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Let 𝑓 : F𝑛 . Then 0, 1 3 → [ ] 𝑓 Λ3( ) − ( E 𝑓 3 ) max 𝑟≠0 | 𝑓 𝑟 ( )| ∥ 𝑓 2 2 . ∥ ≤ Proof. By Proposition 6.1.9 (also see (6.4)), 𝑓 Λ3( ) = 𝑟 ∑︁ , we have . 𝑓 𝑟 ( ) 𝑟≠0 ∑︁ 𝑓 0 ) ( Since E 𝑓 = Λ3( 𝑓 ) − ( E 𝑓 3 ) ≤ The final step is by Parseval. 3 𝑓 𝑟 ( )| | max 𝑟≠0 | 𝑓 𝑟 ( )| · ≤ 𝑟 | ∑︁ 𝑓 𝑟 ( ... |
e proof of Roth’s theorem from F𝑛 3 to Z. Here R Z is the set of reals mod 1. A function 𝑓 : R C is the same as a function / C that is periodic mod 1 (i.e., → / for all 𝑥 R). ∈ Graph Theory and Additive Combinatorics — Yufei Zhao 6.3 Fourier Analysis in the Integers 215 Definition 6.3.1 (Fourier transform in Z) Give... |
ombinatorics — Yufei Zhao 6.4 Roth’s Theorem in the Integers 221 By translation and rescaling, we can identify 𝑃 with 𝑃 𝑁 ′ becomes a subset 𝐴′ . Note that 𝐴′ is 3-AP-free (here we are invoking the important fact that 3-APs are translation and dilation invariant). We can now iterate the argument. (Think .) about w... |
whenever 𝑘 ∈ [ ≤ ] 3 / ≤ 1 + 2𝑛 𝑥 ) + 3, we have ( ≤ / . The sum equals to the coefficients of all the monomials 𝑥 𝑘 with 3 and 𝑛. By deleting contributions 𝑥 𝑘 with 𝑘 > 2𝑛 𝑥2 / 𝑛! 𝑎!𝑏!𝑐! ≤ 1 ( 𝑥2 𝑥 + + 𝑥2𝑛 3 / ) 𝑛 . 𝑎,𝑏,𝑐 0 ∑︁ ≥ 𝑐=𝑛 𝑎 𝑏 + + 2𝑛 2𝑐 𝑏 3 ≤ + / Setting 𝑥 = 0.6 shows that the... |
re 𝑓 = 𝑓str + 𝑓psr + 𝑓sml • • • (structured piece) 𝑓str = 𝑓 ∗ 𝑓psr∥∞ ≤ (pseudorandom piece) ∥ 𝜀0. 𝑓sml∥2 ≤ (small piece) ∥ 𝜀codim 𝑊 ; 𝜇𝑊 for some subspace 𝑊 of codimension at most 𝑀; Graph Theory and Additive Combinatorics — Yufei Zhao 230 Forbidding 3-Term Arithmetic Progressions Remark 6.6.12. It is wo... |
r every 𝑁 𝑁0 and 𝐴 𝑁 ] ⊆ [ with ≥ 𝑥 : 𝑥, 𝑥 |{ 𝑦, 𝑥 2𝑦, 𝑥 3𝑦 + ∈ 𝐴 + + }| ≥ ( 𝛼4 𝑁. 𝜀 ) − Surprisingly, such a statement is false for APs of length 5 or longer. This was shown by Bergelson, Host, and Kra (2005) with an appendix by Ruzsa giving a construction that is a clever modification of the Behrend c... |
= exp (see Exercise 7.11.2). Schoen (2011) ) ( 𝑜 1 = 𝐾 1 further improved the bounds to 𝑑 . Sanders (2012, ( 2013) showed that if we change GAPs to “convex progressions” (see Section 7.12), then an analogous theorem holds with 𝑑 𝐾 ( )) ) and 𝑓 = exp = exp = 𝐾 𝐾 𝐾 log 2𝐾 ( It is easy to see that one cannot do ... |
heorem. The proof only uses the tools introduced so far, and so it is easier than Freiman’s theorem in the integers. Graph Theory and Additive Combinatorics — Yufei Zhao 246 Structure of Set Addition Theorem 7.5.1 (Freiman’s theorem in F𝑛 2 ) 𝐴 If 𝐴 𝐾 𝐾 𝑓 F𝑛 𝐴 2 has | + , where ( | | | is a constant depending o... |
choice of 𝜆 ) 𝑞 1 ] − ∈ [ , we define 𝜙𝜆 𝜙 = 𝜙𝜆 : Z mod 𝑞 −−−−→ Z 𝑞Z · 𝜆 −−−−→ Z 𝑞Z ( 1 mod 𝑞 −−−−−−−−−→ { ) − / / The first map is the mod 𝑞 map. The second map sends 𝑥 to 𝜆𝑥. The last map inverts the mod 𝑞 map Z If 𝜆 𝑞 [ 𝑁. Since there are fewer than 𝑁 nonzero elements in 𝑠 𝐴 is chosen uniforml... |
+ (a) Show that if > 𝑐𝛼3 𝑟 1𝐴( (b) By iterating (a), show that 𝐴 𝛼− 𝑂 ( / (c) Deduce that 2𝐴 )| ) . 2 1 − 2𝐴 contains a subspace of codimension 𝑂 1 𝛼− / 2 . ) ( 7.9 Geometry of Numbers We will need some results concerning lattices and convex bodies belonging to a topic in number theory called the geometry of... |
ry prime 𝑁, every Bohr set of dimension 𝑑 and width 𝜀 𝑑 𝜀 and volume 𝑁, then there | 𝑠 such that 𝐴′ is Freiman 𝑠-isomorphic to a subset of contains a proper GAP with dimension ⊆ 𝛼2 and width 1 𝑁Z with 4. = 𝛼𝑁, 2𝐴 𝐴 with Z and | ≥ | 𝑑 𝑁. 0, 1 | ≤ ∈ ( 2𝐴 𝑠 𝐴 𝑠 𝐴 𝐴 / / / ) | | | | | ≥ ( / ) Now we w... |
of two elements of 𝐴. ) ∈ |{( × + 𝒓 𝑨(𝒙) 𝑎, 𝑏 𝐴 𝐴 : 𝑎 𝑏 = 𝑥 We have the easy bound 2 𝐴 | 2 | − | 𝐴 | ≤ The lower bound is due to trivial solutions 𝑎 is tight for sets without nontrivial solutions to 𝑎 being determined by 𝑎, 𝑏, 𝑐 when 𝑎 𝑏 = 𝑐 + + Here is the main question we explore in this section ... |
ked. See the original paper by Gowers (1998a) on Szemerédi’s theorem for 4-APs as well as excellent lecture notes by Gowers (1998b), Green (2009b), and Soundararajan (2007). Graph Theory and Additive Combinatorics — Yufei Zhao 7.13 Additive Energy and the Balog–Szemerédi–Gowers Theorem 271 Chapter Summary • • • • • • •... |
