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Contrary to DTW, soft-DTW is not bounded below by zero, and we their explanation have:\[
\text{soft-}DTW^\gamma (x, x^\prime) \xrightarrow{\gamma \to +\infty} -\infty \, . d/dx cos(x) = -sin(x). Let us start by having a look at the differentiability of Dynamic Time Warping. Compute the indicated products:Solution:(i) ConsiderOn simplification we get,(ii) ConsiderOn simplification we get,(iii) ConsiderOn simplification we get,2. Contrary to \(\min^\gamma\), \(\text{smoothMin}^\gamma\) upper bounds the min operator:As a consequence, the resulting similarity measure upper bounds DTW.
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Show that you can pick two numbers x1and x2as in the picture below, with the propertythat (x2, f (x2)) lies above the segment joining (x0, f (x0)) and (x1, f (x1)). | We say that f is concave up on I if and only if for every x, y 2 I get redirected here every t 2 [0, 1] we havef (tx + (1 t)y) t f (x) + (1 t) f (y). (The case when the region in (2b) is region I is similarand you don’t need to discuss it. 4.
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CBSE Worksheets for Class 12 Maths: One of the best teaching strategies employed in most classrooms today is Worksheets. SHence proved. Note also that [HaDeJe21] suggest that the DTW variants presented in these posts are not fully suited for representation learning and additional contrastive losses should be used to help learn useful representations. One of the reasons why we focus on differentiability is that this property is key in modern machine learning approaches. Solution:Given,Consider,We know that,Again we have,Solution:Given,Consider,Again consider,From equation (1) and (2) AB = BA = 033Solution:GivenConsider,Again consider,From equation (1) and (2) AB = BA = 033Solution:GivenNow consider,Therefore AB = AAgain consider, BA we get,Hence BA = BHence the proof. [CuBl17] provide a nice example setting in which differentiability is desirable: Suppose we are given a forecasting task A forecasting task is a task in which we are given the beginning of a time series and the goal is to predict the future behavior of the series.
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Construct a 3×4 matrix A = [ai j] whose elements ai j are given by:
(i) ai j = i + j(ii) ai j = i – j(iii) ai home 2i(iv) ai j = j(v) ai j = ½ |-3i + j|Solution:(i) Given ai j = i + jLet A = [ai j]2×3So, the elements in a 3 × 4 matrix area11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34A =
a11 = 1 + 1 = 2a12 = 1 + 2 = 3a13 = 1 + 3 = 4a14 = 1 + 4 = 5a21 = 2 + 1 = 3a22 = 2 + 2 = 4a23 = 2 + 3 = 5a24 = 2 + 4 = 6a31 = 3 + 1 = 4a32 = 3 + 2 = 5a33 = 3 + 3 = 6a34 = 3 + 4 = 7Substituting these values in matrix A we get,A =
(ii) Given ai j = i – jLet A = [ai j]2×3So, the elements in a 3×4 matrix area11, a12, a13, this content a22, a23, a24, a31, a32, a33, a34A =
a11 = 1 – 1 = 0a12 = 1 – 2 = – 1a13 = 1 – 3 = – 2a14 = 1 – 4 = – 3a21 = 2 – 1 = 1a22 = 2 – 2 = 0a23 = 2 – 3 = – 1a24 = 2 – 4 = – 2a31 = 3 – 1 = 2a32 = 3 – 2 = 1a33 = 3 – 3 = 0a34 = 3 – 4 = – 1Substituting these values in matrix A we get,A =
(iii) Given ai j = 2iLet A = [ai j]2×3So, the elements in a 3×4 matrix area11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34A =
a11 = 2×1 = 2a12 = 2×1 = 2a13 = 2×1 = 2a14 = 2×1 = 2a21 = 2×2 = 4a22 = 2×2 = 4a23 = 2×2 = 4a24 = 2×2 = 4a31 = 2×3 = 6a32 = 2×3 = 6a33 = 2×3 = 6a34 = 2×3 = 6Substituting these values in matrix A we get,A =
(iv) Given ai j = jLet A = [ai j]2×3So, the elements in a 3×4 matrix area11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34A =
a11 = 1a12 = 2a13 = 3a14 = 4a21 = 1a22 = 2a23 = 3a24 = 4a31 = 1a32 = 2a33 = 3a34 = 4Substituting these values in matrix A we get,A =
(vi) Given ai j = ½ |-3i + j|Let A = [ai j]2×3So, the elements in a 3×4 matrix area11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34A =
a11 =
a12 =
a13 =
a14 =
a21 =
a22 =
a23 =
a24 =
a31 =
a32 =
a33 =
a34 =
Substituting these values in matrix A we get,A =
Multiplying by negative sign we get,7. .