I wrote down a code to calculate the integral ParEa in terms of a function J as follows:
Laa = 0.01;
ja = 1;
De = 0.001;
J = #(Oma) ja.* Oma .*exp(-Oma/Laa);
ParEaa = #(Oma) pi.^(-1).*J./(Oma.*(Oma + De));
IParEaa = integral(ParEaa, 0, inf)
But it gives the errors:
Undefined function or method 'integral' for input arguments of
type 'function_handle'.
Error in ==> sample at 9
IParEaa = integral(ParEaa, 0, inf)
I want to calculate the integral without substituting the explicit expression of J into ParEa? How should I do it?
I think you're mixing up the symbolic and numeric integration.
I cannot check with Matlab but I would suggest the following:
For numerical simulation:
J = #(Oma) ja.* Oma .*exp(-Oma/Laa);
ParEaa = #(Oma,J) pi.^(-1).*J./(Oma.*(Oma + De));
IParEaa = integral(ParEaa, 0, inf)
For the symbolic integration:
syms oma
expr = pi^(-1)*(ja*Oma*exp(-Oma/Laa))/(Oma*(Oma + De));
int(expr,oma,0,inf)
I hope it works,
Michael
Related
I'm having troubles to define the objective fucntion in a SMT problem with z3py.
Long story, short, I have to optimize the placing of smaller blocks inside a board that has fixed width but variable heigth.
I have an array of coordinates (represented by an array of integers of length 2) and a list of integers (representing the heigth of the block to place).
# [x,y] list of integer variables
P = [[Int("x_%s" % (i + 1)), Int("y_%s" % (i + 1))]
for i in range(blocks)]
y = [int(b) for a, b in data[2:]]
I defined the objective function like this:
obj= Int(max([P[i][1] + y[i] for i in range(blocks)]))
It calculates the max height of the board given the starting coordinate of the blocks and their heights.
I know it could be better, but I think the problem would be the same even with a different definition.
Anyway, if I run my code, the following error occurs on the line of the objective function:
" raise Z3Exception("Symbolic expressions cannot be cast to concrete Boolean values.") "
While debugging I've seen that is P[i][1] that gives an error and I think it's because the program reads "y_i + 3" (for example) and they can't be added togheter.
Point is: it's obvious that the objective function depends on the variables of the problem, so how can I get rid of this error? Is there another place where I should define the objective function so it waits to have the P array instantiated before doing anything?
Full code:
from z3 import *
from math import ceil
width = 8
blocks = 4
x = [3,3,5,5]
y = [3,5,3,5]
height = ceil(sum([x[i] * y[i] for i in range(blocks)]) / width) + 1
# [blocks x 2] list of integer variables
P = [[Int("x_%s" % (i + 1)), Int("y_%s" % (i + 1))]
for i in range(blocks)]
# value/ domain constraint
values = [And(0 <= P[i][0], P[i][0] <= width - 1, 0 <= P[i][1], P[i][1] <= height - 1)
for i in range(blocks)]
obj = Int(max([P[i][1] + y[i] for i in range(blocks)]))
board_problem = values # other constraints I've not included for brevity
o = Optimize()
o.add(board_problem)
o.minimize(obj)
if (o.check == 'unsat'):
print("The problem is unsatisfiable")
else:
print("Solved")
The problem here is that you're calling Python's max on symbolic values, which is not designed to work for symbolic expressions. Instead, define a symbolic version of max and use that:
# Return maximum of a vector; error if empty
def symMax(vs):
m = vs[0]
for v in vs[1:]:
m = If(v > m, v, m)
return m
obj = symMax([P[i][1] + y[i] for i in range(blocks)])
With this change your program will go through and print Solved when run.
The Goal is to format a polynomial with more than 6 parameters into a plot title.
Here is my polynomial parameter to string expression function, inspired by this answer, followed by sym.latex():
def func(p_list):
str_expr = ""
for i in range(len(p_list)-1,-1,-1):
if (i%2 == 0 and i !=len(p_list)):
str_expr =str_expr + " \n "
if p_list[i]>0:
sign = " +"
else:
sign = ""
if i > 1:
str_expr = str_expr+" + %s*x**%s"%(p_list[i],i)
if i == 1:
str_expr = str_expr+" + %s*x"%(p_list[i])
if i == 0:
str_expr = str_expr+sign+" %s"%(p_list[i])
print("str_expr",str_expr)
return sym.sympify(str_expr)
popt = [-2,1,1] # some toy data
tex = sym.latex(func(popt))
print("tex",tex)
Outputs:
str_expr
+ -1*x**2 + 1*x
-2
tex - x^{2} + x - 2
in str_expr the line breaks from \n are visible, yet in the sympy.latex output the are gone.