m Cauchy–Schwarz inequality. Therefore, 𝐼 , (P L) ≤ |P | |L| 2 1 / . + |L| By the same argument with the roles of points and lines swapped (or by applying point-line duality), (P In particular these inequalities tell us that 𝑛 points and 𝑛 lines have 𝑂 L) ≤ |L| |P | + |P | 𝐼 , 1 2 / . 2 𝑛3 / incidences. ( ) / 2 (... |
graph theoretic approach by Conlon, Fox, and Zhao (2014, 2015), which simplified both the hypotheses and the proof of the relative Szemerédi theorem. 9.1 Green–Tao Theorem In this section, we give a high-level overview of the proof strategy of the Green–Tao theorem. Recall Szemerédi’s theorem: 283 Graph Theory and Add... |
3𝑧1, 3𝑧0, 3𝑧1, 3𝑧0, 3𝑧1 ⊆ 𝑆 Here is the statement for 𝑘-APs. (You may wish to skip it and simply imagine how it should generalize based on the above examples.) 𝑘 ≤ = 𝑘𝑥1 + ( Definition 9.2.6 (𝑘-linear forms condition) 𝑟 𝑘, let For each 1 ≤ 𝑥𝑘 −... |
s devoted to proving the above theorem. First, we reformulate the cut norm using convex geometry. 𝑓 ∥ 𝑔 □ ∥ − ≤ 𝜀. Let Φ denote the set of all functions Γ of convolutions of the form 1𝐴 ∗ ⊆ 1𝐵 : 𝐴, 𝐵 1𝐴 ∗ ({ Note that Φ is a centrally symmetric convex set of functions Γ Φ = ConvexHull 1𝐴 ∗ } ∪ {− 1𝐵 or R. ⊆ Γ... |
he problem to another triangle counting problem where 𝜈 is replaced by 1 on one of the edges of the triangle. / This densification strategy reduces a sparse triangle counting problem to a progressively easier triangle counting problem where some of the sparse bipartite graphs among 𝑋, 𝑌 , 𝑍 become dense. Let Sparsi... |
ove 12 linear forms and also change the “12” in 𝑝12 to the number of linear forms remaining. 𝑧0, 2𝑥0 + 𝑧0, 2𝑥1 + 𝑧1, 2𝑥0 + 2𝑥1 + 𝑧1, 𝑥0 − 𝑥1 − 𝑥0 − 𝑥1 − 2𝑧0, 2𝑧0, 2𝑧1, 2𝑧1, 𝑦0, 𝑦0, 𝑦1, 𝑦1 𝜀 ) ± ⊆ 1 𝑆 ( The proof follows the strategy outlined in Section 9.3 on the transference princi... |
national Congress of Mathematicians—Rio de Janeiro 2018. Vol. IV. Invited lectures, World Scientific Publishing, pp. 3059–3092. MR:3966523 Bateman, Michael & Katz, Nets Hawk (2012) New bounds on cap sets, J. Amer. Math. Soc. 25, 585–613. MR:2869028 Behrend, F. A. (1946) On sets of integers which contain no three terms ... |
a Turán type problem of Erdős, Combinatorica 11, 75–79. MR:1112277 Füredi, Zoltan & Gunderson, David S. (2015) Extremal numbers for odd cycles, Combin. Probab. Comput. 24, 641–645. MR:3350026 Füredi, Zoltán & Simonovits, Miklós (2013) The history of degenerate (bipartite) extremal graph problems, Erdős centennial, Jáno... |
58 Lovett, Shachar (2012) Equivalence of polynomial conjectures in additive combinatorics, Combinatorica 32, 607–618. MR:3004811 Lovett, Shachar (2015) An exposition of Sanders’ quasi-polynomial Freiman-Ruzsa theorem, Theory of Computing Library Graduate Surveys, vol. 6, pp. 1–14. Lovett, Shachar & Regev, Oded (2017) A... |
cta Arith. 27, 199–245. MR:369312 Szemerédi, Endre & Trotter, William T., Jr. (1983) Extremal problems in discrete geometry, Combinatorica 3, 381–392. MR:729791 Tao, Terence (2006) A variant of the hypergraph removal lemma, J. Combin. Theory Ser. A 113, 1257–1280. MR:2259060 Tao, Terence (2007a) Structure and randomnes... |
urán density, 19 graph, 14 hypergraph, 20, 191 number, 11 problem, 11 theorem, 14, 188 𝑈3 norm, 214, 262 undecidability, 163 unit distance problem, 24 𝑉 , see graph , 𝑣 𝐺 𝐺 ( ) ( ) Graph Theory and Additive Combinatorics — Yufei Zhao 326 Index van der Waerden’s theorem, 5 vertex-transitive graph, 113 𝑊-random gra... |
e start with an example. Example. Consider the Cauchy-Riemann equation ux = vu, uy = −vx. We know that the pair u, v satisfies the Cauchy-Riemann equations, if and only if both u, v are harmonic, i.e. uxx + uyy = 0 etc. 26 2 Partial Differential Equations II Integrable Systems Now suppose we have managed to find a harmoni... |
can be shown that the function β = β(κ) has only finitely many zeroes χ1 > χ2 > · · · > χN > 0. So we have a finite list of bound-states {ψn}N n=1, written ψn(x) = cnϕχn (x), where cn are normalization constants chosen so that ψn = 1. 3.1.3 Summary of forward scattering problem In summary, we had a spectral problem wher... |
