Spezialfall β = 0 Σ(w) BLi Σ(w + w̃) n(w̃)− n(w) [−2z,−2z,−2z, 0] BL1 [−2z,−2z,−2z,−δ1/2z] −2 [−2z,−2z, 0, 0] BL1 [−2z,−2z,−2z,−δ1/2z] −2 [−2z, 0,−2z, 0] BL1 [−2z,−δ1/2z,−2z,−δ1/2z] −4 [−2z, 0, 0, 0] BL1 [−2z [...] S0 und p−k0 /∈ S0. Nach entsprechender Umbennung bzw. Shift der Bezeichnung, kann man o.B.d.A. immer davon ausgehen, dass die Polkette (pk)k=0,1,2,... ⊂ S0 maximal ist. Proposition 7. Für r0 > 0 hinreichend [...] −1w(hn + hn −1ζ) = (kn − ζ0hn −1)−1w(kn + hn −1(ζ − ζ0)) = 1 kn ( 1 1− (ζ0/(hnkn)) ) w(kn + hn −1(ζ − ζ0)) = 1 kn (1 +O(|kn|−2))w(kn + 1 kn (ζ − ζ0)(1 +O(|kn−2|))) = (1 + o(1))wkn(ζ − ζ0 + o(1)) 15 3.2. Normalität …

2,9 0,0 7,3 7,0 0,3 BY 3,7 3,6 0,1 8,6 8,4 0,2 BE 5,6 5,5 0,1 8,3 8,1 0,2 BB 6,0 6,0 0,0 10,9 10,8 0,1 HB 3,1 3,1 0,0 7,1 7,1 0,0 HH 4,6 4,6 0,0 7,8 7,8 0,0 HE 3,6 3,6 0,0 9,0 9,0 0,0 MV 5,7 5,7 0,0 13 [...] 4 13,2 0,2 NI 3,8 3,7 0,1 7,9 7,7 0,2 NRW 3,3 3,4 -0,1 8,7 8,3 0,4 RP 3,3 3,4 -0,1 8,5 8,2 0,3 SL 3,4 3,4 0,0 8,9 8,8 0,1 SN 6,0 6,0 0,0 12,3 12,2 0,1 ST 6,2 6,0 0,2 11,4 11,2 0,2 SH 3,5 3,5 0,0 8,2 7,9 [...] -1 ,5 – -1 ,0 -1 ,0 – -0 ,5 -0 ,5 – 0, 0 0, 0 – +0 ,5 + 0, 5 – +1 ,0 + 1, 0 – +1 ,5 + 1, 5 – +2 ,0 + 2, 0 – +2 ,5 + 2, 5 – +3 ,0 + 3, 0 – +3 ,5 + 3, 5 – +4 ,0 + 4, 0 – +4 ,5 + 4, 5 – +5 ,0 Prozentuale …

No Title Given

27 0 0 0 1 F b 1 0 1 0 1 0 1 0 m F 0 1 1 0 0 F 0 1 1 0 0 1 0 1 1 0 0 0 1 F 1 0 1 0 0 1 0 0 1 0 1 b 0 m F 0 1 b 0 1 0 0 1 b 1 0 1 0 m F 0 1 0 0 1 b 1 0 1 0 0 F 1 0 1 0 0 1 0 1 0 F b m 0 1 0 0 F b 0 1 0 [...] z16 a 1 b b 0 1 1 1 1 1 1 1 1 1 1 1 1 b 0 0 m 1 0 0 0 0 0 0 0 0 0 0 0 0 c 1 F F F 1 F F 0 0 0 0 0 m 1 1 1 d 0 0 m s 0 0 m 1 1 1 1 F F 0 0 0 e 1 1 1 1 1 1 1 1 F F 0 0 0 0 0 m f 0 0 0 0 0 0 0 0 0 m 1 1 1 1 [...] 0 1 0 b1 0 b s 1 0 0 0 1 F m b 1 0 0 1 F m 0 0 F 0 1 b m F 0 1 b m 1 0 m F b m 0 1 0 m F b 1 0 m F b s 1 0 m F b m 1 0 1 0 0 1 1 0 0 F b m 1 0 0 b s F 0 1 0 1 b m 0 1 0 1 b s 1 0 1 0 b m 0 1 0 0 1 F 1 0 …

