== The problem of differential equations ==
The problem of differential equations is - "cyclic redemption" with the consideration that there weren't any z axis. Or it that when there is x,y axis it is always that z axis is implicit with unit 1.
It is known that point lies on the circle (x,y) represented by (rCosθ,rSinθ).
By that we can say
x=rCosθ
y=rSinθ
In nature, we cannot find anything perfect as circle or sphere, but it is almost to it. Having that we can say,
x=rCosθ+Δx
y=rSinθ+Δy
x/y=rCosθ+Δx/rSinθ+Δy
To avoid abnormalities consider x/y=1 as z=1 so its equality is maintained. See, NP0 = 1. Sinθ/θ = 1 when θ tends to 0.
rCosθ+Δx/rSinθ+Δy=1
rCosθ+Δx/rSinθ+Δy=1
rCosθ+Δx=rSinθ+Δy
Δy=rCosθ+Δx-rSinθ
Δy/Δx=[rCosθ/Δx]+[1]-[rSinθ/Δx]
Δy/Δx=[(rCosθ-rSinθ)/Δx]+1
Δy/Δx=[rCosθ+Δx-rSinθ/Δx]
[Δy=rCosθ+Δx-rSinθ]
'''When you say dy then it is that,'''
dy=y+Δy
dy=y+rCosθ+Δx-rSinθ [as x=rCosθ, y=rSinθ So,] [or logical plane z=1]
dy=y+x+Δx-y
dy=x+Δx
as dx=x+Δx so
[dy=dx]
=== The General Equality Locus of differentiation is ===
dy=dx when plane-z=1, if it is not (z!=1) this equality is accords. So it is not dy/dx gives some ratio rather, make it dy=dx leverage with z-plane analysis like 1,NPr series.
This will avoid cyclic redemption problem of differentiation that it is ignoring that there is a z-axis
~~~~
Ref:
https://en.wikipedia.org/wiki/Group_structure_and_the_axiom_of_choice
http://en.wikipedia.org/wiki/Talk:Derivative#The_problem_of_differential_equations
The problem of differential equations is - "cyclic redemption" with the consideration that there weren't any z axis. Or it that when there is x,y axis it is always that z axis is implicit with unit 1.
It is known that point lies on the circle (x,y) represented by (rCosθ,rSinθ).
By that we can say
x=rCosθ
y=rSinθ
In nature, we cannot find anything perfect as circle or sphere, but it is almost to it. Having that we can say,
x=rCosθ+Δx
y=rSinθ+Δy
x/y=rCosθ+Δx/rSinθ+Δy
To avoid abnormalities consider x/y=1 as z=1 so its equality is maintained. See, NP0 = 1. Sinθ/θ = 1 when θ tends to 0.
rCosθ+Δx/rSinθ+Δy=1
rCosθ+Δx/rSinθ+Δy=1
rCosθ+Δx=rSinθ+Δy
Δy=rCosθ+Δx-rSinθ
Δy/Δx=[rCosθ/Δx]+[1]-[rSinθ/Δx]
Δy/Δx=[(rCosθ-rSinθ)/Δx]+1
Δy/Δx=[rCosθ+Δx-rSinθ/Δx]
[Δy=rCosθ+Δx-rSinθ]
'''When you say dy then it is that,'''
dy=y+Δy
dy=y+rCosθ+Δx-rSinθ [as x=rCosθ, y=rSinθ So,] [or logical plane z=1]
dy=y+x+Δx-y
dy=x+Δx
as dx=x+Δx so
[dy=dx]
=== The General Equality Locus of differentiation is ===
dy=dx when plane-z=1, if it is not (z!=1) this equality is accords. So it is not dy/dx gives some ratio rather, make it dy=dx leverage with z-plane analysis like 1,NPr series.
This will avoid cyclic redemption problem of differentiation that it is ignoring that there is a z-axis
~~~~
Ref:
https://en.wikipedia.org/wiki/Group_structure_and_the_axiom_of_choice
http://en.wikipedia.org/wiki/Talk:Derivative#The_problem_of_differential_equations
