That is also equal to 0. Can you explain why?

[76Ahmad]

Prove of summation claim ??

Hi every one,,

any idea how to prove the following claim

\sum_{i=0}^{n}a_iz^i=(1-z)^{\binom{m}{2}}(1+z)^{\binom{m+1}{2}}




i think we need to use some derivatives, may be the second derivative will help.


please help.

 

[micromass]

Please tell us what a_i is, what n is, what m is and what z is.

When communicating with others it is crucial to give all the information.

 

[76Ahmad]

Yes sure,,

a_i is the coefficient of z.

m is the integer power of z where it goes ++ to n.

Thanks

 

[micromass]

Sigh...

What IS a_i??

Right now you are asking:

Prove that a+b=c, but I won't tell you what a,b and c are.

Please state the FULL problem.

 

[76Ahmad]

OK,, :)

Let


n=m^2 and\,\,\, s=\binom{m}{2}, n-s=\binom{m+1}{2}

prove that the following claim is true

\sum_{i=0}^{n}a_iz^i=(1-z)^{\binom{m}{2}}(1+z)^{\binom{m+1}{2}}

 

[76Ahmad]

prove that this polynomials has only 3 zero coefficient, and they are

a_2, and a_{m^2-2}, and

a_{\frac{m^2}{2}} if m=2 mod 4.

 

[chiro]

I think micromass is asking things like the structure.

For example are they real numbers? complex numbers? integers? prime integers?

Mathematicians usually use sets to describe these numbers and it will probably be beneficial for you to do the same because it will help you both in a) reading other mathematical work and b) get you thinking in the right way to do mathematics.

 

[micromass]

I think micromass is asking things like the structure.

No, I'm just wanting to know how the a_i are defined. Am I the only one who sees a problem with the OP?? Hmmm, I'll let others solve the question since I'm confused...

 

[Mark44]

OK,, :)

Let


n=m^2 and\,\,\, s=\binom{m}{2}, n-s=\binom{m+1}{2}

prove that the following claim is true

\sum_{i=0}^{n}a_iz^i=(1-z)^{\binom{m}{2}}(1+z)^{\binom{m+1}{2}}

No, I'm just wanting to know how the a_i are defined. Am I the only one who sees a problem with the OP?? Hmmm, I'll let others solve the question since I'm confused...

micromass, no, you aren't the only one who sees a problem.

76Ahmed, no one can help you if until you tell us what the values[/color] of the coefficients a_i are.

 

[Norwegian]

The a_i are of course the coefficients of the given polynomial in Z[z], and the question is about which of these are zero.

To prove that a_2 is zero, just use the first 3 terms in each factor, and multiply, and you get the coefficient of z². You need to prove the identity (m 2)(m+1 2) = ((m 2) 2)+((m+1 2) 2) of binomials to get there.

There is also a, not too illuminating, combinatorial way of showing this.

The other zeros follows since the coefficients are (anti-)symmetric.

To prove that no other a_i is zero may be the hard part. It may also be that you are not really asking about that.

 

[76Ahmad]

micromass + Mark44

I do not know what the is difficulty of a_i that is facing you both !
Knowing that I told you before a_i is the coefficient of z.

if you still have a problem with a_i, just put m = 3 for example.

you will get

1 + 3 z - 8 z^3 - 6 z^4 + 6 z^5 + 8 z^6 - 3 z^8 - z^9
then you see
a_0 = 1
a_1 = 3
a_2 = 0
a_3 = -8 ... a_9 = -1DID YOU GET IT.

 

[76Ahmad]

Thanks Norwegian..

I think you are right, can you explain in detail please.

you can see that
|a_2|=|a_{m^2-2}|

so if we prove that a_2 = 0, then a_{m^2-2} is also = 0.

what about
a_{\frac{m^2}{2}}