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ClearAll["Global`*"];
Subscript[i, 0] := {{1 - 1/E^(1/10), 
   5.196153021391857`}, {1 - 1/E^(1/5), 
   5.196159705940993`}, {1 - 1/E^(3/10), 
   5.196180885069776`}, {1 - 1/E^(2/5), 
   5.196222762115215`}, {1 - 1/Sqrt[E], 
   5.196288209096974`}, {1 - 1/E^(3/5), 
   5.196377257424081`}, {1 - 1/E^(7/10), 
   5.196487957277755`}, {1 - 1/E^(4/5), 
   5.196617178572843`}, {1 - 1/E^(9/10), 5.19676123455273`}, {1 - 1/E,
    5.1969163216793515`}}
Subscript[i, 1] := {{1 - 1/E^(1/10), 
   5.19191873287261`}, {1 - 1/E^(1/5), 
   5.191925494281923`}, {1 - 1/E^(3/10), 
   5.1919469304276875`}, {1 - 1/E^(2/5), 
   5.191989341643741`}, {1 - 1/Sqrt[E], 
   5.192055661285492`}, {1 - 1/E^(3/5), 
   5.192145944109194`}, {1 - 1/E^(7/10), 
   5.19225823182757`}, {1 - 1/E^(4/5), 
   5.192389362814114`}, {1 - 1/E^(9/10), 
   5.192535604203823`}, {1 - 1/E, 5.192693098899519`}}
Subscript[i, 2] := {{1 - 1/E^(1/10), 
   5.771775702415989`}, {1 - 1/E^(1/5), 
   5.771779980434482`}, {1 - 1/E^(3/10), 
   5.771793319215499`}, {1 - 1/E^(2/5), 
   5.771819298487611`}, {1 - 1/Sqrt[E], 
   5.771859339949759`}, {1 - 1/E^(3/5), 
   5.771913134732232`}, {1 - 1/E^(7/10), 
   5.771979242055644`}, {1 - 1/E^(4/5), 
   5.772055604738526`}, {1 - 1/E^(9/10), 5.77213992681313`}, {1 - 1/E,
    5.7722299243470445`}}
Subscript[i, 3] := {{1 - 1/E^(1/10), 
   5.201117908780306`}, {1 - 1/E^(1/5), 
   5.201124560147791`}, {1 - 1/E^(3/10), 
   5.201145637585236`}, {1 - 1/E^(2/5), 
   5.2011873188104385`}, {1 - 1/Sqrt[E], 
   5.201252466062782`}, {1 - 1/E^(3/5), 
   5.201341113234828`}, {1 - 1/E^(7/10), 
   5.201451320907582`}, {1 - 1/E^(4/5), 
   5.201579973722657`}, {1 - 1/E^(9/10), 
   5.201723401383306`}, {1 - 1/E, 5.201877816815951`}}
Subscript[i, 4] := {{1 - 1/E^(1/10), 
   5.2510206064477964`}, {1 - 1/E^(1/5), 
   5.251027254946757`}, {1 - 1/E^(3/10), 
   5.251048271461156`}, {1 - 1/E^(2/5), 
   5.251089743481776`}, {1 - 1/Sqrt[E], 
   5.2511544448332`}, {1 - 1/E^(3/5), 
   5.2512423458399775`}, {1 - 1/E^(7/10), 
   5.251351475521756`}, {1 - 1/E^(4/5), 
   5.251478716734289`}, {1 - 1/E^(9/10), 
   5.251620420810767`}, {1 - 1/E, 5.2517728383892575`}}


Plot[{Interpolation[Subscript[i, 0]][a]}, {a, 0, 1}]
Plot[{Interpolation[Subscript[i, 1]][a]}, {a, 0, 1}]
Plot[{Interpolation[Subscript[i, 2]][a]}, {a, 0, 1}]
Plot[{Interpolation[Subscript[i, 3]][a]}, {a, 0, 1}]
Plot[{Interpolation[Subscript[i, 4]][a]}, {a, 0, 1}]
Plot[{Interpolation[Subscript[i, 0]][a], 
  Interpolation[Subscript[i, 1]][a], 
  Interpolation[Subscript[i, 2]][a], 
  Interpolation[Subscript[i, 3]][a], 
  Interpolation[Subscript[i, 4]][a]}, {a, 0, 1}]

I am trying to plot these five curves in one plot. The problem is that when I do that the lines appear as if they are straight horizontal lines and they don't change as $a$ increases. Is there a way or scale to use to plot these five curves in one plot so that it appears they do change with $a$?

