Gauss quadrature approximation by Lanczos algorithm

一、 Gauss Quadrature
- 1.1 Gauss quadrature with weight function
- 1.2 The caculation of the weights and nodes
二、Lanczos algorithm
- 2.1 Krylov subspace
- 2.2 Arnoldi algorithm
- 2.3 Lanczos algorithm
三、Two important theory
- 3.1 Theorem 1
- 3.2 Theorem 2
四、Gauss quadrature by Lanczos algorithm

一、 Gauss Quadrature

1.1 Gauss quadrature with weight function

For
I(f)=∫abρ(x)f(x)dx(1)I(f)=\int_{a}^{b} \rho(x) f(x)dx \tag{1} I(f)=∫abρ(x)f(x)dx(1)
If the following quadrature formula[2,p220]
∫abρ(x)f(x)dx≈∑k=0nAkf(xk)(2)\int_{a}^{b}\rho(x)f(x)dx\approx \sum_{k=0}^{n}A_{k}f(x_{k}) \tag{2} ∫abρ(x)f(x)dx≈k=0∑nAkf(xk)(2)
with the (2n+1)(2n+1)(2n+1) algebraic degree of precision, then it’s called the Gauss-type quadrature formula with weights ρ(x)\rho(x)ρ(x).
The nodes xkx_{k}xk are called Gauss points (nodes), AkA_{k}Aks are called quadrature coefficients about ρ(x)\rho(x)ρ(x) (weights or coefficients).

1.2 The caculation of the weights and nodes

(Note: This section is mainly about how to caculate the Gauss points and coefficents. The theoretical foundation can refer to [2, p223]. )

Orthogonal polynomials method: For a given special weight function ρ(x)\rho(x)ρ(x), find the orthogonal polynomials related to the weight funciton ρ(x)\rho(x)ρ(x).
Let pn+1(x)p_{n+1}(x)pn+1(x) is exact degree nnn, suppose its zeros are t1,⋯,tnt_{1},\cdots,t_{n}t1,⋯,tn, then they are also the Gauss points{xk}k=1n\{x_{k}\}_{k=1}^{n}{xk}k=1n, their coresponding coefficients {Ak}k=1n\{A_{k}\}_{k=1}^{n}{Ak}k=1n are derived by
Ak=∫abρ(x)lk(x)dx,(3)A_{k}=\int_{a}^{b}\rho(x)l_{k}(x)dx, \tag{3}Ak=∫abρ(x)lk(x)dx,(3)
where lk(t)l_{k}(t)lk(t) are base function of interpolation
lk(t)=∏j=1,j≠knt−tjtk−tj.l_{k}(t)=\prod_{j=1,j\neq k}^{n}\frac{t-t_{j}}{t_{k}-t_{j}}.lk(t)=j=1,j=k∏ntk−tjt−tj.
Note: Using the zeros of orthogoanl polynomials to construct the Gauss quadrature formula, which is only effective for some special weight functions, such as 111(Legendre), 11−x2\frac{1}{\sqrt{1-x^{2}}}1−x21(Chebyshev 1), 1−x2\sqrt{1-x^{2}}1−x2(Chebyshev 2) and e−x2e^{-x^{2}}e−x2(Hermite), et.al.
However, for the general weight function, the following undertermined coefficient method is usually adopted.
Undeterminated coefficient method: As equation (2) is accurately established for {1,x,x2,⋯,x2n−1}\{1,x,x^{2},\cdots,x^{2n-1}\}{1,x,x2,⋯,x2n−1}.
Then we can construct the following (2n+2)(2n+2)(2n+2) equations about quadrature nodes {xk}k=0n\{x_{k}\}_{k=0}^{n}{xk}k=0n and quadrature coefficient {Ak}k=0n\{A_{k}\}_{k=0}^{n}{Ak}k=0n :
{∑k=0nAk=∫ab1dx∑k=0nxkAk=∫abxdx∑k=0nxk2Ak=∫abx2dx⋮∑k=0nxk2n+1Ak=∫abx2n+1dx\left\{\begin{array}{l}\sum_{k=0}^{n} A_{k}=\int_{a}^{b} 1 \mathrm{~d} x \\ \sum_{k=0}^{n} x_{k} A_{k}=\int_{a}^{b} x \mathrm{~d} x \\ \sum_{k=0}^{n} x_{k}^{2} A_{k}=\int_{a}^{b} x^{2} \mathrm{~d} x \\ \vdots \\ \sum_{k=0}^{n} x_{k}^{2 n+1} A_{k}=\int_{a}^{b} x^{2 n+1} \mathrm{~d} x\end{array}\right.⎩⎨⎧∑k=0nAk=∫ab1 dx∑k=0nxkAk=∫abx dx∑k=0nxk2Ak=∫abx2 dx⋮∑k=0nxk2n+1Ak=∫abx2n+1 dx