ksets/3dcb310e-f2d1-444e-b991-280e0fff2918","type":"application/linkset"},{"href":"https://eldorado.tu-dortmund.de/signposting/linksets/3dcb310e-f2d1-444e-b991-280e0fff2918/json","type":"application/l [...] dby/3dcb310e-f2d1-444e-b991-280e0fff2918","type":"application/vnd.datacite.datacite+xml"}],"anchor":"https://eldorado.tu-dortmund.de/items/3dcb310e-f2d1-444e-b991-280e0fff2918","cite-as":[{"href":"http://hdl [...] de/items/3dcb310e-f2d1-444e-b991-280e0fff2918","type":"text/html"}],"linkset":[{"href":"https://eldorado.tu-dortmund.de/signposting/linksets/3dcb310e-f2d1-444e-b991-280e0fff2918","type":"application/linkset"} …

ing/linksets/f2c4d393-0711-4a44-bd0b-fc98af5ceb55","type":"application/linkset"},{"href":"https://eldorado.tu-dortmund.de/signposting/linksets/f2c4d393-0711-4a44-bd0b-fc98af5ceb55/json","type":"applic [...] escribedby/f2c4d393-0711-4a44-bd0b-fc98af5ceb55","type":"application/vnd.datacite.datacite+xml"}],"anchor":"https://eldorado.tu-dortmund.de/items/f2c4d393-0711-4a44-bd0b-fc98af5ceb55","cite-as":[{"hre [...] tu-dortmund.de/items/f2c4d393-0711-4a44-bd0b-fc98af5ceb55","type":"text/html"}],"linkset":[{"href":"https://eldorado.tu-dortmund.de/signposting/linksets/f2c4d393-0711-4a44-bd0b-fc98af5ceb55","type":"applicatio …

tu-dortmund.de/signposting/linksets/0ce76a74-2bc6-4259-b2ee-85e2e190acac","type":"application/linkset"},{"href":"https://eldorado.tu-dortmund.de/signposting/linksets/0ce76a74-2bc6-4259-b2ee-85e2e190acac/json" [...] tu-dortmund.de/signposting/describedby/0ce76a74-2bc6-4259-b2ee-85e2e190acac","type":"application/vnd.datacite.datacite+xml"}],"anchor":"https://eldorado.tu-dortmund.de/items/0ce76a74-2bc6-4259-b2ee-85e2e190acac" [...] ://eldorado.tu-dortmund.de/items/0ce76a74-2bc6-4259-b2ee-85e2e190acac","type":"text/html"}],"linkset":[{"href":"https://eldorado.tu-dortmund.de/signposting/linksets/0ce76a74-2bc6-4259-b2ee-85e2e190acac" …

Writing your Thesis with LateX

By replacing J× J×U by J× J×U − {(t0, t0, x0)}, we may assume, without loss of generality, that (t0, t0, x0) = (0, 0, 0) = 0 ∈ Rm+2. Now, proceeding as in our argument for the previous corollary, we obtain [...] corollary, we thus obtain d dt ⃓⃓⃓⃓ 0 PE,ω t,0 (e(γ(t))) = d dt ⃓⃓⃓⃓ 0 [γ(0)] (︁ [γ(t)]−1e(γ(t)) )︁ = [︃ γ(0), d dt ⃓⃓⃓⃓ 0 e(γ(t)) ]︃ = [γ(0), de(γ̇(0))] = (∇ωγ̇(0)e)(γ(0)). We notice that this special case [...] ∇ωγ̇(t)e(γ(t)) = ∇ ω γ̇t(0)e(γt(0)) = d du ⃓⃓⃓⃓ 0 PE,ω u,0 e(γt(u)) = d du ⃓⃓⃓⃓ 0 PE,ω t+u,te(γ(t+ u)) = d du ⃓⃓⃓⃓ 0 PE,ω 0,t ◦ P E,ω t+u,0e(γ(t+ u)) = PE,ω 0,t d dt PE,ω t,0 e(γ(t)). 3.6 Introduction …

DeepLearning on FPGAs - Artificial Neural Networks: Backpropagation and more

= { z if z ≥ 0 0 else = max(0, z) ∂h(z) ∂z = { 1 if z ≥ 0 0 else Note: ReLu is not differentiable in z = 0! But: Usually that is not a problem Practical: z = 0 is pretty rare, just use 0 there. It works [...] = { z if z ≥ 0 0 else = max(0, z) ∂h(z) ∂z = { 1 if z ≥ 0 0 else Note: ReLu is not differentiable in z = 0! But: Usually that is not a problem Practical: z = 0 is pretty rare, just use 0 there. It works [...] subgradient of h(z) at 0 DeepLearning on FPGAs 20 New activation function: ReLu Rectified Linear (ReLu): x y −2 −1 1 2 1 2 h(z) = { z if z ≥ 0 0 else = max(0, z) ∂h(z) ∂z = { 1 if z ≥ 0 0 else Note: ReLu is …