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Classical Physics Doesnt Explain Mirror Reflection by Zig-zag-zug Classical Physics Doesnt Explain Mirror Reflection :iconzig-zag-zug:Zig-zag-zug 0 0
Literature
infinitessimal limited to calculable functions
The cardioid equation is
f(x,y) = (x^2 + y^2 - ax)^2 = a*a(x^2 + y^2)
with c = (x^2 + y^2)^0.5, f(x,y) = g(x,c) = (c^2 - ax)^2 = a^2*c^2
deriving it by variable t
f'(x,y) = d(c^2 - ax)^2/dt = a^2*c^2
f'(x,y) = 2(c^2 - ax)(2c*dc/dt - x*da/dt - a*dx/dt) = 2a^2*c*dc/dt + 2a*c^2*da/dt
f'(x,y) = (c^2 - ax)(2c*dc/dt - x*da/dt - a*dx/dt) = a*c(a*dc/dt + c*da/dt)
f'(x,y) = (2c*dc/dt - x*da/dt - a*dx/dt) = a*c(a*dc/dt + c*da/dt)/(c^2 - ax)
diff. eq.: 1/dt*[2dc/a - x/a/c*da - dx/c = (a*dc + c*da)/(c^2 - ax)]
f'(x,y)dt = 2/a - x/a/c*da/dc - dx/dc/c = (a + c*da/dc)/(c^2 - ax)
deriving with dx/dc = (dc/dx)^-1 and c = (x^2 + y^2)^0.5
f'(x,y)dt = 2/a - x/a/(x^2 + y^2)^0.5(d(x^2 + y^2)^0.5/da)^-1 - 1/(x^2 + y^2)^0.5(d(x^2 + y^2)^0.5/dx)^-1 = (a + (x^2 + y^2)^0.5(d(x^2 + y^2)^0.5/da)^-1)/(x^2 + y^2 - ax)
f'(x,y)dt = 2/a - x(x + y)/a/(x^2 + y^2) - (x + y)/(x^2 + y^2) = [a + (x + y)/(x^2 + y^2)]/(x^2 + y^2 - ax)
f'(x,y)dt = 2/a - x(x + y)/a/(x - y)/(x + y) - (x + y)/(x + y)/(x - y) = [a + (x + y)/(x + y)
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Moebius Strip, Magnetic Monopole, tiny starship by Zig-zag-zug Moebius Strip, Magnetic Monopole, tiny starship :iconzig-zag-zug:Zig-zag-zug 0 0
Literature
Dehydration And Cancer
dehydration and cancer
All cells require water,
all cancer cells require water
cancer cells will continuously send a signal to initiate growth
during dehydration, growth is slowed
rehydration returns water to the organism
cancer cells will continue to replicate and be active during dehydration, while other cells will slow their activities
cancer cells move closer to the blood vessels and neglect large scale signals
they receive "cease" signals during this
they
cease and repair
do not cease and continue sending growth signals
if the former, further body reactions occur, at larger and greater scales. Up to immune response?
Cell repair succesful if the latter
Furthermore, cancer cells internal clock will unsynchronize more as time progresses during dehydration, until they no longer respond to the body's internal clock, and once that happens they could be treated.
As they move more than other cells, these cells will be more stressed, and overall have more markers to distinguish them from h
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Literature
Proof of product rule and integration by parts
d(u*v)/dx = u*dv/dx + v*du/dx {?}
u*v = a + b
d(u*v)/dx = da/dx + db/dx
da = u*m = u*dv
db = v*n = v*du
d(u*v)/dx = u*dv/dx + v*du/dx
proof of product rule. Done. Extends easily!
u*v*h = a + b + c
v*h = l
b + c = f
u*l = a + f
d(u*l)/dx = da/dx + df/dx
da = u*m = u*dl
df = l*n = l*du
d(u*l)/dx = u*dl/dx + l*du/dx
and so,
d(u*v*h)/dx = u*d(v*h)/dx + v*h*du/dx
d(u*v*h)/dx = u*[vdh/dx + hdv/dx] + v*h*du/dx
d(u*v*h)/dx = u*v*dh/dx + u*h*dv/dx + v*h*du/dx
integration by parts
d(u*v)/dx = u*dv/dx + v*du/dx
~d(u*v) = ~u*dv + ~v*du
u*v = ~u*dv + ~v*du
u*v - ~u*dv = ~v*du
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Laser Made With Sugar plug by Zig-zag-zug Laser Made With Sugar plug :iconzig-zag-zug:Zig-zag-zug 0 0 Mirror Placed Within A Conic Well by Zig-zag-zug Mirror Placed Within A Conic Well :iconzig-zag-zug:Zig-zag-zug 0 0 Sorting metallic graphene from semi-conductive by Zig-zag-zug Sorting metallic graphene from semi-conductive :iconzig-zag-zug:Zig-zag-zug 0 0 Holographic chemistry - an exploration! by Zig-zag-zug Holographic chemistry - an exploration! :iconzig-zag-zug:Zig-zag-zug 0 0
Literature
