[QUOTE=blob100;220011]For B a given non-empty set |B| will denote the large of the set (the number of objects in the set).
I have another approach to this proposition's proof (much easier):
We will first prove the next lemma:
For a given set A, |A|=n, and we define k as a natural number 0=/<k</=n.
The number of sequences which contain k objects out of A's objects is:
n(n-1)(n-2)....(n-k+1)=n!/(n-k)!.

Proof:
We see that for the first objects in the sequence we have n options to choose (there are n objects not contained in the sequence).
The next object is able to be n-1 options,
The next is n-2...
When we will choose the object number k (the last object in the sequence) will have n-k+1 options to choose.
This gives n(n-1)...(n-k+1)=n!/(n-k)!.

[/QUOTE]

Correct, so far

[QUOTE]

This last theorem gives the approach to prove our problem:


We firstly see that the number of promotations of every sequence is k!
[/QUOTE]

What is a "promotion"? Stop inventing terminolgy.

I deleted the rest. It was so poorly worded as to make no sense.
Proving that you must now divide by k! is simple, but you first need
to prove that the number of different arrangements of k items is k!

[QUOTE=R.D. Silverman;220014]
What is a "promotion"? Stop inventing terminology.[/QUOTE]

Bob,

I'm sure that he meant "Permutation". I don't know if the error occurred because of a language barrier or careless typing?

[QUOTE=blob100;220011]We firstly see that the number of promotations of every sequence is k!
*I can prove this easily, but it does not seem to be nessecary.[/QUOTE]
necessary
[QUOTE=R.D. Silverman;220014]but you first need to prove that the number of different arrangements of k items is k![/QUOTE]
I will agree with Bob that you really do need to supply this portion of the proof.

[quote=Wacky;220016]Bob,

I'm sure that he meant "Permutation". I don't know if the error occurred because of a language barrier or careless typing?


necessary

I will agree with Bob that you really do need to supply this portion of the proof.[/quote]

Yes, for the proof of our proposition I had problem expressing myself.
Ok, to write the Hebrew taken from English word permutation and what I have written (promotation which does not exist, this made me to think it is an individual mathematical word) are written in Hrebew the same.
For two different words with different pronouncements, same word writing.

[quote=R.D. Silverman;220014]Correct, so far



What is a "promotion"? Stop inventing terminolgy.

I deleted the rest. It was so poorly worded as to make no sense.
Proving that you must now divide by k! is simple, but you first need
to prove that the number of different arrangements of k items is k![/quote]

I'll try again expressing myself.
If the number of permutations of a sequence of k out of n objects is k!,
For the same k objects we can build just one set (these objects are the same as in the sequences).
This set can be found as a combination of k objects out of n objects.
This gives that for every k! sequences there is just one such a combination. This is equivallent to say (the number of sequences)/k!.
The number of sequences was discussed before as n!/(n-k)!,
These two results (the formula and the ratio between the sequneces and combinations) direct us to the next n!/(n-k)!k!=(the number of combinations of k objects out of n objects).

[quote=Wacky;220016]




necessary

.[/quote]

I'll show that the number of permutations of a sequence is k!,
Where the sequence includes k objects.
The number of objects in a given permutation is as we defined k. The first objects in a permutation can be chosen as k different objects,
The next one can be k-1 options,..., and the number k will be able to be just one object.
This gives that the number of permutations is k!.

[quote=blob100;220039]I'll show that the number of permutations of a sequence is k!,
Where the sequence includes k objects.
The number of objects in a given permutation is as we defined k. The first objects in a permutation can be chosen as k different objects,
The next one can be k-1 options,..., and the number k will be able to be just one object.
This gives that the number of permutations is k!.[/quote]
I think, it is not finished:
I was needed to write: "this gives that the number of permutations is:
k(k-1)(k-2)...1=k!."

From now, what to do?

[QUOTE=blob100;220137]From now, what to do?[/QUOTE]

You still haven't applied what you have now learned about combinations ([sub]n[/sub]C[sub]k[/sub]) to the polynomial coefficient problem (#2 of Silverman's polynomial problems). Can you state and prove a simple formula for the coefficient of x^k?

