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Science Forum Index » Math - Symbolic Forum » [Maxima] Newbie question
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| Olivier |
 Posted: Tue Apr 08, 2008 3:44 am |
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Dear all,
I've decided to switch to maxima.
So I will come here with obvious questions --
A first one:
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(%i23) facts();
(%o23) [eps > 0, 1 >= eps, 1 > eps, t > - 1]
(%i24) integrate(1/u,u,1,a);
Is a - 1 positive, negative, or zero?
NEG;
Is a - 1 positive, negative, or zero?
NEG;
Is a - 1 positive, negative, or zero?
NEG;
Is a - 1 positive, negative, or zero?
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I simply want to get Log(a) ...
My aim today:
integrate(1/t*integrate(1/u*(2*u/t-u^2/t^2,u,eps,t),t,eps,1)
with eps in ]0,1[.
Help appreciated :-p
Best,
Amities,
Olivier |
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| Olivier |
| Posted: Tue Apr 08, 2008 3:49 am |
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Olivier a écrit :
[...]
Quote: Is a - 1 positive, negative, or zero?
negative; !!!!!
:-(
A.O. |
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| Peter Pein |
| Posted: Tue Apr 08, 2008 8:08 am |
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Olivier schrieb:
....
Quote: My aim today:
integrate(1/t*integrate(1/u*(2*u/t-u^2/t^2,u,eps,t),t,eps,1)
with eps in ]0,1[.
Help appreciated :-p
Best,
Amities,
Olivier
/*wxMaxima 0.7.4 http://wxmaxima.sourceforge.netMaxima 5.14.0
http://maxima.sourceforge.netUsing Lisp GNU Common Lisp (GCL) GCL 2.6.7 (aka
GCL)Distributed under the GNU Public License. See the file COPYING.Dedicated
to the memory of William Schelter.The function bug_report() provides bug
reporting information.
(%i1) assume(0<a,a<1);
(%o1) [a>0,a<1]
(%i2) integrate(1/u,u,1,a);
(%o2) log(a)
(%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;
(%o4) -(3*log(eps))/2+(8*eps-eps^2)/4-7/4
this is not too difficult ;-)
I guessed where the missing closing paranthesis should be.
Peter |
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| Olivier |
| Posted: Tue Apr 08, 2008 11:32 am |
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Peter Pein a écrit :
[...]
Quote: (%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);
Quote: (%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?
Quote: I guessed where the missing closing parenthesis should be.
That's right 
A.O. |
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