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RecurrenceTable involving inputs from a vector

Prevent Part[] from trying to extract parts of symbolic expressionsHow can I solve a difference-differential equation?Can RecurrenceTable make use of CompiledFunction?RecurrenceTable with vectorCompile issues, scoping and order of evaluationBasis for the intersection of vector spacesNDSolve mixing many scalar and vector equationsmultiplication of vector spacesset of recursive equations with derivativesVector and matrix use in RecurrenceTableAssociate a unique real number to any n-dimensional vector

2

$begingroup$

Directly going to the issue, my problem is how to use a RecurrenceTable when it have inputs from a vector defined previously. For instance I have:

p={1,10,100,1000,10000};

And the recursive relation I want to solve is:

RecurrenceTable[{a[n + 1] == 2 a[n] + p[[n]], a[1] == 1}, a, {n, 1, 5}]

which must have an answer as:

{1,3,16,132,1264}

But I face with this error:

Part::pkspec1: The expression n cannot be used as a part specification.

Part::pkspec1: The expression #1 cannot be used as a part specification.

For avoiding part specification I tried the following:

z = 1;

RecurrenceTable[{a[n + 1] == 2 a[n] + p[[z++]], a[1] == 1}, a, {n, 1, 5}]

But the answer is wrong:

{1, 3, 7, 15, 31}

recursion vector

share|improve this question

asked 5 hours ago

Reza FarajiReza Faraji

132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Related: (14645)
  $endgroup$
  – Mr.Wizard♦
  4 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Directly going to the issue, my problem is how to use a RecurrenceTable when it have inputs from a vector defined previously. For instance I have:

p={1,10,100,1000,10000};

And the recursive relation I want to solve is:

RecurrenceTable[{a[n + 1] == 2 a[n] + p[[n]], a[1] == 1}, a, {n, 1, 5}]

which must have an answer as:

{1,3,16,132,1264}

But I face with this error:

Part::pkspec1: The expression n cannot be used as a part specification.

Part::pkspec1: The expression #1 cannot be used as a part specification.

For avoiding part specification I tried the following:

z = 1;

RecurrenceTable[{a[n + 1] == 2 a[n] + p[[z++]], a[1] == 1}, a, {n, 1, 5}]

But the answer is wrong:

{1, 3, 7, 15, 31}

recursion vector

share|improve this question

asked 5 hours ago

Reza FarajiReza Faraji

132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Related: (14645)
  $endgroup$
  – Mr.Wizard♦
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Directly going to the issue, my problem is how to use a RecurrenceTable when it have inputs from a vector defined previously. For instance I have:

p={1,10,100,1000,10000};

And the recursive relation I want to solve is:

RecurrenceTable[{a[n + 1] == 2 a[n] + p[[n]], a[1] == 1}, a, {n, 1, 5}]

which must have an answer as:

{1,3,16,132,1264}

But I face with this error:

Part::pkspec1: The expression n cannot be used as a part specification.

Part::pkspec1: The expression #1 cannot be used as a part specification.

For avoiding part specification I tried the following:

z = 1;

RecurrenceTable[{a[n + 1] == 2 a[n] + p[[z++]], a[1] == 1}, a, {n, 1, 5}]

But the answer is wrong:

{1, 3, 7, 15, 31}

recursion vector

share|improve this question

asked 5 hours ago

Reza FarajiReza Faraji

132

$endgroup$

p={1,10,100,1000,10000};

RecurrenceTable[{a[n + 1] == 2 a[n] + p[[n]], a[1] == 1}, a, {n, 1, 5}]

Part::pkspec1: The expression n cannot be used as a part specification.

Part::pkspec1: The expression #1 cannot be used as a part specification.

z = 1;

RecurrenceTable[{a[n + 1] == 2 a[n] + p[[z++]], a[1] == 1}, a, {n, 1, 5}]

share|improve this question

asked 5 hours ago

Reza FarajiReza Faraji

132

share|improve this question

asked 5 hours ago

Reza FarajiReza Faraji

132

asked 5 hours ago

Reza FarajiReza Faraji

132

asked 5 hours ago

Reza FarajiReza Faraji

132

1 Answer
1

active

oldest

votes

2

$begingroup$

Try Indexed:

p = {1, 10, 100, 1000, 10000};



RecurrenceTable[{a[n + 1] == 2 a[n] + Indexed[p, n], a[1] == 1}, a, {n, 1, 5}]

{1, 3, 16, 132, 1264}

Alternatives:

RecurrenceTable[{a[n + 1] == 2 a[n] + Unevaluated[p[[n]]], a[1] == 1}, a, {n, 1, 5}]

{1, 3, 16, 132, 1264}

part[x_, n_Integer] := x[[n]]



RecurrenceTable[{a[n + 1] == 2 a[n] + part[p, n], a[1] == 1}, a, {n, 1, 5}]

