sasomao

602 Reputation

7 Badges

12 years, 225 days
Phd.
Paris, France

MaplePrimes Activity


These are replies submitted by sasomao

Hi Axel

I've found the problem, the change of variable implies that the argument of log should be

ln(f0/x)

and not

ln(x*f0)
 

as I've wrongly deduced by the fact that the change of variable was x = t/f0 --> t= x*f0

So now it work with the log, and I'm trying with the lorentzian 1/(1+x^2) that should be good behaved.

I'll tell you

 

thanks

Salvo

this can't work because a has already a numerical values when it comes, so I can't eval it for a=0 and see if nothing changes...

this can't work because a has already a numerical values when it comes, so I can't eval it for a=0 and see if nothing changes...

Hi Axel

 

it doesn't seem to work properly:

 

S_h:=proc(t)
local f_0,T;
f_0:=215.0:
T:=t/f_0:
return simplify((T^(-4.14) -5*T^(-2)+ 111*(1-T^2+T^4/2)/(1+T^2/2)))
end proc:

evalf(Int(f^2*ln(f)/S_h(f),f=20..2000))

7909703.651


f0:=215:
has_ln :=unapply(ln(f0*x),x);

myInt2_DE(has_ln,2,20,2000,21)

5147164.54256176411203

Two different values...

Salvo

 

Hi Axel

 

it doesn't seem to work properly:

 

S_h:=proc(t)
local f_0,T;
f_0:=215.0:
T:=t/f_0:
return simplify((T^(-4.14) -5*T^(-2)+ 111*(1-T^2+T^4/2)/(1+T^2/2)))
end proc:

evalf(Int(f^2*ln(f)/S_h(f),f=20..2000))

7909703.651


f0:=215:
has_ln :=unapply(ln(f0*x),x);

myInt2_DE(has_ln,2,20,2000,21)

5147164.54256176411203

Two different values...

Salvo

 

Hi I've a similar issue with Maple13 on a 64bit linux machine, the main difference is that I can't even see the printer dialog windows. If I push on the "print" icon on maple nothing happens..! I've a couple of cups printers (that work normally with gedit emacs etc) Is that a 64 bit related problem? Thanks. S.

I'm sorry about the name, Axel. I've read it wrongly the first time and it remained in my head.

Sorry

Salvo

I'm sorry about the name, Axel. I've read it wrongly the first time and it remained in my head.

Sorry

Salvo

Hi Alex

I want to thank you because your mws works incredibly well and fastly!!!!!!

I think that this resolve my problems for this particular program.

Today I'll modify the code in order to make it able to split the integrals in a sum of integrals each one of the kind you have so amazingly tamed!!!!

Could you just, in some qualitative words, explain to me how your methods works?

Many thanks to all

salvo

Hi Alex

I want to thank you because your mws works incredibly well and fastly!!!!!!

I think that this resolve my problems for this particular program.

Today I'll modify the code in order to make it able to split the integrals in a sum of integrals each one of the kind you have so amazingly tamed!!!!

Could you just, in some qualitative words, explain to me how your methods works?

Many thanks to all

salvo

Hi pagan

 

unfortunately the command you suggested asks a lot of time too.

By the way, I've a doubt about the meaning of epsilon. The help page says

The specification epsilon = <numeric> specifies the relative error tolerance for the computed result. The routines attempt to achieve a final result with a relative error less than this value. By default, the relative error tolerance which the routines attempt to achieve for the final result is
 

but, how can maple calculate the relative error? In my ingenuous mind  rel_err= (TrueValue - ApproxValue)/TrueValue, but maple don't know what the true value is...

 

Hi pagan

 

unfortunately the command you suggested asks a lot of time too.

By the way, I've a doubt about the meaning of epsilon. The help page says

The specification epsilon = <numeric> specifies the relative error tolerance for the computed result. The routines attempt to achieve a final result with a relative error less than this value. By default, the relative error tolerance which the routines attempt to achieve for the final result is
 

but, how can maple calculate the relative error? In my ingenuous mind  rel_err= (TrueValue - ApproxValue)/TrueValue, but maple don't know what the true value is...

 

Hi Alex

 

this is exaclty my case: the results I plotted were indeed the (1,1) elements of (the same) matrix (up_fisher, in the code I gave earlier in this thread)  divided by a normalization factor!!

S.

Hi Alex

 

this is exaclty my case: the results I plotted were indeed the (1,1) elements of (the same) matrix (up_fisher, in the code I gave earlier in this thread)  divided by a normalization factor!!

S.

Hi Doug

 

thank for your post.

But how'd you do, for exemple, if you were in one or more of these situations (witouth using for cycles)?

1)  You want to check, before the assignment, wether 3^(i-1) (in your ex) is positive or negative, or smaller than 1, etc, and print a string in a file that takes memory of this checks

2) Suppose that your assigment function is composed of many addends, and you want, before the assigment is done for a particular set of indexes, to check for the positiveness of all the addends, and then add all the positives togher, all the negatives together, and finally this two quantities togher

3) You want to assign only one part of the entries, eg only these for whom the second index is smaller that the first.

Maybe I'm wrong, but I've always thought that the direct assigment is possible only in very simple cases...
Salvo

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