[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 agree that it isn't stupidy. He does seem, however, to lack discipline.

He needs to learn how to take and use hints. He was confused because
he kept thinking about problem #2 even though we told him not to use
it. He also needs to practice applying definitions practice in using and
defining his own variables.

He claims to have taken Euclidean geometry. But he does not seem
to be using any formal proof techniques that he might have learned
from that class.

Discipline!

The proof of problem 2 by induction will require careful attention to
the use of variables.

[quote=blob100;219504]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]
.

[QUOTE=blob100;219529].


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).

[/QUOTE]

Stop here. You have already screwed up. (no offense) You have not
defined P. What does "paralleled in f(x)" mean?
I suspect a language difficulty here.

Start by STATING the result that you are trying to prove. Pay careful
attention to the variables and their definition that you use in the statement
of the problem. We are given a polynomial f(x) with coefficients a_i.
(i=0,....n) We are given its roots x_j. (j=0, ....n) State the exact formula
that you are trying to prove using these variables.


You have not defined x1. You have not shown the "acceptance" of the
formula. You have merely declared it to be correct. And, since we do
not [b]have[/b] a well defined formula, declaring it correct is bogus.
And any declaration that such a fomula is correct based on the assumption
that 1 and -1 are the coefficients can [b]not be[/b] correct.

The coefficients are NOT 1 and -1. The free variable of the linear polynomial
is x. x1 is not a variable. It therefore [b]does not have[/b] a coefficient.
Once more you have confused yourself by IMPROPER USE OF VARIABLES.
The linear polynomial you have defined is 1*x^1 - x1 *x^0. The coefficient of x^0 is -x1.

You are not going to be able to do college level math if you keep
failing to make proper use and definitions of your variables!!!!!!

[quote=R.D. Silverman;219533]Stop here. You have already screwed up. (no offense) You have not
defined P. What does "paralleled in f(x)" mean?
I suspect a language difficulty here.

Start by STATING the result that you are trying to prove. Pay careful
attention to the variables and their definition that you use in the statement
of the problem. We are given a polynomial f(x) with coefficients a_i.
(i=0,....n) We are given its roots x_j. (j=0, ....n) State the exact formula
that you are trying to prove using these variables.


You have not defined x1. You have not shown the "acceptance" of the
formula. You have merely declared it to be correct. And, since we do
not [B]have[/B] a well defined formula, declaring it correct is bogus.
And any declaration that such a fomula is correct based on the assumption
that 1 and -1 are the coefficients can [B]not be[/B] correct.

The coefficients are NOT 1 and -1. The free variable of the linear polynomial
is x. x1 is not a variable. It therefore [B]does not have[/B] a coefficient.
Once more you have confused yourself by IMPROPER USE OF VARIABLES.
The linear polynomial you have defined is 1*x^1 - x1 *x^0. The coefficient of x^0 is -x1.

You are not going to be able to do college level math if you keep
failing to make proper use and definitions of your variables!!!!!![/quote]

Sorry for not defining x1.
X={x1,x2,...,xn} the set of all roots of f(x).
P=n-k where k is a given natural number less than n (this is what we defined before as paralleled in f(x)).

[QUOTE=blob100;219536]Sorry for not defining x1.
X={x1,x2,...,xn} the set of all roots of f(x).
P=n-k where k is a given natural number less than n.[/QUOTE]

OK. Now you can state what is to be proved.

[quote=R.D. Silverman;219533]Stop here. You have already screwed up. (no offense) You have not
defined P. What does "paralleled in f(x)" mean?
I suspect a language difficulty here.

Start by STATING the result that you are trying to prove. Pay careful
attention to the variables and their definition that you use in the statement
of the problem. We are given a polynomial f(x) with coefficients a_i.
(i=0,....n) We are given its roots x_j. (j=0, ....n) State the exact formula
that you are trying to prove using these variables.


You have not defined x1. You have not shown the "acceptance" of the
formula. You have merely declared it to be correct. And, since we do
not [B]have[/B] a well defined formula, declaring it correct is bogus.
And any declaration that such a fomula is correct based on the assumption
that 1 and -1 are the coefficients can [B]not be[/B] correct.

