RSolve not reducing for a certain recurrence relation
up vote
4
down vote
favorite
I'm trying to use RSolve as follows to solve a recurrence relation:
RSolve[{a[0] == 1, a[2 n + 1] == a[2 n]*2, a[2 n] == a[2 n - 1] + 2}, a, n]
I think the meaning of the relation is clear: for odd terms in the sequence, multiply the previous term by two, and for even terms, add two to the previous term.
However, when evaluating this, Mathematica simply echos the input:
rather than attempting to solve the recurrence.
I can't find anything in the RSolve documentation which talks about cases where RSolve will do nothing, without any error message.
Have I made a syntax error, or does this mean that Mathematica is not able to solve this type of recurrence relation? How can I change my input so that Mathematica does solve the recurrence, assuming it is possible?
equation-solving symbolic syntax recursion difference-equations
edited Nov 27 at 0:27
bbgodfrey
43.9k857108
asked Nov 26 at 19:34
konsolas
1235
add a comment |
up vote
4
down vote
favorite
I'm trying to use RSolve as follows to solve a recurrence relation:
RSolve[{a[0] == 1, a[2 n + 1] == a[2 n]*2, a[2 n] == a[2 n - 1] + 2}, a, n]
I think the meaning of the relation is clear: for odd terms in the sequence, multiply the previous term by two, and for even terms, add two to the previous term.
However, when evaluating this, Mathematica simply echos the input:
rather than attempting to solve the recurrence.
I can't find anything in the RSolve documentation which talks about cases where RSolve will do nothing, without any error message.
Have I made a syntax error, or does this mean that Mathematica is not able to solve this type of recurrence relation? How can I change my input so that Mathematica does solve the recurrence, assuming it is possible?
equation-solving symbolic syntax recursion difference-equations
edited Nov 27 at 0:27
bbgodfrey
43.9k857108
asked Nov 26 at 19:34
konsolas
1235
Try replacing the next-to-last term inRSolve(a) witha[n].
– yosimitsu kodanuri
Nov 26 at 20:30
It still simply echoes the input:RSolve[{a[0] == 1, a[1 + 2 n] == 2 a[2 n], a[2 n] == 2 + a[-1 + 2 n]}, a[n], n]
– konsolas
Nov 26 at 20:36
1
Welcome to Mathematica.SE! Interesting question(+1). I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise.
– bbgodfrey
Nov 27 at 0:28
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I'm trying to use RSolve as follows to solve a recurrence relation:
RSolve[{a[0] == 1, a[2 n + 1] == a[2 n]*2, a[2 n] == a[2 n - 1] + 2}, a, n]
I think the meaning of the relation is clear: for odd terms in the sequence, multiply the previous term by two, and for even terms, add two to the previous term.
However, when evaluating this, Mathematica simply echos the input:
rather than attempting to solve the recurrence.
I can't find anything in the RSolve documentation which talks about cases where RSolve will do nothing, without any error message.
Have I made a syntax error, or does this mean that Mathematica is not able to solve this type of recurrence relation? How can I change my input so that Mathematica does solve the recurrence, assuming it is possible?
equation-solving symbolic syntax recursion difference-equations
edited Nov 27 at 0:27
bbgodfrey
43.9k857108
asked Nov 26 at 19:34
konsolas
1235
I'm trying to use RSolve as follows to solve a recurrence relation:
RSolve[{a[0] == 1, a[2 n + 1] == a[2 n]*2, a[2 n] == a[2 n - 1] + 2}, a, n]
I think the meaning of the relation is clear: for odd terms in the sequence, multiply the previous term by two, and for even terms, add two to the previous term.
However, when evaluating this, Mathematica simply echos the input:
rather than attempting to solve the recurrence.
I can't find anything in the RSolve documentation which talks about cases where RSolve will do nothing, without any error message.