[quote=jyb;220185]You still haven't applied what you have now learned about combinations ([sub]n[/sub]C[sub]k[/sub]) to the polynomial coefficient problem (#2 of Silverman's polynomial problems). Can you state and prove a simple formula for the coefficient of x^k?[/quote]
Of course I will do so, but first I want Silverman to agree with what I have done till now. This is what tutoring means.

[COLOR=black][FONT=Verdana]Now, for the primary problem:[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]We try to prove that the coefficient a_k of x^k in [/FONT][/COLOR]
[COLOR=black][FONT=Verdana]f(x)=x^n+a_(n-1)x^(n-1)+...+a_0 (a polynomial of degree n) is [/FONT][/COLOR]
[COLOR=black][FONT=Verdana]a_k=((-1)^(n-k))(c1+c2+...+c(n_C_n-k)).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]A={a_0,a_1,...,a_n} the set of the coefficients of the polynomial f(x).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]B={c1,c2,...,c(n_C_k)} is defined as the set of all product combinations of k roots out of n roots.[/FONT][/COLOR]
We of course define n_C_k as the number of combinations of k objects out of n objects (number of sub-sets of a set of n objects).
We will use the result of the value of n_C_k, so we must assume it first:
n_C_k=n!/(n-k)!k!.

[COLOR=black][FONT=Verdana]We will prove this by induction on n:[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]I'll repeat the proposition: [/FONT][/COLOR]
[COLOR=black][FONT=Verdana]a_k=((-1)^(n-k))(c1+c2+...+c(n_C_(n-k))).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]*We define R as R=(c1+c2+...+c(n_C_(n-k)) and we say:[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]If n-k=0 then R=1.[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]*a_k=((-1)^(n-k))(R) is our proposition.[/FONT][/COLOR]


[COLOR=black][FONT=Verdana]Step 1:[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]We show that for a polynomial of n=1 the proposition is true:[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]f(x)=(x-x1)[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]Where D={x1,x2,...,xn} the set of all roots of a polynomial of degree n.[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]We see that the coefficient of x^0 by the formula is -x1=((-1)^(1-0))(x1) which fits the reality.[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]For the coefficient of x^1 by the formula we get ((-1)^(1-1))(1)[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]We see that R=1 just becuase n-k=0 (again, it fits the reality).[/FONT][/COLOR]

[COLOR=black][FONT=Verdana]Step 2:[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]We assume the proposition is true for n.[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]From here we will try proving the proposition is true for n+1.[/FONT][/COLOR]

[COLOR=black][FONT=Verdana]Step 3:[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]If we had a polynomial f(x)=a_nx^n+...+a_0,[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]From here, the approach will be: proving the coefficients of [/FONT][/COLOR]
[COLOR=black][FONT=Verdana]f(x)(x-x(n+1)) fit the proposition.[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]*x(n+1) is a root of the polynomial f(x)(x-x(n+1)) not a product between two factors.[/FONT][/COLOR]
[COLOR=black][FONT=Verdana](a_nx^n+...+a_0)(x-x(n+1))=f(x)(x-x(n+1)).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]We see that f(x)(x-x(n+1))=[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]a_(n+1)x^(n+1)+...+a_(k-1)x^(k-1)*x+((a_k)x^k)(-x(n+1))+...+a_0.[/FONT][/COLOR]

[COLOR=black][FONT=Verdana]We, from now, concern to the phrase a_(k-1)x^(k-1)*x+((a_k)x^k)(-x(n+1)).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]This last phrase means the coefficient of x^k in f(x)(x-x(n+1)).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]And by this last result it is: a_(k-1)+a_k(-x(n+1)).[/FONT][/COLOR]

