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))

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We are given a small 1D dataset with 5 points: X = [-π, -0.5π, 0, 0.5π, π] and corresponding labels y = [1, -1, -1, -1, 1]. We need to evaluate which feature mappings φ(x) make the data linearly separable in the new 2D space (the first component remains x, the second is the chosen transform).

Option A: φ(x) = (x, cos(x))
- Compute cos(x) at the points: cos(-π) = -1, cos(-0.5π) = 0, cos(0) = 1, cos(0.5π) = 0, cos(π) = -1.
- The mapped points are: (-π, -1) with label 1, (-0.5π, 0) with label -1, (0, 1) with label -1, (0.5π, 0) with label -1, (π, -1) with label 1.
- Visually, the two positive examples lie at the leftmost and rightmost x with y2 = -1, while the negative examples occupy the top and middle-left/middle-right regions. It is plausible that a straight line can separate the ends from the interior cluster in this 2D space, particularly since the end points share the same label and the interior points sit at distinct cos values. Therefore A can yield linear separability.

Option B: φ(x) = (x, sin(x))
- Compute sin(x) at the points: sin(-π) = 0, sin(-0.5π) = -1, sin(0) = 0, sin(0.5π) = 1, sin(π) = 0.
- Mapped points: (-π, 0) with label 1, (-0.5π, -1) with label -1, (0, 0) with label -1, (0.5π, 1) with label -1, (π, 0) with label 1.
- The two positive examples lie on the extremes with y2 = 0, while interior points include negative y2 values. A single straight line would struggle to separate the end points (which lie on the same horizontal level as some negatives) from the interior points, making linear separability unlikely. Thus B is unlikely to yield separability.

Option C: φ(x) = (x, cos(0.5x))
- Compute cos(0.5x) for x values: 0.5x = [-0.5π, -0.25π, 0, 0.25π, 0.5π]; cos(-0.5π) = 0, cos(-0.25π) ≈ 0.707, cos(0) = 1, cos(0.25π) ≈ 0.707, cos(0.5π) = 0.
- Mapped points: (-π, 0) with label 1, (-0.5π, 0.707) with label -1, (0, 1) with label -1, (0.5π, 0.707) with label -1, (π, 0) with label 1.
- Here the end points still share label 1 while the interior points have higher cos values. The distribution in this 2D plane creates a pattern where a linear boundary can potentially separate the ends (y=0) from the interior cluster (y≈0.707–1). This setup is conducive to linear separability, so C is plausible.

Option D: φ(x) = (x, sin(0.5x))
- Compute sin(0.5x) for x values: 0.5x = [-0.5π, -0.25π, 0, 0.25π, 0.5π]; sin(-0.5π) = -1, sin(-0.25π) ≈ -0.707, sin(0) = 0, sin(0.25π) ≈ 0.707, sin(0.5π) = 1.
- Mapped points: (-π, -1) with label 1, (-0.5π, -0.707) with label -1, (0, 0) with label -1, (0.5π, 0.707) with label -1, (π, 1) with label 1.
- The end points are labeled 1, while the inner points include negative and near-zero values of sin(0.5x). This arrangement tends to defy a single straight line separating the two classes, because the same class labels appear across a broad range of second-coordinate values, creating overlap in the projection. Consequently, D is unlikely to be linearly separable.

Summarizing the assessments: A provides a configuration that can support a linear decision boundary separating the extremes from the interior, B introduces interior points that collide with the ends in the (x, sin(x)) space, C yields a similar end-vs-interior separation pattern with cos(0.5x), and D tends to mix class labels along the boundary in the transformed space. Based on these reasoned evaluations, the transformations that would make the data linearly separable are A and C.

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