In article <402### [at] hotmailcom>,
 andrel <a_l### [at] hotmailcom> wrote:

> Are you sure? I have used this myself and when I used it
> the tangents were correct. Note also the symmetries in the
> coefficient, the binomial coeeficients and the nice alternating
> signs. I think the original error may be in the line:
>  > P2*3*(2*(1 - t)*(-1)*t + (1 - t)2) -
> In my deriviation the final '-' is a '+'. BTW, I derived the
> equations in the same way as you did. Well, of course, we both
> have the same sort of math training I suspect :).

The one at the very end? That was in the original equation:
P1*(1 - t)^3 +
P2*3*(1 - t)^2*t -
P3*3*t^2*(1 - t) +
P4*t^3

If that's wrong, than the original equation is too. If you're talking 
about something in that term, the Maxima result is: 3*P2*(1 - t)^2 - 
6*P2*(1 - t)*t
=
3*P2*((1 - t)^2 - 2*(1 - t)*t)
=
3*P2*(-2*(1 - t)*t + (1 - t)^2)

Which agrees with my result.
As for training, I'm currently slogging my way through Calculus II, and 
am mainly self-taught. I'm still mainly teaching myself, due to the fact 
that the instructor can't speak clear English or write legibly. I'm also 
taking Numeric Analysis, which may cover splines later this semester.


>  > My results look right, any errors are small ones.
>  >
> What equations did you use in the end?

a():=0.1; b():=0.9; c():=0.2; d():=1;
(Defined as functions because I couldn't figure out how to define them 
as variables in Maxima...damned annoying program, with practically 
useless documentation.)

f(t):=a()*(1 - t)^3 + b()*3*(1 - t)^2*t - c()*3*t^2*(1 - t) + d()*t^3;

fd(t):=3*(b() - a() + (a() - 2*b() - c())*2*t + (b()*3 + c()*3 + d() - 
a())*t^2);

tgt(t, p):= (t - p)*fd(p) + f(p);

plot2d([f(t), tgt(t, 0.2), tgt(t, 0.5), tgt(t, 0.9)], [t, 0, 1]);

-- 
Christopher James Huff <cja### [at] earthlinknet>
http://home.earthlink.net/~cjameshuff/
POV-Ray TAG: <chr### [at] tagpovrayorg>
http://tag.povray.org/

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