Writing polynomial as a product of elementary symmetric polynomials
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Write $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2 $ as a product of elementary symmetric polynomial
I get $E1=x+y+z$, $E2=xy+xz+yz$, $E3=xyz$.
I've tried factoring out E3(xyz) but I can tell that's not right. I know this probably isn't that difficult, think I'm just going about it the wrong way.
Please help!!
abstract-algebra symmetric-polynomials
asked Nov 22 at 19:32
11276
34
|
show 1 more comment
up vote
0
down vote
favorite
Write $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2 $ as a product of elementary symmetric polynomial
I get $E1=x+y+z$, $E2=xy+xz+yz$, $E3=xyz$.
I've tried factoring out E3(xyz) but I can tell that's not right. I know this probably isn't that difficult, think I'm just going about it the wrong way.
Please help!!
abstract-algebra symmetric-polynomials
asked Nov 22 at 19:32
11276
34
That can't be done. The best you can do is to write your polynomial as a sum of products of elementary symmetric polynomials.
– José Carlos Santos
Nov 22 at 19:37
why do you have only five terms? you are missing $x^2 z$
– Will Jagy
Nov 22 at 19:39
@WillJagy You're right, my bad. I've made the edit to include it now. Thanks!
– 11276
Nov 22 at 19:54
After your edit, the sum is symmetric and we find easily that $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2 = E_{2}*E_{1} - 3 E_3$
– Dr. Wolfgang Hintze
Nov 22 at 19:54
@Dr.WolfgangHintze Thanks for the answer but I still can't manipulate it to... (xy+xz+yz)(x+y+z)-3(xyz). Could you show me the workings please?
– 11276
Nov 22 at 20:01
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Write $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2 $ as a product of elementary symmetric polynomial
I get $E1=x+y+z$, $E2=xy+xz+yz$, $E3=xyz$.
I've tried factoring out E3(xyz) but I can tell that's not right. I know this probably isn't that difficult, think I'm just going about it the wrong way.
Please help!!
abstract-algebra symmetric-polynomials
asked Nov 22 at 19:32
11276
34
Write $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2 $ as a product of elementary symmetric polynomial
I get $E1=x+y+z$, $E2=xy+xz+yz$, $E3=xyz$.
I've tried factoring out E3(xyz) but I can tell that's not right. I know this probably isn't that difficult, think I'm just going about it the wrong way.
Please help!!
abstract-algebra symmetric-polynomials
abstract-algebra symmetric-polynomials
asked Nov 22 at 19:32
11276
34
asked Nov 22 at 19:32
11276
34
edited Nov 22 at 19:53
asked Nov 22 at 19:32
11276
34
asked Nov 22 at 19:32
11276
34
asked Nov 22 at 19:32
11276
34
34
That can't be done. The best you can do is to write your polynomial as a sum of products of elementary symmetric polynomials.
– José Carlos Santos
Nov 22 at 19:37
why do you have only five terms? you are missing $x^2 z$
– Will Jagy
Nov 22 at 19:39
@WillJagy You're right, my bad. I've made the edit to include it now. Thanks!
– 11276
Nov 22 at 19:54
After your edit, the sum is symmetric and we find easily that $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2 = E_{2}*E_{1} - 3 E_3$
– Dr. Wolfgang Hintze
Nov 22 at 19:54
@Dr.WolfgangHintze Thanks for the answer but I still can't manipulate it to... (xy+xz+yz)(x+y+z)-3(xyz). Could you show me the workings please?
– 11276
Nov 22 at 20:01
|
show 1 more comment
That can't be done. The best you can do is to write your polynomial as a sum of products of elementary symmetric polynomials.
– José Carlos Santos
Nov 22 at 19:37
why do you have only five terms? you are missing $x^2 z$
– Will Jagy
Nov 22 at 19:39
@WillJagy You're right, my bad. I've made the edit to include it now. Thanks!
– 11276
Nov 22 at 19:54
After your edit, the sum is symmetric and we find easily that $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2 = E_{2}*E_{1} - 3 E_3$
– Dr. Wolfgang Hintze
Nov 22 at 19:54
@Dr.WolfgangHintze Thanks for the answer but I still can't manipulate it to... (xy+xz+yz)(x+y+z)-3(xyz). Could you show me the workings please?
– 11276
Nov 22 at 20:01
That can't be done. The best you can do is to write your polynomial as a sum of products of elementary symmetric polynomials.
– José Carlos Santos
Nov 22 at 19:37
That can't be done. The best you can do is to write your polynomial as a sum of products of elementary symmetric polynomials.
– José Carlos Santos
Nov 22 at 19:37
why do you have only five terms? you are missing $x^2 z$
– Will Jagy
Nov 22 at 19:39
why do you have only five terms? you are missing $x^2 z$
– Will Jagy
Nov 22 at 19:39
@WillJagy You're right, my bad. I've made the edit to include it now. Thanks!
– 11276
Nov 22 at 19:54
@WillJagy You're right, my bad. I've made the edit to include it now. Thanks!