{1, 3, 16, 132, 1264}

Of course you could just Quiet the error and ignore it:

RecurrenceTable[{a[n + 1] == 2 a[n] + p[[n]], a[1] == 1}, a, {n, 1, 5}] // Quiet

{1, 3, 16, 132, 1264}

share|improve this answer

answered 4 hours ago

Mr.Wizard♦Mr.Wizard

233k294791067

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a bunch @Mr.Wizard♦
  $endgroup$
  – Reza Faraji
  4 hours ago
  

  
  
  
  

add a comment | 

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2

$begingroup$

Try Indexed:

p = {1, 10, 100, 1000, 10000};



RecurrenceTable[{a[n + 1] == 2 a[n] + Indexed[p, n], a[1] == 1}, a, {n, 1, 5}]

{1, 3, 16, 132, 1264}

Alternatives:

RecurrenceTable[{a[n + 1] == 2 a[n] + Unevaluated[p[[n]]], a[1] == 1}, a, {n, 1, 5}]

{1, 3, 16, 132, 1264}

part[x_, n_Integer] := x[[n]]



RecurrenceTable[{a[n + 1] == 2 a[n] + part[p, n], a[1] == 1}, a, {n, 1, 5}]

{1, 3, 16, 132, 1264}

Of course you could just Quiet the error and ignore it:

RecurrenceTable[{a[n + 1] == 2 a[n] + p[[n]], a[1] == 1}, a, {n, 1, 5}] // Quiet

{1, 3, 16, 132, 1264}

share|improve this answer

answered 4 hours ago

Mr.Wizard♦Mr.Wizard

233k294791067

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a bunch @Mr.Wizard♦
  $endgroup$
  – Reza Faraji
  4 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Try Indexed:

p = {1, 10, 100, 1000, 10000};



RecurrenceTable[{a[n + 1] == 2 a[n] + Indexed[p, n], a[1] == 1}, a, {n, 1, 5}]

{1, 3, 16, 132, 1264}

Alternatives:

RecurrenceTable[{a[n + 1] == 2 a[n] + Unevaluated[p[[n]]], a[1] == 1}, a, {n, 1, 5}]

{1, 3, 16, 132, 1264}

part[x_, n_Integer] := x[[n]]



RecurrenceTable[{a[n + 1] == 2 a[n] + part[p, n], a[1] == 1}, a, {n, 1, 5}]

{1, 3, 16, 132, 1264}

Of course you could just Quiet the error and ignore it:

RecurrenceTable[{a[n + 1] == 2 a[n] + p[[n]], a[1] == 1}, a, {n, 1, 5}] // Quiet

{1, 3, 16, 132, 1264}

share|improve this answer

answered 4 hours ago

Mr.Wizard♦Mr.Wizard

233k294791067

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a bunch @Mr.Wizard♦
  $endgroup$
  – Reza Faraji
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Try Indexed:

p = {1, 10, 100, 1000, 10000};



RecurrenceTable[{a[n + 1] == 2 a[n] + Indexed[p, n], a[1] == 1}, a, {n, 1, 5}]

{1, 3, 16, 132, 1264}

Alternatives:

RecurrenceTable[{a[n + 1] == 2 a[n] + Unevaluated[p[[n]]], a[1] == 1}, a, {n, 1, 5}]

{1, 3, 16, 132, 1264}

part[x_, n_Integer] := x[[n]]



RecurrenceTable[{a[n + 1] == 2 a[n] + part[p, n], a[1] == 1}, a, {n, 1, 5}]

{1, 3, 16, 132, 1264}

Of course you could just Quiet the error and ignore it:

RecurrenceTable[{a[n + 1] == 2 a[n] + p[[n]], a[1] == 1}, a, {n, 1, 5}] // Quiet

{1, 3, 16, 132, 1264}

share|improve this answer

answered 4 hours ago

Mr.Wizard♦Mr.Wizard

233k294791067

$endgroup$

p = {1, 10, 100, 1000, 10000};



RecurrenceTable[{a[n + 1] == 2 a[n] + Indexed[p, n], a[1] == 1}, a, {n, 1, 5}]

RecurrenceTable[{a[n + 1] == 2 a[n] + Unevaluated[p[[n]]], a[1] == 1}, a, {n, 1, 5}]

part[x_, n_Integer] := x[[n]]



RecurrenceTable[{a[n + 1] == 2 a[n] + part[p, n], a[1] == 1}, a, {n, 1, 5}]

RecurrenceTable[{a[n + 1] == 2 a[n] + p[[n]], a[1] == 1}, a, {n, 1, 5}] // Quiet

share|improve this answer

answered 4 hours ago

Mr.Wizard♦Mr.Wizard

233k294791067

answered 4 hours ago

Mr.Wizard♦Mr.Wizard

233k294791067

answered 4 hours ago

Mr.Wizard♦Mr.Wizard

233k294791067