Have I made a syntax error, or does this mean that Mathematica is not able to solve this type of recurrence relation? How can I change my input so that Mathematica does solve the recurrence, assuming it is possible?
equation-solving symbolic syntax recursion difference-equations
equation-solving symbolic syntax recursion difference-equations
edited Nov 27 at 0:27
bbgodfrey
43.9k857108
asked Nov 26 at 19:34
konsolas
1235
edited Nov 27 at 0:27
bbgodfrey
43.9k857108
asked Nov 26 at 19:34
konsolas
1235
edited Nov 27 at 0:27
bbgodfrey
43.9k857108
edited Nov 27 at 0:27
bbgodfrey
43.9k857108
edited Nov 27 at 0:27
bbgodfrey
43.9k857108
43.9k857108
asked Nov 26 at 19:34
konsolas
1235
asked Nov 26 at 19:34
konsolas
1235
asked Nov 26 at 19:34
konsolas
1235
1235
Try replacing the next-to-last term inRSolve(a) witha[n].
– yosimitsu kodanuri
Nov 26 at 20:30
It still simply echoes the input:RSolve[{a[0] == 1, a[1 + 2 n] == 2 a[2 n], a[2 n] == 2 + a[-1 + 2 n]}, a[n], n]
– konsolas
Nov 26 at 20:36
1
Welcome to Mathematica.SE! Interesting question(+1). I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise.
– bbgodfrey
Nov 27 at 0:28
add a comment |
Try replacing the next-to-last term inRSolve(a) witha[n].
– yosimitsu kodanuri
Nov 26 at 20:30
It still simply echoes the input:RSolve[{a[0] == 1, a[1 + 2 n] == 2 a[2 n], a[2 n] == 2 + a[-1 + 2 n]}, a[n], n]
– konsolas
Nov 26 at 20:36
1
Welcome to Mathematica.SE! Interesting question(+1). I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise.
– bbgodfrey
Nov 27 at 0:28
Try replacing the next-to-last term in
RSolve (a) with a[n].– yosimitsu kodanuri
Nov 26 at 20:30
Try replacing the next-to-last term in
RSolve (a) with a[n].– yosimitsu kodanuri
Nov 26 at 20:30
It still simply echoes the input:
RSolve[{a[0] == 1, a[1 + 2 n] == 2 a[2 n], a[2 n] == 2 + a[-1 + 2 n]}, a[n], n]– konsolas
Nov 26 at 20:36
It still simply echoes the input:
RSolve[{a[0] == 1, a[1 + 2 n] == 2 a[2 n], a[2 n] == 2 + a[-1 + 2 n]}, a[n], n]– konsolas
Nov 26 at 20:36
1
1
Welcome to Mathematica.SE! Interesting question(+1). I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise.
– bbgodfrey
Nov 27 at 0:28
Welcome to Mathematica.SE! Interesting question(+1). I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise.
– bbgodfrey
Nov 27 at 0:28
add a comment |
2 Answers
2
active
oldest
votes
up vote
7
down vote
accepted
It appears that RSolve cannot solve difference equations in which two different equations describe the same variable. If so, a work-around is to represent a[k] with k an even index as c[k], and with k an odd indix as b[k], so that there is only one equation per variable.
FullSimplify[RSolveValue[{c[0] == 1, b[k] == c[k - 1]*2, c[k + 2] == b[k + 1] + 2},
{c[k], b[k]}, k] /. C[1] -> 0]
(* {-2 + 3 2^(-1 + k/2) (1 + (-1)^k), -4 - 3 2^(1/2 (-1 + k)) (-1 + (-1)^k)} *)
Then, construct the desired a[k] as even-index terms of c[k] and odd-index terms of b[k].
sol[k_] := If[EvenQ[k], -2 + 3 2^(k/2) , -4 + 3 2^((1 + k)/2) ]
Table[sol[k], {k, 0, 10}]
(* {1, 2, 4, 8, 10, 20, 22, 44, 46, 92, 94} *)
as desired.
answered Nov 26 at 23:16
bbgodfrey
43.9k857108
add a comment |
up vote
4
down vote
When Mathematica returns the input without any error message, it is unable to evaluate the input.
