Describe a list of numbers that has alternating functions without using trigonometric functions?












1












$begingroup$


The list is: $ {5,6,10,12,15,18,20,24,25...}$



Is there a way to describe this list of numbers mathematically without using $sin(x)
, cos(x), cos(nx)$, etc?



There is a way using the trigonometric function $sin$. Simply use the jumping between $0$ and $1$ of $$frac{1+sin(npi)}{2}$$ and $$frac{1+sinleft(npi+frac{3pi}{2}right)}{2 }$$ which instead oscillates between $1$ and $0$. Then you can get the answer by describing $A = {5, x, 10, y, 15, z, 20}$ as well as $ B = {x, 6, y, 12, z, 18}$. Every other number in the lists $A$ and $B$ will either be numbers or they will be zero, since they're multiplied with said $sin$ function. $A + B$ is therefore a list containing two different methods of sorting, i.e. $A$'s function could be $ x^2 $ and $B$'s function could be $x^3-x^2+2x $ where every item in the list $C$, where $C = A + B$, is based on a different equation than the one before and after that said item.



Now, again, is there a way to create this said list $C$, containing the pattern of two very different functions that create the list $C $ where every item in the slot $ 2n$ is connected to one function but that function is different from the item in the slot $2n + 1$?






sequences-and-series







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Very simply: it's the sequence $(5n,6n)_{nge 1}$.
    $endgroup$
    – Bernard
    Dec 24 '18 at 19:24






  • 2




    $begingroup$
    $f(n)=5n$, when $n equiv 1 pmod{2} $ and $f(n)=3n$, when $n equiv 0 pmod{2}$.
    $endgroup$
    – Anurag A
    Dec 24 '18 at 19:26








  • 1




    $begingroup$
    To give a single formula that works for all $n$, think about $(1+(-1)^n)/2$ and $(1-(-1)^n)/2$.
    $endgroup$
    – David C. Ullrich
    Dec 24 '18 at 19:55


















1












$begingroup$


The list is: $ {5,6,10,12,15,18,20,24,25...}$



Is there a way to describe this list of numbers mathematically without using $sin(x)
, cos(x), cos(nx)$, etc?



There is a way using the trigonometric function $sin$. Simply use the jumping between $0$ and $1$ of $$frac{1+sin(npi)}{2}$$ and $$frac{1+sinleft(npi+frac{3pi}{2}right)}{2 }$$ which instead oscillates between $1$ and $0$. Then you can get the answer by describing $A = {5, x, 10, y, 15, z, 20}$ as well as $ B = {x, 6, y, 12, z, 18}$. Every other number in the lists $A$ and $B$ will either be numbers or they will be zero, since they're multiplied with said $sin$ function. $A + B$ is therefore a list containing two different methods of sorting, i.e. $A$'s function could be $ x^2 $ and $B$'s function could be $x^3-x^2+2x $ where every item in the list $C$, where $C = A + B$, is based on a different equation than the one before and after that said item.



Now, again, is there a way to create this said list $C$, containing the pattern of two very different functions that create the list $C $ where every item in the slot $ 2n$ is connected to one function but that function is different from the item in the slot $2n + 1$?






sequences-and-series







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Very simply: it's the sequence $(5n,6n)_{nge 1}$.
    $endgroup$
    – Bernard
    Dec 24 '18 at 19:24






  • 2




    $begingroup$
    $f(n)=5n$, when $n equiv 1 pmod{2} $ and $f(n)=3n$, when $n equiv 0 pmod{2}$.
    $endgroup$
    – Anurag A
    Dec 24 '18 at 19:26








  • 1




    $begingroup$
    To give a single formula that works for all $n$, think about $(1+(-1)^n)/2$ and $(1-(-1)^n)/2$.
    $endgroup$
    – David C. Ullrich
    Dec 24 '18 at 19:55
















1












1








1





$begingroup$


The list is: $ {5,6,10,12,15,18,20,24,25...}$



Is there a way to describe this list of numbers mathematically without using $sin(x)
, cos(x), cos(nx)$, etc?



There is a way using the trigonometric function $sin$. Simply use the jumping between $0$ and $1$ of $$frac{1+sin(npi)}{2}$$ and $$frac{1+sinleft(npi+frac{3pi}{2}right)}{2 }$$ which instead oscillates between $1$ and $0$. Then you can get the answer by describing $A = {5, x, 10, y, 15, z, 20}$ as well as $ B = {x, 6, y, 12, z, 18}$. Every other number in the lists $A$ and $B$ will either be numbers or they will be zero, since they're multiplied with said $sin$ function. $A + B$ is therefore a list containing two different methods of sorting, i.e. $A$'s function could be $ x^2 $ and $B$'s function could be $x^3-x^2+2x $ where every item in the list $C$, where $C = A + B$, is based on a different equation than the one before and after that said item.



