abdulganiy

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8 years, 318 days

MaplePrimes Activity


These are replies submitted by abdulganiy

@Preben Alsholm 

Your suggestion and correction is appreciated

@vv 

Thank you. Your suggestion is appreciated.

@Kitonum 

Thank you. the correction you made is greatly appreciated.

@Rouben Rostamian  

Thank you sir. it really worked.

Kind regards

@Kitonum

mtaylor command has been used by didnt work out.

 

Thank you for your suggestion. It is appreciated

@tomleslie 

Your comments and suggestions are appreciated.

Thank you and kind regards

@tomleslie 

Thank you for your quick response.

The Idea is to extract the maximum of all the errors in the column 10 and 12 of the displayed table of results. Techically if "N:=solve(h*p = 12*Pi/6, p):" is solved for using the value "h = Pi/6", the desired result should be N=12. However, I dont know why this is not so. 

@acer I am very grateful. Thank you and kind .

@acer

I tried to adapt the code you sent on Legend but still encounter some difficulties. I notice that the one you sent was ploted for the three curves on the same interval. However, the curves I intend to plot are not within the same interval. so I have decided to include two codes here for your perusal before adaptation and after I have adapted your approach.

before adaptation

B:=<<150,200,300,600,800,1600,2400,3200>|<1.23E-2,5.694E-3,3.134E-4,1.259E-7,1.264E-7,4.947E-10,1.931E-11,1.994E-12>|<170,225,381,680,1207,2144,3806,6762>|<2.886E-1,7.846E-3,1.399E-3,1.690E-4,1.846E-5,1.938E-6,1.993E-7,2.021E-8>|<277,496,884,1573,2796,4970,8833,15706>|<2.153E0,1.494E-1,9.359E-3,6.2E-4,4.416E-5,3.412E-6,2.848E-7,2.530E-8>>:

for i from 1 to 8 do
   B[i,2] := log(B[i,2]):
   B[i,4] := log(B[i,4]):
   B[i,6] := log(B[i,6]):
end do:  # computing the log of the max-error
B; # This is the table of values we'll plot.
               Matrix(%id = 18446746337307461022)
plot([B[()..(),[1, 2]], B[()..(), [3, 4]], B[()..(), [5, 6]]], legend = [TSDM, FESDIRK4, ESDIRK4], axis = [gridlines = [colour = green, majorlines = 2]]);


After adapting the approach you sent I came up with this

restart;


B:=<<150,200,300,600,800,1600,2400,3200>|<1.23E-2,5.694E-3,3.134E-4,1.259E-7,1.264E-7,4.947E-10,1.931E-11,1.994E-12>|<170,225,381,680,1207,2144,3806,6762>|<2.886E-1,7.846E-3,1.399E-3,1.690E-4,1.846E-5,1.938E-6,1.993E-7,2.021E-8>|<277,496,884,1573,2796,4970,8833,15706>|<2.153E0,1.494E-1,9.359E-3,6.2E-4,4.416E-5,3.412E-6,2.848E-7,2.530E-8>>:

for i from 1 to 8 do
   B[i,2] := log(B[i,2]):
   B[i,4] := log(B[i,4]):
   B[i,6] := log(B[i,6]):
end do:  # computing the log of the max-error
B; # This is the table of values we'll plot.
               Matrix(%id = 18446746759326532422)

f1 := TSDM:
f2 := FESDIRK4:
f3 := ESDIRK4:
plot([B[()..(),[1, 2]], B[()..(), [3, 4]], B[()..(), [5, 6]]], 
colour = [blue,green,red],style=pointline, symbol=[solidcircle,diamond,box], symbolsize=10, 
adaptive=false, size=[500,500],axis = [gridlines = [colour = green,majorlines = 2,linestyle = dot]], 
thickness=1,legend = [f1, f2, f3]);



Thank you in anticipation of your quick and positive response

@acer Thank you very much for this hint. It is appreciated

I am sorry to have bothered you. Can you please give a prototype for three curves with legends appearing as symbols instead of line when "pointline" is used in the command?

@acer

Thank you.

@rlopez 

Thank you for the hint.

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