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Odd and Even equation |
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| Mar15-07, 11:47 PM | #1 |
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Odd and Even equation
I have been wondering were to put this question but since it is related to algebra, i think i'll post it here.
I was playing with numbers and suddenly i came up with the idea of wanting to make a equation f(x) where all x odd number domain will result of range 1, and all even number domain will result of range 0. so: if x = even, f(x) = 0 if x = odd, f(x) = 1 x can only be whole integers {...,-2,-1,0,1,2...} y can only be 0 or 1 I want only one equation that give me such result. Please lend me a hand. I have been cracking my head all day and i always get it wrong. Thanks. |
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| Mar16-07, 12:00 AM | #2 |
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What about
y=sin2(x*pi/2) |
| Mar16-07, 12:25 AM | #3 |
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Great! Thanks! That is one way to do it. I wonder if you can make it into step functions? Pretty plesase
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| Mar16-07, 03:01 AM | #4 |
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Recognitions:
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Odd and Even equation |
| Mar16-07, 03:21 AM | #5 |
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How about
mod(x,2) |
| Mar16-07, 03:49 AM | #6 |
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-Odd then 3x+1 -Even then x/2 So suppose that i pick 5, then since it is odd it will start: 5 -> 16 -> 8 -> 4 -> 2 -> 1 -> 4 -> 2 -> .... When the number becomes odd, the odd condition is applied. When the number becomes even, the even condition is applied. However, there is also a loop in there. As you can see that from 1 it goes back to 4. The original problem was how to make a loop that doesn't go forever and it was by adding the condition that when x = 1 stop or while x is different to 1 continue to loop. However, the problem was extended to that supposing you pick a very large number. And because of the magnitude of the number, there could be a intermediate loop before reaching x = 1 which makes the program loop for ever even if you have the x = 1 condition. Is there any way to find that there is a intermediate loop inside and stop it? Then the professor explain that it is a little game that mathematicians like to play with and they don't have an answer. But i decided to play a bit as well. I begun to play with real equations and real numbers and i got as far as to be able to write a difference equation which i learned not very long ago and it serves some of my basic purposes to be able to determine the intermediate loop. yk+1=(3yk+1)(sin2(ykpi/2))+(yk/2)(cos2(ykpi/2)) |
| Mar16-07, 03:55 AM | #7 |
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mod(x,2) is a function like any other function. It's perfectly well known by mathematicians.
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| Mar16-07, 03:57 AM | #8 |
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I didn't know that.Well, thanks then. Learned something new. But what are those kind of equations called? Edit: Is that parametric equations? |
| Mar16-07, 03:58 AM | #9 |
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I don't think there's any way to tell in advance if this loop terminates. There may be a proof to that extent.
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| Mar16-07, 03:59 AM | #10 |
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Equations like sin are going to be useless for C++ (only if you have a symbolic computer library will you be OK to use pi - remember to evaluate something like sin the computer will only use approximations, and even then you have to load math.h in C, say). I seem to remember that if(x % 2 ==1) then (do something on odd numbers else do something else on even numbers (though I don't promise to have my precedence of operators correct) would do. Every computer language has a modulo function, and I think it is % in C, C++, python, perl et al. This is modular arithmetic (aka clock arithmetic). |
| Mar16-07, 04:05 AM | #11 |
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| Mar16-07, 10:17 AM | #12 |
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Well, remember us if you succeed and get the Fields medal for math.
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| Mar16-07, 10:59 AM | #13 |
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It is always difficult to decide if one should encourage work like this, or offer strong cautions. I think I would have to edge towards the latter: this problem is very old, very well studied, and no solution has been found. Similar examples are known to be undecidable. It is not something you should start with - I'm sure there are better things to learn from that you can actually make progress with (even if it is already known to someone else).
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| Mar16-07, 01:28 PM | #14 |
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______________ matt grime. I understand what you mean. If it is something that i mathematicians haven't been able to solve, there is a very little change that i would - especially with my limited knowledge in mathematics. But who, knows by the smallest change there still a possibility. Or even if i am not able to solve it at all, i would be able to learn new things on the way. Things that i never thought there were and things i never thought i could use it that way. I mean with the helps of you guys i was able to get the a function where any odd and even integer number will give me only 0 or 1. I learned a new notation (which i haven't been the name of it yet). I even applied difference equations which i hated so much a few days ago because it is too much trouble to get the results analytically. Besides i really feels encouraging and challenging to do things that sot everyone can solve. |
| Mar16-07, 01:31 PM | #15 |
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Doing something hard (impossible even) can be useful, but just as long as you're prepared to accept in advance that, whatever else you may gain, progress towards a solution is unlikely, so don't be disappointed.
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| Mar16-07, 02:49 PM | #16 |
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Well, i have been able to make so progress. I am using matlab to graph the difference equations and i have been able to identify a few patterns.
In the case of: Odd then 3x +1 Even then x/2 -Intermediate loops most likely happen when x is close to 0 -in positive numbers we know that intermediate loop will occur once x reach 1. But in negative integers there are more than one possible -x close to 0 which would lead to intermediate loops (try different -yk) -Most integer numbers inputed (regardless of being positive or negative) would most likely go close to 0 as the process is carried on. Without the graph, there are a few more patterns you can see by analyzing the odd and even conditions. -If odd, then (3x+1) will always transform x to even. -If even, then (x/2) might or might not transform x to odd. -If negative odd, then (3x+1) will lead first father away from 0 then +1 unit closer to 0. -If positive odd, then (3x+1) will always lead father away from 0. -If positive or negative even, then (x/2) will lead closer to 0. Big numbers can possibly have more than one whole division by 2 which will most likely lead the numbers towards 0. Any odd number will always convert it to even, therefore, most likely be reduced again. There are 3 cases where there will be loops which leads to no end. 1-Intermediate loops inside any y interval 2-Continuous increments towards positive infinity. 3-Continuous decrements towards negative infinity. I found case number 2 and 3 by graphing large positive or negative numbers in matlab. I am not sure if it was overflow or if it is going towards large infinity then goes back close to 0 (unless i can do it analytically). Here is my matlab script file so that you can analyze the problem as well if you have the software. Edit: Intermediate loops is my short way to call the loops that repeat the same values |
| Mar16-07, 02:54 PM | #17 |
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That doesn't make sense unless you bother to define what an 'intermediate' loop is.
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