Now, again, is there a way to create this said list $C$, containing the pattern of two very different functions that create the list $C $ where every item in the slot $ 2n$ is connected to one function but that function is different from the item in the slot $2n + 1$?






sequences-and-series







share|cite|improve this question











$endgroup$




The list is: $ {5,6,10,12,15,18,20,24,25...}$



Is there a way to describe this list of numbers mathematically without using $sin(x)
, cos(x), cos(nx)$, etc?



There is a way using the trigonometric function $sin$. Simply use the jumping between $0$ and $1$ of $$frac{1+sin(npi)}{2}$$ and $$frac{1+sinleft(npi+frac{3pi}{2}right)}{2 }$$ which instead oscillates between $1$ and $0$. Then you can get the answer by describing $A = {5, x, 10, y, 15, z, 20}$ as well as $ B = {x, 6, y, 12, z, 18}$. Every other number in the lists $A$ and $B$ will either be numbers or they will be zero, since they're multiplied with said $sin$ function. $A + B$ is therefore a list containing two different methods of sorting, i.e. $A$'s function could be $ x^2 $ and $B$'s function could be $x^3-x^2+2x $ where every item in the list $C$, where $C = A + B$, is based on a different equation than the one before and after that said item.



Now, again, is there a way to create this said list $C$, containing the pattern of two very different functions that create the list $C $ where every item in the slot $ 2n$ is connected to one function but that function is different from the item in the slot $2n + 1$?






sequences-and-series




sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 23:37









Dave

8,78711033




8,78711033










asked Dec 24 '18 at 19:22









lmaolmao

83




83








  • 3




    $begingroup$
    Very simply: it's the sequence $(5n,6n)_{nge 1}$.
    $endgroup$
    – Bernard
    Dec 24 '18 at 19:24






  • 2




    $begingroup$
    $f(n)=5n$, when $n equiv 1 pmod{2} $ and $f(n)=3n$, when $n equiv 0 pmod{2}$.
    $endgroup$
    – Anurag A
    Dec 24 '18 at 19:26








  • 1




    $begingroup$
    To give a single formula that works for all $n$, think about $(1+(-1)^n)/2$ and $(1-(-1)^n)/2$.
    $endgroup$
    – David C. Ullrich
    Dec 24 '18 at 19:55
















  • 3




    $begingroup$
    Very simply: it's the sequence $(5n,6n)_{nge 1}$.
    $endgroup$
    – Bernard
    Dec 24 '18 at 19:24






  • 2




    $begingroup$
    $f(n)=5n$, when $n equiv 1 pmod{2} $ and $f(n)=3n$, when $n equiv 0 pmod{2}$.
    $endgroup$
    – Anurag A
    Dec 24 '18 at 19:26








  • 1




    $begingroup$
    To give a single formula that works for all $n$, think about $(1+(-1)^n)/2$ and $(1-(-1)^n)/2$.
    $endgroup$
    – David C. Ullrich
    Dec 24 '18 at 19:55










3




3




$begingroup$
Very simply: it's the sequence $(5n,6n)_{nge 1}$.
$endgroup$
– Bernard
Dec 24 '18 at 19:24




$begingroup$
Very simply: it's the sequence $(5n,6n)_{nge 1}$.
$endgroup$
– Bernard
Dec 24 '18 at 19:24




2




2




$begingroup$
$f(n)=5n$, when $n equiv 1 pmod{2} $ and $f(n)=3n$, when $n equiv 0 pmod{2}$.
$endgroup$
– Anurag A
Dec 24 '18 at 19:26






$begingroup$
$f(n)=5n$, when $n equiv 1 pmod{2} $ and $f(n)=3n$, when $n equiv 0 pmod{2}$.
$endgroup$
– Anurag A
Dec 24 '18 at 19:26






1




1




$begingroup$
To give a single formula that works for all $n$, think about $(1+(-1)^n)/2$ and $(1-(-1)^n)/2$.
$endgroup$
– David C. Ullrich
Dec 24 '18 at 19:55






$begingroup$
To give a single formula that works for all $n$, think about $(1+(-1)^n)/2$ and $(1-(-1)^n)/2$.
$endgroup$
– David C. Ullrich
Dec 24 '18 at 19:55












