ndomes's profile - activity

What's about: http://sagecell.sagemath.org/?q=oynzya ?

Are you looking for something like this ?

You can resize an image. See PIL or PILLOW documentation.

...
img2 = img2.resize(img.size)
img3 = PIL.Image.blend(img,img2,alpha=0.7)

See an example using PIL here.

Please note: The images 'test.png' and 'test2.png' have to match, (same size, aspect_ratio).

What's about a python function?

def my_piecewise(x0,x1):
    if 0 < x0 < x1 < 1:
        return piecewise([((0,x0),0), ([x0,x1],1), ((x1,1),2)], var=x)

my_piecewise(0.2,0.7).plot()

var('x,y,z,K')
#K=4.48e-13   # use later for substitution
xi=0.75
yi=0
zi=0
eq1 = K==y^2*z/x^2
eq2 = xi+yi==x+y
eq3 = 2*xi+yi+2*zi==2*x+y+2*z

#  x , y dependent on z
solns = solve([eq2,eq3],x,y)
print solns

# z dependent on K
solns2 = solve(eq1.subs(solns),z)
print solns2

print solns2[2].subs(K=4.48e-13)   # using real solution
solns3 = solve([eq2,eq3.subs(z == 0.0000397877574911779)],x,y)
for s in solns3[0]: print s.lhs() == s.rhs().n(30)

My suggestion see http://sagecell.sagemath.org/?q=rmiwzc

Put both plots together with suitable translation and axes.

var('term_0 term_1 term_2 term_3')
equ0 = term_0 == term_1 + term_2 + term_3 

# use the operands method of expressions
print equ0.rhs().operands()

# subtract third term (the last in the list of operands)
print equ0.rhs() - equ0.rhs().operands()[-1]
# or build expression from list of operands
print sum(equ0.rhs().operands()[:2])

sage: M = matrix(2,2,var('a b c d'))
sage: M;  M.parent(); M(a=5)

[a b]
[c d]
Full MatrixSpace of 2 by 2 dense matrices over Symbolic Ring
[5 b]
[c d]

We can use substitution M(a=5) (looking similar to a function call). What happens if we try to use M as variable of a function?

sage: M = matrix(2,2,var('a b c d'))
sage: print M.parent(); print
sage: f(M) = M.transpose()

The error message indicates that the type of M has changed.

Full MatrixSpace of 2 by 2 dense matrices over Symbolic Ring

Traceback (click to the left of this block for traceback)
...
AttributeError: 'sage.symbolic.expression.Expression' object has no attribute 'transpose'

M is no longer a matrix over SR but a symbolic expression.

sage: M.parent()

Symbolic Ring

When defining a symbolic function with a variable M, this M becomes (implicitly) a member of SR overwriting the previous definition of M. After redefining M as matrix:

sage: A = M[0,0] + M[1,1]
sage: A; A(a=5,b=3,c=0,d=-2)

a + d
3

var('t1', 't2', 't3', 'a', 'b')
expr1 = ((t1 + t2 + t3 - 1)*a)/b
equ = t1 + t2 + t3 == 1
expr1.subs(solve(equ,t1))

math.sqrt returns a numerical value, sage sqrt returns a symbolic expression.

See an example: http://sagecell.sagemath.org/?q=bgdeqj

Exact symbolic operations are much slower than numerical algorithms. Converting a symbolic expression to a numerical value takes additional time.

One bottleneck comes from working with the symbolic ring. The following code is some hundred times faster (may be still not fast enough ;-) )

import math
def count_2017(bound=100000):
    cpt = 0
    for n in range (1,bound):
        if int(RR(math.sqrt(n)).frac()*10000) == 2017:
            cpt += 1
    return cpt

Update: Using cython makes the code another ten times faster, for example:

fits_2017 = cython_lambda("double x", "bool(int(10000*(x-int(x))) == 2017)")
f = cython_lambda("int n", "sqrt(n)")

def count_2017(bound=100000):
    cpt = 0
    for n in range(1,bound):
        if fits_2017(f(n)):
            cpt += 1
    return cpt

count_2017(10^7)

A special solution using symbolic functions:

var('t')
k(t) = t*(2-sign(t-1)^2)
f(x)=sin(x)/k(t)
f(t=3).integrate(x, 0, 1)