s is again AMAZING. ˙cn − 4χ3 ncn = 0. cn(t) = cn(0)e4χ3 nt. 3.4.3 Summary of inverse scattering transform So in summary, suppose we are given that u = u(x, t) evolves according to KdV, namely ut + uxxx − 6uux = 0. If we have an initial condition u0(x) = u(x, 0), then we can compute its scattering data n=1, R(k, 0) . T... |
iven function u(x), the total derivative with respect to x is d dx f (x, u(x), ux(x), · · · ) = ∂f ∂x + ux ∂f ∂u + uxx ∂f ∂ux + · · · Example. ∂ ∂x (xu) = u, Dx(xu) = u + xux. 46 4 Structure of integrable PDEs II Integrable Systems Finally, we need to figure out an alternative for J. In the case of a finitedimensional Ha... |
2 0 1 ... 0 0 0 ... un−1 Now differentiate “ψt + Aψ = 0” i times with respect to x to obtain (∂i−1 x ψ)t + ∂i−1 x m−1 j=0 vj(x, t) j ∂ ∂x n j=1 Vij (x,t)∂j−1 x ψ ψ = 0 for some Vij(x, t) depending on vj, ui and their derivatives. We see that this equation then just says ∂ ∂t Ψ = V Ψ. 52 4 Structure... |
ε)T = (I + εa + o(ε))(I + εaT + o(ε)) = I + ε(a + aT ) + o(ε). 58 5 Symmetry methods in PDEs II Integrable Systems So we must have a + aT = 0, i.e. a is anti-symmetric. The other condition says 1 = det A(ε) = det(I + εa + o(ε)) = 1 + ε tr(a) + o(ε). So we need tr(a) = 0, but this is already satisfied since A is antisymm... |
re we need to assume that ∆ is of maximal rank , i.e. the matrix of derivatives ∂∆j ∂yi is of maximal rank, where the yi runs over x, u, and in general all coordinates. So for example, the following theory will not work if, say ∆[x, u] = x2. Assuming ∆ is indeed of maximal rank, it is an exercise on the example sheet t... |
In the wave equation, we can try a solution of the form u(x, t) = f (x − ct), and then the wave equation gives us an ODE (or lack of) in terms of f . Example. Consider the sine–Gordon equation in light cone coordinates This equation admits a Lie-point symmetry uxt = sin u. gε : (x, t, u) → (eεx, e−εt, u), which is gene... |
root of a. However, if w is a complex number, what is the meaning of √w? Let us try to find a reasonable definition of √w. We know that the ei(Arg w)/2. If we equation z2 = w has two solutions, namely z = ± | ei(Arg w)/2. However, want √ w ! | | with this definition, the expression √w√w = √w2 will not hold in general. ! I... |
are metrics: (a). d(z, w) = − z (b). d(z, w) = | 1 + z | (c). d(z, w if z = w = w. 1 if z 4.5. Let the point z = x + iy correspond to the point (ξ, η, ζ ) on the Riemann sphere (see Figure 4.3). Show that ξ = 2 Re z 2 + 1 z | | ,η = 2 Im z 2 + 1 z | | , , and Re z = 1 ξ − , ζ Im = 1 η − . ζ 4.6. Show that if z1 and z2... |
at a point z0 = x0 + iy0 if and only if u(x, y) and v(x, y) are continuous at (x0, y0). Example 5.9. The exponential function f (z) = ez is continuous on the whole complex plane since ex cos y and ex sin y both are continuous for all (x, y) IR2. ∈ The following result is an immediate consequence of Theorem 5.2. Theore... |
z + Thus, d dz zn = lim → ∆z 0 (z +∆ z)n ∆z zn − = nzn − 1. Example 6.2. Clearly, the function f (z) = z is continuous for all z. We shall show that it is nowhere differentiable. Since f (z0 +∆ z) ∆z − f (z0) = (z0 +∆ z) ∆z z0 − = ∆z ∆z . R.P. Agarwal et al., An Introduction to Complex Analysis, DOI 10.1007/978-1-4614-0... |
erywhere in S, then f (z) is a constant in S. If f (z) is analytic in a domain S and if f ′(z) = 0 Proof. Since f ′(z) = 0 in S, all first-order partial derivatives of u and v vanish in S; i.e., ux = uy = vx = vy = 0. Now, since S is connected, we have u = a constant and v = a constant in S. Consequently, f = u + iv is ... |
rentiable everywhere, f ′(z) = (6x + 2) + 2i, i(6y), analytic everywhere (entire). (d). f is differentiable for all z y)3, ux = 3x2, uy = 0, vx = 0, vy = 3, f ′(0 + iy) = 3y2 16 i, f ′( 3 16 i) = 1 16 − 16 − = 0, 4 − 3(1 3(1 − − − − − − 1 ± ̸ 50 Lecture 7 2(z4 + 5z2 + 12)/[z2(z2 + 4)2], analytic in C 0, , analytic every... |