Applied Stuff - the beg. of relativistic applied
mass m, time interval dt, small angle dw, speed of light c
-0.5m*v(dw/dt) + E - mc^2 = 0
-0.5m/c*v(dw/dt) + E/c - mc = 0
-0.5m/c*v(dw/dt) + E/c = mc
E = m*g[r]
-0.5m/c*vdw/dt + m*g[r]/c = mXc
equation still works with kinetic energy formula
-0.5m*v(dx/dt) + E - mc^2 = 0
-0.5m/c*v(dx/dt) + E/c - mc = 0
-0.5m/c*v(dx/dt) + E/c = mc
E = m*g[r]
-0.5m/c*v(dx/dt) + m*g[r]/c = mXc
whatever teeheehee!
speed make by deriving m, or setting m=1.
-0.5v/c*dw/dt + g[r]/c = c
-(1/2)v/c*dx/dt + g[r]/c = c
"Argh!" You say later on! E/c^2 = m, is better however. As we saw previously.
-0.5E/c*v(dx/dt) + E/c*g[r] = E/c
-0.5v(dx/dt) + g[r] = 1
-0.5v(dx/dt)/[r] + g = 1/[r]
g = 1/[r] + 0.5v(dx/dt)/[r]
Where g is constant acceleration on a surface above a point mass. It is noticed
that there is a parabolic equation however!
0 = 0.5v(dx/dt) - g[r] + 1
Therefore! r=x
0.5v^2 - g*x + 1  = 0
(dx/dt)^2 = 2g*x - 2
dx/dt = [2(g*x - 1)]^0.5
time maker
-0.5/c*dw/dt + g[r]/(vXc) = c/v
-0.5/c*dw/dt + g[r]/(vXc) = c(x
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Literature
Mbers, Or Microbic Numbers
It is said that a million monkeys typing randomly for a million years could produce a
play by shakespear. Comparably, a number sufficentely large could contain within it a
unkown "program". When it is read by a particular, individual computer (or person per-
haps, it runs. I propose calling these programs containing numbers: mbers or "embers".
For a particular (or particular group) of computer(s) the mbers would create unknown
errors during the computer(s) activities. This would appear much like a virus, and
could run a program that would appear from thin air, much like people thought bacteria
appeared from thin air before germ theory was developped.
In more practical terms, different computers would produce different results, even if
roughly constructed identically, when using large numbers. This would apply to weather
predicting computers as well. The numbers themselves have propreties that would apply
changes to the program that runs them. Constants are saved numbers, that would the
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Literature
Two thirds as a defined mber (proof mbers exists)
Perimeter of trapeze base plus sides. We must the length of the missing side.
Base of trapeze B = 81 cm is the largest side.
Side length 1 = 15 cm is one side.
Side length 2 = 15 cm is the other side.
The inner angles of the outer pointy things is fourty-five degrees.
x is the length of the missing side.
What is the length of the missing side in centimeters?
15 + 15 + 81 = perimeter measured
111 = perimeter measured
remembered formula for trapezes
B(L+h)/2 = area (wrong!)
where h is the heigth of the trapeze
(x+B)*h/2 = area
2(15)cos(45)+(-81 cm) = x
[30cos(45) - 81 ]cm = x
calculate cosinus of 45 degrees.
a=b
a*a + b*b = c*c
calculating formula
1*1 + 1*1 = 2
2*2 + 2*2 = 8
3*3 + 3*3 = 18
4*4 + 4*4 = 32
Using the above equations we find
(c*c - b*b)/a = b
since a=b
c*c/a - a = a
Length*Length = Area for a rectangle
Length*Length/2 = Area for a triangle with a ninety degree angle
a*b/2 = Area of the triangle with two fourty-five degree angle
calculating area
1*1/2 = 1/2
2*2/2 = 2
3*3/2 =
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Literature
Through the thing theorem and three points problem
x^2 + y^2 = c^2
x=a+a*i, y=v+v*i, c=n+n*i
x=a(1+i), y=v(1+i), c=n(1+i)
a^2(1+i)^2 + v^2(1+i)^2 = n^2(1+i)^2
a^2(1+2i-i^2)^2 + v^2(1+2i-i^2)^2 = n^2(1+2i-i^2)
By respecting the order of operations we obtain:
2a^2(1+i) + 2v^2(1+i)   = n^2 + 2n^(2)i - n^2*i^2
= n^2 + 2n^(2)i + n^2
= 2n^2 + 2n^(2)*i
= 2n^2(1+i)
And the variables are changed.
