How to simplify function handles?
显示 更早的评论
Y =
function_handle with value:
@(s) (F(s)-Q(s))/P(s)
As seen, I have three functions F, Q and P and Y is function made by algebraic operations on these three. Is there a way, to get a simplified expression for Y in terms of the function variable (s) ? When I type in Y in command window, I just get the result as shown above.
As a simplified example:
F = @(s) s;
Q = @(s) s^2;
P = @(s) s^3-3*s^2-1;
Y = @(s) (F(s) - Q(s))/P(s);
When i type in Y in command window, what is see is:
Y =
function_handle with value:
@(s) (F(s)-Q(s))/P(s)
What I wish to see is:
Y = (s-s^2)/(s^3-3*s^2-1)
2 个评论
the cyclist
2021-8-19
It would helpful if you included a small example of what you see, and what you wish to see.
anuj maheshwari
2021-8-19
采纳的回答
Walter Roberson
2021-8-19
You used a tag of symbolic which implies that you might have the Symbolic Toolbox. If so then
syms s
Ys = Y(s)
You might want to simplify() the result.
7 个评论
F = @(s) s;
Q = @(s) s.^2;
P = @(s) s.^3-3*s.^2-1;
Y = @(s) (F(s) - Q(s))./P(s);
syms s real
Ys = Y(s)
dY = diff(Ys, s)
critical = solve(dY, s)
d2Y = diff(dY, s)
crit2 = subs(d2Y, s, critical)
definitely_max = isAlways(crit2 < 0, 'unknown', 'false');
definitely_min = isAlways(crit2 > 0, 'unknown', 'true');
not_sure = ~(definitely_max | definitely_min);
max_locations = critical(definitely_max)
max_values = Y(max_locations)
vpa(max_values)
min_locations = critical(definitely_min)
min_values = Y(min_locations)
vpa(min_values)
other_locations = critical(not_sure)
other_values = Y(other_locations)
anuj maheshwari
2021-8-20
F = @(s) s;
Q = @(s) s.^2;
P = @(s) s.^3-3*s.^2-1;
Y = @(s) (F(s) - Q(s))./P(s);
syms s real
Ys = Y(s)
fplot(Ys, [-10 10])
[N,D] = numden(Ys)
location_of_zeros = solve(N,s)
location_of_poles = solve(D, s, 'maxdegree', 4)
vpa(location_of_poles)
I guess my calculas analysis needed to be more careful as it missed the +/- infinity of the pole.
anuj maheshwari
2021-8-20
F = @(s) s;
Q = @(s) s.^2;
P = @(s) s.^3-3*s.^2-1;
Y = @(s) (F(s) - Q(s))./P(s);
syms s real
Ys = Y(s)
iY = ilaplace(Ys)
matlabFunction(iY)
That is, ilaplace() generated a root() [really a RootOf] and matlabFunction() is not able to deal with those.
In theory as this was a polynomial of degree 3, ilaplace could have generated the exact formula instead of using a root() or RootOf() placeholder... but it didn't, and there is no method provided by Mathworks to "fix-up" the expression afterwards.
A few weeks ago, I posted an attempt to do conversion of these kinds of structures. I will try to dig up the code and see if I can get it to work for your case.
anuj maheshwari
2021-8-20
In the below, eiY is the output to pay attention to: it is the symbolic form of the ilaplace expaned out from sym of roots into closed formula.
The plotting after that is investigative, to try to figure out why the fplot() is showing up with discontinuities.
According to the X / YY values, the matlabFunction() version of eiY is producing values with imaginary component roughly 1e-17 there. Zooming in trying to look for visible infinities or whatever, did not help. But taking some of the locations that produce non-zero imaginary part and evaluating them symbolically to 50 digits shows a mix of pure-rule output and very small imaginary parts; the amplitude of the imaginary parts decreases as the number of digits increases.
So what is happening here is that the matlabFunction() version is suffering from round-off error (and even the symbolic version does too). The root() that is the solution of the ilaplace has a conjugate pair, so in theory the positive and negative imaginary components should cancel, but in practice because of floating point round off, they do not.
In this particular case, at least for positive t, you should be able to just take real() of the values.
format long g
digits(50)
F = @(s) s;
Q = @(s) s.^2;
P = @(s) s.^3-3*s.^2-1;
Y = @(s) (F(s) - Q(s))./P(s);
syms s real
Ys = Y(s)
iY = ilaplace(Ys)
Nth = @(expr,N) expr(N);
resolve_a_root = @(X6) Nth(solve(children(X6,1),children(X6,2),'maxdegree',4),children(X6,3));
resolve_a_summand = @(X5) mapSymType(X5, 'root', resolve_a_root);
resolve_nth_summand = @(X2,N) resolve_a_summand(subs(children(X2,1),children(X2,2),N));
map_symsum = @(X2) sum(arrayfun(resolve_nth_summand, repmat(X2,1,double(children(X2,4)-children(X2,3)+1)), children(X2,3):children(X2,4)));
expand_symsum = @(X1) mapSymType(X1, 'symsum', map_symsum);
eiY = simplify(expand_symsum(iY))
y = matlabFunction(eiY);
fplot(y, [1e-4,10e-4]);
fplot(y, [4.257475e-4 4.25749e-4]); ylim([-1.001 -1]); title('zoom fplot numeric')
fplot(eiY, [4.257475e-4 4.25749e-4]); ylim([-1.001 -1]); title('zoom fplot symbolic')
X = linspace(4.257475e-4, 4.25749e-4,1000);
YY = y(X);
subplot(2,1,1); plot(X, real(YY)); title('fixed numeric evaluation real')
subplot(2,1,2); plot(X, imag(YY)); title('fixed numeric evaluation imag')
X5idx = find(imag(YY), 5);
X5 = X(X5idx)
Y5 = YY(X5idx)
syms t
subs(eiY, t, X5(:))
vpa(ans)
更多回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Symbolic Math Toolbox 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
