# Product of these sums is a sum of the same form...

• Jun 30th 2013, 04:38 PM
Leviathantheesper
Product of these sums is a sum of the same form...
Hi

This time I'd like to know how to show this:

Attachment 28706

Thanks for the help.
(Hi)

I edit: This is my first attempt of proof...but I'm not too happy with it...it seems to work, but it's too vague...

Attachment 28708
• Jun 30th 2013, 05:38 PM
chiro
Re: Product of these sums is a sum of the same form...
Hey Leviathantheesper.

Is y a constant for all k or does it change for a specific k?
• Jul 1st 2013, 06:25 AM
HallsofIvy
Re: Product of these sums is a sum of the same form...
The way you have written it, with y constant, it is not true. (Hence chiro's question.) Rather you need something like
$\displaystyle \sum_{k=0}^n a_kx^k \sum_{k= 0}^n b_kx^k= \sum_{k= 0}^{2n} y_kx^k$

That is, you need that "y" different for each power of x and, obviously, if you are multiplying two nth degree polynomials together, you do NOT get an nthe degree polynomial, you get a 2n degree polynomial:

(a+ bx)(c+ dx)= ac+ (ad+ bc)x+ (bd)x^2.
• Jul 1st 2013, 09:56 AM
Leviathantheesper
Re: Product of these sums is a sum of the same form...
Quote:

Originally Posted by HallsofIvy
The way you have written it, with y constant, it is not true. (Hence chiro's question.) Rather you need something like
$\displaystyle \sum_{k=0}^n a_kx^k \sum_{k= 0}^n b_kx^k= \sum_{k= 0}^{2n} y_kx^k$

That is, you need that "y" different for each power of x and, obviously, if you are multiplying two nth degree polynomials together, you do NOT get an nthe degree polynomial, you get a 2n degree polynomial:

(a+ bx)(c+ dx)= ac+ (ad+ bc)x+ (bd)x^2.

Yes, y is not a constant, that was my mistake.

And...I know i don't get a nth degree polynomial... I edited some time ago to change the picture and replace the n with infinite (I replaced it yesterday)...
And the only thing I have to prove is that the result is a sum of the same form.
This, correcting the $\displaystyle y$ with $\displaystyle y_k$
http://mathhelpforum.com/attachments...e-form-asd.png
• Jul 1st 2013, 10:45 AM
HallsofIvy
Re: Product of these sums is a sum of the same form...
Again, that certainly is NOT true! Unless you mean, as both chiro and I said before, you mean $\displaystyle \sum_{k=0}^\infty y_kx^k$.

And, in that case, it is pretty close to trivial. The product of any two terms, $\displaystyle a_ix^i$ times $\displaystyle b_jx^j$ must be of the form $\displaystyle a_ib_jx^{j+i}$ so the only thing you can have are "numbers times non-negative powers of x" and that is all the term "$\displaystyle \sum_{k=0}^\infty y_kx^k$ means!
• Jul 1st 2013, 10:52 AM
Leviathantheesper
Re: Product of these sums is a sum of the same form...
Quote:

Originally Posted by HallsofIvy
Again, that certainly is NOT true! Unless you mean, as both chiro and I said before, you mean $\displaystyle \sum_{k=0}^\infty y_kx^k$.

And, in that case, it is pretty close to trivial. The product of any two terms, $\displaystyle a_ix^i$ times $\displaystyle b_jx^j$ must be of the form $\displaystyle a_ib_jx^{j+i}$ so the only thing you can have are "numbers times non-negative powers of x" and that is all the term "$\displaystyle \sum_{k=0}^\infty y_kx^k$ means!

I have just said:
This, correcting the $\displaystyle y$ with $\displaystyle y_k$

And thanks, that's it...