For the periodic sequence, is there always an algebraic closed form?












2












$begingroup$


This question is a generalized form of the problem I asked before:



Algebraic Closed Form for $sum_{n=1}^{k}left( n- 3 lfloor frac{n-1}{3} rfloorright)$




Let, look at this periodic sequence:



$$a_n=left{a_1,a_2,a_3,a_4,a_5,cdots a_k ; a_1,a_2,a_3,a_4,a_5, cdots a_k; a_1,a_2,a_3,a_4,a_5,cdots a_k;cdots right}$$, where $left{a_1, a_2, a_3, cdots a_k right} in mathbb{Z^{+}}$ and $a_{k+1}=a_1, a_{k+2}=a_2, a_{k+3}=a_3, cdots a_{2k}=a_k, cdots$




For the sequence $a_k=left{ a_1,a_2,a_3,...,a_k right}$ , $k$ is a finite number. $a_1,a_2,...a_k$ are arbitary numbers.



Verbally, the series $a_n$ consists of an infinite number of periodic repetitions of the finite series $a_k.$




Finally my question is:



a) If there is an exist a algebraic closed form, for finite series $a_k$, in this case, does the $ a_n $ series always have a algebraic closed form?



b) If there is not an exist a algebraic closed form, for finite series $a_k$, in this case, does the $ a_n $ series always have a algebraic closed form?




I mean ,for example:




a)



$a_n=left{ 1,3,5,7,1,3,5,7,1,3,5,7,1,3,5,7,1,3,5,7cdotsright}$



b)



$a_n=left{1,8,2,6,5,9,1,8,2,6,5,9,1,8,2,6,5,9,1,8,2,6,5,9cdotsright}$




Thank you very much.






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$endgroup$












  • $begingroup$
    A finite sequence always does have a closed form, and so does a periodic sequence.
    $endgroup$
    – Robert Israel
    Jan 8 at 12:39










  • $begingroup$
    @RobertIsrael for any arbitary numbers for $a_1, a_2,a_3,...a_k$, There's an always closed form for $a_n$. Do I understand correct?
    $endgroup$
    – Beginner
    Jan 8 at 12:46












  • $begingroup$
    @Beginner, it depends on what you consider a "closed form". That term is not really well-defined. In a sense you have given a "closed form" in your definition of the sequence.
    $endgroup$
    – Mees de Vries
    Jan 8 at 12:53










  • $begingroup$
    @MeesdeVries I tried to give 2 example for a good understanding of the question..
    $endgroup$
    – Beginner
    Jan 8 at 12:56










  • $begingroup$
    This might be a closed form for your first sequence: $$ a_k = begin{cases} 1&text{if $k = 1 mod4$,}\ 3&text{if $k = 2 mod4$,}\ 5&text{if $k = 3 mod4$,}\ 7&text{else.} end{cases} $$ Does that satisfy what you're looking for in a "closed form"?
    $endgroup$
    – Mees de Vries
    Jan 8 at 13:08
















2












$begingroup$


This question is a generalized form of the problem I asked before:



Algebraic Closed Form for $sum_{n=1}^{k}left( n- 3 lfloor frac{n-1}{3} rfloorright)$




Let, look at this periodic sequence:



$$a_n=left{a_1,a_2,a_3,a_4,a_5,cdots a_k ; a_1,a_2,a_3,a_4,a_5, cdots a_k; a_1,a_2,a_3,a_4,a_5,cdots a_k;cdots right}$$, where $left{a_1, a_2, a_3, cdots a_k right} in mathbb{Z^{+}}$ and $a_{k+1}=a_1, a_{k+2}=a_2, a_{k+3}=a_3, cdots a_{2k}=a_k, cdots$




For the sequence $a_k=left{ a_1,a_2,a_3,...,a_k right}$ , $k$ is a finite number. $a_1,a_2,...a_k$ are arbitary numbers.



