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4

$begingroup$

I need to work with a variable $u$ such that $u^2 + u + 1 = 0$. I don't want to find a root of the polynomial $u^2 + u + 1$. Rather, I have to work with $u$ symbolically so that a (polynomial) expression in $u$ gets simplified using the equation $u^2 + u + 1 =0$.

For example, let

y = Series[u + 1 + u*x + x^2, x, 0, 4]
z = Series[u^2 + u^2*x + x^4, x, 0, 4]

Then, I'd expect

SeriesCoefficient[y+z, 0] = 0
SeriesCoefficient[y+z, 1] = -1

Thank you.

symbolic polynomials series-expansion

share|improve this question

asked May 13 at 15:08

MyathMyath

15444 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm a Python/R user, this is unintelligible to me, can you explain without Mathematica jargon please?
  $endgroup$
  – smci
  May 14 at 5:51
  
  

  
  
  
  

add a comment
 | 

4

$begingroup$

I need to work with a variable $u$ such that $u^2 + u + 1 = 0$. I don't want to find a root of the polynomial $u^2 + u + 1$. Rather, I have to work with $u$ symbolically so that a (polynomial) expression in $u$ gets simplified using the equation $u^2 + u + 1 =0$.

For example, let

y = Series[u + 1 + u*x + x^2, x, 0, 4]
z = Series[u^2 + u^2*x + x^4, x, 0, 4]

Then, I'd expect

SeriesCoefficient[y+z, 0] = 0
SeriesCoefficient[y+z, 1] = -1

Thank you.

symbolic polynomials series-expansion

share|improve this question

asked May 13 at 15:08

MyathMyath

15444 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm a Python/R user, this is unintelligible to me, can you explain without Mathematica jargon please?
  $endgroup$
  – smci
  May 14 at 5:51
  
  

  
  
  
  

add a comment
 | 

$begingroup$

I need to work with a variable $u$ such that $u^2 + u + 1 = 0$. I don't want to find a root of the polynomial $u^2 + u + 1$. Rather, I have to work with $u$ symbolically so that a (polynomial) expression in $u$ gets simplified using the equation $u^2 + u + 1 =0$.

For example, let

y = Series[u + 1 + u*x + x^2, x, 0, 4]
z = Series[u^2 + u^2*x + x^4, x, 0, 4]

Then, I'd expect

SeriesCoefficient[y+z, 0] = 0
SeriesCoefficient[y+z, 1] = -1

Thank you.

symbolic polynomials series-expansion

share|improve this question

asked May 13 at 15:08

MyathMyath

15444 bronze badges

$endgroup$

y = Series[u + 1 + u*x + x^2, x, 0, 4]
z = Series[u^2 + u^2*x + x^4, x, 0, 4]

SeriesCoefficient[y+z, 0] = 0
SeriesCoefficient[y+z, 1] = -1

share|improve this question

asked May 13 at 15:08

MyathMyath

15444 bronze badges

share|improve this question

asked May 13 at 15:08

MyathMyath

15444 bronze badges

asked May 13 at 15:08

MyathMyath

15444 bronze badges

asked May 13 at 15:08

MyathMyath

15444 bronze badges

3 Answers
3

active

oldest

votes

4

$begingroup$

You can use Assumptions

assume = u^2 + u + 1 == 0;

y = Series[u + 1 + u*x + x^2, x, 0, 4];
z = Series[u^2 + u^2*x + x^4, x, 0, 4];

Assuming[assume, SeriesCoefficient[y + z, 0] // Simplify]

(* 0 *)

Assuming[assume, SeriesCoefficient[y + z, 1] // Simplify]

(* -1 *)

share|improve this answer

answered May 13 at 18:15

Bob HanlonBob Hanlon

66.6k33 gold badges3737 silver badges102102 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  With y = Series[u + 1 + x^2, x, 0, 4] your method produces u^2 for Assuming[assume, SeriesCoefficient[y + z, 1] // Simplify]. I think the expected result is -1-u.
  $endgroup$
  – Carl Woll
  May 13 at 20:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @CarlWoll - I do not know what is "expected", but LeafCount /@ u^2, -1 - u indicates that u^2 is simpler in the usual sense.
  $endgroup$
  – Bob Hanlon
  May 13 at 20:30
  