With this particular recursion there is another approach. The recursion can be defined by
Clear[a, ar]
ar[0] = 1; ar[n_?OddQ] := ar[n - 1]*2; ar[n_?EvenQ] := ar[n - 1] + 2;
Generating a sequence from this recursion
seq = ar /@ Range[10]
(* {2, 4, 8, 10, 20, 22, 44, 46, 92, 94} *)
Use FindSequenceFunction to find the closed form of the recursion
a[n_] = FindSequenceFunction[seq, n] // FullSimplify
(* -3 + (-1)^n + 3 2^(-1 + n/2) (1 + Sqrt[2] + (-1)^(1 + n) (-1 + Sqrt[2])) *)
Checking equivalence outside of the range of seq
And @@ Table[a[n] == ar[n], {n, 0, 100}]
(* True *)
EDIT: Verifying,
Simplify[{a[2 n + 1] == a[2 n]*2,
a[2 n] == a[2 n - 1] + 2}, {Element[n, Integers], n >= 0}]
(* {True, True} *)
answered Nov 26 at 20:30
Bob Hanlon
58k23593
Hi, thanks for the alternative solution. Do you know why Mathematica was unable to evaluate the input given?
– konsolas
Nov 26 at 20:36
Don’t know. Presumably the algorithms used are not sufficiently robust to cover this case.
– Bob Hanlon
Nov 26 at 20:52
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
It appears that RSolve cannot solve difference equations in which two different equations describe the same variable. If so, a work-around is to represent a[k] with k an even index as c[k], and with k an odd indix as b[k], so that there is only one equation per variable.
FullSimplify[RSolveValue[{c[0] == 1, b[k] == c[k - 1]*2, c[k + 2] == b[k + 1] + 2},
{c[k], b[k]}, k] /. C[1] -> 0]
(* {-2 + 3 2^(-1 + k/2) (1 + (-1)^k), -4 - 3 2^(1/2 (-1 + k)) (-1 + (-1)^k)} *)
Then, construct the desired a[k] as even-index terms of c[k] and odd-index terms of b[k].
sol[k_] := If[EvenQ[k], -2 + 3 2^(k/2) , -4 + 3 2^((1 + k)/2) ]
Table[sol[k], {k, 0, 10}]
(* {1, 2, 4, 8, 10, 20, 22, 44, 46, 92, 94} *)
as desired.
answered Nov 26 at 23:16
bbgodfrey
43.9k857108
add a comment |
up vote
7
down vote
accepted
It appears that RSolve cannot solve difference equations in which two different equations describe the same variable. If so, a work-around is to represent a[k] with k an even index as c[k], and with k an odd indix as b[k], so that there is only one equation per variable.
FullSimplify[RSolveValue[{c[0] == 1, b[k] == c[k - 1]*2, c[k + 2] == b[k + 1] + 2},
{c[k], b[k]}, k] /. C[1] -> 0]
(* {-2 + 3 2^(-1 + k/2) (1 + (-1)^k), -4 - 3 2^(1/2 (-1 + k)) (-1 + (-1)^k)} *)
Then, construct the desired a[k] as even-index terms of c[k] and odd-index terms of b[k].
sol[k_] := If[EvenQ[k], -2 + 3 2^(k/2) , -4 + 3 2^((1 + k)/2) ]
Table[sol[k], {k, 0, 10}]
(* {1, 2, 4, 8, 10, 20, 22, 44, 46, 92, 94} *)
as desired.
answered Nov 26 at 23:16
bbgodfrey
43.9k857108
add a comment |
up vote
7
down vote
accepted
up vote
7
down vote
accepted
It appears that RSolve cannot solve difference equations in which two different equations describe the same variable. If so, a work-around is to represent a[k] with k an even index as c[k], and with k an odd indix as b[k], so that there is only one equation per variable.
FullSimplify[RSolveValue[{c[0] == 1, b[k] == c[k - 1]*2, c[k + 2] == b[k + 1] + 2},
{c[k], b[k]}, k] /. C[1] -> 0]
(* {-2 + 3 2^(-1 + k/2) (1 + (-1)^k), -4 - 3 2^(1/2 (-1 + k)) (-1 + (-1)^k)} *)
Then, construct the desired a[k] as even-index terms of c[k] and odd-index terms of b[k].
sol[k_] := If[EvenQ[k], -2 + 3 2^(k/2) , -4 + 3 2^((1 + k)/2) ]
Table[sol[k], {k, 0, 10}]
(* {1, 2, 4, 8, 10, 20, 22, 44, 46, 92, 94} *)
as desired.
answered Nov 26 at 23:16
bbgodfrey
43.9k857108
It appears that RSolve cannot solve difference equations in which two different equations describe the same variable. If so, a work-around is to represent a[k] with k an even index as c[k], and with k an odd indix as b[k], so that there is only one equation per variable.
