[?] List of elements of a complex sequence

Posted 1 month ago
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 Hi all, I am new to Mathematica but was wondering if it is possible to list complex (a+ib) elements of a sequence as a function of an element. For instance if I define a sequence recursively as x(k+1) = (1-i) x(k), can Mathematica display as output x(1)=(1-i) x(0); x(2)=(-2i) x(0); x(3)= -2(1+i) x(0); x(4)=-4 x(0); etc. ? If it is possible, how ? What would be the best way to plot the result ? Thanks !
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Posted 1 month ago
 sol=RSolve[{x[k] == (1 - I) x[-1 + k], x[0] == x0}, x[k], k]
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Posted 1 month ago
 Thank you very much for your answer. However I was more looking for the following one (your answer did help a little bit...): RecurrenceTable[{x[n + 1] == (1 - I) x[n], x[0] == 1}, x, {n, 1, 12}] or RecurrenceTable[{x[n + 1] == (1 - I) x[n], x[0] == x0}, x, {n, 1, 12}] Now I have to figure out how I can plot those points in the (Re, Im) plane...:)
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Posted 1 month ago
 Hi Al,You can use ComplexListPlot. values = RecurrenceTable[{x[n + 1] == (1 - I) x[n], x[0] == x0}, x, {n, 1, 12}] /. x0 -> 1 ComplexListPlot[Callout[#, #] & /@ values, PlotRange -> All, Joined -> True, ImageSize -> 500]
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Posted 1 month ago
 OKThe RSolve given above yields as answer {{x[k] -> (1 - I)^k x0}} Now define a Function fx fx[k_, x0_] := FullSimplify[(1 - I)^k x0] Look what happens to fx[5, 2 + 3 I] or Table[fx[j, x0], {j, 1, 10}] For a plot you must specify x0. Let us chose x0 = 1 tt = Table[{j, fx[j, 1]}, {j, 1, 15}]; pl1 = ListLinePlot[Re[#[[2]]] & /@ tt,PlotStyle->Red] (*real parts*) pl2 = ListLinePlot[Im[#[[2]]] & /@ tt,PlotStyle->Blue] Look at both plots in one Show[pl1, pl2]
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Posted 1 month ago
 Without having acces to such cool functions like ComplexListPlot one could do it in an old fashioned manner oneself t1 = Table[fx[j, 1], {j, 1, 15}]; t2 = If[Head[#] === Integer, {#, 0}, List[Re[#], Im[#]]] & /@ t1 ListLinePlot[t2]
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Posted 1 month ago
 Thank you.
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