how to know if my ideas didnt tought before?

[R.D. Silverman]

Originally Posted by blob100

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)

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.

[Wacky]

Originally Posted by blob100

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)

Not Responsive! PLEASE try to do exactly 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 show your work.

HINT: There are two canonical forms in which an algebraic polynomial can be expressed. A_nXⁿ + A_n-1X^n-1 + ... + A₁X + A₀ is one of them.

[wblipp]

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.

[R.D. Silverman]

Originally Posted by wblipp

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.

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 given 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 already know, 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.

[R.D. Silverman]

Originally Posted by R.D. Silverman

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

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

<snip>


.

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)

[blob100]

Originally Posted by R.D. Silverman

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 given 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 already know, 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.

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

[R.D. Silverman]

Originally Posted by blob100

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

You were not applying the definition!

[only_human]

Originally Posted by blob100

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

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.

[blob100]

Originally Posted by R.D. Silverman

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)

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)).

[blob100]

Originally Posted by R.D. Silverman

You were not applying the definition!

On my note book. Look, I occure myself, just me.

[blob100]

Originally Posted by only_human

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.

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