# Prove of summation claim ?

1. Jan 19, 2012

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.

2. Jan 19, 2012

### micromass

Re: Prove of summation claim ??

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.

3. Jan 19, 2012

Re: Prove of summation claim ??

Yes sure,,

a_i is the coefficient of z.

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

Thanks

4. Jan 19, 2012

### micromass

Re: Prove of summation claim ??

Sigh...

What IS $a_i$??

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

5. Jan 19, 2012

Re: Prove of summation claim ??

OK,, :)

Let

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

prove that the following claim is true

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

6. Jan 19, 2012

Re: Prove of summation claim ??

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.

7. Jan 19, 2012

### chiro

Re: Prove of summation claim ??

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.

8. Jan 19, 2012

### micromass

Re: Prove of summation claim ??

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...

9. Jan 21, 2012

### Staff: Mentor

Re: Prove of summation claim ??

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 of the coefficients ai are.

10. Jan 22, 2012

### Norwegian

Re: Prove of summation claim ??

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 ai is zero may be the hard part. It may also be that you are not really asking about that.

11. Jan 22, 2012

Re: Prove of summation claim ??

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 = -1

DID YOU GET IT.

12. Jan 22, 2012

Re: Prove of summation claim ??

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.

$a_{\frac{m^2}{2}}$