vie_charlie

63 Reputation

10 Badges

9 years, 98 days

MaplePrimes Activity


These are answers submitted by vie_charlie

Sergey/Acer,

Yes .. i agree with Acer that my codes are not the most efficient.  But the idea is to get it working first.  In that respect, I've tried both methods proposed above.  My findings are that using the DirectSearch package routinely cause the Kernel to crash whereas implementing acer's method was able to provide solutions, albeit long processing time.

Cheers!

How much of the 2*Pi*h*fw*(int(v^3/(exp(hv/kTs)-1), v = 0 .. infinity))/c^2 can we harnest?

http://faculty.uca.edu/weijiul/MATH3331_DiffEq/Arrigo_Maple_Lect1.pdf

good luck!

How much of the 2*Pi*h*fw*(int(v^3/(exp(hv/kTs)-1), v = 0 .. infinity))/c^2 can we harnest?

Anyway .. after reading your question again I kinda figured out what you're asking for.

anyway, here's something i just found.

http://www.math.utah.edu/~zwick/Classes/Summer2008_REU/polydiv.pdf

hope it helps

How much of the 2*Pi*h*fw*(int(v^3/(exp(hv/kTs)-1), v = 0 .. infinity))/c^2 can we harnest?

I've found a way to use single quotes around the expression to pass values of parameters from the specified range into the expressions to be plotted before the expression is evaluated.

REf: http://kb.iu.edu/data/affi.html

I'm trying now with my PEff function.

 

How much of the 2*Pi*h*fw*(int(v^3/(exp(hv/kTs)-1), v = 0 .. infinity))/c^2 can we harnest?

You can define a function say:

 

MC := proc(n)::[some kind of output param] 

....

end proc:

animate( plot, MC(n) .....):

 

hope it helps

How much of the 2*Pi*h*fw*(int(v^3/(exp(hv/kTs)-1), v = 0 .. infinity))/c^2 can we harnest?

Good luck

 

How much of the 2*Pi*h*fw*(int(v^3/(exp(hv/kTs)-1), v = 0 .. infinity))/c^2 can we harnest?

First if you're so new, you really ought to do some reading.  tutorial under help will be useful.

> a := [1, x, 4, y];

                          [1, x, 4, y]
> f := a[1]^4+2*a[2]^3-3*a[3]^2+10*a[4];
                                3       
                       -47 + 2 x  + 10 y
> contourplot3d(f, x = 0 .. 4, y = 0 .. 10, filledregions = true, coloring = [yellow, red]);

 

experiment and have fun.  you may try to do the following

1) define another function g:=

> h:= f + g;

Maple is all about symbolic MATH .--> suppose to be intuitive and user-friendly. 

How much of the 2*Pi*h*fw*(int(v^3/(exp(hv/kTs)-1), v = 0 .. infinity))/c^2 can we harnest?

you can try Ctrl + K and Ctrl +J to add maple input lines.

?package will give you some links about the packages available.

cheers

How much of the 2*Pi*h*fw*(int(v^3/(exp(hv/kTs)-1), v = 0 .. infinity))/c^2 can we harnest?

HI there,

I learned it from school about 15 yrs ago.  For self-learning, you can try the tutorial from the help menu or get a book from the library.  THere is also plenty of resources on the internet after so many years in the market.

Have fun!

 

How much of the 2*Pi*h*fw*(int(v^3/(exp(hv/kTs)-1), v = 0 .. infinity))/c^2 can we harnest?

http://www.mapleprimes.com/files/16982_MJC efficiency calculation.mw

Alright folks.  thanks so far for all the input.  very educational!  Not quite there yet .. hope to get a little more help.

My next task was to plot the real formulas as show in my worksheet above. 

1) My first problem is that I'm trying to plot the function PEffv just like a the f in the previous communication.  On top of that I've had to do a substitution of variables so i can plot the PEffv with the parameter of interest, Eg.  as you can see in Eqn (5).  ran into problem.  So I took a look at the Isc function and try to plot it,  that came out ok as seen in the last plot as Iscv function of Eg.  So what do i have to do to get PEffv to crank out some values?

2) my second problem is that the function Iscv is giving me values that are about 5 order too large.  I've checked the constants multiple times. they should be correct.  Although there are no syntax problem, but i wonder if there are some systematic math problem with my substitution in my evalf function.

MANY THANKS folks! God bless!

Just for the sake of argument:

I've ran this experiment:
> restart;
> E := x^2/(exp(x)-1);
                                      2   
                                     x    
                                 ----------
                                 exp(x) - 1
> f := proc (y) options operator, arrow; evalf(Int(E, x = y .. infinity)) end proc; g := proc (y) options operator, arrow; evalf(Int(E, x = y .. infinity, method = _d01amc)) end proc;
> time(plot(f, 2 .. 10));
                                   14.851
> time(plot(g, 2 .. 10));
                                    0.015
 

I wonder what exactly is the method=_d01amc doing to allow computing so much faster?

Cheers!

THank you!

Page 1 of 1