A live example with nested for loops: http://sagecell.sagemath.org/?q=dzgfjy

To get the rotation axis stay constant, add an enveloping circle making this invisible with opacity = 0. Further set aspect_ratio=(1,1,1) as an option of animate.

r = 2
y_axis = [r*vector((0,-1,0)),r*vector((0,1,0))]

var('alpha')
C = parametric_plot((r*cos(alpha),0,r*sin(alpha)),(alpha,0,2*pi),opacity=0)

n=19
t = (2*pi/n).n()
M=matrix([[cos(t),0,sin(t)],[0,1,0],[-sin(t),0,cos(t)]])
p=polytopes.cube()
vl = [v[:3] for v in p.vertices()]
sf = [] #[p]
for i in range(n):
    vl = [M*v for v in vl]
    sf.append(Polyhedron(vl).plot(color='green',opacity=0.5,frame=False)+ 
              C + line(y_axis,color='gold',thickness=2))
s=animate(sf,aspect_ratio=(1,1,1))
s.gif(savefile="an.gif")

Answer to the second question:

solve returns a list of solutions. So you get something like solve([equ1,[equ2]],k,C).

You have to refer to an equation in the list:

solve([k^3==4*C,equC[0]],k,C)

There are far more simplify methods. expr.simplify? gives a hint:

See also: "simplify_full()", "simplify_trig()", "simplify_rational()", "simplify_rectform()", "simplify_factorial()", "simplify_log()", "simplify_real()", "simplify_hypergeometric()", "canonicalize_radical()"

In this case simplify_trig will do the job.

var('u')
expr = 1-cos(u)^2
expr.simplify_trig()

If solve can't find an explicit solution, you can use find_root or the sympy solver:

var ('N,m,sec,a,t,c')

T = (1.27 * 10^6)
Fg = ((9.81*107000))
a = ((T-Fg)/107000)

v = a*t
P3 = plot(v, (t,0,11))

y = v.integral(t)
P1 = plot(y, (t,0,12))

T1 = 1.27*e^(-(t-11)/2) * 10^6
Fg = 9.81*107000
a1 =(T1-Fg)/107000

v1 = a1.integral(t)
v1 = v1 + 154.299065420560
P4 = plot(v1, (t,11,22))

y1 = v1.integral(t) - 1026.68224299065
P2 = plot (y1, (t,12,22))

P = P1 + P2 

Impact = y1.find_root(21,22)
#print Impact

P5 = P3 + P4
show(P5)

dy1 = diff(y1, t)
#print dy1

print 'solve with maxima solver:', solve(v1,t)
print 'solve with find_root:', v1.find_root(0,20)
import sympy
print 'solve with sympy solver:', sympy.solve(v1,t)

The error "list index out of range" occurs if solve returns an empty list. You can capture this exception with a try except statement:

    try:
        C = circle((sol[1][0].rhs(),sol[1][1].rhs()),sol[1][2].rhs())    
    except:
        pass

This works for me:

print("all {4:{0}.{1}} good {2} eat {3}".format(2,6,"dogs","shoes",x))

I am afraid you are not wrong. If we set z = y + I then conjugate(z.subs(y=5)) gives the result -I + 5 in contradiction to conjugate(z).subs(y=5) which results to I + 5

You can use PIL to blend two images with alpha option. See an example here

Sorry for my misunderstanding. I ignored the 'even' because the function you provided isn't even. The easiest way to build a periodic function is to start with a periodic function. The start interval depends on the function you use and therefore we may need axis transformations to get the desired result. One more suggestion:

f(x) = abs(arccos(cos(x))-pi)

Thanks for the hint. I tried to improve my answer.

EDITED

f(t) = -t + pi        # interval  [0, pi]
g(t) = 1/2*f(t*pi)    # transformed to interval  [-1, 1]
# h = lambda x: 2*g(RR(x/pi).frac())   # including retransformation
h = lambda x: (heaviside(x)-1)*pi + 2*g(RR(x/pi).frac())
plot(h,-3*pi,3*pi)

UPDATE 2 --- a symbolic function:

f(x) = 1/2*(sign(sin(2*x))*arccos(cos(2*x-pi)) + pi)
plot(f,-3*pi,3*pi,aspect_ratio=1)

The function f returns z values. That's all. Do not confuse with implicit_plot3d. There you normally use an expression containing x, y and z.