[QUOTE=blob100;219679]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.[/QUOTE]

Did you read my prior post? You formula must state HOW MANY terms
are in the sum. It is a necessary part of any induction proof. It is not
enough to say "all product combinations". How many combinations are
there?

[quote=R.D. Silverman;219682]Did you read my prior post? You formula must state HOW MANY terms
are in the sum. It is a necessary part of any induction proof. It is not
enough to say "all product combinations". How many combinations are
there?[/quote]

OK,
For a polynomial with n roots,
The number of terms (in the sum.....) is (for P roots combinations);
(n-P)+(n-(P+1))+...+1=((n-P)^2+(n-P))/2.

[QUOTE=blob100;219684]OK,
For a polynomial with n roots,
The number of terms (in the sum.....) is (for P roots combinations);
(n-P)+(n-(P+1))+...+1=((n-P)^2+(n-P))/2.[/QUOTE]

Tomer, did you check this for any reasonable small example? Does it work for the coefficient of the x^2 term? What about for the x^1 term? Does your answer to those two questions tell you anything?

[quote=jyb;219691]Tomer, did you check this for any reasonable small example? Does it work for the coefficient of the x^2 term? What about for the x^1 term? Does your answer to those two questions tell you anything?[/quote]
Sorry, I gave the wrong answer.

[QUOTE=blob100;219738]Sorry, I gave the wrong answer.[/QUOTE]

Check your results before posting.

n-k=P. k>P.
The number of terms for x^k is:
T=F(P)+F(P-1)+...+F(1).
Where F(x) is the Fermat formula taken by x.
f(x)=(x^2+x)/2.

The number of terms for P is equivallent to x^k's.
To open F(P)+F(P-1)+...+F(1)?

[QUOTE=blob100;219744]n-k=P. k>P.
The number of terms for x^k is:
T=F(P)+F(P-1)+...+F(1).
Where F(x) is the Fermat formula taken by x.
f(x)=(x^2+x)/2.

The number of terms for P is equivallent to x^k's.
To open F(P)+F(P-1)+...+F(1)?[/QUOTE]

I have only one word: Huh????
This is total nonsnse.

[quote=R.D. Silverman;219750]I have only one word: Huh????
This is total nonsnse.[/quote]
What is unable to be understood?
I'll try again:
n-k=P. And we (arbitarically) assume k>P.
F(x)=(x^2+x)/2 (known as the Fermat formula).
We would say that the number of terms in the coefficnet of x^k is:
F(P)+F(P-1)+...+F(1)=((P^2+P)+((P-1)^2+(P-1))+...+2)/2.

I'll use a private case (private cases in our situation throw light on the general case):
a,b,c,d,e are the roots.
k=3, which means P=3.
(a,b,c), (a,b,d), (a,b,e)
(a,c,d), (a,c,e)
(a,d,e)

(b,c,d), (b,c,e)
(b,d,e)

(c,d,e)

Now, we see that the number of terms is 6+3+1.
6=F(3)
3=F(2)
1=F(1)

This private case does not come to prove anything, it comes to show from where did I take the formula
F(P)+F(P-1)+...+F(1)=((P^2+P)+((P-1)^2+(P-1))+...+2)/2,
Which you replayed by "Huh????".

[QUOTE=blob100;219755]What is unable to be understood?
I'll try again:
n-k=P. And we (arbitarically) assume k>P.
[/QUOTE]
Why this assumption?
[QUOTE]
F(x)=(x^2+x)/2 (known as the Fermat formula).
[/QUOTE]

(1) You are again being sloppy about notation.

In your prior post you referred to F(x), but you did not define it.
Instead you [b]redefined[/b] f(x) (which was already defined at the
very beginning of the problem set!!!). f(x) is not the same as F(x).
I am also curious. Where did you see that x(x+1)/2 is referred to as
the 'Fermat formula'? I have never seen Fermat's name attached to it.
(that I can recall)

[QUOTE]

We would say that the number of terms in the coefficnet of x^k is:
F(P)+F(P-1)+...+F(1)=((P^2+P)+((P-1)^2+(P-1))+...+2)/2.

[/QUOTE]

We are not discussing the number of terms in the coefficient of x^k.
We [b]are[/b] discussing the the number of terms in the summation
formula that gives the coefficient as a function of the roots. The coefficient
of x^k is a constant. It has one term: itself. You are being sloppy
about definitions here.

And your formula above is not correct. It can not be correct. It is
nonsense.

F(x) is a quadratic in x. When summed over an arithmetic progression
of difference 1, i.e. (P, P-1, P-2, .....1) the result will be a cubic polynomial
in P. But the number of terms in the summation formula is not a cubic
function of P. Indeed. It is not even a polynomial in P.

I will give a hint: The number of different products of roots taken r
at a time from a set of size n is a well known COMBINATORIAL object.

[QUOTE=R.D. Silverman;219757]
\
<snip>


.[/QUOTE]

I will also add:

You wrote "The number of terms for P is equivallent to x^k's.
To open F(P)+F(P-1)+...+F(1)?"

The first sentence is total gibberish. What does "equivalent to x^k's"
mean? It is nonsense.

And what does the word "open" mean?? And "To open F(P)+F(P-1)+...+F(1)?" is not even a sentence.

[quote=R.D. Silverman;219757]Why this assumption?


(1) You are again being sloppy about notation.

In your prior post you referred to F(x), but you did not define it.
Instead you [B]redefined[/B] f(x) (which was already defined at the
very beginning of the problem set!!!). f(x) is not the same as F(x).
I am also curious. Where did you see that x(x+1)/2 is referred to as
the 'Fermat formula'? I have never seen Fermat's name attached to it.
(that I can recall)



We are not discussing the number of terms in the coefficient of x^k.
We [B]are[/B] discussing the the number of terms in the summation
formula that gives the coefficient as a function of the roots. The coefficient
of x^k is a constant. It has one term: itself. You are being sloppy
about definitions here.

And your formula above is not correct. It can not be correct. It is
nonsense.

F(x) is a quadratic in x. When summed over an arithmetic progression
of difference 1, i.e. (P, P-1, P-2, .....1) the result will be a cubic polynomial
in P. But the number of terms in the summation formula is not a cubic
function of P. Indeed. It is not even a polynomial in P.

I will give a hint: The number of different products of roots taken r
at a time from a set of size n is a well known COMBINATORIAL object.[/quote]
I defined F(x) for every x (it does not concern f(x)).
I know, it both concern x (this is my mistake).
I'll define y(y+1)/2=F(y) for any natural y.