How to propagate this linebreak?
Edit: I took # wsdookadr answer and modified it, so that plt.title takes the result of the function as text the argument
def tex_multiline_poly(e, chunk_size=2, separator="\n"):
tex = ""
# split into monomials
print("reversed(e.args)",reversed(e.args))
mono = list(e.args)
print("mono",mono)
mono.reverse()
print("mono",mono)
# we're going split the list of monomials into chunks of chunk_size
# serialize each chunk, and insert separators between the chunks
for i in range(0,len(mono),chunk_size):
chunk = mono[i:i + chunk_size]
print("sum(chunk)",sum(chunk))
print("sym.latex(sum(chunk))",sym.latex(sum(chunk)))
if i == 0:
tex += r'$f(x)= %s$'%(sym.latex(sum(chunk)))+separator
else:
tex += '$%s$'%(sym.latex(sum(chunk))) + separator
return tex
popt = est.params
x = sym.symbols('x')
p = sym.Poly.from_list(reversed(popt),gens=x)
tex = tex_multiline_poly(p.as_expr(),chunk_size=2)
plt.title(text=tex)
In your code, you're inserting a linebreak for every even-power monomial, except for the last one.
if (i%2 == 0 and i !=len(p_list)):
str_expr =str_expr + " \n "
Since you are just building a polynomial from a list of coefficients, your code can be simplified.
Generally what we want is to build/transform/handle things symbolically, and only at the end serialize them and print the result in some specific format
import sympy as sym
x = sym.symbols('x')
def func(p_list):
expr = 0
for i in range(len(p_list)-1,-1,-1):
expr += p_list[i] * (x ** i)
return sym.sympify(expr)
popt = [-2,1,1]
p = func(popt)
p_tex = sym.latex(p)
p_str = str(p)
print("str:", p_str)
print("tex:", p_tex)
Output:
str: x**2 + x - 2
tex: x^{2} + x - 2
We could simplify this even further by using SymPy's built-in functions to build the poly from a list of coefficients:
import sympy as sym
from sympy import symbols
popt = [-2,1,1]
x = symbols('x')
p = sym.Poly.from_list(reversed(popt),gens=x)
p_tex = sym.latex(p.as_expr())
p_str = str(p.as_expr())
print("str:", p_str)
print("tex:", p_tex)
Output:
str: x**2 + x - 2
tex: x^{2} + x - 2
Does the output look like what you would expect?
UPDATE:
After learning more about the use-case, here's a version that inserts separators every N=2 monomials in the latex form of your expression.
import sympy as sym
from sympy import symbols
popt = [-2,1,1]
x = symbols('x')
p = sym.Poly.from_list(reversed(popt),gens=x)
def tex_multiline_poly(e, chunk_size=2, separator="\n"):
tex = ""
# split into monomials
mono = list(reversed(e.args))
# we're going split the list of monomials into chunks of chunk_size
# serialize each chunk, and insert separators between the chunks
for i in range(0,len(mono),chunk_size):
chunk = mono[i:i + chunk_size]
tex += sym.latex(sum(chunk)) + separator
return tex
p_tex = tex_multiline_poly(p.as_expr(),chunk_size=2)
p_str = str(p.as_expr())
print("str:",p_str)
print("tex:",p_tex)
Output:
str: x**2 + x - 2
tex: x^{2} + x
-2
Edit: wrong edit
I am trying to find three parameters (a, b, c) to fit my experimental data using ODE solver and optimization by least squares using Scilab in-built functions.
However, I keep having the message "submatrix incorrectly defined" at line "y_exp(:,1) = [0.135 ..."
When I try another series of data (t, yexp) such as the one used in the original template I get no error messages. The template I use was found here: https://wiki.scilab.org/Non%20linear%20optimization%20for%20parameter%20fitting%20example
function dy = myModel ( t , y , a , b, c )
// The right-hand side of the Ordinary Differential Equation.
dy(1) = -a*y(1) - b*y(1)*y(2)
dy(2) = a*y(1) - b*y(1)*y(2) - c*y(2)
endfunction
function f = myDifferences ( k )
// Returns the difference between the simulated differential
// equation and the experimental data.
global MYDATA
t = MYDATA.t
y_exp = MYDATA.y_exp
a = k(1)
b = k(2)
c = k(3)
y0 = y_exp(1,:)
t0 = 0
y_calc=ode(y0',t0,t,list(myModel,a,b,c))
diffmat = y_calc' - y_exp
// Make a column vector
f = diffmat(:)
MYDATA.funeval = MYDATA.funeval+ 1
endfunction
// Experimental data
t = [0,20,30,45,75,105,135,180,240]';
y_exp(:,1) =
[0.135,0.0924,0.067,0.0527,0.0363,0.02445,0.01668,0.012,0.009]';
y_exp(:,2) =
[0,0.00918,0.0132,0.01835,0.0261,0.03215,0.0366,0.0393,0.0401]';
// Store data for future use
global MYDATA;
MYDATA.t = t;
MYDATA.y_exp = y_exp;
MYDATA.funeval = 0;
function val = L_Squares ( k )
// Computes the sum of squares of the differences.
