[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.