nd cosine functions with those of trigonometric functions, we find cosh(iz) = cos z, cos(iz) = cosh z, sin(iz) = i sinh z. (8.2) Using relations (8.2) and the trigonometric identities, we can show that sinh(iz) = i sin z, the following hyperbolic identities are valid in the complex case: − cosh( z) = sinh z, sinh( − sin... |
e hyperbolic functions can be treated in a corresponding manner. It turns out that sinh− cosh− tanh− 1 z = log[z + (z2 + 1)1/2], 1 z = log[z + (z2 1)1/2], 1 + z 1 z 1 2 1 z = − , log and d dz d dz d dz sinh− 1 z = cosh− 1 z = tanh− 1 z = − 1 (z2 + 1)1/2 , 1 1)1/2 , (z2 − 1 z2 . 1 − 62 Lecture 9 Problems 9.1. Find all v... |
3) Once again, the condition bc function. − = 0 ensures that we do not have a constant − c ad Equation (11.3) reveals that, when c = 0, a linear transformation is a composition of the mappings Z = cz + d, W = 1 Z , w = bc + a c − c ad W, ad bc = 0. − Thus, a linear fractional transformation always transforms circles an... |
ex numbers z1, z2, z3, z4 lie on a line or a circle in the complex plane if and only if their cross ratio is a real number. 11.10. Show that the image of the vertical line x = x0, where π/2 < x0 <π/ 2, under the mapping w = sin z is the right-half of the hyperbola − u2 sin2 x0 − v2 cos2 x0 = 1 if x0 > 0 and the left-ha... |
x domain is simply connected. 12.4. Let S C be an open set, and let f : S C be a continuous function. Show that if S is connected, then f (S) is also connected. → ⊂ Answers or Hints − − − − − ∩ ⇒ ⇒ Aj (a). (b). simply connected, (a). Multiply-connected, (b). Since I(γ) and Lj are convex, so is Ij := I(γ) 12.1. (c). mul... |
.13. Let γR be the circle = R described in the counterclockwise direction, where R > 2. Suppose Log z is the principal branch of the logarithm function. Show that z | Log z2 z2 + z + 1 dz =γR 2πR ≤ # π + 2Log R R2 1 R − − . $ " " " " 13.14. Let S be an open connected set and γ a closed curve in S. Suppose f (z) is anal... |
ess Media, LLC 2011 96 Cauchy-Goursat Theorem 97 dxdy, Theorem, we have f (z)dz = ∂v ∂x − ∂u ∂y − dxdy + i ∂u ∂x − ∂v ∂y =γ = =S′ # $ where S′ is the domain interior to γ. Since f (z) is analytic in S, the first partial derivatives of u and v satisfy the Cauchy-Riemann equations. Hence, it follows that $ = =S′ # f (z)dz... |
point z2 zndz = 1 n + 1 zn+1 2 − zn+1 1 for n = 0, 1, 2, . · · · 16.2. Let γ be the semicircle from 2 3 3i to 3i in anticlockwise direction. − Show that = πi. =γ dz z =γ 16.3. Evaluate each of the following integrals where the path is an arbitrary contour between the limits of integrations i/2 (a). eπz dz, (b). π+2i c... |
we get Λ= g(ξ)dξ =γ (ξ z − − ∆z)(ξ z) − =γ − g(ξ)dξ (ξ z)2 =∆ z − g(ξ)dξ =γ (ξ z − − ∆z)(ξ − z)2 . Cauchy’s Integral Formula 115 ξ · d z d/2 · Figure 17.5 γ z +∆ z Let M = maxξ that z ξ | we may assume that and let d equal the shortest distance from z to γ, so γ | ∈ d > 0 for all ξ on γ. Since we are letting ∆z approac... |
ion is dense in C. 18.18. Suppose f (z, t) is continuous, and continuously differentiable with respect to t, and suppose γ is a smooth curve in the domain of defi- nition of f. Then, the function F (t) = f (z, t)dz is continuously differen- tiable with respect to t, and its derivative is =γ dF (t) dt = ∂f (z, t) ∂t dz. =γ... |
an|− | B z | |− 0, 1 ≥ ̸ 128 Lecture 19 which is the same as then Pn(z) | | Corollary 19.2. zk, k = 1, 2, · · · B an| | > 0; i.e., there is no zero on or outside the circle = R, 1 + |≥ z | = R. z | | 1 , n of Pn(z) lie outside the circle = 0 and C = max ≤ Let a0 ̸ j ≤ . Then, all zeros aj a0| | = r. (19.10) Proof. Set ... |
mum of occur on the boundary of the disk; i.e., on the circle the circle by z(t) = eit, 0 in the disk | 3 is analytic in the disk B(0, 1) and continuous on must = 1. Parameterizing − | z2 +2z 2π, we get z2 + 2z − − 3 3 z t | | | | z2 + 2z | 3 − ≤ ≤ 2 = (z2 + 2z | − e2it + 2eit = = (14 2 − 3)(z2 + 2z 3) 2it + 2e− − − 6 ... |
z0 + z1 + or, equivalently, ∞j=0 zj, where the terms zj are complex numbers. The nth partial sum of the series, denoted by sn, is the sum of the first n + 1 terms; i.e., C n sn = ∞n=0 has a limit s, the j=0 zj. If the sequence of partial sums ∞j=0 zj. A series series is said to converge, or sum, to s, and we write s = ∞... |
