Discussion:
the problem solved
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spiny urchin
2015-05-27 11:23:57 UTC
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This finds the intersection of a curve with the x-axis:

http://www.stonetabernacle.com/intersect-curve.html

The curve is drawn, all points derived and the tangent and normal to the
tangent are determined. Vector algebra is then applied between points on
the graph. From this it is shown that f '(x)=I Newton's Method is derived.
Newton's Method is extrapolated to the nth derivative of the nth degree
polynomial (such as n=1000). From there the smallest increment is used to
reconstruct the equation working backwards until the original is obtained.
All roots to the power series are obtained and the problem solved.

Notice for anyone who visited my web page before: this is new material
added to the end of the page.
spiny urchin
2015-05-27 12:27:39 UTC
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Post by spiny urchin
http://www.stonetabernacle.com/intersect-curve.html
The curve is drawn, all points derived and the tangent and normal to the
tangent are determined. Vector algebra is then applied between points on
the graph. From this it is shown that f '(x)=I Newton's Method is
derived. Newton's Method is extrapolated to the nth derivative of the nth
degree polynomial (such as n=1000). From there the smallest increment is
used to reconstruct the equation working backwards until the original is
obtained. All roots to the power series are obtained and the problem
solved.
Notice for anyone who visited my web page before: this is new material
added to the end of the page.
Case in fact. Consider the 7th degree polynomial. Six derivatives are
taken for the numerator, and seven derivatives are taken for the numerator.
Relatively taken, the result is just the same as a 0th derivative in the
numerator and the 1st derivative in the denominator:

f^6(x) ->f^0(x)
f^7(x) -> f^1(x)

that is,

x_c = x - (f^6)/(f^7) is the same as,
x_c = x - (f^0)/(f^1) per increment

where

f^1(x)=+/- i

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