二、Lanczos algorithm

2.1 Krylov subspace

Krylov subspace: AAA is real matrix (unsymmetric) of order nnn. Let vvv be a given vector and
Kk=(v,Av,⋯,Ak−1v)K_{k}=\left(v, \quad A v, \quad \cdots, \quad A^{k-1} v\right)Kk=(v,Av,⋯,Ak−1v)
be the Krylov matrix of dimension n×kn\times kn×k. The subspace spanned by the columns of matrix KkK_{k}Kk is called a Krylov subspace and denoted by Kk(A,v)K_{k}(A,v)Kk(A,v) or K(A,v)K(A,v)K(A,v).

Note: The natural basis of the Krylov subspace K(A,v)K(A,v)K(A,v) given by the columns of the Krylov matirx KkK_{k}Kk is badly conditioned when kkk is large.

2.2 Arnoldi algorithm

Arnoldi algorithm constructs an orthonormal basis of the Krylov subspace K(A,v)K(A,v)K(A,v) by applying a variant of the Gram-Schmidt orthogonalization process.

Arnoldi algorithm: Set vectors v(j+1)=Av(j)v^{(j+1)}=Av^{(j)}v(j+1)=Av(j) with v(1)=vv^{(1)}=vv(1)=v, then K(A,v)K(A,v)K(A,v) is spanned by the vectors {v(j)}j=1k\{v^{(j)}\}_{j=1}^{k}{v(j)}j=1k.For constucting orthogonal basis vectors vjv^{j}vj, instead of orthogonalizing AjvA^{j}vAjv against the previous vectors, orthogonalize AvjAv^{j}Avj.
Starting from v1=vv^{1}=vv1=v(normalized),the (j+1)(j+1)(j+1)st vector of the basis is computed by using the previous vectors,
Projection: hi,j=(Avj,vi),i=1,⋯,j,h_{i,j}=(Av^{j},v^{i}),i=1,\cdots,j,hi,j=(Avj,vi),i=1,⋯,j,
Orthogonalize: v~j+1=Avj−∑i=1jhi,jvi,\tilde{v}^{j+1}=Av^{j}-\sum_{i=1}^{j}h_{i,j}v^{i},v~j+1=Avj−∑i=1jhi,jvi,
Length: hj+1,j=∥v~j+1∥h_{j+1,j}=\|\tilde{v}^{j+1}\|hj+1,j=∥v~j+1∥,(if hj+1,j=0h_{j+1,j}=0hj+1,j=0 then stop)
Normalize: vj+1=v~j+1hj+1,jv^{j+1}=\frac{\tilde{v}^{j+1}}{h_{j+1,j}}vj+1=hj+1,jv~j+1

As hj+1,jvj+1=Avj−∑i=1jhi,jvih_{j+1,j}v^{j+1}=Av^{j}-\sum_{i=1}^{j}h_{i,j}v^{i}hj+1,jvj+1=Avj−∑i=1jhi,jvi,
If we collect the vectors vj,j=1,⋯,kv^{j},j=1,\cdots,kvj,j=1,⋯,k in a matrix VkV_{k}Vk, that is Vk=[v1,v2,⋯,vk]V_{k}=[v_{1},v_{2},\cdots,v_{k}]Vk=[v1,v2,⋯,vk], the relations defining the vectors vk+1v^{k+1}vk+1 can be written in the matrix form as
AVk=VkHk+hk+1,kvk+1(ek)T,(4)A V_{k}=V_{k} H_{k}+h_{k+1, k} v^{k+1}\left(e^{k}\right)^{T}, \tag{4} AVk=VkHk+hk+1,kvk+1(ek)T,(4)
where HkH_{k}Hk is an upper Hessenberg matrix with elements hi,jh_{i,j}hi,j, hi,j=0,j=1,⋯,i−2,i>2h_{i,j}=0,j=1,\cdots,i-2,i>2hi,j=0,j=1,⋯,i−2,i>2.
Hk=[h1,1h1,2h1,3⋯h1,kh2,1h2,2h2,3⋯h2,k0h3,2h3,3⋯h3,k00h4,3⋯h4,k⋮⋮⋮⋱⋮000hk,k−1hk,k]H_{k}=\left[ \begin{array}{lllll} h_{1,1} & h_{1,2} & h_{1,3} & \cdots & h_{1,k}\\ h_{2,1} & h_{2,2} & h_{2,3} & \cdots & h_{2,k}\\ 0 & h_{3,2} & h_{3,3} & \cdots & h_{3,k}\\ 0 & 0 & h_{4,3} & \cdots & h_{4,k}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & h_{k,k-1} & h_{k,k}\\ \end{array} \right ] Hk=⎣⎡h1,1h2,100⋮0h1,2h2,2h3,20⋮0h1,3h2,3h3,3h4,3⋮0⋯⋯⋯⋯⋱hk,k−1h1,kh2,kh3,kh4,k⋮hk,k⎦⎤

