Ramakrishnan

Dr. Ramakrishnan Vaidyanathan

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4 years, 360 days

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I have retired as Professor-Mechanical in Sri Venkateswara College of Engineering and Technology under Anna University affiliated colleges in Tamil Nadu, India. I have 19 years of Industrial and 20 years of teaching experience. I am learning Maple for the past four and half years hoping to make at least one appreciable maple presentation.

MaplePrimes Activity


These are replies submitted by Ramakrishnan

@acer 

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

Ramakrishnan V
 

``

 

NULL

27.90

(1)

BTE := evalf(2.3)*Import("E:/A_PhD2017/MP1_Data.xlsx", "Brake_Thermal_Efficiency")

BTE := map(proc (u) options operator, arrow; parse(sprintf("%.3f", u)) end proc, Matrix(BTE, datatype = anything))

``

``

NULL

27.90

(2)

NULL


 

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


 

restart
``

z := "Maple1"; x := "Map"; y := "Maple2"; x1 := 9.080787; y1 := 9.98765; z1 := 8.090909

I have done it using the print format directly with variable values. This is fine for me.

Maple:        Maple:     Map

  9.9876:   9.0808       8.1

 

I have done it differently defining a variable for the print format. This is fine for me and required as standard.

y3 := "%-2.5s:        %2.5s:     %2.5s\n  %-2.5g:   %-2.5g       %-3.2g\n", z, y, x, y1, x1, z1

printf(y3)

Maple:        Maple:     Map

  9.9876:   9.0808       8.1

 

 

y4 := sprintf("\n %-2.5s:        %2.5s:     %2.5s   \n %-2.5g:   %-2.5g       %-3.2g\n", z, y, x, y1, x1, z1)

"
 Maple:        Maple:     Map   
 9.9876:   9.0808       8.1
"

(1)

text := GetProperty("printarea0", value)

"Maple:        Maple:     Map  9.9876:   9.0808       8.1
"

(2)

``

y4

"Maple:        Maple:     Map   
 9.9876:   9.0808       8.1
"

(3)

with(DocumentTools); SetProperty("printarea0", value, cat(text, y4))

``

                                 

 

SetProperty("printarea0", Value, "")NULL

 

 

NULL

 

 

NULL


 

Download FormattedOutput_TextBox.mwFormattedOutput_TextBox.mw

 

@John May 

Thank you very much sir.

Ramaakrishnan V

Hope it is not too late:

I attach a checkboxembedded component examples i made which works. Please see if you can its worth for teaching/use.

Thanks for using.

Ramakrishnan V

CheckBox_ExecutionExample_byVRK_for_forPrime.mw

@John Fredsted 

Thanks to my Professor Robert, I got the answer for my question I want a number of calculations under arithmatic ( a+b and a-b)

adding and subtracting in one statement in module
 

 

action := module () export arithmatic, times;  arithmatic := proc (a, b) options operator, arrow; [a+b, a-b] end proc; times := proc (a, b) options operator, arrow; `mod`(a*b, 5) end proc end module

NULLNULL

NULL

action:-arithmatic(2, 5) = [7, -3]NULL

action:-times(2, 6) = times(2, 6)NULL

``

``

``Thank you all.

Ramakrishnan V

 

NULL


 

Download 2_problems_in_module.mw

@John Fredsted 

Dear John Fredsted, I did not know about usage of 'module inside module' also. I understood from your answer about the term export and how to use export with module. I found an app very useful and was tying to make one similar for curve fitting in my project. I enclose herewith the same though it was from clod i got it. It involves startup code where in everything was programmed using module 'action'. Therefore I wanted to know about it. No problem if this could not be taken up further.

Thank you.RequestHelp_regression_curve_fixing_forData.mw

Ishall

@John Fredsted 

Thank. How is it used. What is the answer ? How is the problem framed for this answer?

n(10) = 20 is the answer ok.

will m1 give the same answer.

will m2 also give the same answer 20. To be specific I want a command to plot sin(x) in a plot0 area embedded area using this.

Hope I am not more confusing. Thanks.

Congratulations. MapleCloud improvements should invite more participants and knowledge sharing among the users.

With more translations, Maple has all the power and potential cover a wider world network.

Cheers.

Ramakrishnan v

@Gillee 

Thanks.

Ramakrishnan V

@Rouben Rostamian  

Your 3 line command was very useful in my understanding the logic behind solving coupled odes.

1. Each second order differential equation (for solving),  needs just two known quantities: initial values and initial slope

2. Even coupled equations the above requirements are sufficient.

3. For n number of equations to be solved, 2 n quantities are required.  Here 3 equations and hence 6 initial conditions.

Thank you very much.

Ramakrishnan V

@kambiz1199 

plot command is seen in your doc. Thanks.

rhs is visibly F(t) and not the expression. This is one case where maple (behaves like a human) has rightly nterpretted the middle term as rhs and gives the required plot (y vs t and not F vs t).

Thanks.

Ramakrishnan V.

@kambiz1199 

use of for loop may help.

I am not getting the answer shown in your message. hence i assumed a function (in place of yr rhs of solution
 

y := proc (x) options operator, arrow; x^2 end proc

proc (x) options operator, arrow; x^2 end proc

(1)

y(1)

1

(2)

y(2)

4

(3)

for x from 0 to 8 do y(x) end do

64

(4)

plot(y(x1), x1 = 0 .. 8)

 

``


 

Download UseOfForLoop.mwUseOfForLoop.mw

)and did my code. Enclosed doc for reference. Hope it is useful.

Ramakrishnan V

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