[quote=Wacky;219380]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].[/quote]

This isn't a more direct hint, it is the question itself.

OK, I had hoped that you only were having difficulty in understanding just what was expected.

Find a relationship between [B]X[/B] and [B]Y[/B] that is useful in relating the a[SUB]i[/SUB] and the b[SUB]i[/SUB].

Substitute that in the definition of f. Then regroup the A terms so that they map onto appropriate B terms.

As Bob has said, this exercise is not difficult.

This "change of variable" technique is a fundamental technique in algebraic proofs.

[quote=R.D. Silverman;219374]You need to finish the first problem.......[/quote]
Is this the direction:
Using Wacky's noations.
B[sub]j[/sub]b[sup]j[/sup][sub]i[/sub]=-(g(b[sub]i[/sub])-B[sub]j[/sub]b[sup]j[/sup][sub]i[/sub])+f(a[sup]j[/sup][sub]i[/sub]).
This equalls B[sub]j[/sub]/a[sup]j[/sup][sub]i[/sub].
And we do so for every such b[sub]i[/sub] root.
We get a number of equations of the form:
B[sub]j[/sub]/a[sup]j[/sup][sub]i[/sub]=-(g(b[sub]i[/sub])-B[sub]j[/sub]b[sup]j[/sup][sub]i[/sub])+f(a[sup]j[/sup][sub]i[/sub]).
By these equations we get B[sub]j[/sub].
To show B[sub]j[/sub] as a function of A[sub]i[/sub], the g(b[sub]i[/sub]) part in the equation -(g(b[sub]i[/sub])-B[sub]j[/sub]b[sup]j[/sup][sub]i[/sub])+f(a[sup]j[/sup][sub]i[/sub]) will be reduced by showing the equallity between the equations of the form:
B[sub]j[/sub]=a[sup]j[/sup][sub]i[/sub](-(g(b[sub]i[/sub])-B[sub]j[/sub]b[sup]j[/sup][sub]i[/sub])+f(a[sup]j[/sup][sub]i[/sub])).

[QUOTE=blob100;219390]Is this the direction:
Using Wacky's noations.
B[sub]j[/sub]b[sup]j[/sup][sub]i[/sub]=-(g(b[sub]i[/sub])-B[sub]j[/sub]b[sup]j[/sup][sub]i[/sub])+f(a[sup]j[/sup][sub]i[/sub]).
This equalls B[sub]j[/sub]/a[sup]j[/sup][sub]i[/sub].
And we do so for every such b[sub]i[/sub] root.
We get a number of equations of the form:
B[sub]j[/sub]/a[sup]j[/sup][sub]i[/sub]=-(g(b[sub]i[/sub])-B[sub]j[/sub]b[sup]j[/sup][sub]i[/sub])+f(a[sup]j[/sup][sub]i[/sub]).
By these equations we get B[sub]j[/sub].
To show B[sub]j[/sub] as a function of A[sub]i[/sub], the g(b[sub]i[/sub]) part in the equation -(g(b[sub]i[/sub])-B[sub]j[/sub]b[sup]j[/sup][sub]i[/sub])+f(a[sup]j[/sup][sub]i[/sub]) will be reduced by showing the equallity between the equations of the form:
B[sub]j[/sub]=a[sup]j[/sup][sub]i[/sub](-(g(b[sub]i[/sub])-B[sub]j[/sub]b[sup]j[/sup][sub]i[/sub])+f(a[sup]j[/sup][sub]i[/sub])).[/QUOTE]

Sorry, but no. You are making it much more difficult than it is.

[quote=R.D. Silverman;219397]Sorry, but no. You are making it much more difficult than it is.[/quote]
Can you please explain why not? Every time you disagree with my results it confuses me (things seem to me logical, and you decline them), it will be really helpful for me to understand my mistakes.

Thanks.

[QUOTE=blob100;219407]Can you please explain why not? Every time you disagree with my results it confuses me (things seem to me logical, and you decline them), it will be really helpful for me to understand my mistakes.

Thanks.[/QUOTE]

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

I can't say more without just telling you the answer.

[QUOTE=R.D. Silverman;219429]I can't say more without just telling you the answer.[/QUOTE]

Even if you give the answer, the exercise of _deriving_ the answer might be worthwhile.

I am slightly surprised that, with his habit of doing small examples, Tomer still hasn't latched on to the answer.

[quote=axn;219431]Even if you give the answer, the exercise of _deriving_ the answer might be worthwhile.

I am slightly surprised that, with his habit of doing small examples, Tomer still hasn't latched on to the answer.[/quote]
It is just becuase of my stupidity.
By the examples I found:
b_i=(((-1)^(|i-P|+1))a_P)/a_0 And proved it by the result of the second problem.
b_i the coefficients of g(y) and a_i the coefficients of f(x).
P=n-i.

[quote=R.D. Silverman;219429]You are not using the definition of what it means to be a root of a polynomial.

I can't say more without just telling you the answer.[/quote]

Can you just give me the answer?

Let's try another approach:

Suppose that I want to create a polynomial that has the rational roots (5) and (2/3).

How would you create that polynomial?

First express it in a form that [B]shows[/B] that it has the required roots.
Then express it in the form: A[SUB]n[/SUB][B]X[/B][SUP]n[/SUP]+ ...

How about another polynomial with roots (1/5) and (3/2)?

What do you observe?
Can you generalize the result?

(Be sure to "show your work")

[quote=Wacky;219444]Let's try another approach:

Suppose that I want to create a polynomial that has the rational roots (5) and (2/3).

How would you create that polynomial?

First express it in a form that [B]shows[/B] that it has the required roots.
Then express it in the form: A[sub]n[/sub][B]X[/B][sup]n[/sup]+ ...

How about another polynomial with roots (1/5) and (3/2)?

What do you observe?
Can you generalize the result?

(Be sure to "show your work")[/quote]

This approach is what I tried on the begining.
From here, we can prove just by the second result.

a_P=((-1)^P)(R) where R is the sum of product combinations between i roots of f(x).
b_i=((-1)^i)(G)
G=R/-a_0.
b_i=((-1)^i)(a_P/((-1)^P))/-a_0=((-1)^(i+1-P))(a_P)/(a_0)