Expressing polynomial including a indefinite parameter

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Cagas Akalin
Cagas Akalin 2021-10-21
编辑： Matt J 2021-10-21
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

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Matt J
Matt J 2021-10-21
编辑：Matt J 2021-10-21
Ran in:
One way.
p = @(x,a) polyval([a,-4,2-a],x);
Usage:
x=1; a=2;
p(x,a)
ans = -2
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Matt J
Matt J 2021-10-21
编辑：Matt J 2021-10-21
Ran in:
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)
ans = 3
a=2; p(a)
ans = 0
a=0; p(a)
ans = -2

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更多回答（1 个）

Matt J
Matt J 2021-10-21
编辑：Matt J 2021-10-21
Ran in:
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 = 
p=subs(P,x,1)
p = 
coefficients = sym2poly(p)
coefficients = 1×2
1 -2
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Cagas Akalin
Cagas Akalin 2021-10-21
Dear Matt,
Since my question covers an different topic in the bigger picture I will pose it in an another question page.
Bests,
Cagdas
Matt J
Matt J 2021-10-21
编辑：Matt J 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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