@Ramakrishnan I have uploaded a wrong doc. Enclosed is doc that gives the answer you wanted in specified widths. Ali

gnment??
 

Download Real_Re_part.mw

Real_Re_part.mw

Excellent presentation of the document. Examples are easy to understand though not in english.

Congratulations. Thank you for letting us know the way of teaching.

Ramakrishnan V

@bouhafsmaple 

Thank you. I shall use embedded components to make this work more attractive and easy to follow. Concept understanding can be made easy with Maple.Unfortunately, I did not do anything further on it for the past two years . Your appreciation motivates me to work on it further. Thanks very much.

Sincerely 

Ramakrishnan. V

@Kitonum 

Dear Sir,
 

Download Two_equnsThreeUnknowns.mwTwo_equnsThreeUnknowns.mw

I want to draw a vertical line in the graph in my document which gives pair of x and z values for a given y. Could you find a way out? Thanks.

@acer 

I think I did not make the variable name changed inside the map argument. I have made a perfect workbook and enclosed now with your codes..

Can you help me in removing the y label from column graph in this document. Thanks again.

Ramakrishnan V

DataAnalysis-FindCurveandOptimumPoints.maple

Expt1_Data.xlsx

@acer 

Dear Sir, Early in the morning now, after posting the full example to you, I tried and surprisingly found it working!. But I donot know how it happened yesterday. It gave an error once. I was working with few other programmes at that time.In case you happened to diagnose I will be very thankful.In case you also found it working without problem, please let me know. Thanks.

Ramakrishnan V

@acer 

Hope you would pursuee the attached workbook for help.

Thanks.There is a ComoBox ChCB where in I have called the module to import the excel data. The data table on top of the doc displays the data in the format i need. The rest of the repeated codes also may be commented for my improvement. This doc gives you option to select parameters and report analysis.

Thanks for your help.

DataAnalysis-FindCurveandOptimumPoints.maple

Ramakrishnan V

@Carl Love .

Thank you very much. It works.

@acer 

Dear sir, Though the following codes given by you , work well in a document directly when specified, I have found difficulties in using the same codes for my use inside procedure or module. The arrow operator does not work. Should I add anything like with(DocumentTools) or Statistics like that. The arrow oerator is copied as an operator  inside the procedurewhen I attempted to copy iin a sprocedure.

BTE := map(u->parse(sprintf("%.3f",u)),
           Matrix(BTE, datatype = anything)):

@tomleslie 

Please find attached doc with proc call which works andwhich does not where there is an error 0 and %1 are not evaluated in module. Hope these two documents help you find the issue and solution.

Thanks.

Ramakrishnan V

Download readstatExample1.mw

readstatExample1.mw

readstatExampleNW.mw

@acer 

Dear Professor, Thank you so much. It works fine.

Ramakrishnan V
 

Download Doubt_NumberFormat_in_DataTable.mwDoubt_NumberFormat_in_DataTable.mw

@acer 

I have referred only the digits after the decimal point.

I considered that A decimal refers to any digit after the decimal separator, such as in "3.14 is the approximation of π to two decimals".

Sorry for not very clear in my usaget.

%2.3g is what normally i use. What is f stands for? I do not understand the following from help page.

The object is formatted as a fixed point number. The number of digits specified by the precision appears after the decimal point (the six digits, if no precision is specified).

Thank you very much.

Ramakrishnan V
You have considered 6 digits total with 3 after decimal pointer. [099.999 or 99.999 or 999.99 or 099.990 or 99.99 as per data].Am I correctly understood?

@adel-00 

Contour is a locus of points that satisfies all the three expressions here. The expressions given here are lamda1 = numerical values, lamda2 = numerical values and third is a function of Result, lamda1 and lamda2 which should be zero since contours = 0. Since this as an equation, i changed it in to an expression for Results  in terms of lamda1 and lamda2.Then it is simple coding to plot the three expressions as contour. The commands are

Result := (lambda__1*Pi*(lambda__2+1)^2/(4*lambda__2))^.5;
with(plots);
contourplot(Result, lambda__1 = 0 .. 1, lambda__2 = 0 .. 1);
 

@John May 

Dear Professor, It worked very well. I attach the document for just reference. Thanks. Ramakrishnan V

Download FormattedOutput_TextBox.mwFormattedOutput_TextBox.mw

@John May 

Thank you very much sir.

Ramaakrishnan V