FullSimplify[RSolveValue[{c[0] == 1, b[k] == c[k - 1]*2, c[k + 2] == b[k + 1] + 2},
{c[k], b[k]}, k] /. C[1] -> 0]
(* {-2 + 3 2^(-1 + k/2) (1 + (-1)^k), -4 - 3 2^(1/2 (-1 + k)) (-1 + (-1)^k)} *)
Then, construct the desired a[k] as even-index terms of c[k] and odd-index terms of b[k].
sol[k_] := If[EvenQ[k], -2 + 3 2^(k/2) , -4 + 3 2^((1 + k)/2) ]
Table[sol[k], {k, 0, 10}]
(* {1, 2, 4, 8, 10, 20, 22, 44, 46, 92, 94} *)
as desired.
answered Nov 26 at 23:16
bbgodfrey
43.9k857108
edited Nov 27 at 11:29
answered Nov 26 at 23:16
bbgodfrey
43.9k857108
answered Nov 26 at 23:16
bbgodfrey
43.9k857108
answered Nov 26 at 23:16
bbgodfrey
43.9k857108
43.9k857108
add a comment |
add a comment |
up vote
4
down vote
When Mathematica returns the input without any error message, it is unable to evaluate the input.
With this particular recursion there is another approach. The recursion can be defined by
Clear[a, ar]
ar[0] = 1; ar[n_?OddQ] := ar[n - 1]*2; ar[n_?EvenQ] := ar[n - 1] + 2;
Generating a sequence from this recursion
seq = ar /@ Range[10]
(* {2, 4, 8, 10, 20, 22, 44, 46, 92, 94} *)
Use FindSequenceFunction to find the closed form of the recursion
a[n_] = FindSequenceFunction[seq, n] // FullSimplify
(* -3 + (-1)^n + 3 2^(-1 + n/2) (1 + Sqrt[2] + (-1)^(1 + n) (-1 + Sqrt[2])) *)
Checking equivalence outside of the range of seq
And @@ Table[a[n] == ar[n], {n, 0, 100}]
(* True *)
EDIT: Verifying,
Simplify[{a[2 n + 1] == a[2 n]*2,
a[2 n] == a[2 n - 1] + 2}, {Element[n, Integers], n >= 0}]
(* {True, True} *)
answered Nov 26 at 20:30
Bob Hanlon
58k23593
Hi, thanks for the alternative solution. Do you know why Mathematica was unable to evaluate the input given?
– konsolas
Nov 26 at 20:36
Don’t know. Presumably the algorithms used are not sufficiently robust to cover this case.
– Bob Hanlon
Nov 26 at 20:52
add a comment |
up vote
4
down vote
When Mathematica returns the input without any error message, it is unable to evaluate the input.
With this particular recursion there is another approach. The recursion can be defined by
Clear[a, ar]
ar[0] = 1; ar[n_?OddQ] := ar[n - 1]*2; ar[n_?EvenQ] := ar[n - 1] + 2;
Generating a sequence from this recursion
seq = ar /@ Range[10]
(* {2, 4, 8, 10, 20, 22, 44, 46, 92, 94} *)
Use FindSequenceFunction to find the closed form of the recursion
a[n_] = FindSequenceFunction[seq, n] // FullSimplify
(* -3 + (-1)^n + 3 2^(-1 + n/2) (1 + Sqrt[2] + (-1)^(1 + n) (-1 + Sqrt[2])) *)
Checking equivalence outside of the range of seq
And @@ Table[a[n] == ar[n], {n, 0, 100}]
(* True *)
EDIT: Verifying,
Simplify[{a[2 n + 1] == a[2 n]*2,
a[2 n] == a[2 n - 1] + 2}, {Element[n, Integers], n >= 0}]
(* {True, True} *)
answered Nov 26 at 20:30
Bob Hanlon
58k23593
Hi, thanks for the alternative solution. Do you know why Mathematica was unable to evaluate the input given?
– konsolas
Nov 26 at 20:36
Don’t know. Presumably the algorithms used are not sufficiently robust to cover this case.
– Bob Hanlon
Nov 26 at 20:52
add a comment |
up vote
4
down vote
up vote
4
down vote
When Mathematica returns the input without any error message, it is unable to evaluate the input.