[COLOR=black][FONT=Verdana]We will show every part of this coefficient in it's most helpful form:[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]a_(k-1)=((-1)^(n-k+1))(C1+C2+...+C(n_C_(n-k+1)))=[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]((-1)^(n-k+1))(G).[/FONT][/COLOR]
Where E={C1,C2,...,C(n_C_(n-k+1))} the set of all product combinations of n-k+1 roots out of n roots.
[COLOR=black][FONT=Verdana]*We define G as C1+C2+...+C(n_C_(n-k+1)).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]a_k=((-1)^(n-k))(c1+c2+...+c(n_C_(n-k)))=((-1)^(n-k))(R).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]*We will again discuss the value of R: R=(c1+c2+...+c(n_C_(n-k)).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]We must see the difference between G and R, by the number of such product combinations and their number of root factors.[/FONT][/COLOR]

[COLOR=black][FONT=Verdana]a_k(-x(n+1))=((-1)^(n-k))(R)(-x(n+1))=((-1)^(n-k+1))(R(x(n+1)))).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]R(-x(n+1)) is of course a sum of product combinations, each n-k+1 roots (similar to G's product combinations).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]a_(k-1)+a_k(-x(n+1))=((-1)^(n-k+1))(G+R(x(n+1))).[/FONT][/COLOR]


[COLOR=black][FONT=Verdana]Now, we will try to verify ((-1)^(n-k+1))(G+R(x(n+1))) fits the proposition.[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]By the proposition:[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]The coefficient of x^k in the new polynomial is:[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]((-1)^(n+1-k))(M)[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]Where M is the sum of the n_C_(n+1-k) product combinations of n+1-k roots.[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]We now try to prove M=G+R(x(n+1)).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]G and R(x(n+1)) are two sums of different product combinations of n-k+1 [/FONT][/COLOR]
[COLOR=black][FONT=Verdana]roots.[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]From here, things seem pretty trivial (by the definitions of R,M,G).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]R(x(n+1))+G is the sum of the n!/(k-1)!(n-k+1)!+n!/(n-k)!k! product combinations of the n+1 roots. [/FONT][/COLOR]
[COLOR=black][FONT=Verdana]M is the sum of the [U]same[/U] n!/(n-k+1)!k! product combinations of the n+1 roots.[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]By Pascal's rule (or simple algebra) [/FONT][/COLOR]
[COLOR=black][FONT=Verdana]n!/(n-k+1)!k!=n!/(k-1)!(n-k+1)!+n!/(n-k)!k! is true,[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]And this completes the proof of M=G+R(x(n+1)).[/FONT][/COLOR]

[quote=blob100;220216][COLOR=black][FONT=Verdana]Now, for the primary problem:[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]We try to prove that the coefficient a_k of x^k in [/FONT][/COLOR]
[COLOR=black][FONT=Verdana]f(x)=x^n+a_(n-1)x^(n-1)+...+a_0 (a polynomial of degree n) is [/FONT][/COLOR]
[COLOR=black][FONT=Verdana]a_k=((-1)^(n-k))(c1+c2+...+c(n_C_n-k)).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]A={a_0,a_1,...,a_n} the set of the coefficients of the polynomial f(x).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]B={c1,c2,...,c(n_C_k)} is defined as the set of all product combinations of k roots out of n roots.[/FONT][/COLOR]
We of course define n_C_k as the number of combinations of k objects out of n objects (number of sub-sets of a set of n objects).
We will use the result of the value of n_C_k, so we must assume it first:
n_C_k=n!/(n-k)!k!.

[COLOR=black][FONT=Verdana]We will prove this by induction on n:[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]I'll repeat the proposition: [/FONT][/COLOR]
[COLOR=black][FONT=Verdana]a_k=((-1)^(n-k))(c1+c2+...+c(n_C_(n-k))).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]*We define R as R=(c1+c2+...+c(n_C_(n-k)) and we say:[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]If n-k=0 then R=1.[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]*a_k=((-1)^(n-k))(R) is our proposition.[/FONT][/COLOR]


[COLOR=black][FONT=Verdana]Step 1:[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]We show that for a polynomial of n=1 the proposition is true:[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]f(x)=(x-x1)[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]Where D={x1,x2,...,xn} the set of all roots of a polynomial of degree n.[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]We see that the coefficient of x^0 by the formula is -x1=((-1)^(1-0))(x1) which fits the reality.[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]For the coefficient of x^1 by the formula we get ((-1)^(1-1))(1)[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]We see that R=1 just becuase n-k=0 (again, it fits the reality).[/FONT][/COLOR]

