Tamas Papp <tkpapp at gmail.com> writes:
> Hi,
>
> If I do
>
> texput(%lambda,"\\lambda");
> assume(%lambda > 0);
> integrate(1/(1+A*exp(%lambda*(p-s))),p,x,inf);
>
> I get an expression where x-s always occurs together, so it would be
> convenient to replace them by a single parameter y=x-s. I tried
>
> ratsubst(y,x-s,%);
>
> but the substitution is not performed inside the logarithm. Is there
> a way I could make Maxima simplify this? I guess I need to cancel the
> exp(something)*exp(-something) terms first inside the log, but I don't
> know how.
There's a slight work-around that I generally use in this sort of
situation. Instead substitute y+s for x:
(%i1) assume(? > 0);
(%o1) [? > 0]
(%i2) assume(A > 0);
(%o2) [A > 0]
(%i3) integrate(1/(1+A*exp(?*(p-s))),p,x,inf);
Is s ? an integer?
no;
- s ? x ? s ?
log(%e (A %e + %e )) log(A)
(%o3) ------------------------------ - ------ - x + s
? ?
(%i4) subst(y+s, x, %);
- s ? (y + s) ? s ?
log(%e (A %e + %e )) log(A)
(%o4) ------------------------------------ - ------ - y
? ?
(%i5) factor(%);
y ?
log(A %e + 1) - y ? - log(A)
(%o5) -------------------------------
?
Tada!
I'm not sure how to convince ratsubst to deal with the exponentials
correctly: my attempts with map and scanmap failed miserably. Anyone got
any suggestions?
Rupert
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