Verbally, the series $a_n$ consists of an infinite number of periodic repetitions of the finite series $a_k.$




Finally my question is:



a) If there is an exist a algebraic closed form, for finite series $a_k$, in this case, does the $ a_n $ series always have a algebraic closed form?



b) If there is not an exist a algebraic closed form, for finite series $a_k$, in this case, does the $ a_n $ series always have a algebraic closed form?




I mean ,for example:




a)



$a_n=left{ 1,3,5,7,1,3,5,7,1,3,5,7,1,3,5,7,1,3,5,7cdotsright}$



b)



$a_n=left{1,8,2,6,5,9,1,8,2,6,5,9,1,8,2,6,5,9,1,8,2,6,5,9cdotsright}$




Thank you very much.






sequences-and-series soft-question closed-form







share|cite|improve this question











$endgroup$












  • $begingroup$
    A finite sequence always does have a closed form, and so does a periodic sequence.
    $endgroup$
    – Robert Israel
    Jan 8 at 12:39










  • $begingroup$
    @RobertIsrael for any arbitary numbers for $a_1, a_2,a_3,...a_k$, There's an always closed form for $a_n$. Do I understand correct?
    $endgroup$
    – Beginner
    Jan 8 at 12:46












  • $begingroup$
    @Beginner, it depends on what you consider a "closed form". That term is not really well-defined. In a sense you have given a "closed form" in your definition of the sequence.
    $endgroup$
    – Mees de Vries
    Jan 8 at 12:53










  • $begingroup$
    @MeesdeVries I tried to give 2 example for a good understanding of the question..
    $endgroup$
    – Beginner
    Jan 8 at 12:56










  • $begingroup$
    This might be a closed form for your first sequence: $$ a_k = begin{cases} 1&text{if $k = 1 mod4$,}\ 3&text{if $k = 2 mod4$,}\ 5&text{if $k = 3 mod4$,}\ 7&text{else.} end{cases} $$ Does that satisfy what you're looking for in a "closed form"?
    $endgroup$
    – Mees de Vries
    Jan 8 at 13:08














2












2








2





$begingroup$


This question is a generalized form of the problem I asked before:



Algebraic Closed Form for $sum_{n=1}^{k}left( n- 3 lfloor frac{n-1}{3} rfloorright)$




Let, look at this periodic sequence:



$$a_n=left{a_1,a_2,a_3,a_4,a_5,cdots a_k ; a_1,a_2,a_3,a_4,a_5, cdots a_k; a_1,a_2,a_3,a_4,a_5,cdots a_k;cdots right}$$, where $left{a_1, a_2, a_3, cdots a_k right} in mathbb{Z^{+}}$ and $a_{k+1}=a_1, a_{k+2}=a_2, a_{k+3}=a_3, cdots a_{2k}=a_k, cdots$




For the sequence $a_k=left{ a_1,a_2,a_3,...,a_k right}$ , $k$ is a finite number. $a_1,a_2,...a_k$ are arbitary numbers.



Verbally, the series $a_n$ consists of an infinite number of periodic repetitions of the finite series $a_k.$




Finally my question is:



a) If there is an exist a algebraic closed form, for finite series $a_k$, in this case, does the $ a_n $ series always have a algebraic closed form?



b) If there is not an exist a algebraic closed form, for finite series $a_k$, in this case, does the $ a_n $ series always have a algebraic closed form?




I mean ,for example:




a)



$a_n=left{ 1,3,5,7,1,3,5,7,1,3,5,7,1,3,5,7,1,3,5,7cdotsright}$



b)



$a_n=left{1,8,2,6,5,9,1,8,2,6,5,9,1,8,2,6,5,9,1,8,2,6,5,9cdotsright}$




Thank you very much.






sequences-and-series soft-question closed-form







share|cite|improve this question











$endgroup$




This question is a generalized form of the problem I asked before:



Algebraic Closed Form for $sum_{n=1}^{k}left( n- 3 lfloor frac{n-1}{3} rfloorright)$




