Slow computation of recursive sequencesCan one identify the design patterns of Mathematica?What does the...
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Slow computation of recursive sequences
Can one identify the design patterns of Mathematica?What does the construct f[x_] := f[x] = … mean?Creating Recursive SequencesDefining a recursive integral sequenceTangled up in sequences and recurrence relationsEvaluate recursive funtion as a whole before computing a recursive stepMimic a procedural, recursive clustering algorithm for site percolation using functional programmingRecursive calculation is very slowImproving my method for solving iterative algebraic equations
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$begingroup$
I want to investigate the asymptotic behavior of the following recursive system:
a[n_] := (a[n-1] - 0.45 a[n-1]) + 0.3 b[n-1]
b[n_] := (b[n-1] - 0.3 b[n-1]) + 0.45 a[n-1]
a[0] = 450;
b[0] = 450;
Grid[Table[{n, a[n], b[n]}, {n, 0, 12}]]
It is working, however it is very slow. It seems that it doesn't memorize the calculated value, so it repeats the computation from the beginning.
Is there a way to make this computations faster?
recursion difference-equations sequence markov-chains
edited 8 hours ago
WReach
55.7k2 gold badges122 silver badges222 bronze badges
asked 9 hours ago
David LingardDavid Lingard
82 bronze badges
New contributor
David Lingard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment
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$begingroup$
I want to investigate the asymptotic behavior of the following recursive system:
a[n_] := (a[n-1] - 0.45 a[n-1]) + 0.3 b[n-1]
b[n_] := (b[n-1] - 0.3 b[n-1]) + 0.45 a[n-1]
a[0] = 450;
b[0] = 450;
Grid[Table[{n, a[n], b[n]}, {n, 0, 12}]]
It is working, however it is very slow. It seems that it doesn't memorize the calculated value, so it repeats the computation from the beginning.
Is there a way to make this computations faster?
recursion difference-equations sequence markov-chains
edited 8 hours ago
WReach
55.7k2 gold badges122 silver badges222 bronze badges
asked 9 hours ago
David LingardDavid Lingard
82 bronze badges
New contributor
David Lingard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Hi David. Please get into the habit of posting code directly in copyable form (with the appropriate markup). Towards your question: You probably want to learn about memoization, i.e. the art of defining Functions That Remember Values They Have Found.
$endgroup$
– Henrik Schumacher
8 hours ago
$begingroup$
@HenrikSchumacher Hi Henrik, I am sorry I first typed it as a copiable code, but I get the persisting error: "your post appears to contain code that is not properly formatted as code". That's why I inserted it as a picture.
$endgroup$
– David Lingard
8 hours ago
$begingroup$
related: Basic/Advanced Memoization, What does the construct f(x_) := f(x) = … mean?
$endgroup$
– WReach
8 hours ago
add a comment
|
$begingroup$
I want to investigate the asymptotic behavior of the following recursive system:
a[n_] := (a[n-1] - 0.45 a[n-1]) + 0.3 b[n-1]
b[n_] := (b[n-1] - 0.3 b[n-1]) + 0.45 a[n-1]
a[0] = 450;
b[0] = 450;
Grid[Table[{n, a[n], b[n]}, {n, 0, 12}]]
It is working, however it is very slow. It seems that it doesn't memorize the calculated value, so it repeats the computation from the beginning.
Is there a way to make this computations faster?
recursion difference-equations sequence markov-chains
edited 8 hours ago
WReach
55.7k2 gold badges122 silver badges222 bronze badges
asked 9 hours ago
David LingardDavid Lingard
82 bronze badges
New contributor
David Lingard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I want to investigate the asymptotic behavior of the following recursive system:
a[n_] := (a[n-1] - 0.45 a[n-1]) + 0.3 b[n-1]
b[n_] := (b[n-1] - 0.3 b[n-1]) + 0.45 a[n-1]
a[0] = 450;
b[0] = 450;
Grid[Table[{n, a[n], b[n]}, {n, 0, 12}]]
It is working, however it is very slow. It seems that it doesn't memorize the calculated value, so it repeats the computation from the beginning.
