Complicated linear recursion relations
$begingroup$
Is it possible to obtain a solution for $a_n$ for the following recursion relation?
$$a_n = -frac{1}{epsilon}(n+1)a_{n+1}+Rleft(frac{n+2}{n+1}right)a_{n+2}+frac{R}{epsilon}left(frac{(n+3)(n+2)}{n+1}right)a_{n+3}$$
If not, any other methods and hints on how to get rid of the recursion will be appreciated.
recurrence-relations recursion
asked Dec 16 '18 at 16:29
jarheadjarhead
1308
$endgroup$
add a comment |
$begingroup$
Is it possible to obtain a solution for $a_n$ for the following recursion relation?
$$a_n = -frac{1}{epsilon}(n+1)a_{n+1}+Rleft(frac{n+2}{n+1}right)a_{n+2}+frac{R}{epsilon}left(frac{(n+3)(n+2)}{n+1}right)a_{n+3}$$
If not, any other methods and hints on how to get rid of the recursion will be appreciated.
recurrence-relations recursion
asked Dec 16 '18 at 16:29
jarheadjarhead
1308
$endgroup$
add a comment |
$begingroup$
Is it possible to obtain a solution for $a_n$ for the following recursion relation?
$$a_n = -frac{1}{epsilon}(n+1)a_{n+1}+Rleft(frac{n+2}{n+1}right)a_{n+2}+frac{R}{epsilon}left(frac{(n+3)(n+2)}{n+1}right)a_{n+3}$$
If not, any other methods and hints on how to get rid of the recursion will be appreciated.
recurrence-relations recursion
asked Dec 16 '18 at 16:29
jarheadjarhead
1308
$endgroup$
Is it possible to obtain a solution for $a_n$ for the following recursion relation?
$$a_n = -frac{1}{epsilon}(n+1)a_{n+1}+Rleft(frac{n+2}{n+1}right)a_{n+2}+frac{R}{epsilon}left(frac{(n+3)(n+2)}{n+1}right)a_{n+3}$$
If not, any other methods and hints on how to get rid of the recursion will be appreciated.
recurrence-relations recursion
recurrence-relations recursion
asked Dec 16 '18 at 16:29
jarheadjarhead
1308
asked Dec 16 '18 at 16:29
jarheadjarhead
1308
asked Dec 16 '18 at 16:29
jarheadjarhead
1308
asked Dec 16 '18 at 16:29
jarheadjarhead
1308
asked Dec 16 '18 at 16:29
jarheadjarhead
1308
1308
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Maple gives:
$$a_n=frac{(-epsilon)^n}{Gamma(n+1)}left(a_0+sum_{i=0}^{n-1}tright)$$
$$t=-frac{1}{piepsilon}frac{4}{epsilon^2R}^frac{i}{2}(a_0epsilon+a_1)Gammaleft(frac{i+1}{2}right)^2 icolon even$$
$$t=frac{4}{epsilon^2R}^frac{i-1}{2}Gammaleft(frac{i+1}{2}right)^2left(frac{2a_2}{epsilon^2}+frac{a_1}{epsilon}right) icolon odd$$
answered Dec 16 '18 at 19:03
IV_IV_
1,511525
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1 Answer
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1 Answer
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active
oldest
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active
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$begingroup$
Maple gives:
$$a_n=frac{(-epsilon)^n}{Gamma(n+1)}left(a_0+sum_{i=0}^{n-1}tright)$$
$$t=-frac{1}{piepsilon}frac{4}{epsilon^2R}^frac{i}{2}(a_0epsilon+a_1)Gammaleft(frac{i+1}{2}right)^2 icolon even$$
$$t=frac{4}{epsilon^2R}^frac{i-1}{2}Gammaleft(frac{i+1}{2}right)^2left(frac{2a_2}{epsilon^2}+frac{a_1}{epsilon}right) icolon odd$$
answered Dec 16 '18 at 19:03
IV_IV_
1,511525
$endgroup$
add a comment |
$begingroup$
Maple gives:
$$a_n=frac{(-epsilon)^n}{Gamma(n+1)}left(a_0+sum_{i=0}^{n-1}tright)$$
$$t=-frac{1}{piepsilon}frac{4}{epsilon^2R}^frac{i}{2}(a_0epsilon+a_1)Gammaleft(frac{i+1}{2}right)^2 icolon even$$
$$t=frac{4}{epsilon^2R}^frac{i-1}{2}Gammaleft(frac{i+1}{2}right)^2left(frac{2a_2}{epsilon^2}+frac{a_1}{epsilon}right) icolon odd$$
answered Dec 16 '18 at 19:03
IV_IV_
1,511525
$endgroup$
add a comment |
$begingroup$
Maple gives:
$$a_n=frac{(-epsilon)^n}{Gamma(n+1)}left(a_0+sum_{i=0}^{n-1}tright)$$
$$t=-frac{1}{piepsilon}frac{4}{epsilon^2R}^frac{i}{2}(a_0epsilon+a_1)Gammaleft(frac{i+1}{2}right)^2 icolon even$$
$$t=frac{4}{epsilon^2R}^frac{i-1}{2}Gammaleft(frac{i+1}{2}right)^2left(frac{2a_2}{epsilon^2}+frac{a_1}{epsilon}right) icolon odd$$
answered Dec 16 '18 at 19:03
IV_IV_
1,511525
$endgroup$
Maple gives:
$$a_n=frac{(-epsilon)^n}{Gamma(n+1)}left(a_0+sum_{i=0}^{n-1}tright)$$
$$t=-frac{1}{piepsilon}frac{4}{epsilon^2R}^frac{i}{2}(a_0epsilon+a_1)Gammaleft(frac{i+1}{2}right)^2 icolon even$$
$$t=frac{4}{epsilon^2R}^frac{i-1}{2}Gammaleft(frac{i+1}{2}right)^2left(frac{2a_2}{epsilon^2}+frac{a_1}{epsilon}right) icolon odd$$
answered Dec 16 '18 at 19:03
IV_IV_
1,511525
answered Dec 16 '18 at 19:03
IV_IV_
1,511525
answered Dec 16 '18 at 19:03
IV_IV_
1,511525
answered Dec 16 '18 at 19:03
IV_IV_
1,511525
1,511525
add a comment |
add a comment |
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