Is there any way to express an expression in terms of other expressions in MATLAB?
For example, the following expressions have been written as sum (X + Y) and product (XY)
1/X + 1/Y = (X + Y)/XY
1/X^2 + 1/Y^2 + 2/(XY) = (X + Y)^2/(XY)
2*X/Y + 2*Y/X = 2*((X + Y)^2 - 2*X*Y)/(XY)
I know about the rewrite() function but I couldn't find how it can be used to do what I want?
There are a few different functions you can try to change the format of your symbolic expression:
collect: collects coefficients (can specify an expression to collect powers of):
>> collect(1/X + 1/Y)
ans =
(X + Y)/(Y*X)
simplify: perform algebraic simplification:
>> simplify(1/X^2 + 1/Y^2 + 2/(X*Y))
ans =
(X + Y)^2/(X^2*Y^2)
numden: convert to a rational form, with a numerator and denominator:
>> [n, d] = numden(2*X/Y + 2*Y/X)
n =
2*X^2 + 2*Y^2
d =
X*Y
>> n/d
ans =
(2*X^2 + 2*Y^2)/(X*Y)
Related
I am solving a fourth order equation in matlab using the solve function.My script looks like this:
syms m M I L Bp Bc g x
m = 0.127
M = 1.206
I = 0.001
L = 0.178
Bc = 5.4
Bp = 0.002
g = 9.8
eqn = ((m + M)*(I + m*L^2) - m^2*L^2)*x^4 + ((m + M)*Bp + (I + m*L^2)*Bc)*x^3 + ((m + M)*m*g*L + Bc*Bp)*x^2 + m*g*L*Bc*x == 0
S = solve(eqn, x)
In the answer, I should get 4 roots, but instead I get such strange expressions:
S =
0
root(z^3 + (34351166180215288*z^2)/7131724328013535 + (352922208800606144*z)/7131724328013535 + 1379250971773894912/7131724328013535, z, 1)
root(z^3 + (34351166180215288*z^2)/7131724328013535 + (352922208800606144*z)/7131724328013535 + 1379250971773894912/7131724328013535, z, 2)
root(z^3 + (34351166180215288*z^2)/7131724328013535 + (352922208800606144*z)/7131724328013535 + 1379250971773894912/7131724328013535, z, 3)
The first root, which is 0, is displayed clearly. Is it possible to make the other three roots appear as numbers as well? I looked for something about this in the documentation for the solve function, but did not find it.
a(x) here is a type of equation used in signal
Let
y(x) = x^3 + x^2 + x + 1
a(x) is inputed by the user so for example if user inputs
a(x) = D2 + D + 6
y(x)*a(x) = D2(x^3 + x^2 + x + 1) + D(x^3 + x^2 + x + 1) + 6(x^3 + x^2 + x + 1)
here D2(x^3 + x^2 + x + 1) = 6x + 2 and D(x^3 + x^2 + x + 1) = 3x^2 + 2x + 1
So D is differentiation and D2 is double differential
So i want to know how i would do something like this on MATLAB
Use Matlab symbolic computation toolbox https://www.mathworks.com/products/symbolic.html
Calculus
Evaluate exact analytical solutions for definite or indefinite integral,
calculate derivatives of symbolic expressions or functions,
and approximate functions using series expansions.
How does one extract specific parts of an expression in Matlab/ Octave symbolic package? In XCAS, one can use indexing expressions, but I can't find anything similar in Octave/ Matlab.
For instance, with X = C*L*s**2 + C*R*s + 1, is there a way to get C*R*s by X(2) or the like?
It would be nice to do this with factors too. X = (alpha + s)*(beta**2 + s**2)*(C*R*s + 1), and have X(2) give (beta**2 + s**2).
Thanks!
children (MATLAB doc, Octave doc) does this but the order in which you write the expressions will not necessarily be the same. The order is also different in MATLAB and Octave.
