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Thread: function point

  1. #1
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    function point

    see the attachment
    Attached Files Attached Files
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  2. #2
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    Re: function point

    Hey point1967.

    What exactly are you trying to convey in your PDF?
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  3. #3
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    Re: function point

    Quote Originally Posted by chiro View Post
    Hey point1967.

    What exactly are you trying to convey in your PDF?
    that there are different forms of functions, which is based on the current functions and analytic geometry and geometry, I told you he presented the simplest form, I hope you understand the essence

    the current form of the function - y=f(x) one variable , y=f(x1 ,x2 , ..., xn) more variable , functional form that I he presented not exist in the current mathematics
    A=x , B=x1=2x+1 , f( x , x1=2x+1 , y ) , Read :
    Function point ( y ) of the independent variables x and the dependent variable
    x1 ( function , which is dependent on the independent variables x )


    no hybridization functions
    x-(2x+1)=0 , x=-1 - zero for the hybridization of two function point
    y=A-B ( y=x-x1 ) , y=x-( 2x+1) x>-1 , y=B-A ( y=x1 -x) , y=( 2x+1)-x x<-1
    - positive hybridization of the first and second functions points
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  4. #4
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    Re: function point

    How exactly is this idea different or new? I'm not quite understanding how it is different.

    As you long as you are specifying functions that take some inputs and map them to a unique output then everything has already been considered.

    The decomposition of the function can be whatever you want to be but as long as you can actually get an output value that is unique then nothing you stated is actually new.
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  5. #5
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    Re: function point

    extension Function point.pdf

    Quote Originally Posted by chiro View Post
    How exactly is this idea different or new?

    current function y=3x+5 , Independent variables (3x) , operation ( +) constant ( 5) y dependent variable .

    function point , given independent variable ( B=x ) and constant ( A=6 ) , should find the conditions for operation(?) for straihgt line ( AB)
    1.Number line
    2.Plane , Descartes Coordinate System , 6 ( y coordinate ) , x ( x coordinate )

    Number line , in the picture below ( x=7 ) , operation (-) , has two solutions y(straight line)=x(point B)-6(point A) and y(straight line)=6(point A)-x(point B)
    function point-ww.png

    Plane , in the picture below ( x=7 ) , operation (+ , 62 , B2 , \sqrt ) , solution y(straight line)= \sqrt{6^2(point A)+x^2(point B)}
    function point-w.png
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  6. #6
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    Re: function point

    solution of the task -


    we have the numerical line , on it is straight line AB , point A is located on a number of numerical line , point B anywhere of numerical line , how to describe this as a function ?
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  7. #7
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    Re: function point

    A=a , B=x , AB=y , y=|a-x| or y=|x-a|

    example
    a=5 , x=(2,5,10)
    y=|5-2|=3 or y=|2-5|=3 , length straight AB=3
    y=|5-5|=0 or y=|5-5|=0 , no straight AB
    y=|5-10|=5 or y=|10-5|=5 , length straight AB=5

    we have the number line , on it is straight line AB , point A anywhere of number line , point B anywhere of number line , how to describe this as a function ?
    Last edited by point1967; Feb 15th 2015 at 01:10 AM.
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  8. #8
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    Re: function point

    first solution
    A=x$_1$ , B=x$_2$ , AB=y , y=|x$_1$ - x$_2$| or y=|x$_2$ - x$_1$|
    example
    x$_1$ =(2,8) , x$_2$=(-10,-20)
    y=|2-(-10)|=12 , length straight AB=12
    y=|2-(-20)|=22 , length straight AB=22
    y=|8-(-10)|=18 , length straight AB=18
    y=|8-(-20)|=28 , length straight AB=28

    second solution
    A=x , B=y$_1$=f(x) , AB=y$_2$ , y$_2$=|x-f(x)| or y$_2$=|f(x)-x|
    example
    x=(3,7) , f(x)=3$x^2$-2
    y=|3-(27-2)|=22 , length straight AB=22
    y=|7-(147-2)|=138 , length straight AB=138
    Comment - the structure of the current function: the dependent variable (y) of n independent variables (x$_n$).
    here we have a new structure functions: x independent variable, dependent variable y$_1$ (depending on x), y$_2$ dependent variable (depending on x and depends on y$_1$ (f (x))

    continuation - dynamic graphics, static graphics, partial graph y=|a-x| ?
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  9. #9
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    Re: function point

    All of this is pretty much secondary school algebra. Again, what is your point?
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  10. #10
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    Re: function point

    Quote Originally Posted by HallsofIvy View Post
    Again, what is your point?
    y = a-x
    The graph of the current solution:
    x-coordinate represents all real numbers, when solved function we have two numbers (y, x) , introduces the new coordinates y perpendicular to the x-coordinate and cut the number 0 (plane), the number of y is transferred to the y-coordinate , line (which is parallel to the y-coordinate, and on it is a point that is the number x) is cut from the line (which is parallel to the x-coordinate and it is a point that is the number y) gets the point in the plane (x, y)
    which means that the point (x, y) on the x-coordinate of the mapped into a point in the plane (x, y) points are merged to obtain a graph

    y = | a-x |
    Graph of my solution:
    x-coordinate represents all real numbers, when solved function we have three numbers (a, y, x), introduces the new coordinates y perpendicular to the x-coordinate and cut the number 0 (plane), the number of y is transferred to the y-coordinates, lines (the first parallel to the y-coordinate, and on it is a point that is the number a , the second is parallel to the y-coordinate, and on it is a point that is the number x) is cut from the line (which is parallel to the x-coordinate and it is the point which is also the number y) gave the points in the plane (x, y) and (a, y) of the connecting point is obtained straight line
    which means that the points (a, x, y) on the x-coordinates are mapped onto the straight line AB in the plane (A (x, y) B (s, y)), the straight line are merged to obtain the graph of

