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How's this? range = {1, 3, 5, 7, 9}; DeleteCases[Flatten[Table[i/j, {i, range}, {j, range}]], 1] (* {1/3, 1/5, 1/7, 1/9, 3, 3/5, 3/7, 1/3, 5, 5/3, 5/7, 5/9, 7, 7/3, 7/5, 7/9, 9, 3, 9/5, 9/7} *)

Here's one way, using Replace to wrap your points x with {r,x} and Table to iterate over r. res = Flatten[Table[ list = RecurrenceTable[{x[n] == r (x[n - 1] - x[n - 1]^3), x[1] == 0.5}, x, {n, 1, 2 …

I think Sequence does the job for you: a[[Sequence @@ l[[1]]]] (* a11 *) Equivalently, without (most) brackets, call it Part: Part[a, Sequence @@ l[[1]]] (* a11 *)

I ran into a strange problem, where using ConstantArray inside FindRoot results in a FindRoot::jsing error. Here's a minimal example, with the first instance working and the second failing: FindRoot …

Seems to be the same underlying issue as here: by default, Table compiles its argument when the number of values is 250 or more. Evidently Internal`Bag doesn't like this! If all Internal`Bags are th …

I need a lot of Internal`Bags so I made them with Table. It works fine for 249 or fewer Bags but goes horribly wrong for 250 or more: Table[Internal`Bag[0], {i, 249}] (* {Internal`Bag["<" 1 ">"], .. …

This could be made nicer, but how about something like this? GraphicsColumn[{ ListLogPlot[Select[xp, #[[2]] > 0 &], PlotRange -> {{0, 10}, All}, PlotRangePadding -> Scaled[.05]], ListLogPlot[A …

Something like this? f[x_] := x^2; With[{x := 10^xp}, Table[{x, f[x]}, {xp, 0, 5}]] {{1, 1}, {10, 100}, {100, 10000}, {1000, 1000000}, {10000, 100000000}, {100000, 10000000000}}

Another solution: increase the PlotPoints in testPlot2: testPlot1 = Plot[Log[HarmonicNumber[n]], {n, 1, 5}]; testPlot2 = Plot[Log[HarmonicNumber[n]], {n, 1, 43}, PlotPoints -> 100]; Show[testPlot1 …

My idea is to make a new function g that builds up a list of values, then use FixedPoint on g with a SameTest that looks for repeat values. First, a helper (since MemberQ wasn't checking for numerica …

For this particular example you can find the bifurcations analytically by also setting the slopes of the isoclines to be equal. eq1 := x^2 + y + b; eq2 := x + y^2 - a; bif = Solve[{eq1 == 0, eq2 == 0 …

What's the fastest way to find the local maxima of a 2D list? E.g. nx = ny = 100; dat = Table[Sin[2. \[Pi] x/nx] (0.1 + Cos[2. \[Pi] y/ny]), {y, 0, ny}, {x, 0, nx}]; ListPlot3D[dat] This (updat …

Not sure you know this, but if you use Plot you don't need to manually define the x-coordinates. Plot[Evaluate[Table[f[x, t], {t, 0, 0.9, 0.1}]], {x, -1, 1}] The Evaluate is only necessary to get …

I think the easiest way to get what you want is with Table: z0 = 1; ta = 0.01; ytotal[ta] = Table[z0 = y[z0, ta][T] /. sol, {i, 100}]; For those parameter values, it looks like the population quick …

Here's a solution using my EcoEvo package, which is designed for just this kind of problem. First, install the package (only need to do this once): PacletInstall["EcoEvo", "Site" -> "http://raw.githu …