2.3 Lanczos algorithm

Multiplying equation (4) by VkTV_{k}^{T}VkT and using orthogonality, we have
Hk=VkTAVk.H_{k}=V_{k}^{T}A V_{k}.Hk=VkTAVk.
If matrix AAA is symmetric, then the Hessenberg matrix HkH_{k}Hk is tridiagonal and denoted by JkJ_{k}Jk, that is Ji,j=0,j=i+2,⋯,k.J_{i,j}=0,j=i+2,\cdots,k.Ji,j=0,j=i+2,⋯,k. (or hi,j=0,i=1,⋯,j−2,j+2,⋯,kh_{i,j}=0,i=1,\cdots,j-2,j+2,\cdots,khi,j=0,i=1,⋯,j−2,j+2,⋯,k). This implies that the new vector vk+1v^{k+1}vk+1 can be computed by using only the two previous vkv^{k}vk and vk−1v^{k-1}vk−1.
hj+1,jvj+1=Avj−∑i=1jhi,jvi=Avj−hj−1,jvj−1−hj,jvj.h_{j+1,j}v^{j+1}=Av^{j}-\sum_{i=1}^{j}h_{i,j}v^{i}=Av^{j}-h_{j-1,j}v^{j-1}-h_{j,j}v^{j}.hj+1,jvj+1=Avj−i=1∑jhi,jvi=Avj−hj−1,jvj−1−hj,jvj.
In matrix form, set ηi\eta_{i}ηi denote the nonzero off-diagonal entries of JkJ_{k}Jk,
AVk=VkJk+ηkvk+1(ek)T.(5)A V_{k}=V_{k} J_{k}+\eta_{k} v^{k+1}\left(e^{k}\right)^{T}. \tag{5} AVk=VkJk+ηkvk+1(ek)T.(5)

Equation (5) describes in matrix form the elgant Lanczos algorithm. For simple notation system, starting from a nonzero vector v1=v/∥v∥,α1=(Av1,v1),v~2=Av1−α1v1v^{1}=v/\|v\|,\alpha_{1}=(Av^{1},v^{1}),\tilde{v}^{2}=Av^{1}-\alpha_{1}v^{1}v1=v/∥v∥,α1=(Av1,v1),v~2=Av1−α1v1, and then for k=2,3,⋯,k=2,3,\cdots,k=2,3,⋯,
ηk−1=∥v~k∥,vk=v~kηk−1,αk=(vk,Avk)=(vk)TAvk,v~k+1=Avk−αkvk−ηk−1vk−1.\begin{array}{c} \eta_{k-1}=\|\tilde{v}^{k}\|,\\ v^{k}=\frac{\tilde{v}^{k}}{\eta_{k-1}},\\ \alpha_{k}=\left(v^{k}, A v^{k}\right)=\left(v^{k}\right)^{T} A v^{k},\\ \tilde{v}^{k+1}=A v^{k}-\alpha_{k} v^{k}-\eta_{k-1} v^{k-1}.\\ \end{array}ηk−1=∥v~k∥,vk=ηk−1v~k,αk=(vk,Avk)=(vk)TAvk,v~k+1=Avk−αkvk−ηk−1vk−1.
Jk=[α1η10⋯0η1α2η2⋯00η2α3⋯000η3⋯0⋮⋮⋮⋱⋮000ηk−1αk]J_{k}=\left[ \begin{array}{lllll} \alpha_{1} & \eta_{1} & 0 & \cdots & 0\\ \eta_{1} & \alpha_{2} & \eta_{2} & \cdots & 0\\ 0 & \eta_{2} & \alpha_{3} & \cdots & 0\\ 0 & 0 & \eta_{3} & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \eta_{k-1} & \alpha_{k}\\ \end{array} \right ] Jk=⎣⎡α1η100⋮0η1α2η20⋮00η2α3η3⋮0⋯⋯⋯⋯⋱ηk−10000⋮αk⎦⎤

