Is there a way in sage to construct and plot a smooth function which has a given set of data:
OK, the third and last time ;-)
L_roots = [-1,4]
L_max = [(-3,1)]
L_min = [(1,1)]
L_other = []
distinguish_min_max = False # set to True or False
N = len(L_roots)+2*len(L_min)+2*len(L_max)+len(L_other)-1
param_list = [var("a%d" % k) for k in [0..N]]
f(x) = sum([ param_list[k]*x^k for k in [0..N]])
df = diff(f,x)
ddf = diff(df,x)
show(f)
eqns = [ f(x0)==0 for x0 in L_roots ]
eqns += [ f(P[0])==P[1] for P in L_min ]
eqns += [ df(P[0])==0 for P in L_min]
eqns += [ f(P[0])==P[1] for P in L_max ]
eqns += [ df(P[0])==0 for P in L_max ]
eqns += [ f(P[0])==P[1] for P in L_other ]
if distinguish_min_max:
eqns += [ ddf(P[0])> 0 for P in L_min]
eqns += [ ddf(P[0])<0 for P in L_max]
#for eq in eqns:
# print eq
sol = solve(eqns,param_list,solution_dict=True)
show(sol)
try:
g = f.substitute(sol[0])
show(g)
G = Graphics()
G += g.plot(xmin=-10,xmax=10,ymin=-10,ymax=10)
G += points(L_min,color='red')
G += points(L_max,color='red')
G += points(L_other,color='black')
G += points([(x0,0) for x0 in L_roots],color='green')
G.show()
except:
print 'no solution'
Here an example where you don't have to rewrite the code. I do not distinguish between minima and maxima. Controlling behavior for x to infinity: set some additional points.
L_roots = [-1,0]
L_min_max = [(1,1),(5,1)]
L_other = [(10,10)]
N = len(L_roots)+2*len(L_min_max)+len(L_other) -1
param_str = reduce(lambda x,y:x+y,['a%s,'%(k) for k in [0..N]])
param_list = var(param_str[:-1])
f(x) = sum([ param_list[k]*x^k for k in [0..N]])
df = diff(f,x)
show(f)
eqns = [ f(x0)==0 for x0 in L_roots ]
eqns += [ f(P[0])==P[1] for P in L_min_max ]
eqns += [ df(P[0])==0 for P in L_min_max ]
eqns += [ f(P[0])==P[1] for P in L_other ]
sol=solve(eqns,param_list,solution_dict=True)
show(f.substitute(sol[0]))
What kind of function do you want? Here a basic example for beginners. (http://www.hr.shuttle.de:9000/home/pu...)
# Minima / Maxima
A = (-1 , 1)
B = ( 3 , 3)
# Root
C = (1 , 0)
var("a0 a1 a2 a3 a4 a5 a6")
# f(x) = a5*x^5+ a4*x^4+a3*x^3 + a2*x^2+ a1*x + a0
f(x) = a4*x^4+a3*x^3 + a2*x^2+ a1*x + a0
# f(x) = a3*x^3 + a2*x^2+ a1*x + a0
df = diff(f,x);
ddf = diff(df,x);
xa,ya = A; xb,yb = B; xc,yc = C
# equations
eq1 = ( f(xa) == ya )
eq2 = ( df(xa) == 0 )
eq3 = ( f(xb) == yb )
eq4 = ( df(xb) == 0 )
eq5 = ( f(xc) == yc )
sol = solve([eq1,eq2,eq3,eq4,eq5],a0,a1,a2,a3,a4, solution_dict=True)
show(sol)
g = f.substitute(sol[0])
show(g)
plot(g,-4,4,ymin=-5,ymax=5)
Thanks, that's nice. Polynomial functions are fine. The problem is that if I want to add two roots or a minima to your example, I have to rewrite essentially the whole thing. How do you control the behaviour for $x\to \pm \infinity$? How do you distinguish between minima and maxima in your example?
Minima will be at points where the second derivative is positive and maxima where it is negative. See http://en.wikipedia.org/wiki/Second_derivative_test.
Asked: 2012-06-08 17:45:23 +0200
Seen: 383 times
Last updated: Jun 10 '12
Copyright Sage, 2010. Some rights reserved under creative commons license. Content on this site is licensed under a Creative Commons Attribution Share Alike 3.0 license.
I don't think so (this would be a strange feature to have), but you can do it analytically through calculus and other methods. Do you need help doing that?
If you have a good idea how to do it, it would be nice if you could explain it.