0
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I don't understand why, but, with this code, I can't get a graph. Nothing is happening in output. And I have no errors.. Here is the code:

ListPlot[RecurrenceTable[{a[n + 1] == (a[n] + b[n])/2, 
   b[n + 1] == (2 / (1 / a[n] + 1/b[n])), a[0] == 4, b[0] == 3}, {a, 
   b}, {n, 0, 30}],
 PlotRange -> {0, 100},
 ImageSize -> 400,
 AxesLabel -> {"n", "anbn"},
 PlotLabel -> "Premiers Termes :-) "
 ]

Do you know what is the problem?

Thanks for your time!!!

Edit: Now, changing a[0], b[0] and n, I have this:

enter image description here

New edit:

enter image description here

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9
  • $\begingroup$ Exact calculations are too time consumption in this case. Is using a[0] == 4.0 and b[0] == 3.0 an option for you? I think you will have to do this numerically. $\endgroup$ Commented Feb 23, 2018 at 16:08
  • $\begingroup$ What do you mean? Its not an obligation to put a[0] and b[0] with these values. But, It must be superior to 0 for both, thats all .. Have you got any suggestions? $\endgroup$
    – nolwww
    Commented Feb 23, 2018 at 16:32
  • 1
    $\begingroup$ When you replace the 4 with 4.0 and the 3 with 3.0 you trigger numerical evaluation which is much faster. To see this, evaluate your RecurrenceTable (remove the plot stuff) replacing the 30 with much smaller values and increment them gradually. You will see some huge fractions that take increasingly more time to calculate. $\endgroup$ Commented Feb 23, 2018 at 16:37
  • $\begingroup$ Why is there only two points and not a wide range of points? is it not supposed to be points for a[0] a[1] ....... a[10] and same for b ? I put a new edit with your advices. $\endgroup$
    – nolwww
    Commented Feb 23, 2018 at 16:39
  • $\begingroup$ I dont understand why n dont have values higher than 3.55 $\endgroup$
    – nolwww
    Commented Feb 23, 2018 at 16:45

1 Answer 1

1
$\begingroup$
        I think you can see the dependence of a[0]=s, b[0]=q given below plot.

         Manipulate[
          ListLinePlot[
         RecurrenceTable[{a[n + 1] == (a[n] + b[n])/2, 
       b[n + 1] == (2/((1/a[n]) + (1/b[n]))), a[0] == s, b[0] == q}, {a, 
      b}, {n, 0, 20, 1}], PlotRange -> All, ImageSize -> 400, 
     PerformanceGoal -> "Speed", Frame -> True, 
         PlotMarkers -> {Automatic, 16}, PlotStyle -> Blue], {s, 1, 
      30}, {q, .001, 1}]

plot of data

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1
  • $\begingroup$ Thanks !!! But we don't see b values, right? $\endgroup$
    – nolwww
    Commented Feb 23, 2018 at 17:10

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