how to know if my ideas didnt tought before?

[R.D. Silverman]

Originally Posted by blob100

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.

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?

[blob100]

Originally Posted by R.D. Silverman

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?

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.

[jyb]

Originally Posted by blob100

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.

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?

[blob100]

Originally Posted by jyb

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?

Sorry, I gave the wrong answer.

[R.D. Silverman]

Originally Posted by blob100

Sorry, I gave the wrong answer.

Check your results before posting.

[blob100]

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)?

[R.D. Silverman]

Originally Posted by blob100

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)?

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

[blob100]

Originally Posted by R.D. Silverman

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

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????".

[R.D. Silverman]

Originally Posted by blob100

What is unable to be understood?
I'll try again:
n-k=P. And we (arbitarically) assume k>P.

Why this assumption?

F(x)=(x^2+x)/2 (known as the Fermat formula).

(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 redefined 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 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.

We are not discussing the number of terms in the coefficient of x^k.
We are 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.

[R.D. Silverman]

Originally Posted by R.D. Silverman

\
<snip>


.

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.

[blob100]

Originally Posted by R.D. Silverman

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 redefined 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 are 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.

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.