Returning to the initial equation.
4a^2(i+i^3) + 4b^2(i+i^3) = 4c^2(i+i^3)
a^2(i+i^3) + b^2(i+i^3) = c^2(i+i^3)
a^2 + v^2 = n^2
And we know that x=a(1+i), y=b(1+i), and that c=n(1+i), therefore
c^2 = 2n^2(1+i)
(c^2)/(1+i) = 2n^2
c/[2(1+i)]^(1/2) = n
And angle is cos(m)= c/x, or cos(w)= n/a
cos^2(m) = c^2/x^2 = 2n^2(1+i)/2a^2(1+i)
cos^2(m) = n^2/a^2
cos(m) = n/a
My crazy thought, and as seen in "euleur's identity remix - update":
So, Pythagore theorem holds true for imaginary numbers.
cos(m) = n/a
x=a(1+i)
x/a=1+i
x/a-i=1
nx/cos(m) -i=1
x(a^2+b^2)/cos(m) -i=1
Let's just relax and put x^2-1=0, x=i, so
-x+x(a^2+b^2)/cos(m) = 1
x(-1+(a^2+b^2)/cos(m))
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Literature
euleur's identity remix - update
x^2 + y^2 = c^2
x=a+b*i, y=u+v*i, c=n+m*i
(a+b*i)^2 + (u+v*i)^2 = (n+m*i)^2
(a^2 + abi + abi - b^2) + (u^2 + uvi + uvi -v^2) = n^2 + nmi + nmi - m^2
a^2 - b^2 + u^2 - v^2 - n^2 + m^2 = 2nmi - 2abi - 2uvi
a^2 - b^2 + u^2 - v^2 - n^2 + m^2 = 2i(nm - ab - uv)
a^2 - b^2 + u^2 - v^2 - n^2 + m^2
___________________________________ = i
2(nm - ab - uv)
With Euler's identity
e^(i*pi) + 1 = 0
We find that:
 a^2 - b^2 + u^2 - v^2 - n^2 + m^2  
 ___________________________________*pi  
e^     2(nm - ab - uv)                   + 1 = 0
Update:
By selecting square vectors initialy, another form of the theorem can be obtained.
x^2 + y^2 = c^2
x=a+a*i, y=v+v*i, c=n+n*i
x=a(1+i), y=v(1+i), c=n(1+i)
a^2(1+i)^2 + v^2(1+i)^2 = n^2(1+i)^2
a^2(1+2i-i^2)^2 + v^2(1+2i-i^2)^2 = n^2(1+2i-i^2)^2
By respecting the order of operations we obtain:
(1+2i-i^2)*(1+2i-i^2) = 1+2i-i^2 + 2i + 4i^2 + 2i^3 -i^2 - 2i^3 - i^4
(1+2i-i^2)*(1+2i-i^2) = 1+4
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Literature
Mandelbrot set remix
Zn^2 + C = Zn+1
Zn+1^2 + C = Zn+2
is
(An+Bn*i)^2 + C = (An+1 + Bn+1*i)
We can make the mandelbrot real at any point by using the remix theorem.
a^2 - b^2 + u^2 - v^2 - n^2 + m^2
___________________________________ = i
2(nm - ab - uv)
Such that
(An+1 + Bn+1 (a^2 - b^2 + u^2 - v^2 - n^2 + m^2))^2  = Z
    ___________________________________
2(nm - ab - uv)
At any point in the series, whether it diverges or not.
As an aside, the mandelbrot could be modified further by Euleur's number itself.
Here I simply divide C by e for a new form of Mandelbrot set.
Zn^2 + C/e = Zn+1
It can be done for any constant, including physical constants, to obtain a fractal
of that constant.
For x smaller than 2
Zn^2 + C*x = Zn+1
For x greater than 2
Zn^2 + C/x = Zn+1
Where x is the chosen constant, such as the golden ration, the speed of light, etc.
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Literature
Relativistic kinetic energy of rigid bodies remix
x^2 + y^2 = c^2
x=a+b*i, y=u+v*i, c=n+m*i
(a+b*i)^2 + (u+v*i)^2 = (n+m*i)^2
(a^2 + abi + abi - b^2) + (u^2 + uvi + uvi -v^2) = n^2 + nmi + nmi - m^2
a^2 - b^2 + u^2 - v^2 - n^2 + m^2 = 2nmi - 2abi - 2uvi
a^2 - b^2 + u^2 - v^2 - n^2 + m^2 = 2i(nm - ab - uv)
a^2 - b^2 + u^2 - v^2 - n^2 + m^2
___________________________________ = i
                2(nm - ab - uv)
There are three cases of special relativity that remain untreated, first, when the
square root itself is negative, secondaly when the number beneath the root is imaginary.
Thirdly, when the root is negative and the lorentz factor imaginary.
First imaginary numbers result when (v^2/c^2) > 1
Using Relativistic_kinetic_energy_of_rigid_bodies from
https://en.wikipedia.org/wiki/Kinetic_energy#Relativistic_kinetic_energy_of_rigid_bodies
M(1-v^2/c^2)^-1/2*v^2*i + Mc^2(1-v^2/c^2)^1/2*i ± E0  (the constant of integration can be negative)
Where M is mass, c is speed of light. It simplifi
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Viva Jinx, aie-ya yeai!

