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Derivative by infinitesimal multiplication
Newton's derivative by infinitesimal addition
The limit when a --> 0 is used to check the slope of a point on a graph.
Calculus uses the addition to derive the slope at a point.
a line f(x)=x, with two points y1, y2
x=y1
x+a=y2
for a number n
if
f(x+n) = f(x)+f(n)
then derivative is
f(x+a)-f(x) f(x)+f(a)-f(x) f(a)
----------- = ------------------ = -----------
(x+a) - x a a
which when evaluated creates indeterminate answer at x=0
lim f(0) infnity 0
a-->0 ------- = or ------
0 0
We could also check a point on a graph with multiplication of tiny a.
if
f(a*x) = a*f(x)
lim f(a*x) - f(x) f(x)*(a - 1) f(x)
a-->0 --------------- = ------------ = ------
a*x - x x*(a-1) x
So any function where a constant can be pulled out will have the above
form if multiplication happens.
Also, calling multiplicative derivative "merivative"
because first come first served. :3
(Actually, someone probably thought of this at some point already.)
Right, because for merivitave, f(x)=x (a line) is gonna be wonky.
f(x)=x
Mx(f(x)) = ?
lim ax - x lim x(a-1) lim 1
a-->0 ------- = a-->0 --------- = a-->0 ------
ax-x x(a-1) 1
That limit makes no sense, so we can't have it go like that, so, for f(x)=x
we gotta have an exception. Oh hey, isn't that how Newton's derivitive works?
Gotta have both! So, for a equation with x*f(x), how's it going to work?
Keep it stupid, simple! Better to have an error at the start then the end, eh?
Mx[f(x)]= (x+a)*f(a*x)
1) Mx[f(x)]=x^2
lim ax^2 - x^2 x^2(a-1) x^2
a-->0 ---------------- = ------------ = ---- = x
ax-x x(a-1) x
2) Mx[f(x)]=x^n
lim ax^n - x^n x^n(a-1) x^n
a-->0 ---------- = ------------ = ------- = x^(n-1)
ax-x x(a-1) x
Similar, but different than calculus where Dx(x^n)= n*x^(n-1)
f(x)=cos(a*x)
using half-angle formula
lim cos(a*x)-cos(x) (0.5a*cos(x)^2 - 1) - cos(x)
a-->0 --------------- = ----------------------------
ax-x x(a-1)
lim cos(x)(0.5a*cos(x)-1) - 1
a-->0 -------------------------
x(a-1)
lim cos(x)(0.5a*cos(x)-1) lim -1
a-->0 --------------------- + a-->0 -----
x(a-1) x(a-1)
cos(x)(0 - 1) -1
----------------- + --------
x(0-1) x(0-1)
cos(x) 1
------ + ---
x x
f(x)=sin(a*x)
lim sin(a*x)-sin(x) lim 0.5*a*sin(x)cos(x) - sin(x)
a-->0 --------------- = a-->0 --------------------------
a*x-x x(a-1)
lim sin(x)(0.5*a*cos(x)-1) sin(x)(0-1) sin(x)
a-->0 ---------------------- = ---------------- = ------
x(a-1) x(0-1) x
??
y2-y2 sin(x)
----- = ------
x2-x1 x
??
if sin(x)=1
then is related to merivative of cos(x) ?
cos(x) (1)
Mx(cos(x)) = ----- + ----
x x
if cos(x)=0
then it too could be related to merivative of sin(x)
(0) sin(x)
Mx(sin(x) = ---- + -----
x x
f(x)=tan(x)
tan(a*x)-tan(x) -sec(x)*sin(x-a*x)*sec(a*x) 0
-------------- = --------------------------------- = ---
a*x-x a*x-x x
Here tan(x)=0
is due to simplification?
is due to f1(x)-f2(x)=0?
needs fiddling!