– 11276
Nov 22 at 19:54
After your edit, the sum is symmetric and we find easily that $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2 = E_{2}*E_{1} - 3 E_3$
– Dr. Wolfgang Hintze
Nov 22 at 19:54
After your edit, the sum is symmetric and we find easily that $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2 = E_{2}*E_{1} - 3 E_3$
– Dr. Wolfgang Hintze
Nov 22 at 19:54
@Dr.WolfgangHintze Thanks for the answer but I still can't manipulate it to... (xy+xz+yz)(x+y+z)-3(xyz). Could you show me the workings please?
– 11276
Nov 22 at 20:01
@Dr.WolfgangHintze Thanks for the answer but I still can't manipulate it to... (xy+xz+yz)(x+y+z)-3(xyz). Could you show me the workings please?
– 11276
Nov 22 at 20:01
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
There's a unique way to write $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2$ as a polynomial in $E_1,E_2,E_3$. Note that $x^2y$ could nicely be obtained by expanding the product $(x+ldots)(xy+ldots)$, and indeed after expanding $E_1E_2$, we obtain the desired expression - plus $3xyz$. Hence
$$x^2y+xy^2+x^2z+xz^2+y^2z+yz^2=E_1E_2-3E_3.$$
answered Nov 22 at 20:08
Hagen von Eitzen
274k21266494
Thanks @Hagen. Should of noticedthat and expanded E1.E2 straight away. Appreciate the help!
– 11276
Nov 23 at 9:42
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
There's a unique way to write $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2$ as a polynomial in $E_1,E_2,E_3$. Note that $x^2y$ could nicely be obtained by expanding the product $(x+ldots)(xy+ldots)$, and indeed after expanding $E_1E_2$, we obtain the desired expression - plus $3xyz$. Hence
$$x^2y+xy^2+x^2z+xz^2+y^2z+yz^2=E_1E_2-3E_3.$$
answered Nov 22 at 20:08
Hagen von Eitzen
274k21266494
Thanks @Hagen. Should of noticedthat and expanded E1.E2 straight away. Appreciate the help!
– 11276
Nov 23 at 9:42
add a comment |
up vote
0
down vote
accepted
There's a unique way to write $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2$ as a polynomial in $E_1,E_2,E_3$. Note that $x^2y$ could nicely be obtained by expanding the product $(x+ldots)(xy+ldots)$, and indeed after expanding $E_1E_2$, we obtain the desired expression - plus $3xyz$. Hence
$$x^2y+xy^2+x^2z+xz^2+y^2z+yz^2=E_1E_2-3E_3.$$
answered Nov 22 at 20:08
Hagen von Eitzen
274k21266494
Thanks @Hagen. Should of noticedthat and expanded E1.E2 straight away. Appreciate the help!
– 11276
Nov 23 at 9:42
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
There's a unique way to write $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2$ as a polynomial in $E_1,E_2,E_3$. Note that $x^2y$ could nicely be obtained by expanding the product $(x+ldots)(xy+ldots)$, and indeed after expanding $E_1E_2$, we obtain the desired expression - plus $3xyz$. Hence
$$x^2y+xy^2+x^2z+xz^2+y^2z+yz^2=E_1E_2-3E_3.$$
answered Nov 22 at 20:08
Hagen von Eitzen
274k21266494
There's a unique way to write $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2$ as a polynomial in $E_1,E_2,E_3$. Note that $x^2y$ could nicely be obtained by expanding the product $(x+ldots)(xy+ldots)$, and indeed after expanding $E_1E_2$, we obtain the desired expression - plus $3xyz$. Hence
$$x^2y+xy^2+x^2z+xz^2+y^2z+yz^2=E_1E_2-3E_3.$$
answered Nov 22 at 20:08
Hagen von Eitzen
274k21266494
answered Nov 22 at 20:08
Hagen von Eitzen
274k21266494
answered Nov 22 at 20:08
Hagen von Eitzen
274k21266494
answered Nov 22 at 20:08
Hagen von Eitzen
274k21266494
274k21266494
Thanks @Hagen. Should of noticedthat and expanded E1.E2 straight away. Appreciate the help!
– 11276
Nov 23 at 9:42
add a comment |
Thanks @Hagen. Should of noticedthat and expanded E1.E2 straight away. Appreciate the help!
– 11276
Nov 23 at 9:42
Thanks @Hagen. Should of noticedthat and expanded E1.E2 straight away. Appreciate the help!
– 11276
Nov 23 at 9:42
Thanks @Hagen. Should of noticedthat and expanded E1.E2 straight away. Appreciate the help!
– 11276
Nov 23 at 9:42
add a comment |
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That can't be done. The best you can do is to write your polynomial as a sum of products of elementary symmetric polynomials.
– José Carlos Santos
Nov 22 at 19:37
why do you have only five terms? you are missing $x^2 z$
– Will Jagy
Nov 22 at 19:39
@WillJagy You're right, my bad. I've made the edit to include it now. Thanks!
– 11276
Nov 22 at 19:54
After your edit, the sum is symmetric and we find easily that $x^2y+xy^2+x^2z+xz^2+y^2z+yz^2 = E_{2}*E_{1} - 3 E_3$
– Dr. Wolfgang Hintze
Nov 22 at 19:54
@Dr.WolfgangHintze Thanks for the answer but I still can't manipulate it to... (xy+xz+yz)(x+y+z)-3(xyz). Could you show me the workings please?
– 11276
Nov 22 at 20:01