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Direct sum factorization of polynomials

Direct sum and subspacesFinding irreducible polynomials and factorizationLinear Algebra: Direct SumFactorization of Polynomials. Irreducible polynomial (basic question)On factorization of polynomialsFactorization of a PolynomialsUnderstanding Direct SumFactorization of polynomialsFactorization of polynomials.Factorization of polynomials

1

$begingroup$

• I have been recently reading the paper "Mixed finite elements for second order elliptic problems in three variables" by Brezzi et. al.


• I noticed the claim in the proof of Lemma $2.1$, which basically boils down to given a polynomial
  $mathrmpleft(x,yright) in P_k$, we can write it as $$
  mathrmpleft(x,yright) =
  cleft(1 - x - yright) + x,mathrmp_1left(x,yright) +
  y,mathrmp_2left(x,yright)
  $$
  for some $c$ constant and $mathrmp_1left(x,yright), mathrmp_2left(x,yright) in P_k - 1$. This is equivalent to the claim that
  $$
  P_k = xP_k - 1oplus yP_k - 1 oplus
  textSpanleft1 - x - yright
  $$ as far as I understand.


• It sounds plausible but I am not sure if it is true. Does anyone know more about this ?.

polynomials irreducible-polynomials direct-sum finite-element-method

share|cite|improve this question

edited Mar 22 at 5:03

Felix Marin

68.9k7110147

asked Mar 22 at 3:37

Abdullah Ali SivasAbdullah Ali Sivas

162

$endgroup$

add a comment | 

1

$begingroup$

• I have been recently reading the paper "Mixed finite elements for second order elliptic problems in three variables" by Brezzi et. al.


• I noticed the claim in the proof of Lemma $2.1$, which basically boils down to given a polynomial
  $mathrmpleft(x,yright) in P_k$, we can write it as $$
  mathrmpleft(x,yright) =
  cleft(1 - x - yright) + x,mathrmp_1left(x,yright) +
  y,mathrmp_2left(x,yright)
  $$
  for some $c$ constant and $mathrmp_1left(x,yright), mathrmp_2left(x,yright) in P_k - 1$. This is equivalent to the claim that
  $$
  P_k = xP_k - 1oplus yP_k - 1 oplus
  textSpanleft1 - x - yright
  $$ as far as I understand.


• It sounds plausible but I am not sure if it is true. Does anyone know more about this ?.

polynomials irreducible-polynomials direct-sum finite-element-method

share|cite|improve this question

edited Mar 22 at 5:03

Felix Marin

68.9k7110147

asked Mar 22 at 3:37

Abdullah Ali SivasAbdullah Ali Sivas

162

$endgroup$

add a comment | 

$begingroup$

• I have been recently reading the paper "Mixed finite elements for second order elliptic problems in three variables" by Brezzi et. al.


• I noticed the claim in the proof of Lemma $2.1$, which basically boils down to given a polynomial
  $mathrmpleft(x,yright) in P_k$, we can write it as $$
  mathrmpleft(x,yright) =
  cleft(1 - x - yright) + x,mathrmp_1left(x,yright) +
  y,mathrmp_2left(x,yright)
  $$
  for some $c$ constant and $mathrmp_1left(x,yright), mathrmp_2left(x,yright) in P_k - 1$. This is equivalent to the claim that
  $$
  P_k = xP_k - 1oplus yP_k - 1 oplus
  textSpanleft1 - x - yright
  $$ as far as I understand.


• It sounds plausible but I am not sure if it is true. Does anyone know more about this ?.

polynomials irreducible-polynomials direct-sum finite-element-method

share|cite|improve this question

edited Mar 22 at 5:03

Felix Marin

68.9k7110147

asked Mar 22 at 3:37

Abdullah Ali SivasAbdullah Ali Sivas

162

$endgroup$

share|cite|improve this question

edited Mar 22 at 5:03

Felix Marin

68.9k7110147

asked Mar 22 at 3:37

Abdullah Ali SivasAbdullah Ali Sivas

162

share|cite|improve this question

edited Mar 22 at 5:03

Felix Marin

68.9k7110147

asked Mar 22 at 3:37

Abdullah Ali SivasAbdullah Ali Sivas

162

edited Mar 22 at 5:03

Felix Marin

68.9k7110147

edited Mar 22 at 5:03

Felix Marin

68.9k7110147

asked Mar 22 at 3:37

Abdullah Ali SivasAbdullah Ali Sivas

162

asked Mar 22 at 3:37

Abdullah Ali SivasAbdullah Ali Sivas

162

1 Answer
1

active

oldest

votes

1

$begingroup$

Here, few days after I came up with this answer.

