Derivatives of spectral functions. 4.Use the residue theorem to compute Z C g(z)dz. Vivek R. Cauchy's integral formula helps you to determine the value of a function at a point inside a simple closed curve, if the function is analytic at all points inside and on the curve. If a proof under general preconditions ais needed, it should be learned after studenrs get a good knowledge of topology. 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 $$K(x,y) = \frac{1}{2a} e^{-|x-y|/a}$$. Note that similar constructions can be used to take into account several poles. Join the initiative for modernizing math education. See the detailed computation at the end of the post. sur les intégrales définies, prises entre des limites imaginaires, Polynomial magic III : Hermite polynomials, The many faces of integration by parts – II : Randomized smoothing and score functions, The many faces of integration by parts – I : Abel transformation. This representation can be used to compute derivatives of $$F$$, by simple derivations, to obtain the same result as . Cauchy’s Residue Theorem Dan Sloughter Furman University Mathematics 39 May 24, 2004 45.1 Cauchy’s residue theorem The following result, Cauchy’s residue theorem, follows from our previous work on integrals. • The residue theorem relates a contour integral around some of a function's poles to the sum of their residuals \sum_{ \lambda \in {\rm poles}(f)} {\rm Res}\big( f(z) \pi \frac{1}{\sin \pi z} ,\lambda\big).\) See [7, Section 11.2] for more details. Before going to the spectral analysis of matrices, let us explore some cool choices of contours and integrands, and (again!) The desired integral is then equal to $$2i\pi$$ times the sum of all residues of $$f$$ within the unit disk. Academic Press, 1990. for the cauchy’s integration theorem proved with them to be used for the proof of other theorems of complex analysis (for example, residue theorem.) Now that you are all experts in residue calculus, we can move on to spectral analysis. I have been working on machine learning since 2000, with a focus on algorithmic and theoretical contributions, in particular in optimization. This will allow us to compute the integrals in Examples 5.3.3-5.3.5 in … If the function $$f$$ is holomorphic and has no poles at integer real values, and satisfies some basic boundedness conditions, then $$\sum_{n \in \mathbb{Z}} f(n) = \ – \!\!\! X is holomorphic, and z0 2 U, then the function g(z)=f (z)/(z z0) is holomorphic on U \{z0},soforanysimple closed curve in U enclosing z0 the Residue Theorem gives 1 2⇡i ‰ f (z) z z0 dz = 1 2⇡i ‰ g(z) dz = Res(g, z0)I (,z0); Required fields are marked *. 5.Combine the previous steps to deduce the value of the integral we want. Explanation of Cauchy residue formula If around λ, f(z) has a series expansions in powers of (z − λ), that is, f(z) = + ∞ ∑ k = − ∞ak(z − λ)k, then Res(f, λ) = a − 1. (1) seems unsettling. SEE: Residue Theorem. Spline models for observational data. I have just scratched the surface of spectral analysis, and what I presented extends to many interesting situations, for example, to more general linear operators in infinite-dimensional spaces , or to the analysis fo the eigenvalue distribution of random matrices (see a nice and reasonably simple derivation of the semi-circular law from Terry Tao’s blog).  Grace Wahba. Matrix Perturbation Theory.  Jan R. Magnus. Because residues rely on the understanding of a host of topics such as the nature of the logarithmic function, integration in the complex plane, and Laurent series, it is recommended that you be familiar with all of these topics before proceeding. Cauchy's Residue Theorem is as follows: Let be a simple closed contour, described positively.$$ Note here that the asymptotic remainder $$o(\| \Delta\|_2)$$ can be made explicit. Explore anything with the first computational knowledge engine. The contour $$\gamma$$ is defined as a differentiable function $$\gamma: [0,1] \to \mathbb{C}$$, and the integral is equal to $$\oint_\gamma f(z) dz = \int_0^1 \!\!f(\gamma(t)) \gamma'(t) dt = \int_0^1 \!\! 