abdulganiy

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3 years, 335 days

MaplePrimes Activity


These are replies submitted by abdulganiy

@Rouben Rostamian  

Your Suggestion is  highly apprciated. Thank you and kind regards

@acer Thank you very much. I really appreciate your suggestion and value the time you spent on it.

@Kitonum I really appreciate your suggestion. However, I have applied the "allvalues" command before settling for the line b the results dont just make any sense to me (too long and some contain imaginary) which shouldnt be.

@acer Thank you and kind regards

@acer 

Your suggestion and correction is appreciated . I have view questions as I am new to some of the command used in the correction

1. What is the function of

 

kernelopts(version);

 

2. Is it possible to display the final resuts " all2 " one after the other as contained in the original code. i.e alpha[1],beta[0],beta[1],beta[2],gamma0,gamma1,gamma2 separately and on separate lines. Your suggestion will be highly appreciated.

Thank you and kind regards.

@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

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