How can one solve the recurrence relation $a(n+3) = Ba(n)/n^2$?












0














As the title suggests, I am looking for the solutions to the recurrence relation



$a(n+3) = B frac{a(n)}{n^2}$.



In particular, I am attempting to solve a differential equation using the power series method and came across this formula for the coefficients. If you're curious, the differential equation is something like



$frac{y'(x)}{x^2} - frac{y''(x)}{x} - By(x) = 0. $



I was wondering if this were solvable in closed-form. Since I'm not an expert on recurrence relations, I didn't really know where to start (though the internet tells me nonlinear recurrence relations are usually not easy to solve).



Nevertheless, I note that if one defines $b(n) equiv 1/Gamma(n)$, one gets the relation $b(n+2) = frac{b(n)}{n^2}$. This gives me hope it's solvable in terms of (scaled/shifted) gamma functions and exponentials.



Any advice?



Note: Not homework or anything, I'm just playing around with trying to find functions orthogonal to each other given the weight function w(x) = 1/x.










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  • $e^{-x/2}y(x)$ is a solution of Airy's differential equation: mathworld.wolfram.com/AiryDifferentialEquation.html
    – Jack D'Aurizio
    Jan 3 at 22:18










  • About orthogonal functions with respect to $langle f,grangle=int_{0}^{+infty}f(x)g(x)frac{dx}{x}$, Bessel functions $J_{2n}(x)$ do the job.
    – Jack D'Aurizio
    Jan 3 at 22:22






  • 1




    wolframalpha.com/input/?i=a(n%2B3)%3DB+a(n)%2Fn%5E2
    – R. Burton
    Jan 3 at 22:44










  • Woah, WolframAlpha is stronger than I thought, Mr. Burton. Thanks to you and Jack! Though Jack, Wikipedia tells me the Bessel functions are orthogonal with respect to the weight function x, not 1/x. Do you have a source I could read for it?
    – Adam Robert Denchfield
    2 days ago


















0














As the title suggests, I am looking for the solutions to the recurrence relation



$a(n+3) = B frac{a(n)}{n^2}$.



In particular, I am attempting to solve a differential equation using the power series method and came across this formula for the coefficients. If you're curious, the differential equation is something like



$frac{y'(x)}{x^2} - frac{y''(x)}{x} - By(x) = 0. $



I was wondering if this were solvable in closed-form. Since I'm not an expert on recurrence relations, I didn't really know where to start (though the internet tells me nonlinear recurrence relations are usually not easy to solve).



Nevertheless, I note that if one defines $b(n) equiv 1/Gamma(n)$, one gets the relation $b(n+2) = frac{b(n)}{n^2}$. This gives me hope it's solvable in terms of (scaled/shifted) gamma functions and exponentials.



Any advice?



Note: Not homework or anything, I'm just playing around with trying to find functions orthogonal to each other given the weight function w(x) = 1/x.










share|cite|improve this question







New contributor




Adam Robert Denchfield is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • $e^{-x/2}y(x)$ is a solution of Airy's differential equation: mathworld.wolfram.com/AiryDifferentialEquation.html
    – Jack D'Aurizio
    Jan 3 at 22:18










  • About orthogonal functions with respect to $langle f,grangle=int_{0}^{+infty}f(x)g(x)frac{dx}{x}$, Bessel functions $J_{2n}(x)$ do the job.
    – Jack D'Aurizio
    Jan 3 at 22:22






  • 1




    wolframalpha.com/input/?i=a(n%2B3)%3DB+a(n)%2Fn%5E2
    – R. Burton
    Jan 3 at 22:44










  • Woah, WolframAlpha is stronger than I thought, Mr. Burton. Thanks to you and Jack! Though Jack, Wikipedia tells me the Bessel functions are orthogonal with respect to the weight function x, not 1/x. Do you have a source I could read for it?
    – Adam Robert Denchfield
    2 days ago
















0












0








0







As the title suggests, I am looking for the solutions to the recurrence relation



$a(n+3) = B frac{a(n)}{n^2}$.



