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comment Solve this equation symbolically
@user71346 For does not accumulate output, and is also inefficient here. (Read this.) Use Table instead: Table[vals, {a, -1, 1, 0.02}]
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revised Why don't these styling rules work when viewed in TableForm?
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answered Why don't these styling rules work when viewed in TableForm?
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comment How to combine SparseArray and If
@Mencia I have updated my answer.
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revised How to combine SparseArray and If
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comment Beginner's question: Gap between FrameLabel and BarChart
@Stefan For what it's worth the output (from this) looks a bit different on version 7, and the longer code was required to match your output. Also, I believe you should be able to use the slightly shorter PlotRangePadding -> 0 to produce the same output as given by PlotRangePadding -> {0, 0}.
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comment vectorized operations is not efficient?
@explorer I believe you are asking: is UnitStep[1 - r] equivalent to 1 - UnitStep[r - 1]? Not precisely, because of the case of 1: {UnitStep[1 - r], 1 - UnitStep[r - 1]} /. r -> 1. It is unlikely to make a difference in your application however, IMHO, but to be rigorous I used the longer form.
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reviewed Leave Open Generating a list of all factorizations
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answered Beginner's question: Gap between FrameLabel and BarChart
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comment Beginner's question: Gap between FrameLabel and BarChart
Welcome, Stefan. I embedded the images for you. The image/link limitation is an anti-spam measure that will be lifted once you start using the site more.
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revised Beginner's question: Gap between FrameLabel and BarChart
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revised Variable scoping problem when mapping over delayed replacement
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comment How could I find the correct values for every array that would lead me to unique summation number for every unique chain?
@m.rizeg somehow missed your comment that you aren't using Mathematica. In light of this, regarding our earlier discussion, I'll let the StackOverflow folks try answering this to your satisfaction (after all there are a lot more of them). You're still welcome in the Mathematica.SE chat room of course.
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answered Applying a function with the HoldAll attribute inside NestList
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revised vectorized operations is not efficient?
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comment Solve this equation symbolically
@user71346 you could substitute an arbitrary real value for a and convert to numeric values with N. This will not prove that certain solutions are always real (I think), but it can help you eliminate the ones that return complex values. Try: sols /. a -> 1 // N
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answered vectorized operations is not efficient?
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comment Solve this equation symbolically
@user71346 I see I missed a major aspect of your updated question: the Root objects! Please see this fine answer.
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