On Friday, November 2, 2018 at 5:44:34 PM UTC-7, Emmanuel Charpentier wrote:
>
> One way to define (something almost the same as) what you want is :
> sage: f=piecewise([((-oo,0),x^3),((0,oo),x^2)],var=x)
> sage: f
> piecewise(x|-->x^3 on (-oo, 0), x|-->x^2 on (0, +oo); x)
> [...] Except for the po
One way to define (something almost the same as) what you want is :
sage: f=piecewise([((-oo,0),x^3),((0,oo),x^2)],var=x)
sage: f
piecewise(x|-->x^3 on (-oo, 0), x|-->x^2 on (0, +oo); x)
An indeed; you can do
sage: plot(f(x),(x,-1,1), figsize=3)
which seems correct. Except for the point 0, for wh
Il giorno mercoledì 31 ottobre 2018 10:11:34 UTC+1, Francesco ha scritto:
>
> Hello; I have installed sage 8.4 and I have problem with the derivatives
> ...
> I have defined a function in sage of this type:
>
> x=var('x')
> def funz(x):
>if x >= 0:
> return x^2
>else:
> retu
I'm really naive on this one: the problem I'm trying to solve is to
write a recurrence for the Legendre Q(n,x) polynomials / Q(n,m,x)
functions. Numeric results can be easily conjugated but a symbolic
expression with conjugated log functions is tedious to use and, as it
seems, impossible to dfiffer
Hi!
Be careful! conjugate(·) is not complex differentiable! The Example you
gave in your link had not conjugate(log(x)) as function but
conjugate(log(conjugate(x)). Which exists.
You can alos show it easily by using the facts, if log(·) is differentialbe
in an neighbourhood of x, it has a seri
Le samedi 2 mars 2013 17:00:15 UTC+1, Emmanuel Charpentier a écrit :
[ An idiocy ... ]
[ Snip... ]
> Since s1 is irst-degree equation in t, this is the only real
> nonnegative maximum (it is easy to show that there is no negative
> maximum).
> Now, try brute force via the to_poly_solve solver
Dear Jose,
Le mercredi 27 février 2013 14:18:45 UTC+1, Jose Guzman a écrit :
>
> I am trying to find the maximun of an exponential expression of the form:
>
> sage: t=var('t')
> sage: g(t) = e**(-t/10)-e^(-t/2)
>
> between 0 and say 50. My idea is to get the maximun to normalize the
> function
A: Yes.
> Q: Are you sure?
>> A: Because it reverses the logical flow of conversation.
>>> Q: Why is top posting annoying in email?
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For mo
Thanks. Will do.
On Apr 3, 12:19 pm, Alec Mihailovs wrote:
> On Apr 3, 2:32 pm, pallab wrote:
>
> > Is there any way to check whether a symbolic expression is a
> > derivative. Like,
>
> > isinstance(diff(f(x),x),"what to put?")
>
> > gives "True"
>
> > and
>
> > isinstance(f(x),"what to put?")
On Apr 3, 2:32 pm, pallab wrote:
> Is there any way to check whether a symbolic expression is a
> derivative. Like,
>
> isinstance(diff(f(x),x),"what to put?")
>
> gives "True"
>
> and
>
> isinstance(f(x),"what to put?")
>
> gives false, assuming f is not a derivative itself.
One can do the follo
I mean...
I do not know. If it is already evaluated to cos(x) then it should be
false (not *true*). current "sympy" gives false, I think I would go
with it.
On Apr 3, 11:51 am, pallab wrote:
>
>
> On Apr 3, 11:36 am, "ma...@mendelu.cz" wrote:
>
> > What kind of behavior do you expect from the
I do not know. If it is already evaluated to cos(x) then it should be
true. current "sympy" gives false, I think I
would go with it.
On Apr 3, 11:36 am, "ma...@mendelu.cz" wrote:
> What kind of behavior do you expect from the following command?
>
> isinstance(diff(sin(x),x),"what to put?")
>
> Ro
What kind of behavior do you expect from the following command?
isinstance(diff(sin(x),x),"what to put?")
Robert
On 3 dub, 20:32, pallab wrote:
> Is there any way to check whether a symbolic expression is a
> derivative. Like,
>
> isinstance(diff(f(x),x),"what to put?")
