留学生作业代写 Quadrature

Quadrature

Interpolation and Gaussian quadrature¶
Polynomial interpolation is the process of finding a polynomial that equals data at a precise set of points.

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Quadrature is the act of approximating an integral by a weighted sum:
\int_a^b f(x) w(x) {\rm d}x ≈ \sum_{j=1}^n w_j f(x_j).
In these notes we see that the two concepts are intrinsically linked: interpolation leads naturally
to quadrature rules. Moreover, by using a set of points $x_j$ linked to orthogonal polynomials we get
significantly more accurate rules, and in fact can use quadrature to compute expansions in orthogonal polynomials that
interpolate. That is, we can mirror the link between the Trapezium rule, Fourier series, and interpolation but
now for polynomials.

Polynomial Interpolation: we describe how to interpolate a function by a polynomial.
Truncated Jacobi matrices: we see that truncated Jacobi matrices are diagonalisable
in terms of orthogonal polynomials and their zeros.
Interpolatory quadrature rule: polynomial interpolation leads naturally to ways to integrate
functions, but onely realisable in the simplest cases.
Gaussian quadrature: Using roots of orthogonal polynomials and truncated Jacobi matrices
leads naturally to an efficiently
computable interpolatory quadrature rule. The miracle is its exact for twice as many polynomials as

1. Polynomial Interpolation¶
We already saw a special case of polynomial interpolation, where we saw that the polynomial
f(z) ≈ ∑_{k=0}^{n-1} f̂_k^n z^k
equaled $f$ at evenly spaced points on the unit circle: ${\rm e}^{{\rm i} 2π j/n}$.
But here we consider the following:

Definition (interpolatory polynomial) Given $n$ distinct points $x_1,…,x_n ∈ ℝ$
and $n$ samples $f_1,…,f_n ∈ ℝ$, a degree $n-1$
interpolatory polynomial $p(x)$ satisfies
p(x_j) = f_j

The easiest way to solve this problem is to invert the Vandermonde system:

Definition (Vandermonde) The Vandermonde matrix associated with $n$ distinct points $x_1,…,x_n ∈ ℝ$
is the matrix
V := \begin{bmatrix} 1 & x_1 & ⋯ & x_1^{n-1} \\
⋮ & ⋮ & ⋱ & ⋮ \\
1 & x_n & ⋯ & x_n^{n-1}
\end{bmatrix}

Proposition (interpolatory polynomial uniqueness)
The interpolatory polynomial is unique, and the Vandermonde matrix is invertible.

Suppose $p$ and $p̃$ are both interpolatory polynomials. Then $p(x) – p̃(x)$ vanishes at $n$ distinct points $x_j$. By the fundamental theorem of
algebra it must be zero, i.e., $p = p̃$.

For the second part, if $V 𝐜 = 0$ for $𝐜 ∈ ℝ$ then for $q(x) = c_1 + ⋯ + c_n x^{n-1}$ we have
q(x_j) = 𝐞_j^⊤ V 𝐜 = 0
hence $q$ vanishes at $n$ distinct points and is therefore 0, i.e., $𝐜 = 0$.

Thus a quick-and-dirty way to to do interpolation is to invert the Vandermonde matrix
(which we saw in the least squares setting with more samples then coefficients):

using Plots, LinearAlgebra
f = x -> cos(10x)

x = range(0, 1; length=n)# evenly spaced points (BAD for interpolation)
V = x .^ (0:n-1)’ # Vandermonde matrix
c = V \ f.(x) # coefficients of interpolatory polynomial
p = x -> dot(c, x .^ (0:n-1))

g = range(0,1; length=1000) # plotting grid
plot(g, f.(g); label=”function”)
plot!(g, p.(g); label=”interpolation”)
scatter!(x, f.(x); label=”samples”)

But it turns out we can also construct the interpolatory polynomial directly.
We will use the following with equal $1$ at one grid point
and are zero at the others:

Definition (Lagrange basis polynomial) The Lagrange basis polynomial is defined as
ℓ_k(x) := ∏_{j ≠ k} {x-x_j \over x_k – x_j} = {(x-x_1) ⋯(x-x_{k-1})(x-x_{k+1}) ⋯ (x-x_n) \over (x_k – x_1) ⋯ (x_k – x_{k-1}) (x_k – x_{k+1}) ⋯ (x_k – x_n)}

Plugging in the grid points verifies the following:

Proposition (delta interpolation)
ℓ_k(x_j) = δ_{kj}

We can use these to construct the interpolatory polynomial:

Theorem (Lagrange interpolation)
The unique polynomial of degree at most $n-1$ that interpolates $f$ at $x_j$ is:
p(x) = f(x_1) ℓ_1(x) + ⋯ + f(x_n) ℓ_n(x)

p(x_j) = ∑_{k=1}^n f(x_k) ℓ_k(x_j) = f(x_k)
so we just need to show it is unique. Suppose $p̃(x)$ is a polynomial
of degree at most $n-1$
that also interpolates $f$.

