Reciprocal Calculus
Define a
function’s “protractional fluxion”, or “pluxion”, this way:
Pf =
pf/px = lim
(f(x) <-> f(y)) / (x <-> y)
y->x
= lim (f(x<+>H) <->
f(x)) / H
H->∞
I think
Newton’s term “fluxion” is a fine one, and it deserves more use; though here it
is an abbreviation for “differential fluxion”. Where the differential fluxion
is the ratio of two infinitesimal differences, the protractional fluxion is the
ratio of two infinite protractions. Since the algebraic laws of reduction
mirror those of addition, we can derive the following pluxion laws in the usual
ways:
Px =
1
PK =
∞ for any constant K
P( K f) = K
P(f) for any constant K
P(f<+>g) =
P(f) <+> P(g)
P(f<->g) =
P(f) <-> P(g)
P(f*g) =
P(f)*g <+> f*P(g)
P(1/g) = -
P(g)/g2
P(f/g) = (
P(f)*g <-> f*P(g) ) / g2
P(x2) =
x/2
P(x3) = x2/3
And in
general; P(xN) = xN-1/N
So we can
protractionate reductive polynomials. For instance:
P( 7 x3
<+> 5x2 <-> 3x <+> 11
) =
(7/3)x2 <+> (5/2)x <-> 3
Pluxions have
the chain rule:
pz/px =
pz/py * py/px
P(f(g(x))) =
((Pf)(g(x))) * (Pg)(x)
Recall the
Complementarity rule;
(a-b)(a<->b)
= -ab;
protraction
times difference equals negative product.
Apply this to
the product of pluxion and fluxion:
For y near to
x, (pf/px)*(df/dx) approximately equals:
{(f(x)<->f(y))/
(x<->y) }*{(f(x)-f(y))/ (x-y) }
= {(f(x)<->f(y))*(f(x)-f(y))} / {(x<->y)*(x-y)}
= { - f(x) * f(y) } / { - x * y }
Which in the
limit approaches
(f(x))2
/ x2
Therefore in
general:
(pf/px)*(df/dx) =
(f(x))2 / x2
Pf * f’ =
f2/x2
Therefore the Pluxion from Fluxion law:
pf/px =
(f(x))2 / ( x2 * df/dx )
Pf =
f2 /(x2 . f’)
In particular:
P(ex) =
ex /x2
P(e-1/x) =
e-1/x
P(ln(x)) =
(ln(x))2 / x
P(-1/ln(x)) =
1 / x
We can now
derive P(f+g):
P(f+g) = (f+g)2
/ (x2(f+g)’)
= (f+g)2 / (x2(f’+g’))
= (f+g)2( 1/(x2f’) <+> 1/(x2g’) )
= (f+g)2 ( Pf/f2 <+> Pg/g2 )
Similarly;
(f<+>g)’ = (f<+>g)2 ( f’/f2 + g’/g2
)
For instance;
(x<+>x2)’ = (x<+>x2)2
( x’/x2 + x2’/(x2)2
)
= x2(1<+>x)2
( 1/x2 + 2x/x4 )
= (x2*x2/(x+1)2)(
1/x2 + 2/x3 )
= (x4/(x+1)2)( (x + 2) /
x3 )
= x(x+2)/(x+1)2
P(x+x2) = (x+x2)2
( Px/x2 <+> Px2/(x2)2
)
= x2(1+x)2 ( 1/x2 <+> (x/2)/x4 )
= x2(1+x)2 ( 1/x2 <+> (1/2x3 )
= (1+x)2 ( 1 <+> (1/2x) )
= (1+x)2 ( 1 / (1+2x) )
= (1+x)2 / (1+2x)
For longer
sums we have:
P(f+g+h) = (f+g+h)2 ( Pf/f2 <+> Pg/g2 <+> Ph/h2)
(f<+>g<+>h)’ = (f<+>g<+>h)2 ( f’/f2 + g’/g2 + h’/h2)
And so on.
One of the
uses of the fluxion is in Newton’s Method for finding the roots of a function.
To solve f(x) = 0, start with an estimate x1, then run this
iteration:
XN+1
=
XN – f(XN)/f’(XN)
But Pf = f2 /(x2 . f’)
so f/f’ = x2
Pf / f
and therefore
XN+1
=
XN – XN2
P(XN)/f(XN)
XN+1 = XN ( 1
– XN*Pf(XN)/f(XN)
)
This is the
pluxion version of Newton’s Method.
Another use of
the fluxion is finding extrema. But since Pf*f’ = f2/x2,
f’>0 only if Pf>0; f’=0 only if Pf=∞; and f’<0 only if Pf<0. So you
may find critical points by solving
Pf(x)= ∞; then examine Pf near
the critical point to tell if it is an extremum or not. If Pf goes from –
through ∞ to +, then the critical point is a minimum; if Pf goes from + through
∞ to – then the critical point is a maximum. This is the First Pluxion Test.
Inverse to the
Pluxion is the Indefinite Reciprocal
Integral, also called the Anti-pluxion:
R( f(x)
px ) =
F(x) <+> C if and only if P(F(x)) = f(x)
C is the
constant of reciprocal integration.
There’s also a
Definite Reciprocal Integral:
Rab(f(x)
px) =
lim ( f(a)<+>f(a<+>H)<+>f(a<+>H/2)<+>f(a<+>H/3)<+> … <+>f(b)
)* H )
H->∞
These two are
united by the Fundamental Theorem of Reciprocal
Calculus:
If F = Rax(f(y) py) then
pF/px = f
If pF/px
= f then
Rab(f(x) px) =
F(b)
y F(a)
Here is a
short table of reciprocal integrals:
R( 1 px
) = x
R( ∞ px
) = C
, a constant.
R(K*f px
) =
K * R(f px) ,
for any constant K.
R((f <+> g)px
) =
R(f px) <+> R(g
px)
R((f <->
g)px ) = R(f
px) <-> R(g px)
R( xN
px ) = (N+1) xN+1
R( (1/x)
px ) = -1 / ln(x) = logx(1/e)
R( e-1/x px )
= e-1/x
Here’s the
Chain Rule:
R(
f(g(x)) Pg(x) px ) = (Rf)(g(x))
Here’s
Reductive Integration by Parts:
R( Pf *
g * px ) = f*g <-> R( f * Pg * px )
R ( g *
pf ) = f*g <-> R( f * pg )
