These can be obtained by other means [5], but using contour integrals shows that this is simply done by looking at the differential of \((z I – A)^{-1}\) and integrating it. Mathematics of Operations Research, 21(3):576–588, 1996. Thus holomorphic functions correspond to differentiable functions on \(\mathbb{R}^2\) with some equal partial derivatives. It consists in finding \(r\) pairs \((u_j,v_j) \in \mathbb{R}^{n} \times \mathbb{R}^d\), \(j=1,\dots,r\), of singular vectors and \(r\) positive singular values \(\sigma_1 \geqslant \cdots \geqslant \sigma_r > 0\) such that \(W = \sum_{j=1}^r \sigma_j u_j v_j^\top\) and \((u_1,\dots,u_r)\) and \((v_1,\dots,v_r)\) are orthonormal families. The Cauchy residue trick: spectral analysis made “easy”. 1 Residue theorem problems Important note. Complex analysis. Since the zeros of sinπz occur at the integers and are all simple zeros (see Example 1, Section 4.6), it follows that cscπz has simple poles at the integers. The result above can be naturally extended to vector-valued functions (and thus to any matrix-valued function), by applying the identity to all components of the vector. [u(x(t),y(t)) +i v(x(t),y(t))] [ x'(t) + i y'(t)] dt,$$ where \(x(t) = {\rm Re}(\gamma(t))\) and \(y(t) = {\rm Im}(\gamma(t))\). This leads to \(2i \pi\) times the sum of all residues of the function \(z \mapsto f(z) e^{ i \omega z}\) in the upper half plane. Vol. \end{array}\right.$$ This leads to $$\left\{ \begin{array}{l} \displaystyle \frac{\partial u}{\partial x}(x,y) = {\rm Re}(f'(z)) \\ \displaystyle \frac{\partial u}{\partial y}(x,y) = \ – {\rm Im}(f'(z)) \\ \displaystyle \frac{\partial v}{\partial x}(x,y) = {\rm Im}(f'(z)) \\ \displaystyle \frac{\partial v}{\partial y}(x,y) = {\rm Re}(f'(z)). Fourier transforms. Expert Answer The Cauchy Residue Theorem states as- Ifis analytic within a closed contour C except some finite number of poles at C view the full answer This in turn leads to the Cauchy-Riemann equations \(\displaystyle \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\) and \(\displaystyle \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}\), which are essentially necessary and sufficient conditions to be holomorphic. Because residues rely on the understanding of a host of topics such as the nature of the logarithmic function, integration in the complex plane, and Laurent series, it is recommended that you be familiar with all of these topics before proceeding. 1.The Cauchy-Goursat Theorem says that if a function is analytic on and in a closed contour C, then the integral over the closed ... that show how widely applicable the Residue Theorem is. \end{array}\right.$$. Wolfram Web Resources. Note that this result can be simply obtained by the simple (rough) calculation: if \(x\) is a unit eigenvector of \(A\), then \(Ax =\lambda x\), and \(x^\top x = 1\), leading to \(x^\top dx = 0\) and \(dA\ x + A dx = d\lambda \ x + \lambda dx\), and by taking the dot product with \(x\), \(d\lambda = x^\top dA\ x + x^\top A dx = x^\top dA \ x + \lambda x^\top dx = x^\top dA \ x\), which is the same result. The residue theorem is effectively a generalization of Cauchy's integral formula. If around λ, f(z) has a series expansions in powers of (z − λ), that is, f(z) = + ∞ ∑ k = − ∞ak(z − λ)k, then Res(f, λ) = a − 1. Many classical functions are holomorphic on \(\mathbb{C}\) or portions thereof, such as the exponential, sines, cosines and their hyperbolic counterparts, rational functions, portions of the logarithm. $$ This leads to, by contour integration:$$ \lambda_{k}(A+\Delta) -\lambda_k(A) = \frac{1}{2i \pi} \oint_\gamma \Big[ \sum_{j=1}^n \frac{ z \cdot u_j^\top \Delta u_j}{(z-\lambda_j)^2} \Big] dz + o(\| \Delta \|_2). If f(z) has a pole of order m at z = a, then the residue of f(z) at z = a is given by . Cauchy’s residue theorem is a consequence of Cauchy’s integral formula f(z. We consider a function which is holomorphic in a region of \(\mathbb{C}\) except in \(m\) values \(\lambda_1,\dots,\lambda_m \in \mathbb{C}\), which are usually referred to as poles. if m > 1. Walk through homework problems step-by-step from beginning to end. Matrix Perturbation Theory. Join the initiative for modernizing math education. Question on evaluating $\int_{C}\frac{e^{iz}}{z(z-\pi)}dz$ without the residue theorem. I have been working on machine learning since 2000, with a focus on algorithmic and theoretical contributions, in particular in optimization. Note that several eigenvalues may be summed up by selecting a contour englobing more than one eigenvalues. The Cauchy Residue Theorem Before we develop integration theory for general functions, we observe the following useful fact. Here I derive a perturbation result for the projector \(\Pi_k(A)=u_k u_k^\top\), when \(\lambda_k\) is a simple eigenvalue. Et quand le lien “expert” est T.Tao tout va bien , Your email address will not be published. $$ Taking the trace, the cross-product terms \({\rm tr}(u_j u_\ell^\top) = u_\ell^\top u_j\) disappear for \(j \neq \ell\), and we get: $$ {\rm tr} \big[ z (z I – A – \Delta)^{-1} \big] – {\rm tr} \big[ z (z I – A)^{-1} \big]= \sum_{j=1}^n \frac{ z \cdot u_j^\top \Delta u_j}{(z-\lambda_j)^2} + o(\| \Delta \|_2). $$ The dependence on \(z\) of the form \( \displaystyle \frac{1}{z- \lambda_j}\) leads to a nice application of Cauchy residue formula. Understanding how the spectral decomposition of a matrix changes as a function of a matrix is thus of primary importance, both algorithmically and theoretically. If a proof under general preconditions ais needed, it should be learned after studenrs get a good knowledge of topology. The Cauchy residue formula gives an explicit formula for the contour integral along γ: ∮γf(z)dz = 2iπ m ∑ j = 1Res(f, λj), where Res(f, λ) is called the residue of f at λ. This “shows” that the integral does not depend on the contour, and so in applications we can be quite liberal in the choice of contour. Using residue theorem to compute an integral. Here it is more direct to consider the so-called Jordan-Wielandt matrix, defined by blocks as $$ \bar{W} = \left( \begin{array}{cc}0 & W \\W^\top & 0 \end{array} \right). The central component is the following expansion, which is a classical result in matrix differentiable calculus, with \(\|\Delta\|_2\) the operator norm of \(\Delta\) (i.e., its largest singular value): $$ (z I- A – \Delta)^{-1} = (z I – A)^{-1} + (z I- A)^{-1} \Delta (z I- A)^{-1} + o(\| \Delta\|_2). Singular value decompositions are also often used, for a rectangular matrix \(W \in \mathbb{R}^{n \times d}\). Cauchy’s integral formula is worth repeating several times. We can then extend by \(1\)-periodicity to all \(x-y\). No dependence on the contour. Proposition 1.1. We first consider a contour integral over a contour \(\gamma\) enclosing a region \(\mathcal{D}\) where the function \(f\) is holomorphic everywhere. When we consider eigenvalues as functions of \(A\), we use the notation \(\lambda_j(A)\), \(j=1,\dots,n\). $$ The matrix \(\bar{W}\) is symmetric, and its non zero eigenvalues are \(+\sigma_i\) and \(-\sigma_i\), \(i=1,\dots,r\), associated with the eigenvectors \(\frac{1}{\sqrt{2}} \left( \begin{array}{cc}u_i \\ v_i \end{array} \right)\) and \(\frac{1}{\sqrt{2}} \left( \begin{array}{cc}u_i \\ -v_i \end{array} \right)\). Indeed, letting \(f(z) = \frac{1}{iz} Q\big( \frac{z+z^{-1}}{2}, \frac{z-z^{-1}}{2i} \big)\), it is exactly equal to the integral on the unit circle. Here is a very partial and non rigorous account (go to the experts for more rigor!). Do not simply evaluate the real integral – you must use complex methods. Definition Let f ∈ Cω(D\{a}) and a ∈ D with simply connected D ⊂ C with boundary γ. Define the residue of f at a as Res(f,a) := 1 2πi Z 9. See an example below related to kernel methods. 1. [12] Adrian S. Lewis, and Hristo S. Sendov. The desired integral is then equal to \(2i\pi\) times the sum of all residues of \(f\) within the unit disk. Note that this extends to piecewise smooth contours \(\gamma\). Springer Science & Business Media, 2011. Trigonometric integrals. The Residue Theorem has Cauchy’s Integral formula also as special case. Springer, 2013. These properties can be obtained from many angles, but a generic tool can be used for all of these: it is a surprising and elegant application of Cauchy’s residue formula, which is due to Kato [3]. A function \(f : \mathbb{C} \to \mathbb{C}\) is said holomorphic in \(\lambda \in \mathbb{C}\) with derivative \(f'(\lambda) \in \mathbb{C}\), if is differentiable in \(\lambda\), that is if \(\displaystyle \frac{f(z)-f(\lambda)}{z-\lambda}\) tends to \(f'(\lambda)\) when \(z\) tends to \(\lambda\). Complex Analysis, volume 103. Spectral functions are functions on symmetric matrices defined as \(F(A) = \sum_{k=1}^n f(\lambda_k(A))\), for any real-valued function \(f\). [1] Gilbert W. Stewart and Sun Ji-Huang. If the function \(f\) is holomorphic and has no poles at integer real values, and satisfies some basic boundedness conditions, then $$\sum_{n \in \mathbb{Z}} f(n) = \ – \!