On Friday, June 27, 2014 4:03:27 AM UTC-4, Camille Chambon wrote:
Hello Chris,
Thanks a lot for your answer! It's very interesting. By now I just need
the bisection method: nsolve(eq,c2,(1,1000),solver='bisect'). But I will
keep in mind the continuation method for the future.
Cheers,
Hello Chris,
Thanks a lot for your answer! It's very interesting. By now I just need the
bisection method: nsolve(eq,c2,(1,1000),solver='bisect'). But I will keep
in mind the continuation method for the future.
Cheers,
Camille
Le jeudi 26 juin 2014 18:36:58 UTC+2, Chris Smith a écrit :
Above
Above 0, the function is monotonic but is ill-behaved in that it is very
flat; this can cause troubles for different solver routines. A slow but
robust method is bisection
nsolve(eq,c2,(1,1000),solver='bisect')
mpf('22.964256014441664')
Using that, you don't need a very precise range for the
I want to find the root for various given coefficients. My parameters are
fixed. Isn't there any other solution than trying different starting points
in an interval and merge the results? I thought that if the function was
bijective and monotonic, I wouldn't need to set a starting point,
Hello Vinzent,
It's exactly what I was looking for. Thanks!
How do you choose the starting point? The coefficients (0.66, 0.34 and 0.7)
may vary. Is there a way to choose the right starting point automatically?
Cheers,
Camille
P.S. scipy.optimize.root works too, but with a starting point of
On Monday, June 23, 2014 3:55:51 AM UTC-4, Camille Chambon wrote:
Hello Vinzent,
It's exactly what I was looking for. Thanks!
How do you choose the starting point? The coefficients (0.66, 0.34 and
0.7) may vary. Is there a way to choose the right starting point
automatically?
I'm not
On Tuesday, June 17, 2014 4:01:05 AM UTC-4, Camille Chambon wrote:
Hello Christophe,
Thanks for your answer.
OK. So if my equation is not solvable symbolically, I can't use SymPy?
Then I will use a numeric solver, like SciPy.
Have a look at `nsolve`, it solves equations numerically given
Hello.
Does your equation solvable symbolically ? At a first glance, I don't think.
C.
2014-06-16 10:39 GMT+02:00 Camille Chambon camille.cham...@gmail.com:
Hello,
I would like to find
c2
as
0.66 + 0.34 * (1 + 1.0 / c2) - 1.0 / ln(1 + c2) - 0.7 == 0
Thus I used the function
solve
Hello Christophe,
Thanks for your answer.
OK. So if my equation is not solvable symbolically, I can't use SymPy? Then
I will use a numeric solver, like SciPy.
And what does mean
PolynomialError: 1/log(c2 + 1) contains an element of the generators set
when using the function
roots
?
Cheers,
Hello,
I would like to find
c2
as
0.66 + 0.34 * (1 + 1.0 / c2) - 1.0 / ln(1 + c2) - 0.7 == 0
Thus I used the function
solve
like so:
c2 = Symbol('c2')
solve(0.66 + 0.34 * (1 + 1.0 / c2) - 1.0 / ln(1 + c2) - 0.7, c2)
But I got the following error:
NotImplementedError
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