Week 1: Preparation

Week 1: Preparation#

Reading Material#

We recommend that you read the textbook. Watching YouTube videos on the week’s topics can be useful, but it should not replace proper preparation for the week’s program and is not recommended as a standalone approach.

Read and study the following:

Key Concepts#

After reading, you should be able to explain the following key concepts:

  • Scalar functions: particularly quadratic forms

  • Vector functions

  • Visualization of functions: Graphs and level curves/sets

  • Continuity

  • The standard inner product (dot product) and norm in \(\mathbb{R}^n\)

  • Partial derivatives and the Gradient vector

This week, we will explore these key concepts in great detail. We expect you to have familiarized yourself with these topics before lectures.

Preparatory Exercises#

I: The Function Value at a Point#

Question a#

Substitute the values \(x = 2\) and \(y = -1\) into the functional expression \(f(x, y) = x^2 + 3xy + 4y^2\) and calculate \(f(2, -1)\).

Hint

Substitute \(x\) with \(2\) and \(y\) with \(-1\) in the expression \(x^2 + 3xy + 4y^2\).

Answer

\(f(2, -1) = 2^2 + 3(2)(-1) + 4(-1)^2 = 4 - 6 + 4 = 2\)

Question b#

Let \(g: \mathbb{R}^2 \to \mathbb{R}\) be given by the functional expression \(g(x_1, x_2) = x_1^2 + 3x_1 x_2 + 4 x_2^2\). Calculate \(g(2, -1)\).

Hint

This is the same functional expression as in the previous question, so method and answer are the same.

Question c#

Let \(\alpha \in \mathbb{R}\). Find \(g(2 \alpha, \alpha)\) and \(g(\alpha, 2 \alpha)\), where \(g\) is defined in the previous question. Calculate the derivative of \(g(2 \alpha, \alpha)\) with respect to \(\alpha\).

Hint

For \(g(2 \alpha, \alpha)\), substitute \(x_1 = 2 \alpha\) and \(x_2 = \alpha\) into \(g(x_1, x_2) = x_1^2 + 3x_1x_2 + 4x_2^2\). Repeat for \(g(\alpha, 2 \alpha)\). For the derivative, differentiate the expression for \(g(2 \alpha, \alpha)\) with respect to \(\alpha\).

Answer

Substituting gives:
\(g(2 \alpha, \alpha) = (2\alpha)^2 + 3(2 \alpha)(\alpha) + 4(\alpha)^2 = (4+6+4)\alpha^2 = 14 \alpha^2\) and \(g(\alpha, 2\alpha) = (\alpha)^2 + 3(\alpha)(2\alpha) + 4(2\alpha)^2 = (1+6+8) \alpha^2 = 15 \alpha^2\).

The derivative of \(g(2 \alpha, \alpha) = 14 \alpha^2\) with respect to \(\alpha\) is \(\frac{d}{d\alpha} \big( 14 \alpha^2 \big) = 28 \alpha\).

II: Limit of a Function \(f: \mathbb{R}^2 \to \mathbb{R}\)#

Let \(f: \mathbb{R}^2 \to \mathbb{R}\) be given by:

\[\begin{equation*} f(x,y) = \begin{cases} \displaystyle \frac{x\,y + x^3 + xy^2}{x^2 + y^2} & \text{if } (x,y)\neq (0,0),\\ 0, & \text{if } (x,y) = (0,0). \end{cases} \end{equation*}\]

Question a#

Find \(f(x, x)\) for \(x \neq 0\). Then find \(f(y, y)\) for \(y \neq 0\).

Hint

Substitute \(y = x\) into the functional expression for \(f(x,y)\). Be sure to simplify the fraction.

Hint

After finding \(f(x, x)\), you can find \(f(y, y)\) by simply replacing \(x\) with \(y\). It’s after all the same expression, just with a different variable name.

Answer

For \(x \neq 0\):

\[\begin{equation*} f(x,x) \;=\; \frac{x\,x + x^3 + xx^2}{x^2 + x^2} \;=\; \frac{x^2}{2x^2} + \frac{2x^3}{2x^2} \;=\; \tfrac{1}{2} + x. \end{equation*}\]

Question b#

Determine \(\lim_{x \to 0} f(x,x)\).

Answer

From the expression in the previous question, we get:

\[\begin{equation*} \lim_{x \to 0} f(x,x) \;=\; \lim_{x\to 0} \Bigl(\tfrac12 + x\Bigr) \;=\; \tfrac12. \end{equation*}\]

Question c#

Determine \( \lim_{x \to 0} f(x,2x) \).

