how to know if my ideas didnt tought before?

[blob100]

Originally Posted by 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.

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

[Wacky]

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

Find a relationship between X and Y that is useful in relating the a_i and the b_i.

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.

[blob100]

Originally Posted by R.D. Silverman

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

Is this the direction:
Using Wacky's noations.
B_jb^j_i=-(g(b_i)-B_jb^j_i)+f(a^j_i).
This equalls B_j/a^j_i.
And we do so for every such b_i root.
We get a number of equations of the form:
B_j/a^j_i=-(g(b_i)-B_jb^j_i)+f(a^j_i).
By these equations we get B_j.
To show B_j as a function of A_i, the g(b_i) part in the equation -(g(b_i)-B_jb^j_i)+f(a^j_i) will be reduced by showing the equallity between the equations of the form:
B_j=a^j_i(-(g(b_i)-B_jb^j_i)+f(a^j_i)).

[R.D. Silverman]

Originally Posted by blob100

Is this the direction:
Using Wacky's noations.
B_jb^j_i=-(g(b_i)-B_jb^j_i)+f(a^j_i).
This equalls B_j/a^j_i.
And we do so for every such b_i root.
We get a number of equations of the form:
B_j/a^j_i=-(g(b_i)-B_jb^j_i)+f(a^j_i).
By these equations we get B_j.
To show B_j as a function of A_i, the g(b_i) part in the equation -(g(b_i)-B_jb^j_i)+f(a^j_i) will be reduced by showing the equallity between the equations of the form:
B_j=a^j_i(-(g(b_i)-B_jb^j_i)+f(a^j_i)).

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

[blob100]

Originally Posted by R.D. Silverman

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

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.

[R.D. Silverman]

Originally Posted by blob100

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.

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.

[axn]

Originally Posted by R.D. Silverman

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

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.

[blob100]

Originally Posted by axn

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.

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.

[blob100]

Originally Posted by R.D. Silverman

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.

Can you just give me the answer?

[Wacky]

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 shows that it has the required roots.
Then express it in the form: A_nXⁿ+ ...

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")

[blob100]

Originally Posted by Wacky

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 shows that it has the required roots.
Then express it in the form: A_nXⁿ+ ...

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")

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)