Newton-raphson method with vectors as functions, and derivative

I am during the last stage of an assignment, in which i have to use the newton-raphson iterative method to find the root of a polynom, where the polynom is given by a derivative of a vector where n+1= the length of the original vector. To make things more clear, i have something like this-

vector[a1 a2 a3 a4 a5 a6]

i need to derive it once to get the polynom i need to work with, so the first deriving function i thought of is-

for i=2:1:n+1
der1(i-1)=(vector(i)*n-1*(x^(n-2)))

and to use newton raphson i will need to derive again, so-

for i=2:1:n
der2(i-1)=(der1(i)*n-1*(x^(n-2)))

These two loops will give vectors, and summing these vectors would give the actual functions f (der1) and it's derivative (der2). From this point the newton raphson method itself is obvious, just choose an initial guess and let it start running.

On paper it all looks good, but i am far from sure that this is actually a correct process. If anyone can give his opinion if this is ok, or if anyone has a better idea- i will appreciate any input. Thanks.

(just to be clear about it- i am a complete matlab noob, and we were given a few assignments without instructions on how to work with matlab, so i have to learn as i go)

Answer 1

[Torsten]

help polyval

help polyder

Best wishes

Torsten.

[Rob Dunham]

while i appreciate pitching the idea, i have both a deadline to meet and the assignment is to purposely use weird ineffective ways. It's a numerical analysis course, not a programming course. This is why i'm asking very specific questions....

[Torsten]

And I gave you a very specific answer:

If your polynomial is given by a vector

p = [3 2 1];

you can get its first and second derivatives by

derp = polyder(p);
der2p = polyder(derp);

and you can evaluate the polynomial and its first and second derivatives with polyval (e.g. at x=1.0):

y=polyval(p,1.0);
yder=polyval(derp,1.0);
yder2=poyval(der2p,1.0);

Best wishes

Torsten.

[Rob Dunham]

i might not have been specific enough, my bad- I'm not allowed to use these functions, and have to make my own for deriving. I actually solved the problem, my first guess was close enough. Here are the functions i ended up with-

      for i=2:1:n+1;
       fun(i-1)=(vector(i)*(i-1)*(x^(i-2)));
      end    

      for i=2:1:n
        ftag(i-1)=((fun(i)*(i-1))/x);
      end

Second function is repeatable for higher order derivatives, it's also possible to loop in order to correct n for each order of derivative. Hope this helps if anyone gets an assignment as function-restrictive as mine.