Since x+x+x+... = n*x
A multiplicative derivative can either be random, or some constant number of additions.
a = dx+dy+dz+dw+...
b = dx+dx+dx+dx+... = n*dx
x+a = x+a = x+dx+dy+dz+dw+... !~ x+dx
x+b = x+n*dx ~ x+dx
a*x = x*(dx+dy+dz+dw+...) !~ x+dx
b*x = n*dx*x ~ lim dx->0 x+dx
I should check when lim 0 and inf, as they are both indeterminate when additive derivation.
f(x) = x*cos(x)
(x+a)*cos(a*x) - x*cos(x)
Mx(f(x)) = -------------------------
a
x*cos(a*x)+a*cos(a*x) - x*cos(x)
= --------------------------------
a
cos(x) 1
Mx(cos(a*x) = ----- + ---
x x
x*(cos(x)/x + 1/x) +a*(cos(x)/x + 1/x) - x*cos(x)
= ----------------------------------------------------
a
cos(x) +1 +a*(cos(x)/x + 1/x) - x*cos(x)
= -------------------------------------------------
a
cos(x) +1 +a*(cos(x)/x + 1/x) - x*cos(x)
= ------ ---- ------------- -------
a a a a
cos(x) +1 . +(cos(x)/x + 1/x) - x*cos(x)
= ------ ---- ------
a a a
Here, the question of "how small or big is 'a'" is interesting!
If 0, 3 indeterminate terms.
If large, 3 zeroes.
Kinda interesting, since there are 3 ways x*cos(x)=0.
1 linear x=0
2 trig circle cos(x) for x={-1,1}
0*cos(n)=0
1*cos(1)=0
-1*cos(-1)=0
Mx(x*sin(x)) = (x*a)*sin(a*x) - x*sin(x)
x*sin(a*x) + a*sin(x) - x*sin(x)
= --------------------------------
a
x*sin(x)/x + a*sin(x) - x*sin(x)
= --------------------------------
a
x*sin(x)*(1/x -1) + a*sin(x)
= ------------------- --------
a a
x*sin(x)*(1/x -1) + sin(x)
= --------------------
a
if
lim a
--> 0
= f(x)/0 + sin(x)
yet if
lim a
--> infinity
= f(x)/infinity + sin(x)
= sin(x)
Newton's derivative by infinitesimal addition
The limit when a --> 0 is used to check the slope of a point on a graph.
Calculus uses the addition to derive the slope at a point.
a line f(x)=x, with two points y1, y2
x=y1
x+a=y2
for a number n
if
f(x+n) = f(x)+f(n)
then derivative is
f(x+a)-f(x) f(x)+f(a)-f(x) f(a)
----------- = ------------------ = -----------
(x+a) - x a a
which when evaluated creates indeterminate answer at x=0
lim f(0) infnity 0
a-->0 ------- = or ------
0 0
We could also check a point on a graph with multiplication of tiny a.
if
f(a*x) = a*f(x)
lim f(a*x) - f(x) f(x)*(a - 1) f(x)
a-->0 --------------- = ------------ = ------
a*x - x x*(a-1) x
So any function where a constant can be pulled out will have the above
form if multiplication happens.
Also, calling multiplicative derivative "merivative"
because first come first served. :3
(Actually, someone probably thought of this at some point already.)
Right, because for merivitave, f(x)=x (a line) is gonna be wonky.
f(x)=x
Mx(f(x)) = ?
lim ax - x lim x(a-1) lim 1
a-->0 ------- = a-->0 --------- = a-->0 ------
ax-x x(a-1) 1
That limit makes no sense, so we can't have it go like that, so, for f(x)=x
we gotta have an exception. Oh hey, isn't that how Newton's derivitive works?
Gotta have both! So, for a equation with x*f(x), how's it going to work?
Keep it stupid, simple! Better to have an error at the start then the end, eh?
