I have the following equation and want to simplify it using Matlab's MuPAD.
So, I used this in MuPAD:
Simplify(Gb = Gm*((2*(Gd+t)+3*Gm+3*P*(Gd+t-Gm))/(2*(Gd+t)+3*Gm-2*P*(Gd+t-Gm))))
and I get this:
Gb*(3*Gm + 2*Gd + 2*t - 2*P*(Gd - Gm + t)) = Gm*(3*Gm + 2*Gd + 2*t + 3*P*(Gd - Gm + t)) and 3*Gm + 2*Gd + 2*t <> 2*P*(Gd - Gm + t)
I cannot understand the first and second part (after "and"), what are these?
As per MuPAD's documentation, the relational operator <> denotes inequality, i.e., the left- and right-hand side are not equal. In other words, the simplification (before "and") is only valid if 3*Gm + 2*Gd + 2*t is not equal to 2*P*(Gd - Gm + t). This is the same thing as requiring that the denominator of the original expression not be zero (which would make it undefined).
Note that in MuPAD for Matlab R2015b, the <> operator is displayed as ≠ and "and" is rendered as ∧ (logical and):
Related
I am trying to take the derivative of a function including a boolean variable with sympy.
My expected result:
Two different derivatives, depending on the boolean being either True or False (i.e. 1 or 0).
Example:
import sympy as sy
c, x = sy.symbols("c x", positive=True, real=True)
bo = sy.Function("bo")
fct1 = sy.Function("fct1")
fct2 = sy.Function("fct2")
FOC2 = sy.Function("FOC2")
y = 5
a = 2
b = 4
def fct1(x):
return -0.004*x**2 + 0.25*x + 4
# the following gives the smaller positive intercept with the x-axis)
# this intercept is the threshold value for the boolean function, bo
min(sy.solve(fct1(x)-y, x))
def bo(x):
if fct1(x) <= y:
return 1
else:
return 0
def fct2(c, x):
return a + b*c + bo(x)*c
def FOC2(c, x):
return sy.diff(fct2(c, x), c)
print(FOC2(c, x))
The min-function after the comments shows me the threshold of x for bo being True or False would be 4.29..., thus positive and real.
Output:
TypeError: cannot determine truth value of Relation
I understand that the truth value depends on x, which is a symbol. Thus, without knowing x one cannot determine bo.
But how would I get my expected result, where bo is symbolic?
First off, I would advise you to carefully consider what is going on in your code the way it is pasted above. You first define a few sympy functions, e.g.
fct1 = sy.Function("fct1")
So after this, fct1 is an undefined sympy.Function - undefined in the sense that it is neither specified what its arguments are, nor what the function looks like.
However, then you define same-named functions explicitly, as in
def fct1(x):
return -0.004*x**2 + 0.25*x + 4
Note however, that at this point, fct1 ceases to be a sympy.Function, or any sympy object for that matter: you overwrite the old definition, and it is now just a regular python function!
This is also the reason that you get the error: when you call bo(x), python tries to evaluate
-0.004*x**2 + 0.25*x + 4 <= 5
and return a value according to your definition of bo(). But python does not know whether the above is true (or how to make that comparison), so it complains.
I would suggest 2 changes:
Instead of python functions, as in the code, you could simply use sympy expressions, e.g.
fct1 = -0.004*x**2 + 0.25*x + 4
To get the truth value of your condition, I would suggest to use the Heaviside function (wiki), which evaluates to 0 for a negative argument, and to 1 for positive. Its implementation in sympy is sympy.Heaviside.
Your code could then look as follows:
import sympy as sy
c, x = sy.symbols("c x", positive=True, real=True)
y = 5
a = 2
b = 4
fct1 = -0.004*x**2 + 0.25*x + 4
bo = sy.Heaviside(y - fct1)
fct2 = a + b*c + bo * c
FOC2 = sy.diff(fct2, c)
print(FOC2)
Two comments on the line
bo = sy.Heaviside(y - fct1)
(1) The current implementation does not evaluate sympy.Heaviside(0)by default; this is beacause there's differing definitions around (some define it to be 1, others 1/2). You'd want it to be 1, to be in accordance with the (weak) inequality in the OP. In sympy 1.1, this can be achieved by passing an additional argument to Heaviside, namely whatever you want Heaviside(0) to evaluate to:
bo = sy.Heaviside(y - fct1, 1)
This is not supported in older versions of sympy.
(2) You will get your FOC2, again involving a Heaviside term. What I like about this, is that you could keep working with this expression, say if you wanted to take a second derivative and so on. If, for the sake of readability, you would prefer a piecewise expression - no problem. Just replace the according line with
bo = sy.Heaviside(y - fct1)._eval_rewrite_as_Piecewise(y-fct1)
Which will translate to a piecewise function automatically. (note that under older versions, this automatically implicitly uses Heaviside(0) = 0.5 - best to use (1) and (2) together:
bo = sy.Heaviside(y - fct1, 1)._eval_rewrite_as_Piecewise(y-fct1)
Unfortunately, I don't have a working sympy 1.1 at my hands right now and can only test the old code.
