Expressing polynomial including a indefinite parameter

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

Answer 1

[Matt J]

One way.

p = @(x,a) polyval([a,-4,2-a],x);

Usage:

x=1; a=2;

p(x,a)

ans = -2

[Cagas Akalin]

Hi Matt,

Thank you for your answer. But what I want to do is to keep the "a" as indefinite parameter. I do not want to evaluate the value of the polynomial for any value of "a".

Actually my final goal is to evaluate "a" as p(1) = a*x^2 - 4x + 2-a = 1 @ x = 1

Bests,

Cagdas

[Matt J]

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

ans = -5

p0(10)%a=10,x=0

ans = -8

[Matt J]

Comment by Cagas Akalin moved here

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  

[Matt J]

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

[Cagas Akalin]

Thank you Matt,

It works.

Bests,

Cagdas

Answer 2

[Matt J]

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

[Cagas Akalin]

Yes, what I was trying to implement was exactly that you guessed.

[Cagas Akalin]

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]

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.