>
> gives "True"
>
> and
Hi Jason,
On Fri, 18 Sep 2009 13:15:46 -0500
Jason Grout wrote:
>
> On alpha.sagenb.org, I get the following:
>
> sage: t=var('t')
> sage: diff(cot(t),t)
> D[0](cot)(t)
> sage: diff(cos(t)/sin(t),t)
> -cos(t)^2/sin(t)^2 - 1
>
>
> Does Sage not know that cot(t) is cos(t)/sin(t)? Or am I jus
On Thu, 17 Sep 2009 13:29:01 -0400
Dan Aldrich wrote:
> y=|x|
> 1st derivative should be +1 x>0 and -1,X<0
>
> f(x) = abs(x)
> Dx = x.derivative()
This should be
Dx = f.derivative(x)
In your example Dx is 1, and the graph is correct.
> p1 = plot (f(x),(-5,5),color='black')
> p2 = plot (Dx,
Hi all:
Thanks to those who worked on closing ticket 6243 regarding
derivatives as dictionary keys. It appears that there are still some
bugs, though (see below).
Alex
--
| Sage Version 4.1.1, Release Date: 2009-08-14
Hi Burcin:
When using Sage for my work i makes heavy use of dictionaries to
substitute values for derivatives of symbolic functions. Thanks for
your help on fixing this bug for Sage 4.1.1.
Alex
On Jun 8, 7:12 am, Burcin Erocal wrote:
> Hi Alex,
>
> On Sun, 7 Jun 2009 11:31:26 -0700 (PDT)
>
>
Hi Alex,
On Sun, 7 Jun 2009 11:31:26 -0700 (PDT)
Alex Raichev wrote:
>
> Hi all:
>
> Upon upgrading to Sage 4.0, i can no longer make a dictionary with
> derivatives as keys (see below). Can someone please fix this?
> --
>
Thanks for the news, William. I will hold off on this chain rule
business till the new symbolics arrive.
Alex
On Apr 24, 3:43 pm, William Stein wrote:
> On Thu, Apr 23, 2009 at 7:18 PM, Alex Raichev wrote:
>
> > Hmm, implementing the chain rule is trickier than i thought. My
> > straightforw
On Thu, Apr 23, 2009 at 7:18 PM, Alex Raichev wrote:
>
> Hmm, implementing the chain rule is trickier than i thought. My
> straightforward plan of attack was to write a function that
> differentiates a symbolic expression as usual but when it comes to a
> composition f o g, it uses the chain rul
Hmm, implementing the chain rule is trickier than i thought. My
straightforward plan of attack was to write a function that
differentiates a symbolic expression as usual but when it comes to a
composition f o g, it uses the chain rule and returns the appropriate
entry of the matrix (Df o g)Dg. P
Woops, that ain't right: 'write'.
On Apr 23, 1:43 pm, Alex Raichev wrote:
> Never mind. I'll just right a short recursive function. It's easy
> enough.
>
> Alex
>
> On Apr 23, 11:10 am, Alex Raichev wrote:
>
> > Hi all:
>
> > Do any of you know how to get Sage to use the chain rule and expand
Never mind. I'll just right a short recursive function. It's easy
enough.
Alex
On Apr 23, 11:10 am, Alex Raichev wrote:
> Hi all:
>
> Do any of you know how to get Sage to use the chain rule and expand
> the derivative of a composition involving one or two callable symbolic
> functions? Here
Thank you very much !
Loïc
Le samedi 28 février 2009 à 02:37 -0800, Rolandb a écrit :
> This works:
>
> f=arccos((1-x^2)/(1+x^2))
> g=f.diff(x)
> g.simplify_full()
> 2*x/((x^2 + 1)*abs(x))
>
> In general: type g. TAB and you will find all kind of handy functions.
>
> Roland
>
> On 28 feb, 10
This works:
f=arccos((1-x^2)/(1+x^2))
g=f.diff(x)
g.simplify_full()
2*x/((x^2 + 1)*abs(x))
In general: type g. TAB and you will find all kind of handy functions.
Roland
On 28 feb, 10:58, Loïc wrote:
> Hello
>
> Another problem:
> I want the derivative for the function arccos((1-x^2)/(1+x^2))
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