7:42 pm   Post subject: (No subject)

nope, still another fault.

[1/(-10 + 1)] = -1, [1/(-1 + 1)] = undefined

still not every single number.
Andy




PostPosted: Mon Dec 08, 2003 8:48 pm   Post subject: (No subject)

okok i think i actually got it...
f(x)=[(x-(|x|/x))/x]

how about that?
Mazer




PostPosted: Mon Dec 08, 2003 9:07 pm   Post subject: (No subject)

hey dodge, besides the fact that f(x) != f(10x) with your function, what is f(0)?
Andy




PostPosted: Mon Dec 08, 2003 9:20 pm   Post subject: (No subject)

yes it does equal... sub in any value... and undefined=undefined rite?
Mazer




PostPosted: Mon Dec 08, 2003 9:41 pm   Post subject: (No subject)

uh... as we all know, i really suck at math, so feel free to point out my stupidity if i do something that seems incredibly n00bishly stupid.

WW wrote:
f(x)=[(x-(|x|/x))/x]


f(5) = (5 - (|5|/5))/5
= (5 - 1) / 5
= 4/5
= 0.8

right?

f(50) = (50 - (|50|/50))/50
= (50 - 1)/50
= 49/50
= 0.98
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