newton.html

Jason Newton

MA 502

Dr. Carl Eberhart

October 7, 1999

Project 1

1. Show that the set of polynomials , , , ..., , ... where satisfy the identity (***) . In semigroup parlance, this says that the function is an isomorphism between the semigroup of all polynomials under composition and the semigroup of non-negative integers under multiplication. Note: if you show this directly, great! You don't have to take the exam. Otherwise, you can show first that and then that the identity (***) is satified. But, you still have to take the exam in that case.

Generate for = 0 to 6

> for n from 0 to 6 do p[n]:=cos(n*arccos(x)) od;

expand the above

> for n from 0 to 6 do p[n]:=expand(cos(n*arccos(x))) od;

generate for = 0 to 6

> p[0]:=1;

> p[1]:= x;

> for m from 2 to 6 do p[m]:=expand(2*x*p[m-1]-p[m-2]) od;

From comparing and we see that when

Therefore for all (positive integers)

no. This only does it for n = 0 to 6. Use induction to complete the proof.

Now I will show that

= =

> p[6];

> p2p3:=expand(subs(x=p[3],p[2]));)

> p3p2:=expand(subs(x=p[2],p[3]));

As you can see from above = =

Below is the proof for the general

> p[s]:=cos(s*arccos(x));

> p[t]:=cos(t*arccos(x));

Therefore, for all and

correct

This is what I have finished I was wondering if you could email me, and let me know if this id sufficient to answer the corresponding part of problem 1.

Yes, except you need to complete the argument that p[n] is cos(n*arccos(x)).

Here, are the changes to my proof. All changes are labeled or marked in green

1. From above we know it is true for the case.