## Tuesday, November 19, 2013

### Conjugates of Fields, 2 of 6; Reciprocal Addition

Reciprocal Arithmetic

Reciprocal arithmetic is arithmetic conjugated by the reciprocal function:

x   1/[+]   y     =     1 / (1/x  + 1/y)    =    xy/(x+y)   =   "x with y"   =  "reduction"
x   1/[-]   y     =      x  1/+ (-y)          =    xy/(y-x)    =   "x without y"  =  "retraction"
x   1/[*]  y     =      1 /  ((1/x)(1/y))   =       xy

Reduction is a field under *, with identities 1 and 1/0. Infinity is the zero of this field; and zero, in return, is its infinity - that is, its attractor:

x  1/+        =    x
x  1/-  x       =
x  1/+  0       =    0

Reduction and retraction obey these "common numerator" and "general numerator" laws:

c/a   1/+  c/b     =    c/(a+b)
c/a   1/-   c/b     =    c/(a-b)
a/b   1/+  c/d     =    ac / (ad+bc)   =   (ad 1/+ bc) / bd
a/b   1/-   c/d     =    ac / (ad-bc)    =   (ad 1/- bc) / bd

Reciprocal arithmetic is just like ordinary arithmetic, but "standing on its head" - i.e. with the roles of numerator and denominator reversed. For instance, these reciprocal sums apply:

1  1/+  1                                      =   1/2
1  1/+  1  1/+  1                              =   1/3
1  1/+  1  1/+  1  1/+  1                      =   1/4
1  1/+  1  1/+  1  1/+  1  1/+  1              =   1/5  ...
2  1/+  2                                      =    1
3  1/+  3  1/+  3                              =    1
4  1/+  4  1/+  4  1/+  4                      =    1
5  1/+  5  1/+  5  1/+  5  1/+  5              =    1  ...

These formulas connect linear and reciprocal sums:

a(a+b)    1/+   b(a+b)    =     ab
a(a-b)     1/-   b(a-b)      =     ab
a(a+1)    1/+   (a+1)      =      a

So   a+b = c   implies that     ac  1/+  bc   =   ab .   Therefore:

2+3=5         implies      10  1/+  15   =    6 ;
3+4=7         implies      21  1/+  28   =  12 ;
3+5=8         implies      24  1/+  40   =  15 ;
5+7=12       implies      60  1/+  84   =  35 ;
23+17=40   implies     920 1/+ 680  =  391;  and so on.

If a whole number is partitioned into a sum of its divisors, then that implies a whole-number reciprocal partition of unity. For instance:
6   =   1+2+3;     therefore  6/6  =  6/(1+2+3) ;     therefore   1  =   6  1/+ 3 1/+  2
12  =  1+2+3+6                      ;   so  1  =   12  1/+  6  1/+ 4  1/+  2
20  =  1+4+5+10                    ;   so  1  =   20  1/+  5  1/+  4  1/+  2
30  =  2+3+10+15                  ;   so  1  =   15  1/+ 10 1/+  3  1/+  2
36  =  2+3+4+6+9+12            ;   so  1  =   18  1/+  12  1/+  9  1/+  6  1/+ 4  1/+ 3
60  =  4+5+6+10+15+20        ;   so  1  =  15  1/+  12  1/+ 10  1/+ 6  1/+  4  1/+  3
84  =  3+6+12+14+21+28      ;   so  1  =  28  1/+  14  1/+  7  1/+ 6   1/+  4  1/+  3
90  =  3+5+9+10+15+18+30  ;   so  1  =  30  1/+  18  1/+  10 1/+  9 1/+  6 1/+  5  1/+  3

Reduction, like addition, obeys the transposition law:
X  1/+  A      =     B             implies      X   =    B  1/-  A
Therefore we can solve "reciprocal-linear" algebraic equations:
AX  1/+  B   =   CX  1/+  D    implies    X    =   (D  1/-  B) / (A  1/-  C)

The reciprocal version of the quadratic equation:
AX2  1/+  BX  1/+  C    =
has this solution:
X  =  ( - 2B  1/+  ± \/{4B2 1/-  AC}  )  /  A

