What would be the correct input in the wolfram website to give u(n) for a given n, where u is a sequence ?
Something like:
evaluate [u(n+1)=27*u(n)-14, u(0) = 1] at n = 3
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Welcome to MSE! – Culver Kwan Aug 16 '19 at 09:41
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1Most of the work can be done with " recurrance a_(n+1) = 27*a_(n)-14 " – Matti P. Aug 16 '19 at 09:50
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Entering "a(0)=1, a(n+1) = 27*a(n)-14" outputs a closed form and first nine values. You can enter the closed form and specify n if you want other values "n=3, 1/13 (7 + 2 3^(1 + 3 n))". – Vepir Aug 16 '19 at 10:10
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@MattiP. @ Vepir I had figured that out, but then 2 steps are required. I want to be able to change u(0) quickly, in one step... Is that not possible ? – MFX Aug 16 '19 at 13:23
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There is a Mathematica Stack Exchange. – Jair Taylor Aug 16 '19 at 20:27
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@JairTaylor I gave my answer to OP's question here - but I'm not sure If I understand the "wolfram-alpha" tag on this site. What is the borderline between wolfram-alpha queries and Mathematica interpreted (related) queries? – Vepir Aug 16 '19 at 20:49
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@vepir Not sure myself. Just thought the other site might be helpful. – Jair Taylor Aug 16 '19 at 20:53
1 Answers
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The comments already discussed how you can manipulate the recurrence. - But if you are not satisfied with that, why not use the Mathematica cloud interpreter in the wolfram lab?
- There are two ways to do this. The direct one is to open the code of your wolfram query:
$1.$
$2.$
- Which will lead you to the computable notebook, and after you scroll down a bit, you'll find the code:
$3.$
- Where for example I told it to get me the terms from $4$th to $6$th. After running that code section (Either clicking Run or pressing Shift + Enter), you will get your results instantly:
What if "Open code" is not available?
However, if you can't directly open the code of your wolfram query, you can go and open a new notebook and paste that code there.
You navigate to programming lab and click "Start programming now". Then click File and New notebook.
Click on (+) sign and select "Wolfram language input". Now type in the code from the lab:
RecurrenceTable[{a[0] == 1, a[1 + n] == -14 + 27 a[n]}, a, {n, 4, 6}]
Modify the {n, 4, 6} part to list the terms you want, and the {a[0] == 1, a[1 + n] == -14 + 27 a[n]} part to setup your recurrence.
Then either click the gear icon on the right and select "Evaluate", or press Shift+Enter.
You'll get your output:
{245281,6622573,178809457}
Bonus points, if you want to quickly change the starting condition, you can make a table:
A code snippet to help with your example
You can use the following code:
ClearAll[recurrence,a]
recurrenceName = a;
recurrence[x_] := {a[0] == x, a[1 + n] == -14 + 27 a[n]}
a0min = 0;
a0max = 3;
nmin = 0;
nmax = 5;
Grid[Prepend[Transpose[Prepend[Transpose[Table[RecurrenceTable[recurrence[x], recurrenceName, {n, nmin, nmax}],{x,a0min,a0max}]],Table[StringForm["a(0) = ``", a],{a,a0min,a0max}]]],Table[If[n<nmin,"",StringForm["n = ``", n]],{n,nmin-1,nmax}]],Frame->All]
To instantly generate a table for all values you need:
Vepir
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