\!\! $$ The key benefit of these representations is that when the matrix \(A\) is slightly perturbed, then the same contour \(\gamma\) can be used to enclose the corresponding eigenvalues of the perturbed matrix, and perturbation results are simply obtained by taking gradients within the contour integral. This leads to, for \(x-y \in [0,1]\), \(K(x,y) = \frac{1}{2a} \frac{ \cosh (\frac{1-2(x-y)}{2a})}{\sinh (\frac{1}{2a})}\). SIAM Journal on Matrix Analysis and Applications 23.2: 368-386, 2001. I have just scratched the surface of spectral analysis, and what I presented extends to many interesting situations, for example, to more general linear operators in infinite-dimensional spaces [3], or to the analysis fo the eigenvalue distribution of random matrices (see a nice and reasonably simple derivation of the semi-circular law from Terry Tao’s blog). By expanding the product of complex numbers, it is thus equal to $$\int_0^1 [ u(x(t),y(t)) x'(t) \ – v(x(t),y(t))y'(t)] dt +i \int_0^1 [ v(x(t),y(t)) x'(t) +u (x(t),y(t))y'(t)] dt,$$ which we can rewrite in compact form as (with \(dx = x'(t) dt\) and \(dy = y'(t)dt\)): $$\oint_\gamma ( u \, dx\ – v \, dy ) + i \oint_\gamma ( v \, dx + u \, dy ).$$ We can then use Green’s theorem because our functions are differentiable on the entire region \(\mathcal{D}\) (the set “inside” the contour), to get $$\oint_\gamma ( u \, dx\ – v \, dy ) + i \oint_\gamma ( v \, dx + u \, dy ) =\ – \int\!\!\!\!\int_\mathcal{D} \! Indeed, we have: $$ \oint_\gamma (z I- A)^{-1} z dz = \sum_{j=1}^m \Big( \oint_\gamma \frac{z}{z – \lambda_j} dz \Big) u_j u_j^\top = 2 i \pi \lambda_k u_k u_k^\top, $$ and by taking the trace, we obtain $$ \oint_\gamma {\rm tr} \big[ z (z I- A)^{-1} \big] dz = \lambda_k. So, now we give it for all derivatives ( ) ( ) of . Cauchy Residue Formula. This can be done considering two contours \(\gamma_1\) and \(\gamma_2\) below with no poles inside, and thus with zero contour integrals, and for which the integrals along the added lines cancel. SEE: Residue Theorem. \int\!\!\!\!\int_\mathcal{D} \!\Big( \frac{\partial u}{\partial x} – \frac{\partial v}{\partial y} \Big) dx dy.$$ Thus, because of the Cauchy-Riemann equations, the contour integral is always zero within the domain of differentiability of \(f\). The contour \(\gamma\) is defined as a differentiable function \(\gamma: [0,1] \to \mathbb{C}\), and the integral is equal to $$\oint_\gamma f(z) dz = \int_0^1 \!\!f(\gamma(t)) \gamma'(t) dt = \int_0^1 \!\! All of my papers can be downloaded from my web page or my Google Scholar page. For example, for \(\alpha_0=1\) and \(\alpha_1=a^2\), we get for \(x-y>0\), one pole \(i/a\) in the upper half plane for the function \(\frac{1}{1+a^2 z^2} = \frac{1}{(1+iaz)(1-iaz)}\), with residue \(-\frac{i}{2a} e^{-(x-y)/a}\), leading to the familiar exponential kernel \(K(x,y) = \frac{1}{2a} e^{-|x-y|/a}\). Complex-valued functions on \(\mathbb{C}\) can be seen as functions from \(\mathbb{R}^2\) to itself, by writing $$ f(x+iy) = u(x,y) + i v(x,y),$$ where \(u\) and \(v\) are real-valued functions. (7.13) Note that we could have obtained the residue without partial fractioning by evaluating the coefficient of 1/(z −p) at z = p: 1 1−pz z=p = 1 1−p2. \sum_{ \lambda \in {\rm poles}(f)} {\rm Res}\big( f(z) \pi \frac{\cos \pi z}{\sin \pi z} ,\lambda\big).$$ This is a simple consequence of the fact that the function \(z \mapsto \pi \frac{\cos \pi z}{\sin \pi z}\) has all integers \(n \in \mathbb{Z}\) as poles, with corresponding residue equal to \(1\). This will allow us to compute the integrals in Examples 5.3.3-5.3.5 in … The Cauchy residue theorem can be used to compute integrals, by choosing the appropriate contour, looking for poles and computing the associated residues. Get the free "Residue Calculator" widget for your website, blog, Wordpress, Blogger, or iGoogle.