i_ swore i'd never science again!

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Zig-zag-zug

Artist | Hobbyist | Literature
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Ground Up Shark teeth protein shakes can sup_ly Christian with their fish requirements. 



Viva Jinx, aie-ya yeai!

i_ swore i'd never science again!
Classical Physics Doesnt Explain Mirror Reflection
A straight line path couldn't produce the image in the mirror, as light from the object must travel through itself to reach the mirror.

Mirror reflection cannot be explained by bouncy balls.  :D 

(I think...!)
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The cardioid equation is
f(x,y) = (x^2 + y^2 - ax)^2 = a*a(x^2 + y^2)
with c = (x^2 + y^2)^0.5, f(x,y) = g(x,c) = (c^2 - ax)^2 = a^2*c^2
deriving it by variable t
f'(x,y) = d(c^2 - ax)^2/dt = a^2*c^2
f'(x,y) = 2(c^2 - ax)(2c*dc/dt - x*da/dt - a*dx/dt) = 2a^2*c*dc/dt + 2a*c^2*da/dt
f'(x,y) = (c^2 - ax)(2c*dc/dt - x*da/dt - a*dx/dt) = a*c(a*dc/dt + c*da/dt)
f'(x,y) = (2c*dc/dt - x*da/dt - a*dx/dt) = a*c(a*dc/dt + c*da/dt)/(c^2 - ax)
diff. eq.: 1/dt*[2dc/a - x/a/c*da - dx/c = (a*dc + c*da)/(c^2 - ax)]
f'(x,y)dt = 2/a - x/a/c*da/dc - dx/dc/c = (a + c*da/dc)/(c^2 - ax)
deriving with dx/dc = (dc/dx)^-1 and c = (x^2 + y^2)^0.5
f'(x,y)dt = 2/a - x/a/(x^2 + y^2)^0.5(d(x^2 + y^2)^0.5/da)^-1 - 1/(x^2 + y^2)^0.5(d(x^2 + y^2)^0.5/dx)^-1 = (a + (x^2 + y^2)^0.5(d(x^2 + y^2)^0.5/da)^-1)/(x^2 + y^2 - ax)
f'(x,y)dt = 2/a - x(x + y)/a/(x^2 + y^2) - (x + y)/(x^2 + y^2) = [a + (x + y)/(x^2 + y^2)]/(x^2 + y^2 - ax)
f'(x,y)dt = 2/a - x(x + y)/a/(x - y)/(x + y) - (x + y)/(x + y)/(x - y) = [a + (x + y)/(x + y)(x - y)]/(x^2 + y^2 - ax)
f'(x,y)dt = 2/a - x/a/(x - y) - 1/(x - y) = [a + 1/(x - y)]/(x^2 + y^2 - ax)
f'(x,y)dt = -a[a + 1/(x - y)]/(x^2 + y^2 - ax) - a/(x - y) + 2 = x/(x + y)

