TENTAMENSSKRIVNING Funktionsteori 20161222 kl 8 13 LUNDS TEKNISKA HÖGSKOLA MATEMATIK HJÄLPMEDEL: Utdelat formelblad. Lösningarna ska vara försedda med ordentliga motiveringar. 1. Låt talföljden (an )n∈N deneras genom an+2 − 3an+1 + 2an = 2n , a0 = 1, a1 = 1. a) Bestäm lösningen till rekursionsekvationen. b) Konvergerar serien ∞ X an an+1 (0.8) n ? (0.2) 2. a) Bestäm alla (komplexa) lösningar till ekvationen cos(z + 23 π) = −2. (0.7) n=0 ∞ X 1 an z n , i en potensserie f (z) = b) Om man utvecklar funktionen f (z) = cos(z + 32 π) + 2 n=0 z ∈ C, vilken konvergensskiva har den potenserien? (Du behöver inte beräkna koecienterna an !) (0.3) 3.a) Beräkna den exponentiella och trigonometriska Fourierserien för den 2π -periodiska funktionen f (t) = | sin t| och rita f i intervallet [−π, π]. (0.7) b) Beräkna seriesumman ∞ X n=1 1 (2n − 1)2 (2n + 1)2 . (0.3) 4. Låt f (x+iy) = u(x, y)+iv(x, y) vara en hel funktion. Antag att u(x, y) = g(x) cos y för en två gånger kontinuerligt deriverbar funktion g . Bestäm f (z) som uppfyller f (0) = 1 + i och f (i π2 ) = 2i. 5. a) Beräkna Z 0 ∞ cos(2x) dx. 1 + x2 (0.6) b) Låt γ vara en sluten slät kurva i C \ {±i}. Vilka värden kan integralen Z γ anta? 6. a) För vilka komplexa tal z konvergerar/divergerar serien ∞ X n=1 zn ? 1 − zn cos(2z) dz 1 + z2 (0.4) (0.7) b) För ett naturligt tal n deneras funktionsvärdet ψ(n) som antalet positiva delare till n. Exempelvis delar 1, 2, 3, 6 talet 6 och ψ(6) = 4. Visa att ∞ X n=1 ∞ X zn = ψ(n)z n 1 − zn n=1 för |z| < 1. (0.3) Lycka till! LUNDS TEKNISKA HÖGSKOLA MATEMATIK LÖSNINGAR Funktionsteorie 20111222 1.a) Characteristic equation for the homogeneous solution is r2 − 3r + 2 = 0, r1 = 1, r2 = 2. The homogeneous solution is therefore ahn = c1 + c2 2n . Ansatz for the particular solution (2 is a root of the characteristic equation): apn = cn2n gives c = 21 . Fitting the initial conditions, i.e. adapt the constants c1 , c2 gives an = 2 − 2n + 2n−1 . an n : b) We apply the root crterion to bn = an+1 lim n→∞ b1/n n 1 n2n−1 − n1 + 12 1 2 − 2n + n2n−1 = < 1. = lim 1 = lim 1 2 n→∞ n→∞ 2 − 2n+1 + (n + 1)2n − + 1 n−1 n2 n Therefore the series converges. 2. We will use Euler's formula and set w = eiz+i 2 π . We get 3 1 = −4 w w+ or w2 + 4w + 1 = 0. √ Hence w1,2 = −2 ± 3 och √ √ 3 i(z + π) = log(−2 ± 3) = ln(2 ∓ 3) + (2k + 1)πi, 2 Finally, z = −i ln(2 ∓ √ 1 3) + (2k − )π, 2 k ∈ Z. k ∈ Z. b) The radius of convergence equals the distance from the origin (the series is centered at z = 0) √ to the closest singularity, i.e. to the closest solution of the equation from a). We have √ ln(2 + 3) = − ln(2 − 3) and therefore (with k = 0) the series converges for ( z∈ w ∈ C : |w| < r ) √ π2 + ln2 (2 + 3) . 4 3.a) f is even, so bk = 0. For the exponential series we have 1 π 1 ck = ∈−π | sin t|e−ikt dt = 2π π Z 0 π sin te−ikt dt. Since e−ik(t−π) = (−1)k eikt follows ck = 0 for odd k. c2k 1 = π Z π −2kit sin te 0 1 dt = 2πi Z π 0 1 4 eit − e−it e−2kit dt = − 2 2π 4k − 1 1 4 =− . 