0) = 1 2ˇi I. C. f(z) z z. The first pivotal theorem proved by Cauchy, now known as Cauchy's integral theorem , was the following: where f ( z ) is a complex-valued function analytic on and within the non-self-intersecting closed curve C (contour) lying in the complex plane . Let Note. We thus obtain an expression for projectors on the one-dimensional eigen-subspace associated with the eigenvalue $$\lambda_k$$. Cauchy’s integral formula is worth repeating several times. Et quand le lien “expert” est T.Tao tout va bien , Your email address will not be published. The key property that we will use below is that we can express the so-called resolvent matrix $$(z I – A)^{-1} \in \mathbb{C}^{n \times n}$$, for $$z \in \mathbb{C}$$, as:$$ (z I- A)^{-1} = \sum_{j=1}^n \frac{1}{z-\lambda_j} u_j u_j^\top. f(x) e^{ i \omega x} dx\) for holomorphic functions $$f$$ by integrating on the real line and a big upper circle as shown below, with $$R$$ tending to infinity (so that the contribution of the half-circle goes to zero because of the exponential term). With simple manipulations, we can also access the eigenvalues. Here are classical examples, before I show applications to kernel methods. The Cauchy residue formula gives an explicit formula for the contour integral along $$\gamma$$: $$\oint_\gamma f(z) dz = 2 i \pi \sum_{j=1}^m {\rm Res}(f,\lambda_j), \tag{1}$$ where $${\rm Res}(f,\lambda)$$ is called the residue of $$f$$ at $$\lambda$$ . In an upcoming topic we will formulate the Cauchy residue theorem. At first, the formula in Eq. Section 5.1 Cauchy’s Residue Theorem 103 Coeﬃcient of 1 z: a−1 = 1 5!,so Z C1(0) sinz z6 dz =2πiRes(0) = 2πi 5!. Here it is more direct to consider the so-called Jordan-Wielandt matrix, defined by blocks as $$\bar{W} = \left( \begin{array}{cc}0 & W \\W^\top & 0 \end{array} \right). Indeed, letting $$f(z) = \frac{1}{iz} Q\big( \frac{z+z^{-1}}{2}, \frac{z-z^{-1}}{2i} \big)$$, it is exactly equal to the integral on the unit circle. Assuming the $$k$$-th eigenvalue $$\lambda_k$$ is simple, we consider the contour $$\gamma$$ going strictly around $$\lambda_k$$ like below (for $$k=5$$). A classical question is: given the norm defined above, how to compute $$K$$? Theorem 4.5. The Cauchy residue formula gives an explicit formula for the contour integral along γ: ∮γf(z)dz = 2iπ m ∑ j = 1Res(f, λj), where Res(f, λ) is called the residue of f at λ. Suppose C is a positively oriented, simple closed contour. \int\!\!\!\!\int_\mathcal{D} \!\Big( \frac{\partial u}{\partial x} – \frac{\partial v}{\partial y} \Big) dx dy.$$ Thus, because of the Cauchy-Riemann equations, the contour integral is always zero within the domain of differentiability of $$f$$. 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. We first consider a contour integral over a contour $$\gamma$$ enclosing a region $$\mathcal{D}$$ where the function $$f$$ is holomorphic everywhere. Vol. Hot Network Questions 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): $$(z I- A – \Delta)^{-1} = (z I – A)^{-1} + (z I- A)^{-1} \Delta (z I- A)^{-1} + o(\| \Delta\|_2). For more details on complex analysis, see . All possible errors are my faults. Complex variables and applications.Boston, MA: McGraw-Hill Higher Education.  Alain Berlinet, and Christine Thomas-Agnan.$$ 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 $$\frac{1}{\sqrt{2}} \left( \begin{array}{cc}u_i \\ -v_i \end{array} \right)$$. An analytic function whose Laurent series is given by(1)can be integrated term by term using a closed contour encircling ,(2)(3)The Cauchy integral theorem requires thatthe first and last terms vanish, so we have(4)where is the complex residue. By expanding the expression on the basis of eigenvectors of $$A$$, we get $$z (z I- A – \Delta)^{-1} – z (z I- A)^{-1} = \sum_{j=1}^n \sum_{\ell=1}^n u_j u_\ell^\top \frac{ z \cdot u_j^\top \Delta u_\ell}{(z-\lambda_j)(z-\lambda_\ell)} + o(\| \Delta \|_2). The #1 tool for creating Demonstrations and anything technical. Suppose that f(z) has an isolated singularity at z0 and f(z) = X∞ k=−∞ ak(z − z0)k is its Laurent expansion in a deleted neighbourhood of z0. We thus need a perturbation analysis or more generally some differentiability properties for eigenvalues or eigenvectors , or any spectral function . I am Francis Bach, a researcher at INRIA in the Computer Science department of Ecole Normale Supérieure, in Paris, France. Deﬁnition Let f ∈ Cω(D\{a}) and a ∈ D with simply connected D ⊂ C with boundary γ. Deﬁne the residue of f at a as Res(f,a) := 1 2πi Z In non-parametric estimation, regularization penalties are used to constrain real-values functions to be smooth. Indeed, we have:$$ \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. Complex Analysis, volume 103. New York: Springer, 2010. 