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2021-08-29 23:35:35 +0200	received badge	● Popular Question (source)
2017-10-10 22:15:46 +0200	asked a question	an integral with arccos wrong The value of the integral `integrate(exp(arccos(x)),x,0,1)` is badly return by Sage. However, `sympy_integrator` or `giac_integrator` give the exact value (and also the primitive), maxima giving the same wrong value than Sage (the both unable to give a primitive) `sage: integrate(exp(arccos(x)),x,0,1) -1/2e^(-1/2pi) + 1/2` Calculus gives through the change of variable x=cos t the equality \int_0^1 exp(arccos(x))dx= \int_0^{pi/2}e^t sin(t)dt and the second integral is easily computed : `sage: integrate(exp(t)sin(t),t,0,pi/2) 1/2e^(1/2pi) + 1/2 from sage.symbolic.integration.external import sympy_integrator sympy_integrator(exp(arccos(x)), x, 0, 1) 1/2e^(1/2pi) + 1/2 # GOOD from sage.symbolic.integration.external import maxima_integrator maxima_integrator(exp(arccos(x)), x, 0, 1) -1/2e^(-1/2pi) + 1/2 # BAD from sage.symbolic.integration.external import giac_integrator giac_integrator(exp(arccos(x)), x, 0, 1) 1/2e^(1/2*pi) + 1/2 # GOOD`
2017-10-10 22:12:10 +0200	asked a question	an integral with arccos wrongly evaluated The value of the integral `integrate(exp(arccos(x)),x,0,1)` is badly return by Sage. However, `sympy_integrator` or `giac_integrator` give the exact value (and also the primitive), maxima giving the same wrong value than Sage `sage: integrate(exp(arccos(x)),x,0,1) -1/2e^(-1/2pi) + 1/2` Calculus gives through the change of variable x=cos t the equality \int_0^1 exp(arccos(x))dx= \int_0^{pi/2}e^t sin(t)dt and the second integral is easily computed. A primitive is also given by sympy and giac, but nothing by sage, nor maxima `from sage.symbolic.integration.external import sympy_integrator sympy_integrator(exp(arccos(x)), x, 0, 1) 1/2e^(1/2pi) + 1/2 # GOOD from sage.symbolic.integration.external import maxima_integrator maxima_integrator(exp(arccos(x)), x, 0, 1) -1/2e^(-1/2pi) + 1/2 # BAD from sage.symbolic.integration.external import giac_integrator giac_integrator(exp(arccos(x)), x, 0, 1) 1/2e^(1/2pi) + 1/2 # GOOD`
2017-10-01 12:06:10 +0200	commented answer	t_span in ode_solver Thanks for explanations ! In fact, I was also puzzled by the ode_solver feature giving solution only for positive times : t_1>t_0 if t_span=t_0,t_1] (for negative times (t_1<t_0) ,="" consider="" the="" ode="" y'="-F(-t,y)).</p">
2017-10-01 12:03:55 +0200	received badge	● Editor (source)
2017-09-30 19:27:30 +0200	received badge	● Student (source)
2017-09-30 19:24:58 +0200	asked a question	t_span in ode_solver The definition of the first time in t_span has no effect : [t_0,t_1] has the same effect than [0,t_1]. See the code below. What to do ? `def f(t,y): return[y[0],y[1]] T = ode_solver() T.function=f T.ode_solve(y_0=[1,1],t_span=[-1,0.5],num_points=100) sx=[j[1][0] for j in T.solution] sy=[j[1][1] for j in T.solution] p=line(zip(sx,sy)) p.show(xmin=-3,xmax=3,ymax=3,ymin=-3)`
2017-09-04 22:31:00 +0200	received badge	● Popular Question (source)
2015-05-15 08:49:41 +0200	commented answer	hyphen in subscript It doesn't work in my full text : [dis]appearance of the hyphen subscript seems to be at random !
2013-02-22 07:45:47 +0200	received badge	● Taxonomist
2012-03-11 13:18:22 +0200	asked a question	contour and function definition Hello Let us consider this piece of code var('x,y') px=(x,-1,1) py=(y,-1,1) def V(x,y) : if (x2+y2 <1) : return(x+y+(abs(1-x2-y2))(1/2)) else: return(20) W(x,y)=x+y+(abs(1-x2-y2))(1/2) r=Graphics() r+=contour_plot(W(x,y),px,py,contours=[1.70],fill=false,cmap=["red"]) r+=contour_plot(V(x,y),px,py,contours=[1.50],fill=false,cmap=["magenta"]) plot(r) In 4.7, the level lines in magenta and red are displayed, but in 4.8 only the red one appears. is this change expected ? how to let sage knowing the definition of the V function ?