Week 8: Closure

Week 8: Closure#

Continue working on the the preparatory exercises and the in-class exercises that you have not yet completed.

Key Concepts#

  • The Riemann Integral of Scalar Functions of \(n\) variables

  • The Riemann Integral of Vector Functions

  • The Change-of-Variables Theorem: Coordinate Change in \(\mathbb{R}^n\)

  • The Jacobian Determinant

  • Typical coordinates:

    • In \(\mathbb{R}^2\): Cartesian and Polar Coordinates

    • In \(\mathbb{R}^3\): Cartesian, Spherical, Cylindrical/Semi-Polar Coordinates

If there are still concepts you are unsure about, you should reread the relevant chapters in the textbook or revisit the exercises of the week.

Extra Exercises#

We do not expect you to complete more exercises than those from from the week’s program. The following additional exercises are purely an optional offer for those who want extra practice and challenge.

1: Integration as a “Smoothener”#

In this exercise we aim to study the following rather loosely stated proposition:

Integration makes functions more smooth (i.e., they smoothen sharp corners), while differentiation does the opposite.

Remember that the ReLU function is continuous but not differentiable.

Question a#

Argue (without calculations!) that \(h_1(x) = \int_{0}^x \text{ReLU}(t) \mathrm dt\) is a differentiable function with respect to \(x \in \mathbb{R}\).

Hint

It is an antiderivative of ReLU according to the fundamental theorem of calculus.

Question b#

Find a functional expression for \(h_1(x)\). Argue that \(h \in C^1(\mathbb{R})\).

Answer

\[\begin{equation*} h_1(x) = \begin{cases} 0, & x < 0, \\ \frac{1}{2}x^2, & x \geq 0. \end{cases} \end{equation*}\]

As ReLU is continuous, its antiderivative \(h_1\) is differentiable with \(h_1'(x) = \text{ReLU}(x)\). Hence, \(h_1\) is of the type \(C^1(\mathbb{R})\).

Question c#

Does \(\tilde{h}_1(x) = \int_{1}^x \text{ReLU}(t) \mathrm dt\) give the same functional expression as \(h_1\)?

Answer

No, they do not have the same functional expression, but they are both antiderivatives of ReLU. We see this because

\[\begin{equation*} \tilde{h}_1(x) = \int_{1}^{x} \operatorname{ReLU}(t)\,\mathrm dt = \int_{0}^{x} \operatorname{ReLU}(t)\,\mathrm dt - \int_{0}^{1} \operatorname{ReLU}(t)\,\mathrm dt = h_1(x) - h_1(1). \end{equation*}\]

As \(h_1(1) = \frac{1}{2}\), then \(\tilde{h}_1\) differs from \(h_1\) by the constant \(-\frac{1}{2}\).

Question d#

Find a functional expression for \(h_2(x) := \int_{0}^x h(t) \mathrm dt\). Argue that \(h_2 \in C^2(\mathbb{R})\).

Answer

\[\begin{equation*} h_1(x) = \begin{cases} 0, & x < 0, \\ \frac{1}{6}x^3, & x \geq 0. \end{cases} \end{equation*}\]

Question e#

If we keep on integrating like this, will the function then become more and more smooth? Meaning, will \(h_n\) become a \(C^n\) function?

Answer

Yes.

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