Steffensen's method: Difference between revisions
julia example
(julia example) |
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t0 = 0.9: intersection at -0.6810249675905098, 2.6810249675906883
t0 = 1: no answer
</pre>
=={{header|Julia}}==
{{trans|C}}
<syntaxhighlight lang="julia">""" Aitken's extrapolation """
function aitken(f, p0)
p1 = f(p0)
p2 = f(p1)
return p0 - (p1 - p0)^2 / (p2 - 2 * p1 + p0)
end
""" Steffensen's method using Aitken """
function steffensen_aitken(f, pinit, tol, maxiter)
p0 = pinit
p = aitken(f, p0)
iter = 1
while abs(p - p0) > tol && iter < maxiter
p0 = p
p = aitken(f, p0)
iter += 1
end
return abs(p - p0) > tol ? NaN : p
end
""" deCasteljau function """
function deCasteljau(c0, c1, c2, t)
s = 1.0 - t
return s * (s * c0 + t * c1) + t * (s * c1 + t * c2)
end
xConvexLeftParabola(t) = deCasteljau(2, -8, 2, t)
yConvexRightParabola(t) = deCasteljau(1, 2, 3, t)
implicit_equation(x, y) = 5 * x^2 + y - 5
""" may return NaN on overflow """
function f(t)
return t in [nothing, NaN, Inf, -Inf] ? NaN :
implicit_equation(xConvexLeftParabola(t), yConvexRightParabola(t)) + t
end
""" test the example """
function test_steffensen(tol = 0.00000001, iters = 1000, stepsize = 0.1)
for t0 in 0:stepsize:1.1
print("t0 = $t0 : ")
t = steffensen_aitken(f, t0, tol, iters)
if isnan(t)
println("no answer")
else
x = xConvexLeftParabola(t)
y = yConvexRightParabola(t)
if abs(implicit_equation(x, y)) <= tol
println("intersection at ($(Float32(x)), $(Float32(y)))")
else
println("spurious solution")
end
end
end
return 0
end
test_steffensen()
</syntaxhighlight>{{out}}
<pre>
t0 = 0.0 : no answer
t0 = 0.1 : intersection at (0.88102496, 1.118975)
t0 = 0.2 : no answer
t0 = 0.3 : no answer
t0 = 0.4 : no answer
t0 = 0.5 : no answer
t0 = 0.6 : no answer
t0 = 0.7 : no answer
t0 = 0.8 : no answer
t0 = 0.9 : intersection at (-0.68102497, 2.681025)
t0 = 1.0 : no answer
t0 = 1.1 : no answer
</pre>
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