see the attachment
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
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.
extension Function point.pdf
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)
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)
Plane , in the picture below ( x=7 ) , operation (+ , 62 , B2 , ) , solution y(straight line)=
A=a , B=x , AB=y , y=|a-x| or y=|x-a|
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 ?
A=x$_1$ , B=x$_2$ , AB=y , y=|x$_1$ - x$_2$| or y=|x$_2$ - x$_1$|
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
A=x , B=y$_1$=f(x) , AB=y$_2$ , y$_2$=|x-f(x)| or y$_2$=|f(x)-x|
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| ?
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>0- straight line
0<x( )- straight line
reads : semi-line ( ) , passes into straight line (( )>x>0) reduces the length , exceeds the point ( x=0 ) , goes straight line and changes in direction and increases the length (0<x( ), straight line the semi-line passes into ( )
y=|3-x| , x=(1 ,2,3 ,4.5 ) , red color to the solution
Dynamic graph: y solution
y=0 - point
0<y<( ) - straight line
reads : point ( y=0) passes into straight line ( 0<y<( ) and increases the length of the , passes straight line the line ( )
y=( 0 , 1 , 2 ) , red color to the solution
Ap and Aq semi-line and surface between them
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
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
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 .
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
spajaju se tačke AC ( BD ) duži AB i CD
tačke ABDC čine površinu za 4≥x≥3.5
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.
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
connect the dots AC (BD) straight lines AB and CD
ABDC points form the surface of 4≥x≥3.5