Mx[f(x)]= (x+a)*f(a*x)
1) Mx[f(x)]=x^2
lim ax^2 - x^2 x^2(a-1) x^2
a-->0 ---------------- = ------------ = ---- = x
ax-x x(a-1) x
2) Mx[f(x)]=x^n
lim ax^n - x^n x^n(a-1) x^n
a-->0 ---------- = ------------ = ------- = x^(n-1)
ax-x x(a-1) x
Similar, but different than calculus where Dx(x^n)= n*x^(n-1)
f(x)=cos(a*x)
using half-angle formula
lim cos(a*x)-cos(x) (0.5a*cos(x)^2 - 1) - cos(x)
a-->0 --------------- = ----------------------------
ax-x x(a-1)
lim cos(x)(0.5a*cos(x)-1) - 1
a-->0 -------------------------
x(a-1)
lim cos(x)(0.5a*cos(x)-1) lim -1
a-->0 --------------------- + a-->0 -----
x(a-1) x(a-1)
cos(x)(0 - 1) -1
----------------- + --------
x(0-1) x(0-1)
cos(x) 1
------ + ---
x x
f(x)=sin(a*x)
lim sin(a*x)-sin(x) lim 0.5*a*sin(x)cos(x) - sin(x)
a-->0 --------------- = a-->0 --------------------------
a*x-x x(a-1)
lim sin(x)(0.5*a*cos(x)-1) sin(x)(0-1) sin(x)
a-->0 ---------------------- = ---------------- = ------
x(a-1) x(0-1) x
??
y2-y2 sin(x)
----- = ------
x2-x1 x
??
if sin(x)=1
then is related to merivative of cos(x) ?
cos(x) (1)
Mx(cos(x)) = ----- + ----
x x
if cos(x)=0
then it too could be related to merivative of sin(x)
(0) sin(x)
Mx(sin(x) = ---- + -----
x x
f(x)=tan(x)
tan(a*x)-tan(x) -sec(x)*sin(x-a*x)*sec(a*x) 0
-------------- = --------------------------------- = ---
a*x-x a*x-x x
Here tan(x)=0
is due to simplification?
is due to f1(x)-f2(x)=0?
needs fiddling!
Since x+x+x+... = n*x
A multiplicative derivative can either be random, or some constant number of additions.
a = dx+dy+dz+dw+...
b = dx+dx+dx+dx+... = n*dx
x+a = x+a = x+dx+dy+dz+dw+... !~ x+dx
x+b = x+n*dx ~ x+dx
a*x = x*(dx+dy+dz+dw+...) !~ x+dx
b*x = n*dx*x ~ lim dx->0 x+dx
I should check when lim 0 and inf, as they are both indeterminate when additive derivation.
f(x) = x*cos(x)
(x+a)*cos(a*x) - x*cos(x)
Mx(f(x)) = -------------------------
a
x*cos(a*x)+a*cos(a*x) - x*cos(x)
= --------------------------------
a
cos(x) 1
Mx(cos(a*x) = ----- + ---
x x
x*(cos(x)/x + 1/x) +a*(cos(x)/x + 1/x) - x*cos(x)
= ----------------------------------------------------
a
cos(x) +1 +a*(cos(x)/x + 1/x) - x*cos(x)
= -------------------------------------------------
a
cos(x) +1 +a*(cos(x)/x + 1/x) - x*cos(x)
= ------ ---- ------------- -------
a a a a
cos(x) +1 . +(cos(x)/x + 1/x) - x*cos(x)
= ------ ---- ------
a a a
Here, the question of "how small or big is 'a'" is interesting!
If 0, 3 indeterminate terms.
If large, 3 zeroes.
Kinda interesting, since there are 3 ways x*cos(x)=0.
1 linear x=0
2 trig circle cos(x) for x={-1,1}
0*cos(n)=0
1*cos(1)=0
-1*cos(-1)=0
Mx(x*sin(x)) = (x*a)*sin(a*x) - x*sin(x)
x*sin(a*x) + a*sin(x) - x*sin(x)
= --------------------------------
a
x*sin(x)/x + a*sin(x) - x*sin(x)
= --------------------------------
a
x*sin(x)*(1/x -1) + a*sin(x)
= ------------------- --------
a a
x*sin(x)*(1/x -1) + sin(x)
= --------------------
a
if
lim a
--> 0
= f(x)/0 + sin(x)
yet if
lim a
--> infinity
= f(x)/infinity + sin(x)
= sin(x)
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I try to math or something. Me having flunked Calc 3 should be apparent from reading this. Eh. Live and learn!
Needs more thinking!
Needs more thinking!
Comments1
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Darn it, DA edited the spacing.