I share your doubts, but if mathematics doesn't forsake me, it has to be
right.
When a curve f(x) crosses the x-axis at point t, for small distances close
to t, the curve approximates a straight line, right? NO, IT DOESN'T.
That's one flaw in my method. If I take the derivative of f(x) at t, the
tangent line is a straight line.
One area is between f(x) and the x-axis from t to (t + delta t), and the
other area is between f(x) and the x-axis from (t - delta t) to t. I simply
say that when f(x) crosses the x-axis at f(x)=0, THE TWO AREAS CANCEL TO
ZERO.
A better strategy is to use the areas above and below the tangent line when
f(x) crosses the x-axis.
g(x) = f '(x) x + b equation of tangent line
INT (f '(x) x + b) dx from (t - delta t) to (t + delta t) = 0
For instance, using the example suggested by François and letting delta t =
c,
f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1,
f '(x) = 4x^3 - 12x^2 + 12x - 4
INT (4x^4 - 12x^3 + 12x^2 - 4x + bx) from (t - c) to (t + c) = 0
= [(4/5)x^5 - 3x^4 + 4x^3 - 2x^2 + (b/2)x^2] from (t-c) to (t+c)
= 0
cancel x^2
= [(4/5)x^3 - 3x^2 + 4x - 2 + (b/2)] from (t-c) to (t+c) = 0
h(t)=
+(4/5)[(t+c)^3 - (t-c)^3]
-3[(t+c)^2 - (t-c)^2]
+4[(t+c) - (t-c)]
- 2 + b/2
=0
f '(t)=
4t^3 - 12t^2 + 12t - 4=
-b/t=
[f(t+c) - f(t-c)]/(2c)
b = -t f '(t)
h(t)=
+(4/5)[(t+c)^3 - (t-c)^3]
-3[(t+c)^2 - (t-c)^2]
+4[(t+c) - (t-c)]
- 2 - (1/2)t f '(t)
=0
Express h(t) as a Maclaurin series, take the limit as c approaches zero and
WA LA! There's the solution to the quartic in terms of the known solution
to the cubic.
Notice, however, in the original equation,
f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1,
The lonely 1 has been entirely omitted out of the solution. This is a
heinous injustice to the lonlely 1.
Maybe one of you can FIX IT.
--
Jon G.
***@peoplepc.com
http://mypeoplepc.com/members/jon8338/math/index.html
Post by Jon G.roots to 4th, 5th, 6th & 7th degree polynomials
http://mypeoplepc.com/members/jon8338/math/id19.html
The best I can promise is all denominators are nonzero, and dimensional
analysis passes.
See if you can understand the concept. If it works, it can find the roots
to any degree polynomial. If it doesn't, then at best it may give a rough
estimate.
--
Jon Giffen
Have you tried the formulae for a variety of polynomials.
Your cubic might be right; the quartic is probably wrong since a
resolvent cubic is needed and you do not have one; for higher degrees,
Abel's theorem says "Nope!"