-- Sep 13 In-Class Exercise

δ= 1/1000 ε = 1/(n + 1) m=max(144lnδ/2, m1) m1=r+s=144(n + 1)^2/25(kε)^2 = 144(n+1)^2/25(k(1/(n + 1))^2 = 144(n + 1)^4/25k^2 Above, one can observe that m1 is always positive, whereas the ln(1/1000) is negative. So, m = 144(n + 1)^4/25k^2. Since the number of examples used = 2m/ε = 2m(n + 1), one can next say, # ex = 2(n + 1)( 144(n + 1)^4/25k^2 = (288/25)(n + 1)^5/k^2 For a gradual threshold function, one can choose c = 3. Recall, ∣∣{x∈{0,1}n:∣∣∣∣w∣∣∣∣−1∣∣w⃗ ⋅x⃗ −θ∣∣≤τ}∣∣=∣∣{x∈{0,1}n:∣∣w⃗ ⋅x⃗ −θ∣∣≤τ∣∣∣∣w∣∣∣∣}∣∣≤2τ∣∣∣∣w∣∣∣∣+1<2τ2n+1<3τ2n and we can take c in the definition of gradual to be 3. By definition of k, k = 2*c = 2*3 = 6. So, the # ex = (288/25)(n + 1)^5/k^2 = (8/25)(n + 1)^5

-- Sep 13 In-Class Exercise

number of examples= 2m /epsilon m = max(144lnδ/2,mi) m1 = 144(n+1)4 / 25 k2 so, number of examples = 288 (n+1)5 / 25(1/6)2 (Repost)