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R. Brown
Posts:
54
From:
texas
Registered:
12/29/05


Real & Imaginary Parabolas
Posted:
Feb 28, 2007 10:26 AM


Parabolas of Real numbers AND Parabolas of Imaginaries by Robt. W. Brown
Let's assume that we have a Quadratic equation in which neither of the constants A or C = zero. While the third constant (B) does equal zero: B= 0
Under these special conditions, a form of the quadratic solution should reduce to:
x= [sqrt] [4A(C y)]/(2A)
Now let's designate (y) as the Independent variable and make another special condition: Namely, the following form of the discriminant will always be GREATER than zero. B^{2}[4A(C y)] = always greater than zero (in function #1)
Now we finally set the domain of (y) as being only numbers from the 'real number set', which comply to the 'discriminant condition' above.
Now, under the above conditions we should always have a parabola that falls within the XY plane. But let's assume that a third coordinate is implied and always equals zero . z= 0
******************** Now let's move to a second similiar function (function #2).
Let's assume that we have a Quadratic equation in which neither of the constants A or C = zero. While the third constant (B) does equal zero: B= 0
Under these special conditions, a form of the quadratic solution should reduce to:
x= [sqrt] [4A(C y)]/(2A)
Now let's designate (y) as the Independent variable and make this special condition: Namely, the following form of the discriminant will always be LESS than zero.
B^{2}[4A(C y)] = always less than zero (in function #2)
Now we finally set the domain of (y) as being only numbers from the 'real number set', which comply to the 'discriminant condition' above.
Now, under the above conditions we should always have a parabola that falls within the YZ plane. But let's assume that a third coordinate is implied and always equals zero (in this #2 function): x= 0 ***********
Since the discriminant in function #2 is always 'less than zero', we know that (z) will always be an 'imaginary number' and contain (i). Such as:
z= 1.4142135i
So, we see a pattern developing. In function #1, the depedent variable (x) was a number from the 'real number set'. Therefore it was 'implied' that z= 0.
While in function #2, the dependent variable (z) is an imaginary number and x= 0. So we can make this general connection to both functions:
A*[(x + z)^{2}] + [B(x +z)] + C = y .............eq. 101
When z= 0, eq. 101 becomes:
Ax^{2} + Bx + C = y
When x= 0, eq. 101 becomes:
Az^{2} + Bz + C = y
Now let's restate eq. 101:
A*[(x + z)^{2}] + [B(x +z)] + C = y .............eq. 101
Finally we see this connection between "complex numbers" and eq. 101: assume: x= a z= bi
Then eq. 101 becomes:
A*[(a + bi)^{2}] + [B (a + bi)] + C = y .............
Now to alleviate any confusion that might arise from graphing these 2 functions on the same 3 dimensional graph, I propose the following 3 axis system:
X..... axis Y... axis Z/i...axis
I will also state that the independent variable (y) should be from the REAL number set in the system I am proposing, because it relates to the 'power of a log base'. which can be summarized as:
N^{1/y}= B B^{Y} = N Log_{B}N= y
Now, if N is a term in a 'geometric sequence' we also have this relationship: B^{Y} = N_{y}
In my system of Terms and Ranks, we then see a connection between the related Logarithm and the Rank of the number (N):
logarithm= y rank of number= y Therefore: logarithm= rank of number/term
I will further state that it is possible to consolidate Function #1 and function #2 into a single function where y= the parameter of x & z This will require the graphing of the parameter (y), as the 3rd coordinate also, so as to complete a 3dimensional configuration. There will no longer be any implied coordinates, when this is done. Furthermore the above discriminant conditions could be relaxed. Because: y.. = the real number set
Compliments of Robt. W. Brown / Barnyard Physicist of Texas
NOTE: This same 'posting' appears on MMB.(less confusing)