With this particular recursion there is another approach. The recursion can be defined by
Clear[a, ar]
ar[0] = 1; ar[n_?OddQ] := ar[n - 1]*2; ar[n_?EvenQ] := ar[n - 1] + 2;
Generating a sequence from this recursion
seq = ar /@ Range[10]
(* {2, 4, 8, 10, 20, 22, 44, 46, 92, 94} *)
Use FindSequenceFunction to find the closed form of the recursion
a[n_] = FindSequenceFunction[seq, n] // FullSimplify
(* -3 + (-1)^n + 3 2^(-1 + n/2) (1 + Sqrt[2] + (-1)^(1 + n) (-1 + Sqrt[2])) *)
Checking equivalence outside of the range of seq
And @@ Table[a[n] == ar[n], {n, 0, 100}]
(* True *)
EDIT: Verifying,
Simplify[{a[2 n + 1] == a[2 n]*2,
a[2 n] == a[2 n - 1] + 2}, {Element[n, Integers], n >= 0}]
(* {True, True} *)
answered Nov 26 at 20:30
Bob Hanlon
58k23593
When Mathematica returns the input without any error message, it is unable to evaluate the input.
With this particular recursion there is another approach. The recursion can be defined by
Clear[a, ar]
ar[0] = 1; ar[n_?OddQ] := ar[n - 1]*2; ar[n_?EvenQ] := ar[n - 1] + 2;
Generating a sequence from this recursion
seq = ar /@ Range[10]
(* {2, 4, 8, 10, 20, 22, 44, 46, 92, 94} *)
Use FindSequenceFunction to find the closed form of the recursion
a[n_] = FindSequenceFunction[seq, n] // FullSimplify
(* -3 + (-1)^n + 3 2^(-1 + n/2) (1 + Sqrt[2] + (-1)^(1 + n) (-1 + Sqrt[2])) *)
Checking equivalence outside of the range of seq
And @@ Table[a[n] == ar[n], {n, 0, 100}]
(* True *)
EDIT: Verifying,
Simplify[{a[2 n + 1] == a[2 n]*2,
a[2 n] == a[2 n - 1] + 2}, {Element[n, Integers], n >= 0}]
(* {True, True} *)
answered Nov 26 at 20:30
Bob Hanlon
58k23593
edited Nov 27 at 14:35
answered Nov 26 at 20:30
Bob Hanlon
58k23593
answered Nov 26 at 20:30
Bob Hanlon
58k23593
answered Nov 26 at 20:30
Bob Hanlon
58k23593
58k23593
Hi, thanks for the alternative solution. Do you know why Mathematica was unable to evaluate the input given?
– konsolas
Nov 26 at 20:36
Don’t know. Presumably the algorithms used are not sufficiently robust to cover this case.
– Bob Hanlon
Nov 26 at 20:52
add a comment |
Hi, thanks for the alternative solution. Do you know why Mathematica was unable to evaluate the input given?
– konsolas
Nov 26 at 20:36
Don’t know. Presumably the algorithms used are not sufficiently robust to cover this case.
– Bob Hanlon
Nov 26 at 20:52
Hi, thanks for the alternative solution. Do you know why Mathematica was unable to evaluate the input given?
– konsolas
Nov 26 at 20:36
Hi, thanks for the alternative solution. Do you know why Mathematica was unable to evaluate the input given?
– konsolas
Nov 26 at 20:36
Don’t know. Presumably the algorithms used are not sufficiently robust to cover this case.
– Bob Hanlon
Nov 26 at 20:52
Don’t know. Presumably the algorithms used are not sufficiently robust to cover this case.
– Bob Hanlon
Nov 26 at 20:52
add a comment |
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Try replacing the next-to-last term in
RSolve(a) witha[n].– yosimitsu kodanuri
Nov 26 at 20:30
It still simply echoes the input:
RSolve[{a[0] == 1, a[1 + 2 n] == 2 a[2 n], a[2 n] == 2 + a[-1 + 2 n]}, a[n], n]– konsolas
Nov 26 at 20:36
1
Welcome to Mathematica.SE! Interesting question(+1). I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise.
– bbgodfrey
Nov 27 at 0:28