Is there a way to make this computations faster?
recursion difference-equations sequence markov-chains
recursion difference-equations sequence markov-chains
edited 8 hours ago
WReach
55.7k2 gold badges122 silver badges222 bronze badges
asked 9 hours ago
David LingardDavid Lingard
82 bronze badges
New contributor
David Lingard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 8 hours ago
WReach
55.7k2 gold badges122 silver badges222 bronze badges
asked 9 hours ago
David LingardDavid Lingard
82 bronze badges
New contributor
David Lingard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 8 hours ago
WReach
55.7k2 gold badges122 silver badges222 bronze badges
edited 8 hours ago
WReach
55.7k2 gold badges122 silver badges222 bronze badges
edited 8 hours ago
WReach
55.7k2 gold badges122 silver badges222 bronze badges
55.7k2 gold badges122 silver badges222 bronze badges
asked 9 hours ago
David LingardDavid Lingard
82 bronze badges
New contributor
David Lingard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
David LingardDavid Lingard
82 bronze badges
asked 9 hours ago
David LingardDavid Lingard
82 bronze badges
82 bronze badges
New contributor
David Lingard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
David Lingard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Hi David. Please get into the habit of posting code directly in copyable form (with the appropriate markup). Towards your question: You probably want to learn about memoization, i.e. the art of defining Functions That Remember Values They Have Found.
$endgroup$
– Henrik Schumacher
8 hours ago
$begingroup$
@HenrikSchumacher Hi Henrik, I am sorry I first typed it as a copiable code, but I get the persisting error: "your post appears to contain code that is not properly formatted as code". That's why I inserted it as a picture.
$endgroup$
– David Lingard
8 hours ago
$begingroup$
related: Basic/Advanced Memoization, What does the construct f(x_) := f(x) = … mean?
$endgroup$
– WReach
8 hours ago
add a comment
|
$begingroup$
Hi David. Please get into the habit of posting code directly in copyable form (with the appropriate markup). Towards your question: You probably want to learn about memoization, i.e. the art of defining Functions That Remember Values They Have Found.
$endgroup$
– Henrik Schumacher
8 hours ago
$begingroup$
@HenrikSchumacher Hi Henrik, I am sorry I first typed it as a copiable code, but I get the persisting error: "your post appears to contain code that is not properly formatted as code". That's why I inserted it as a picture.
$endgroup$
– David Lingard
8 hours ago
$begingroup$
related: Basic/Advanced Memoization, What does the construct f(x_) := f(x) = … mean?
$endgroup$
– WReach
8 hours ago
$begingroup$
Hi David. Please get into the habit of posting code directly in copyable form (with the appropriate markup). Towards your question: You probably want to learn about memoization, i.e. the art of defining Functions That Remember Values They Have Found.
$endgroup$
– Henrik Schumacher
8 hours ago
$begingroup$
Hi David. Please get into the habit of posting code directly in copyable form (with the appropriate markup). Towards your question: You probably want to learn about memoization, i.e. the art of defining Functions That Remember Values They Have Found.
$endgroup$
– Henrik Schumacher
8 hours ago
$begingroup$
@HenrikSchumacher Hi Henrik, I am sorry I first typed it as a copiable code, but I get the persisting error: "your post appears to contain code that is not properly formatted as code". That's why I inserted it as a picture.
$endgroup$
– David Lingard
8 hours ago
$begingroup$
@HenrikSchumacher Hi Henrik, I am sorry I first typed it as a copiable code, but I get the persisting error: "your post appears to contain code that is not properly formatted as code". That's why I inserted it as a picture.
$endgroup$
– David Lingard
8 hours ago
$begingroup$
related: Basic/Advanced Memoization, What does the construct f(x_) := f(x) = … mean?
$endgroup$
– WReach
8 hours ago
$begingroup$
related: Basic/Advanced Memoization, What does the construct f(x_) := f(x) = … mean?