I hope you understand what I mean. Any help is appreciated.

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4
  • $\begingroup$ Plotting Interpolation[Subscript[i, 3]][a] is straightening all the plots because the y-axes range increases. Perhaps, you can try a double vertical axes for the Interpolation[Subscript[i, 3]][a] mathematica.stackexchange.com/a/629/19742 $\endgroup$
    – Anjan Kumar
    Commented Feb 6, 2017 at 8:10
  • $\begingroup$ It is the problem of the scale. Indeed, the difference between the last and the first terms within the same list is Last[Subscript[i, 1]][[2]] - First[Subscript[i, 1]][[2]] gives 0.000774, while the difference between any terms of different lists is, say, First[Subscript[i, 4]][[2]] - First[Subscript[i, 1]][[2]] giving 0.0591. This is two orders of magnitude greater. For this reason, if you show all the lists on the same plot you cannot visualize the tiny variations within each of the lists. If you show them on different plots, you will see the individual variations. $\endgroup$
    – Alexei Boulbitch
    Commented Feb 6, 2017 at 8:18
  • $\begingroup$ @AlexeiBoulbitch, I have already noticed that but I am asking is there a way to solve this like using some sort of scaling or something. $\endgroup$
    – MrDi
    Commented Feb 6, 2017 at 8:21
  • $\begingroup$ @MrDi If you had two lists, you might make different axis to the left and to the right. If you are interested, there has been such a question/answer in the past. You may, of course, make four axis, but to my taste it will be a mess. If you notice the origin of your problem I encourage you to edit the question and formulate it. This will enable us to help you more efficiently. $\endgroup$
    – Alexei Boulbitch
    Commented Feb 6, 2017 at 8:27

2 Answers 2

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This is rather kludgy and very manual, i.e. the values have to be optimized by hand, but I think that it expresses the idea you are hoping to convey:

interpolations = {Interpolation[Subscript[i, 0]][a], 
   Interpolation[Subscript[i, 1]][a], 
   Interpolation[Subscript[i, 2]][a],
   Interpolation[Subscript[i, 3]][a], 
   Interpolation[Subscript[i, 4]][a]};

MapIndexed[
  With[{multi = 0.1},
   LogPlot[#1,
     {a, -multi First@#2, 1},
     PlotRange -> {{0, 1}, All}, PlotRangePadding -> None,
     AxesOrigin -> {multi (4 - First@#2), Automatic},
     ImageSize -> 400 , ImagePadding -> {{30, 10}, {20, 10}},
     Ticks -> {Automatic, {5.194, 5.195, 5.198, 5.199, 5.202, 5.203, 
        5.772, 5.773}},
     PlotStyle -> ColorData[97][First@#2], 
     AxesStyle -> {Automatic, ColorData[97][First@#2]}
     ] &
   ],
  interpolations
  ] // Overlay

Mathematica graphics

Essentially, it generates four different plots, with a vertical axis positioned farther and farther to the left, all the way to the leftmost border. The plots are color-coded for readability. They are then overlaid to generate the final plot.

Note that this requires careful control of all dimensions of the generated graphic object, i.e. padding, size, etc.

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0
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Another way to go would be to rescale them all to start at zero by subtracting their minimum value, and have the list with the largest range in y-values go to 1. Then print out their actual ranges as a legend. 

scales = MinMax[Last /@ #] & /@ {i0, i1, i2, i3, i4};
maxScale = Max[Subtract @@@ Reverse /@ scales];
rescaledLists = Thread[{First /@ #,
      ((Last /@ #) - Min[Last /@ #])/maxScale}] & /@
   {i0, i1, i2, 
    i3, i4};
legends = MapIndexed[
   Row[{Subscript["i", First@#2], ":", #1}] &, scales];
ListLinePlot[
 rescaledLists,
 PlotLegends -> legends,
 PlotRange -> All,
 InterpolationOrder -> 2]

Mathematica graphics

I renamed all the lists so they aren't subscripts, and just used the interpolation built in to ListLinePlot because I think that's better practice, but you could easily change it back to suit your needs.

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