Re: [Maxima] Newbie question

Peter Pein a écrit :
[...]
(%i3) assume(eps>0,eps<1);
(%o3) [eps>0,eps<1]
(%i4) integrate(1/t*integrate(1/u*(2*u/t-u^2/t^2),u,eps,t),t,eps,1);
Is t-eps positive, negative, or zero?
p;

Yep, but I did not stumble on the manual where this answer was
described :-p I tried: POS; NEG; t>eps; t-eps>0; assume(t>eps);

(%o4) -(3*log(eps))/2+(8*eps-eps^2)/4-7/4

this is not too difficult ;-)

Newbies are newbies :-p

By the way,

(1) do you know a way to avoid the above question?
("Is t-eps positive, negative, or zero?")
(2) I discovered lambda in between and want to
replace the 1/u and 1/t above by f:lambda([t],1/t).
So far, so good. But my next lambda reads :
f:lambda([t],if (t>2*eps) then (t-eps)/t) else eps/t);
Any idea?


I guessed where the missing closing parenthesis should be.

That's right :-)
A.O.
.