How can I tell if a sequence can be a recursive sequence












2














Obviously 1, 1, 2, 3, 5, 8, etc.. is possibly a recursive sequence because we know it's the Fibonacci sequence and because it's really easy to see the pattern. It's recursive "function" for the form $a(n_{i - 1}) + b (n_{i - 2})$ where $a = 1$ and $b = 1$.



But how do we know for sure that 1, 2, 3, 4, _, _, etc.. can't have a recursive function? where there is a $a$ or $b$ that works for that sequence?






linear-algebra







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  • Argh I don't know how to make more than one character go in a subscript can someone help me out? Thanks!
    – ming
    Dec 3 '18 at 8:52










  • to make multiple subscript charachters you can do : normal text_{your subscript text} which gives: $$normal text_{your subscript text}$$ Place all characters inside brackets
    – TheD0ubleT
    Dec 3 '18 at 8:59












  • Fancy, thanks!!
    – ming
    Dec 3 '18 at 9:40
















2














Obviously 1, 1, 2, 3, 5, 8, etc.. is possibly a recursive sequence because we know it's the Fibonacci sequence and because it's really easy to see the pattern. It's recursive "function" for the form $a(n_{i - 1}) + b (n_{i - 2})$ where $a = 1$ and $b = 1$.



But how do we know for sure that 1, 2, 3, 4, _, _, etc.. can't have a recursive function? where there is a $a$ or $b$ that works for that sequence?






linear-algebra







share|cite|improve this question
























  • Argh I don't know how to make more than one character go in a subscript can someone help me out? Thanks!
    – ming
    Dec 3 '18 at 8:52










  • to make multiple subscript charachters you can do : normal text_{your subscript text} which gives: $$normal text_{your subscript text}$$ Place all characters inside brackets
    – TheD0ubleT
    Dec 3 '18 at 8:59












  • Fancy, thanks!!
    – ming
    Dec 3 '18 at 9:40














2












2








2







Obviously 1, 1, 2, 3, 5, 8, etc.. is possibly a recursive sequence because we know it's the Fibonacci sequence and because it's really easy to see the pattern. It's recursive "function" for the form $a(n_{i - 1}) + b (n_{i - 2})$ where $a = 1$ and $b = 1$.



But how do we know for sure that 1, 2, 3, 4, _, _, etc.. can't have a recursive function? where there is a $a$ or $b$ that works for that sequence?






linear-algebra







share|cite|improve this question















Obviously 1, 1, 2, 3, 5, 8, etc.. is possibly a recursive sequence because we know it's the Fibonacci sequence and because it's really easy to see the pattern. It's recursive "function" for the form $a(n_{i - 1}) + b (n_{i - 2})$ where $a = 1$ and $b = 1$.



But how do we know for sure that 1, 2, 3, 4, _, _, etc.. can't have a recursive function? where there is a $a$ or $b$ that works for that sequence?






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 3 '18 at 9:40

























asked Dec 3 '18 at 8:51









ming

3165




3165












  • Argh I don't know how to make more than one character go in a subscript can someone help me out? Thanks!
    – ming
    Dec 3 '18 at 8:52










  • to make multiple subscript charachters you can do : normal text_{your subscript text} which gives: $$normal text_{your subscript text}$$ Place all characters inside brackets
    – TheD0ubleT
    Dec 3 '18 at 8:59












  • Fancy, thanks!!
    – ming
    Dec 3 '18 at 9:40


















  • Argh I don't know how to make more than one character go in a subscript can someone help me out? Thanks!
    – ming
    Dec 3 '18 at 8:52










  • to make multiple subscript charachters you can do : normal text_{your subscript text} which gives: $$normal text_{your subscript text}$$ Place all characters inside brackets
    – TheD0ubleT
    Dec 3 '18 at 8:59












  • Fancy, thanks!!
    – ming
    Dec 3 '18 at 9:40
















Argh I don't know how to make more than one character go in a subscript can someone help me out? Thanks!
– ming
Dec 3 '18 at 8:52




Argh I don't know how to make more than one character go in a subscript can someone help me out? Thanks!
– ming
Dec 3 '18 at 8:52












to make multiple subscript charachters you can do : normal text_{your subscript text} which gives: $$normal text_{your subscript text}$$ Place all characters inside brackets
– TheD0ubleT
Dec 3 '18 at 8:59






to make multiple subscript charachters you can do : normal text_{your subscript text} which gives: $$normal text_{your subscript text}$$ Place all characters inside brackets
– TheD0ubleT
Dec 3 '18 at 8:59














Fancy, thanks!!
– ming
Dec 3 '18 at 9:40




Fancy, thanks!!
– ming
Dec 3 '18 at 9:40










1 Answer
1






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oldest

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1














You might want to look into : Solving homogeneous linear recurrence relations with constant coefficients.



