(01-30-2015 05:20 PM)Snorre Wrote:  Hello,

why not define it by yourself:
a(f,k):=int(f(t)*cos(k*t),t=-π..π)/π
b(f,k):=int(f(t)*sin(k*t),t=-π..π)/π
or something similiar.
(Beware: this may suffer from a little issue which will be fixed in next software version.)

Greetings

(01-30-2015 06:48 PM)salvomic Wrote:  define the functions it works, but I get first a lot of warnings, like those claiming the use of Taylor or subs() and so on...
ex.: "Warning, evalueting univariate expression(value) like if expression was a function. You should write subst(t^2,t,taylorx21" if I try with t^2 as function

(01-30-2015 07:23 PM)Han Wrote:  This may have to do with using an expression as a function. Are you passing a CAS function as your argument? Or are you passing a CAS expression?

(01-30-2015 07:32 PM)salvomic Wrote:  

(01-30-2015 07:23 PM)Han Wrote:  This may have to do with using an expression as a function. Are you passing a CAS function as your argument? Or are you passing a CAS expression?

I tried this a(t^2,1)

***
then some variables were a mess, until I purge some of them: d, g, v, used in another programs...

(01-30-2015 07:54 PM)Han Wrote:  If you look at how a(f,k) was defined, the f is assumed to be a CAS function (notice the f(t) reference in how a(f,k) is defined). This means you need to pass a CAS function.

f(t):=t^2;
a(f,1);

(01-30-2015 09:06 PM)salvomic Wrote:  

(01-30-2015 07:54 PM)Han Wrote:  If you look at how a(f,k) was defined, the f is assumed to be a CAS function (notice the f(t) reference in how a(f,k) is defined). This means you need to pass a CAS function.

f(t):=t^2;
a(f,1);

in this way it works (I used also variant with Int(...,0, 2π) ), but I tried this:
create a0 (first term), ak, bk variable with expressions
f(t):=|sin(t)|
I got warning about the absolute value...
then I got value for a0 (as a0:=(1/2π)Int(f(t),t,0,2π)
After this value for a0, ak, bk and also f=f(t) are reset and I must rewrite all three functions...
There is a reason? Is is possible to maintain the function

Perhaps is better a little program that give a list with (a0, ak, bk)?

(01-31-2015 12:05 AM)Han Wrote:  I'm not sure I understand what you did; perhaps you could copy/paste the actual commands you used? I am not having any trouble with respect to things "disappearing"

(01-30-2015 06:07 PM)salvomic Wrote:  However, it would be likely to have such a function included in the CAS, if it's possible ;-)

(01-31-2015 09:29 PM)Tim Wessman Wrote:  XCAS has 3 commands for this. When we was going through the list of commands at some point in the past ...

Those commands are fourier_an, fourier_bn, fourier_cn ...

I assume those will do what you are looking for?

(01-31-2015 09:11 PM)Snorre Wrote:  Hello Salvo,

a program is always a good idea ;-)

Just two tips:

...

(01-31-2015 09:53 PM)salvomic Wrote:  Yes, however, I would export a list of three arguments: a0 (however this one is always the same), ak, bk...

Last: doing so:

Code:


a0:=int(fun(t),t,0,2*PI)/(2*PI);
ak:=int(fun(t)*cos(k*t),t,0,2*PI)/PI;
bk:=int(fun(t)*sin(k*t),t,0,2*PI)/PI;

How can we avoid this error?
I set g:=x^2 then input fourcoeff(g(x),1) or fourcoeff(x^2,1) and I get "warning, evaluating invariable expression(value) like if expression was a function. You should write subst(x^2,x,t)"...
Sure I miss something else
I would better have the possibility to input both functions or expressions :-)

(02-01-2015 03:10 AM)Han Wrote:  Again, you need to understand the difference between a function and an expression.

g:=x^2; <-- this creates an expression whose name is g
g(x):=x^2; <-- this creates a function whose name is g

a0:=int(fun(t),t,0,2*PI)/(2*PI); <-- this creates a variable named a0 whose value is the result of integrating a function named 'fun' (which hopefully you've already pre-defined); if 'fun(t)' is not defined, then a0 is an expression (otherwise a0 is presumably a numerical value)

ak:=int(fun(t)*cos(k*t),t,0,2*PI)/PI; <-- this creates a variable named ak whose value is the result of integrating the product of the function 'fun(t)' and 'cos(k*t)'; if 'fun' and 'k' were predefined prior to creating ak, then ak is just a numerical value; otherwise ak is an expression (same as a0)

Neither a0 nor ak are functions when defined this way. However, the CAS will let you get away with using expressions as if they were functions -- it will just always throw warnings/complaints about using expressions as if they were functions. So don't do it if you don't want to see the warnings.

ak(f,k):=int(f(t)*cos(k*t),t,0,2*PI)/(2*PI); <-- this creates a function named a0, whose arguments are a CAS function f and a constant k; that means 'f' must be a function (see the g(x):=x^2 example), and NOT an expression (see the g:=x^2 example)

Once you understand the difference between the two, your warnings will go away.

(02-01-2015 03:22 AM)rprosperi Wrote:  fyi - For your final example, in the explanation 'a0' should be 'ak', that is if I understood it correctly.

(02-01-2015 03:10 AM)Han Wrote:  ...
ak(f,k):=int(f(t)*cos(k*t),t,0,2*PI)/(2*PI); <-- this creates a function named ak, whose arguments are a CAS function f and a constant k; that means 'f' must be a function (see the g(x):=x^2 example), and NOT an expression (see the g:=x^2 example)

Once you understand the difference between the two, your warnings will go away.

(01-31-2015 09:11 PM)Snorre Wrote:  a program is always a good idea ;-)

Just two tips: