[sage-support] Re: Why is f.inverse() so slow?
> > Maybe try to alter that via: > f = prod((1 - x^n + O(x^size))^a(n) for n in (1..size)) > Yes, that is the solution. Thank you so much! -- 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 view this discussion on the web visit https://groups.google.com/d/msgid/sage-support/d4b1447f-4d4e-49e7-9a06-30e9606b3965n%40googlegroups.com.
[sage-support] Re: Why is f.inverse() so slow?
Are you sure it's the inverse function which is slow? Just looking at the command f = prod((1 - x^n)^a(n) for n in (1..size)) + O(x^size) I would guess that first the gigantic product is computed, and only then would the resulting polynomial be truncated by the O(x^size) part. Maybe try to alter that via: f = prod((1 - x^n + O(x^size))^a(n) for n in (1..size)) On Tuesday, November 10, 2020 at 4:07:12 PM UTC-5 Peter Luschny wrote: > Hi all, > > I wanted to implement the Euler Transformation. > https://oeis.org/wiki/Euler_transform > My first attempt was: > > def EulerTransform(size, a): > R. = ZZ[[]] > f = prod((1 - x^n)^a(n) for n in (1..size)) + O(x^size) > return f.inverse().list() > > print(EulerTransform(6, lambda n: factorial(n))) > > Now try this with size = 30. I have no idea how long this > takes because after 3 years I turned the computer off. > > Am I doing something fundamentally wrong, or is f.inverse() > hopelessly slow? > > You can of course implement it differently, but that's not my > question. For those who are interested in the transformation > here the fast variant: > > def EulerTrans(a): > @cached_function > def b(n): > if n == 0: return 1 > s = sum(sum(d*a(d) for d in divisors(j))*b(n-j) for j in (1..n)) > return s//n > return b > > b = EulerTrans(lambda n: factorial(n)) > print([b(n) for n in range(30)]) > > Thanks > > > -- 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 view this discussion on the web visit https://groups.google.com/d/msgid/sage-support/eb26f117-fe6c-4cbd-94c4-5b63f374500en%40googlegroups.com.
[sage-support] Re: why is this so slow?
On Thu, Oct 23, 2008 at 7:56 AM, John H Palmieri [EMAIL PROTECTED] wrote:
The setup:
sage: var('x y')
sage: F = sin(x^2 + y^2) * cos(y) * exp(-0.5*(x^2+y^2))
sage: G=F.derivative(x,x); G
-3.00*x^2*e^(-(0.500*(y^2 +
x^2)))*cos(y)*sin(y^2 + x^2) -
1.00*e^(-(0.500*(y^2 + x^2)))*cos(y)*sin(y^2 +
x^2) - 4.00*x^2*e^(-(0.500*(y^2 +
x^2)))*cos(y)*cos(y^2 + x^2) + 2*e^(-(0.500*(y^2 +
x^2)))*cos(y)*cos(y^2 + x^2)
Then the following takes a long time, between 5 and 10 seconds on my
2.4 GHz Intel iMac.
sage: G(1.2, 1.2)
0.193703001636676
(I tried using 'timeit(G(1.2, 1.2))' but got: AttributeError:
'SymbolicArithmetic' object has no attribute 'find'.)
Why is it so slow, and are there any tricks I can use to speed it up?
Because (the way Sage uses!?) Maxima sucks.
For two months now several of us have been working on completely
replacing Sage's use of Maxima for symbolic manipulation with reliance
on a C++ library called Ginac. This will make its first appearance as
a non-default option in Sage-3.2.
William
I'd like to use these calculations in a class (graphing the degree 2
Taylor approximation to a function in two variables), but if it takes
minutes to compute all of the relevant first and second derivatives,
it's annoying. (If a student says, Let's see what happens if we
compute the Taylor polynomial at a different point, I'd like to be
able to oblige them.)
John
--
William Stein
Associate Professor of Mathematics
University of Washington
http://wstein.org
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[sage-support] Re: why is this so slow?
On Oct 23, 8:05 am, William Stein [EMAIL PROTECTED] wrote:
On Thu, Oct 23, 2008 at 7:56 AM, John H Palmieri [EMAIL PROTECTED] wrote:
The setup:
sage: var('x y')
sage: F = sin(x^2 + y^2) * cos(y) * exp(-0.5*(x^2+y^2))
sage: G=F.derivative(x,x); G
-3.00*x^2*e^(-(0.500*(y^2 +
x^2)))*cos(y)*sin(y^2 + x^2) -
1.00*e^(-(0.500*(y^2 + x^2)))*cos(y)*sin(y^2 +
x^2) - 4.00*x^2*e^(-(0.500*(y^2 +
x^2)))*cos(y)*cos(y^2 + x^2) + 2*e^(-(0.500*(y^2 +
x^2)))*cos(y)*cos(y^2 + x^2)
Then the following takes a long time, between 5 and 10 seconds on my
2.4 GHz Intel iMac.
sage: G(1.2, 1.2)
0.193703001636676
(I tried using 'timeit(G(1.2, 1.2))' but got: AttributeError:
'SymbolicArithmetic' object has no attribute 'find'.)
Why is it so slow, and are there any tricks I can use to speed it up?
Because (the way Sage uses!?) Maxima sucks.
For two months now several of us have been working on completely
replacing Sage's use of Maxima for symbolic manipulation with reliance
on a C++ library called Ginac. This will make its first appearance as
a non-default option in Sage-3.2.
William
My impression is that Ginac is in 3.2-alpha0, which I just installed.
Can I use it to do these computations? If so, how?
John
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[sage-support] Re: why is this so slow?
On Thu, Oct 23, 2008 at 7:56 AM, John H Palmieri [EMAIL PROTECTED] wrote:
The setup:
sage: var('x y')
sage: F = sin(x^2 + y^2) * cos(y) * exp(-0.5*(x^2+y^2))
sage: G=F.derivative(x,x); G
-3.00*x^2*e^(-(0.500*(y^2 +
x^2)))*cos(y)*sin(y^2 + x^2) -
1.00*e^(-(0.500*(y^2 + x^2)))*cos(y)*sin(y^2 +
x^2) - 4.00*x^2*e^(-(0.500*(y^2 +
x^2)))*cos(y)*cos(y^2 + x^2) + 2*e^(-(0.500*(y^2 +
x^2)))*cos(y)*cos(y^2 + x^2)
Then the following takes a long time, between 5 and 10 seconds on my
2.4 GHz Intel iMac.
sage: G(1.2, 1.2)
0.193703001636676
(I tried using 'timeit(G(1.2, 1.2))' but got: AttributeError:
'SymbolicArithmetic' object has no attribute 'find'.)
Quick note: You should have done
timeit(G(1.2,1.2))
since timeit takes a string as input.
My impression is that Ginac is in 3.2-alpha0, which I just installed.
Can I use it to do these computations? If so, how?
Here's me doing this in sage-3.2.alpha0. Note that not all the same
commands and conventions as with Sage's usual symbolic variables
are yet supported. However, the good thing is that the final substitution
takes 1.81 ms on sage.math instead of 10 seconds.
[EMAIL PROTECTED]:~/build/sage-3.2.alpha0$ ./sage
--
| SAGE Version 3.2.alpha0, Release Date: 2008-10-20 |
| Type notebook() for the GUI, and license() for information.|
--
sage: var('x,y',ns=1)#note the ns=1, which will go away in the future...
(x, y)
sage: F = sin(x^2 + y^2) * cos(y) * exp(-0.5*(x^2+y^2))
sage: G=F.diff(x,1).diff(x,1); G
-exp(-(0.500)*x^2 - (0.500)*y^2)*sin(x^2 +
y^2)*cos(y) + 2*exp(-(0.500)*x^2 -
(0.500)*y^2)*cos(x^2 + y^2)*cos(y) -
(3.00)*exp(-(0.500)*x^2 -
(0.500)*y^2)*sin(x^2 + y^2)*cos(y)*x^2 -
(4.00)*exp(-(0.500)*x^2 -
(0.500)*y^2)*cos(x^2 + y^2)*cos(y)*x^2
sage: G.subs(x=1.2).subs(y=1.2)
0.193703001636676
sage: timeit('G.subs(x=1.2).subs(y=1.2)')
125 loops, best of 3: 1.81 ms per loop
William
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[sage-support] Re: why is this so slow?
On Oct 23, 8:56 am, William Stein [EMAIL PROTECTED] wrote:
Here's me doing this in sage-3.2.alpha0. Note that not all the same
commands and conventions as with Sage's usual symbolic variables
are yet supported. However, the good thing is that the final substitution
takes 1.81 ms on sage.math instead of 10 seconds.
[EMAIL PROTECTED]:~/build/sage-3.2.alpha0$ ./sage
--
| SAGE Version 3.2.alpha0, Release Date: 2008-10-20 |
| Type notebook() for the GUI, and license() for information. |
--
sage: var('x,y',ns=1) #note the ns=1, which will go away in the future...
(x, y)
sage: F = sin(x^2 + y^2) * cos(y) * exp(-0.5*(x^2+y^2))
sage: G=F.diff(x,1).diff(x,1); G
or G=F.diff(x,2)
-exp(-(0.500)*x^2 - (0.500)*y^2)*sin(x^2 +
y^2)*cos(y) + 2*exp(-(0.500)*x^2 -
(0.500)*y^2)*cos(x^2 + y^2)*cos(y) -
(3.00)*exp(-(0.500)*x^2 -
(0.500)*y^2)*sin(x^2 + y^2)*cos(y)*x^2 -
(4.00)*exp(-(0.500)*x^2 -
(0.500)*y^2)*cos(x^2 + y^2)*cos(y)*x^2
sage: G.subs(x=1.2).subs(y=1.2)
0.193703001636676
sage: timeit('G.subs(x=1.2).subs(y=1.2)')
125 loops, best of 3: 1.81 ms per loop
William
Great! That's much better.
John
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[sage-support] Re: why is this so slow?
Okay, using what William said, I produced this an 'interact' example,
showing the linear and quadratic Taylor approximations to a certain
function in two variables. Right now it's using a combo of the old
and new symbolic stuff, because that seems to be the best way to get
things done...
var('x y')
F = sin(x^2 + y^2) * cos(y) * exp(-0.5*(x^2+y^2))
plotF = plot3d(F, (x, 0.4, 2), (y, 0.4, 2), adaptive=True,
color='blue')
var('xx yy', ns=1)
G = sin(xx^2 + yy^2) * cos(yy) * exp(-0.5*(xx^2+yy^2))
@interact
def _(x0=(0.8,1.6), y0=(0.8, 1.6), show_plane=(Show linear
approximation, False), show_quad=(Show quadratic approximation,
False)):
F0 = float(G.subs(xx=x0).subs(yy=y0))
F1 = float(G.diff(xx).subs(xx=x0).subs(yy=y0))
F2 = float(G.diff(yy).subs(xx=x0).subs(yy=y0))
F11 = float(G.diff(xx, 2).subs(xx=x0).subs(yy=y0))
F12 = float(G.diff(xx).diff(yy).subs(xx=x0).subs(yy=y0))
F22 = float(G.diff(yy, 2).subs(xx=x0).subs(yy=y0))
P = (x0, y0, F0)
dot = point3d(P, size=15, color='red')
plane = F0 + F1 * (x-x0) + F2 * (y-y0)
quad = plane + 1/2 * (F11 * (x-x0)^2 + 2 * F12 * (x-x0) * (y-y0) +
F22 * (y-y0)^2)
plane_string = 'z = %.2f + %.2f(x-x_0)+%.2f(y-y_0)'%(F0, F1, F2)
quad_string = plane_string + '+ \\frac{1}{2} \\left((%.2f) (x-
x_0)^2 + 2(%.2f) (x-x_0)(y-y_0) + (%.2f)(y-y_0)^2\\right)'%(F11, F12,
F22)
plot1 = plot3d(plane, (x, 0.4, 2), (y, 0.4, 2), color='green')
plot2 = plot3d(quad, (x, max(0.4, x0-0.5), min(2.0, x0+0.5)),
(y, max(0.4, y0-0.5), min(2.0, y0+0.5)), color='purple')
plot = dot + plotF
if show_plane: plot += plot1
if show_quad: plot += plot2
html('$F(x,y) = e^{-(x^2+y^2)/2} \\cos(y) \\sin(x^2+y^2)$')
if show_plane: html('tangent plane: $%s$'%plane_string)
if show_quad: html('quad approx: $%s$'%quad_string)
show(plot)
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