In particular, I am attempting to solve a differential equation using the power series method and came across this formula for the coefficients. If you're curious, the differential equation is something like



$frac{y'(x)}{x^2} - frac{y''(x)}{x} - By(x) = 0. $



I was wondering if this were solvable in closed-form. Since I'm not an expert on recurrence relations, I didn't really know where to start (though the internet tells me nonlinear recurrence relations are usually not easy to solve).



Nevertheless, I note that if one defines $b(n) equiv 1/Gamma(n)$, one gets the relation $b(n+2) = frac{b(n)}{n^2}$. This gives me hope it's solvable in terms of (scaled/shifted) gamma functions and exponentials.



Any advice?



Note: Not homework or anything, I'm just playing around with trying to find functions orthogonal to each other given the weight function w(x) = 1/x.










share|cite|improve this question







New contributor




Adam Robert Denchfield is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











As the title suggests, I am looking for the solutions to the recurrence relation



$a(n+3) = B frac{a(n)}{n^2}$.



In particular, I am attempting to solve a differential equation using the power series method and came across this formula for the coefficients. If you're curious, the differential equation is something like



$frac{y'(x)}{x^2} - frac{y''(x)}{x} - By(x) = 0. $



I was wondering if this were solvable in closed-form. Since I'm not an expert on recurrence relations, I didn't really know where to start (though the internet tells me nonlinear recurrence relations are usually not easy to solve).



Nevertheless, I note that if one defines $b(n) equiv 1/Gamma(n)$, one gets the relation $b(n+2) = frac{b(n)}{n^2}$. This gives me hope it's solvable in terms of (scaled/shifted) gamma functions and exponentials.



Any advice?



Note: Not homework or anything, I'm just playing around with trying to find functions orthogonal to each other given the weight function w(x) = 1/x.







differential-equations recurrence-relations frobenius-method






share|cite|improve this question







New contributor




Adam Robert Denchfield is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Adam Robert Denchfield is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




Adam Robert Denchfield is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked Jan 3 at 22:13









Adam Robert Denchfield

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81




New contributor




Adam Robert Denchfield is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Adam Robert Denchfield is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Adam Robert Denchfield is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $e^{-x/2}y(x)$ is a solution of Airy's differential equation: mathworld.wolfram.com/AiryDifferentialEquation.html
    – Jack D'Aurizio
    Jan 3 at 22:18










  • About orthogonal functions with respect to $langle f,grangle=int_{0}^{+infty}f(x)g(x)frac{dx}{x}$, Bessel functions $J_{2n}(x)$ do the job.
    – Jack D'Aurizio
    Jan 3 at 22:22






  • 1




    wolframalpha.com/input/?i=a(n%2B3)%3DB+a(n)%2Fn%5E2
    – R. Burton
    Jan 3 at 22:44










  • Woah, WolframAlpha is stronger than I thought, Mr. Burton. Thanks to you and Jack! Though Jack, Wikipedia tells me the Bessel functions are orthogonal with respect to the weight function x, not 1/x. Do you have a source I could read for it?
    – Adam Robert Denchfield
    2 days ago




















  • $e^{-x/2}y(x)$ is a solution of Airy's differential equation: mathworld.wolfram.com/AiryDifferentialEquation.html
    – Jack D'Aurizio
    Jan 3 at 22:18










  • About orthogonal functions with respect to $langle f,grangle=int_{0}^{+infty}f(x)g(x)frac{dx}{x}$, Bessel functions $J_{2n}(x)$ do the job.
    – Jack D'Aurizio
    Jan 3 at 22:22






  • 1




    wolframalpha.com/input/?i=a(n%2B3)%3DB+a(n)%2Fn%5E2
    – R. Burton
    Jan 3 at 22:44










  • Woah, WolframAlpha is stronger than I thought, Mr. Burton. Thanks to you and Jack! Though Jack, Wikipedia tells me the Bessel functions are orthogonal with respect to the weight function x, not 1/x. Do you have a source I could read for it?
    – Adam Robert Denchfield
    2 days ago


















