Expressing polynomial including a indefinite parameter
显示 更早的评论
Hi All,
I know that the expression p=[1 -4 2] corresponds to p(x) = x^2 - 4x + 2.
I want to express p(x) = a*x^2 - 4x + 2-a
I would appreciate it if you told me how to do that.
Bests,
Cagdas
采纳的回答
One way.
p = @(x,a) polyval([a,-4,2-a],x);
Usage:
x=1; a=2;
p(x,a)
5 个评论
Cagas Akalin
2021-10-21
编辑：Cagas Akalin
2021-10-21
Maybe you mean,
p = @(x) @(a) polyval([a,-4,2-a],x);
This means p(x) will return a function of a for fixed x. Example,
p0=p(0); %returns a function of a at x=0
p0(7) %a=7, x=0
p0(10)%a=10,x=0
Matt J
2021-10-21
编辑：Walter Roberson
2021-10-21
I think I made a confision but let me correct myself and ask my question as follows.
I understood that I can express a polynomial either by
p = @(x,a) polyval([a,-4,2-a],x);
or by
p = @(x) @(a) polyval([a,-4,2-a],x);
But what I am aiming to do is to give it a value only for x which returns a polynomial including the constant(indefinite) "a". Namely,
Let's assume my polynomial is
p(x,a) = 2*a*x^2 - 4x + 2-a
and my input is x = 1
and this returns
p(1,a) = 2*a*1^2 - 4*1 + 2-a = 2*a - 4 + 2 - a = a - 2
But what I am aiming to do is to give it a value only for x which returns a polynomial including the constant(indefinite) "a".
I maintain that that is what is achieved by doing as follows (updated for your new example).
Pxa = @(x) @(a) polyval([2*a,-4,2-a],x);
Now if we invoke px() at x=1,
p=Pxa(1); %x=1
then I claim that p(a) is the polynomial function a-2. We can check various inputs a to verify that this is true:
a=5; p(a)
a=2; p(a)
a=0; p(a)
Cagas Akalin
2021-10-21
更多回答（1 个）
Possibly you mean that you want a coefficient vector as the output, rather than a polynomial in functional form.
syms a x
P = 2*a*x^2 - 4*x + 2-a
p=subs(P,x,1)
coefficients = sym2poly(p)
3 个评论
Cagas Akalin
2021-10-21
Cagas Akalin
2021-10-21
Yes, what I was trying to implement was exactly that you guessed.
I'm glad we converged, but please Accept-click the answer to indicate that your question was resolved.
Since my question covers an different topic in the bigger picture I will pose it in an another question page.
Yes, by all means.
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