s B(0, 1). In fact, since sn(z) − ∞j=0 fj(z) converges uniformly on a do∞j=0 fj(z) = s(z) if and only if for any ϵ> 0 there N, and any natural number m, C S, n ∈ ≥ <ϵ holds. ∞j=0 zj does not converge uniformly on C sn+m(z) | sn(z) | − = | = | · · · zn+1(1 + z + zm 1 z || − n+1 | − z | 1 | | ≥ + zm z | 1) − | n+1 | (1 |... |
An Introduction to Complex Analysis, DOI 10.1007/978-1-4614-0195-7_23, © Springer Science+Business Media, LLC 2011 151 ̸ 152 Lecture 23 Corollary 23.1. diverges at all points z that satisfy the inequality If (23.1) diverges at some point z = z1, then it z | z0| − > z1 − | . z0| Corollary 23.2. Let R be the radius of co... |
. Clearly, 1)j/2/[2j+n (j/2)! (j/2 + n)!] aj| 1/[(j/2)!]2/j. Now use [(k)!]1/k ( z = zn − 2 1)j 2n(j+n) j!(j+n)! aj| | 2j+2n − 1 − →∞ | ≤ ∞j=0 1/j if j is odd if j is even. We will 1/j = 0 if j is odd, and for j → ∞ ∞j=0 . as k ( − j!(j+n 1)j → ∞ 1)! − 2 3 C C j)/(j + 1a C C C − →∞ | →∞ | = limj = 1. (b). Compute · · ·... |
22j − 1)j 2)B2j − (2j)! z2j, <π. z | | ( − ∞ j=1 / 24.11. Euler’s numbers Ek are defined as E0 = 1, E2j+1 = 0, j = 0, 1, 2, , · · · 2j 0 # E2j + $ 2j 2 # $ E2j 2 + − · · · + 2j 2j # E0 = 0, j = 1, 2, $ . · · · Extend the method of Example 24.6 to show that sech z = E0 + E2j (2j)! z2j, ∞ j=1 / <π/ 2. z | | 24.12. The Leg... |
series for f (z). 3. We can write where f (z) = ∞ j= / −∞ aj(z − z0)j , A, z ∈ aj = 1 2πi f (ξ) z0)j+1 dξ − (ξ =γ for all j Z. ∈ − ∞j= −∞ C aj(z z0)j converges to f (z) in some annular domain, 4. If a series then it must be the Laurent series for f (z) at z0; i.e., Laurent’s series for f (z) in a given annulus is uniqu... |
such that am ̸ = 0. Then, f (z) has a zero of order m at z0, and so the representation z0)mg(z) by Theorem 26.1 is valid. Since g(z0) f (z) = (z = 0 and g(z) is continuous at z0, there exists a disk B(z0,δ ) throughout which g(z) is nonzero. Consequently, f (z) = 0 for 0 < <δ. ≥ − z z0| | − Thus, for a function f (z) t... |
n f (z) is also called an analytic continuation of the function f (x) of the real variable into the complex domain S. Various such analytic continuations were given in Lecture 26. ⊆ ∈ ⊂ ∈ Example 27.2. Consider the function f1(z) = ∞ zj. /j=0 Clearly, this function converges to 1/(1 z) when z | | − < 1. Hence, f1(z) = ... |
he definition of symmetry above is independent of the points z1, z2, z3. For this, we need the following elementary lemma. Lemma 28.1. (z1, z2, z3, z′′) = (z1, z2, z3, z′) if and only if z′′ = z′. Let z1, z2, z3 be distinct complex numbers. Then, Theorem 28.1. The points z′ and z′′ are symmetric with respect to a line γ... |
0 | z0| − < R | z − z0| < R, and hence bounded in every closed neighboris analytic on hood of z0. This in turn implies that f (z) is bounded in some punctured neighborhood of z0. Conversely, suppose there is a punctured neighborz0| hood 0 < < r < R and a finite M such that on this neighbor− hood M. Then, in Problem 25.1... |
ple 30.3. ble singularities are z = kπ, k = 0, 1, We claim that z = kπ, k k be even. Then, since limz , it follows that limz For the function f (z) = ez/ sin z, the only possi0(z/ sin z) = ± 0 f (z) = e, and hence f (z) has a removable singularity at z = 0. = 0 are essential singularities of f (z). For this, let kπ− (z... |
/2] + 2nπ. Then < R. Thus, h(z) = ∞j=0 ajzj = z | | < 1, we have aj aj+1 + h(z), where h(z) is analytic in < R. Clearly, bj → cos− Z C since the quantity inside the brackets is never zero. is defined for all w ∈ Fix w and set ζn = . So − we can take zn = 1/ζn. The result for cosh(1/z) = cos(1/(iz)) also follows from thi... |
31.7. Suppose that f (z) = ∞j= z0)j are valid in an annulus around z0. Show that C aj(z z0)j and g(z) = −∞ − ∞j= −∞ bj(z − C R[f g, z0] = ∞ j= / −∞ ajb 1. − j − 31.8. Suppose that f (z) and g(z) are analytic functions with zeros of respective orders k and k + 1 at z0. Show that R f g - , z0 = (k + 1) . f (k)(z0) g(k+1)... |
ow, in view of (33.2), we have π/4 exp iρ2e2iθ ρieiθdθ ρ2 e− ρ − π/4 exp[iρ2(cos 2θ + i sin 2θ)] dθ π/4 " " e− ρ2 sin 2θdθ " " π/4 e− ρ2(4θ/π )dθ ρ ≤ 0 = 0 as ρ , → ∞ → (33.5) whereas (33.3) gives 0 lim ρ →∞ = ρ exp ir2eπi/2 eπi/4dr = & ’ Combining (33.4)-(33.6), we get eπi/4 ∞ r2 e− dr = 0 = − 1 + i √2 − . √π 2 (33.6)... |
c), then the improper integral of f (x) over [a, c] and its Cauchy principal value are defined by ∈ c a = f (x)dx := lim r → b− = r a f (x)dx + lim b+ s → f (x)dx s = c c b t − f (x)dx + f (x)dx , b+t = 1 and c p.v. f (x)dx := lim a = 0+ 0= provided the appropriate limits exist. on the interval ( value of its integral i... |
, we e2πiαI1 as r − , it has the expansion f (z) = a0 + (a ∞ f (z). This leads to I5 = 2πia0eπiα . Thus, as r · · · 0 and ρ 1/z) + → − → → ∞ and I3 = at z = limz get →∞ 2πi R[zα 1(1 − − z)− αf (z), zj] = (1 − e2πiα )I + 2πia0eπiα , which gives / 1 xα 1(1 − − 0 = x)− αf (x)dx = πa0 sin πα + (1 2πi e2πiα) − / R[zα − 1(1 ... |
2πi [(m1 + + mℓ) (n1 + + np)] = 2πi[Zf − · · · Pf ], − · · · R.P. Agarwal et al., An Introduction to Complex Analysis, DOI 10.1007/978-1-4614-0195-7_37, © Springer Science+Business Media, LLC 2011 247 ̸ 248 Lecture 37 which is the same as (37.1). Remark 37.1. The nomenclature argument principle for (37.1) comes from t... |
z0| (38.2), for z | − ≥ 1, in view of (38.1), we can choose r sufficiently small so rm/2 and, for a w. From z0| set g(z) to be the constant w0 − < r. Now let ρ = z = 0 for all 0 < | − w0| w <ρ, | = r, we have am| − | g(z) | | = w0 − | w | < | rm am| 2 < f (z) | − . w0| R.P. Agarwal et al., An Introduction to Complex Anal... |
onformal Mappings 259 Furthermore, if w = f (z) maps S onto a domain S ′ and is conformal at every point in S, then w = f (z) is called a conformal mapping of S onto S′. Moreover, S′ is called a conformal image of S. The following result provides an easier test for the conformality of the mapping w = f (z) at the point... |
aps the circle = 1 onto the real interval [ = 1) onto the ellipse = r (r > 0, r 1, 1]. = 1 − z z z | | | | | | u2 r + 1 r 1 2 2 + v2 1 2 r 1 r − 2 = 1, 4 1. 2 35 4 2 35 ± which has foci at (f). J is a one-to-one continuous mapping of both the interior and the z exterior of the unit circle | (g). w = J(z) is conformal e... |
∞ j=0 & / 1 1 θ) − θ) r R ei(t − r − t) + − eij(θ − R ei(t 3 2 − r R ei(t − ∞ 1 − t) + r R r R e− i(t θ) − 3 θ) j eij(t − θ) ’ j=1 & / t) + eij(t − θ) − j eij(θ ( % j ’ j ’ cos j(θ t) − cos jθ cos jt sin jθ sin jt, r R ∞ = 1 + j=1 & / ∞ = 1 + 2 j=1 & / ∞ = 1 + 2 +2 /j=1 & r ∞ R j=1 & / ’ r R r R j ’ from (40.3) it fol... |
t, we generalize the discussion above for the derivative f ′(w) of a function f (w) that maps the half-plane v 0 onto a polygonal region with any number of sides. ≥ ≥ Theorem 41.1 (The Schwarz-Christoffel Transformation). Let f (w) be analytic in the domain v > 0 and have the derivative f ′(w) = A(w u1)(α1/π) − 1(w u2)(... |
ctions bn(z) are analytic Example 42.3. For < 1, uniformly we have z | | ∞ 1 + z2j Kj=0 & ’ = (1 + z)(1 + z2)(1 + z41 z − z2)(1 + z2)(1 + z4) · · · 1 z − lim 1 n →∞ & − z2n = . z 1 − 1 1. z | |≥ ’ We also note that this infinite product diverges for Example 42.4. Since (1 + z/√j)(1 zi/√j) = (1 + z2/j), j (1 + zi/√j)(1 z... |
ince r > 0 is arbitrary, in (43.2) the limit function f (z) is entire and has the prescribed zeros. ∞j=1(Eℓj (z/zj) 1 for n 1 z/zj| |≤| C − Remark 43.2. chosen differently, the representation (43.2) is not unique. Since the sequence of positive integers ℓn} { can be Remark 43.3. The general entire function with simple z... |
ire, and since φ(z)ψ(z) = lim f (zn) = lim zn → z zn it satisfies the conditions (44.5). → z φ(z) z − − φ(zn) zn ψ(z)(z = − zn) . φ′(zn)An φ′(zn) , Finally, we state the following result, which provides an explicit partial fraction expansion of a meromorphic function f (z). Theorem 44.2. Let f (z) be a meromorphic funct... |