  
  
  
  

add a comment
 | 

5

$begingroup$

You can give u an UpValues for Power:

u /: u^n_Integer := Block[u,
 If[n<0,
 PolynomialMod[(-u-1)^-n, 1+u+u^2],
 PolynomialMod[u^n,1+u+u^2]
 ]
]

Then:

y = Series[u + 1 + u x + x^2, x, 0, 4];
z = Series[u^2 + u^2 x + x^4,x, 0, 4];

and:

y + z //TeXForm

$-x+x^2+x^4+Oleft(x^5right)$

share|improve this answer

edited May 13 at 17:49

answered May 13 at 15:34

Carl WollCarl Woll

92.1k33 gold badges121121 silver badges233233 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Carl, what does the Block[u, ...] do here? I think I'm still confused about the usage of Block.
  $endgroup$
  – Roman
  May 13 at 15:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Roman The Block is needed so that recursion is avoided (preventing evaluation of u^n on the right hand side)
  $endgroup$
  – Carl Woll
  May 13 at 16:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I get Iteration limit exceeded with $1/u$.
  $endgroup$
  – Myath
  May 13 at 17:40
  

  
  
  
  

add a comment
 | 

2

$begingroup$

The simplest methods are usually the best. I suggest

rule = u^n_ :> 1, u, -1 - u[[Mod[n, 3] + 1]];
y + z /. rule

which will do what you want. Also, the following code

Table[u^n, n, 0, 6] /. rule

demonstrates that $u^3 = 1$ and the powers of $u$ are periodic with period $3$.

share|improve this answer

edited May 13 at 21:58

AccidentalFourierTransform

7,06311 gold badge1212 silver badges4545 bronze badges

answered May 13 at 18:09

SomosSomos

3,04011 gold badge22 silver badges1111 bronze badges

$endgroup$

add a comment
 | 

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4

$begingroup$

You can use Assumptions

assume = u^2 + u + 1 == 0;

y = Series[u + 1 + u*x + x^2, x, 0, 4];
z = Series[u^2 + u^2*x + x^4, x, 0, 4];

Assuming[assume, SeriesCoefficient[y + z, 0] // Simplify]

(* 0 *)

Assuming[assume, SeriesCoefficient[y + z, 1] // Simplify]

(* -1 *)

share|improve this answer

answered May 13 at 18:15

Bob HanlonBob Hanlon

66.6k33 gold badges3737 silver badges102102 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  With y = Series[u + 1 + x^2, x, 0, 4] your method produces u^2 for Assuming[assume, SeriesCoefficient[y + z, 1] // Simplify]. I think the expected result is -1-u.
  $endgroup$
  – Carl Woll
  May 13 at 20:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @CarlWoll - I do not know what is "expected", but LeafCount /@ u^2, -1 - u indicates that u^2 is simpler in the usual sense.
  $endgroup$
  – Bob Hanlon
  May 13 at 20:30
  

  
  
  
  

add a comment
 | 

4

$begingroup$

You can use Assumptions

assume = u^2 + u + 1 == 0;

y = Series[u + 1 + u*x + x^2, x, 0, 4];
z = Series[u^2 + u^2*x + x^4, x, 0, 4];

Assuming[assume, SeriesCoefficient[y + z, 0] // Simplify]

(* 0 *)

Assuming[assume, SeriesCoefficient[y + z, 1] // Simplify]

(* -1 *)

share|improve this answer

answered May 13 at 18:15

Bob HanlonBob Hanlon

66.6k33 gold badges3737 silver badges102102 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  With y = Series[u + 1 + x^2, x, 0, 4] your method produces u^2 for Assuming[assume, SeriesCoefficient[y + z, 1] // Simplify]. I think the expected result is -1-u.
  $endgroup$
  – Carl Woll
  May 13 at 20:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @CarlWoll - I do not know what is "expected", but LeafCount /@ u^2, -1 - u indicates that u^2 is simpler in the usual sense.
  $endgroup$
  – Bob Hanlon
  May 13 at 20:30
  

  
  
  
  

add a comment
 | 

$begingroup$

You can use Assumptions

assume = u^2 + u + 1 == 0;

y = Series[u + 1 + u*x + x^2, x, 0, 4];
z = Series[u^2 + u^2*x + x^4, x, 0, 4];

Assuming[assume, SeriesCoefficient[y + z, 0] // Simplify]

(* 0 *)

Assuming[assume, SeriesCoefficient[y + z, 1] // Simplify]

(* -1 *)

share|improve this answer

answered May 13 at 18:15

Bob HanlonBob Hanlon

66.6k33 gold badges3737 silver badges102102 bronze badges

$endgroup$

assume = u^2 + u + 1 == 0;

y = Series[u + 1 + u*x + x^2, x, 0, 4];
z = Series[u^2 + u^2*x + x^4, x, 0, 4];

Assuming[assume, SeriesCoefficient[y + z, 0] // Simplify]

(* 0 *)

Assuming[assume, SeriesCoefficient[y + z, 1] // Simplify]

(* -1 *)

share|improve this answer

answered May 13 at 18:15

Bob HanlonBob Hanlon

66.6k33 gold badges3737 silver badges102102 bronze badges

answered May 13 at 18:15

Bob HanlonBob Hanlon

66.6k33 gold badges3737 silver badges102102 bronze badges

answered May 13 at 18:15

Bob HanlonBob Hanlon

66.6k33 gold badges3737 silver badges102102 bronze badges

5

$begingroup$

You can give u an UpValues for Power:

u /: u^n_Integer := Block[u,
 If[n<0,
 PolynomialMod[(-u-1)^-n, 1+u+u^2],
 PolynomialMod[u^n,1+u+u^2]
 ]
]

Then:

y = Series[u + 1 + u x + x^2, x, 0, 4];
z = Series[u^2 + u^2 x + x^4,x, 0, 4];

and:

y + z //TeXForm

$-x+x^2+x^4+Oleft(x^5right)$

share|improve this answer

edited May 13 at 17:49

answered May 13 at 15:34

Carl WollCarl Woll

92.1k33 gold badges121121 silver badges233233 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Carl, what does the Block[u, ...] do here? I think I'm still confused about the usage of Block.
  $endgroup$
  – Roman
  May 13 at 15:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Roman The Block is needed so that recursion is avoided (preventing evaluation of u^n on the right hand side)
  $endgroup$
  – Carl Woll
  May 13 at 16:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I get Iteration limit exceeded with $1/u$.
  $endgroup$
  – Myath
  May 13 at 17:40
  

  
  
  
  

add a comment
 | 

5

$begingroup$

You can give u an UpValues for Power:

u /: u^n_Integer := Block[u,
 If[n<0,
 PolynomialMod[(-u-1)^-n, 1+u+u^2],
 PolynomialMod[u^n,1+u+u^2]
 ]
]

Then:

y = Series[u + 1 + u x + x^2, x, 0, 4];
z = Series[u^2 + u^2 x + x^4,x, 0, 4];

and:

y + z //TeXForm

$-x+x^2+x^4+Oleft(x^5right)$

share|improve this answer

edited May 13 at 17:49

answered May 13 at 15:34

Carl WollCarl Woll

92.1k33 gold badges121121 silver badges233233 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Carl, what does the Block[u, ...] do here? I think I'm still confused about the usage of Block.
  $endgroup$
  – Roman
  May 13 at 15:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Roman The Block is needed so that recursion is avoided (preventing evaluation of u^n on the right hand side)
  $endgroup$
  – Carl Woll
  May 13 at 16:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I get Iteration limit exceeded with $1/u$.
  $endgroup$
  – Myath
  May 13 at 17:40
  

  
  
