[quote=wblipp;219091]
William[/quote]

I presume you have met Paul by now.
I haven't, but as you will understand, this would be superfluous.

Will check out that Tom Lehrer ditty in due course.

Visited the Flask the other day. Met a really nice bloke
from Norfolk.
OTOH met that same bloke I told you about - opening line
"I'm going to knock your ******* head off".

Happy Days,

David

[quote=R.D. Silverman;219296]No. Not even close. You are not taking my hints:

(1) This problems does NO depend on any later problems.
(2) Apply the definition of what it means to be a root.[/quote]

A root r of the polynomial f(x) such that for x=r, f(r)=f(x)=0.
It is really hard to prove what I don't know.
Is my formula on #636 right?
My problem here is how not to use the result of the second problem.
Showing the coefficients of g(x) as a function of the coefficients of f(x) include understanding the coefficients of f(x).
The common sense we can find between those coefficients (of f and g) is the roots which with those we can show the coefficients (by the result of the second problem), I can't see another way to find the coefficients as a function of the roots.

[quote=blob100;219344]A root r of the polynomial f(x) such that for x=r, f(r)=f(x)=0.
It is really hard to prove what I don't know.
Is my formula on #636 right?
My problem here is how not to use the result of the second problem.
Showing the coefficients of g(x) as a function of the coefficients of f(x) include understanding the coefficients of f(x).
The common sense we can find between those coefficients (of f and g) is the roots which with those we can show the coefficients (by the result of the second problem), I can't see another way to find the coefficients as a function of the roots.[/quote]

Here is a mathematical way to show my problem:
f(x)=(x-x1)(x-x2)...(x-xn)
g(x)=(x-X1)(x-X2)...(x-Xn)
Where Xi (given root of g(x))=xi^(-1) (the inverse of f(x)'s root).
Let R={x1,x2,...,xn}, M={X1,X2,...,Xn}={1/x1,1/x2,...,1/xn}.
a_i=F(R) a function of the roots of f(x) (this function is what problem 2 relate to (the result)).
b_i=F(M) a function of the roots of g(x).
a_i and b_i are the coefficients of x^i in f(x) and g(x) order wise.
To show b_i as a function of a_i, we must understand the function F.

[QUOTE=blob100;219347]Here is a mathematical way to show my problem:
f(x)=(x-x1)(x-x2)...(x-xn)
g(x)=(x-X1)(x-X2)...(x-Xn)
Where Xi (given root of g(x))=xi^(-1) (the inverse of f(x)'s root).
Let R={x1,x2,...,xn}, M={X1,X2,...,Xn}={1/x1,1/x2,...,1/xn}.
a_i=F(R) a function of the roots of f(x) (this function is what problem 2 relate to (the result)).
b_i=F(M) a function of the roots of g(x).
a_i and b_i are the coefficients of x^i in f(x) and g(x) order wise.
To show b_i as a function of a_i, we must understand the function F.[/QUOTE]

You are not using the definition of what it means to be a root.

Is this the direction?:
f(x1)=0
g(1/x1)=0.
f(x1)=g(1/x1)=0.
It does not seem easy to find the coefficients of g(x) as a function of the ones of f(x) by this equation.

[QUOTE=blob100;219364]Is this the direction?:
f(x1)=0
g(1/x1)=0.
f(x1)=g(1/x1)=0.
It does not seem easy to find the coefficients of g(x) as a function of the ones of f(x) by this equation.[/QUOTE]

This is correct.
And finding the coefficients [b]IS[/b] trivial by these equations.

[quote=R.D. Silverman;219368]This is correct.
And finding the coefficients [B]IS[/B] trivial by these equations.[/quote]

But hard. To do it?

[QUOTE=blob100;219369]But hard. To do it?[/QUOTE]

No, it is not hard. It is trivial.

[quote=R.D. Silverman;219372]No, it is not hard. It is trivial.[/quote]

Ok, To prove the result of the secondary problem?
Or to try the third?

[QUOTE=blob100;219373]Ok, To prove the result of the secondary problem?
Or to try the third?[/QUOTE]

You need to finish the first problem.......

Perhaps you can use a more direct hint:

Given
f([B]X[/B]) = A[SUB]n[/SUB][B]X[/B][SUP]n[/SUP] + A[SUB]n-1[/SUB][B]X[/B][SUP]n-1[/SUP] + ... + A[SUB]1[/SUB][B]X[/B] + A[SUB]0[/SUB]
with roots a[SUB]1[/SUB], a[SUB]2[/SUB], ..., a[SUB]n[/SUB]
and
g([B]Y[/B]) = B[SUB]n[/SUB][B]Y[/B][SUP]n[/SUP] + B[SUB]n-1[/SUB][B]Y[/B][SUP]n-1[/SUP] + ... + B[SUB]1[/SUB][B]Y[/B] + B[SUB]0[/SUB]
with roots b[SUB]1[/SUB], b[SUB]2[/SUB], ..., b[SUB]n[/SUB]
where, for i in 1 ... n,
b[SUB]i[/SUB] = 1/a[SUB]i[/SUB]

You need to express B[SUB]j[/SUB] in terms of the A[SUB]i[/SUB].