Arclength Calculation

E^8
===

ax^8+bx^7+cx^6+dx^5+ex^4+fx^3+gx^2+hx+i = 0

2{2{2{2{2{2{2{Ax^8}^7/8-Bx^7}^6/7-Cx^6}^5/6-Dx^5}^4/5-Ex^4}^3/4-Fx^3}^2/3-Gx
^2}^1/2-Hx - I = 0

x = I / [ 2{2{2{2{2{2{2{A}^7/8-B}^6/7-C}^5/6-D}^4/5-E}^3/4-F}^2/3-G}^1/2 -
H]

Let J = I / [
2{2{2{2{2{2{2{A}^7/8-B}^6/7-C}^5/6-D}^4/5-E}^3/4-F}^2/3-G}^1/2 - H]

Suppose

f(x) = (x - J)^8 = 0
f '(x) = 8(x - J)^7
f ''(x) = 8*7(x - J)^6
f '''(x) = 8*7*6(x - J)^5
f ''''(x) = 8*7*6*5(x - J)^4
f^{5}(x) = 8*7*6*5*4(x - J)^3
f^{6}(x) = 8*7*6*5*4*3(x - J)^2
f^{7}(x) = 8!(x - J)
f^{8}(x) = 8!

f(0) = (- J)^8
f '(0) = 8(- J)^7
f ''(0) = 8*7(- J)^6
f '''(0) = 8*7*6(- J)^5
f ''''(0) = 8*7*6*5(- J)^4
f^{5}(0) = 8*7*6*5*4(- J)^3
f^{6}(0) = 8*7*6*5*4*3(- J)^2
f^{7}(0) = 8!(- J)
f^{8}(0) = 8!

(1/0!)f(0) = (- J)^8 = i/a
(1/1!)f '(0) = 8(- J)^7 = h/a
(1/2!)f ''(0) = (1/2!)8*7(- J)^6 = g/a
(1/3!)f '''(0) = (1/3!)8*7*6(- J)^5 = f/a
(1/4!)f ''''(0) = (1/4!)8*7*6*5(- J)^4 = e/a
(1/5!)f^{5}(0) = (1/5!)8*7*6*5*4(- J)^3 = d/a
(1/6!)f^{6}(0) = (1/6!)8*7*6*5*4*3(- J)^2 = c/a
(1/7!)f^{7}(0) = (1/7!)8!(- J) = b/a
(1/8!)f^{8}(0) = 1 = a/a

{-J}^8 ={i/a}
{-J}^7 ={(1/8)(h/a)}
{-J}^6 ={(2!/[8*7])(g/a)}
{-J}^5 = {(3!/[8*7*6])(f/a)}
{-J}^4 = {(4!/[8*7*6*5])(e/a)}
{-J}^3 = {(5!/[8*7*6*5*4])(d/a)}
{-J}^2 = {(6!/[8*7*6*5*4*3])(c/a)}
{-J}^1 = {(1/8)(b/a)}

J^8 = {i/a} eqns i.
J^7 = -{(1/8)(h/a)}
J^6 = {(2!/[8*7])(g/a)}
J^5 = -{(3!/[8*7*6])(f/a)}
J^4 = {(4!/[8*7*6*5])(e/a)}
J^3 = -{(5!/[8*7*6*5*4])(d/a)}
J^2 = {(6!/[8*7*6*5*4*3])(c/a)}
J^1 = -{(1/8)(b/a)}


Where
J = I / [ 2{2{2{2{2{2{2{A}^7/8-B}^6/7-C}^5/6-D}^4/5-E}^3/4-F}^2/3-G}^1/2 -H]

I = J[ 2{2{2{2{2{2{2{A}^7/8-B}^6/7-C}^5/6-D}^4/5-E}^3/4-F}^2/3-G}^1/2 - H]

2{2{2{2{2{2{2{AJ^8}^7/8-BJ^7}^6/7-CJ^6}^5/6-DJ^5}^4/5-EJ^4}^3/4-FJ^3}^2/3-GJ
^2}^1/2-HJ - I = 0

2{2{2{2{2{2{2{A{i/a}^8}^7/8+B{(1/8)(h/a)}^7}^6/7-C{(2!/[8*7])(g/a)}^6}^5/6+D
{(3!/[8*7*6])(f/a)}^5}^4/5-E{(4!/[8*7*6*5])(e/a)}^4}^3/4+F{(5!/[8*7*6*5*4])(
d/a)}^3}^2/3-G{(6!/[8*7*6*5*4*3])(c/a)}^2}^1/2+H{(1/8)(b/a)} -I = 0