Expanded Expression:
syms R L C s;
X1 = C*L*s^2 + C*R*s + 1;
partsX1 = children(X1);
In MATLAB:
>> X1
X1 =
C*L*s^2 + C*R*s + 1
>> partsX1
partsX1 =
[ C*R*s, C*L*s^2, 1]
In Octave:
octave:1> X1
X1 = (sym)
2
C⋅L⋅s + C⋅R⋅s + 1
octave:2> partsX1
partsX1 = (sym 1×3 matrix)
⎡ 2 ⎤
⎣1 C⋅L⋅s C⋅R⋅s⎦
Factorised Expression:
syms R C a beta s; %alpha is also a MATLAB function so don't shadow it with your variable
X2 = (a + s) * (beta^2 + s^2) * (C*R*s + 1);
partsX2 = children(X2);
In MATLAB:
>> X2
X2 =
(a + s)*(C*R*s + 1)*(beta^2 + s^2)
>> partsX2
partsX2 =
[ a + s, C*R*s + 1, beta^2 + s^2]
In Octave:
octave:3> X2
X2 = (sym)
⎛ 2 2⎞
(a + s)⋅⎝β + s ⎠⋅(C⋅R⋅s + 1)
octave:4> partsX2
partsX2 = (sym 1×3 matrix)
⎡ 2 2⎤
⎣C⋅R⋅s + 1 a + s β + s ⎦
I need to rewrite a symbolic expression in terms of a specific subexpression.
Consider the following scenario:
expression f with 2 variables a, b
subexpression c = a / b
syms a b c
f = b / (a + b) % = 1 / (1 + a/b) = 1 / (1 + c) <- what I need
Is there a way to achieve this?
Edit:
The step from 1 / (1 + a/b) to 1 / (1 + c) can be achieved by calling
subs(1 / (1 + a/b),a/b,c)
So a better formulated question is:
Is there a way to tell MATLAB to 'simplify' b / (a + b) into 1 / (1 + a/b)?
Just calling simplify(b / (a + b) makes no difference.
Simplification to your desired form is not automatically guaranteed, and in my experience, isn't likely to be achieved directly through simplify-ing as I've noticed simplification rules prefer rational polynomial functions. However,
if you know the proper reducing ratio, you can substitute and simplify
>> syms a b c
>> f = b / (a + b);
>> simplify(subs(f,a,c*b))
ans =
1/(c + 1)
>> simplify(subs(f,b,a/c))
ans =
1/(c + 1)
And then re-substitute without simplification, if desired:
>> subs(simplify(subs(f,a,c*b)),c,a/b)
ans =
1/(a/b + 1)
>> subs(simplify(subs(f,b,a/c)),c,a/b)
ans =
1/(a/b + 1)
I think I missed reading a theorem or postulate or something..
The equation is: wy + wxz + xyz
According to my professor, the simplification is (which she didn't explain how):
wy + xz(w'y + wy')
= wy + xz (w XOR y)
Where did that (w'y + wy') came from??
I tried to calculate it but so far I only got: (w+x)(w+y)(w+z)(x+y)(y+z)
In a Boolean expression + is XOR and * is AND. This is the same as identifying true with 1 and false with 0, with the only special convention that 1 + 1 = 0 (or 2 = 0 if you wish.)
With these definitions both + and * are commutative, i.e., a + b = b + a and a * b = b * a. In addition, the distributive law is also valid a * (b + c) = a * b + a * c. Note also that the * operator is usually implicit in the sense that we write ab instead of a * b.
Applying these properties to the expression wy + wxz + xyz, there are some few obvious transformations you can do:
wy + wxz + yxz (commute x with y)
wy + (w + y)xz (prev plus distribute xz)
wy + xz(w + y) (prev plus commute (w + y) with xz
Note that the last one is wy + xz(w XOR y) because + is nothing but XOR.
ADDENDUM
Regarding the expression of your professor, let's recall that a' = 1 - a by definition. So
w'y + wy' = (1 - w)y + w(1 - y) - def
= y - wy + w - wy - distribute
= y + w - simplify (a + a = 0 always)
= w + y - commute
which shows that s/he was right.