Esercizi svolti di analisi reale e complessa - Dipartimento di ...

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Res √ i 2 = i√2 Res √ i 3 = 2 −i√3 . 2 γR f dz = CR f dz + σR f dz = π( √ 3 − √ 2) . R → ∞ z 2 CR z4 + 5z2 + 6 dz σR ≤ f(z) dz R→∞ −→ ≤ CR CR = πR ∞ −∞ R2 |z4 + 5z2 |dz| ≤ + 6| R2 R4 − 5R2 |dz| = − 6 R 2 R 4 − 5R 2 − 6 x2 x4 + 5x2 dx . + 6 R→∞ −→ 0 ; π( √ 3 − √ ∞ 2) = lim f(z) dz = f(x) dx R→∞ γR −∞ π 2 (√3 − √ ∞ 2) = 0 x 2 x 4 + 5x 2 + 6 . a = 0 0 f(z) = eiz z 2 + a 2 γR z = ±i|a| R z = i|a| Resi|a| = e−|a| 2i|a| . • lim f(z) dz = R→∞ γR π |a| .
• • CR e iz z 2 + a +∞ f(x) dx = −∞ = dz 2 ≤ +∞ −∞ +∞ −∞ ≤ 1 R2 − a2 |e Cr iz | |dz| ≤ πR R2 − a2 R→0 −→ 0 . cos x x2 +∞ sin x dx + i + a2 −∞ x2 dx = + a2 cos x x2 dx . + a2 +∞ 0 cos x x2 1 dx = + a2 2 +∞ −∞ cos x x2 π dx = + a2 2|a| e−|a| . |z| = 1 π 0 cos θ = 1 2 (eiθ + e −iθ ) = 1 2 dθ a + cos θ z + 1 z . 2π 1 dθ = 2 a + cos θ = = 1 2 = 1 i 0 |z|=1 |z|=1 dz (iz) −1 a + 1 2 1 = (z + z ) dz z 2 + 2az + 1 α± = −a± √ a 2 − 1 Resα+ = 1 α+ − α− = −Resα− . α+ = π √ a 2 − 1 .
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Rn Rn Rn
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Rn Rn Rn
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ℓ 1 ℓ ∞ x = {xn}n ∈ R N
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P ≡ {(x, y) ∈ R 2 y = x 2 (x
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∂ 5 f (1, 1, .., 1) n > 5 ∂x1
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z = z(x, y) (1, 1) z 3 −
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K ⊂ R2 \ y = 1 2 0 < δ < β
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(i) ∞ n=0 n 3 2 n (ii) ∞ n=1 n
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v ∈ R n 1 f(x) ≡ x +
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X ⊂ R n o X = ∅ X Qx ,
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E = {(x, y) ∈ R 2 : x ≥ 0, (x 2
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R 3 Γ ≡ {(u(t), v(t)) : t ∈ (
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F F (x, y, z) = (x, y, z) ;
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Im {(1 + i) n + (1 − i) n } Re
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x dz σ 1 + i σ x dz
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f Imz ≥ 0 ∃ lim f(z) <
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∞ n=2 1 − 1 n2 = 1 2 . a
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Ω ∂
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a) b) c) d) e) f) g) h) i) 2π 0
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Λ = {λ ∈ R n : λ = λ(t), a
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x, y ∈ E k > N0 |(fk(x) −
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x ∈ ℓ 1 k > N0 x1 ≤ x
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X
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limk→+∞ x (k) = limk→+∞ y (
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(⇐=) i = 1, . . . , m fi
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|f(x) − f(y)| ≤ 2 1 1
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L 4m3 m→+∞ −→ +∞ 3
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F ′ (t) = ∂ f ∂ x (g(t), 1
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∂f ∂y = ∂f ∂x = ∂f ∂t =
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• M1 = ( 2 √ 3 , 1 √ 3 ) M
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s(x, y) ≡ x2s(x) c(x, y) ≡ y2s
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f ∂xf(x, y) = 2x − y 2 = 0
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2 3 4 4 i=1 xi i x2 4 i=1 xii x3 4
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dist lim dist (x(t), y(t)), y = t
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• y(t, β) = e βt .