$endgroup$
– WReach
8 hours ago
add a comment
|
3 Answers
3
active
oldest
votes
$begingroup$
These are all equivalent
RecurrenceTable[{a[n] == a[n - 1] - 0.45 a[n - 1] + 0.3 b[n - 1],
b[n] == b[n - 1] - 0.3 b[n - 1] + 0.45 a[n - 1], a[0] == 450,
b[0] == 450}, {a[n], b[n]}, {n, 10}]
Clear[a, b]
{a[n], b[n]} /. RSolve[{a[n] == a[n - 1] - 0.45 a[n - 1] + 0.3 b[n - 1],
b[n] == b[n - 1] - 0.3 b[n - 1] + 0.45 a[n - 1], a[0] == 450,
b[0] == 450}, {a[n], b[n]}, n];
Table[%, {n, 0, 10}]
Clear[a, b]
a[n_] := a[n] = a[n - 1] - 0.45 a[n - 1] + 0.3 b[n - 1]
b[n_] := b[n] = b[n - 1] - 0.3 b[n - 1] + 0.45 a[n - 1]
a[0] = 450;
b[0] = 450;
Table[{a[n], b[n]}, {n, 0, 10}]
(* {{450, 450}, {382.5, 517.5}, {365.625, 534.375}, {361.406,
538.594}, {360.352, 539.648}, {360.088, 539.912}, {360.022,
539.978}, {360.005, 539.995}, {360.001, 539.999}, {360.,
540.}, {360., 540.}}*)
answered 8 hours ago
Suba ThomasSuba Thomas
4,28411 silver badges20 bronze badges
$endgroup$
add a comment
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$begingroup$
The system can be solved in closed-form using RSolve
Clear["Global`*"]
eqns = {a[n] == (a[n - 1] - 0.45 a[n - 1]) + 0.3 b[n - 1],
b[n] == (b[n - 1] - 0.3 b[n - 1]) + 0.45 a[n - 1],
a[0] == 450, b[0] == 450} // Rationalize;
sol = RSolve[eqns, {a, b}, n][[1]]
(* {a -> Function[{n}, 45 2^(1 - 2 n) (1 + 2^(2 + 2 n))],
b -> Function[{n}, 45 2^(1 - 2 n) (-1 + 3 2^(1 + 2 n))]} *)
Verifying that the solutions satisfy the equations
And @@ (eqns /. sol // Simplify)
(* True *)
The functions asymptotically approach
Limit[{a[n], b[n]} /. sol, n -> Infinity]
(* {360, 540} *)
N[{a[#], b[#]} /. sol] & /@ Range [0, 10]
(* {{450., 450.}, {382.5, 517.5}, {365.625, 534.375}, {361.406,
538.594}, {360.352, 539.648}, {360.088, 539.912}, {360.022,
539.978}, {360.005, 539.995}, {360.001, 539.999}, {360., 540.}, {360.,
540.}} *)
Plot[Evaluate[{b[n], a[n]} /. sol], {n, 0, 15},
PlotLegends -> Placed[{b[n], a[n]}, {0.5, 0.5}]]
Alternatively, using FixedPointList
FixedPointList[
{(#[[1]] - 0.45 #[[1]]) + 0.3 #[[2]],
(#[[2]] - 0.3 #[[2]]) + 0.45 #[[1]]} &,
{450, 450}]
(* {{450, 450}, {382.5, 517.5}, {365.625, 534.375}, {361.406, 538.594}, {360.352,
539.648}, {360.088, 539.912}, {360.022, 539.978}, {360.005,
539.995}, {360.001, 539.999}, {360., 540.}, {360., 540.}, {360.,
540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360.,
540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360.,
540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360., 540.}} *)
answered 6 hours ago
Bob HanlonBob Hanlon
66.4k3 gold badges37 silver badges102 bronze badges
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The basic idea is
a[n_]:=(a[n]=(a[n-1)0...)