), e3 = ℘((ω1 + ω2)/2), and consider the function g(z) = (℘(z) e1)(℘(z) e2)(℘(z) e3). − − − Since ω1/2 and ω2/2, (ω1 + ω2)/2 are simple zeros of ℘′(z), they are double zeros of ℘(z) e3, respectively, so g(z) also has zeros of order 2 at ω1/2,ω 2/2, and (ω1 + ω2)/2 and a pole of order 6 at z = 0. Then, f (z)/g(z) is an ... |
one in infinity. Then, { c1 z z : + z | | c2 z2 + > 1 } , · · · (47.2) except for a pole at (47.3) 2 j cj| | ≤ 1. ∞ j=1 / R.P. Agarwal et al., An Introduction to Complex Analysis, DOI 10.1007/978-1-4614-0195-7_47, © Springer Science+Business Media, LLC 2011 308 Bieberbach’s Conjecture 309 Proof. We assume that c0 = 0; o... |
surface has infinitely many sheets with two logarithmic branch points above w = i. It is constructed by joining an infinite number of w-planes with a cut u = 0, 1, corresponding to the vertical strips kπ < x < (k + 1)π. ± v | |≤ Riemann surface for w = cot z. Since cot z = tan(π/2 Riemann surface for w = cot z is the sa... |
s were not conceived in the Hellenistic world). We also have the following quotation from Bhaskara Acharya (working in 486 AD), a Hindu mathematician: “The square of a positive number, also that of a negative number, is positive: and the square root of a positive number is two-fold, positive and negative; there is no s... |
es of mystery and distrust of complex numbers could be said to have disappeared, although some resistance continued among a few textbook writers well into the 20th century. Nowadays, complex numbers are viewed in the following different ways: − 1. points or vectors in the plane; 2. ordered pairs of real numbers; 3. oper... |
O){cos¢ + i sin¢) =(cosO cos ¢-sinO sin¢)+ i{sinO cos¢ + cosO sin¢) = cos(O + ¢) + i sin{O + ¢). {2.9) ahead Looking that, if we extend numbers, we have, to a notation the series for any real 0, that we shall definition in Chapter 3, we note function to complex of the exponential justify properly We may therefore write... |
such that l z11 > l z2l· Show that, for 2.13. that, if z1, z2 Prove C, then E Deduce that, for all c, d in C, 2.14. Sum the series cos 8 +cos 38 + + cos(2n + 1)8. · · · 2.15. Let 1 = peifi (¢�)be a root of P(z) = 0, where and where deduce a0, a1, . . • , an are real. Show that that z2 -2p cos 8 + p2 is a factor of P( ... |
(a, b) is 2} is open. Describe its P relude to Complex Analysis 3. 41 3 .3 Functions and Continuity a In complex "process" areas of mathema number analysis, as in other one complex transf orming descript a complicated may involve use of a formula. Frequently we talk of the domain of definition, or simply Thus the func... |
(1+ z)2 _ (1-2z)l = �1 -(1-2z)(1 + 2 z + z2) I = (1+ z)2 l3z2 + 2z31 3lzl2 + 2lzl3 11 + zl2 - (1-lzl)2 4lzl2 < ---'-:--'______,---,-:....,,.:- 16lzl2 -1/4 , < = that ( 1 -lzl)2 since lzl ::; � implies 1 ( )2 1 + z follows. and the result � i and 3lzl2 + 2lzl3 ::; 4lzl2. Thus 2 O(z ) , -(1 -2z ) = 0 Example 3 . 12 Show... |
- i ax ax ay ay ' au and so f'(z) = that 0 implies av = au av au ax ax {}y ay = = = 0 at every Let point in U. = a + bi and q = c p + di be points in U. Then 58 Complex Analysis at least one of r = a + di and Suppose, without = c + bi lies in U. (See Exercise 4.2 U. Then both below.) essential loss of generality, that... |
, and the open disc N(a, R Then f is holomorphic ) = L ncn(z- at-1. n=1 f'(z) 00 4.17 we know that g is defined 0. this for the case where for lzl Proof* to prove Again, it will be sufficient R. We shall a Let g(z) =:::: L::'=1 ncnzn-1. From Theorem , show that, within the disc N(a, R ) f(z + h -f(z) -g(z)l-+ 0 ash--+... |
the answer but we cannot For example, the principal be sure that using then log( e if z = 5irr/2, ) .: = In fact all we can say is that ezw is a value of the multifunc tion ( ez)w. 0 as but in practice it is usually sur or in All this prisingly unction multif may seem somewhat easy to sort out whether confusing, or no... |
lix Edouard Justin Emile Borel, 1871-1956. 2 80 Complex Analysis Theorem 5 . 1 (The Heine-Borel Let S be a closed, a finite bounded subset subcovering Theorem) of C. of S. Then every open covering of S contains Proof* Since side l: S is bounded, we may suppose that it is enclosed within a square Q with is no parts fini... |
