[sage-support] Re: numerical approximation with units of measurement?
Thank you! I have a lot to learn about Sage, I can see. Will study &
experiment with recn.
In most cases, an integer exponent should look like "2" and not
"2.000", don't you think? So I guess I would not n() the exponents.
I have just started using Sage and SMC with some online classes. It's great
to:
- check work with units of measure
- take notes that look like math rather than COBOL (+1 typeset_mode)
- do symbolic and numerical calculations in those same typeset notes
I'm hooked. :-)
On Friday, June 6, 2014 6:53:18 PM UTC-5, Nils Bruin wrote:
>
> On Thursday, June 5, 2014 6:32:42 PM UTC-7, Hal Snyder wrote:
>>
>> IIs there a simple way to take n() of things without getting into the
>> following?
>>
> You could automate the application, but you'll quickly see you need to be
> a bit careful:
>
> #unfortunately, the operators returned for sums and products of multiple
> #arguments are callable, but don't accept multiple arguments, so we need to
> #do a little surgery ourselves (borrow the functionality from elsewhere):
>
> opdict = {
> operator.mul : sage.interfaces.maxima_lib.mul_vararg,
> operator.add : sage.interfaces.maxima_lib.add_vararg,
> }
> def recn(e):
> try:
> return n(e)
> except TypeError:
> pass
> op=e.operator()
> if op:
> if op in opdict:
> op = opdict[op]
> return op(*[recn(c) for c in e.operands()])
> else:
> return e
> --
>
> This now works, a little bit:
>
> sage: recn(area)
> 21.5161409036487*meter^2.00
>
> As you can see, the exponent in meter^2 was also numerified. Perhaps you
> didn't want that?
>
> Nonetheless, a recursive n(..) method seems eminently reasonable and
> desirable to implement.
>
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[sage-support] Re: numerical approximation with units of measurement?
On Thursday, June 5, 2014 6:32:42 PM UTC-7, Hal Snyder wrote:
>
> IIs there a simple way to take n() of things without getting into the
> following?
>
You could automate the application, but you'll quickly see you need to be a
bit careful:
#unfortunately, the operators returned for sums and products of multiple
#arguments are callable, but don't accept multiple arguments, so we need to
#do a little surgery ourselves (borrow the functionality from elsewhere):
opdict = {
operator.mul : sage.interfaces.maxima_lib.mul_vararg,
operator.add : sage.interfaces.maxima_lib.add_vararg,
}
def recn(e):
try:
return n(e)
except TypeError:
pass
op=e.operator()
if op:
if op in opdict:
op = opdict[op]
return op(*[recn(c) for c in e.operands()])
else:
return e
--
This now works, a little bit:
sage: recn(area)
21.5161409036487*meter^2.00
As you can see, the exponent in meter^2 was also numerified. Perhaps you
didn't want that?
Nonetheless, a recursive n(..) method seems eminently reasonable and
desirable to implement.
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[sage-support] Re: Numerical approximation of ceiling functions?
On 04/21/2013 09:50 AM, Kenneth Lin wrote:
Hi Sage,
I'm not sure if it's that I'm not doing this right, but I have this
function that has a ceiling in it. I defined it like so:
|
botrk(h0_prime,a0,s0,c0)=h0_prime /ceil(log(20*(a0 +25)/(h0_prime
+20*(a0 +25)),0.95))*(s0 +0.4)*(1+c0)
|
But it won't do approximations of the ceiling, only returning another
symbolic expression that can't be approximated.
|
sage: botrk(3000, 10, 1, 0.1)
4620.000/ceil(-19.4957257462237*log(7/37))
sage: botrk(3000, 10, 1, 0.1).n()
---
TypeError Traceback (most recent call last)
in ()
> 1 botrk(Integer(3000), Integer(10), Integer(1), RealNumber('0.1')).n()
/usr/lib/sagemath/local/lib/python2.7/site-packages/sage/symbolic/expression.so
in sage.symbolic.expression.Expression._numerical_approx
(sage/symbolic/expression.cpp:21011)()
TypeError: cannot evaluate symbolic expression numerically
|
Does anyone know how to approximate ceilings? For my purpose, I could
just plug this in again and get a result, but I was hoping for a better,
cleaner way of doing it.
Don't know why it stumbles on the ceiling. A function works on the other
hand, since it avoids symbolics.
sage: def botrk(h0_prime, a0, s0, c0): return h0_prime / ceil(log(20 *
(a0 + 25) / (h0_prime + 20 * (a0 + 25)), 0.95)) * (s0 + 0.4) * (1 + c0)
:
sage: botrk(3000, 10, 1, 0.1)
140.
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[sage-support] Re: numerical approximation in sage.
Hi, This is an unavoidable consequence of using Maxima's solve commands, I think - with multiple equations, Maxima's solve uses things like algsys, if I'm not mistaken, and those return "real" solutions if they can't find "symbolic" ones. With one equation the (new) behavior is to not do this automatically (the to_poly_solve=True option would allow this), but I don't think there is any way to avoid the possibility of this with more than one equation. The documentation should be clear on this in the newest version of Sage - I hope! - kcrisman -- To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to sage-support-unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-support URL: http://www.sagemath.org
Re: [sage-support] Re: numerical approximation in sage.
My question is about the syntax and why does this syntax give a numerical
approximation.