This will set downvalues for a, for each n to prevent re-computation
answered 8 hours ago
John McGeeJohn McGee
1,8387 silver badges13 bronze badges
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
These are all equivalent
RecurrenceTable[{a[n] == a[n - 1] - 0.45 a[n - 1] + 0.3 b[n - 1],
b[n] == b[n - 1] - 0.3 b[n - 1] + 0.45 a[n - 1], a[0] == 450,
b[0] == 450}, {a[n], b[n]}, {n, 10}]
Clear[a, b]
{a[n], b[n]} /. RSolve[{a[n] == a[n - 1] - 0.45 a[n - 1] + 0.3 b[n - 1],
b[n] == b[n - 1] - 0.3 b[n - 1] + 0.45 a[n - 1], a[0] == 450,
b[0] == 450}, {a[n], b[n]}, n];
Table[%, {n, 0, 10}]
Clear[a, b]
a[n_] := a[n] = a[n - 1] - 0.45 a[n - 1] + 0.3 b[n - 1]
b[n_] := b[n] = b[n - 1] - 0.3 b[n - 1] + 0.45 a[n - 1]
a[0] = 450;
b[0] = 450;
Table[{a[n], b[n]}, {n, 0, 10}]
(* {{450, 450}, {382.5, 517.5}, {365.625, 534.375}, {361.406,
538.594}, {360.352, 539.648}, {360.088, 539.912}, {360.022,
539.978}, {360.005, 539.995}, {360.001, 539.999}, {360.,
540.}, {360., 540.}}*)
answered 8 hours ago
Suba ThomasSuba Thomas
4,28411 silver badges20 bronze badges
$endgroup$
add a comment
|
$begingroup$
These are all equivalent
RecurrenceTable[{a[n] == a[n - 1] - 0.45 a[n - 1] + 0.3 b[n - 1],
b[n] == b[n - 1] - 0.3 b[n - 1] + 0.45 a[n - 1], a[0] == 450,
b[0] == 450}, {a[n], b[n]}, {n, 10}]
Clear[a, b]
{a[n], b[n]} /. RSolve[{a[n] == a[n - 1] - 0.45 a[n - 1] + 0.3 b[n - 1],
b[n] == b[n - 1] - 0.3 b[n - 1] + 0.45 a[n - 1], a[0] == 450,
b[0] == 450}, {a[n], b[n]}, n];
Table[%, {n, 0, 10}]
Clear[a, b]
a[n_] := a[n] = a[n - 1] - 0.45 a[n - 1] + 0.3 b[n - 1]
b[n_] := b[n] = b[n - 1] - 0.3 b[n - 1] + 0.45 a[n - 1]
a[0] = 450;
b[0] = 450;
Table[{a[n], b[n]}, {n, 0, 10}]
(* {{450, 450}, {382.5, 517.5}, {365.625, 534.375}, {361.406,
538.594}, {360.352, 539.648}, {360.088, 539.912}, {360.022,
539.978}, {360.005, 539.995}, {360.001, 539.999}, {360.,
540.}, {360., 540.}}*)
answered 8 hours ago
Suba ThomasSuba Thomas
4,28411 silver badges20 bronze badges
$endgroup$
add a comment
|
$begingroup$
These are all equivalent
RecurrenceTable[{a[n] == a[n - 1] - 0.45 a[n - 1] + 0.3 b[n - 1],
b[n] == b[n - 1] - 0.3 b[n - 1] + 0.45 a[n - 1], a[0] == 450,
b[0] == 450}, {a[n], b[n]}, {n, 10}]
Clear[a, b]
{a[n], b[n]} /. RSolve[{a[n] == a[n - 1] - 0.45 a[n - 1] + 0.3 b[n - 1],
b[n] == b[n - 1] - 0.3 b[n - 1] + 0.45 a[n - 1], a[0] == 450,
b[0] == 450}, {a[n], b[n]}, n];
Table[%, {n, 0, 10}]
Clear[a, b]
a[n_] := a[n] = a[n - 1] - 0.45 a[n - 1] + 0.3 b[n - 1]
b[n_] := b[n] = b[n - 1] - 0.3 b[n - 1] + 0.45 a[n - 1]
a[0] = 450;
b[0] = 450;
Table[{a[n], b[n]}, {n, 0, 10}]
(* {{450, 450}, {382.5, 517.5}, {365.625, 534.375}, {361.406,
538.594}, {360.352, 539.648}, {360.088, 539.912}, {360.022,
539.978}, {360.005, 539.995}, {360.001, 539.999}, {360.,
540.}, {360., 540.}}*)
answered 8 hours ago
Suba ThomasSuba Thomas
4,28411 silver badges20 bronze badges
$endgroup$
These are all equivalent
RecurrenceTable[{a[n] == a[n - 1] - 0.45 a[n - 1] + 0.3 b[n - 1],
b[n] == b[n - 1] - 0.3 b[n - 1] + 0.45 a[n - 1], a[0] == 450,
b[0] == 450}, {a[n], b[n]}, {n, 10}]
Clear[a, b]
{a[n], b[n]} /. RSolve[{a[n] == a[n - 1] - 0.45 a[n - 1] + 0.3 b[n - 1],
b[n] == b[n - 1] - 0.3 b[n - 1] + 0.45 a[n - 1], a[0] == 450,
b[0] == 450}, {a[n], b[n]}, n];
Table[%, {n, 0, 10}]
Clear[a, b]
