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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                      Management

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                      Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
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                      Has there ever been an instance of an active nuclear power plant within or near a war zone?

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                      36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
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