.3. Sketch the curves a) { (acost,bsint) : 0 S t S 27r} (a,b > 0); b) {(acosht,bsinht) : t E [O,oo)} (a,b > 0); c) {(at,aft): t E(O,oo)} (a>O). curve { teit : t E with a < b. Determine [a,b] }. Determine 5.4. Sketch the 5.5. Let a, b E JR., {et+it : t E lima-t-oo A(a,b). the length (0, 21r] } , and determine of A( a, i... |
ille Jordan, 1838-1922. 3 98 Complex A n alysis EXERCISES 5.6. Let 1(t) 5.7. Evaluate z-a-t (0:::; t :::; h). Show that 1 = 1 (�:1 = � Cz-a -h)n-(z! a)n) J, f(z) dz, where a) f(z) = Rez, 1(t) = t2 +it, t b) f(z) z2, 1(t) = eit, t E c) f(z) 1/z, 1(t) d) f ( z) cos z, where by the straight [0, 11']; eit, t E [0,611']; 1*... |
mple, a strong later, but many for a piecewise of the Funda 'Y . when "reasonable" Corollary 'Y , (Simply observe The following general (-cos)'(z), cosz occupies result that sinz i sin z dz = i cos z dz = i exp z dz 0 . (sin)'(z), expz position a central = = = = {exp)'(z).) analysis: in complex Theorem 6 . 1 (Cauchy's... |
Cauchy ' s Integral Formula 7 .1 We have already then observed in Theorem 5.13 that if is a circle with centre 0 a if �t( a, r) is a circle with centre More generally, r 1-dz (a,r) z -a J... = 21l"i j a, then for, with z a+ rei9, (7.1) = . a re' dz - .9 -21l"l. which shows that the value of a holomorphic 0 will play a ... |
lzl > R. By whenever bounded set closed entire function, that the 5.3, the function 1/p(z) is also bounded we deduce, as required, 0 must have at least one root. as lzl -+ oo -+ oo 0 = It is to prove now straightforward Theorem 7 . 1 1 (The Fundamental Theorem of Algebra) Let p(z) be a polynomial of degree /3, a1, a2,... |
), has a Taylor series tan z = a1 z + a3z3 + asz5 + · · · sin z = tan z cos z . and the result of the previous Use the exercise identity to show that (- 1)n (2n + 1)! Use this identity =a2n+l-2! +� - ·· ·+ - 1 to calculate a2n-l a2n-3 ( )n a1 ar, a3 and as. ( ) (2n)! n� O. 136 Complex Analysis 7.14. The odd function ta... |
tial singularity the assumption that f has an essen As a consequence we have the following result close arbitrarily says that a non in complex number to every function comes polynomial entire { z : lzl > R}: any region 0 , which Theorem 8 . 1 1 Let f be an entire exists Then there not a polynomial function, z such t... |
-1 + i 3) + 2 2iV3 res(!, - 1 + iV3) = V3 156 Complex Analysis In the upper half-plane , for all sufficiently lzl, large x= Re zandy =Imz) izf(z)i = < zei"'e-Y It follows I z2 :�;: + 41 -z2+ 2z+ 4 (where I I lzl (since lei"' I = 1 and le-Y I -lzl2 - 2lzl - 4 uniformly that lzf(z)l tends All the conditions of Theorem !o... |
uate X 100 1 + x --4dx. 0 Solution is no point Here there grand is an odd function, Jq(O,R)[z/(1 + z4)] dz, where in finding from the integral to is trivially and this integral equal since the inte to 0. We consider the first quadrant. q(O, R) is the quarter circle in oo , - oo iR R 168 Complex Analysas ___ has a simp... |
the interval 0 since x 1 -,.---..,-,---:---.-: 178 Complex Analysis It now follows, O'n} min{lzl : z since n +! ,that E = the contour O'n is of total length 4(2n + 1), and since g(z)dz $; 4(2n+ 1)11" sup 2 2 $; 4(2n+ 1) ( 1)2 2, cot 1rz I 1r Hence, letting 11 I I as n and this tends to 0 ZECTn CTn in (9.17), we find t... |
er, a contour that, for f(iy) oo = = = = = ..:11'2 (arg /) ..:11'2 (arg z4) --+ 2rr as R --+ oo . is one root of f(z) largeR, and so there 0 2rr for sufficiently = Thus ..:1-r (arg!) in the first quadrant . = = D 188 Complex Analysis EXERCISES z8 + 3z3 + 7z + 5 = 0 10.1 . Show that the equation first quadrant ? distinc... |
ssible 0. For example, Remark It is of course w (z-1)2(2z-3)(z-3), with zeros ("t would give (�; dz = 27ri[(2 r f ;f 20 1)] = 107r . X Conformal Mappings 1 1 P reservation of Angles = u = geometric of a remarkable the consequence at Figures property of 3 . 1 and 3.2 on page 42. For l in the w-plane are of course that t... |
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