[quote=R.D. Silverman;218858]Sorry, but this isn't even close.

One can express the coefficients of g(x) very simply in terms of the coefficients of f(x). You are making the problem much more difficult than
it is.

And once again, you are introducing extraneous variables. The entire
problem can be solved using ONLY the variables presented in the problem
itself. You are also trying to solve the general case by presenting a
specific example . Don't.





This is irrelevant to the problem.[/quote]

I tought again about the problem (1) and I think I understood what it means:
There is a polynomial f(x) such that it's roots are:
x1, x2,...., xn
And also:
1/x1, 1/x2,...., 1/xn.
I'm I right?

[quote=R.D. Silverman;218858][quote=blob100;218855]For the first question:
As you said, there is a missunderstanding about the problem.
So, I will give a private case (n=3) to show what I have understood:
f(x)=(x+a)(x+b)(x+c)= x^3+(a+b+c)x^2+(ca+cb+ab)x+abc (By simple algebra).
The roots are -a,-b,-c.
In your problem asks to find the coefficients of the polynomial g(x) where it's roots are the inverses of f(x)'s roots.
g(x)=(x+1/a)(x+1/b)(x+1/c)=
x^2+((ca+cb+ab)/abc)x^2+((a+b+c)/abc)x+1/abc
[/quote]
Sorry, but this isn't even close.[/quote]The language problem here may be obscuring your view of Tomer's meaning.

[quote] You are also trying to solve the general case by presenting a specific example.[/quote]No, that is [I]not[/I] what he is trying to do. What he has done is present an example to illustrate [I]only[/I] that he has properly understood what you are [I]asking[/I], not an example of a solution to the problem.

He was asking you whether his n=3 example is [U]a correct interpretation of what the problem asks him to examine[/U]. That is, he was asking whether his example is a correct example of the problem for the case n=3. He was not yet[I] (at that point)[/I] trying to present a solution to the problem.

[quote=blob100;218942]I tought again about the problem (1) and I think I understood what it means:
There is a polynomial f(x) such that it's roots are:
x1, x2,...., xn
And also:
1/x1, 1/x2,...., 1/xn.
I'm I right?[/quote]No, that is not what it means. You were correct earlier (#603), but Silverman's misunderstanding led him to post comments (#604) that confused you.

The roots of f(x) are x[sub]1[/sub], x[sub]2[/sub], ... x[sub]n[/sub].

1/x[sub]1[/sub], 1/x[sub]2[/sub], ... 1/x[sub]n[/sub] are roots of some different polynomial g(x), not f(x).

Your task is to express the coefficients of g(x) in terms of the coefficients of f(x) (which are a[sub]n[/sub], a[sub]n-1[/sub], ... a[sub]0[/sub]) .... and determine whether g(x) is unique.

In your n=3 example, the coefficients of f(x) are (from highest power of x to lowest power) 1, (a+b+c), (ca+cb+ab), and abc.

However, there you specified the roots as negative (-a, -b, -c). I recommend that you rewrite your n=3 example with the f(x) roots given in positive form (a, b, c).

Proof of the rational roots theorem:
alpha=a/b a rational root of f(x),
* we must note that a,b are co-primes.
f(x)=a_nx^n+.....+a_0.
a_i={the set of all coefficients of f(x)}.
We say a_i are all integers.
f(a/b)=0 (a/b a root of f(x)).
f(a/b)*b^(n-1)=a_n(a^n/b)+(.....)=0
We verify (.....) is a sum of integers:
It is a sum of products between integers.
a and b are co-primes, then a_n is devided by b.
With analogy to what was related till now, we will prove a is deviding a_0.
f(a/b)*(b^n/a)=(b^n*f(a/b))/a=(sum of integers)+a_0(b^n/a)=0.
b,a are co-primes which gives a|a_0.

[quote=cheesehead;218948]
The roots of f(x) are x[sub]1[/sub], x[sub]2[/sub], ... x[sub]n[/sub].

1/x[sub]1[/sub], 1/x[sub]2[/sub], ... 1/x[sub]n[/sub] are roots of some different polynomial g(x), not f(x).