2 Answers
2






active

oldest

votes


















2












$begingroup$

What you have is three sequences



$$ a_1,a_2,a_3,cdots $$



$$ b_1,b_2,b_3,cdots$$



with a third sequence which is a 'shuffle' of the first two



$$ a_1,b_1,a_2,b_2,a_3,b_3,cdots=c_1,c_2,c_3,c_4,c_5,c_6cdots$$



Define two functions on $mathbb{N}$:




  1. $k(n)=leftlfloordfrac{n+1}{2}rightrfloor$

  2. $h(n)=dfrac{1+(-1)^{n+1}}{2}$


Then $k$ defines a sequence $1,1,2,2,3,3,cdots$ and



$h$ defines a sequence $1,0,1,0,1,0,cdots$



Then the sequence $c_n$ can be defined



$$ c_n=a_{k(n)}h(n)+b_{k(n)}h(n+1) $$



Then



begin{eqnarray}
c_1&=&a_{k(1)}h(1)+b_{k(1)}h(2)\
&=&a_1cdot1+b_1cdot0\
&=&a_1
end{eqnarray}



begin{eqnarray}
c_2&=&a_{k(2)}h(2)+b_{k(2)}h(3)\
&=&a_1cdot0+b_1cdot1\
&=&b_1
end{eqnarray}



etc.



Substituting in your particular sequences will result in



$$c_n=left(frac{11+(-1)^n}{2}right)k_n$$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    If you just want to define a sequence so that the terms come from one function and then another, alternately, a simple way is:



    $$
    c_n = begin{cases}
    f(n) & text{if $n equiv 0 pmod 2$}, \
    g(n) & text{if $n equiv 1 pmod 2$}.
    end{cases}
    $$






    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












      $begingroup$

      What you have is three sequences



      $$ a_1,a_2,a_3,cdots $$



      $$ b_1,b_2,b_3,cdots$$



      with a third sequence which is a 'shuffle' of the first two



      $$ a_1,b_1,a_2,b_2,a_3,b_3,cdots=c_1,c_2,c_3,c_4,c_5,c_6cdots$$



      Define two functions on $mathbb{N}$:




      1. $k(n)=leftlfloordfrac{n+1}{2}rightrfloor$

      2. $h(n)=dfrac{1+(-1)^{n+1}}{2}$


      Then $k$ defines a sequence $1,1,2,2,3,3,cdots$ and



      $h$ defines a sequence $1,0,1,0,1,0,cdots$



      Then the sequence $c_n$ can be defined



      $$ c_n=a_{k(n)}h(n)+b_{k(n)}h(n+1) $$



      Then



      begin{eqnarray}
      c_1&=&a_{k(1)}h(1)+b_{k(1)}h(2)\
      &=&a_1cdot1+b_1cdot0\
      &=&a_1
      end{eqnarray}



      begin{eqnarray}
      c_2&=&a_{k(2)}h(2)+b_{k(2)}h(3)\
      &=&a_1cdot0+b_1cdot1\
      &=&b_1
      end{eqnarray}



      etc.



      Substituting in your particular sequences will result in



      $$c_n=left(frac{11+(-1)^n}{2}right)k_n$$






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        What you have is three sequences



        $$ a_1,a_2,a_3,cdots $$



        $$ b_1,b_2,b_3,cdots$$



        with a third sequence which is a 'shuffle' of the first two



        $$ a_1,b_1,a_2,b_2,a_3,b_3,cdots=c_1,c_2,c_3,c_4,c_5,c_6cdots$$



        Define two functions on $mathbb{N}$:




        1. $k(n)=leftlfloordfrac{n+1}{2}rightrfloor$

        2. $h(n)=dfrac{1+(-1)^{n+1}}{2}$


        Then $k$ defines a sequence $1,1,2,2,3,3,cdots$ and



        $h$ defines a sequence $1,0,1,0,1,0,cdots$



        Then the sequence $c_n$ can be defined



        $$ c_n=a_{k(n)}h(n)+b_{k(n)}h(n+1) $$



        Then



        begin{eqnarray}
        c_1&=&a_{k(1)}h(1)+b_{k(1)}h(2)\
        &=&a_1cdot1+b_1cdot0\
        &=&a_1
        end{eqnarray}



        begin{eqnarray}
        c_2&=&a_{k(2)}h(2)+b_{k(2)}h(3)\
        &=&a_1cdot0+b_1cdot1\
        &=&b_1
        end{eqnarray}



        etc.



