[QUOTE=blob100;219474]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)[/QUOTE]

Stop fixating on the second problem. You are not allowed to
use its results. And you are not using the DEFINITION of what it
means to be a root.

[QUOTE=blob100;219474]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)[/QUOTE]

[B]Not Responsive! PLEASE[/B] try to do [B]exactly[/B] what has been requested.

In this instance, I requested that you provide a series of steps which should lead to a result. And I specifically ask that you [B]show your work[/B].

HINT: There are two canonical forms in which an algebraic polynomial can be expressed. 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] is one of them.

Tomer,

I'm trying to figure out if you are so focused on Problem #2 that you just can't see the simple solution to Problem #1, or if you are missing a fundamental skill that needs to be taught first. To help me figure that out, let's try something simpler as a warm up exercise.

Let f(x) = x^2 + 3x + 2

If it helps, you can think of f as the weight of yeast of some biological experiment on day x.

Let x = 2y+1

If it helps, you can think of y as the amount of water that has been added to the experiment by day x.

Let g(y) be the weight of yeast that after y units of water have been added to the experiment.

Find g(y)

William

PS - It will help us if you show your work.

[QUOTE=wblipp;219485]Tomer,

I'm trying to figure out if you are so focused on Problem #2 that you just can't see the simple solution to Problem #1, or if you are missing a fundamental skill that needs to be taught first. To help me figure that out, let's try something simpler as a warm up exercise.

Let f(x) = x^2 + 3x + 2

If it helps, you can think of f as the weight of yeast of some biological experiment on day x.

Let x = 2y+1

If it helps, you can think of y as the amount of water that has been added to the experiment by day x.

Let g(y) be the weight of yeast that after y units of water have been added to the experiment.

Find g(y)

William

PS - It will help us if you show your work.[/QUOTE]

Enough already! It is clear that he is not going to solve this.

Here is the full derivation, with careful attention to notation.

We are [b]given[/b] the following:

f(x) = a_nx^n + .... + a0

A = {x1, x2, .... xn} is the set of roots of f(x). x_i are integers.

We are asked to find the coefficients of a polynomial
whose roots are B = {1/x1, 1/x2, .... 1/xn}

Let use call it g(y) = b_n y^n + .... + b_0

Step 1:

Use the definition of root!

We have, for some x \in A

(1) a_nx^n + a_{n-1}x^n-1 + .... a_0 = 0

AND

(2) b_n(1/x)^n + b_{n-1}(1/x)^n-1 + ..... b_0 = 0

from the definition of root.


Step 2:

But the LHS of equation (2) is not a polynomial in x. So
make it one. Multiply both sides by x^n (!!!!!!!)

Whence

(3) b_n + b_{n-1}x + ...... b_0 x^n = 0

Step 3:

By the DEFINITION OF ROOT, x is a root of the polynomial on the
left hand side of (3). But we [b]already know[/b], from the statement of
the problem a polynomial for which x is a root! It is f(x).

Hence f(x) = b_n + b_{n-1}x + ..... b_0 x^n = a_n x^n + a_{n-1} x^n-1 + .... a_0

Thus:

b_n = a_0
b_{n-1} = a_1
.
.
.
b_0 = a_n

Thus the polynomial which has 1/x as a root is the same as the one that
has x as a root WITH THE COEFFICIENTS REVERSED.

Now what was so bloody difficult? Use the definition of (1/x) as a root,
clear the denominators by multiplying by x^n, and then recognize the
result as the original polynomial with coefficients revesed.

I fail to understand why you found this so difficult. Despite repeated
hints, you were not using the definition of root. And multiplying by x^n
to clear denominators in equation (2) should have come to mind immediately,
since the left hand side of (2) was not a polynomial.

[QUOTE=R.D. Silverman;219489]Enough already! It is clear that he is not going to solve this.

Here is the full derivation, with careful attention to notation.

<snip>


.[/QUOTE]

Now Tomer can do problem 2. He already knows the answer.
The problem is to PROVE IT.

I will offer a hint: INDUCTION. (on the degree of the polynomial)

[quote=R.D. Silverman;219489]Enough already! It is clear that he is not going to solve this.

Here is the full derivation, with careful attention to notation.

We are [B]given[/B] the following:

f(x) = a_nx^n + .... + a0

A = {x1, x2, .... xn} is the set of roots of f(x). x_i are integers.

We are asked to find the coefficients of a polynomial
whose roots are B = {1/x1, 1/x2, .... 1/xn}

Let use call it g(y) = b_n y^n + .... + b_0

Step 1:

Use the definition of root!

We have, for some x \in A

(1) a_nx^n + a_{n-1}x^n-1 + .... a_0 = 0

AND

(2) b_n(1/x)^n + b_{n-1}(1/x)^n-1 + ..... b_0 = 0

from the definition of root.


Step 2:

But the LHS of equation (2) is not a polynomial in x. So
make it one. Multiply both sides by x^n (!!!!!!!)

Whence

(3) b_n + b_{n-1}x + ...... b_0 x^n = 0

Step 3:

By the DEFINITION OF ROOT, x is a root of the polynomial on the
left hand side of (3). But we [B]already know[/B], from the statement of
the problem a polynomial for which x is a root! It is f(x).

Hence f(x) = b_n + b_{n-1}x + ..... b_0 x^n = a_n x^n + a_{n-1} x^n-1 + .... a_0

Thus:

b_n = a_0
b_{n-1} = a_1
.
.
.
b_0 = a_n

Thus the polynomial which has 1/x as a root is the same as the one that
has x as a root WITH THE COEFFICIENTS REVERSED.

Now what was so bloody difficult? Use the definition of (1/x) as a root,
clear the denominators by multiplying by x^n, and then recognize the
result as the original polynomial with coefficients revesed.

I fail to understand why you found this so difficult. Despite repeated
hints, you were not using the definition of root. And multiplying by x^n
to clear denominators in equation (2) should have come to mind immediately,
since the left hand side of (2) was not a polynomial.[/quote]

I'm so stupid.
I got till step 3 and then got confused.
I'm sorry.

[QUOTE=blob100;219498]I'm so stupid.
I got till step 3 and then got confused.
I'm sorry.[/QUOTE]

You were not applying the definition!

[quote=blob100;219498]I'm so stupid.
I got till step 3 and then got confused.
I'm sorry.[/quote]Don't call yourself stupid. It's not true and doesn't help.

Take a break. Relax. Try to do something else for a little bit so that you can take a fresh look at it.

[quote=R.D. Silverman;219490]Now Tomer can do problem 2. He already knows the answer.
The problem is to PROVE IT.

I will offer a hint: INDUCTION. (on the degree of the polynomial)[/quote]

The induction steps:
1) to show it is true for n=1.
2) assume it is true for n.
3) show that if it is true for n, it is true for n+1.

Proposition:
The coefficient of x^k is:
((-1)^(P))(R_k) where R_k is the sum of the product combinations of P roots, P and k are paralleled in f(x).
1) We see that for n=1, f(x)=(x-x1), and we see that the coefficents are 1,-1 (this accepts the formula).
2) let's assume it is true for n, f(x)=(x-x1)...(x-xn)=x^n+...+a0.
ak=((-1)^(P))(R_k).
3) f(x)=(x-x1)...(x-xn)(x-x(n+1))=x^(n+1)+....+a0.
Now we see: (x-x1)...(x-xn)(x-x(n+1))=(x^n+...+a0)(x-x_(n+1)).
Now, by the last equation, if we had:
akx^k=((-1)^(P))(R_k)x^k the next x^k's (in the next degree polynomial)
will have a coefficient:
a(k-1)+(ak)(-x(n+1))=((-1)^(P+1))(R_(k-1))+((-1)^(P))(R_k)(-x(n+1)).
((-1)^(P+1))(R_(k-1)+(R_k)(x(n+1))). P and k are paralleled in
f(x)(x-x(n+1)).
Now we will consider how ((-1)^(P+1))(R_(k-1)+(R_k)(x(n+1))) is of the form ak=((-1)^(P))(R_k):
We see that R_(k-1) is a product combination of more roots then R_k is (by one root), then, by producting R_k with x(n+1) it earns the next roots (and then it is as R_k by root devisors).
P and k are paralleled in f(x)(x-x(n+1)).

[quote=R.D. Silverman;219501]You were not applying the definition![/quote]
On my note book. Look, I occure myself, just me.

[quote=only_human;219502]Don't call yourself stupid. It's not true and doesn't help.

Take a break. Relax. Try to do something else for a little bit so that you can take a fresh look at it.[/quote]

I'm just upset that I left the right result behind.