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Math Help - function's formula

  1. #1
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    function's formula

    Hello.

    How could I make a description of the function which relate an argument and a value this way:

    x y
    0 0
    1 1
    2 2
    ...
    6 6
    7 5
    8 4
    ...
    12 0
    13 1
    14 2

    etc...

    This is obviously periodic function, so I can write that f(x)=f(x+12) but how could I give a comprehensive formula for order of y? Should I write piecewise definition or rather recursive, or...?

    Thanks for all help.
    Last edited by yosso; October 7th 2010 at 02:34 PM.
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  2. #2
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    Hello, yosso!

    I don't have a complete solution yet,
    . . but I have some ideas . . .


    How could I construct a function which relate an argument and a value this way?

    . . \begin{array}{c|ccccccccccccccccc}<br />
x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & \cdots \\ \hline<br />
y & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 5 & 4 & 3 & 2 & 1 & 0 & 1 & 2 & 3 & \cdots \end{array}


    This is obviously periodic function, so I can write: . f(x)\:=\:f(x+12)
    but how could I give a comprehensive formula for \,y?
    Should I write piecewise definition, or recursive, or ... ?

    We can describe the first "cycle" with this function:

    . . f(x) \;=\;6-|x-6|



    We can also describe the first cycle with a piecewise function:

    . . f(x) \;=\;\begin{Bmatrix} x & \text{if }0 \le x < 6\;\; \\ 6-x & \text{if }6 < x \le 12 \end{array}



    Either way, we must make it cyclic with a period of 12.

    Hmmm . . .
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  3. #3
    MHF Contributor

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    That looks like a complete solution to me! It is a "sawtooth" function.
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  4. #4
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    If it is a sawtooth function, there should be x=y (or x=|y|, )isn't it? So, it is true for arguments from 0 to 6, but it is not true in refference to numbers following last one. Maybe it could be better to define values in connection of multiples of arguments? For x=4 there is true, that y=4 and for 2x=4, 4x=4, 5x=4 etc.
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