We can easily find the solutions for this kind of linear recurrence relation:

If $x^2-ax-b = 0$ has 2 real solutions $r_1$ and $r_2$ then the sequence looks like : $u_n=A*r_1^n + B*r_2^n$



If $x^2-ax-b = 0$ has 1 real solution $r$ then the sequence looks like : $u_n=(n*A+B)*r^n $



If $x^2-ax-b = 0$ has 2 complex solutions $r_1$ and $r_2$ then the sequence looks like : $u_n=|r|^n*(Acos(ntheta)+Bsin(ntheta))$ with $theta = arg(r)$



As you can see there is always r^n which prevents linear sequences like $1,2,3,4,5...$ (except if $r = 1$)

In that case your characteristic polynomial looks like $(r-1)^2=0$, thus your recursive relation looks like $u_n=2u_{n-1}-u_{n-2}$

Also, you'll need $A = 1$ and $B = 0$.

that means $u_0=0$ and $u_1=1$



Note: I can make a more in depth answer but is is very well described in the article i have linked






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    You might want to look into : Solving homogeneous linear recurrence relations with constant coefficients.



    We can easily find the solutions for this kind of linear recurrence relation:

    If $x^2-ax-b = 0$ has 2 real solutions $r_1$ and $r_2$ then the sequence looks like : $u_n=A*r_1^n + B*r_2^n$



    If $x^2-ax-b = 0$ has 1 real solution $r$ then the sequence looks like : $u_n=(n*A+B)*r^n $



    If $x^2-ax-b = 0$ has 2 complex solutions $r_1$ and $r_2$ then the sequence looks like : $u_n=|r|^n*(Acos(ntheta)+Bsin(ntheta))$ with $theta = arg(r)$



    As you can see there is always r^n which prevents linear sequences like $1,2,3,4,5...$ (except if $r = 1$)

    In that case your characteristic polynomial looks like $(r-1)^2=0$, thus your recursive relation looks like $u_n=2u_{n-1}-u_{n-2}$

    Also, you'll need $A = 1$ and $B = 0$.

    that means $u_0=0$ and $u_1=1$



    Note: I can make a more in depth answer but is is very well described in the article i have linked






    share|cite|improve this answer


























      1














      You might want to look into : Solving homogeneous linear recurrence relations with constant coefficients.



      We can easily find the solutions for this kind of linear recurrence relation:

      If $x^2-ax-b = 0$ has 2 real solutions $r_1$ and $r_2$ then the sequence looks like : $u_n=A*r_1^n + B*r_2^n$



      If $x^2-ax-b = 0$ has 1 real solution $r$ then the sequence looks like : $u_n=(n*A+B)*r^n $



      If $x^2-ax-b = 0$ has 2 complex solutions $r_1$ and $r_2$ then the sequence looks like : $u_n=|r|^n*(Acos(ntheta)+Bsin(ntheta))$ with $theta = arg(r)$



      As you can see there is always r^n which prevents linear sequences like $1,2,3,4,5...$ (except if $r = 1$)

      In that case your characteristic polynomial looks like $(r-1)^2=0$, thus your recursive relation looks like $u_n=2u_{n-1}-u_{n-2}$

      Also, you'll need $A = 1$ and $B = 0$.

      that means $u_0=0$ and $u_1=1$



      Note: I can make a more in depth answer but is is very well described in the article i have linked






      share|cite|improve this answer
























        1












        1








        1






        You might want to look into : Solving homogeneous linear recurrence relations with constant coefficients.



        We can easily find the solutions for this kind of linear recurrence relation:

        If $x^2-ax-b = 0$ has 2 real solutions $r_1$ and $r_2$ then the sequence looks like : $u_n=A*r_1^n + B*r_2^n$



        If $x^2-ax-b = 0$ has 1 real solution $r$ then the sequence looks like : $u_n=(n*A+B)*r^n $



        If $x^2-ax-b = 0$ has 2 complex solutions $r_1$ and $r_2$ then the sequence looks like : $u_n=|r|^n*(Acos(ntheta)+Bsin(ntheta))$ with $theta = arg(r)$



        As you can see there is always r^n which prevents linear sequences like $1,2,3,4,5...$ (except if $r = 1$)

        In that case your characteristic polynomial looks like $(r-1)^2=0$, thus your recursive relation looks like $u_n=2u_{n-1}-u_{n-2}$

        Also, you'll need $A = 1$ and $B = 0$.

        that means $u_0=0$ and $u_1=1$



        Note: I can make a more in depth answer but is is very well described in the article i have linked






        share|cite|improve this answer












        You might want to look into : Solving homogeneous linear recurrence relations with constant coefficients.



        We can easily find the solutions for this kind of linear recurrence relation:

        If $x^2-ax-b = 0$ has 2 real solutions $r_1$ and $r_2$ then the sequence looks like : $u_n=A*r_1^n + B*r_2^n$



        If $x^2-ax-b = 0$ has 1 real solution $r$ then the sequence looks like : $u_n=(n*A+B)*r^n $



        If $x^2-ax-b = 0$ has 2 complex solutions $r_1$ and $r_2$ then the sequence looks like : $u_n=|r|^n*(Acos(ntheta)+Bsin(ntheta))$ with $theta = arg(r)$



        As you can see there is always r^n which prevents linear sequences like $1,2,3,4,5...$ (except if $r = 1$)

        In that case your characteristic polynomial looks like $(r-1)^2=0$, thus your recursive relation looks like $u_n=2u_{n-1}-u_{n-2}$

        Also, you'll need $A = 1$ and $B = 0$.

        that means $u_0=0$ and $u_1=1$



        Note: I can make a more in depth answer but is is very well described in the article i have linked







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 9:44









        TheD0ubleT

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