To my understanding, solns is contracted from two arrays with p,q,x,y being
the keys (because there are two solutions to the equations set). The part "for
s in solns" is putting in s ab array, and the part s[blah].n(30) asks sage
to give s[blah] in 30 bits precision.
Here is what I don't understand:
- Is this syntax equivalent to:
for s in solns
s[q].n(30)
s[p].n(30)
s[x].n(30)
s[y].n(30)
- Why does this method give a numerical solution.
Thanks.
On Fri, Nov 27, 2009 at 10:46 PM, John H Palmieri wrote:
> On Nov 27, 10:03 am, Yotam Avital wrote:
> > Hello.
> >
> > In the tutorials there is an example for numerical approximation:
> >
> > var('x y p q')
> > (x, y, p, q)
> > eq1 = p+q==9
> > eq2 = q*y+p*x==-6
> > eq3 = q*y^2+p*x^2==24
> > solns = solve([eq1,eq2,eq3,p==1],p,q,x,y, solution_dict=True)
> > [[s[p].n(30), s[q].n(30), s[x].n(30), s[y].n(30)] for s in solns]
> > [[1.000, 8.000, -4.8830369, -0.13962039],
> > [1.000, 8.000, 3.5497035, -1.1937129]]
> >
> > As I far as I can understand, solution_dict tells sage that I want the
> > output to be in dictionary form(that is, {x:1, y:8 ...})
> > I also know that the .n(30) tell sage I want the answer to have 30
> > digits accuracy. I can't understand though the logic of the last
> > command. Can any of you explain it to me?
>
> If you're asking about the command
>
> [[s[p].n(30), s[q].n(30), s[x].n(30), s[y].n(30)] for s in solns]
>
> then note first that "solns" is a list, and a construction like [blah
> for s in solns] evaluates "blah" for each entry s in solns. If you
> just print solns at this point, you should get
>
> [{q: 8, x: -4/3*sqrt(10) - 2/3, p: 1, y: 1/6*sqrt(2)*sqrt(5) - 2/3},
> {q: 8, x: 4/3*sqrt(10) - 2/3, p: 1, y: -1/6*sqrt(2)*sqrt(5) - 2/3}]
>
> Each entry s in solns is a dictionary with keys the variables p, q, x,
> y. For the first entry, s[p] is 1, s[q] is 8, etc. So the command
> that I think you were asking about prints s[p], s[q], s[x], and s[y],
> each with 30 bits of precision, for each of the two solutions.
>
> --
> John
>
> --
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[sage-support] Re: numerical approximation in sage.
On Nov 27, 10:03 am, Yotam Avital wrote:
> Hello.
>
> In the tutorials there is an example for numerical approximation:
>
> var('x y p q')
> (x, y, p, q)
> eq1 = p+q==9
> eq2 = q*y+p*x==-6
> eq3 = q*y^2+p*x^2==24
> solns = solve([eq1,eq2,eq3,p==1],p,q,x,y, solution_dict=True)
> [[s[p].n(30), s[q].n(30), s[x].n(30), s[y].n(30)] for s in solns]
> [[1.000, 8.000, -4.8830369, -0.13962039],
> [1.000, 8.000, 3.5497035, -1.1937129]]
>
> As I far as I can understand, solution_dict tells sage that I want the
> output to be in dictionary form(that is, {x:1, y:8 ...})
> I also know that the .n(30) tell sage I want the answer to have 30
> digits accuracy. I can't understand though the logic of the last
> command. Can any of you explain it to me?
If you're asking about the command
[[s[p].n(30), s[q].n(30), s[x].n(30), s[y].n(30)] for s in solns]
then note first that "solns" is a list, and a construction like [blah
for s in solns] evaluates "blah" for each entry s in solns. If you
just print solns at this point, you should get
[{q: 8, x: -4/3*sqrt(10) - 2/3, p: 1, y: 1/6*sqrt(2)*sqrt(5) - 2/3},
{q: 8, x: 4/3*sqrt(10) - 2/3, p: 1, y: -1/6*sqrt(2)*sqrt(5) - 2/3}]
Each entry s in solns is a dictionary with keys the variables p, q, x,
y. For the first entry, s[p] is 1, s[q] is 8, etc. So the command
that I think you were asking about prints s[p], s[q], s[x], and s[y],
each with 30 bits of precision, for each of the two solutions.
--
John
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[sage-support] Re: Numerical approximation
sage: round(x,2) 0.0 To show it really does what you want: sage: round(1.2345,2) 1.23 On May 2, 8:35 pm, Fidel wrote: > Hello, > > I think I tried to post this about an hour ago, but the discussion > didn't show up. So I'm doing it again, sorry in case it is repeated. > > I am working in sage 3.4.1 > > I am trying to define a function to get the LaTeX string of a graph, > so I am trying to convert a number to string. > > I have x=6.1230317691118863e-17, which I got from > > sage: x=graphs.PetersenGraph().get_pos()[0][0] > > I would like a two decimal approximation of x, that is 0.00. > > I have tried > > sage: numerical_approx(x,digits=2) > 6.1e-17 > sage: n(x,digits=2) > 6.1e-17 > > Is there a way to get 0.00 as output? > > Thanks in advance for your attention and help. > > Greetings, > Fidel --~--~-~--~~~---~--~~ To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to sage-support-unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-support URLs: http://www.sagemath.org -~--~~~~--~~--~--~---