    Dynamic graph: x solution
    x\rightarrow-\infty - semi-line
    ( x\rightarrow-\infty)>x>0- straight line
    x=0 -point
    0<x( x\rightarrow+\infty)- straight line
    x\rightarrow+\infty semi-line
    reads : semi-line ( x\rightarrow-\infty) , passes into straight line (( x\rightarrow-\infty)>x>0) reduces the length , exceeds the point ( x=0 ) , goes straight line and changes in direction and increases the length (0<x( x\rightarrow+\infty), straight line the semi-line passes into ( x\rightarrow+\infty)
    y=|3-x| , x=(1 ,2,3 ,4.5 ) , red color to the solution
    https://d6pdkg.bn1302.livefilestore....sse.png?psid=1

    Dynamic graph: y solution
    y=0 - point
    0<y<( y\rightarrow+\infty) - straight line
    y\rightarrow+\infty - line
    reads : point ( y=0) passes into straight line ( 0<y<( y\rightarrow+\infty) and increases the length of the , passes straight line the line ( y\rightarrow+\infty)
    y=( 0 , 1 , 2 ) , red color to the solution
    https://qhdrnq.bn1302.livefilestore....ssp.png?psid=1
    Last edited by point1967; Feb 20th 2015 at 08:16 AM.
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  11. #11
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    Re: function point

    Again, this is all secondary school material. And you still haven't said why you are posting it although you have been asked twice.
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  12. #12
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    Re: function point

    Quote Originally Posted by HallsofIvy View Post
    And you still haven't said why you are posting it although you have been asked twice.
    I discovered a new form of function, probably from the previous demonstrated ...

    static graphics
    Ap and Aq semi-line and surface between them
    https://o9alca.bn1302.livefilestore..../ss.png?psid=1
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  13. #13
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    Re: function point

    The problem seems to be that you are using a lot of words whose meanings you do not know!

    In particular, while talking about a "new form of function", you do not appear to know what "function" means.
    Thanks from zzephod
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  14. #14
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    Re: function point

    SERBIAN
    Pošto sam otkrio nove mogućnosti , moguća rešenja

    Na x-koordinati , postoji duž AB , tačka A je nepokretna na x-koordinati , tačka B se nalazi na bilo kojem mestu x-koordinate , opisati ovo funkcijom .

    Rešenja : A=a , B=x , AB=y

    a) y=|a-x|

    b) y=-|a-x|

    c) y=a-x

    d) y=x-a

    GOOGLE TRANSLATOR
    Since I discovered new possibilities, possible solutions

    On the x-coordinate, there is straight AB, point A is fixed on the x-coordinate, point B is located at any point x-coordinates, to describe this function.

    Solutions : A=a , B=x , AB=y

    a) y=|a-x|

    b) y=-|a-x|

    c) y=a-x

    d) y=x-a
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  15. #15
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    Re: function point

    SERBIAN
    Preslikavanje funkcije iz x-koordinate u ravan ( dekartov koordinatni sistem )
    y=x-a , x i a ostaju na x-koordinatu , y ide na y-koordinatu .
    prati sliku
    https://pkxnqg.bn1302.livefilestore..../ii.png?psid=1
    prave iz x i a paralelne sa y-koordinate
    prava iz y paralelna sa x-koordinate
    u preseku pravih nastaju tačke A i B
    tačke A i B se spajaju i dobija se duž AB
    dato je za x=4 , a=2 , y=2
    ponovimo postupak za x=3.5 , a=2 , y=1.5 , prati sliku
    u preseku pravih nastaju tačke C i D
    tačke C i D se spajaju i dobija se duž CD

    https://befwwg.bn1302.livefilestore....M/i.png?psid=1
    spajaju se tačke AC ( BD ) duži AB i CD
    tačke ABDC čine površinu za 4≥x≥3.5

    GOOGLE TRANSLATOR
    The mapping function from the x-coordinates of the plane (Cartesian coordinate system)
    y = x-a, x and a remain on the x-coordinate, y goes to the y-coordinate.
    view photo
    https://pkxnqg.bn1302.livefilestore..../ii.png?psid=1
    the lines of x and a parallel to the y-coordinates
    line of y parallel to the x-coordinate
    formed at the intersection of real points A and B
    points A and B are combined and gets straight line AB
    is given by x = 4, a = 2, y = 2
    Repeat for x = 3.5, a = 2, y = 1.5, view photo
    formed at the intersection of real points C and D
    points C and D are combined and received straight line CD

    https://befwwg.bn1302.livefilestore....M/i.png?psid=1
    connect the dots AC (BD) straight lines AB and CD
    ABDC points form the surface of 4≥x≥3.5
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