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







share|improve this question









$endgroup$












  • $begingroup$
    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


















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









$endgroup$












  • $begingroup$
    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
















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







share|improve this question









$endgroup$




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






share|improve this question













share|improve this question











share|improve this question




share|improve this question










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




















  • $begingroup$
    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












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









$endgroup$













  • $begingroup$
    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



















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











$endgroup$













  • $begingroup$
    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



















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











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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 *)





    share|improve this answer









    $endgroup$













    • $begingroup$
      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
















    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









    $endgroup$













    • $begingroup$
      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














    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 *)





    share|improve this answer









    $endgroup$



    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 *)






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered May 13 at 18:15









    Bob HanlonBob Hanlon

    62.8k33599




    62.8k33599












    • $begingroup$
      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


















    • $begingroup$
      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
















    $begingroup$
    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











    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











    $endgroup$













    • $begingroup$
      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
















    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











    $endgroup$













    • $begingroup$
      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














    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



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







    share|improve this answer











    $endgroup$



    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



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








    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited May 13 at 17:49

























    answered May 13 at 15:34









    Carl WollCarl Woll

    81.4k3105209




    81.4k3105209












    • $begingroup$
      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


















    • $begingroup$
      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
















    $begingroup$
    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











    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











    $endgroup$


















      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











      $endgroup$
















        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 $3$.






        share|improve this answer











        $endgroup$



        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 $3$.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited May 13 at 21:58









        AccidentalFourierTransform

        5,45511142




        5,45511142










        answered May 13 at 18:09









        SomosSomos

        2,4151111




        2,4151111






























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            Bunad

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            Tradisjonelle folkedrakter frå Aust-Telemark, truleg frå 1880-åra. Drakter som dette er grunnlaget for moderne bunadar. Foto: Axel Lindahl Bunad frå Eggedal i Buskerud fotografert i 1906. Bunad (frå kledebunad , der norrønt «búnaðr» har same rot som (føre)bu) er ei samnemning brukt om klesdrakter i tradisjonell nordisk stil — ofte utvikla frå folkedrakter, men likevel i ein eigen tradisjon. Ein bunad er typisk ei forseggjort og kostbar drakt, med mykje broderi, og med handsmidde smykke i gull eller sølv, særleg dei særeigne søljene, som obligatorisk prynad i tillegg. Bunad og bunadbruk er eit debattemne med mange ulike oppfatningar, til dømes om kva som er ein bunad, og om ein kan endra på utforminga av draktene eller ikkje. Den norske bunadrørsla går tilbake til folkedansrørsla og føregangsfolk som Hulda Garborg og Klara Semb. I dag blir bunad brukt som nasjonaldrakt. Festbunad er likestilt med galla-antrekk, og ofte nytta ved særskilde høve, som konfirmasjonar eller d
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            Færeyskur hestur Heimild | Tengill | Tilvísanir | LeiðsagnarvalRossið - síða um færeyska hrossið á færeyskuGott ár hjá færeyska hestinum

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            FæreyjarHestar færeyskahjaltlandshestinumíslenski hesturinnBretlandskolanámumenskuWikipediafæreyskuWikipediaRossið - síða um færeyska hrossið á færeysku Færeyskur hestur Úr Wikipediu, frjálsa alfræðiritinu Jump to navigation Jump to search Rani og Grani. Grani í vetrarfeldi. Færeyski hesturinn (færeyska: Føroyska rossið ) er hestur sem hefur lifað í Færeyjum í hundruði ára. Hann er smágerður, 120-132 cm á hæð, og helst skyldur hjaltlandshestinum. Hlutverk hans var að draga vagna og plóg og bera klyfjar. Hann hefur fjórar gangtegundir eins og íslenski hesturinn (þar með talið tölt) og er með fjölda litaafbrigði. Nú til dags er hann notaður til frístunda. Í lok 19. aldar voru til um 800 hross. Mörg þeirra voru seld til Bretlands til að vinna í kolanámum og um 1960 voru aðeins nokkrir hestar eftir. Átak var gert í að varðveita hestinn og eru nú til 70-80 hross í Færeyjum. [1] Heimild |   Wikimedia Commons er með margmiðlunarefni sem te
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