Let, look at this periodic sequence:



$$a_n=left{a_1,a_2,a_3,a_4,a_5,cdots a_k ; a_1,a_2,a_3,a_4,a_5, cdots a_k; a_1,a_2,a_3,a_4,a_5,cdots a_k;cdots right}$$, where $left{a_1, a_2, a_3, cdots a_k right} in mathbb{Z^{+}}$ and $a_{k+1}=a_1, a_{k+2}=a_2, a_{k+3}=a_3, cdots a_{2k}=a_k, cdots$




For the sequence $a_k=left{ a_1,a_2,a_3,...,a_k right}$ , $k$ is a finite number. $a_1,a_2,...a_k$ are arbitary numbers.



Verbally, the series $a_n$ consists of an infinite number of periodic repetitions of the finite series $a_k.$




Finally my question is:



a) If there is an exist a algebraic closed form, for finite series $a_k$, in this case, does the $ a_n $ series always have a algebraic closed form?



b) If there is not an exist a algebraic closed form, for finite series $a_k$, in this case, does the $ a_n $ series always have a algebraic closed form?




I mean ,for example:




a)



$a_n=left{ 1,3,5,7,1,3,5,7,1,3,5,7,1,3,5,7,1,3,5,7cdotsright}$



b)



$a_n=left{1,8,2,6,5,9,1,8,2,6,5,9,1,8,2,6,5,9,1,8,2,6,5,9cdotsright}$




Thank you very much.






sequences-and-series soft-question closed-form




sequences-and-series soft-question closed-form






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 16:50







Beginner

















asked Jan 8 at 12:36









BeginnerBeginner

341110




341110












  • $begingroup$
    A finite sequence always does have a closed form, and so does a periodic sequence.
    $endgroup$
    – Robert Israel
    Jan 8 at 12:39










  • $begingroup$
    @RobertIsrael for any arbitary numbers for $a_1, a_2,a_3,...a_k$, There's an always closed form for $a_n$. Do I understand correct?
    $endgroup$
    – Beginner
    Jan 8 at 12:46












  • $begingroup$
    @Beginner, it depends on what you consider a "closed form". That term is not really well-defined. In a sense you have given a "closed form" in your definition of the sequence.
    $endgroup$
    – Mees de Vries
    Jan 8 at 12:53










  • $begingroup$
    @MeesdeVries I tried to give 2 example for a good understanding of the question..
    $endgroup$
    – Beginner
    Jan 8 at 12:56










  • $begingroup$
    This might be a closed form for your first sequence: $$ a_k = begin{cases} 1&text{if $k = 1 mod4$,}\ 3&text{if $k = 2 mod4$,}\ 5&text{if $k = 3 mod4$,}\ 7&text{else.} end{cases} $$ Does that satisfy what you're looking for in a "closed form"?
    $endgroup$
    – Mees de Vries
    Jan 8 at 13:08


















  • $begingroup$
    A finite sequence always does have a closed form, and so does a periodic sequence.
    $endgroup$
    – Robert Israel
    Jan 8 at 12:39










  • $begingroup$
    @RobertIsrael for any arbitary numbers for $a_1, a_2,a_3,...a_k$, There's an always closed form for $a_n$. Do I understand correct?
    $endgroup$
    – Beginner
    Jan 8 at 12:46












  • $begingroup$
    @Beginner, it depends on what you consider a "closed form". That term is not really well-defined. In a sense you have given a "closed form" in your definition of the sequence.
    $endgroup$
    – Mees de Vries
    Jan 8 at 12:53










  • $begingroup$
    @MeesdeVries I tried to give 2 example for a good understanding of the question..
    $endgroup$
    – Beginner
    Jan 8 at 12:56










  • $begingroup$
    This might be a closed form for your first sequence: $$ a_k = begin{cases} 1&text{if $k = 1 mod4$,}\ 3&text{if $k = 2 mod4$,}\ 5&text{if $k = 3 mod4$,}\ 7&text{else.} end{cases} $$ Does that satisfy what you're looking for in a "closed form"?
    $endgroup$
    – Mees de Vries
    Jan 8 at 13:08
















