[sage-support] Re: How solve simple log equation?
@Chris : sometimes we will get the career we have not expected at all ... -) Just watched last sunday one video interview with Simon (famous math, philantropic, NY city market and so on .. millionnaire businessman) as guest : he had never predicted that his math work at young age will be included into the main core of modern ... physics theory. The question "is it normal for simple non linear equations to be unsolvable systematically ?" is debatable. Since long time. Try to think about the length of arc of an ellipse. What is simple ? Sometimes we forget, the "closed form" of solution of equation doesn't exist, at least with current known formulas. Before my post reply : I went to URL about maxima (which is running under the scene of sagemath) ... and of course, my POV,.., not solving everything...especially some quadratic equations or in the p-adic fields (where is a basic Hensel lemma algorithm?) Dominique. -- You received this message because you are subscribed to the Google Groups "sage-support" group. To unsubscribe from this group and stop receiving emails from it, send an email to sage-support+unsubscr...@googlegroups.com. To post to this group, send email to sage-support@googlegroups.com. Visit this group at https://groups.google.com/group/sage-support. For more options, visit https://groups.google.com/d/optout.
[sage-support] Re: How solve simple log equation?
Is it always a coin toss whether a computer algebra system can solve a log equation? Should I not expect to make a career out of using Sage to solve nonlinear equations? cs On Sunday, July 16, 2017 at 3:41:42 PM UTC-5, Emmanuel Charpentier wrote: > > Wups... My bad : I wasn't really awake, it seems... > > Anyway, as suggested by Dominique, you can do : > > sage: E=log(y) == C + log(x) + log(y-1);E > log(y) == C + log(x) + log(y - 1) > sage: S=E.solve(x)[0].solve(y);S > [y == x*e^C/(x*e^C - 1)] > sage: bool(E.subs(S).expand_log()) > True > > which checks. > > Again, sorry for the noise... > > -- > Emmanuel Charpentier > > > Le dimanche 16 juillet 2017 18:29:46 UTC+2, Chris Seberino a écrit : >> >> Emmanuel >> >> Thank you for your reply but you solved a DIFFERENT equation. Notice >> mine has an x variable in it. >> I can get your's to work but not mine. >> >> cs >> > -- You received this message because you are subscribed to the Google Groups "sage-support" group. To unsubscribe from this group and stop receiving emails from it, send an email to sage-support+unsubscr...@googlegroups.com. To post to this group, send email to sage-support@googlegroups.com. Visit this group at https://groups.google.com/group/sage-support. For more options, visit https://groups.google.com/d/optout.
[sage-support] Re: How solve simple log equation?
Wups... My bad : I wasn't really awake, it seems... Anyway, as suggested by Dominique, you can do : sage: E=log(y) == C + log(x) + log(y-1);E log(y) == C + log(x) + log(y - 1) sage: S=E.solve(x)[0].solve(y);S [y == x*e^C/(x*e^C - 1)] sage: bool(E.subs(S).expand_log()) True which checks. Again, sorry for the noise... -- Emmanuel Charpentier Le dimanche 16 juillet 2017 18:29:46 UTC+2, Chris Seberino a écrit : > > Emmanuel > > Thank you for your reply but you solved a DIFFERENT equation. Notice mine > has an x variable in it. > I can get your's to work but not mine. > > cs > -- You received this message because you are subscribed to the Google Groups "sage-support" group. To unsubscribe from this group and stop receiving emails from it, send an email to sage-support+unsubscr...@googlegroups.com. To post to this group, send email to sage-support@googlegroups.com. Visit this group at https://groups.google.com/group/sage-support. For more options, visit https://groups.google.com/d/optout.
Re: [sage-support] Re: How solve simple log equation?
Dominique
THANK YOU! Without or without declaring x your way works
This...
var("y C")
solve( log(y) == C + log(x) + log(y-1),x)
solve( x == y/(y*e^C - e^C), y)
Gives...
[y == x*e^C/(x*e^C - 1)]
What is amazing is that simply having y appear in 2 places makes it
unsolvable directly without solving for x first.
What if my equation did not have the option of solving for x first? Is it
normal for "simple" nonlinear equations
to be unsolvable systematically? I wouldn't be surprised if it wasn't
Sage's fault and there simply isn't
a foolproof systematic way to solve log equations but not sure.
cs
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[sage-support] Re: How solve simple log equation?
Why not adding "x" ? (and of cause declaring x in the same way than y and C)
Because
var("x y C")
solve( log(y) == C + log(x) + log(y-1),x,y)
returns
([x == y/(y*e^C - e^C)], [1])
Dominique
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[sage-support] Re: How solve simple log equation?
Emmanuel Thank you for your reply but you solved a DIFFERENT equation. Notice mine has an x variable in it. I can get your's to work but not mine. cs -- You received this message because you are subscribed to the Google Groups "sage-support" group. To unsubscribe from this group and stop receiving emails from it, send an email to sage-support+unsubscr...@googlegroups.com. To post to this group, send email to sage-support@googlegroups.com. Visit this group at https://groups.google.com/group/sage-support. For more options, visit https://groups.google.com/d/optout.
[sage-support] Re: How solve simple log equation?
Works for me :
sage: reset()
sage: var("y,C")
(y, C)
sage: E=log(y)==C+log(y)+log(1-y);E
log(y) == C + log(y) + log(-y + 1)
sage: S=solve(E,y);S
[y == (e^C - 1)*e^(-C)]
Let's check this unique solution :
sage: y0=S[0].rhs()
sage: E.subs(y==y0)
log((e^C - 1)*e^(-C)) == C + log((e^C - 1)*e^(-C)) + log(-(e^C - 1)*e^(-C)
+ 1)
Not nice...
sage: E.subs(y==y0).expand().simplify()
log(-e^(-C) + 1) == log(-e^(-C) + 1)
Nicer. And indeed :
sage: bool(E.subs(y==y0).expand().simplify())
True
Whereas :
sage: bool(E.subs(y==y0))
False
Sometime, sage needs a little help : systematically simplifying
intermediate results leads to dead ends, and that's why sage doesn't do
that.
HTH,
--
Emmanuel Charpentier
Le dimanche 16 juillet 2017 00:44:03 UTC+2, Chris Seberino a écrit :
>
> This does not solve...
>
> var("y C")
> solve( log(y) == C + log(x) + log(y-1),y)
>
> It returns
>
> [log(y) == C + log(x) + log(y - 1)]
>
> Any ideas?
>
> Thanks!
>
> Chris
>
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