Example We can interpolate $\exp(x)$ at the points $0,1,2$:
\begin{align*}
p(x) &= ℓ_1(x) + {\rm e} ℓ_2(x) + {\rm e}^2 ℓ_3(x) =
{(x – 1) (x-2) \over (-1)(-2)} + {\rm e} {x (x-2) \over (-1)} +
{\rm e}^2 {x (x-1) \over 2} \
&= (1/2 – {\rm e} +{\rm e}^2/2)x^2

(-3/2 + 2 {\rm e} – {\rm e}^2 /2) x + 1
\end{align*}

Remark Interpolating at evenly spaced points is a really bad idea:
interpolation is inheritely ill-conditioned.
The problem sheet asks you to explore
this experimentally.

2. Roots of orthogonal polynomials and truncated Jacobi matrices¶
We now consider roots (zeros) of orthogonal polynomials $q_n(x)$. This is important as we shall
see they are useful for interpolation and quadrature. For interpolation to be well-defined we
first need to guarantee that the roots are distinct.

Lemma $q_n(x)$ has exactly $n$ distinct roots.

Suppose $x_1, …,x_j$ are the roots where $q_n(x)$ changes sign, that is,
q_n(x) = c_j (x-x_j) + O((x-x_j)^2)
for $c_j ≠ 0$. Then
q_n(x) (x-x_1) ⋯(x-x_j)
does not change sign.
In other words:
⟨q_n,(x-x_1) ⋯(x-x_j) ⟩ = \int_a^b q_n(x) (x-x_1) ⋯(x-x_j) w(x) {\rm d} x ≠ 0.
This is only possible if $j = n$ as $q_n(x)$ is orthogonal w.r.t. all lower degree
polynomials.

Definition (truncated Jacobi matrix) Given a symmetric Jacobi matrix $X$,
(or the weight $w(x)$ whose orthonormal polynomials are associated with $X$)
the truncated Jacobi matrix is
X_n := \begin{bmatrix} a_0 & b_0 \\
b_0 & ⋱ & ⋱ \\
& ⋱ & a_{n-2} & b_{n-2} \\
&& b_{n-2} & a_{n-1} \end{bmatrix} ∈ ℝ^{n × n}

Lemma (zeros) The zeros $x_1, …,x_n$ of $q_n(x)$ are the eigenvalues of the truncated Jacobi matrix $X_n$.
More precisely,
X_n Q_n = Q_n \begin{bmatrix} x_1 \\ & ⋱ \\ && x_n \end{bmatrix}
for the orthogonal matrix
Q_n = \begin{bmatrix}
q_0(x_1) & ⋯ & q_0(x_n) \\
⋮ & ⋯ & ⋮ \\
q_{n-1}(x_1) & ⋯ & q_{n-1}(x_n)
\end{bmatrix} \begin{bmatrix} α_1^{-1} \\ & ⋱ \\ && α_n^{-1} \end{bmatrix}
where $α_k = \sqrt{q_0(x_k)^2 + ⋯ + q_{n-1}(x_k)^2}$.

We construct the eigenvector (noting $b_{n-1} p_n(x_j) = 0$):
X_n \begin{bmatrix} p_0(x_j) \\ ⋮ \\ p_{n-1}(x_j) \end{bmatrix} =
\begin{bmatrix} a_0 p_0(x_j) + b_0 p_1(x_j) \\
b_0 p_0(x_j) + a_1 p_1(x_j) + b_1 p_2(x_j) \\
b_{n-3} p_{n-3}(x_j) + a_{n-2} p_{n-2}(x_j) + b_{n-2} p_{n-1}(x_j) \\
b_{n-2} p_{n-2}(x_j) + a_{n-1} p_{n-1}(x_j) + b_{n-1} p_n(x_j)
\end{bmatrix} = x_j \begin{bmatrix} p_0(x_j) \\
p_1(x_j) \\
\end{bmatrix}
The result follows from normalising the eigenvectors. Since $X_n$ is symmetric
the eigenvector matrix is orthogonal.

3. Interpolatory quadrature rules¶
Definition (interpolatory quadrature rule) Given a set of points $x_1,…,x_n$
the interpolatory quadrature rule is:
Σ_n^{\rm i}[f] := ∑_{k=1}^n w_k f(x_k)
w_k := ∫_a^b ℓ_k(x) w(x) {\rm d} x

Proposition (interpolatory quadrature is exact for polynomials)
Interpolatory quadrature is exact for all degree $n-1$ polynomials $p$:
∫_a^b p(x) w(x) {\rm d}x = Σ_n^{\rm i}[f]

The result follows since, by uniqueness of interpolatory polynomial:
p(x) = ∑_{k=1}^n p(x_k) ℓ_k(x)

4. Gaussian quadrature¶
Gaussian quadrature is the interpolatory quadrature rule corresponding
to the grid $x_k$ defined as the roots of the orthogonal polynomial $q_n(x)$.
We shall see that it is exact for polynomials up to degree $2n-1$, i.e., double
the degree of other interpolatory quadrature rules from other grids.

Definition (Gauss quadrature) Given a weight $w(x)$, the Gauss quadrature rule is:
∫_a^b f(x)w(x) {\rm d}x ≈ \underbrace{∑_{k=1}^n w_k f(x_k)}_{Σ_n^w[f]}
where $x_1,…,x_n$ are the eigenvalues of $X_n$ and
w_k = Q_n[k,1]^2 = {1 \over α_k^2}

In analogy to how Fourier series are orthogonal with respect to Trapezium rule,
Orthogonal polynomials are orthogonal with respect to Gaussian quadrature:

Lemma (Discrete orthogonality)
For $0 ≤ ℓ,m ≤ n-1$,
Σ_n^w[q_ℓ q_m] = δ_{ℓm}

Σ_n^w[q_ℓ q_m] = ∑_{k=1}^n {q_ℓ(x_k) q_m(x_k) \over α_k^2}
= \left[q_ℓ(x_1)/ α_1 | ⋯ | {q_ℓ(x_n)/ α_n}\right]
\begin{bmatrix}
q_m(x_1)/α_1 \\
q_m(x_n)/α_n \end{bmatrix} = 𝐞_ℓ Q_n Q_n^⊤ 𝐞_m = δ_{ℓm}

Just as approximating Fourier coefficients using Trapezium rule gives a way of
interpolating at the grid, so does Gaussian quadrature:

Theorem (interpolation via quadrature)
f_n(x) = ∑_{k=0}^{n-1} c_k^n q_k(x)\hbox{ for } c_k^n := Σ_n^w[f q_k]
interpolates $f(x)$ at the Gaussian quadrature points $x_1,…,x_n$.

Note that we can write:
\begin{bmatrix}
c_{n-1}^n \end{bmatrix} = Q_n^w \begin{bmatrix}
f_n \end{bmatrix}.

Consider the Vandermonde-like matrix:
Ṽ = \begin{bmatrix} q_0(x_1) & ⋯ & q_{n-1}(x_1) \\
⋮ & ⋱ & ⋮ \\
q_0(x_n) & ⋯ & q_{n-1}(x_n) \end{bmatrix}
Note that if $p(x) = [q_0(x) | ⋯ | q_{n-1}(x)] 𝐜$ then
\begin{bmatrix}
\end{bmatrix} = Ṽ 𝐜
But we see that (similar to the Fourier case)
Q_n^w Ṽ = \begin{bmatrix} Σ_n^w[q_0 q_0] & ⋯ & Σ_n^w[q_0 q_{n-1}]\\
⋮ & ⋱ & ⋮ \\
Σ_n^w[q_{n-1} q_0] & ⋯ & Σ_n^w[q_{n-1} q_{n-1}]
\end{bmatrix} = I_n

Corollary Gaussian quadrature is an interpolatory quadrature rule.

A consequence of being an interpolatory quadrature rule is that it is exact for all
polynomials of degree $n-1$. The miracle of Gaussian quadrature is it is exact for twice

Theorem (Exactness of Gauss quadrature) If $p(x)$ is a degree $2n-1$ polynomial then
Gauss quadrature is exact:
∫_a^b p(x)w(x) {\rm d}x = Σ_n^w[p].

Using polynomial division algorithm (e.g. by matching terms) we can write
p(x) = q_n(x) s(x) + r(x)
where $s$ and $r$ are degree $n-1$. Then we have:
\begin{align*}
Σ_n^w[p] &= \underbrace{Σ_n^w[q_n s]}_{\hbox{$0$ since evaluating $q_n$ at zeros}} + Σ_n^w[r] = ∫_a^b r(x) w(x) {\rm d}x
= \underbrace{∫_a^b q_n(x)s(x) w(x) {\rm d}x}_{\hbox{$0$ since $s$ is degree $CS代考 加微信: powcoder QQ: 1823890830 Email: powcoder@163.com