problem with symbolic factorization with two symbols
显示 更早的评论
The output is not a factorized one. I want this form : (s+s1)*(s+s2)*(s+s3). What is the problem?
syms s K
factor(s^3 + 10*s^2 + (21+K)*s + 4*K, [s, K],'FactorMode','full')
The output is
4*K + 21*s + K*s + 10*s^2 + s^3
1 个评论
Dyuman Joshi
2023-10-21
It is factorized according to the inputs given.
From the documentation - F = factor(x,vars) returns an array of factors F, where vars specifies the variables of interest. All factors not containing a variable in vars are separated into the first entry F(1). The other entries are irreducible factors of x that contain one or more variables from vars.
Note the last sentence.
What is the expected output from the vectorization?
采纳的回答
That you want to see a simple set of factors is not relevant. You essentially need to obtain the roots of this cubic polynomial in s, where K is a parameter, which would give you that factorization.
syms s K
sroots = solve(s^3 + 10*s^2 + (21+K)*s + 4*K,s,'maxdegree',3)
As you can see, they are fairly messy, and depending on the value of K, they may well be complex even if K is real. This is expected for any cubic polynomial, that we may find exactly two conjugate complex roots.
The factors now are just
prod(s-sroots)
We can expand that to show the result recovers your original polynomial.
simplify(expand(ans))
Finally, why did not factor understand what you wanted to see? You gave it two variables s and K. How should factor possibly know that you wanted it to do what I just did, essentially, that s is the variable of interest, and K a symbolic parameter in the problem? Software cannot read your mind, at least, not yet.
7 个评论
It also doesn't seem to work if you only specify s instead of [s,K] in
syms s K
factor(s^3 + 10*s^2 + (21+K)*s + 4*K, s,'FactorMode','full')
rustem
2023-10-21
I think the problem is, even though factor seems like it should solve this, it looks for only more obvious factors. Can factor handle a simpler problem? Well, yes.
syms x
factor(x^2 + 4*x +4,x)
How about this one? It handled that one.
syms K
factor(x^2 - 4*K*x + 4*K^2,x)
But what if I give it something slightly less obvious?
factor(x^2 + 4*K^2 - 4*K,x)
So factor was unable to solve even that one, even though we can write it simply enough in terms of sqrt(-4*K^2+4*K). Yet solve would handle it with not even a sweat.
Dyuman Joshi
2023-10-21
I feel that the question should be - Why is factor() not able to do so?
Maybe I should raise a query to TMW about this.
Factorization of expressions with real components (or components of unknown reality) traditionally only proceeds to the point where any further factoring would require use of imaginary components.
syms x
factor(x^2 + 1)
factor(x^2 - 1)
P1 = expand((x-2) * (x+7) * (x-7))
factor(P1)
P2 = expand((x-2) * (x+7i) * (x-7i))
factor(P2)
P3 = expand((x-2i) * (x+7) * (x-7))
factor(P3)
I cannot see a rule when "factor" works and when if does not work.
syms x
P2 = expand((x-2) * (x+7i) * (x-7i))
factor(P2,x,'FactorMode','complex')
更多回答（1 个）
Here are the factors:
syms s K
eqn = s^3 + 10*s^2 + (21+K)*s + 4*K == 0;
factors = solve(eqn,s,'MaxDegree',3)
2 个评论
Dyuman Joshi
2023-10-21
Shouldn't the factors be
s - roots_of_equation
Torsten
2023-10-21
Yes, I should have said: -s1, -s2 and -s3 from the factorization (s+s1)*(s+s2)*(s+s3).
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