This eq. f'(x,y)dt  cannot be integrated.
f'(x,y) is a number producing equation
yet is multiplied by an infinitessimal
a number so small it is near zero
It can be derived or integrated again if the following is done:
since u*v = u + c there is a constant of derivation that here is called n
such that u*dt = u + n

Let's consider f'(x,y)dt.

f'(x,y)*dt = f'(x,y) + n

when lim dt --> 0 then f'(x,y) + n must tend towards 0

so f'(x,y) + n ≈ 0

For algebraic function, n --> infinity as

x^2 + n ≈ 0
(5)^2 + (-24.999...) ≈ 0
(25)^2 + (-625.0001) ≈ 0
and so on

For some functions, this is not the case. One example is cos(x) = f(x), which oscillates between -1 and 1.

cos(x) + n ≈ 0

and so, cos(x)*dt = cos(x) + n ≈ 0

here, n is necessarily between -1, 1 for all real numbers

f(x) = g(u)
sin(x) = u^2 + u
f'(x) = g'(u)
cos(x)*dx/dt = 2u*du/dt + 1*du/dt
cos(x)*(1) = (2u + 1)du/dt
diff. eq. f'(x)dt = cos(x)*dt = (2u + 1)du
cos(x)*dt = cos(x) + n ≈ 0
cos(x) + n = 2u + 1 + m
n = ± 1
cos(x) ± 1 = 2u + 1 + m
cos(x) - 2u - 1 ± 1 = m
let's put u = tan(x)
cos(x) - 2tan(x) - 1 ± 1  = m
checking the bounds with cos(x) = ±1
-2tan(x) - 1 ± 2 = m

m is between ±infinity ± 2

Okay, tan(x) was a toughie.
Let's make it something not oscillating so bigly.

cos(x) - 2u - 1 ± 1 = m
with u = sin(x)cos(x)
cos(x) - 2sin(x)cos(x) - 1 ± 1 = m
at the utmost
± 1 - 2(±0.5) - 1 ± 1 = m
± 3 - 1 = m
The thing (m) bounces between [-4,2]

Trying with u = sin(x)
cos(x) - 2sin(x) - 1 ± 1 = m
± 1 ± 2 - 1 ± 1 = m
± 4 - 1 = m

m is between [-5,3]
infinitessimal limited to calculable functions
There are functions which can be treated mathematically but do not produce numbers!

x^3 + y^3 + n = {?}
d(x^3)/dt + d(y^3)/dt + dn/dt = f'(x,y)
3x^2*dx/dt + 3y^2*dy/dt + 0 = f'(x,y)
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Moebius Strip, Magnetic Monopole, tiny starship
Not sure how it would come about initially.

I'm not saying it's magnetic... I am saying that but it could spin around a whole bunch.

Somewhere out there, far away in time, is a guy with a moëbius strip and some particles spinning around real good!

Not sure it's naturally occuring, but who knows? It could. oooo, maybe a giant ring world built by crazies to harness a blackhole jet. Yeah.

Yeah!!!

Or a tiny starship trying to harness the super charged jet of stuff coming outta there!
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Artoveli Featured By Owner Apr 29, 2017  Hobbyist General Artist
Hey, thanks for stopping by! :wave:
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TheN00bie Featured By Owner Apr 1, 2017  Hobbyist Traditional Artist
Thanks for the Favorites Icon 3D Llama Emoji-02 (Blush) [V1] 
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BlueDebut Featured By Owner Mar 21, 2017  Hobbyist Digital Artist
Interesting profile :D! Mind if I watch?

I like your maths submissions but they are fairly hard to read as Journals don't have a maths text display feature Llama Emoji-10 (Shy) [V1] .