The "reciprocal continued fraction"     x  =  a  1/+  1 / x
solves the equation       x2  1/-  ax  1/-  1    =
and therefore equals     2a  1/+   \/{4a2 1/+ 1}

Reduction connects to multiplication via logarithmic bases:

If C = ab,     then      logC(x)   =    loga(x)   1/+   logb(x)
If C = a/b,    then      logC(x)   =    loga(x)   1/-   logb(x)
If C = ab,      then      logC(x)   =    loga(x)   /   b
If C = b\/a,    then      logC(x)   =    loga(x)   *  b

Applications of Reduction

Reduction has a geometric construction; the "Fence", or “Fishtail”:

*     *
|\   /|
| \ / |
x |  *  | y
| /|\ |
|/ |z\|
*--*--*

In this diagram the middle line, z, has length equal to x 1/+ y. This is a construction in affine geometry; it needs no metrics, only parallels. If we double the length of the middle line, then we get the "reciprocal mean"  2(x 1/+ y):

"Kirkhoff's law" for resistors involves reduction: if resistors R1 and R2  are wired in parallel, then their combined resistance Rc  equals   R1  1/+  R2 .
Thus, for instance, if  R1 = (a break) then    Rc   =    ∞ 1/+ R2   =   R2 ;
and  if   R1 = 0  (a short)  then  Rc  =  0 1/+  R2   =   0  .

If two lenses with focal lengths F1 and F2  are combined,
then the resulting lens has focal length  F1 1/+  F2 .

Given the time-distance-rate relations  D  =  R1T1 =  R2T2 , then
D / (R1+R2)   =    T1 1/+ T2
D / (T1+T2)   =    R1 1/+ R2
D / (R1-R2)   =    T1 1/- T2
D / (T1-T2)    =    R1 1/- R2
When rates add, times reduce, and vice versa. "Many hands make light work".

These relations also apply to work-rates. If Alice and Bob can do a job (say, mow the lawn) in, respectively,  T1 and T2  minutes of work, then if they work together and their rates add, they will finish the job in  T1 1/+ T2  minutes. If Bob works against Alice, then the time will be T1 1/- T2. If Bob, that lazy bum, would take forever to do the job, then together they finish in time T1 1/+   =  T1; and if Bob, that brilliant inventor, builds a gizmo that can do the job in no time at all, then together they finish in time T1 1/+ 0  =  0 .

If a particle moves at velocity V1 over a distance D, then at velocity V2 over the same distance D, then its average velocity over the distance 2D is  2(V1 1/+ V2) ; the reciprocal mean.

If Alice's and Bob's cars are one mile apart, and if Alice's car can drive a mile in X minutes, and Bob's car can drive a mile in Y minutes, and if they drive toward each other, then they will meet in (X 1/+ Y) minutes; whereas if Alice chases after Bob, then she will overtake him in (X 1/- Y) minutes. (This is assuming that Alice's car is faster; i.e. that X<Y.)
If Alice and Bob meet in E minutes, and overtake in F minutes, then we obtain this reciprocal-linear system:
X   1/+  Y      =      E
X   1/-   Y      =      F
We can solve this reciprocal-linear system by methods similar to solving a linear system; i.e. by substitution, or by a reciprocal version of Gauss-Jordan elimination. This yields:
X     =     2 ( E  1/+  F )   ;    the reciprocal mean
Y     =     2 ( E  1/-  F )    ;    the reciprocal radius.

In general, the reciprocal-linear system:
aX   1/+  bY      =      E
cX   1/+  dY      =      F
has this reciprocal-Cramer's-rule solution:
X     =     ( Ed  1/-  bF )  /   ( ad  1/-  bc )
Y     =     ( aF  1/-  Ec )  /   ( ad  1/-  bc )
That solution can be phrased in terms of reciprocal determinants. We can also define reciprocal matrices, vectors, etc. Reciprocal-linear algebra is just like linear algebra, except with 1/+ standing in for +.