Let's consider the basis of the $P_k$, e.g. $ i,jgeq 0 text and i+jleq k $. Observe that we can write each function in this basis as given above

$ 1 = 1times (1-x-y) + xtimes 1 + ytimes 1 $

$ x^i = 0times (1-x-y) + xtimes x^i-1 + ytimes 0$, for $i>0$

$ y^j = 0times (1-x-y) + xtimes 0 + ytimes y^j-1$, for $j>0$

$ x^iy^j = 0times (1-x-y) + xtimes tfrac12x^i-1y^j + ytimes tfrac12x^i-1y^j-1$, for $j>0$ and $i>0$. Since sum of $k$-th degree polynomials is a $k$-th degree polynomial and $1$, $x^i-1$, $y^j-1$, $tfrac12x^i-1y^j$ and $tfrac12x^i-1y^j-1$ are $(k-1)$-st degree polynomials, we have the answer.

I am going to leave this here in case someone else (or future me) needs it.

share|cite|improve this answer

answered Mar 24 at 18:51

Abdullah Ali SivasAbdullah Ali Sivas

162

$endgroup$

add a comment | 

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1

$begingroup$

Here, few days after I came up with this answer.

Let's consider the basis of the $P_k$, e.g. $ i,jgeq 0 text and i+jleq k $. Observe that we can write each function in this basis as given above

$ 1 = 1times (1-x-y) + xtimes 1 + ytimes 1 $

$ x^i = 0times (1-x-y) + xtimes x^i-1 + ytimes 0$, for $i>0$

$ y^j = 0times (1-x-y) + xtimes 0 + ytimes y^j-1$, for $j>0$

$ x^iy^j = 0times (1-x-y) + xtimes tfrac12x^i-1y^j + ytimes tfrac12x^i-1y^j-1$, for $j>0$ and $i>0$. Since sum of $k$-th degree polynomials is a $k$-th degree polynomial and $1$, $x^i-1$, $y^j-1$, $tfrac12x^i-1y^j$ and $tfrac12x^i-1y^j-1$ are $(k-1)$-st degree polynomials, we have the answer.

I am going to leave this here in case someone else (or future me) needs it.

share|cite|improve this answer

answered Mar 24 at 18:51

Abdullah Ali SivasAbdullah Ali Sivas

162

$endgroup$

add a comment | 

1

$begingroup$

Here, few days after I came up with this answer.

Let's consider the basis of the $P_k$, e.g. $ i,jgeq 0 text and i+jleq k $. Observe that we can write each function in this basis as given above

$ 1 = 1times (1-x-y) + xtimes 1 + ytimes 1 $

$ x^i = 0times (1-x-y) + xtimes x^i-1 + ytimes 0$, for $i>0$

$ y^j = 0times (1-x-y) + xtimes 0 + ytimes y^j-1$, for $j>0$

$ x^iy^j = 0times (1-x-y) + xtimes tfrac12x^i-1y^j + ytimes tfrac12x^i-1y^j-1$, for $j>0$ and $i>0$. Since sum of $k$-th degree polynomials is a $k$-th degree polynomial and $1$, $x^i-1$, $y^j-1$, $tfrac12x^i-1y^j$ and $tfrac12x^i-1y^j-1$ are $(k-1)$-st degree polynomials, we have the answer.

I am going to leave this here in case someone else (or future me) needs it.

share|cite|improve this answer

answered Mar 24 at 18:51

Abdullah Ali SivasAbdullah Ali Sivas

162

$endgroup$

add a comment | 

$begingroup$

Here, few days after I came up with this answer.

Let's consider the basis of the $P_k$, e.g. $ i,jgeq 0 text and i+jleq k $. Observe that we can write each function in this basis as given above

$ 1 = 1times (1-x-y) + xtimes 1 + ytimes 1 $

$ x^i = 0times (1-x-y) + xtimes x^i-1 + ytimes 0$, for $i>0$

$ y^j = 0times (1-x-y) + xtimes 0 + ytimes y^j-1$, for $j>0$

$ x^iy^j = 0times (1-x-y) + xtimes tfrac12x^i-1y^j + ytimes tfrac12x^i-1y^j-1$, for $j>0$ and $i>0$. Since sum of $k$-th degree polynomials is a $k$-th degree polynomial and $1$, $x^i-1$, $y^j-1$, $tfrac12x^i-1y^j$ and $tfrac12x^i-1y^j-1$ are $(k-1)$-st degree polynomials, we have the answer.

I am going to leave this here in case someone else (or future me) needs it.

share|cite|improve this answer

answered Mar 24 at 18:51

Abdullah Ali SivasAbdullah Ali Sivas

162

$endgroup$

share|cite|improve this answer

answered Mar 24 at 18:51

Abdullah Ali SivasAbdullah Ali Sivas

162

answered Mar 24 at 18:51

Abdullah Ali SivasAbdullah Ali Sivas

162

answered Mar 24 at 18:51

Abdullah Ali SivasAbdullah Ali Sivas

162