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I am trying to find the reflection function. Here is my function and its graph.

eq[n_, \[Beta]_, a_] :=  Hypergeometric1F1[1/2 - 1/4 a, n + 1, \[Beta]]
ED[n_, a_, k_Integer: 1] := \[Beta] /.   FindRoot[eq[n, \[Beta], a] == 0, {\[Beta], 0}]
rootslist[n_Integer, k_Integer, a_Integer] :=  Rest@FoldList[
FindRoot[eq[n, \[Beta], #2] == 0, {\[Beta], #1}][[1, 2]] &, 0,Range@a]
firstroot = Table[rootslist[n, 1, 50], {n, 0, 5}]
gr = ListLinePlot[firstroot, PlotRange -> Automatic, AxesLabel -> {(2 - 4 a), \[Beta]},
frame -> False]

If I tried this I can get the function of a graph, (x-axis:a and y-axis:$\beta$). I am wondering how to get the reflection function of this? (x-axis:$\beta$ and y-axis:a)? I have tried this code, but it didn't work.

Show[gr, Plot[a, {a, Automatic}, PlotStyle -> Black], 
gr /. L_Line -> {Red, GeometricTransformation[L, ReflectionTransform[{-1, 1}]]}, PlotRange   
-> All]
share|improve this question

marked as duplicate by whuber, Jens, J. M. Apr 25 '13 at 5:14

This question was marked as an exact duplicate of an existing question.

    
Welcome to our site, Keith! You probably didn't know or guess, but searching for keywords like listlineplot axes would have spared you the effort of re-asking this question. – whuber Apr 25 '13 at 3:07

Just swap the Axes:

ListLinePlot[((Partition[#, 2] & /@ (Riffle[#, Range@50] & /@ firstroot))[[All, 2 ;;]])]

enter image description here

share|improve this answer
    
Belisarius, I myself have used a combination of Riffle and Partition so often that I made myself remember to use that. However, I was very glad to hear Thread did exactly this. Perhaps interesting: ListLinePlot[(Thread[{#, Range@50}] & /@ firstroot)[[All, 2 ;;]]]. But maybe I'm being annoying :P. – Jacob Akkerboom Jun 7 '13 at 11:28

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