2π (2k + 1)(2k − 1) Using ak = ck + c−k (and the continuity of f ) we get ∞ ∞ 4 2 1X 4 1 X 2kit e = − cos(2kt). | sin t| = − 2π k=−∞ (2k + 1)(2k − 1) π π k=1 (2k + 1)(2k − 1) b) We will use Parseval's formula. ∞ π Z 1 1 = 2 2π | sin t|2 dt = −π Therefore ∞ X k=1 4 1 1 16 X + . π 2 2 π 2 k=1 (2k − 1)2 (2k + 1)2 π2 1 1 = − . (2k − 1)2 (2k + 1)2 16 2 4.a) u has to be harmonic, so or 0 ≡ u00xx + u00yy = (g 00 (x) − g(x)) cos y g 00 (x) = g(x). The latter dierential equation has the general solution g(x) = c1 ex + c2 e−x . We also have f (0) = 1 + i =⇒ u(0, 0) = 1 =⇒ g(0) = 1 =⇒ c1 + c2 = 1. We seperate u(x, y) = (c1 ex + (1 − c1 )e−x ) cos y = c1 ex cos y + (1 − c1 )e−x cos y = u1 + u2 . The harmonic conjugates are v1 (x, y) = c1 ex sin y + d1 and v2 (x, y) = −(1 − c1 )e−x sin y + d2 , d1,2 ∈ R, respectively. Setting z = x, y = 0 and using the identity theorem we get f (z) = c1 ez + (1 − c1 )e−z + id d ∈ R. Finally, f (0) = 1 + i =⇒ d = 1 and f (i π2 ) = 2i =⇒ c1 = 1. Hence, f (z) = ez + i. 5.a) We use cos(2x) = Re(e2ix ) and consider Z C e2iz dz = 1 + z2 Z R −R e2ix dx + 1 + x2 Z CR e2iz dz 1 + z2 where CR = {z ∈ C : |z| = R, Im(z) > 0}. By Jordan's lemma the second integral on the right-hand-side vanishes as R → ∞. Using the residue theorem (z = i is the only singularity surrounded by C ) we get Z ∞ −∞ cos(2x) dx = Re 1 + x2 Z ∞ −∞ 2iz e2ix e dx = Re 2πi · Resz=i = πe−2 . 2 2 1+x 1+z 2 cos(2z) b) The function f (z) = has two singularities z1,2 = ±i. Let γ be a closed smooth 1 + z2 curve. By Wi (γ), W−i (γ) ∈ Z we denote the winding number (with orientation) of γ around i, respectively −i. That is we count how often the curve γ encircles each singularity taking into account the direction. The residue theorem says Z f (z) dz = 2πi (Wi (γ)Resz=i (f (z)) + W−i (γ)Resz=−i (f (z))) = Wi (γ)π cos(2i) − W−i (γ)π cos(2i) γ = mπ cos(2i), where m = m(γ) = Wi (γ) − W−i (γ) is an arbitrary integer (depending on the curve γ ). 6.a) Let 0 < r < 1 and |z| ≤ r then ∞ ∞ ∞ X zn X r rn 1 X n ≤ r = <∞ ≤ 1 − zn n 1−r 1 − r n=1 1−r n=1 n=1 and the series converges uniformly on {z ∈ C : |z| ≤ r} and hence converges for |z| < 1. If |z| > 1 then zn 1 ≥ lim lim n n→∞ 1 − z n→∞ 1 − 1 |z|n =1>0 and the series diverges since the terms do not tend to zero. If |z| = 1 then |1 − z n | ≤ 2 and zn 1 ≥ >0 lim n→∞ 1 − z n 2 and again the series does not converge. To summarize the series diverges for |z| ≥ 1, converges for |z| < 1 (and converges uniformly on each disc |z| ≤ r < 1). p Remark: The series is not dened for z = ei q π , p, q ∈ Z. b) Let |z| ≤ r < 1. By the view of a) the left-hand-side series converges uniformly and we can ∞ X 1 (z k )l for |z| < 1, change the order of the terms in an arbitrary manner. By using = 1 − zk l=0 we conclude ∞ X n=1 ∞ ∞ X zk X zn = = zk n k 1−z 1−z k=1 k=1 = ∞ X X n=1 k|n zn = ∞ X ∞ X (z k )l l=0 ψ(n)z n . n=1 3 ! = ∞ X ∞ X k=1 l=0 z (l+1)k