[COLOR=black][FONT=Verdana]Step 2:[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]We assume the proposition is true for n.[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]From here we will try proving the proposition is true for n+1.[/FONT][/COLOR]

[COLOR=black][FONT=Verdana]Step 3:[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]If we had a polynomial f(x)=a_nx^n+...+a_0,[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]From here, the approach will be: proving the coefficients of [/FONT][/COLOR]
[COLOR=black][FONT=Verdana]f(x)(x-x(n+1)) fit the proposition.[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]*x(n+1) is a root of the polynomial f(x)(x-x(n+1)) not a product between two factors.[/FONT][/COLOR]
[COLOR=black][FONT=Verdana](a_nx^n+...+a_0)(x-x(n+1))=f(x)(x-x(n+1)).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]We see that f(x)(x-x(n+1))=[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]a_(n+1)x^(n+1)+...+a_(k-1)x^(k-1)*x+((a_k)x^k)(-x(n+1))+...+a_0.[/FONT][/COLOR]

[COLOR=black][FONT=Verdana]We, from now, concern to the phrase a_(k-1)x^(k-1)*x+((a_k)x^k)(-x(n+1)).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]This last phrase means the coefficient of x^k in f(x)(x-x(n+1)).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]And by this last result it is: a_(k-1)+a_k(-x(n+1)).[/FONT][/COLOR]

[COLOR=black][FONT=Verdana]We will show every part of this coefficient in it's most helpful form:[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]a_(k-1)=((-1)^(n-k+1))(C1+C2+...+C(n_C_(n-k+1)))=[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]((-1)^(n-k+1))(G).[/FONT][/COLOR]
Where E={C1,C2,...,C(n_C_(n-k+1))} the set of all product combinations of n-k+1 roots out of n roots.
[COLOR=black][FONT=Verdana]*We define G as C1+C2+...+C(n_C_(n-k+1)).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]a_k=((-1)^(n-k))(c1+c2+...+c(n_C_(n-k)))=((-1)^(n-k))(R).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]*We will again discuss the value of R: R=(c1+c2+...+c(n_C_(n-k)).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]We must see the difference between G and R, by the number of such product combinations and their number of root factors.[/FONT][/COLOR]

[COLOR=black][FONT=Verdana]a_k(-x(n+1))=((-1)^(n-k))(R)(-x(n+1))=((-1)^(n-k+1))(R(x(n+1)))).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]R(-x(n+1)) is of course a sum of product combinations, each n-k+1 roots (similar to G's product combinations).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]a_(k-1)+a_k(-x(n+1))=((-1)^(n-k+1))(G+R(x(n+1))).[/FONT][/COLOR]


[COLOR=black][FONT=Verdana]Now, we will try to verify ((-1)^(n-k+1))(G+R(x(n+1))) fits the proposition.[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]By the proposition:[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]The coefficient of x^k in the new polynomial is:[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]((-1)^(n+1-k))(M)[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]Where M is the sum of the n_C_(n+1-k) product combinations of n+1-k roots.[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]We now try to prove M=G+R(x(n+1)).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]G and R(x(n+1)) are two sums of different product combinations of n-k+1 [/FONT][/COLOR]
[COLOR=black][FONT=Verdana]roots.[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]From here, things seem pretty trivial (by the definitions of R,M,G).[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]R(x(n+1))+G is the sum of the n!/(k-1)!(n-k+1)!+n!/(n-k)!k! product combinations of the n+1 roots. [/FONT][/COLOR]
[COLOR=black][FONT=Verdana]M is the sum of the [U]same[/U] n!/(n-k+1)!k! product combinations of the n+1 roots.[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]By Pascal's rule (or simple algebra) [/FONT][/COLOR]
[COLOR=black][FONT=Verdana]n!/(n-k+1)!k!=n!/(k-1)!(n-k+1)!+n!/(n-k)!k! is true,[/FONT][/COLOR]
[COLOR=black][FONT=Verdana]And this completes the proof of M=G+R(x(n+1)).[/FONT][/COLOR][/quote]

Anyone would like to help here?