Double integral of a function (Beginners Question)

Doktor Bernd <doktorbernd at hotmail.com> writes:
<snip>
> rho_0(k,L,Cn2) := block([k:k,L:L,Cn2:Cn2],(1.46*k^2*integrate(Cn2(l),l,0,L))^(-3/5)) $
<snip more>
> My questions are:
>
> Do I have to use the blocks?
> I read at some places that I have to and at others that all passed variables are treated as local ones.
>
> Why do I not get a result from the last statement?
> I guess I am using ' in a wrong way.

Well, I'll answer the easy one at least :-) What block does is set up
local bindings for a variable. So

foo(x) := block([a], a: x);

doesn't do anything (it sets up a local binding for a, sets a to x and
then forgets about a again). On the other hand,

bar(x) := a: x;

affects the bindings for a

Transcript:

(%i1) foo(x) := block([a], a: x);
(%o1)                     foo(x) := block([a], a : x)
(%i2) foo(3);
(%o2)                                  3
(%i3) a;
(%o3)                                  a
(%i4) bar(x) := a: x;
(%o4)                           bar(x) := a : x
(%i5) bar(3);
(%o5)                                  3
(%i6) a;
(%o6)                                  3

So your definition for rho_0 could be

rho_0(k,L,Cn2) := (1.46*k^2*integrate(Cn2(l),l,0,L))^(-3/5)$

instead of

rho_0(k,L,Cn2) := block([k:k,L:L,Cn2:Cn2],(1.46*k^2*integrate(Cn2(l),l,0,L))^(-3/5)) $

and similarly for all of the other functions, since none of them
actually bind anything.

Transcript:

(%i12) rho_0(k,L,Cn2) := (1.46*k^2*integrate(Cn2(l),l,0,L))^(-3/5)$

(%i13) rho_0(1,2,lambda([x], 1));
(%o13)                         .5257391142650575
(%i14) rho_0(k,L,Cn2) := block([k:k,L:L,Cn2:Cn2],(1.46*k^2*integrate(Cn2(l),l,0,L))^(-3/5)) $

(%i15) rho_0(1,2,lambda([x], 1));
(%o15)                         .5257391142650575


As for the second question, you probably don't want any of the
integrate calls quoted, but maybe it'll become more obvious once you've
cleaned up the blocks everywhere.


Rupert
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