Alias for root of a polynomial
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
asked May 13 at 15:08
MyathMyath
1423
$endgroup$
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
asked May 13 at 15:08
MyathMyath
1423
$endgroup$
$begingroup$
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
4
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
asked May 13 at 15:08
MyathMyath
1423
$endgroup$
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
symbolic polynomials series-expansion
asked May 13 at 15:08
MyathMyath
1423
asked May 13 at 15:08
MyathMyath
1423
asked May 13 at 15:08
MyathMyath
1423
asked May 13 at 15:08
MyathMyath
1423
asked May 13 at 15:08
MyathMyath
1423
1423
$begingroup$
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'm a Python/R user, this is unintelligible to me, can you explain without Mathematica jargon please?
$endgroup$
– smci
May 14 at 5:51
$begingroup$
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
$begingroup$
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 |
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 *)
answered May 13 at 18:15
Bob HanlonBob Hanlon
62.8k33599
$endgroup$
$begingroup$
Withy = Series[u + 1 + x^2, {x, 0, 4}]your method producesu^2forAssuming[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", butLeafCount /@ {u^2, -1 - u}indicates thatu^2is 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
$-x+x^2+x^4+Oleft(x^5right)$
answered May 13 at 15:34
Carl WollCarl Woll
81.4k3105209
$endgroup$
$begingroup$
Carl, what does theBlock[{u}, ...]do here? I think I'm still confused about the usage ofBlock.
$endgroup$
– 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 $3$.
edited May 13 at 21:58
AccidentalFourierTransform
5,45511142
answered May 13 at 18:09
SomosSomos
2,4151111
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
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 *)
answered May 13 at 18:15
Bob HanlonBob Hanlon
62.8k33599
$endgroup$
$begingroup$
Withy = Series[u + 1 + x^2, {x, 0, 4}]your method producesu^2forAssuming[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", butLeafCount /@ {u^2, -1 - u}indicates thatu^2is 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 *)
answered May 13 at 18:15
Bob HanlonBob Hanlon
62.8k33599
$endgroup$
$begingroup$
Withy = Series[u + 1 + x^2, {x, 0, 4}]your method producesu^2forAssuming[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", butLeafCount /@ {u^2, -1 - u}indicates thatu^2is simpler in the usual sense.
$endgroup$
– Bob Hanlon
May 13 at 20:30
add a comment |
4
4
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 *)
answered May 13 at 18:15
Bob HanlonBob Hanlon
62.8k33599
$endgroup$
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 *)
answered May 13 at 18:15
Bob HanlonBob Hanlon
62.8k33599
answered May 13 at 18:15
Bob HanlonBob Hanlon
62.8k33599
answered May 13 at 18:15
Bob HanlonBob Hanlon
62.8k33599
answered May 13 at 18:15
Bob HanlonBob Hanlon
62.8k33599
62.8k33599
$begingroup$
Withy = Series[u + 1 + x^2, {x, 0, 4}]your method producesu^2forAssuming[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", butLeafCount /@ {u^2, -1 - u}indicates thatu^2is simpler in the usual sense.
$endgroup$
– Bob Hanlon
May 13 at 20:30
add a comment |
$begingroup$
Withy = Series[u + 1 + x^2, {x, 0, 4}]your method producesu^2forAssuming[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", butLeafCount /@ {u^2, -1 - u}indicates thatu^2is simpler in the usual sense.
$endgroup$
– Bob Hanlon
May 13 at 20:30
$begingroup$
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
$begingroup$
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
$begingroup$
@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
$-x+x^2+x^4+Oleft(x^5right)$
answered May 13 at 15:34
Carl WollCarl Woll
81.4k3105209
$endgroup$
$begingroup$
Carl, what does theBlock[{u}, ...]do here? I think I'm still confused about the usage ofBlock.
$endgroup$
– 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
$-x+x^2+x^4+Oleft(x^5right)$
answered May 13 at 15:34
Carl WollCarl Woll
81.4k3105209
$endgroup$
$begingroup$
Carl, what does theBlock[{u}, ...]do here? I think I'm still confused about the usage ofBlock.
$endgroup$
– 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
5
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
$-x+x^2+x^4+Oleft(x^5right)$
answered May 13 at 15:34
Carl WollCarl Woll
81.4k3105209
$endgroup$
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
$-x+x^2+x^4+Oleft(x^5right)$
answered May 13 at 15:34
Carl WollCarl Woll
81.4k3105209
edited May 13 at 17:49
answered May 13 at 15:34
Carl WollCarl Woll
81.4k3105209
answered May 13 at 15:34
Carl WollCarl Woll
81.4k3105209
answered May 13 at 15:34
Carl WollCarl Woll
81.4k3105209
81.4k3105209
$begingroup$
Carl, what does theBlock[{u}, ...]do here? I think I'm still confused about the usage ofBlock.
$endgroup$
– 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 |
$begingroup$
Carl, what does theBlock[{u}, ...]do here? I think I'm still confused about the usage ofBlock.
$endgroup$
– 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
$begingroup$
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
$begingroup$
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
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
$begingroup$
@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
$begingroup$
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 $3$.
edited May 13 at 21:58
AccidentalFourierTransform
5,45511142
answered May 13 at 18:09
SomosSomos
2,4151111
$endgroup$
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 $3$.
edited May 13 at 21:58
AccidentalFourierTransform
5,45511142
answered May 13 at 18:09
SomosSomos
2,4151111
$endgroup$
add a comment |
2
2
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 $3$.
edited May 13 at 21:58
AccidentalFourierTransform
5,45511142
answered May 13 at 18:09
SomosSomos
2,4151111
$endgroup$
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 $3$.
edited May 13 at 21:58
AccidentalFourierTransform
5,45511142
answered May 13 at 18:09
SomosSomos
2,4151111
edited May 13 at 21:58
AccidentalFourierTransform
5,45511142
edited May 13 at 21:58
AccidentalFourierTransform
5,45511142
edited May 13 at 21:58
AccidentalFourierTransform
5,45511142
5,45511142
answered May 13 at 18:09
SomosSomos
2,4151111
answered May 13 at 18:09
SomosSomos
2,4151111
answered May 13 at 18:09
SomosSomos
2,4151111
2,4151111
add a comment |
add a comment |
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$begingroup$
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