三、Two important theory

3.1 Theorem 1

Theorem1 [1,Th.4.1]: Let χk(λ)\chi_{k}(\lambda)χk(λ) be the determinant of Jk−λIJ_{k}-\lambda IJk−λI(which is a monic polynomial); then
vk=pk(A)v1,pk(λ)=(−1)k−1χk−1(λ)η1⋯ηk−1,k>1,p1≡1v^{k}=p_{k}(A) v^{1}, \quad p_{k}(\lambda)=(-1)^{k-1} \frac{\chi_{k-1}(\lambda)}{\eta_{1} \cdots \eta_{k-1}}, k>1, \quad p_{1} \equiv 1vk=pk(A)v1,pk(λ)=(−1)k−1η1⋯ηk−1χk−1(λ),k>1,p1≡1
The polynomials pkp_{k}pk of degree k−1k-1k−1 are called the normalized Lanczos polynomials.

This theorem describes the most important property of Lanczos algorithm:The Lanczos vectors vkv^{k}vk are given as a polynomial in matrix AAA applied to the initial vector v1v^{1}v1.

It is easy to verify that
det⁡(Jk+1)=αk+1det⁡(Jk)−ηkdet⁡(Jk−1),(6)\det(J_{k+1}) = \alpha_{k+1}\det(J_{k})-\eta_{k}\det(J_{k-1}), \tag{6} det(Jk+1)=αk+1det(Jk)−ηkdet(Jk−1),(6)
and
det⁡(Jk+1−λI)=(αk+1−λ)det⁡(Jk−λI)−ηkdet⁡(Jk−1−λI).(7)\det(J_{k+1}-\lambda I) = (\alpha_{k+1}-\lambda)\det(J_{k}-\lambda I)-\eta_{k}\det(J_{k-1}-\lambda I). \tag{7} det(Jk+1−λI)=(αk+1−λ)det(Jk−λI)−ηkdet(Jk−1−λI).(7)
From the expression of polynomial pk(λ)p_{k}(\lambda)pk(λ) in Theorem 1, we know that the Lanczos polynomials satisfy a scalar three-term recurrence,
ηkpk+1(λ)=(λ−αk)pk(λ)−ηk−1pk−1(λ),k=1,2,…,(8)\eta_{k} p_{k+1}(\lambda)=\left(\lambda-\alpha_{k}\right) p_{k}(\lambda)-\eta_{k-1} p_{k-1}(\lambda), k=1,2, \ldots, \tag{8} ηkpk+1(λ)=(λ−αk)pk(λ)−ηk−1pk−1(λ),k=1,2,…,(8)
with initial conditions p0≡0,p1≡1p_{0} \equiv 0, p_{1} \equiv 1p0≡0,p1≡1.

3.2 Theorem 2

Theorem2 [1, Th.4.2]:
Consider the Lanczos vectors vkv^{k}vk. There exists a measure α(λ)\alpha(\lambda)α(λ) such that
(vk,vl)=⟨pk,pl⟩=∫abpk(λ)pl(λ)dα(λ),(9)\left(v^{k}, v^{l}\right)=\left\langle p_{k}, p_{l}\right\rangle=\int_{a}^{b} p_{k}(\lambda) p_{l}(\lambda) d \alpha(\lambda), \tag{9} (vk,vl)=⟨pk,pl⟩=∫abpk(λ)pl(λ)dα(λ),(9)
where α≤λ1=λmin\alpha\leq \lambda_{1}=\lambda_{min}α≤λ1=λmin and b≥λn=λmaxb\geq \lambda_{n}=\lambda_{max}b≥λn=λmax, λmin\lambda_{min}λmin and λmax\lambda_{max}λmax being the smallest and largest eigenvalues of AAA, and pip_{i}pi are the Lanczos polynomials associated with AAA and v1v^{1}v1.
α(λ)={0,if λ<λ1∑j=1i[v^j]2,if λi≤λ<λi+1∑j=1n[v^j]2,if λn≤λ(10)\alpha(\lambda)=\left\{\begin{array}{ll}0, & \text { if } \lambda<\lambda_{1} \\ \sum_{j=1}^{i}\left[\hat{v}_{j}\right]^{2}, & \text { if } \lambda_{i} \leq \lambda<\lambda_{i+1} \\ \sum_{j=1}^{n}\left[\hat{v}_{j}\right]^{2}, & \text { if } \lambda_{n} \leq \lambda\end{array}\right. \tag{10} α(λ)=⎩⎨⎧0,∑j=1i[v^j]2,∑j=1n[v^j]2, if λ<λ1 if λi≤λ<λi+1 if λn≤λ(10)

Note: For the sake of simplicity, here suppose that the eigenvalues of AAA are distinct.