        Substituting in your particular sequences will result in



        $$c_n=left(frac{11+(-1)^n}{2}right)k_n$$






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          What you have is three sequences



          $$ a_1,a_2,a_3,cdots $$



          $$ b_1,b_2,b_3,cdots$$



          with a third sequence which is a 'shuffle' of the first two



          $$ a_1,b_1,a_2,b_2,a_3,b_3,cdots=c_1,c_2,c_3,c_4,c_5,c_6cdots$$



          Define two functions on $mathbb{N}$:




          1. $k(n)=leftlfloordfrac{n+1}{2}rightrfloor$

          2. $h(n)=dfrac{1+(-1)^{n+1}}{2}$


          Then $k$ defines a sequence $1,1,2,2,3,3,cdots$ and



          $h$ defines a sequence $1,0,1,0,1,0,cdots$



          Then the sequence $c_n$ can be defined



          $$ c_n=a_{k(n)}h(n)+b_{k(n)}h(n+1) $$



          Then



          begin{eqnarray}
          c_1&=&a_{k(1)}h(1)+b_{k(1)}h(2)\
          &=&a_1cdot1+b_1cdot0\
          &=&a_1
          end{eqnarray}



          begin{eqnarray}
          c_2&=&a_{k(2)}h(2)+b_{k(2)}h(3)\
          &=&a_1cdot0+b_1cdot1\
          &=&b_1
          end{eqnarray}



          etc.



          Substituting in your particular sequences will result in



          $$c_n=left(frac{11+(-1)^n}{2}right)k_n$$






          share|cite|improve this answer











          $endgroup$



          What you have is three sequences



          $$ a_1,a_2,a_3,cdots $$



          $$ b_1,b_2,b_3,cdots$$



          with a third sequence which is a 'shuffle' of the first two



          $$ a_1,b_1,a_2,b_2,a_3,b_3,cdots=c_1,c_2,c_3,c_4,c_5,c_6cdots$$



          Define two functions on $mathbb{N}$:




          1. $k(n)=leftlfloordfrac{n+1}{2}rightrfloor$

          2. $h(n)=dfrac{1+(-1)^{n+1}}{2}$


          Then $k$ defines a sequence $1,1,2,2,3,3,cdots$ and



          $h$ defines a sequence $1,0,1,0,1,0,cdots$



          Then the sequence $c_n$ can be defined



          $$ c_n=a_{k(n)}h(n)+b_{k(n)}h(n+1) $$



          Then



          begin{eqnarray}
          c_1&=&a_{k(1)}h(1)+b_{k(1)}h(2)\
          &=&a_1cdot1+b_1cdot0\
          &=&a_1
          end{eqnarray}



          begin{eqnarray}
          c_2&=&a_{k(2)}h(2)+b_{k(2)}h(3)\
          &=&a_1cdot0+b_1cdot1\
          &=&b_1
          end{eqnarray}



          etc.



          Substituting in your particular sequences will result in



          $$c_n=left(frac{11+(-1)^n}{2}right)k_n$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 24 '18 at 21:59

























          answered Dec 24 '18 at 21:47









          John Wayland BalesJohn Wayland Bales

          14.1k21238




          14.1k21238























              0












              $begingroup$

              If you just want to define a sequence so that the terms come from one function and then another, alternately, a simple way is:



              $$
              c_n = begin{cases}
              f(n) & text{if $n equiv 0 pmod 2$}, \
              g(n) & text{if $n equiv 1 pmod 2$}.
              end{cases}
              $$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                If you just want to define a sequence so that the terms come from one function and then another, alternately, a simple way is:



                $$
                c_n = begin{cases}
                f(n) & text{if $n equiv 0 pmod 2$}, \
                g(n) & text{if $n equiv 1 pmod 2$}.
                end{cases}
                $$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  If you just want to define a sequence so that the terms come from one function and then another, alternately, a simple way is:



                  $$
                  c_n = begin{cases}
                  f(n) & text{if $n equiv 0 pmod 2$}, \
                  g(n) & text{if $n equiv 1 pmod 2$}.
                  end{cases}
                  $$






                  share|cite|improve this answer









                  $endgroup$



                  If you just want to define a sequence so that the terms come from one function and then another, alternately, a simple way is:



                  $$
                  c_n = begin{cases}
                  f(n) & text{if $n equiv 0 pmod 2$}, \
                  g(n) & text{if $n equiv 1 pmod 2$}.
                  end{cases}
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 24 '18 at 22:03









                  David KDavid K

                  54k342116




                  54k342116






























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                      .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 .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
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