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When we consider eigenvalues as functions of $$A$$, we use the notation $$\lambda_j(A)$$, $$j=1,\dots,n$$. 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 $$\lambda$$ . These equations are key to obtaining the Cauchy residue formula. Using residue theorem to compute an integral. 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. Suppose C is a positively oriented, simple closed contour. 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 $$\displaystyle \frac{e^{ (2q-1)/(2a)}}{4ia\pi} \frac{\pi}{\sin (i/(2a))} =\ – \frac{e^{ (2q-1)/(2a)}}{4a} \frac{1}{\sinh (1/(2a))}$$. Cauchy Residue Formula. The Cauchy residue trick: spectral analysis made “easy”. Perturbation Theory for Linear Operators, volume 132. 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. We thus obtain an expression for projectors on the one-dimensional eigen-subspace associated with the eigenvalue $$\lambda_k$$. Residue theorem. Do not simply evaluate the real integral – you must use complex methods. See the detailed computation at the end of the post.  Tosio Kato. Proof. If a proof under general preconditions ais needed, it should be learned after studenrs get a good knowledge of topology. The contour $$\gamma$$ is defined as a differentiable function $$\gamma: [0,1] \to \mathbb{C}$$, and the integral is equal to $$\oint_\gamma f(z) dz = \int_0^1 \!\!f(\gamma(t)) \gamma'(t) dt = \int_0^1 \!\! Thus the gradient of $$\lambda_k$$ at a matrix $$A$$ where the $$k$$-th eigenvalue is simple is simply $$u_k u_k^\top$$, where $$u_k$$ is a corresponding eigenvector. If the function $$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) = \ – \!\!\! \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$$. Springer Science & Business Media, 2011. Unlimited random practice problems and answers with built-in Step-by-step solutions. We also consider a simple closed directed contour $$\gamma$$ in $$\mathbb{C}$$ that goes strictly around the $$m$$ values above. 0. §33 in Theory of Functions Parts I … Required fields are marked *. Series.  Gilbert W. Stewart and Sun Ji-Huang. We thus need a perturbation analysis or more generally some differentiability properties for eigenvalues or eigenvectors , or any spectral function . So, now we give it for all derivatives f(n)(z) of f. This will include the formula for functions as a special case. Note that several eigenvalues may be summed up by selecting a contour englobing more than one eigenvalues. Find more Mathematics widgets in Wolfram|Alpha. Proposition 1.1. Your email address will not be published. The desired integral is then equal to $$2i\pi$$ times the sum of all residues of $$f$$ within the unit disk. The function $$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$$ encloses all eigenvalues (as shown below). The Cauchy residue theoremgeneralizes both the Cauchy integral theorem(because analytic functionshave no poles) and the Cauchy integral formula(because f⁢(x)/(x-a)nfor analytic fhas exactly one pole at x=awith residue Res(f(x)/(x-a)n,a)=f(n)(a)/n! If you are already familiar with complex residues, you can skip the next section. [u(x(t),y(t)) +i v(x(t),y(t))] [ x'(t) + i y'(t)] dt,$$ where $$x(t) = {\rm Re}(\gamma(t))$$ and $$y(t) = {\rm Im}(\gamma(t))$$. (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. Consistency of trace norm minimization. Matrix Perturbation Theory. Here is a very partial and non rigorous account (go to the experts for more rigor!). For holomorphic functions $$Q$$, we can compute the integral $$\displaystyle \int_0^{2\pi} \!\!\!  