a[n_] := a[n] = a[n - 1] - 0.45 a[n - 1] + 0.3 b[n - 1]
b[n_] := b[n] = b[n - 1] - 0.3 b[n - 1] + 0.45 a[n - 1]
a[0] = 450;
b[0] = 450;
Table[{a[n], b[n]}, {n, 0, 10}]
(* {{450, 450}, {382.5, 517.5}, {365.625, 534.375}, {361.406,
538.594}, {360.352, 539.648}, {360.088, 539.912}, {360.022,
539.978}, {360.005, 539.995}, {360.001, 539.999}, {360.,
540.}, {360., 540.}}*)
answered 8 hours ago
Suba ThomasSuba Thomas
4,28411 silver badges20 bronze badges
answered 8 hours ago
Suba ThomasSuba Thomas
4,28411 silver badges20 bronze badges
answered 8 hours ago
Suba ThomasSuba Thomas
4,28411 silver badges20 bronze badges
answered 8 hours ago
Suba ThomasSuba Thomas
4,28411 silver badges20 bronze badges
4,28411 silver badges20 bronze badges
add a comment
|
add a comment
|
$begingroup$
The system can be solved in closed-form using RSolve
Clear["Global`*"]
eqns = {a[n] == (a[n - 1] - 0.45 a[n - 1]) + 0.3 b[n - 1],
b[n] == (b[n - 1] - 0.3 b[n - 1]) + 0.45 a[n - 1],
a[0] == 450, b[0] == 450} // Rationalize;
sol = RSolve[eqns, {a, b}, n][[1]]
(* {a -> Function[{n}, 45 2^(1 - 2 n) (1 + 2^(2 + 2 n))],
b -> Function[{n}, 45 2^(1 - 2 n) (-1 + 3 2^(1 + 2 n))]} *)
Verifying that the solutions satisfy the equations
And @@ (eqns /. sol // Simplify)
(* True *)
The functions asymptotically approach
Limit[{a[n], b[n]} /. sol, n -> Infinity]
(* {360, 540} *)
N[{a[#], b[#]} /. sol] & /@ Range [0, 10]
(* {{450., 450.}, {382.5, 517.5}, {365.625, 534.375}, {361.406,
538.594}, {360.352, 539.648}, {360.088, 539.912}, {360.022,
539.978}, {360.005, 539.995}, {360.001, 539.999}, {360., 540.}, {360.,
540.}} *)
Plot[Evaluate[{b[n], a[n]} /. sol], {n, 0, 15},
PlotLegends -> Placed[{b[n], a[n]}, {0.5, 0.5}]]
Alternatively, using FixedPointList
FixedPointList[
{(#[[1]] - 0.45 #[[1]]) + 0.3 #[[2]],
(#[[2]] - 0.3 #[[2]]) + 0.45 #[[1]]} &,
{450, 450}]
(* {{450, 450}, {382.5, 517.5}, {365.625, 534.375}, {361.406, 538.594}, {360.352,
539.648}, {360.088, 539.912}, {360.022, 539.978}, {360.005,
539.995}, {360.001, 539.999}, {360., 540.}, {360., 540.}, {360.,
540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360.,
540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360.,
540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360., 540.}} *)
answered 6 hours ago
Bob HanlonBob Hanlon
66.4k3 gold badges37 silver badges102 bronze badges
$endgroup$
add a comment
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$begingroup$
The system can be solved in closed-form using RSolve
Clear["Global`*"]
eqns = {a[n] == (a[n - 1] - 0.45 a[n - 1]) + 0.3 b[n - 1],
b[n] == (b[n - 1] - 0.3 b[n - 1]) + 0.45 a[n - 1],
a[0] == 450, b[0] == 450} // Rationalize;
sol = RSolve[eqns, {a, b}, n][[1]]
(* {a -> Function[{n}, 45 2^(1 - 2 n) (1 + 2^(2 + 2 n))],
b -> Function[{n}, 45 2^(1 - 2 n) (-1 + 3 2^(1 + 2 n))]} *)
Verifying that the solutions satisfy the equations
And @@ (eqns /. sol // Simplify)
(* True *)
The functions asymptotically approach
Limit[{a[n], b[n]} /. sol, n -> Infinity]
(* {360, 540} *)
N[{a[#], b[#]} /. sol] & /@ Range [0, 10]
(* {{450., 450.}, {382.5, 517.5}, {365.625, 534.375}, {361.406,
538.594}, {360.352, 539.648}, {360.088, 539.912}, {360.022,
539.978}, {360.005, 539.995}, {360.001, 539.999}, {360., 540.}, {360.,
540.}} *)
Plot[Evaluate[{b[n], a[n]} /. sol], {n, 0, 15},
PlotLegends -> Placed[{b[n], a[n]}, {0.5, 0.5}]]