Your task is to express the coefficients of g(x) in terms of the coefficients of f(x) (which are a[sub]n[/sub], a[sub]n-1[/sub], ... a[sub]0[/sub]).

[/quote]

If so, I gave the solution on the begining (much earlier then #603, and on #603 itself).

The coefficient of x^k, k<n in g(x) is (a_P*a_n)/(a_0).
f(x)=a_nx^n+...+a_0.
a_i={the set of all coefficients of f(x)}, are all integers.
a_P is the coefficient of x^P in f(x) (where P,k are paralleled in f(x), parallelism defined before).

[QUOTE=blob100;218949]Proof of the rational roots theorem:
alpha=a/b a rational root of f(x),
* we must note that a,b are co-primes.
f(x)=a_nx^n+.....+a_0.
a_i={the set of all coefficients of f(x)}.
[/QUOTE]

What is i? Once again you are adding extraneous, unnecessary
variables and not defining them. And you are redefining a_i
to be a set. Do you not understand set notation?

[QUOTE]
We say a_i are all integers.
[/QUOTE]

I will allow for language difficulty here, but this illustrates
why the extraneous 'i' is bad. As written, this sentence
implies that there are multiple a_i, when in fact a_i is a SINGLE
set. Or do you mean 'elements of a_i'?? If so, you need to define:

A = {a_0, a_1, ..... a_n} == {a_i, i=0,n}

But defining a_i to be a the entire set is bad.

[QUOTE]
f(a/b)=0 (a/b a root of f(x)).
f(a/b)*b^(n-1)=a_n(a^n/b)+(.....)=0
We verify (.....) is a sum of integers:
It is a sum of products between integers
[/QUOTE]

This last sentence is a mere assertion. You have not proved
that a_n * a^n/b is an integer. Or that any other term
must be an integer.


The coefficients of f(x) are each the sum of products of ROOTS,
and each is an integer, but this does not prove that a_n * a^n/b
is an integer.

[QUOTE=blob100;218950]If so, I gave the solution on the begining (much earlier then #603, and on #603 itself).

The coefficient of x^k, k<n in g(x) is (a_P*a_n)/(a_0).
f(x)=a_nx^n+...+a_0.
a_i={the set of all coefficients of f(x)}, are all integers.
a_P is the coefficient of x^P in f(x) (where P,k are paralleled in f(x), parallelism defined before).[/QUOTE]

This is gibberish.
"paralleled in f(x)" is meaningless.

And the answer: (a_P*a_n)/(a_0) is wrong.

[QUOTE=R.D. Silverman;218953]This is gibberish.
"paralleled in f(x)" is meaningless.

And the answer: (a_P*a_n)/(a_0) is wrong.[/QUOTE]

Here is a hint:

Do the problems IN ORDER.

I assume that you know/understand synthetic division of polynomials?
If not, let me know and I will put together an exercize that will lead
you through it.

[quote=R.D. Silverman;218959]Here is a hint:

Do the problems IN ORDER.

I assume that you know/understand synthetic division of polynomials?
If not, let me know and I will put together an exercize that will lead
you through it.[/quote]

A synthetic division of a polynomial is the phrase (x-r) where r is a root of the polynomial.

[QUOTE=blob100;218973]A synthetic division of a polynomial is the phrase (x-r) where r is a root of the polynomial.[/QUOTE]

No. For any polynomial f(x) of arbitrary degree k, and any polynomial q(x)
of degree r < k, we can write

f(x) = q(x) * g(x) + r(x) where g(x) is a polynomial of degree
r-k and r(x) has degree less than q(x).

In words: we can divide any polynomial by any other polynomial of
lesser degree leaving a polynomial remainder of even lesser degree than
the divisor.

It is not specific to division by just linear polynomials and has nothing to
do with the roots of the polynomial.

It is analagous to the division algorithm for integers. Any integer N in Z+
can be written as N = d*q + r where r < d and d is the divisor.
The same applies for polynomials.

Do you know how to prove this?

Bob,

I'm pretty sure you intend for polynomials to always have integer coefficients and that Tomer doesn't know this.

I also think he may benefit from some guidance about how substitution of variables works.

William