The dependence on \(z$$ of the form $$\displaystyle \frac{1}{z- \lambda_j}$$ leads to a nice application of Cauchy residue formula. Wolfram Web Resources. \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. Singular value decompositions are also often used, for a rectangular matrix $$W \in \mathbb{R}^{n \times d}$$. The Cauchy residue theorem can be used to compute integrals, by choosing the appropriate contour, looking for poles and computing the associated residues. Join the initiative for modernizing math education. $$Taking the trace, the cross-product terms $${\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 Residue Theorem has Cauchy’s Integral formula also as special case. Question on evaluating $\int_{C}\frac{e^{iz}}{z(z-\pi)}dz$ without the residue theorem. Journal of Machine Learning Research, 9:1019-1048, 2008. Note that similar constructions can be used to take into account several poles.  Jan R. Magnus. Note that this extends to piecewise smooth contours $$\gamma$$. Cauchy's Residue Theorem contradiction? By expanding the expression on the basis of eigenvectors of $$A$$, we get $$z (z I- A – \Delta)^{-1} – z (z I- A)^{-1} = \sum_{j=1}^n \sum_{\ell=1}^n u_j u_\ell^\top \frac{ z \cdot u_j^\top \Delta u_\ell}{(z-\lambda_j)(z-\lambda_\ell)} + o(\| \Delta \|_2). The Cauchy Residue Theorem Before we develop integration theory for general functions, we observe the following useful fact. By expanding the product of complex numbers, it is thus equal to$$\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} \! 1 Residue theorem problems Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science The Residue Theorem. Hints help you try the next step on your own. If around $$\lambda$$, $$f(z)$$ has a series expansions in powers of $$(z − \lambda)$$, that is, $$\displaystyle f(z) = \sum_{k=-\infty}^{+\infty}a_k (z −\lambda)^k$$, then $${\rm Res}(f,\lambda)=a_{-1}$$. Econometric Theory, 1(2):179–191, 1985. Complex integration: Cauchy integral theorem and Cauchy integral formulas Deﬁnite integral of a complex-valued function of a real variable Consider a complex valued function f(t) of a real variable t: f(t) = u(t) + iv(t), which is assumed to be a piecewise continuous function deﬁned in the closed interval a ≤ t … 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. See an example below related to kernel methods. SIAM Journal on Matrix Analysis and Applications 23.2: 368-386, 2001. 0) = 1 2ˇi I. C. f(z) z z. 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 11.7 The Residue Theorem The Residue Theorem is the premier computational tool for contour integrals. SEE ALSO: Cauchy Integral Formula, Cauchy Integral Theorem, Complex Residue, Contour, Contour Integral, Contour Integration, Group Residue Theorem, Laurent Series, Pole. Assuming the $$k$$-th eigenvalue $$\lambda_k$$ is simple, we consider the contour $$\gamma$$ going strictly around $$\lambda_k$$ like below (for $$k=5$$).  Adrian Stephen Lewis. The same trick can be applied to $$\displaystyle \sum_{n \in \mathbb{Z}} (-1)^n f(n) =\ – \!\!\! Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. This will allow us to compute the integrals in Examples 5.3.3-5.3.5 in … 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$$. Then if C is Where does the multiplicative term $${2i\pi}$$ come from?  Francis Bach. (7.14) This observation is generalized in the following. For a circle contour of center $$\lambda \in \mathbb{C}$$ and radius $$r$$, we have, with $$\gamma(t) = \lambda + re^{ 2i \pi t}$$: $$\oint_{\gamma} \frac{dz}{(z-\lambda)^k} =\int_0^{1} \frac{ 2r i \pi e^{2i\pi t}}{ r^k e^{2i\pi kt}}dt= \int_0^{1} r^{1-k} i e^{2i\pi (1-k)t} dt,$$ which is equal to zero if $$k \neq 1$$, and to $$\int_0^{1} 2i\pi dt = 2 i \pi$$ for $$k =1$$. [ 7 ] Joseph Bak, Donald J. Newman the Cauchy-Goursat Theorem Cauchy! Preconditions ais needed, it should be learned after studenrs get a knowledge...  note here that the asymptotic remainder \ ( \omega > 0\ ), we can \. For more rigor! ) Engineering and Science the Residue Theorem. Cauchy-Goursat Theorem and Cauchy s... ( \lambda_k\ ) 7 ) if we define 0 constrain real-values functions to be smooth value! 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