One more noteconcerning sympy's piecewise functions: they are much more readable if using sympy's latex printing, by inserting
sy.init_printing()
early in the code.
(Disclaimer: I am by no means an expert in sympy, and there might be other, preferable solutions out there. Just trying to make a suggestion!)
So, I have a really long symbolic exrpression in Matlab that I want to copy/paste into a JavaScript-code to animate a numerical solution for it. The problem is that some places in my code i get exponents (mostly ^2), when I'd rather have Matlab express it as A*A.
I have mulitple (and different) expressions like
cos(th2t)^2
That I would rather have expressed as
cos(th2t)*cos(th2t)
Any way I can do this? The alternative is to use a text editor afterward and search for powers of 2 and replace it, but there are multiple different expressions, so that would take some time...
This is an example of one of the exrpressions I end up with:
(J31*(2*ddth2t*cos(th1t) - 2*w12*w13 - 4*dth1t*dth2t*sin(th1t) - 4*dth2t*w13*sin(th1t) + dth1t^2*sin(2*th2t)*cos(th1t) - w12^2*sin(2*th2t)*cos(th1t) + w13^2*sin(2*th2t)*cos(th1t) + 2*dth2t*w11*sin(2*th2t) + 2*w12*w13*cos(th2t)^2 + ddth1t*sin(2*th2t)*sin(th1t) + 4*dth1t*dth2t*cos(th2t)^2*sin(th1t) + 2*dth1t*w13*sin(2*th2t)*cos(th1t) + 4*dth2t*w13*cos(th2t)^2*sin(th1t) + w11*w12*sin(2*th2t)*sin(th1t) + 4*dth1t*w12*cos(th1t)^2*cos(th2t)^2 - 2*dth1t*w11*sin(2*th1t)*cos(th2t)^2 - 2*dth2t*w11*sin(2*th2t)*cos(th1t)^2 + 2*w12*w13*cos(th1t)^2*cos(th2t)^2 - w11*w13*sin(2*th1t)*cos(th2t)^2 - 4*dth2t*w12*cos(th1t)*cos(th2t)*sin(th1t)*sin(th2t)))/(2*(J11 + J31))
Steve's answer should work fine if you want to rely on the ** operator. However, since that operator is not officially supported, and that solution doesn't directly answer the OP's question, here is a function that can expand out the exponents in a symbolic expression.
function [ text ] = remove_powers( exp )
%Cleans the powers from T, and return expanded text representation
% Define functions
enclose =#(t) ['(' t ')'];
expand_pow=#(b,p) strjoin(strcat(cell(1,p),enclose(char(b))),'*');
count_pow=#(s) arrayfun(#(k) count(char(s(k)),'^'), 1:length(s));
sym2str = #(s) strrep(char(s), ' ', '');
% Check for fractions
[N,D]=numden(exp);
if ~isequal(D,sym(1))
% pass up the num and den
text = [remove_powers(N) '/' enclose(remove_powers(D))];
else
% Split a into subterms
Ts = children(exp);
% Clean children
has_pow=count_pow(Ts)>0;
text = sym2str(exp);
if sum(count_pow(Ts))<count_pow(exp)
% We have removed a power, clean it, expand it, and pass up
text = expand_pow(remove_powers(Ts(1)),Ts(2));
else
% Just clean the unclean children and pass up
for t=Ts(has_pow)
text = strrep(text,sym2str(t),remove_powers(t));
end
end
end
end
The function uses the children function in Matlab to recursively clean each subexpression and replace it (as text) in the parent. This method is better than using regex because it avoids the issue of parsing the syntax, as Steve mentioned in the comment with respect to cos(x^2+2*y)^2.
Which makes for a good example:
syms x y real
exp = cos(x^2+2*y)^2;
cleaned_exp = remove_powers(exp)
Outputs: (cos(2*y+(x)*(x)))*(cos(2*y+(x)*(x)))
Notice that since Matlab is doing the parsing, there was no need to parse the order of precedence for the '^' operators, which could be difficult to accomplish with regex.