Alternatively, using FixedPointList
FixedPointList[
{(#[[1]] - 0.45 #[[1]]) + 0.3 #[[2]],
(#[[2]] - 0.3 #[[2]]) + 0.45 #[[1]]} &,
{450, 450}]
(* {{450, 450}, {382.5, 517.5}, {365.625, 534.375}, {361.406, 538.594}, {360.352,
539.648}, {360.088, 539.912}, {360.022, 539.978}, {360.005,
539.995}, {360.001, 539.999}, {360., 540.}, {360., 540.}, {360.,
540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360.,
540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360.,
540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360., 540.}} *)
answered 6 hours ago
Bob HanlonBob Hanlon
66.4k3 gold badges37 silver badges102 bronze badges
$endgroup$
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$begingroup$
The system can be solved in closed-form using RSolve
Clear["Global`*"]
eqns = {a[n] == (a[n - 1] - 0.45 a[n - 1]) + 0.3 b[n - 1],
b[n] == (b[n - 1] - 0.3 b[n - 1]) + 0.45 a[n - 1],
a[0] == 450, b[0] == 450} // Rationalize;
sol = RSolve[eqns, {a, b}, n][[1]]
(* {a -> Function[{n}, 45 2^(1 - 2 n) (1 + 2^(2 + 2 n))],
b -> Function[{n}, 45 2^(1 - 2 n) (-1 + 3 2^(1 + 2 n))]} *)
Verifying that the solutions satisfy the equations
And @@ (eqns /. sol // Simplify)
(* True *)
The functions asymptotically approach
Limit[{a[n], b[n]} /. sol, n -> Infinity]
(* {360, 540} *)
N[{a[#], b[#]} /. sol] & /@ Range [0, 10]
(* {{450., 450.}, {382.5, 517.5}, {365.625, 534.375}, {361.406,
538.594}, {360.352, 539.648}, {360.088, 539.912}, {360.022,
539.978}, {360.005, 539.995}, {360.001, 539.999}, {360., 540.}, {360.,
540.}} *)
Plot[Evaluate[{b[n], a[n]} /. sol], {n, 0, 15},
PlotLegends -> Placed[{b[n], a[n]}, {0.5, 0.5}]]
Alternatively, using FixedPointList
FixedPointList[
{(#[[1]] - 0.45 #[[1]]) + 0.3 #[[2]],
(#[[2]] - 0.3 #[[2]]) + 0.45 #[[1]]} &,
{450, 450}]
(* {{450, 450}, {382.5, 517.5}, {365.625, 534.375}, {361.406, 538.594}, {360.352,
539.648}, {360.088, 539.912}, {360.022, 539.978}, {360.005,
539.995}, {360.001, 539.999}, {360., 540.}, {360., 540.}, {360.,
540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360.,
540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360.,
540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360., 540.}} *)
answered 6 hours ago
Bob HanlonBob Hanlon
66.4k3 gold badges37 silver badges102 bronze badges
$endgroup$
The system can be solved in closed-form using RSolve
Clear["Global`*"]
eqns = {a[n] == (a[n - 1] - 0.45 a[n - 1]) + 0.3 b[n - 1],
b[n] == (b[n - 1] - 0.3 b[n - 1]) + 0.45 a[n - 1],
a[0] == 450, b[0] == 450} // Rationalize;
sol = RSolve[eqns, {a, b}, n][[1]]
(* {a -> Function[{n}, 45 2^(1 - 2 n) (1 + 2^(2 + 2 n))],
b -> Function[{n}, 45 2^(1 - 2 n) (-1 + 3 2^(1 + 2 n))]} *)
Verifying that the solutions satisfy the equations
And @@ (eqns /. sol // Simplify)
(* True *)
The functions asymptotically approach
Limit[{a[n], b[n]} /. sol, n -> Infinity]
(* {360, 540} *)
N[{a[#], b[#]} /. sol] & /@ Range [0, 10]
(* {{450., 450.}, {382.5, 517.5}, {365.625, 534.375}, {361.406,
538.594}, {360.352, 539.648}, {360.088, 539.912}, {360.022,
539.978}, {360.005, 539.995}, {360.001, 539.999}, {360., 540.}, {360.,
540.}} *)
Plot[Evaluate[{b[n], a[n]} /. sol], {n, 0, 15},
PlotLegends -> Placed[{b[n], a[n]}, {0.5, 0.5}]]
Alternatively, using FixedPointList
FixedPointList[
{(#[[1]] - 0.45 #[[1]]) + 0.3 #[[2]],