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α > 0 • [0, 1) • [0, a
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N ≥ N0 |(1+x) α − N−1 k=0 |
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• [0, +∞) • [0, +∞)
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k 2k + 1 2n = (−1)k (2k + 1)! n=
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I − T f ′ (y)∞,∞ = max {|
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g Br(0) r ≡ ρ ρ = 2 In 2 .
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g(x, y) = e x2 +y 2 − x 2 − 2y
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R n R n • q = m
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∂A ∂A
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Q = {0} × R y x = 0 Q0 = R
Page 92 and 93: D xy dx dy = = = = 1 dx √ 1−x
Page 94 and 95: x ′ xi i = 1−|x4| i = 1, 2, 3
Page 96 and 97: Area(E) = = = 1 2 = 1 4 E π 4
Page 98 and 99: (∆) = 2π dx dy dz = dθ ρ d
Page 100 and 101: Kn ⊂ Kn+1 n > 1 n Fp
Page 102 and 103: x ′2 + y ′2 = (ρ ′ (θ) cos
Page 104 and 105: (2, 0) A f dx dy = = = A A π =
Page 106 and 107: T R3 ·1 (
Page 108 and 109: xy z ∈ [0, 1] z C C(z) ≡
Page 110 and 111: (ρ, ϕɛ, θ) P = (x, y,
Page 112 and 113: Area (∂E2) = = = = = = ∂E2 π
Page 114 and 115: Φ + F (∂E1) Ψ 1 u
Page 116 and 117: F Φ + F (∂E) = = = Vol (E) =
Page 118 and 119: 0 ≤ α < 2π 2π ω = γα =
Page 120 and 121: ω 1 R 2 \ {0} f(x, y) ω =
Page 122 and 123: • • • ∞ n=1 1 n 2 = ∞ n=1
Page 124 and 125: X ′′ (x) X(x) = −µ X(0) = X(
Page 126 and 127: {sin nx}n ∆v = 0 ⇐⇒ c ′
Page 128 and 129: • inf | sin z| = 0 4√ i( i = {
Page 130 and 131: 2z + 3 z + 1 1 = 2 = ((z − 1) + 5
Page 132 and 133: 1 = |z − a| 2 1 ρ 2 − |a|
Page 134 and 135: f R = {aω1 + bω2 : 0
Page 136 and 137: •
Page 138 and 139: c > 1 c = 1 Rez = 1 c < 1
Page 140 and 141: Resπ = −1 . Reskπ = (−1) k
Page 144 and 145: sin 2 z (0, π 2 ( π 2 ) ,
Page 146 and 147: R → ∞ z 2 − z + 2 CR
Page 148 and 149: π 2 i 2 ∞ = lim f(z) dz =
Page 150 and 151: f(0) = 0 g(z) = f(z) − f(0)
Page 152 and 153: 1 n −n ∞ sin πz = π
Page 154 and 155: g = f ◦ S−1 f(z0
Page 156 and 157: un [a, b] un (a, b)
Page 158 and 159: f(x) = sin 2 (xy) x 2 +y 2 (x, y)
Page 160 and 161: 2 y(x) (0, 1 2 ) {(x,
Page 162 and 163: Ac 1 + 3x2 dxdy + y2 Ac ≡ {(x, y
Page 164 and 165: ϕ : R ↦→ R 1 1 ϕ f(x)d
Page 166 and 167: P x P A, B, C, D ∈
Page 168 and 169: f : R 3 → R µf (x, r) = 1 4
Page 170 and 171: D = {(x, y, z) : x 2 + y 2 + z 2
Page 172 and 173: az + b T z = . T (R∪∞)
Page 174 and 175: cos z f(z) = z f(z) = e 1 z f(z)
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