4. (a) The Order of a pole of csc(πz)= 1sin πz is the order of the zero of 1 csc(πz)= sinπz. if m =1, and by . $$The dependence on $$z$$ of the form $$\displaystyle \frac{1}{z- \lambda_j}$$ leads to a nice application of Cauchy residue formula. For more mathematical details see Cauchy's integral formula and residue theorem. Proof. When f : U ! Note that this result can be simply obtained by the simple (rough) calculation: if $$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$$, which is the same result. No dependence on the contour. The following theorem gives a simple procedure for the calculation of residues at poles. REFERENCES: Knopp, K. "The Residue Theorem." For functions $$f$$ defined on an interval $$I$$ of the real line, penalties are typically of the form $$\int_I \sum_{k=0}^s \alpha_k | f^{(k)}(x)|^2 dx$$, for non-negative weights $$\alpha_0,\dots,\alpha_k$$. Theorem 9.1. Understanding how the spectral decomposition of a matrix changes as a function of a matrix is thus of primary importance, both algorithmically and theoretically. Springer, 2013. By expanding the product of complex numbers, it is thus equal to$$\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} \! $$This leads to, by contour integration:$$ \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). The Cauchy residue theorem can be used to compute integrals, by choosing the appropriate contour, looking for poles and computing the associated residues. However, this reasoning is more cumbersome, and does not lead to neat approximation guarantees, in particular in the extensions below. Your email address will not be published. Wolfram Web Resources.  Francis Bach. Journal of Machine Learning Research, 9:1019-1048, 2008. (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. Spectral functions are functions on symmetric matrices defined as $$F(A) = \sum_{k=1}^n f(\lambda_k(A))$$, for any real-valued function $$f$$. Springer, 2013.  Tosio Kato. On differentiating eigenvalues and eigenvectors. 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. This leads to, for $$x-y \in [0,1]$$, $$K(x,y) = \frac{1}{2a} \frac{ \cosh (\frac{1-2(x-y)}{2a})}{\sinh (\frac{1}{2a})}$$. Trigonometric integrals. Using residue theorem to compute an integral. This will include the formula for functions as a special case. We consider the function $$f(z) = \frac{e^{i\pi (2q-1) z}}{1+(2a \pi z)^2} \frac{\pi}{\sin (\pi z)}.$$ It is holomorphic on $$\mathbb{C}$$ except at all integers $$n \in \mathbb{Z}$$, where it has a simple pole with residue $$\displaystyle \frac{e^{i\pi (2q-1) n}}{1+(2a \pi n)^2} (-1)^n = \frac{e^{i\pi 2q n}}{1+(2a \pi n)^2}$$, at $$z = i/(2a\pi)$$ where it has a residue equal to $$\displaystyle \frac{e^{ – (2q-1)/(2a)}}{4ia\pi} \frac{\pi}{\sin (i/(2a))} = \ – \frac{e^{ – (2q-1)/(2a)}}{4a} \frac{1}{\sinh (1/(2a))}$$, and at $$z = -i/(2a\pi)$$ where it has a residue equal to $$\displaystyle \frac{e^{ (2q-1)/(2a)}}{4ia\pi} \frac{\pi}{\sin (i/(2a))} =\ – \frac{e^{ (2q-1)/(2a)}}{4a} \frac{1}{\sinh (1/(2a))}$$. 4.3 Cauchy’s integral formula for derivatives.  Gilbert W. Stewart and Sun Ji-Huang. Hints help you try the next step on your own. Cauchy's Residue Theorem contradiction? The function $$F$$ can be represented as $$F(A) = \sum_{k=1}^n f(\lambda_k(A)) = \frac{1}{2i \pi} \oint_\gamma f(z) {\rm tr} \big[ (z I – A)^{-1} \big] dz,$$ where the contour $$\gamma$$ encloses all eigenvalues (as shown below). Get the free "Residue Calculator" widget for your website, blog, Wordpress, Blogger, or iGoogle.  Joseph Bak, Donald J. Newman. Experts will see an interesting link with the Euler-MacLaurin formula and Bernoulli polynomials. (7.14) This observation is generalized in the following. This result is due to Cauchy  in 1825. Reproducing kernel Hilbert spaces in probability and statistics. In this post, I have shown various applications of the Cauchy residue formula, for computing integrals and for the spectral analysis of matrices. Society for Industrial and Applied Mathematics, 1990. The Residue Theorem has Cauchy’s Integral formula also as special case. Q(\cos \theta, \sin \theta) d\theta\). Singular value decompositions are also often used, for a rectangular matrix $$W \in \mathbb{R}^{n \times d}$$. One such examples are combinations of squared $$L_2$$ norms of derivatives. We can also consider the same penalty on the unit interval $$[0,1]$$ with periodic functions, leading to the kernel (see  for more details): $$K(x,y) = \sum_{n \in \mathbb{Z}} \frac{ e^{2in\pi(x-y)}}{\sum_{k=0}^s \alpha_k( 2n\pi)^s}.$$ For the same example as above, that is, $$\alpha_0=1$$ and $$\alpha_1=a^2$$, this leads to an infinite series on which we can apply the Cauchy residue formula as explained above. 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. 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. Ca permet de se décrasser les neurones. 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. Do not simply evaluate the real integral – you must use complex methods. Applications to kernel methods. We can then extend by $$1$$-periodicity to all $$x-y$$.  Serge Lang. 0. dz; where fis an analytic function and Cis a simple closed contour in the complex plane enclosing the point z. SEE ALSO: Cauchy Integral Formula, Cauchy Integral Theorem, Complex Residue, Contour, Contour Integral, Contour Integration, Group Residue Theorem, Laurent Series, Pole. When we consider eigenvalues as functions of $$A$$, we use the notation $$\lambda_j(A)$$, $$j=1,\dots,n$$. For example, if $$\displaystyle f(z) = \frac{g(z)}{z-\lambda}$$ with $$g$$ holomorphic around $$\lambda$$, then $${\rm Res}(f,\lambda) = g(\lambda)$$, and more generally, if $$\displaystyle f(z) = \frac{g(z)}{(z-\lambda)^k}$$ for $$k \geqslant 1$$, then $$\displaystyle {\rm Res}(f,\lambda) = \frac{g^{(k-1)}(\lambda) }{(k-1)!}$$. Before diving into spectral analysis, I will first present the Cauchy residue theorem and some nice applications in computing integrals that are needed in machine learning and kernel methods. For $$I = \mathbb{R}$$, then this can be done using Fourier transforms as: $$K(x,y) = \frac{1}{2\pi} \int_\mathbb{R} \frac{e^{i\omega(x-y)}}{\sum_{k=0}^s \alpha_k \omega^{2k}} d\omega.$$ This is exactly an integral of the form above, for which we can use the contour integration technique. Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. 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 . This is obtained from the contour below with $$m$$ tending to infinity. It is the consequence of the fact that the value of the function can determined by … We also consider a simple closed directed contour $$\gamma$$ in $$\mathbb{C}$$ that goes strictly around the $$m$$ values above. If a function is analytic inside except for a finite number of singular points inside , then Brown, J. W., & Churchill, R. V. (2009). Then if C is 2. Thus holomorphic functions correspond to differentiable functions on $$\mathbb{R}^2$$ with some equal partial derivatives. For holomorphic functions $$Q$$, we can compute the integral $$\displaystyle \int_0^{2\pi} \!\!\! [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))$$. Since the zeros of sinπz occur at the integers and are all simple zeros (see Example 1, Section 4.6), it follows that cscπz has simple poles at the integers. Find out information about Cauchy residue formula. Fourier transforms. With all residues summing to zero (note that this fact requires a precise analysis of limits when $$m$$ tends to infinity for the contour defined in the main text), we get: $$\sum_{n \in \mathbb{Z}} \frac{e^{2i\pi q \cdot n}}{1+(2a \pi n)^2} =\frac{e^{ – (2q-1)/(2a)}}{4a} \frac{1}{\sinh (1/(2a))}+ \frac{e^{ (2q-1)/(2a)}}{4a} \frac{1}{\sinh (1/(2a))} = \frac{1}{2a} \frac{ \cosh (\frac{2q-1}{2a})}{\sinh (\frac{1}{2a})}.