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Re: FindRoot + Compile = incompatible (?)
*To*: mathgroup at smc.vnet.net
*Subject*: [mg111314] Re: FindRoot + Compile = incompatible (?)
*From*: Bob Hanlon <hanlonr at cox.net>
*Date*: Tue, 27 Jul 2010 04:17:47 -0400 (EDT)
func = Compile[{{x, _Real}}, x^2];
FindRoot[func[y] == 1,
{y, 0.5, 1.5}] // Quiet
{y -> 0.9999999999999993}
FindRoot[func[y] == 1,
{y, #}] & /@
{-.5, .5} // Quiet
{{y -> -1.000000000000001},
{y -> 1.000000000000001}}
Solve[func[y] == 1, y] //
Quiet
{{y -> -1}, {y -> 1}}
Reduce[func[y] == 1, y] //
Quiet
y == -1 || y == 1
Bob Hanlon
---- "Scot T. Martin" <smartin at seas.harvard.edu> wrote:
=============
Has anyone ever run into the problem of trying to use Compile'd functions
within FindRoot?
The error message is:
"CompiledFunction::cfsa: Argument y at position 1 should be a machine-size
real number."
[Yes, the function still evaluates but in doing so the purpose of compile
& speed gets defeated.]
The explanation for this error messages is found in ref/FindRoot:
"FindRoot first localizes the values of all variables, then evaluates f
with the variables being symbolic [1], and then repeatedly evaluates the
result numerically." [1] That's why Compile[] fails.
Here is a specific example of code that shows the problem:
*****
func = Compile[{x}, x^2];
FindRoot[func[y] == 1, {y, 0.5, 1.5}]
*****
Anyone have thoughts on how to work around this aspect so that Compile'd
functions can work within FindRoot and thus accelerate evaluation thereof?
[Or is there an inherent reason why they shouldn't?]
[My actual function is not "func"; my actual func is 3x faster as a
Compile'd, hence my motivation to get FindRoot to work.]
Thanks!
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