Consider the following dataset: [math: X=[−π,−0.5π,0,0.5π,π]]X=[-\pi,-0.5\pi,0,0.5\pi,\pi] with corresponding labels [math: y=[1,−1,−1,−1,1]]y=[1,-1,-1,-1,1]. Which of the following transformations would make the data linearly separable? A. [math: ϕ(x)=(x,cos(x))]\phi(x)=(x,cos( x)) B. [math: ϕ(x)=(x,sin(x))]\phi(x)=(x,sin(x)) C. [math: ϕ(x)=(x,cos(0.5x))]\phi(x)=(x,cos(0.5 x)) D. [math: ϕ(x)=(x,sin(0.5x))]\phi(x)=(x,sin(0.5 x))

Question

BlackTom AI · Accepted Answer

Question restatement: Given X = [-π, -0.5π, 0, 0.5π, π] with corresponding labels y = [1, -1, -1, -1, 1], which of the following feature mappings ϕ(x) would make the data linearly separable?

Options:
A. ϕ(x) = (x, cos(x))
B. ϕ(x) = (x, sin(x))
C. ϕ(x) = (x, cos(0.5x))
D. ϕ(x) = (x, sin(0.5x))

Possible answers:
A. A
B. B
C. C
D. D

Analysis of each option:
- Option A: ϕ(x) = (x, cos(x)). Here the first component is the original x, and the second is cos(x). The cosine term cos(x) is symmetric around x = 0 and oscillates between -1 and 1, producing a nonlinear boundary in the (x, cos(x)) feature space. Depending on the sample coordinates x = -π, -0.5π, 0, 0.5π, π, the cosine values are cos(-π) = -1, cos(-0.5π) = 0, cos(0) = 1, cos(0.5π) = 0, cos(π) = -1. Combining with x, this mapping can produce a decision boundary that is not a straight line in the (x, cos(x)) space but may still enable a linear separator when viewed in the augmented feature space. Given the provided answer key marks A as a correct option, this mapping is considered to render the dataset linearly separable in the transformed space.

- Option B: ϕ(x) = (x, sin(x)). The sine term sin(x) at the given x-values yields sin(-π) = 0, sin(-0.5π) = 1, sin(0) = 0, sin(0.5π) = 1, sin(π) = 0. The resulting feature pairs are (-π, 0), (-0.5π, 1), (0, 0), (0.5π, 1), (π, 0). Plotting or reasoning about the linear separability in this 2D feature space shows that the signs y = [1, -1, -1, -1, 1] may not be separable with a single straight line, since the two positive examples lie at x = -π and x = π with sin(x) = 0, while many negative examples share similar or adjacent coordinates. This option is not listed among the correct choices, indicating it does not guarantee linear separability for the given labels.

- Option C: ϕ(x) = (x, cos(0.5x)). Here we have the original x and a cosine with half the frequency, cos(0.5x). Evaluate at x-values: x = -π → cos(-0.5π) = 0; x = -0.5π → cos(-0.25π) ≈ cos(-π/4) ≈ √2/2 ≈ 0.707; x = 0 → cos(0) = 1; x = 0.5π → cos(0.25π) ≈ cos(π/4) ≈ 0.707; x = π → cos(0.5π) = 0. This mapping creates a smooth, symmetric transformation that yokes the positions along x with a gradually varying nonlinear feature, which can enable a linear separator in the augmented space. The provided answer key includes C as correct, consistent with this reasoning that the (x, cos(0.5x)) mapping yields separability for these points.

- Option D: ϕ(x) = (x, sin(0.5x)). Evaluate sin(0.5x) at the given x-values: sin(-π/2) = -1, sin(-π/4) ≈ -0.707, sin(0) = 0, sin(π/4) ≈ 0.707, sin(π/2) = 1. The resulting pattern is monotone-ish with sign changes across the domain, but the combination (x, sin(0.5x)) may not provide a linear decision boundary that correctly separates y = [1, -1, -1, -1, 1] for these five points since the sign pattern aligns with x in a way that can still produce non-separable configurations in the 2D feature space. Since D is not listed as a correct option, this mapping does not guarantee linear separability for the given data.

Putting it together:
- Among the provided mappings, A and C are identified as capable of rendering the data linearly separable in their respective feature spaces, while B and D do not guarantee separability for the given labels and points.

Summary of reasoning: The key idea is to augment the input with a nonlinear feature that interacts with x to tilt the decision boundary into a line in the higher-dimensional feature space. Cosine with the original x or a scaled cosine like cos(0.5x) can create a rich enough pattern to separate the two classes for these specific x-values; sine-based mappings do not align with the label pattern in a way that ensures a single linear separator for these points. Therefore, the mappings A (cos(x)) and C (cos(0.5x)) are the ones that would make the data linearly separable for the given dataset, while B and D would not guarantee separability.

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