$begingroup$
A finite sequence always does have a closed form, and so does a periodic sequence.
$endgroup$
– Robert Israel
Jan 8 at 12:39




$begingroup$
A finite sequence always does have a closed form, and so does a periodic sequence.
$endgroup$
– Robert Israel
Jan 8 at 12:39












$begingroup$
@RobertIsrael for any arbitary numbers for $a_1, a_2,a_3,...a_k$, There's an always closed form for $a_n$. Do I understand correct?
$endgroup$
– Beginner
Jan 8 at 12:46






$begingroup$
@RobertIsrael for any arbitary numbers for $a_1, a_2,a_3,...a_k$, There's an always closed form for $a_n$. Do I understand correct?
$endgroup$
– Beginner
Jan 8 at 12:46














$begingroup$
@Beginner, it depends on what you consider a "closed form". That term is not really well-defined. In a sense you have given a "closed form" in your definition of the sequence.
$endgroup$
– Mees de Vries
Jan 8 at 12:53




$begingroup$
@Beginner, it depends on what you consider a "closed form". That term is not really well-defined. In a sense you have given a "closed form" in your definition of the sequence.
$endgroup$
– Mees de Vries
Jan 8 at 12:53












$begingroup$
@MeesdeVries I tried to give 2 example for a good understanding of the question..
$endgroup$
– Beginner
Jan 8 at 12:56




$begingroup$
@MeesdeVries I tried to give 2 example for a good understanding of the question..
$endgroup$
– Beginner
Jan 8 at 12:56












$begingroup$
This might be a closed form for your first sequence: $$ a_k = begin{cases} 1&text{if $k = 1 mod4$,}\ 3&text{if $k = 2 mod4$,}\ 5&text{if $k = 3 mod4$,}\ 7&text{else.} end{cases} $$ Does that satisfy what you're looking for in a "closed form"?
$endgroup$
– Mees de Vries
Jan 8 at 13:08




$begingroup$
This might be a closed form for your first sequence: $$ a_k = begin{cases} 1&text{if $k = 1 mod4$,}\ 3&text{if $k = 2 mod4$,}\ 5&text{if $k = 3 mod4$,}\ 7&text{else.} end{cases} $$ Does that satisfy what you're looking for in a "closed form"?
$endgroup$
– Mees de Vries
Jan 8 at 13:08










2 Answers
2






active

oldest

votes


















2












$begingroup$

A periodic sequence with period $P$ can always be written as a trigonometric polynomial $$a_n = sum_{j=0}^{P-1} b_j cos(2 pi j n/P) + sum_{j=1}^{P-1} c_j sin(2 pi j n/P)$$
for some coefficients $b_j$ and $c_j$ (look up Finite Fourier Transform).



Thus your first example can be written as
$$a_n = 4-cos left( pi,j/2 right) -cos left( pi,j right) -cos
left( 3,pi,j/2 right) -sin left( pi,j/2 right) +sin
left( 3,pi,j/2 right)
$$






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Let $f(x)$ be a polynomial interpolating $a_1,a_2,a_3,dots,a_k$.



    Then the sequence is given by $a_n = f(1+((n-1) bmod k))$.



    If you can't use mod directly, but can use floor, then note that
    $$
    a bmod b = a - b leftlfloor dfrac{a}{b} rightrfloor
    $$
    for $a,b in mathbb N$.






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      A periodic sequence with period $P$ can always be written as a trigonometric polynomial $$a_n = sum_{j=0}^{P-1} b_j cos(2 pi j n/P) + sum_{j=1}^{P-1} c_j sin(2 pi j n/P)$$
      for some coefficients $b_j$ and $c_j$ (look up Finite Fourier Transform).



