Answer:Newton-Cotes Formula: Proving $\omega_j=\omega_{n-j}$ & $(b-a)$ Sum

[akerman]

I have been given this question and I have no idea how to answer it. I know that the answer will contain two small proofs where one of them uses quadrature formulae.
So I have been ask to show Show it holds that $\omega_j=\omega_{n-j}$ and that $\sum_j{\omega_j}=(b-a)$. Knowing that Newton cotes formulae is in the equi spaced points $x_i=a+ih$ with $h=(b-a)/n$ and $i=0,\dots,n$

Any idea how to properly solve this?

 

[akerman]

First proof should consider odd functions in the interval from -1 to 1, and I would like to show that ωj=ωn−j The second one should consider the quadrature formulae (constant funciton 1) based on that I would like to prove ∑jωj=(b−a) where ω is the weights. However I don't know how...

 

[I like Serena]

I haven't answered before because I don't quite know what to make of your question.

Which formulae do you mean?
What do you mean by $\omega_j$?

 

[akerman]

So in this context ωj are weights, actually I think its called a quadrature weight. So I would like to show that for corresponding weights it is true or it holds that ωj = ωj-1. I am sure for this proof odd functions need to be considered in the interval of -1 to 1. Does that explain it any better?

 

[akerman]

I haven't answered before because I don't quite know what to make of your question.

Which formulae do you mean?
What do you mean by $\omega_j$?

We interpolate $f$ using a Lagrange interpolation polynomial of the form

$$
p_n(x)=\sum_{k=0}^nL_k(x)f(x_k).
$$

We obtain

$$
\int_a^bf(x)dx\approx \int_a^b\sum_{k=0}^nL_k(x)f(x_k)=\sum_{k=0}^nf(x_k)\int_a^bL_k(x)dx:=\sum_{k=0}^nf(x_k)\omega_k,
$$
where the $\omega_k:=\int_a^bL_k(x)dx$ are called integration weights.

So based on that I would like to prove that $ω_j=ω_{n−j}$ and the second proof should be proving that $\sum\limits_j ω_j=(b−a)$.

I like Serena do you know how it can be achieved?

 

[I like Serena]

We interpolate $f$ using a Lagrange interpolation polynomial of the form

$$
p_n(x)=\sum_{k=0}^nL_k(x)f(x_k).
$$

We obtain

$$
\int_a^bf(x)dx\approx \int_a^b\sum_{k=0}^nL_k(x)f(x_k)=\sum_{k=0}^nf(x_k)\int_a^bL_k(x)dx:=\sum_{k=0}^nf(x_k)\omega_k,
$$
where the $\omega_k:=\int_a^bL_k(x)dx$ are called integration weights.

So based on that I would like to prove that $ω_j=ω_{n−j}$ and the second proof should be proving that $\sum\limits_j ω_j=(b−a)$.

I like Serena do you know how it can be achieved?

The Lagrange polynomials $L_k(x)$ are completely independent of $f(x)$.
Since the interval has been split in equal subintervals, the construction of $L_k(x)$ is completely symmetric.
Therefore $L_k(x) = L_{n-k}(x)$, which in turn implies that $ω_j=ω_{n−j}$.

 

[akerman]

The Lagrange polynomials $L_k(x)$ are completely independent of $f(x)$.
Since the interval has been split in equal subintervals, the construction of $L_k(x)$ is completely symmetric.
Therefore $L_k(x) = L_{n-k}(x)$, which in turn implies that $ω_j=ω_{n−j}$.

OK that makes sense.
What about the second proof which is ∑jωj=(b−a). I need to prove it considering the quadrature formulae (constant function 1)

 

[I like Serena]

OK that makes sense.
What about the second proof which is ∑jωj=(b−a). I need to prove it considering the quadrature formulae (constant function 1)

If we fill in $f(x)=1$, we get in the left hand side:
$$\int_a^b 1\ dx = b-a$$
and on the right hand side:
$$\sum_{k=0}^n 1 \cdot \omega_k = \sum_j \omega_j$$

In other words:
$$b-a \approx \sum_j \omega_j$$

Furthermore, with $f(x)=1$ the Lagrange interpolation is a perfect interpolation.
That is:
$$f(x) = p_n(x)$$
So the approximation will actually be an equality.

So:
$$b-a = \sum_j \omega_j$$
which concludes the proof.