  
  

add a comment
 | 

$begingroup$

You can give u an UpValues for Power:

u /: u^n_Integer := Block[u,
 If[n<0,
 PolynomialMod[(-u-1)^-n, 1+u+u^2],
 PolynomialMod[u^n,1+u+u^2]
 ]
]

Then:

y = Series[u + 1 + u x + x^2, x, 0, 4];
z = Series[u^2 + u^2 x + x^4,x, 0, 4];

and:

y + z //TeXForm

$-x+x^2+x^4+Oleft(x^5right)$

share|improve this answer

edited May 13 at 17:49

answered May 13 at 15:34

Carl WollCarl Woll

92.1k33 gold badges121121 silver badges233233 bronze badges

$endgroup$

u /: u^n_Integer := Block[u,
 If[n<0,
 PolynomialMod[(-u-1)^-n, 1+u+u^2],
 PolynomialMod[u^n,1+u+u^2]
 ]
]

y = Series[u + 1 + u x + x^2, x, 0, 4];
z = Series[u^2 + u^2 x + x^4,x, 0, 4];

y + z //TeXForm

share|improve this answer

edited May 13 at 17:49

answered May 13 at 15:34

Carl WollCarl Woll

92.1k33 gold badges121121 silver badges233233 bronze badges

answered May 13 at 15:34

Carl WollCarl Woll

92.1k33 gold badges121121 silver badges233233 bronze badges

answered May 13 at 15:34

Carl WollCarl Woll

92.1k33 gold badges121121 silver badges233233 bronze badges

2

$begingroup$

The simplest methods are usually the best. I suggest

rule = u^n_ :> 1, u, -1 - u[[Mod[n, 3] + 1]];
y + z /. rule

which will do what you want. Also, the following code

Table[u^n, n, 0, 6] /. rule

demonstrates that $u^3 = 1$ and the powers of $u$ are periodic with period $3$.

share|improve this answer

edited May 13 at 21:58

AccidentalFourierTransform

7,06311 gold badge1212 silver badges4545 bronze badges

answered May 13 at 18:09

SomosSomos

3,04011 gold badge22 silver badges1111 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

The simplest methods are usually the best. I suggest

rule = u^n_ :> 1, u, -1 - u[[Mod[n, 3] + 1]];
y + z /. rule

which will do what you want. Also, the following code

Table[u^n, n, 0, 6] /. rule

demonstrates that $u^3 = 1$ and the powers of $u$ are periodic with period $3$.

share|improve this answer

edited May 13 at 21:58

AccidentalFourierTransform

7,06311 gold badge1212 silver badges4545 bronze badges

answered May 13 at 18:09

SomosSomos

3,04011 gold badge22 silver badges1111 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

The simplest methods are usually the best. I suggest

rule = u^n_ :> 1, u, -1 - u[[Mod[n, 3] + 1]];
y + z /. rule

which will do what you want. Also, the following code

Table[u^n, n, 0, 6] /. rule

demonstrates that $u^3 = 1$ and the powers of $u$ are periodic with period $3$.

share|improve this answer

edited May 13 at 21:58

AccidentalFourierTransform

7,06311 gold badge1212 silver badges4545 bronze badges

answered May 13 at 18:09

SomosSomos

3,04011 gold badge22 silver badges1111 bronze badges

$endgroup$

rule = u^n_ :> 1, u, -1 - u[[Mod[n, 3] + 1]];
y + z /. rule

Table[u^n, n, 0, 6] /. rule

share|improve this answer

edited May 13 at 21:58

AccidentalFourierTransform

7,06311 gold badge1212 silver badges4545 bronze badges

answered May 13 at 18:09

SomosSomos

3,04011 gold badge22 silver badges1111 bronze badges

edited May 13 at 21:58

AccidentalFourierTransform

7,06311 gold badge1212 silver badges4545 bronze badges

edited May 13 at 21:58

AccidentalFourierTransform

7,06311 gold badge1212 silver badges4545 bronze badges

answered May 13 at 18:09

SomosSomos

3,04011 gold badge22 silver badges1111 bronze badges

answered May 13 at 18:09

SomosSomos

3,04011 gold badge22 silver badges1111 bronze badges