      Thus your first example can be written as
      $$a_n = 4-cos left( pi,j/2 right) -cos left( pi,j right) -cos
      left( 3,pi,j/2 right) -sin left( pi,j/2 right) +sin
      left( 3,pi,j/2 right)
      $$






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        A periodic sequence with period $P$ can always be written as a trigonometric polynomial $$a_n = sum_{j=0}^{P-1} b_j cos(2 pi j n/P) + sum_{j=1}^{P-1} c_j sin(2 pi j n/P)$$
        for some coefficients $b_j$ and $c_j$ (look up Finite Fourier Transform).



        Thus your first example can be written as
        $$a_n = 4-cos left( pi,j/2 right) -cos left( pi,j right) -cos
        left( 3,pi,j/2 right) -sin left( pi,j/2 right) +sin
        left( 3,pi,j/2 right)
        $$






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          A periodic sequence with period $P$ can always be written as a trigonometric polynomial $$a_n = sum_{j=0}^{P-1} b_j cos(2 pi j n/P) + sum_{j=1}^{P-1} c_j sin(2 pi j n/P)$$
          for some coefficients $b_j$ and $c_j$ (look up Finite Fourier Transform).



          Thus your first example can be written as
          $$a_n = 4-cos left( pi,j/2 right) -cos left( pi,j right) -cos
          left( 3,pi,j/2 right) -sin left( pi,j/2 right) +sin
          left( 3,pi,j/2 right)
          $$






          share|cite|improve this answer











          $endgroup$



          A periodic sequence with period $P$ can always be written as a trigonometric polynomial $$a_n = sum_{j=0}^{P-1} b_j cos(2 pi j n/P) + sum_{j=1}^{P-1} c_j sin(2 pi j n/P)$$
          for some coefficients $b_j$ and $c_j$ (look up Finite Fourier Transform).



          Thus your first example can be written as
          $$a_n = 4-cos left( pi,j/2 right) -cos left( pi,j right) -cos
          left( 3,pi,j/2 right) -sin left( pi,j/2 right) +sin
          left( 3,pi,j/2 right)
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 8 at 16:59

























          answered Jan 8 at 16:39









          Robert IsraelRobert Israel

          321k23210462




          321k23210462























              2












              $begingroup$

              Let $f(x)$ be a polynomial interpolating $a_1,a_2,a_3,dots,a_k$.



              Then the sequence is given by $a_n = f(1+((n-1) bmod k))$.



              If you can't use mod directly, but can use floor, then note that
              $$
              a bmod b = a - b leftlfloor dfrac{a}{b} rightrfloor
              $$
              for $a,b in mathbb N$.






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                Let $f(x)$ be a polynomial interpolating $a_1,a_2,a_3,dots,a_k$.



                Then the sequence is given by $a_n = f(1+((n-1) bmod k))$.



                If you can't use mod directly, but can use floor, then note that
                $$
                a bmod b = a - b leftlfloor dfrac{a}{b} rightrfloor
                $$
                for $a,b in mathbb N$.






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Let $f(x)$ be a polynomial interpolating $a_1,a_2,a_3,dots,a_k$.



                  Then the sequence is given by $a_n = f(1+((n-1) bmod k))$.



                  If you can't use mod directly, but can use floor, then note that
                  $$
                  a bmod b = a - b leftlfloor dfrac{a}{b} rightrfloor
                  $$
                  for $a,b in mathbb N$.






                  share|cite|improve this answer











                  $endgroup$



                  Let $f(x)$ be a polynomial interpolating $a_1,a_2,a_3,dots,a_k$.



                  Then the sequence is given by $a_n = f(1+((n-1) bmod k))$.



                  If you can't use mod directly, but can use floor, then note that
                  $$
                  a bmod b = a - b leftlfloor dfrac{a}{b} rightrfloor
                  $$
                  for $a,b in mathbb N$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 8 at 15:13

























                  answered Jan 8 at 13:08









                  lhflhf

                  164k10170395




                  164k10170395






























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                      Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
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                      0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
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