$e^{-x/2}y(x)$ is a solution of Airy's differential equation: mathworld.wolfram.com/AiryDifferentialEquation.html
– Jack D'Aurizio
Jan 3 at 22:18




$e^{-x/2}y(x)$ is a solution of Airy's differential equation: mathworld.wolfram.com/AiryDifferentialEquation.html
– Jack D'Aurizio
Jan 3 at 22:18












About orthogonal functions with respect to $langle f,grangle=int_{0}^{+infty}f(x)g(x)frac{dx}{x}$, Bessel functions $J_{2n}(x)$ do the job.
– Jack D'Aurizio
Jan 3 at 22:22




About orthogonal functions with respect to $langle f,grangle=int_{0}^{+infty}f(x)g(x)frac{dx}{x}$, Bessel functions $J_{2n}(x)$ do the job.
– Jack D'Aurizio
Jan 3 at 22:22




1




1




wolframalpha.com/input/?i=a(n%2B3)%3DB+a(n)%2Fn%5E2
– R. Burton
Jan 3 at 22:44




wolframalpha.com/input/?i=a(n%2B3)%3DB+a(n)%2Fn%5E2
– R. Burton
Jan 3 at 22:44












Woah, WolframAlpha is stronger than I thought, Mr. Burton. Thanks to you and Jack! Though Jack, Wikipedia tells me the Bessel functions are orthogonal with respect to the weight function x, not 1/x. Do you have a source I could read for it?
– Adam Robert Denchfield
2 days ago






Woah, WolframAlpha is stronger than I thought, Mr. Burton. Thanks to you and Jack! Though Jack, Wikipedia tells me the Bessel functions are orthogonal with respect to the weight function x, not 1/x. Do you have a source I could read for it?
– Adam Robert Denchfield
2 days ago












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Check again your Gamma function understanding. If you set
$$
b_n=Γ(n/3)^2a_n,
$$

then
$$
b_{n+3}=Γ(n/3+1)^2a_{n+3}=(n/3)^2Γ(n/3)^2frac{Ba_n}{n^2}=frac{B}{9}b_n
$$

which is a simple geometric sequence for the sub-sequences of every third sequence element.






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    1 Answer
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    Check again your Gamma function understanding. If you set
    $$
    b_n=Γ(n/3)^2a_n,
    $$

    then
    $$
    b_{n+3}=Γ(n/3+1)^2a_{n+3}=(n/3)^2Γ(n/3)^2frac{Ba_n}{n^2}=frac{B}{9}b_n
    $$

    which is a simple geometric sequence for the sub-sequences of every third sequence element.






    share|cite|improve this answer


























      0














      Check again your Gamma function understanding. If you set
      $$
      b_n=Γ(n/3)^2a_n,
      $$

      then
      $$
      b_{n+3}=Γ(n/3+1)^2a_{n+3}=(n/3)^2Γ(n/3)^2frac{Ba_n}{n^2}=frac{B}{9}b_n
      $$

      which is a simple geometric sequence for the sub-sequences of every third sequence element.






      share|cite|improve this answer
























        0












        0








        0






        Check again your Gamma function understanding. If you set
        $$
        b_n=Γ(n/3)^2a_n,
        $$

        then
        $$
        b_{n+3}=Γ(n/3+1)^2a_{n+3}=(n/3)^2Γ(n/3)^2frac{Ba_n}{n^2}=frac{B}{9}b_n
        $$

        which is a simple geometric sequence for the sub-sequences of every third sequence element.






        share|cite|improve this answer












        Check again your Gamma function understanding. If you set
        $$
        b_n=Γ(n/3)^2a_n,
        $$

        then
        $$
        b_{n+3}=Γ(n/3+1)^2a_{n+3}=(n/3)^2Γ(n/3)^2frac{Ba_n}{n^2}=frac{B}{9}b_n
        $$

        which is a simple geometric sequence for the sub-sequences of every third sequence element.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 3 at 23:27









        LutzL

        56.4k42054




        56.4k42054






















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