$$, Excellent billet ! See more examples in http://residuetheorem.com/, and many in . So, now we give it for all derivatives ( ) ( ) of . 9. \Big( \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y} \Big) dx dy \ – i \!\! The result above can be naturally extended to vector-valued functions (and thus to any matrix-valued function), by applying the identity to all components of the vector. Formula 6) can be considered a special case of 7) if we define 0! This can be done considering two contours $$\gamma_1$$ and $$\gamma_2$$ below with no poles inside, and thus with zero contour integrals, and for which the integrals along the added lines cancel. SIAM Journal on Matrix Analysis and Applications 23.2: 368-386, 2001.  Adrian Stephen Lewis. 9.2 Integrals of functions that decay The theorems in this section will guide us in choosing the closed contour Cdescribed in the introduction. \end{array}\right.$$This leads to$$\left\{ \begin{array}{l} \displaystyle \frac{\partial u}{\partial x}(x,y) = {\rm Re}(f'(z)) \\ \displaystyle \frac{\partial u}{\partial y}(x,y) = \ – {\rm Im}(f'(z)) \\ \displaystyle \frac{\partial v}{\partial x}(x,y) = {\rm Im}(f'(z)) \\ \displaystyle \frac{\partial v}{\partial y}(x,y) = {\rm Re}(f'(z)). To state the Residue Theorem we rst need to understand isolated singularities of holomorphic functions and quantities called winding numbers. derive the Residue Theorem for meromorphic functions from the Cauchy Integral Formula. $$By keeping only the pole $$\lambda_k$$ which is inside the contour $$\gamma$$, we get$$ \lambda_{k}(A+\Delta) -\lambda_k(A) = \frac{1}{2i \pi} \oint_\gamma \Big[ \frac{ z \cdot u_k^\top \Delta u_k}{(z-\lambda_k)^2} \Big] dz + o(| \Delta |_2) \ = u_k^\top \Delta u_k + o(\| \Delta \|_2), $$using the identity $$\displaystyle \oint_\gamma \frac{z dz}{(z – \lambda_k)^2} dz = \oint_\gamma \Big( \frac{\lambda_k}{(z – \lambda_k)^2} + \frac{1}{z – \lambda_k} \Big) dz = 1$$. See an example below related to kernel methods. Contour integral with no poles. Residue theorem. Why doesn’t the result depend more explicitly on the contour $$\gamma$$? Just diﬀerentiate Cauchy’s integral formula n times. Given the gradient, other more classical perturbation results could de derived, such as Hessians of eigenvalues, or gradient of the projectors $$u_k u_k^\top$$. The Cauchy residue theoremgeneralizes both the Cauchy integral theorem(because analytic functionshave no poles) and the Cauchy integral formula(because f⁢(x)/(x-a)nfor analytic fhas exactly one pole at x=awith residue Res(f(x)/(x-a)n,a)=f(n)(a)/n!$$ Taking the trace, the cross-product terms $${\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). If you are already familiar with complex residues, you can skip the next section. Here is a very partial and non rigorous account (go to the experts for more rigor!). Econometric Theory, 1(2):179–191, 1985. Now that the Cauchy formula is true for the circle around a single pole, we can “deform” the contour below to a circle. Proposition 1.1. 1 Residue theorem problems 2 2 Zero Sum theorem for residues problems 76 3 Power series problems 157 Acknowledgement.The following problems were solved using my own procedure in a program Maple V, release 5. Important note. The rst theorem is for functions that decay faster than 1=z. 1. Question on evaluating \int_{C}\frac{e^{iz}}{z(z-\pi)}dz without the residue theorem. \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$$. Knowledge-based programming for everyone. Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science The Residue Theorem. The residue theorem is effectively a generalization of Cauchy's integral formula. There are two natural ways to relate the singular value decomposition to the classical eigenvalue decomposition of a symmetric matrix, first through $$WW^\top$$ (or similarly $$W^\top W$$). This in turn leads to the Cauchy-Riemann equations $$\displaystyle \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$$ and $$\displaystyle \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$$, which are essentially necessary and sufficient conditions to be holomorphic. some positive definite kernels. So, now we give it for all derivatives f(n)(z) of f. This will include the formula for functions as a special case. A function $$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$$. Theorem 2. Find more Mathematics widgets in Wolfram|Alpha. Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. We consider integrating the matrix above, which leads to: $$\oint_\gamma (z I- A)^{-1} dz = \sum_{j=1}^m \Big( \oint_\gamma \frac{1}{z – \lambda_j} dz \Big) u_j u_j^\top = 2 i \pi \ u_k u_k^\top$$ using the identity $$\displaystyle \oint_\gamma \frac{1}{z – \lambda_j} dz = 1$$ if $$j=k$$ and $$0$$ otherwise (because the pole is outside of $$\gamma$$). Springer Science & Business Media, 2011. \end{array}\right.$$. These equations are key to obtaining the Cauchy residue formula. The Cauchy method of residues: theory and applications. Cauchy Residue Formula. In many areas of machine learning, statistics and signal processing, eigenvalue decompositions are commonly used, e.g., in principal component analysis, spectral clustering, convergence analysis of Markov chains, convergence analysis of optimization algorithms, low-rank inducing regularizers, community detection, seriation, etc. Looking for Cauchy residue formula? For $$\omega>0$$, we can compute $$\displaystyle \int_{-\infty}^\infty \!\! we have from the residue theorem I = 2πi 1 i 1 1−p2 = 2π 1−p2. Springer Science & Business Media, 1984. 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. Practice online or make a printable study sheet. The Cauchy Residue Theorem Before we develop integration theory for general functions, we observe the following useful fact. Cauchy’s integral formula for derivatives. If f(z) has a pole of order m at z = a, then the residue of f(z) at z = a is given by . The expression with contour integrals allows to derive simple formulas for gradients of eigenvalues. If ( ) and satisfy the same hypotheses Expanding \(f(z+dz) = f(z) + f'(z) dz$$, which is the definition of complex differentiability, into real and imaginary parts, we get (using $$i^2 = -1$$):$$\left\{ \begin{array}{l} u(x+dx,y+dy) = u(x,y) + {\rm Re}(f'(z)) dx\ – {\rm Im}(f'(z)) dy \\ v(x+dx,y+dy) = v(x,y) + {\rm Re}(f'(z)) dy + {\rm Im}(f'(z)) dx. It will turn out that $$A = f_1 (2i)$$ and $$B = f_2(-2i)$$. 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.  Augustin Louis Cauchy, Mémoire sur les intégrales définies, prises entre des limites imaginaires, 1825, re-published in Bulletin des Sciences Mathématiques et Astronomiques, Tome 7, 265-304, 1874. The original paper where this is presented is a nice read in French where you can find some pepits like “la fonction s’évanouit pour $$x = \infty$$”. 1. f(z) z 2 dz+ Z. C. 2. f(z) z 2 dz= 2ˇif(2) 2ˇif(2) = 4ˇif(2): 4.3 Cauchy’s integral formula for derivatives. It follows that f ∈ Cω(D) is arbitrary often diﬀerentiable. Theorem 45.1. Using the same technique as above, we get: $$\Pi_k(A+\Delta )\ – \Pi_k(A) = \frac{1}{2i \pi} \oint_\gamma (z I- A)^{-1} \Delta (z I – A)^{-1}dz + o(\| \Delta\|_2),$$ which we can expand to the basis of eigenvectors as $$\frac{1}{2i \pi} \oint_\gamma \sum_{j=1}^n \sum_{\ell=1}^n u_j u_j^\top \Delta u_\ell u_\ell^\top \frac{ dz}{(z-\lambda_\ell) (z-\lambda_j) } + o(\| \Delta\|_2).$$ We can then split in two, with the two terms (all others are equal to zero by lack of poles within $$\gamma$$): $$\frac{1}{2i \pi} \oint_\gamma \sum_{j \neq k} u_j^\top \Delta u_k ( u_j u_k^\top + u_k u_j^\top) \frac{ dz}{(z-\lambda_k) (z-\lambda_j) }= \sum_{j \neq k} u_j^\top \Delta u_k ( u_j u_k^\top + u_k u_j^\top) \frac{1}{\lambda_k – \lambda_j}$$ and $$\frac{1}{2i \pi} \oint_\gamma u_k^\top \Delta u_k u_k u_k^\top \frac{ dz}{(z-\lambda_k)^2 } = 0 ,$$ finally leading to $$\Pi_k(A+\Delta ) \ – \Pi_k(A) = \sum_{j \neq k} \frac{u_j^\top \Delta u_k}{\lambda_k – \lambda_j} ( u_j u_k^\top + u_k u_j^\top) + o(\| \Delta\|_2),$$ from which we can compute the Jacobian of $$\Pi_k$$. Note that this extends to piecewise smooth contours $$\gamma$$. Unlimited random practice problems and answers with built-in Step-by-step solutions.  Dragoslav S. Mitrinovic, and Jovan D. Keckic. 1. Note that several eigenvalues may be summed up by selecting a contour englobing more than one eigenvalues.