To test the OP's example:
syms ddth1t dth1t th1t ddth2t dth2t th2t w11 w12 w13 J11 J31 real
exp = (J31*(2*ddth2t*cos(th1t) - 2*w12*w13 - 4*dth1t*dth2t*sin(th1t) - 4*dth2t*w13*sin(th1t) + dth1t^2*sin(2*th2t)*cos(th1t) - w12^2*sin(2*th2t)*cos(th1t) + w13^2*sin(2*th2t)*cos(th1t) + 2*dth2t*w11*sin(2*th2t) + 2*w12*w13*cos(th2t)^2 + ddth1t*sin(2*th2t)*sin(th1t) + 4*dth1t*dth2t*cos(th2t)^2*sin(th1t) + 2*dth1t*w13*sin(2*th2t)*cos(th1t) + 4*dth2t*w13*cos(th2t)^2*sin(th1t) + w11*w12*sin(2*th2t)*sin(th1t) + 4*dth1t*w12*cos(th1t)^2*cos(th2t)^2 - 2*dth1t*w11*sin(2*th1t)*cos(th2t)^2 - 2*dth2t*w11*sin(2*th2t)*cos(th1t)^2 + 2*w12*w13*cos(th1t)^2*cos(th2t)^2 - w11*w13*sin(2*th1t)*cos(th2t)^2 - 4*dth2t*w12*cos(th1t)*cos(th2t)*sin(th1t)*sin(th2t)))/(2*(J11 + J31));
cleaned_exp = remove_powers(exp);
isequal(sym(cleaned_exp),exp) % This should be 1
count(cleaned_exp,'^') % This should be 0
As expected, the new expression is equivalent to the original, but has no '^' symbols.
This answer suggests that you could call the exponential operator in Javascript with e.g. A**2. As such you could replace all instances of ^ with **.
(Solution from comments)
I need to compute some fairly long expressions that contain common subexpressions. For example, consider the following two expressions:
double dfdx1 = 2 * (-x2 + x1 - sin(b2)*n34 + cos(b2)*sin(c2)*n24 - cos(b2)*cos(c2)*n14 + sin(b1)*m34 - cos(b1)*sin(c1)*m24 + cos(b1)*cos(c1)*m14);
double dfdx2 = -2 * (-x2 + x1 - sin(b2)*n34 + cos(b2)*sin(c2)*n24 - cos(b2)*cos(c2)*n14 + sin(b1)*m34 - cos(b1)*sin(c1)*m24 + cos(b1)*cos(c1)*m14);
Aside from eliminating all the trigonometric functions, one obvious elimination is dfdx2 = -dfdx1. The question is whether the compiler will recognise this. I found that using MATLAB's MuPad generate::optimize() function does not, which rather surprised me.
More generally, will the compiler recognise that f2 = -f1 in the example below:
double f1 = a*a + b*b - c*a - c*b;
double f2 = c*a + c*b - a*a - b*b;
Or will it just eliminate the terms a*a, b*b, c*a and c*b?
I am using the MSVC compiler, but I guess that they all do pretty much the same thing.
Normally compilers should recognize this and carry out the asked transformation if you enable "fast math" (-ffast-math for gcc). The reason is that floating point operations are not perfectly exact and the order of the evaluation of an expression might matter.
Example (for doubles, all constants are actually considered being results of other operations):
"1e100"+"1.0"-"1e100" results in 0.0
"1e100"-"1e100"+"1.0" results in 1.0
So the compiler will only reorder expressions if you explicitly allow such transformations.
If I start with the following symbolic expression:
a^2 + 2*a*b + b^2
Then run simplify (or factor), I get the expected result:
>> simplify(a^2 + 2*a*b + b^2)
(a + b)^2
Now when I run the same example, but adding another term, no factorisation occurs:
>> simplify(a^2 + 2*a*b + b^2 + 1)
a^2 + 2*a*b + b^2 + 1
How can I get these functions to return the more practical version of this expression ((a + b)^2 + 1)? I have tried all of the obvious options with these functions (like 'Steps', 'IgnoreAnalyticConstraints', etc.) but to no avail.
Context: I have the expression ax^2 - 2*ax*bx + bx^2 + ay^2 - 2*ay*by + by^2 which I need to convert back into (ax - bx)^2 + (ay - by)^2 so it can then be treated correctly as r^2. I know I could use some blunt substitution rules, but for something so simple I feel like I'm missing an obvious 'non-hack' solution.
you can run simplify on the two terms separately.
simplify(ax^2 - 2*ax*bx + bx^2) + simplify(ay^2 - 2*ay*by + by^2)
It seems like you already know how it should be simplified anyway.
Also, you eventually want to write it as r^2. This is not generally possible for all second-order expressions, so don't bother trying to find a general solution.
My question is quite similar to this one, but with a difference. I want to create a macro (or whatever) that behaves this way:
julia> #my-macro x + 2
:(x + 2)
(note that x + 2 is not enclosed in quotes). Is there something like that in Julia? And if there is not, how do I do it? (Please, give a detailed explanation about why it works.)
The input expression to the macro needs to be quoted because a macro returns an expression, which are evaluated, while you would like to get the expression itself, hence you need an extra quoting. The quoting can be done as:
macro mymacro(ex)
Expr(:quote,ex) # this creates an expression that looks like :(:(x + 2))
end
e=#mymacro x + 2 #returns :(x + 2)
Another shorter possibility:
macro mymacro(ex)
QuoteNode(ex)
end
e = #mymacro x + 2 #returns :(x + 2)