### Autor Tema: Polynomials

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09 Febrero, 2020, 12:29 pm
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#### jacks

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##### Polynomials
Finding coefficient of $x^{97}$ in $\displaystyle \bigg(x-\binom{99}{0}\bigg)\bigg(x-\binom{99}{1}\bigg)\cdots\cdots\bigg(x-\binom{99}{98}\bigg)\bigg(x-\binom{99}{99}\bigg)$

10 Febrero, 2020, 10:05 am
Respuesta #1

#### Luis Fuentes

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##### Re: Polynomials
Hi

Finding coefficient of $x^{97}$ in $\displaystyle \bigg(x-\binom{99}{0}\bigg)\bigg(x-\binom{99}{1}\bigg)\cdots\cdots\bigg(x-\binom{99}{98}\bigg)\bigg(x-\binom{99}{99}\bigg)$

Call $a_k=\displaystyle\binom{n}{k}$.

The coefficiente of degrer $n-2$ of the polynomial of degree $n+1$:

$p(x)=\displaystyle\prod_{k=0}^n(x-a_k)$

is the 3rd elementary symmetric polynomial on $a_0,a_1,\ldots,a_n$:

$\displaystyle\sum_{i=0}^n{}\displaystyle\sum_{j=i+1}^n{}\displaystyle\sum_{k=j+1}^n{}a_ia_ja_k$

But:

$\displaystyle\sum_{i=0}^n{}\displaystyle\sum_{j=i+1}^n{}\displaystyle\sum_{k=j+1}^n{}a_ia_ja_k=\dfrac{1}{6}\left((a_0+a_1+\ldots+a_n)^3-3(a_0^2+a_1^2+\ldots+a_n^2)(a_0+a_1+\ldots+a_n)+2(a_0^3+a_1^3+\ldots+a_n^3)\right)$   (*)

$a_0+a_1+\ldots+a_n=\displaystyle\sum_{k=0}^n{}\displaystyle\binom{n}{k}=2^n$
$a_0^2+a_1^2+\ldots+a_n^2=\displaystyle\sum_{k=0}^n{}\displaystyle\binom{n}{k}^2=\displaystyle\binom{2n}{n}$

But to my knowledge there is no a closed form for (look here):

$a_0^3+a_1^3+\ldots+a_n^3=\displaystyle\sum_{k=0}^n{}\displaystyle\binom{n}{k}^3$

Best regards.

P.S. jack: Where did you find this problem?

12 Febrero, 2020, 02:21 pm
Respuesta #2

#### jacks

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##### Re: Polynomials

It was given by my friend

i still have a doubt i have seems he mean Coefficients of $x^{98}$ in given expression

12 Febrero, 2020, 05:02 pm
Respuesta #3

#### Luis Fuentes

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##### Re: Polynomials
Hola

i still have a doubt i have seems he mean Coefficients of $x^{98}$ in given expression

¡Ah! This can be done.

Now the coefficiente of degrer $n-1$ of the polynomial of degree $n+1$:

$p(x)=\displaystyle\prod_{k=0}^n(x-a_k)$

is the 2rd elementary symmetric polynomial on $a_0,a_1,\ldots,a_n$:

$\displaystyle\sum_{i=0}^n{}\displaystyle\sum_{j=i+1}^n{}a_ia_j$

But:

$\displaystyle\sum_{i=0}^n{}\displaystyle\sum_{j=i+1}^n{}a_ia_j=\dfrac{1}{2}\left((a_0+a_1+\ldots+a_n)^2-(a_0^2+a_1^2+\ldots+a_n^2)\right)$

and

$a_0+a_1+\ldots+a_n=\displaystyle\sum_{k=0}^n{}\displaystyle\binom{n}{k}=2^n$
$a_0^2+a_1^2+\ldots+a_n^2=\displaystyle\sum_{k=0}^n{}\displaystyle\binom{n}{k}^2=\displaystyle\binom{2n}{n}$

Best regards.

14 Febrero, 2020, 03:53 pm
Respuesta #4

#### jacks

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##### Re: Polynomials
Thanks so much admin got it.