J = x

I = I

2{2{2{2{2{2{2{A{i/a}^8}^7/8+B{(1/8)(h/a)}^7}^6/7-C{(2!/[8*7])(g/a)}^6}^5/6+D
{(3!/[8*7*6])(f/a)}^5}^4/5-E{(4!/[8*7*6*5])(e/a)}^4}^3/4+F{(5!/[8*7*6*5*4])(
d/a)}^3}^2/3-G{(6!/[8*7*6*5*4*3])(c/a)}^2}^1/2+H{(1/8)(b/a)}=x[
2{2{2{2{2{2{2{A}^7/8-B}^6/7-C}^5/6-D}^4/5-E}^3/4-F}^2/3-G}^1/2 - H]

A= R^8
B = R^7
C = R^6
D = R^5
E = R^4
F = R^3
G = R^2
H = R R cancels and

x =
2{2{2{2{2{2{2{{i/a}^8}^7/8+{(1/8)(h/a)}^7}^6/7-{(2!/[8*7])(g/a)}^6}^5/6+{(3!
/[8*7*6])(f/a)}^5}^4/5-{(4!/[8*7*6*5])(e/a)}^4}^3/4+{(5!/[8*7*6*5*4])(d/a)}^
3}^2/3-{(6!/[8*7*6*5*4*3])(c/a)}^2}^1/2+{(1/8)(b/a)}

If r[k] are the roots of
(x-r[8])(x-r[7])(x-r[6]) ... (x-r[2])(x-r[1]) = 0
then r[k] are also the roots of
(x-r[8])^8 (x-r[7]) ^8 (x-r[6])^8 ... (x-r[2])^8 (x-r[1])^8 = 0
making the solution to
f(x) = (x - J)^8 a valid root to ax^8+bx^7+cx^6+dx^5+ex^4+fx^3+gx^2+hx+i
= 0

ARCLENGTH CALCULATION

Known arc A and known chord B subtend unknown angle z on a circle. Find z
in terms
of A and B.

Bz/(2A) = sin(z/2)

sin(z/2) = (z/2) - ((z/2)^3)/3! + ((z/2)^5)/5! - ((z/2)^7)/7! +
((z/2)^9)/9! - ((z/2)^11)/11! + ...

B/A = 1 - ((z/2)^2)/3! + ((z/2)^4)/5! - ((z/2)^6)/7! + ...

Letting x = (z/2)^2, and B/A = q 0 = q = 1

(1-q) - x/3! +(x^2)/5! - (x^3)/7! + (x^4)/9! - (x^5)/11! + (x^6)/13! -
(x^7)/15! + (x^8)/17! = 0

ax^8+bx^7+cx^6+dx^5+ex^4+fx^3+gx^2+hx+i

a = 1/17!
b = -1/15! b/a = -17!/15!
c = 1/13! c/a = 17!/13!
d = -1/11! d/a = -17!/11!
e = 1/9! e/a = 17!/9!
f = -1/7! f/a = -17!/7!
g = 1/5! g/a = 17!/5!
h = -1/3! h/a = -17!/3!
i = (1-q) i/a = 17!(1-q)

z =
2[2{2{2{2{2{2{2{{17!(1-q)}^8}^7/8+{(1/8)(-17!/3!)}^7}^6/7-{(2!/[8*7])(17!/5!
)}^6}^5/6+{(3!/[8*7*6])(-17!/7!)}^5}^4/5-{(4!/[8*7*6*5])(17!/9!)}^4}^3/4+{(5
!/[8*7*6*5*4])(-17!/11!)}^3}^2/3-{(6!/[8*7*6*5*4*3])(17!/13!)}^2}^1/2+{(1/8)
(-17!/15!)}]^1/2

or

z =
2[2{2{2{2{2{2{2{{-17!(1-B/A)}^8}^7/8+{(1/8)(17!/3!)}^7}^6/7-{(2!/[8*7])(-17!
/5!)}^6}^5/6+{(3!/[8*7*6])(17!/7!)}^5}^4/5-{(4!/[8*7*6*5])(-17!/9!)}^4}^3/4+
{(5!/[8*7*6*5*4])(17!/11!)}^3}^2/3-{(6!/[8*7*6*5*4*3])(-17!/13!)}^2}^1/2+{(1
/8)(17!/15!)}]^1/2

"Jon" wrote in message
ink.net...

Nice spew.