四、Gauss quadrature by Lanczos algorithm

For symetric matrix AAA with eigenvalue decomposition A=QΛQTA=Q\Lambda Q^{T}A=QΛQT, fff is smooth funciton, vvv is random unit vector, then
vTf(A)v=vTQf(Λ)QTvv^{T}f(A)v=v^{T}Qf(\Lambda)Q^{T}vvTf(A)v=vTQf(Λ)QTv
set v^=QTv\hat{v}=Q^{T}vv^=QTv, then
vTf(A)v=v^Tf(Λ)v=∑j=1nf(λi)[v^j]2v^{T}f(A)v=\hat{v}^{T}f(\Lambda)v=\sum_{j=1}^{n}f(\lambda_{i})[\hat{v}_{j}]^{2}vTf(A)v=v^Tf(Λ)v=j=1∑nf(λi)[v^j]2
Consider the above sum as Riemann-Stieltjes integral
∑j=1nf(λi)[v^j]2=∫abf(t)dα(t).(11)\sum_{j=1}^{n}f(\lambda_{i})[\hat{v}_{j}]^{2}=\int_{a}^{b}f(t)d\alpha(t). \tag{11} j=1∑nf(λi)[v^j]2=∫abf(t)dα(t).(11)
where the measure α(t)\alpha(t)α(t) is defined as:
α(t)={0,if t<λ1=a∑j=1i[v^j]2,if λi≤t<λi+1∑j=1n[v^j]2,if b=λn≤t.\alpha(t)=\left\{\begin{array}{ll}0, & \text { if } t<\lambda_{1}=a \\ \sum_{j=1}^{i}\left[\hat{v}_{j}\right]^{2}, & \text { if } \lambda_{i} \leq t<\lambda_{i+1} \\ \sum_{j=1}^{n}\left[\hat{v}_{j}\right]^{2}, & \text { if } b=\lambda_{n} \leq t.\end{array}\right. α(t)=⎩⎨⎧0,∑j=1i[v^j]2,∑j=1n[v^j]2, if t<λ1=a if λi≤t<λi+1 if b=λn≤t.

The integral in Equation (11) can be estimated using the Gauss quadrature,
∫abf(t)dα(t)≈∑k=1nAkf(xk).(12)\int_{a}^{b}f(t)d\alpha(t)\approx \sum_{k=1}^{n}A_{k}f(x_{k}).\tag{12} ∫abf(t)dα(t)≈k=1∑nAkf(xk).(12)
As stated in the Theorem2, {pk}k=1n\{p_{k}\}_{k=1}^{n}{pk}k=1n are orthonormal polynomial and the measure funciton is α(t)\alpha(t)α(t), the same with Equation (10).
We can rewirte the three-term recurrence in Equation (8) in matrix form. Let p(λ)=[p1(λ),p2(λ),⋯,pk(λ)]T\mathbf{p}(\lambda)=[p_{1}(\lambda),p_{2}(\lambda),\cdots,p_{k}(\lambda)]^{T}p(λ)=[p1(λ),p2(λ),⋯,pk(λ)]T and p0(λ)=0p_{0}(\lambda)=0p0(λ)=0, then
λp(λ)=p(λ)Jk+ηkpk+1(λ)ekT.(13)\lambda \mathbf{p}(\lambda)=\mathbf{p}(\lambda) J_{k}+\eta_{k}p_{k+1}(\lambda)e_{k}^{T}. \tag{13} λp(λ)=p(λ)Jk+ηkpk+1(λ)ekT.(13)
Then the zeros of pk+1(λ)p_{k+1}(\lambda)pk+1(λ) are precisely the eigenvalues of JkJ_{k}Jk.

Note:pk+1(λ)p_{k+1}(\lambda)pk+1(λ) is polynomial of degree k.

So, the Gauss point prolbem in Equation(12) can be get by the eigenvalues of JkJ_{k}Jk which derived by Lanczos algorithm for matrix AAA.

References:
[1] Golub, G. H. and Meurant, G. Matrices, Moments and Quadrature with Applications, Princeton University Press, 2010.
[2] 孙志忠，袁慰平，闻震初. 数值分析(第3版)，东南大学出版社.

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Gauss quadrature approximation by Lanczos algorithm

Gauss quadrature approximation by Lanczos algorithm

一、 Gauss Quadrature

1.1 Gauss quadrature with weight function

1.2 The caculation of the weights and nodes

二、Lanczos algorithm

2.1 Krylov subspace

2.2 Arnoldi algorithm

2.3 Lanczos algorithm

三、Two important theory

3.1 Theorem 1

3.2 Theorem 2

四、Gauss quadrature by Lanczos algorithm

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