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Topic: find the function (Read 4747 times) |
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inexorable
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find the function
« on: Jan 4th, 2006, 12:17pm » |
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Construct a function F from integers to integers such that f(f(n))=-n for any integer n.
Also find a function g from positive rational numbers to positive rational numbers such that g(g(q))=1/q for any positive rational q?
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JocK
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Re: find the function
« Reply #1 on: Jan 4th, 2006, 12:40pm » |
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on Jan 4th, 2006, 12:17pm, inexorable wrote:| Construct a function F from integers to integers such that f(f(n))=-n for any integer n. |
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Any function f obtained from pairing distinct positive integers k and n such that
f(-k) = -n
f(-n) = +k
f(+n) = -k
f(+k) = +n
suffices (if complemented with f(0) = 0).
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| « Last Edit: Jan 4th, 2006, 12:42pm by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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SMQ
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Re: find the function
« Reply #2 on: Jan 4th, 2006, 12:41pm » |
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Didn't we just see g somewhere recently? 
f looks a bit more interesting, though...
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inexorable
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Re: find the function
« Reply #3 on: Jan 4th, 2006, 8:17pm » |
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on Jan 4th, 2006, 12:41pm, SMQ wrote:
but g here is a function from positive rational numbers to positive rational numbers so it can't be g(q)=q^i
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SMQ
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Re: find the function
« Reply #4 on: Jan 5th, 2006, 6:10am » |
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You mean to tell me not all complex numbers are rational?  
Ah well, so not every f() over there is a g() over here; the questions are still related, as any g() here is also an f() there. 
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Eigenray
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Re: find the function
« Reply #5 on: Jan 5th, 2006, 10:49am » |
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Jock's trick for f will also work for g. Partition the rationals greater than 1 into ordered pairs (r,s), and let f act as the cycle r -> s -> 1/r -> 1/s -> r.
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Barukh
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Re: find the function
« Reply #6 on: Jan 7th, 2006, 1:45am » |
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Ilike your argument, Eigenray! But it seems a bit existential, don't you think? Iwas trying to come with some specific pairing but didn't suceed so far. One way of partitioning that I tried was the parity of p+q for some rational r = p/q.
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Eigenray
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Re: find the function
« Reply #7 on: Jan 7th, 2006, 11:23pm » |
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You're right, there's a constructive way using f:
g([prod] pia_i) = [prod] pif(a_i)
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srn437
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Re: find the function
« Reply #8 on: Sep 2nd, 2007, 11:26am » |
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F(n)=n(i)
g(q)=q^i
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towr
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Re: find the function
« Reply #9 on: Sep 2nd, 2007, 12:20pm » |
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on Sep 2nd, 2007, 11:26am, srn347 wrote:
on Jan 4th, 2006, 8:17pm, inexorable wrote:| but g here is a function from positive rational numbers to positive rational numbers so it can't be g(q)=q^i |
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Wikipedia, Google, Mathworld, Integer sequence DB
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srn437
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Re: find the function
« Reply #10 on: Sep 17th, 2007, 6:34am » |
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It still works for f. It is for integers and i is a guasion integer. It could be division by i also.
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towr
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Re: find the function
« Reply #11 on: Sep 17th, 2007, 7:58am » |
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on Sep 17th, 2007, 6:34am, srn347 wrote:| It still works for f. It is for integers and i is a guasion integer. It could be division by i also. |
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A gaussian integer isn't an integer in the normal sense, and so unfortunately doesn't qualify as a solution.
You can compare it to how an imaginary friend, despite having the noun friend in it, isn't really a friend, because it doesn't exist (courtesy of the adjective imaginary). A gaussian integer, despite containing the noun integer, isn't an integer (because the adjective gaussian modifies it to something else).
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srn437
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Re: find the function
« Reply #12 on: Sep 18th, 2007, 7:19am » |
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Though the definition of integer is having a representation that is neither fraction nor decimal. It is a subset of reals though.
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rmsgrey
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Re: find the function
« Reply #13 on: Sep 18th, 2007, 8:01am » |
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on Sep 18th, 2007, 7:19am, srn347 wrote:| Though the definition of integer is having a representation that is neither fraction nor decimal. It is a subset of reals though. |
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The definition of the integers I usually use is "the closure of the natural numbers under subtraction"
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towr
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Re: find the function
« Reply #14 on: Sep 18th, 2007, 9:03am » |
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on Sep 18th, 2007, 7:19am, srn347 wrote:| Though the definition of integer is having a representation that is neither fraction nor decimal. It is a subset of reals though. |
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 is neither a rational, nor a decimal (which btw isn't a type of number, but a number representation. You can't write as decimal, as it would have infinite length. But please do try; but mind you, not here.)
