I Derivation of second order Adams-Bashforth

[goggles31]

My notes state that the method is constructed based on the idea:
y_k+1=y_k+∫f(x,y)dx where the integral is taken from x_k to x_k+1
We can estimate the integral by considering
∫f(x)dx (from x_k to x_k+1) =c₀f_k+c₁f_k-1
To simplify the equation, we move x_k to the origin such that
∫f(x)dx (from 0 to h) =c₀f(0)+c₁f(-h)

Starting from below, I start to get confused.
It says "Replace f(x) with the polynomials, we have"
f(x)=1 : h=c₀(1)+c₁(1)
f(x)=x : h²/2=c₀(0)+c₁(-h)
Solving for c₀ and c₁,
c₀=3h/2 and c₁=-h/2 giving
y_k+1=y_k+h/2*(3f_k-f_k-1) + O(h³)

I've gone through the working and understand where the numbers come from, but I have no idea why they replace f(x) with 1 and x. Why not x² or x³? I've gone through some books but they derive this with the Taylor's series and I also understand that. I just don't understand the part I mentioned.

 

[eys_physics]

My notes state that the method is constructed based on the idea:
y_k+1=y_k+∫f(x,y)dx where the integral is taken from x_k to x_k+1
We can estimate the integral by considering
∫f(x)dx (from x_k to x_k+1) =c₀f_k+c₁f_k-1
To simplify the equation, we move x_k to the origin such that
∫f(x)dx (from 0 to h) =c₀f(0)+c₁f(-h)

Starting from below, I start to get confused.
It says "Replace f(x) with the polynomials, we have"
f(x)=1 : h=c₀(1)+c₁(1)
f(x)=x : h²/2=c₀(0)+c₁(-h)
Solving for c₀ and c₁,
c₀=3h/2 and c₁=-h/2 giving
y_k+1=y_k+h/2*(3f_k-f_k-1) + O(h³)

I've gone through the working and understand where the numbers come from, but I have no idea why they replace f(x) with 1 and x. Why not x² or x³? I've gone through some books but they derive this with the Taylor's series and I also understand that. I just don't understand the part I mentioned.

As I understand the point is to choose the constants $c_0$ and $c_1$ such that the integral $\int_0^h f(x) dx$ is exact if f(x) is a polynomial of degree $n\leq 1$, i.e. for all constants and linear polynomials. All such polynomials can be written in the form $p_1(x)=a\cdot 1+b\cdot x$, for some constants $a$ and $b$. We can thus consider $1$ and $x$ as basis functions. If we determine $c_0$ and $c_1$ such that $\int_0^h 1 dx$ and $\int_0^h x dx$ are reproduced exactly, $\int_0^h p_1(x) dx$ can be expressed in terms of the aforementioned integrals over $1$ and $x$ .

 

[FactChecker]

Those are the two simplest linearly independent examples of f(x) that the method needs to work for. So they are the easiest ones to use to see what c₀ and c₁ should be. It would be more complicated to use x² or x³. I haven't thought it through, but those may be so complicated that they won't give you the solution for c₀ and c₁, but I think they will work also.

PS. Even if x² and x³ work, they may give a different answer and not be the Adams-Bashforth method.

 

[FactChecker]

As I understand the point is to choose the constants $c_0$ and $c_1$ such that the integral $\int_0^h f(x) dx$ is exact if f(x) is a polynomial of degree $n\leq 1$, i.e. for all constants and linear polynomials. All such polynomials can be written in the form $p_1(x)=a\cdot 1+b\cdot x$, for some constants $a$ and $b$. We can thus consider $1$ and $x$ as basis functions. If we determine $c_0$ and $c_1$ such that $\int_0^h 1 dx$ and $\int_0^h x dx$ are reproduced exactly, $\int_0^h p_1(x) dx$ can be expressed in terms of the aforementioned integrals over $1$ and $x$ .

Yes. And it is wise to match the lower order polynomials before worrying about matching higher orders. So that makes good sense.