Mixing
How can one solve the recurrence relation $a(n+3) = Ba(n)/n^2$?
$begingroup$
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.
ordinary-differential-equations recurrence-relations frobenius-method
asked Jan 3 at 22:13
Adam Robert DenchfieldAdam Robert Denchfield
81
$endgroup$
add a comment |
$begingroup$
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.
ordinary-differential-equations recurrence-relations frobenius-method
asked Jan 3 at 22:13
Adam Robert DenchfieldAdam Robert Denchfield
81
$endgroup$
$begingroup$
$e^{-x/2}y(x)$ is a solution of Airy's differential equation: mathworld.wolfram.com/AiryDifferentialEquation.html
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:18
$begingroup$
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.
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:22
1
$begingroup$
wolframalpha.com/input/?i=a(n%2B3)%3DB+a(n)%2Fn%5E2
$endgroup$
– R. Burton
Jan 3 at 22:44
$begingroup$
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?
$endgroup$
– Adam Robert Denchfield
Jan 5 at 2:30
add a comment |
$begingroup$
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.
ordinary-differential-equations recurrence-relations frobenius-method
asked Jan 3 at 22:13
Adam Robert DenchfieldAdam Robert Denchfield
81
$endgroup$
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.
ordinary-differential-equations recurrence-relations frobenius-method
ordinary-differential-equations recurrence-relations frobenius-method
asked Jan 3 at 22:13
Adam Robert DenchfieldAdam Robert Denchfield
81
asked Jan 3 at 22:13
Adam Robert DenchfieldAdam Robert Denchfield
81
asked Jan 3 at 22:13
Adam Robert DenchfieldAdam Robert Denchfield
81
asked Jan 3 at 22:13
Adam Robert DenchfieldAdam Robert Denchfield
81
asked Jan 3 at 22:13
Adam Robert DenchfieldAdam Robert Denchfield
81
81
$begingroup$
$e^{-x/2}y(x)$ is a solution of Airy's differential equation: mathworld.wolfram.com/AiryDifferentialEquation.html
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:18
$begingroup$
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.
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:22
1
$begingroup$
wolframalpha.com/input/?i=a(n%2B3)%3DB+a(n)%2Fn%5E2
$endgroup$
– R. Burton
Jan 3 at 22:44
$begingroup$
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?
$endgroup$
– Adam Robert Denchfield
Jan 5 at 2:30
add a comment |
$begingroup$
$e^{-x/2}y(x)$ is a solution of Airy's differential equation: mathworld.wolfram.com/AiryDifferentialEquation.html
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:18
$begingroup$
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.
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:22
1
$begingroup$
wolframalpha.com/input/?i=a(n%2B3)%3DB+a(n)%2Fn%5E2
$endgroup$
– R. Burton
Jan 3 at 22:44
$begingroup$
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?
$endgroup$
– Adam Robert Denchfield
Jan 5 at 2:30
$begingroup$
$e^{-x/2}y(x)$ is a solution of Airy's differential equation: mathworld.wolfram.com/AiryDifferentialEquation.html
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:18
$begingroup$
$e^{-x/2}y(x)$ is a solution of Airy's differential equation: mathworld.wolfram.com/AiryDifferentialEquation.html
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:18
$begingroup$
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.
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:22
$begingroup$
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.
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:22
1
1
$begingroup$
wolframalpha.com/input/?i=a(n%2B3)%3DB+a(n)%2Fn%5E2
$endgroup$
– R. Burton
Jan 3 at 22:44
$begingroup$
wolframalpha.com/input/?i=a(n%2B3)%3DB+a(n)%2Fn%5E2
$endgroup$
– R. Burton
Jan 3 at 22:44
$begingroup$
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?
$endgroup$
– Adam Robert Denchfield
Jan 5 at 2:30
$begingroup$
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?
$endgroup$
– Adam Robert Denchfield
Jan 5 at 2:30
add a comment |
1 Answer
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$begingroup$
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.
answered Jan 3 at 23:27
LutzLLutzL
57.2k42054
$endgroup$
add a comment |
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$begingroup$
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.
answered Jan 3 at 23:27
LutzLLutzL
57.2k42054
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 3 at 23:27
LutzLLutzL
57.2k42054
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 3 at 23:27
LutzLLutzL
57.2k42054
$endgroup$
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.
answered Jan 3 at 23:27
LutzLLutzL
57.2k42054
answered Jan 3 at 23:27
LutzLLutzL
57.2k42054
answered Jan 3 at 23:27
LutzLLutzL
57.2k42054
answered Jan 3 at 23:27
LutzLLutzL
57.2k42054
57.2k42054
add a comment |
add a comment |
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$begingroup$
$e^{-x/2}y(x)$ is a solution of Airy's differential equation: mathworld.wolfram.com/AiryDifferentialEquation.html
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:18
$begingroup$
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.
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:22
1
$begingroup$
wolframalpha.com/input/?i=a(n%2B3)%3DB+a(n)%2Fn%5E2
$endgroup$
– R. Burton
Jan 3 at 22:44
$begingroup$
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?
$endgroup$
– Adam Robert Denchfield
Jan 5 at 2:30