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| « Last Edit: Sep 18th, 2007, 9:03am by towr » |
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Grimbal
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Re: find the function
« Reply #15 on: Sep 18th, 2007, 9:46am » |
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And in fact, integers do have a representation as a fraction and as a decimal.
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Hippo
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Re: find the function
« Reply #16 on: Sep 18th, 2007, 1:12pm » |
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just some comments:
Eigenray: ... the construction of q using f and factorisation should be applied on both parts of fraction.... oops you are right ... you use negative ai's.
The cycles for f can be defined for example in the following way:
Write nonzero integer n as +/- 2ko where o is odd. If k is odd f(n)=n/2 otherwise f(n)=-2n.
f(0)=0.
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| « Last Edit: Sep 18th, 2007, 1:14pm by Hippo » |
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srn437
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Re: find the function
« Reply #17 on: Sep 18th, 2007, 7:13pm » |
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Pi is a decimal, though infinite. It also has some fraction representations. Check it on wikipedia.
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JP05
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on Sep 18th, 2007, 7:13pm, srn347 wrote:| It also has some fraction representations. Check it on wikipedia. |
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You are incorrect. pi does not have any fractional (rational) representations. It has as many rational "approximations" as you wish to exploit via truncation.
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| « Last Edit: Sep 18th, 2007, 7:19pm by JP05 » |
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srn437
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Re: find the function
« Reply #19 on: Sep 18th, 2007, 8:38pm » |
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No offence, but saying pi has no fraction representation is not even wrong(finally I get to use that phrase). Regardless, let's get back on topic before the spammers come.
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JP05
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Of course, no offense. I think you need to discern the difference between an approximation and a representation in this case. I believe mathematicians regard a representation as being precise.
Again, there are no rational representations of pi. There are as many rational approximations of pi as you wish to make from truncation.
And last, I have reviewed the forums tonight and so have noticed I should not even be responding here with this. After all, there are plenty of cartoons that I would otherwise be missing myself.
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srn437
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Re: find the function
« Reply #21 on: Sep 18th, 2007, 9:14pm » |
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Try multitasking. Anyway, f2(f2-f0)=2n. Hopefully this helps. Try graphing it with x, y, and z(x, f(x), and f(f(x))).
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| « Last Edit: Sep 18th, 2007, 11:29pm by srn437 » |
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ima1trkpny
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Re: find the function
« Reply #22 on: Sep 18th, 2007, 9:22pm » |
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on Sep 18th, 2007, 9:14pm, srn347 wrote:
Ok, I have yet to see you prove you know anything! Mostly it is a bunch of bogus, irrelevant claims from you that you never actually prove! You just go "Oh I think it must involve this and that and that..." (and whatever other theorems name you think throwing around will draw respect) and I must say it truly is pathetic.
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"The pessimist sees difficulty in every opportunity. The optimist sees the opportunity in every difficulty." -Churchill
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JP05
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That's nonsense. Don't you realize that this is a good place to learn mathematics and computer algos? Why are you against the grain here? Don't you see how generous people are here with explanations and knowledge?
You are using all these benefits to your disadvantage. We don't care if you are a dummy. Well, at least no one had to test you to see if you were a dummy, because, well, you showed it on your own.
Bye.
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ima1trkpny
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Re: find the function
« Reply #24 on: Sep 18th, 2007, 9:49pm » |
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on Sep 18th, 2007, 9:43pm, srn347 wrote:| if you don't believe those functions being used, wait until I find the answer. |
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It doesn't matter what I believe or not! My point is that you just spew whatever comes to the top of your head as an answer without actually doing any work! Guessing at an answer means NOTHING if you have no idea how you got there and how to produce the same results! As it is you just spew a bunch of bogus and then provide absolutely NO PROOF or SUPPORT for your arguments! You lack sound logic and the humility to realize there is a possibility you are wrong.
Let's pretend for a moment you are an engineer or someone with some responsibility. Let's say you are in charge of a the design of a building and you pull this same guessing game. Maybe once you will get lucky and come out with the right measurements, etc. for the building to stand. But what the hell happens when you aren't? The building is completely faulty and potentially could kill people and you would have no blueprints or anything for someone else to either notice your mistake or be able to correct the failures to make it work. The difference here is that most of these people have been doing this for years and years (some long before you were even conceived) and can recognise by instinct poor foundations. They are trying to save you from big mistakes and bad habits that will be a road block for you later in life.(And they are completely right) And you have the guts to call them audacious? Your sillyness is beyond words!
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| « Last Edit: Sep 18th, 2007, 9:56pm by ima1trkpny » |
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"The pessimist sees difficulty in every opportunity. The optimist sees the opportunity in every difficulty." -Churchill
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