Autor Tema: Polynomials

0 Usuarios y 1 Visitante están viendo este tema.

09 Febrero, 2020, 12:29 pm
Leído 545 veces

jacks

  • Experto
  • Mensajes: 672
  • País: in
  • Karma: +0/-0
  • Sexo: Masculino
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

  • el_manco
  • Administrador
  • Mensajes: 46,785
  • País: es
  • Karma: +1/-0
  • Sexo: Masculino
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

  • Experto
  • Mensajes: 672
  • País: in
  • Karma: +0/-0
  • Sexo: Masculino
Thanks Admin got.

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

  • el_manco
  • Administrador
  • Mensajes: 46,785
  • País: es
  • Karma: +1/-0
  • Sexo: Masculino
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

  • Experto
  • Mensajes: 672
  • País: in
  • Karma: +0/-0
  • Sexo: Masculino
Thanks so much admin got it.