f = myDifferences ( k )
val = sum(f.^2)
endfunction
// Initial guess
a = 0;
b = 0;
c = 0;
x0 = [a;b;c];
[fopt ,xopt]=leastsq(myDifferences, x0)
Does anyone know how to approach this problem?
Just rewrite lines 28,29 as
y_exp = [0.135,0.0924,0.067,0.0527,0.0363,0.02445,0.01668,0.012,0.009
0,0.00918,0.0132,0.01835,0.0261,0.03215,0.0366,0.0393,0.0401]';
or insert a clear at line 1 (you may have defined y_exp before with a different size).
I have tried to write a function that takes in notes in MIDI form (C2,A4,Bb6) and returns their respective frequencies in hertz. I'm not sure what the best method of doing this should be. I am torn between two approaches. 1) a list based one where I can switch on an input and return hard-coded frequency values given that I may only have to do this for 88 notes (in the grand piano case). 2) a simple mathematical approach however my math skills are a limitation as well as converting the input string into a numerical value. Ultimately I've been working on this for a while and could use some direction.
You can use a function based on this formula:
The basic formula for the frequencies of the notes of the equal
tempered scale is given by
fn = f0 * (a)n
where
f0 = the frequency of one fixed note which must be defined. A common choice is setting the A above middle C (A4) at f0 = 440 Hz.
n = the number of half steps away from the fixed note you are. If you are at a higher note, n is positive. If you are on a lower note, n is negative.
fn = the frequency of the note n half steps away. a = (2)1/12 = the twelth root of 2 = the number which when multiplied by itself 12 times equals 2 = 1.059463094359...
http://www.phy.mtu.edu/~suits/NoteFreqCalcs.html
In Objective-C, this would be:
+ (double)frequencyForNote:(Note)note withModifier:(Modifier)modifier inOctave:(int)octave {
int halfStepsFromA4 = note - A;
halfStepsFromA4 += 12 * (octave - 4);
halfStepsFromA4 += modifier;
double frequencyOfA4 = 440.0;
double a = 1.059463094359;
return frequencyOfA4 * pow(a, halfStepsFromA4);
}
With the following enums defined:
typedef enum : int {
C = 0,
D = 2,
E = 4,
F = 5,
G = 7,
A = 9,
B = 11,
} Note;
typedef enum : int {
None = 0,
Sharp = 1,
Flat = -1,
} Modifier;
https://gist.github.com/NickEntin/32c37e3d31724b229696
Why don't you use a MIDI pitch?
where f is the frequency, and d the MIDI data.
HI! I create a function to get a grayscale version of an image, but i have problem trying to pass by reference the destination of the bit(s) generated by this function:
void grayscale (const unsigned char *source, unsigned char **dest, int data_size) {
for (int i=0; i < data_size; i= i+4) {
int gray = (source[i] + source[i+1] + source[i+2]) / 3;
gray = 255 - (int)cos(source[i])*255;
*dest[i] = (char)gray;
*dest[i + 1] = (char)gray;
*dest[i + 2] = (char)gray; //HERE AN ERROR
*dest[i + 3] = (char)255;
}
}
I call this function with:
grayscale(source, &destination, width*height*4 );
Is there something wrong with pointers ?
(i'm working on objective C and i obtain a EXC_BAD_ACCESS).
Thank you
It's not clear why you're using an additional level of indirection for dest, since *dest is not being modified but most likely you need to do the following - change:
*dest[i] = (char)source[i];
*dest[i + 1] = (char)gray;
*dest[i + 2] = (char)gray;
*dest[i + 3] = (char)255;
to:
(*dest)[i] = (char)source[i];
(*dest)[i + 1] = (char)gray;
(*dest)[i + 2] = (char)gray;
(*dest)[i + 3] = (char)255;
The reason for this is operator precedence/associativity.
Also it's hard to tell without seeing the calling code, but I'm guessing you may need to change:
(*dest)[i] = (char)source[i];
to:
(*dest)[i] = (char)source[i / 4];
if you're trying to do something like convert a single plane image to RGBA.
Your call is fine, you have an illegal memory access. Debug your indices and check the size of the array your passing in, also check that it is still in scope. Most likely i+2 is out of bounds.