Re: TIB: Problematic equation in ti-86
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Re: TIB: Problematic equation in ti-86
On Fri, 26 Dec 1997 20:52:00 +0200 (EET) Janne M{kel{ <maja@otol.fi>
writes:
>I have a little problem with finding root(s) of a
>equation I've created. Actually it's a bunch of
>equations that my program creates, but some pointers
>would certainly be appreciated here.
...
If the equation you are solving roots for has only integer powers, then
you could use the command poly.
If not then you will need to use Newton's method:
Let f(c) = 0, where f is differentiable on an open interval containing c.
Then, to approximate c, use the following steps.
1. Make an initial estimate x1 that is "close" to c.
2. Determine a new approximation
x2 = x1 - f(x1)/f'(x1)
3. if abs(x1 - x2) is less than desired accuracy, let x2 serve as the
final approximation. Otherwise, return to step 2 and calculate a new
approximation(using value of x2 in x1).
If you don't understand anything I just said, here's an example program
from my AP Calculus book(it's for the 81, but you'll get the picture).
Prgm:NEWTON
:(Xmax-Xmin)/100-->D
:1-->I
:Lbl 1
:X-Y1/NDeriv(Y1,D)-->R
:If abs(X-R)<=abs(X/1E10)
:Goto 2
:R-->X
:I+1-->I
:Goto 1
:Lbl 2
:Disp "ROOT="
:Disp R
:Disp "ITER="
:Disp I
To run the program, enter the function as Y1 and enter the initial guess
as X.
Sincerely,
The Unibomer
Jared Ivey
Unibomer@Juno.com
http://www.geocities.com/SiliconValley/Vista/7342
"It is a puzzling thing. The truth knocks on the door and you say, 'Go
away, I'm looking for the truth.' and so it goes away. Puzzling." --
Robert M. Pirsig
References: