[QUOTE=blob100;218575]I have seen these stuff on these years of secondary school, but it does not say these exersizes are on the same level with the secondary school's level (there are many levels for everything, you can teach a theory on a verry deep way and on the same time the opposite).
[/QUOTE]

But you keep claiming that you are ready for more advanced material!!
(calculus, college algebra, etc).

The polynomial exercizes are typical of ones that I encountered when taking
pre-calculus in high school. If you can not handle them, then you are
not, IMO, ready for more advanced material.

Would anyone else like to render an opinion?

The exercizes I gave should be typical for a high level honors high
school class.

[QUOTE=R.D. Silverman;218598]But you keep claiming that you are ready for more advanced material!!
(calculus, college algebra, etc).

The polynomial exercizes are typical of ones that I encountered when taking
pre-calculus in high school. If you can not handle them, then you are
not, IMO, ready for more advanced material.

Would anyone else like to render an opinion?

The exercizes I gave should be typical for a high level honors high
school class.[/QUOTE]

Since we haven't really seen the level of material with which Tomer [I]is[/I] comfortable, I can't really say whether I think he's ready for e.g. calculus. But I will say that just because he is having some difficulty with these problems, it's not necessarily a sign that he [I]isn't[/I] ready.

It appears that your mental model for math education is that a student needs to show true mastery of the material at any given level before attempting to tackle a more advanced level. That sounds like a reasonable--maybe even an ideal--model to have. Unfortunately it doesn't particularly accord with the way that most people actually learn. The truth is that exposure to more advanced material can be one of the fastest ways to solidify understanding of the basic material. In my experience, both as a student and a teacher, it is a frequent occurrence that after being exposed to more advanced topics, the less advanced ones can be revisited with greater understanding; sometimes understanding that the student didn't even realize was missing.

And sure, true mastery of the basics would make learning the more advanced stuff easier. But it is far more common that students learn well by moving back and forth through the levels of difficulty, such that the new, more difficult topics help to reinforce the basics, and that reinforcement in turn helps make the difficult topics easier as they go. And that back-and-forth process ends up taking less total time than trying to achieve complete mastery of the basics and then moving on.

As an example, I would guess that a fair number of people in this forum have taken courses in both number theory and group theory, in that order. How many of those people discovered that many of the concepts in number theory just made more sense after learning group theory?

I think the exercises you are giving to Tomer are good ones for him, but IMHO the fact that he finds them difficult does not definitely mean that he isn't ready to learn the basics of calculus. Indeed, I would guess that he could learn the basics of calculus without too much trouble, and that after doing so he would find these problems to be easier.

[QUOTE=R.D. Silverman;218562]Throughout these exercizes let f(x) = a_n x^n + a_{n-1} x^(n-1) + .... a_0.

Suppose that its roots are x_0, x_1, .... x_n.
[/QUOTE]

Wait a sec, that doesn't sound right. :wink:

[quote=R.D. Silverman;218596]Huh? The first question is clear. You are given a polynomial and its
roots and asked to derive the polynomial whose roots are the reciprocals
of the given roots. Why isn't this understandable?????

The second question is somewhat related to the binomial theorem.[/quote]

For the first question,
Did you mean?:
For f(x)'s roots are x_0, x_1, ..., x_n, we have a polynomial with roots based on f(x)'s roots, which are 1/x_i.

[quote=R.D. Silverman;218598]But you keep claiming that you are ready for more advanced material!!
(calculus, college algebra, etc).

The polynomial exercizes are typical of ones that I encountered when taking
pre-calculus in high school. If you can not handle them, then you are
not, IMO, ready for more advanced material.

Would anyone else like to render an opinion?

The exercizes I gave should be typical for a high level honors high
school class.[/quote]

I did not claim such a thing (maybe three months ago, but not now).
And more then it, didn't claim I can't solve these...
Just, as you gave me Shanks' book (which was great) I wanted to make my solvements much more intelligent ( I mean, less "OUT OF IGNORANCE").
On my age (in school) things as Newton's binom, The remainder theorem, etc, are not taught (not in my country).
Why did you write "this is easy with calculus", I don't know anything in calculus.

[QUOTE=jyb;218629]Wait a sec, that doesn't sound right. :wink:[/QUOTE]

Oops. Wrong indices. One too many roots.......

[QUOTE=blob100;218632]For the first question,
Did you mean?:
For f(x)'s roots are x_0, x_1, ..., x_n, we have a polynomial with roots based on f(x)'s roots, which are 1/x_i.[/QUOTE]

Can someone else help? We seem to have a language problem.
I do not know how to express the problem any more clearly.

[quote=R.D. Silverman;218660]Can someone else help? We seem to have a language problem.
I do not know how to express the problem any more clearly.[/quote]
There is no language problem.
I meant:
The problem concerns a polynomial g(x) with roots of the form 1/x_i right?
When you denote x_i you mean: a root of f(x) (defined on the exercizes post before the first problem)?
Or, a given polynomial g(x) with roots of the form 1/n for any n (which is nothing interesting because every number is an inverse of another number).

[QUOTE=jyb;218627]
As an example, I would guess that a fair number of people in this forum have taken courses in both number theory and group theory, in that order. How many of those people discovered that many of the concepts in number theory just made more sense after learning group theory?
[/QUOTE]

I did it in reverse. I took algebra as an undergrad, but no number theory.
I did not learn any number theory until later.

[QUOTE]

I think the exercises you are giving to Tomer are good ones for him, but IMHO the fact that he finds them difficult does not definitely mean that he isn't ready to learn the basics of calculus. Indeed, I would guess that he could learn the basics of calculus without too much trouble, and that after doing so he would find these problems to be easier.[/QUOTE]

I mean no offense to Tomer, but from what I have seen, I can't
imagine that he has the ability right now to perform a epsilon-delta
proof for a limit problem. Even a simple one.

I talk with colleagues (University professors) all the time. Their biggest
complaint is that too many students do not have sufficient mastery
of high school level algebra/geometry/trig to tackle college calculus.
There are too many students taking too many bonehead "remedial"
courses. And they are even failing at those.

Have you read Andrei Toom's essay on the pseudo-education that is
taking place with math education today [in the US]? It is a terrific
essay.

[QUOTE=blob100;218661]There is no language problem.
I meant:
The problem concerns a polynomial g(x) with roots of the form 1/x_i right?
When you denote x_i you mean: a root of f(x) (defined on the exercizes post before the first problem)?
Or, a given polynomial g(x) with roots of the form 1/n for any n (which is nothing interesting because every number is an inverse of another number).[/QUOTE]

Huh? Where did you get this last bit of nonsense? I defined x_i.
Having defined x_i, 1/x_i also becomes well defined. Where are you pulling
'n' from?? What relation does it have to x_i???

I have commented before that you have a bad habit of introducing
extraneous variables without properly defining them.

We do have a language problem. x_i were [b]defined[/b] as the roots of a
certain polynomial. 1/x_i are therefore well defined.

Someone else needs to help.

[quote=R.D. Silverman;218665]
. x_i were [B]defined[/B] as the roots of a
certain polynomial. 1/x_i are therefore well defined.

.[/quote]

Ho, Great! This is what I wanted to hear.
f(x)=a_nx^n+a_(n-1)x^(n-1)+....+a_0.
g(x) have the roots 1/x_i.
g(x)=(a_na_0/a_o)x^n+(a_na_1/a_0)x^(n-1)+....+
(a_na_(n/2+1)/a_0)x^(n/2)+....+(a_na_n/a_0).
For even n.

[QUOTE=blob100;218667]Ho, Great! This is what I wanted to hear.
f(x)=a_nx^n+a_(n-1)x^(n-1)+....+a_0.
g(x) have the roots 1/x_i.
g(x)=(a_na_0/a_o)x^n+(a_na_1/a_0)x^(n-1)+....+
(a_na_(n/2+1)/a_0)x^(n/2)+....+(a_na_n/a_0).
For even n.[/QUOTE]

What is "a_na_0/a_o" etc. What is "a_na_n/a_0" ????? .......

And why for "even n"???

And the coefficients that you have written are not integers........

Note: The correct answer is quite simple.

[quote=R.D. Silverman;218687]What is "a_na_[B]0[/B]/a_[B]o[/B]" etc. What is "a_na_n/a_0" ????? .......

And why for "even n"???

And the coefficients that you have written are not integers........

Note: The correct answer is quite simple.[/quote]
a_na_n/a_0 is the product of a_n and itself (a_n)^2 devided by a_0
(you defined these by yourself).
"a_na_[B]0[/B]/a_[B]o[/B]"--> a_na_0/a_0.
These does not need to be integers.

For f(x)=a.nx^n+a.(n-1)x^(n-1)+....+a.0
For x^k the coefficient is ((a.n * a.P)/ a.0) where P means the "paralleled" for k, defined: if n is the order of the polynomial, |n-k|=P-0=P.
Where k is any of the powers of x in the polynomial (1,2...,n).
Example:
(x-1)(x+2)(x-2)=f(x)
=x^3-x^2-4x+4
(x-1)(x+1/2)(x-1/2)=g(x)
=x^3-x^2-(1/4)X+1/4
The paralleled for 3 is 0, 2-->1, etc.

The second problem:
Again using P as a "paralleled" of k, x^k's coefficient is (where the order of the polynomial is N):
-/+(The sum of the product combinations between the roots, where every factor in the sum is a product of P roots),
For order 3, the coefficient of x^2 is -(x3+x2+x1).
The value is minus or plus depends on N-k, If it is odd, minus, if it is even, plus.

[QUOTE=R.D. Silverman;218663]I mean no offense to Tomer, but from what I have seen, I can't
imagine that he has the ability right now to perform a epsilon-delta
proof for a limit problem. Even a simple one.[/QUOTE]

I agree that he probably doesn't have that ability right now. But no one is born knowing how to perform epsilon-delta proofs. It has to be taught--and learned. And it's hard to say based on what we've seen, but I would guess that it's something he would be able to learn if given good instruction. And perhaps you're thinking that his track record of learning things doesn't look too good in this forum. But let us agree that this forum is a [I]lousy[/I] medium for learning mathematics.

[QUOTE=R.D. Silverman;218663]I talk with colleagues (University professors) all the time. Their biggest
complaint is that too many students do not have sufficient mastery
of high school level algebra/geometry/trig to tackle college calculus.
There are too many students taking too many bonehead "remedial"
courses. And they are even failing at those.

Have you read Andrei Toom's essay on the pseudo-education that is
taking place with math education today [in the US]? It is a terrific
essay.[/QUOTE]

Yes, I agree with the vast majority of what Toom has to say. (Some of his specific examples, like having to snatch calculators out of the hands of students so they will actually [I]think[/I], hit very close to home. I have had some of those exact experiences.) You won't find me defending the vague state of primary school mathematics education in this country. And indeed, Toom even makes a point about students trying to learn material which is simply too advanced for their abilities, which is pretty much what you're talking about.

Look, what I said above can certainly be taken too far. I don't advocate trying to force advanced material on people with minimal mathematical background (even if they're requesting it). It's just that I see time and time again, in mathematics and in other disciplines, this phenomenon where exposure to advanced topics, methods and mindsets can really help understanding of the more basic topics, even when the student thought they already understood the basic topics. So there's some level of problem difficulty beyond which a failure to solve the problem is not necessarily a barrier to learning more advanced topics. That difficulty level is probably different for every student, but I don't think it's infinitely high, as you seem to.

[QUOTE=blob100;218706]a_na_n/a_0 is the product of a_n and itself (a_n)^2 devided by a_0
(you defined these by yourself).
"a_na_[B]0[/B]/a_[B]o[/B]"--> a_na_0/a_0.
These does not need to be integers.[/QUOTE]

But your last expression a_na_0/a_0 is just a_n........

Your answer is not correct. This problem is quite simple, only takes
two steps, and yields a very simple answer that does not involve
ratios of coefficients. Indeed, the derivation does not even involve
arithmetic operations performed on the coefficients.......

[QUOTE=blob100;218716]The second problem:
Again using P as a "paralleled" of k, x^k's coefficient is (where the order of the polynomial is N):
-/+(The sum of the product combinations between the roots, where every factor in the sum is a product of P roots),
For order 3, the coefficient of x^2 is -(x3+x2+x1).
The value is minus or plus depends on N-k, If it is odd, minus, if it is even, plus.[/QUOTE]

Without quibbling about the wording, this is essentially correct.
The coefficient is (-1)^k times the sum of all products of the coefficients taken k at a time.

[quote=R.D. Silverman;218732]But your last expression a_na_0/a_0 is just a_n........

Your answer is not correct. This problem is quite simple, only takes
two steps, and yields a very simple answer that does not involve
ratios of coefficients. Indeed, the derivation does not even involve
arithmetic operations performed on the coefficients.......[/quote]

a_na_0/a_0 is just a_n........ true, but I wanted to show the structure, not the best and most comfortable expression.

[QUOTE=jyb;218731]I agree that he probably doesn't have that ability right now. But no one is born knowing how to perform epsilon-delta proofs. It has to be taught--and learned. And it's hard to say based on what we've seen, but I would guess that it's something he would be able to learn if given good instruction. And perhaps you're thinking that his track record of learning things doesn't look too good in this forum. But let us agree that this forum is a [I]lousy[/I] medium for learning mathematics.
[/QUOTE]

Agreed. But I wasn't thinking so much about the aspect of learning how
to do epsilon-delta proofs specificallty. Instead, based upon what I have
seen, he has, so far, been unable to put together a rigorous derivation of
anything. He fails to heed my advice about defining variables. He fails
to heed my advice about introducing superfluous variables. (which his posts
have shown only confuses things). He ignores my advice when I give hints.
I have suggested that he try to present a semi-formal proof where he gives a
reason for each step based upon known results and previously derived
results. He doesn't seem to be able to do this either. He is very very
sloppy. He resorts too much to "handwaving".

Epsilon-delta proofs are usually the first instance where students must
show something that depends on more than one quantifier. One must be
very organized to put together such proofs. I don't think he is ready.

He doesn't seem able to handle the first polynomial algebra exercize that
I gave, and this one is totally trivial.

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

For f(x)=(a.n)x^n+a.(n-1)x^(n-1)....a.0.
a.i denotes the coefficient (integers) of x^i, n integer (the order of the polynom).
Do you remember how I defined P on the last posts?
The proposition:
The coefficient of x^k in the polynom g(x) (the polynom with the inverse roots of f(x)) are:
(a.P*a.n)/(a.0).
For k integer<n, P the paralleled of k.

[QUOTE=blob100;218855]For the first question:
As you said, there is a missunderstandment about the problem.
So, I will give a private case shows what I understand you want me to do in the problem.
f(x)=(x+a)(x+b)(x+c)
(x^2+(a+b)x+ab)(x+c)=x^3+cx^2+(a+b)x^2+c(a+b)x+abx+abc=
x^3+(a+b+c)x^2+(ca+cb+ab)x+abc.
The roots are -a,-b,-c.
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.

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.



[QUOTE]

For f(x)=a.nx^n+a.(n-1)x^(n-1)....a.0.
a.n integers, n integer (the order of the polynom).
Do you remember how I defined P on the last posts?
The proposition:
The coefficient of x^k in the polynom g(x) (the polynom with the inverse roots of f(x)) are:
(a.P*a.n)/(a.0).
For k integer<n, P the paralleled of k.[/QUOTE]

This is irrelevant to the problem.

[QUOTE=blob100;218855]The coefficient of x^k in the polynom g(x) (the polynom with the inverse roots of f(x)) are:
(a.P*a.n)/(a.0).
For k integer<n, P the paralleled of k.[/QUOTE]

I think it's standard form to have integer coefficients.

[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

[QUOTE=wblipp;219036]Bob,

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

Huh? I wrote originally:
"Suppose that its roots are x_0, x_1, .... x_n.
a_i are integers."

This doesn't leave much room for doubt that the coefficients are
integers.
[QUOTE]

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

He [b]claims[/b] to be ready for calculus and other university level
courses. If he needs help with first year algebra (i.e. use of variables as
you suggest) then it is clear that he is nowhere near ready.

I have repeatedly told him that he introduces too many 'free variables'
in his problems; often without defining them. He doesn't seem to get
that message.

To be fair to both of us: this all would be better in front of a blackboard.

[QUOTE=R.D. Silverman;219020]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).

[/QUOTE]

Ooops! Clearly r-k should be k-r.

[quote=R.D. Silverman;219041]Huh? I wrote originally:
"Suppose that its roots are x_0, x_1, .... x_n.
a_i are integers."

This doesn't leave much room for doubt that the coefficients are
integers.
.[/quote]
I know f(x)'s coefficients are integers.
But what about g(x)'s coefficients? are these integers too?
g(x) is defined as a polynomial with roots x_0^-1, x_1^-1, .... x_n^-1.

[quote=R.D. Silverman;219020]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?[/quote]

In wolfram alpha it is called a polynomial remainder.

[quote=R.D. Silverman;219020]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?[/quote]

I think you are able to find infinitely many synthetic devisions for a given polynomial.

[QUOTE=blob100;219069]I know f(x)'s coefficients are integers.
But what about g(x)'s coefficients? are these integers too?
g(x) is defined as a polynomial with roots x_0^-1, x_1^-1, .... x_n^-1.[/QUOTE]

Yes, they are integers. Try expressing them in terms of the coefficients of f.

[QUOTE=blob100;219071]I think you are able to find infinitely many synthetic devisions for a given polynomial.[/QUOTE]


Synthetic division involves TWO given polynomials, not 'a given polynomial'.
The divisor polynomial is also given.

[quote=R.D. Silverman;219085]Synthetic division involves TWO given polynomials, not 'a given polynomial'.
The divisor polynomial is also given.[/quote]
OK.

[QUOTE=R.D. Silverman;219020]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?[/QUOTE]


Bob,

I think the process you have described is usually called polynomial division. I think "synthetic division" is usually used for the special problem of checking for divisibility by a linear factor by evaluating the polynomial at the root of the linear polynomial. That usage of "synthetic division" [B]does[/B] have something do with the roots of the polynomial.

Also, I can't tell if the final "this" refers to the result for integers or the result for polynomials.

William

[QUOTE=wblipp;219091]Bob,

I think the process you have described is usually called polynomial division. I think "synthetic division" is usually used for the special problem of checking for divisibility by a linear factor by evaluating the polynomial at the root of the linear polynomial. That usage of "synthetic division" [B]does[/B] have something do with the roots of the polynomial.

Also, I can't tell if the final "this" refers to the result for integers or the result for polynomials.

William[/QUOTE]

'This' refers to the result for polynomials. We had already proved the
result for integers in an earlier post.

The Wikipaedia says of synthetic division:

"It is mostly taught for division by binomials of the form
but the method generalizes to division by any monic polynomial. "

I think the words "mostly" and "generalizes" are relevant.

[QUOTE=R.D. Silverman;219101]"It is mostly taught for division by binomials of the form but the method generalizes to division by any monic polynomial."[/QUOTE]

Hmm. The thing Wikipedia calls synthetic division is VASTLY different from the thing I was taught as synthetic division. I was taught the thing at the first Google link,

[url]http://www.purplemath.com/modules/synthdiv.htm[/url]

which says

"Synthetic division is a shorthand, or shortcut, method of polynomial division in the special case of dividing by a linear factor -- and it [I]only[/I] works in this case."

I guess there must be multiple definitions of synthetic division.

[QUOTE=wblipp;219104]Hmm. The thing Wikipedia calls synthetic division is VASTLY different from the thing I was taught as synthetic division. I was taught the thing at the first Google link,

[url]http://www.purplemath.com/modules/synthdiv.htm[/url]

which says

"Synthetic division is a shorthand, or shortcut, method of polynomial division in the special case of dividing by a linear factor -- and it [I]only[/I] works in this case."

I guess there must be multiple definitions of synthetic division.[/QUOTE]

Just like the definition of N. To some, it means "postive integers".
To others it means "non-negative integers".

f(x)=a_nx^n+....+a_0.
roots={x1,x2,....,xn}.
g(x)=b_nx^n+...+b_0.
roots={1/x1,1/x2,....,1/xn}.

a_i=(-1)^i(g1+....+gn)
b_i=(-1)^i(1/g1+....+1/gn).
Where g={g1,g2,....,gn}
g is the set of all product combinations of P roots of f(x).
a_i is the coefficient of x^i in f(x), similarily with b_i.
n-i=P.

The phrase 1/g1+....+1/gn we may write as:
Denominator=-1(a_0)=x1x2...xn.
The numerator is (-1)^i(the sum of product combinations of i=n-P roots of f(x) which is of course a_P)
The whole phrase is ((-1)^(i-1))(a_P/a_0).
*this assumes x^n's coefficient is 1.

The result: "the sum of product combinations of i=n-P roots of f(x) which is of course a_P" comes from the common denominator between the factors 1/g.
We may find that to make the denominator of 1/gj (given factor of the sum) we are needed to product 1/gj by an combination of i roots product,
Same with every such 1/g.
In conclusion, we get the sum of combinations of such products of i roots, which is similarily |a_P|.

[QUOTE=blob100;219263]f(x)=a_nx^n+....+a_0.
roots={x1,x2,....,xn}.
g(x)=b_nx^n+...+b_0.
roots={1/x1,1/x2,....,1/xn}.

a_i=(-1)^i(g1+....+gn)
b_i=(-1)^i(1/g1+....+1/gn).
Where g={g1,g2,....,gn}

[/QUOTE]

You are doing it again. Introducing extraneous variables and
not defining them. What are g1, g2, ........ gn??????

[QUOTE]
g is the set of all product combinations of P roots of f(x).

[/QUOTE]

Not from what you have written here.
[QUOTE]
a_i is the coefficient of x^i in f(x), similarily with b_i.
n-i=P.
[/QUOTE]

b_i is given in terms of g1, g2.... which you have not even defined.

<snip>

rest deleted....



You are also ignoring my advice once again. I told you to do the
problems in order. You should not be using results from a later problem
to solve this one. In fact, no results from later problems are needed
in this one!

[quote=R.D. Silverman;219269]You are doing it again. Introducing extraneous variables and
not defining them. What are g1, g2, ........ gn??????



Not from what you have written here.


b_i is given in terms of g1, g2.... which you have not even defined.

<snip>

rest deleted....



You are also ignoring my advice once again. I told you to do the
problems in order. You should not be using results from a later problem
to solve this one. In fact, no results from later problems are needed
in this one![/quote]

I'm sorry for not marking g={g1,....,gr}.
b_i=(-1)^i(1/g1+1/g2+...+1/gr)
I'm sorry for writing n instead of r.
r is the number of product combinations for P roots.

We can show this by the identifies:
f(x)=(x-x1)(x-x2)...(x-xn)
g(x)=(x-1/x1)(x-1/x2)...(x-1/xn)

If we calculate these as sums we get again:
By calculating f(x) we get the result of the combinations,
And by calculating g(x) too we get the coefficient of x^i is (-1)^(i+1)(a_P/a_0).

[QUOTE=blob100;219271]I'm sorry for not marking g={g1,....,gr}.
b_i=(-1)^i(1/g1+1/g2+...+1/gr)
I'm sorry for writing n instead of r.
r is the number of product combinations for P roots.
I can't see a way to prove this without the secondary problem's result.[/QUOTE]

You defined g. You did not, and still have not defined g1, g2, ....


I will offer a hint. Go back to the basic definition of what it means
to be a root.

[quote=R.D. Silverman;219274]You defined g. You did not, and still have not defined g1, g2, ....


I will offer a hint. Go back to the basic definition of what it means
to be a root.[/quote]
gi is a given unit of the set if {g1,g2,....} is a sum of product combinations of P roots.

[quote=R.D. Silverman;219274]You defined g. You did not, and still have not defined g1, g2, ....


I will offer a hint. Go back to the basic definition of what it means
to be a root.[/quote]
I know what a root means, it is a number such that putting it in x in the polynomial, f(x)=0.

[quote=blob100;219263]f(x)=a_nx^n+....+a_0.
roots={x1,x2,....,xn}.
g(x)=b_nx^n+...+b_0.
roots={1/x1,1/x2,....,1/xn}.

a_i=(-1)^i(g1+....+gn)
b_i=(-1)^i(1/g1+....+1/gn).
Where g={g1,g2,....,gn}
g is the set of all product combinations of P roots of f(x).
a_i is the coefficient of x^i in f(x), similarily with b_i.
n-i=P.

The phrase 1/g1+....+1/gn we may write as:
Denominator=-1(a_0)=x1x2...xn.
The numerator is (-1)^i(the sum of product combinations of i=n-P roots of f(x) which is of course a_P)
The whole phrase is ((-1)^(i-1))(a_P/a_0).
*this assumes x^n's coefficient is 1.

The result: "the sum of product combinations of i=n-P roots of f(x) which is of course a_P" comes from the common denominator between the factors 1/g.
We may find that to make the denominator of 1/gj (given factor of the sum) we are needed to product 1/gj by an combination of i roots product,
Same with every such 1/g.
In conclusion, we get the sum of combinations of such products of i roots, which is similarily |a_P|.[/quote]

I'll write everything again:
f(x)=a_nx^n+....+a_0.
roots={x1,x2,....,xn}.
g(x)=b_nx^n+...+b_0.
roots={1/x1,1/x2,....,1/xn}.

a_i=((-1)^(i+1))(g1+....+gr)
b_i=((-1)^(i+1))(1/g1+....+1/gr).
G={g1,g2,...,gr} is the set of all product combinations of P roots of f(x).
a_i is the coefficient of x^i in f(x), similarily with b_i.
n-i=P, r is the number of such combinations can be written in terms of G.


The phrase 1/g1+....+1/gn can be written as:
Denominator: x1x2....xn.
Numerator: ((-1)^(i+1))(the sum of product combinations of i=n-P roots of f(x) which is of course a_P)
The numerator is ((-1)^(i+1))(a_P).

And the whole phrase (numberator/denominator) is:
(((-1)^(i+1))(a_P))/(x1x2...xn)=((-1)^(i+1))(a_P)/(a_0) by the result of the second problem (and simple algebra).

[quote=blob100;219280]I'll write everything again:
f(x)=a_nx^n+....+a_0.
roots={x1,x2,....,xn}.
g(x)=b_nx^n+...+b_0.
roots={1/x1,1/x2,....,1/xn}.

a_i=((-1)^(i+1))(g1+....+gr)
b_i=((-1)^(i+1))(1/g1+....+1/gr).
G={g1,g2,...,gr} is the set of all product combinations of P roots of f(x).
a_i is the coefficient of x^i in f(x), similarily with b_i.
n-i=P, r is the number of such combinations can be written in terms of G.


The phrase 1/g1+....+1/gn can be written as:
Denominator: x1x2....xn.
Numerator: ((-1)^(i+1))(the sum of product combinations of i=n-P roots of f(x) which is of course a_P)
The numerator is ((-1)^(i+1))(a_P).

And the whole phrase (numberator/denominator) is:
(((-1)^(i+1))(a_P))/(x1x2...xn)=((-1)^(i+1))(a_P)/(a_0) by the result of the second problem (and simple algebra).[/quote]

Didn't have time to edit:
[quote=blob100;219263]f(x)=a_nx^n+....+a_0.
roots={x1,x2,....,xn}.
g(x)=b_nx^n+...+b_0.
roots={1/x1,1/x2,....,1/xn}.

a_i=(-1)^i(g1+....+gn)
b_i=(-1)^i(1/g1+....+1/gn).
Where g={g1,g2,....,gn}
g is the set of all product combinations of P roots of f(x).
a_i is the coefficient of x^i in f(x), similarily with b_i.
n-i=P.

The phrase 1/g1+....+1/gn we may write as:
Denominator=-1(a_0)=x1x2...xn.
The numerator is (-1)^i(the sum of product combinations of i=n-P roots of f(x) which is of course a_P)
The whole phrase is ((-1)^(i-1))(a_P/a_0).
*this assumes x^n's coefficient is 1.

The result: "the sum of product combinations of i=n-P roots of f(x) which is of course a_P" comes from the common denominator between the factors 1/g.
We may find that to make the denominator of 1/gj (given factor of the sum) we are needed to product 1/gj by an combination of i roots product,
Same with every such 1/g.
In conclusion, we get the sum of combinations of such products of i roots, which is similarily |a_P|.[/quote]

I'll write everything again:
f(x)=a_nx^n+....+a_0.
roots={x1,x2,....,xn}.
g(x)=b_nx^n+...+b_0.
roots={1/x1,1/x2,....,1/xn}.

a_i=((-1)^(i+1))(g1+....+gr)
b_i=((-1)^(i+1))(1/g1+....+1/gr).
G={g1,g2,...,gr} is the set of all product combinations of P roots of f(x).
a_i is the coefficient of x^i in f(x), similarily with b_i.
n-i=P, r is the number of such combinations can be written in terms of G.


The phrase 1/g1+....+1/gn can be written as:
Denominator: x1x2....xn.
Numerator: the sum of product combinations of i=n-P roots of f(x) which will be defined as R.
a_P=((-1)^(P+1))R


And the whole phrase (numberator/denominator) is:
(The sum of product combinations of i=n-P roots of f(x))/( x1x2....xn)=R/(x1x2...xn).
b_i=(((-1)^(i+1))(R)/(x1x2...xn)
Which equals ((-1)^(|i-P|))(a_P)/(-a_0)=((-1)^(|i-P|+1))(a_P)/(a_0).

[QUOTE=blob100;219282]Didn't have time to edit:


I'll write everything again:
f(x)=a_nx^n+....+a_0.
roots={x1,x2,....,xn}.
g(x)=b_nx^n+...+b_0.
roots={1/x1,1/x2,....,1/xn}.

a_i=((-1)^(i+1))(g1+....+gr)
b_i=((-1)^(i+1))(1/g1+....+1/gr).
G={g1,g2,...,gr} is the set of all product combinations of P roots of f(x).
a_i is the coefficient of x^i in f(x), similarily with b_i.
n-i=P, r is the number of such combinations can be written in terms of G.


The phrase 1/g1+....+1/gn can be written as:
Denominator: x1x2....xn.
Numerator: the sum of product combinations of i=n-P roots of f(x) which will be defined as R.
a_P=((-1)^(P+1))R


And the whole phrase (numberator/denominator) is:
(The sum of product combinations of i=n-P roots of f(x))/( x1x2....xn)=R/(x1x2...xn).
b_i=(((-1)^(i+1))(R)/(x1x2...xn)
Which equals ((-1)^(|i-P|))(a_P)/(-a_0)=((-1)^(|i-P|+1))(a_P)/(a_0).[/QUOTE]

No. Not even close. You are not taking my hints:

(1) This problems does NO depend on any later problems.
(2) Apply the definition of what it means to be a root.

Tomer,

1. In my opinion, "solve the problems in order" means that you have solved #2 before #1 and are now trying to use the results of #2 to solve #1. It is, in my opinion, not actually wrong but it misses a much better solution.

2. When using the hint "Go back to the basic definition of what it means to be a root," (hmmm - giving an additional hint here without giving the game away is harder than I anticipated) think about g as well f.

[quote=wblipp;219091]
William[/quote]

I presume you have met Paul by now.
I haven't, but as you will understand, this would be superfluous.

Will check out that Tom Lehrer ditty in due course.

Visited the Flask the other day. Met a really nice bloke
from Norfolk.
OTOH met that same bloke I told you about - opening line
"I'm going to knock your ******* head off".

Happy Days,

David

[quote=R.D. Silverman;219296]No. Not even close. You are not taking my hints:

(1) This problems does NO depend on any later problems.
(2) Apply the definition of what it means to be a root.[/quote]

A root r of the polynomial f(x) such that for x=r, f(r)=f(x)=0.
It is really hard to prove what I don't know.
Is my formula on #636 right?
My problem here is how not to use the result of the second problem.
Showing the coefficients of g(x) as a function of the coefficients of f(x) include understanding the coefficients of f(x).
The common sense we can find between those coefficients (of f and g) is the roots which with those we can show the coefficients (by the result of the second problem), I can't see another way to find the coefficients as a function of the roots.

[quote=blob100;219344]A root r of the polynomial f(x) such that for x=r, f(r)=f(x)=0.
It is really hard to prove what I don't know.
Is my formula on #636 right?
My problem here is how not to use the result of the second problem.
Showing the coefficients of g(x) as a function of the coefficients of f(x) include understanding the coefficients of f(x).
The common sense we can find between those coefficients (of f and g) is the roots which with those we can show the coefficients (by the result of the second problem), I can't see another way to find the coefficients as a function of the roots.[/quote]

Here is a mathematical way to show my problem:
f(x)=(x-x1)(x-x2)...(x-xn)
g(x)=(x-X1)(x-X2)...(x-Xn)
Where Xi (given root of g(x))=xi^(-1) (the inverse of f(x)'s root).
Let R={x1,x2,...,xn}, M={X1,X2,...,Xn}={1/x1,1/x2,...,1/xn}.
a_i=F(R) a function of the roots of f(x) (this function is what problem 2 relate to (the result)).
b_i=F(M) a function of the roots of g(x).
a_i and b_i are the coefficients of x^i in f(x) and g(x) order wise.
To show b_i as a function of a_i, we must understand the function F.

[QUOTE=blob100;219347]Here is a mathematical way to show my problem:
f(x)=(x-x1)(x-x2)...(x-xn)
g(x)=(x-X1)(x-X2)...(x-Xn)
Where Xi (given root of g(x))=xi^(-1) (the inverse of f(x)'s root).
Let R={x1,x2,...,xn}, M={X1,X2,...,Xn}={1/x1,1/x2,...,1/xn}.
a_i=F(R) a function of the roots of f(x) (this function is what problem 2 relate to (the result)).
b_i=F(M) a function of the roots of g(x).
a_i and b_i are the coefficients of x^i in f(x) and g(x) order wise.
To show b_i as a function of a_i, we must understand the function F.[/QUOTE]

You are not using the definition of what it means to be a root.

Is this the direction?:
f(x1)=0
g(1/x1)=0.
f(x1)=g(1/x1)=0.
It does not seem easy to find the coefficients of g(x) as a function of the ones of f(x) by this equation.

[QUOTE=blob100;219364]Is this the direction?:
f(x1)=0
g(1/x1)=0.
f(x1)=g(1/x1)=0.
It does not seem easy to find the coefficients of g(x) as a function of the ones of f(x) by this equation.[/QUOTE]

This is correct.
And finding the coefficients [b]IS[/b] trivial by these equations.

[quote=R.D. Silverman;219368]This is correct.
And finding the coefficients [B]IS[/B] trivial by these equations.[/quote]

But hard. To do it?

[QUOTE=blob100;219369]But hard. To do it?[/QUOTE]

No, it is not hard. It is trivial.

[quote=R.D. Silverman;219372]No, it is not hard. It is trivial.[/quote]

Ok, To prove the result of the secondary problem?
Or to try the third?

[QUOTE=blob100;219373]Ok, To prove the result of the secondary problem?
Or to try the third?[/QUOTE]

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

Perhaps you can use a more direct hint:

Given
f([B]X[/B]) = 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]
with roots a[SUB]1[/SUB], a[SUB]2[/SUB], ..., a[SUB]n[/SUB]
and
g([B]Y[/B]) = B[SUB]n[/SUB][B]Y[/B][SUP]n[/SUP] + B[SUB]n-1[/SUB][B]Y[/B][SUP]n-1[/SUP] + ... + B[SUB]1[/SUB][B]Y[/B] + B[SUB]0[/SUB]
with roots b[SUB]1[/SUB], b[SUB]2[/SUB], ..., b[SUB]n[/SUB]
where, for i in 1 ... n,
b[SUB]i[/SUB] = 1/a[SUB]i[/SUB]

You need to express B[SUB]j[/SUB] in terms of the A[SUB]i[/SUB].

[quote=Wacky;219380]Perhaps you can use a more direct hint:

Given
f([B]X[/B]) = 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]
with roots a[sub]1[/sub], a[sub]2[/sub], ..., a[sub]n[/sub]
and
g([B]Y[/B]) = B[sub]n[/sub][B]Y[/B][sup]n[/sup] + B[sub]n-1[/sub][B]Y[/B][sup]n-1[/sup] + ... + B[sub]1[/sub][B]Y[/B] + B[sub]0[/sub]
with roots b[sub]1[/sub], b[sub]2[/sub], ..., b[sub]n[/sub]
where, for i in 1 ... n,
b[sub]i[/sub] = 1/a[sub]i[/sub]

You need to express B[sub]j[/sub] in terms of the A[sub]i[/sub].[/quote]

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

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

Find a relationship between [B]X[/B] and [B]Y[/B] that is useful in relating the a[SUB]i[/SUB] and the b[SUB]i[/SUB].

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.

[quote=R.D. Silverman;219374]You need to finish the first problem.......[/quote]
Is this the direction:
Using Wacky's noations.
B[sub]j[/sub]b[sup]j[/sup][sub]i[/sub]=-(g(b[sub]i[/sub])-B[sub]j[/sub]b[sup]j[/sup][sub]i[/sub])+f(a[sup]j[/sup][sub]i[/sub]).
This equalls B[sub]j[/sub]/a[sup]j[/sup][sub]i[/sub].
And we do so for every such b[sub]i[/sub] root.
We get a number of equations of the form:
B[sub]j[/sub]/a[sup]j[/sup][sub]i[/sub]=-(g(b[sub]i[/sub])-B[sub]j[/sub]b[sup]j[/sup][sub]i[/sub])+f(a[sup]j[/sup][sub]i[/sub]).
By these equations we get B[sub]j[/sub].
To show B[sub]j[/sub] as a function of A[sub]i[/sub], the g(b[sub]i[/sub]) part in the equation -(g(b[sub]i[/sub])-B[sub]j[/sub]b[sup]j[/sup][sub]i[/sub])+f(a[sup]j[/sup][sub]i[/sub]) will be reduced by showing the equallity between the equations of the form:
B[sub]j[/sub]=a[sup]j[/sup][sub]i[/sub](-(g(b[sub]i[/sub])-B[sub]j[/sub]b[sup]j[/sup][sub]i[/sub])+f(a[sup]j[/sup][sub]i[/sub])).

[QUOTE=blob100;219390]Is this the direction:
Using Wacky's noations.
B[sub]j[/sub]b[sup]j[/sup][sub]i[/sub]=-(g(b[sub]i[/sub])-B[sub]j[/sub]b[sup]j[/sup][sub]i[/sub])+f(a[sup]j[/sup][sub]i[/sub]).
This equalls B[sub]j[/sub]/a[sup]j[/sup][sub]i[/sub].
And we do so for every such b[sub]i[/sub] root.
We get a number of equations of the form:
B[sub]j[/sub]/a[sup]j[/sup][sub]i[/sub]=-(g(b[sub]i[/sub])-B[sub]j[/sub]b[sup]j[/sup][sub]i[/sub])+f(a[sup]j[/sup][sub]i[/sub]).
By these equations we get B[sub]j[/sub].
To show B[sub]j[/sub] as a function of A[sub]i[/sub], the g(b[sub]i[/sub]) part in the equation -(g(b[sub]i[/sub])-B[sub]j[/sub]b[sup]j[/sup][sub]i[/sub])+f(a[sup]j[/sup][sub]i[/sub]) will be reduced by showing the equallity between the equations of the form:
B[sub]j[/sub]=a[sup]j[/sup][sub]i[/sub](-(g(b[sub]i[/sub])-B[sub]j[/sub]b[sup]j[/sup][sub]i[/sub])+f(a[sup]j[/sup][sub]i[/sub])).[/QUOTE]

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

[quote=R.D. Silverman;219397]Sorry, but no. You are making it much more difficult than it is.[/quote]
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.

[QUOTE=blob100;219407]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.[/QUOTE]

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.

[QUOTE=R.D. Silverman;219429]I can't say more without just telling you the answer.[/QUOTE]

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.

[quote=axn;219431]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.[/quote]
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.

[quote=R.D. Silverman;219429]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.[/quote]

Can you just give me the answer?

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 [B]shows[/B] that it has the required roots.
Then express it in the form: A[SUB]n[/SUB][B]X[/B][SUP]n[/SUP]+ ...

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")

[quote=Wacky;219444]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 [B]shows[/B] that it has the required roots.
Then express it in the form: A[sub]n[/sub][B]X[/B][sup]n[/sup]+ ...

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")[/quote]

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=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.

[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.