Calculate The Gradient Of The Function F(x) = 1x2x3 + 2x5 +

Calculate The Gradient Of The Functionfx1x2x3 2x51x32x43x1x2x2

Calculate the gradient of the function f(x₁, x₂, x₃) = 2x₁ + x₂ + x₃ + 2x₁x₂ + x₂x₃ at the point (1, 1, 1). Additionally, analyze the function f(x₁, x₂) = (x₁ + x₂)⁴ – 8(x₁ + x₂)² to identify all local minima and maxima, and hypothesize its graph shape. Explore the function f(x) = xᵗAx, where A is invertible, demonstrating it has only one stationary point at 0 and providing specific matrices A for which 0 is a minimizer or neither a maximizer nor a minimizer. Complete the proof of Lemma 7 from the lecture notes. Find the gradient of f₁(x) = (xᵗA x)² and f₂(x) = (xᵗA x)ⁿ for A being an n×n matrix. For a fat matrix A ∈ R^{n×p} with n

Paper For Above instruction

The comprehensive analysis of gradients and optimization functions forms the backbone of many fields in mathematics and engineering, including calculus, machine learning, and control systems. This paper begins by calculating the gradient of a multivariable function, proceeds to analyze a quadratic form's stationarity and minima, and extends to discussing solutions of underdetermined systems, culminating with graphical visualization in R.

Gradient of the Function f(x₁, x₂, x₃)

The function in question is f(x₁, x₂, x₃) = 2x₁ + x₂ + x₃ + 2x₁x₂ + x₂x₃. To compute its gradient, we derive the partial derivatives with respect to each variable:

  • ∂f/∂x₁ = 2 + 2x₂
  • ∂f/∂x₂ = 1 + 2x₁ + x₃
  • ∂f/∂x₃ = 1 + x₂

At the point (1, 1, 1), these derivatives evaluate to:

  • ∂f/∂x₁ = 2 + 2(1) = 4
  • ∂f/∂x₂ = 1 + 2(1) + 1 = 4
  • ∂f/∂x₃ = 1 + 1 = 2

Thus, the gradient vector at (1,1,1) is ∇f = (4, 4, 2).

Analysis of the Function f(x₁, x₂) = (x₁ + x₂)⁴ – 8(x₁ + x₂)²

This function depends solely on the variable s = x₁ + x₂. Rewriting as f(s) = s⁴ – 8s², critical points occur where its derivative equals zero:

f'(s) = 4s³ – 16s = 4s(s² – 4) = 0

Solutions are s = 0, s = ±2. The second derivative is:

f''(s) = 12s² – 16

  • At s = 0, f''(0) = –16
  • At s = ±2, f''(±2) = 12×4 – 16 = 48 – 16 = 32 > 0, indicating local minima

Returning to the original variables, the minima occur along lines where x₁ + x₂ = ±2, and the maximum along x₁ + x₂ = 0. The graph resembles a 'double-well' shape with peaks at zero and valleys at ±2.

Stationary Points of quadratic form f(x) = xᵗ A x

Considering A invertible ensures that the gradient of f is:

∇f(x) = 2A x

Setting ∇f(x) = 0 gives x = 0, the only stationary point. To classify whether this point is a min or max, analyze the definiteness of A:

  • If A is positive definite, then f(x) ≥ 0 with equality only at x = 0, making 0 a global minimum.
  • If A has negative eigenvalues, 0 is not a minimizer or maximizer; instead, the nature depends on A's definiteness.

For a specific A where 0 is a minimizer, choose A as the identity matrix I. For A where 0 is neither, pick A with mixed eigenvalues, for example, diagonal with entries {1, -1}.

The completeness of the proof in Lemma 7 involves showing that under invertibility, the only stationary point is at 0 because the gradient's zero corresponds solely to x = 0.

Gradients of Functions involving Quadratic Forms

(a) For f₁(x) = (xᵗ A x)², the gradient is derived via chain rule:

∇f₁(x) = 2(xᵗ A x) · ∇(xᵗ A x) = 2(xᵗ A x) · 2A x = 4(xᵗ A x) A x

(b) For f₂(x) = (xᵗ A x)ⁿ, its gradient is:

∇f₂(x) = n(xᵗ A x)^{n – 1} · 2A x = 2n(xᵗ A x)^{n – 1} A x

Solutions in Under-Determined Systems

For A ∈ R^{n×p} with n

minimize ‖x‖₂ subject to A x = y

The solution is given by the Moore-Penrose pseudoinverse:

x = A† y = Aᵗ (A Aᵗ)^{–1} y

This minimizes ‖x‖₂, as proven by convexity of the norm and properties of the pseudoinverse.

Graphing in R

To visualize the function f(x₁, x₂) = (x₁ – 1)² + (x₂ – 1)² + x₁ x₂ over specified ranges, use R commands:

outer_vals_x1 

outer_vals_x2 

z 

persp(outer_vals_x1, outer_vals_x2, z, main="Surface plot of the function", xlab="x1", ylab="x2", zlab="f(x1,x2)")

This code generates a 3D surface plot, providing insights into the function's topology.

In conclusion, the mathematical exploration encompasses gradient calculations, critical point analysis, solving underdetermined systems, and visualization techniques essential for a comprehensive understanding of multivariable calculus applications.

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