The coefficients are NOT 1 and -1. The free variable of the linear polynomial
is x. x1 is not a variable. It therefore [B]does not have[/B] a coefficient.
Once more you have confused yourself by IMPROPER USE OF VARIABLES.
The linear polynomial you have defined is 1*x^1 - x1 *x^0. The coefficient of x^0 is -x1.

You are not going to be able to do college level math if you keep
failing to make proper use and definitions of your variables!!!!!![/quote]
I'll try writing this post again.
Leave it for now.

[quote=R.D. Silverman;219537]OK. Now you can state what is to be proved.[/quote]

f(x)=(x-x1)...(x-xn)=anx^n+...+a0.
A={x1,x2,...,xn} the set of all roots of f(x).
B={a0,a1,...,an} the set of all coefficients of the polynomial f(x).

Proposition:
a_k (the coefficient of x^k in f(x)) is of the form:
((-1)^P)(R) where P=n-k and R is the sum of product combinations between P roots of f(x).
*if P=0 then R=1.

We will prove this statment by induction:
1) show it is true for n degree =1.
2) assume it for n.
3) prove that if it is true for n, it is true for n+1.

Proof:
1) f(x)=(x-x1), degree(n)=1.
This is a sum between two factors, x and x1.
The coefficent of x is 1 and of x1 is -1 which is equivallent to:
a1=1 and a0=-x1.
This fits the proposition:
a1=((-1)^(1-1=0))(1)=1,
a0=(-1)^(1-0=1)(x1)=-x1.

2) Let's assume that if f(x)=(x-x1)...(x-xn)=x^n+...+a0.
The coefficient of x^k is ((-1)^P)(R).

3) g(x) is the polynomial given by producting f(x) and (x-x(n+1)) (which means: a polynomial of degree n+1).
*we must assume x(n+1) is not a product, it is a noation to define "the root number n+1".
g(x)=f(x)(x-x(n+1))=(x^n+...+a0)(x-x(n+1)).
The coefficent of x^k in f(x) is (as we assumed) ((-1)^P)(R),
And we now try to find what is it for g(x).
Two coefficients of this kind will be obtained by producting
akx^k and -x(n+1), and the product (a(k-1)x^(k-1)) and x.
ak(-x(n+1))+(a(k-1)) is the coefficient of x^k in g(x).
ak=((-1)^P)(R) (as we assumed), and then:
ak(-x(n+1))=((-1)^(P+1))(R)(x(n+1)).
a(k-1)=((-1)^(P+1))(G) where G is the sum of the product combinations between P+1 roots.
ak(-x(n+1))+(a(k-1))=((-1)^(P+1))(R(x(n+1))+G).
We assume that P+1=n+1-k (P+1 and k are paralleled in g(x)).
R(x(n+1)) and G are two sums of product combinations of P+1 roots,
Moreover the sum of R(x(n+1)) and G is F the sum of the whole product combinations of P+1 roots of g(x).
This proves:
ak=((-1)^(P+1))(R(x(n+1))+G)=((-1)^(P+1))(F) in g(x),
The proposition itself.

[QUOTE=blob100;219550]f(x)=(x-x1)...(x-xn)=anx^n+...+a0.
A={x1,x2,...,xn} the set of all roots of f(x).
B={a0,a1,...,an} the set of all coefficients of the polynomial f(x).
[/QUOTE]

By this definition a_n equals 1. This is fine.
In fact, in order to prove the theorem you will need
to work with a monic polynomial. Therefore if a_n
isn't one you will need to do something first. This requires
dividing the polynomial by a_n (making it monic). This will
clearly not affect the roots. Now you will need to express
a_i/a_n in terms of the roots. However, we will allow
taking a_n = 1 throughout.

[QUOTE]
Proposition:
a_k (the coefficient of x^k in f(x)) is of the form:
((-1)^P)(R) where P=n-k and R is the sum of product combinations between P roots of f(x).
*if P=0 then R=1.
[/QUOTE]

This is insufficient. What is lacking is the specification of
how [b]many[/b] terms are in this sum. And the expression "product
combinations between P roots of f(x)" is better expressed as "product of
roots of f(x) taken P at a time". You need the sum of ALL products of roots
of f(x) taken P at a time.

The binomial symbol will/must appear in your proof. I will write
m_C_k for the number of different combinations of k items chosen
from a set of m items. This is also known as m choose k.

[QUOTE]
We will prove this statment by induction:
1) show it is true for n degree =1.
2) assume it for n.
3) prove that if it is true for n, it is true for n+1.
[/QUOTE]

Good so far.

[QUOTE]
Proof:
1) f(x)=(x-x1), degree(n)=1.
[/QUOTE]


Go back to the original definition of f(x). Its coefficients were given
as a_i. Thus f(x) = a1 x + a0. Now express the a_i
in terms of the x_i. You can (as stated above) take a1 = 1
[or replace a_i with a_i/a_n for all i; the proof will be the same]

You have replaced a0 with -x1. The coefficients are a_i. They need
to be expressed in terms of the x_i. Thus, you may take

f(x) = x + a0 := (x-x1) whence a0 = -x1 expresses a coefficient of
the polynomial as a function of the roots (the only root in this case)

[QUOTE]
This is a sum between two factors, x and x1

The coefficent of x is 1 and of x1 is -1 which is equivallent to:
a1=1 and a0=-x1.
[/QUOTE]

No! No! No!

The first statement just above is gibberish, but I will attribute it to
English being a second language.

You are again misusing variables. And you did not read what I wrote in my
earlier post. x1 [b]DOES NOT HAVE A COEFFICIENT!!!!
[/b] x1 is NOT a variable!!

The only VARIABLE(s) here are x and its POWERS. The a_i and the x_i
are given CONSTANTS. For f(x) = a_1 x + a_0 the coefficients are
a_1 and a_0. The root is x1. Roots do not have coefficients!


rest deleted.....

A hint for this proof: You will need to establish the well-known combinatorial
identity m_C_k = {m-1}_C_{k-1} + {m-1}_C_{k} in the course of the
proof. The proof can not succeed without this identity.

[quote=R.D. Silverman;219568]



No! No! No!

The first statement just above is gibberish, but I will attribute it to
English being a second language.

You are again misusing variables. And you did not read what I wrote in my
earlier post. x1 [B]DOES NOT HAVE A COEFFICIENT!!!!
[/B] x1 is NOT a variable!!

The only VARIABLE(s) here are x and its POWERS. The a_i and the x_i
are given CONSTANTS. For f(x) = a_1 x + a_0 the coefficients are
a_1 and a_0. The root is x1. Roots do not have coefficients!
[/quote]

Look, on the paragraph:
"This is a sum between two factors, x and x1

The coefficent of x is 1 and of x1 is -1 which is equivallent to:
a1=1 and a0=-x1."
I didn't mean x1 has a coefficient, that's why I showed a0=-x1.
For n=1, f(x)=a1x+a0=(x-x1)---> a1=1, a0=-x1.
I myself don't know why I have written "and of x1 is -1"
But I didn't mean it (x1 has a coefficient).

[QUOTE=blob100;219614]Look, on the paragraph:
"This is a sum between two factors, x and x1

The coefficent of x is 1 and of x1 is -1 which is equivallent to:
a1=1 and a0=-x1."
I didn't mean x1 has a coefficient, that's why I showed a0=-x1.
For n=1, f(x)=a1x+a0=(x-x1)---> a1=1, a0=-x1.
I myself don't know why I have written "and of x1 is -1"
But I didn't mean it (x1 has a coefficient).[/QUOTE]

So start over. Give an exact statement of the formula to be proven.

[quote=R.D. Silverman;219615]So start over. Give an exact statement of the formula to be proven.[/quote]

If I understand, you want me just to show the formula (on this post).

ak=((-1)^P)(n_C_P).
m_C_P is the sum of all product combinations of P roots taken from n roots of f(x) (which has n roots).
ak is the coefficent of x^k in f(x).
P=n-k.