# Tiny question about Indexed[x,n]

Maybe I should apologize for this silly question, but the documentation for Indexed is very short and not explicit.

I have an equation with variables Indexed[x,n] and construct a sparse matrix through CoefficientArrays.

How can I choose all $x$ with an index value $n≥|b|$ as variables in CoefficientArrays?

And how can I set a rule of substitution for an indexed variable? Construction of form Indexed[x,1]=a^2 doesn't work, so what's the correct variant?

Thank you in advance for your help

• I can't quite figure out what you are asking. It would help if you could give a small snippet of code that shows the kind of thing you have and what you expect to see as output. – bill s Jun 5 '15 at 7:48
• You could index your variables using x[n]; you can freely assign values to such variables. – MarcoB Jun 5 '15 at 7:56
• @bills code to the first question: eqnsList = {5*Indexed[x, 1] - f^2 Indexed[x, -3] + Sqrt[2] Indexed[x, 4] == 0, 1/5 Indexed[x, -4] - g Indexed[x, 2] == 0}; linsystem = CoefficientArrays[eqns, Indexed[x, 1]]. This code is good when I get only factors of Indexed[x, 1], but I don't know how to choose as variables all $x$ with an index factor $n\geq \left| 3\right|$ About the second one: if I want to substitute smth as a^2 in all expr. for Indexed[x,1], I do it like Indexed[x,1]=a^2, but I get only an error >Set::write: Tag Indexed in Subscript[coefA, 1] is Protected. >> – Hedin Jun 5 '15 at 9:15
• @Hedin -- you should place your code in the question itself so that others can easily find it (you can use the "edit button" to edit your question). – bill s Jun 5 '15 at 9:42

I'm not sure I know what you are trying to accomplish, but you can do the same thing for any set of indices that you are doing for Index[x,1] using Map (shortcut /@) as follows:

eqnsList = {5*Indexed[x, 1] - f^2 Indexed[x, -3] + Sqrt[2] Indexed[x, 4] == 0,
1/5 Indexed[x, -4] - g Indexed[x, 2] == 0};
linsystem = CoefficientArrays[eqnsList, Indexed[x, #]] & /@ {-3, 1, 2, 4}


This does the same thing for indices {-3, 1, 2, 4}. Of course you can stick any set you wish in their place.

For the second question, you can use a replacement rule, for instance:

eqnsList /. Indexed[x, 1] -> a^2


replaces all occurrences of Indexed[x, 1] with a^2.

• glad to help. I didn't feel easy on pure functions yet, but I'll try to fix that drawback. – Hedin Jun 5 '15 at 10:15