(#[[2]] - 0.3 #[[2]]) + 0.45 #[[1]]} &,
{450, 450}]
(* {{450, 450}, {382.5, 517.5}, {365.625, 534.375}, {361.406, 538.594}, {360.352,
539.648}, {360.088, 539.912}, {360.022, 539.978}, {360.005,
539.995}, {360.001, 539.999}, {360., 540.}, {360., 540.}, {360.,
540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360.,
540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360.,
540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360., 540.}, {360., 540.}} *)
answered 6 hours ago
Bob HanlonBob Hanlon
66.4k3 gold badges37 silver badges102 bronze badges
answered 6 hours ago
Bob HanlonBob Hanlon
66.4k3 gold badges37 silver badges102 bronze badges
answered 6 hours ago
Bob HanlonBob Hanlon
66.4k3 gold badges37 silver badges102 bronze badges
answered 6 hours ago
Bob HanlonBob Hanlon
66.4k3 gold badges37 silver badges102 bronze badges
66.4k3 gold badges37 silver badges102 bronze badges
add a comment
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add a comment
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$begingroup$
The basic idea is
a[n_]:=(a[n]=(a[n-1)0...)
This will set downvalues for a, for each n to prevent re-computation
answered 8 hours ago
John McGeeJohn McGee
1,8387 silver badges13 bronze badges
$endgroup$
add a comment
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$begingroup$
The basic idea is
a[n_]:=(a[n]=(a[n-1)0...)
This will set downvalues for a, for each n to prevent re-computation
answered 8 hours ago
John McGeeJohn McGee
1,8387 silver badges13 bronze badges
$endgroup$
add a comment
|
$begingroup$
The basic idea is
a[n_]:=(a[n]=(a[n-1)0...)
This will set downvalues for a, for each n to prevent re-computation
answered 8 hours ago
John McGeeJohn McGee
1,8387 silver badges13 bronze badges
$endgroup$
The basic idea is
a[n_]:=(a[n]=(a[n-1)0...)
This will set downvalues for a, for each n to prevent re-computation
answered 8 hours ago
John McGeeJohn McGee
1,8387 silver badges13 bronze badges
answered 8 hours ago
John McGeeJohn McGee
1,8387 silver badges13 bronze badges
answered 8 hours ago
John McGeeJohn McGee
1,8387 silver badges13 bronze badges
answered 8 hours ago
John McGeeJohn McGee
1,8387 silver badges13 bronze badges
1,8387 silver badges13 bronze badges
add a comment
|
add a comment
|
David Lingard is a new contributor. Be nice, and check out our Code of Conduct.
David Lingard is a new contributor. Be nice, and check out our Code of Conduct.
David Lingard is a new contributor. Be nice, and check out our Code of Conduct.
David Lingard is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Hi David. Please get into the habit of posting code directly in copyable form (with the appropriate markup). Towards your question: You probably want to learn about memoization, i.e. the art of defining Functions That Remember Values They Have Found.
$endgroup$
– Henrik Schumacher
8 hours ago
$begingroup$
@HenrikSchumacher Hi Henrik, I am sorry I first typed it as a copiable code, but I get the persisting error: "your post appears to contain code that is not properly formatted as code". That's why I inserted it as a picture.
$endgroup$
– David Lingard
8 hours ago
$begingroup$
related: Basic/Advanced Memoization, What does the construct f(x_) := f(x) = … mean?
$endgroup$
– WReach
8 hours ago