how to know if my ideas didnt tought before?

[davieddy]

Originally Posted by wblipp

William

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

[blob100]

Originally Posted by R.D. Silverman

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.

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.

[blob100]

Originally Posted by blob100

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.

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.

[R.D. Silverman]

Originally Posted by blob100

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.

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

[blob100]

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.

[R.D. Silverman]

Originally Posted by blob100

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.

This is correct.
And finding the coefficients IS trivial by these equations.

[blob100]

Originally Posted by R.D. Silverman

This is correct.
And finding the coefficients IS trivial by these equations.

But hard. To do it?

[R.D. Silverman]

Originally Posted by blob100

But hard. To do it?

No, it is not hard. It is trivial.

[blob100]

Originally Posted by R.D. Silverman

No, it is not hard. It is trivial.

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

[R.D. Silverman]

Originally Posted by blob100

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

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

[Wacky]

Perhaps you can use a more direct hint:

Given
f(X) = A_nXⁿ + A_n-1X^n-1 + ... + A₁X + A₀
with roots a₁, a₂, ..., a_n
and
g(Y) = B_nYⁿ + B_n-1Y^n-1 + ... + B₁Y + B₀
with roots b₁, b₂, ..., b_n
where, for i in 1 ... n,
b_i = 1/a_i

You need to express B_j in terms of the A_i.