Have you tried using LaTeX and then taking a screenshot? That'll be much more visually pleasing and easier to read.
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Zig-zag-zug Featured By Owner Apr 5, 2017  Hobbyist Writer
I think I've fainted. begin_rant LaTex has done more to stiful arthimatic than even the worst of teachers, the kind of teacher that says you're bad, and then hangs out in the teacher lounge smoking doobies!

Type text math is concise and dense, works (sorta... ;-; ) in compilers and the sensuality of LaTex is approprietely avoided! /end_rant

:la: It's stuff! Math is stuff! And the more stuff you move the better! And in this way, I can move more stuff. Ohm... Ohm...  You may watch, if you want to. omfg Hiyo from canada! Canada 
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BlueDebut Featured By Owner Apr 6, 2017  Hobbyist Digital Artist
I really don't see how LaTeX has stifled arithmetic :/ why do you say this is the case? It is one of, if not the most, widely used document preparation systems used in publishing mathematical teaching and publication. All I suggested was to submit journals using it or a similar program so that it is easier to read for your audience. For example, It's easier to read int(int(d^2y_1/dy_2^2))dy_2dy_2 as how we would normally write it, else we would write the former. From experience, people generally prefer to read maths displayed the same way they would write it. Another issue that springs to mind with plain text is representing exotic symbols such the direct product of modules or fraktur letters. Indeed math display output from LaTeX code takes up far less space than its plain text representation.

If it works for you, then that's a good thing! I only made the suggestion because it is easier for others to read.
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Zig-zag-zug Featured By Owner Apr 9, 2017  Hobbyist Writer
(I'm not very good at math.)
Ah, you're probably right. The wisdom of the crowd is not readily dismissed! There are stylistic commands required by LaTex sometimes, but overall it copies and pastes easily enough. Exotic symbols are okay, there are more than a few. I suppose I simply prefer clarity over intensity. Once aclimated to LaTex's bevy of curlycues and odd slant I suppose I'd be content.

But enough of my uncouth meadering! The Euleur's remix theorem struck you has interesting!? AH! AAAH! It's been there an eon or two! Thank you! :D Many thanks for your benevolent visit! ^_^ I stuck in it in Snell's theorem, and found it interesting enough, but I've yet to devine it's entire meaning. Pythagore's theorem perhaps holds true for imaginary and real numbers... and perhaps that means negative logarithms must be treated with one day! Oooh Oh!

At the moment, however, I am consumed by functions that make sense mathematically but cannot produce numbers as output when you plug input in. In plain english, a normal function... and then you add a mystery number! Derive, and the constant goes away, and the expression produces numbers once more.

x + y + ç = {?}
dx/dt + dy/dt + (0) = z

Numbers make no sense in the first expression!
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(2 Replies)
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ReclusiveChicken Featured By Owner Nov 28, 2015  Hobbyist General Artist
You are a genius!

I honestly cannot think of any more to say.
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Zig-zag-zug Featured By Owner Dec 9, 2015  Hobbyist Writer
No way! It's about fiddling and learning, and trying to connect things. You have much, much better art than me!
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ReclusiveChicken Featured By Owner Dec 9, 2015  Hobbyist General Artist
Your theories and my art...

...put them together...
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:iconzig-zag-zug:
Zig-zag-zug Featured By Owner Dec 11, 2015  Hobbyist Writer
*blushes* Ah, um, okay, I'll see what I can do my friend! And, how are you doing today?

!
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(1 Reply)
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Kevrekidis Featured By Owner May 19, 2013  Hobbyist Photographer
Thanks for the fav!
See also my new image: [link]
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Zig-zag-zug Featured By Owner Oct 29, 2015  Hobbyist Writer
Interesting perspectives! 
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jaeTanaka Featured By Owner Jan 2, 2013  Professional Digital Artist
Thanks for the fav.
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Zig-zag-zug Featured By Owner Edited Oct 29, 2015  Hobbyist Writer
I really like the ice kream one! And the pretty ladies, that's good too. :)
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Zig-zag-zug Featured By Owner Oct 29, 2015  Hobbyist Writer
My pleasure! :)
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kimsol Featured By Owner Dec 14, 2012
:iconsanta-glompplz: Thanks so much for faving!
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Zig-zag-zug Featured By Owner Dec 18, 2012  Hobbyist Writer
My pleasure! :)
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