Hint

Substitute \(y = 2x\) into the formula for \(f(x, y)\) and simplify the expression.

Answer

For \( x \neq 0 \), we get \(f(x,2x) \;=\; \tfrac{2}{5} + x\), and therefore:

\[\begin{equation*} \lim_{x \to 0} f(x,2x) \;=\; \lim_{x\to 0} \Bigl(\tfrac{2}{5} + x\Bigr) \;=\; \tfrac{2}{5}. \end{equation*}\]

Question d (extra, optional)#

Consider whether the limit \(\lim_{(x,y)\to (0,0)} f(x,y)\) exists.

Answer

As we approach \((0,0)\) along the lines \(y=x\) and \(y=2x\), we find two different values. Since the limit depends on the path taken, \(\lim_{(x,y)\to (0,0)} f(x,y)\) does not exist. This shows that \(f\) is not continuous at \((0,0)\) (even if we were allowed to change the function value \(f(0,0)\)).

III: Level Curves#

Describe the level curves (contour lines) for the function \(f: \mathbb{R}^2 \to \mathbb{R}\) givet ved \(f(x, y) = x^2 + y^2 - 5\).

Hint

For example, use Python. The level curve for \(f(x, y) = 9\) can be plotted as follows:

x, y  = symbols("x y", real=True)   
f = x**2 + y**2 - 5
dtuplot.plot_implicit(Eq(f,9), x,y)

IV: Graph or Level Curve?#

Below are shown the graph of a function \(f_1\) of one variable and a level curve of a function \(f_2\) of two variables. Which plot is the graph, and which is the level curve?

../_images/094a64368e8ee424f1937062ab8ae8aa850a93ad7c780acc3021e30f148a4958.png

V: Discontinuous at One Point#

Draw, describe, or define a function \(h: \mathbb{R} \to \mathbb{R}\) that is continuous at all points except for a single point.

Answer

One possible function is \(h(x) = \begin{cases} x & \text{if } x \neq 1 \\ 0 & \text{if } x = 1 \end{cases}\). This function is continuous everywhere except at \(x = 1\).

VI: Discontinuity of the Heaviside Step Function#

Question a#

Plot the Heaviside step function given here in Python.

Hint

See this week’s Python demo.

Question b#

Indicate the points where the function is discontinuous.

Answer

The function is discontinuous only at \(0\).

Question c#

Can you prove that the function is discontinuous at \(x_0 = 0\)?

Hint

Discontinuity at a point intuitively means that the function “jumps” at this point. However, this is not a definition we can work with. For example, there are many functions that seemingly “jump” everywhere, but are still continuous. Thomae’s function is an example of a function that seemingly “jumps” at every point on the real line but is continuous at all irrational numbers (and discontinuous at all rational numbers!).

Hint

To prove discontinuity, we need to use the precise meaning of this property. Find the \(\epsilon-\delta\) technique in the textbook.

Hint

A function \(f\) is discontinuous at \(x_0\) if we can find an \(\epsilon > 0\) such that no \(\delta > 0\) can make \(|x-x_0| < \delta \Rightarrow |f(x)-f(x_0)| < \epsilon\) true.
Written with quantifiers: \(f\) is discontinuous at \(x_0\) if and only if \(\exists \epsilon >0 \forall \delta >0 \exists x : |x-x_0| < \delta \text{ and } |f(x)-f(x_0)| \ge \epsilon\).

Hint

The function is discontinuous at \(0\), so we need to consider \(x_0 = 0\). Remember that the Heaviside function satisfies \(f(0)=1\).

Hint

Choose \(\epsilon = 1/2\) (the exact value is unimportant as long as it’s \(<1\)).

Answer

Let \(\delta>0\) be an arbitrarily small number, for example \(\delta = 10^{-100}\) (just remember that while \(\delta\) can be arbitrarily small, it must be greater than zero). Choose \(x = -\delta/2\). Then \(|x-x_0| = |-\delta/2 - 0| = \delta/2 < \delta\) and \(|f(x)-f(x_0)| = |0 - 1| = 1 \ge 1/2 = \epsilon\). This shows that the Heaviside function is discontinuous at \(x_0 = 0\). Alternatively, we can think of this as no \(\delta > 0\) making \(|x-0| < \delta \Rightarrow |f(x)-1| < 1/2\) true.

Above, you could also choose \(x = -\delta/3\) (why is that?). But the argument fails if you choose, for instance, \(x = \delta/2\).