Week 5: Preparation

Week 5: Preparation#

Reading Material#

  • Reading: Chapter 4

  • Python: Demo 5

Key Concepts#

Note

Note that degree and order of polynomials sometimes are used synonymously. We use “degree” in this course.

Orders of magnitude and Big O notation are not part of the curriculum, but it is useful to become familiar with “Big O” notation, as it appears, for example, in Sympy outputs (see today’s Python demo).

Long Day will cover:

  • Tangent lines and tangent planes

  • Taylor polynomials in one variable

  • Taylor polynomials in \(n\) variables

  • Taylor polynomials of vector functions

  • Remainder term and error estimation

Short Day will cover:

  • Diverging and converging Taylor series

  • Taylor’s theorem

  • Taylor’s limit formula

Preparatory Exercises#

I: Tangent Lines and Tangent Planes#

Question a#

A function \(f : \mathbb{R} \to \mathbb{R}\) is given by:

\[\begin{equation*} f(x) = x^2. \end{equation*}\]

Find the equation of the tangent line at \(x_0 = 1\).

Hint

Calculate \(f(1)\) and \(f'(1)\), and then use the well-known formula for the slope of the line:

\[\begin{equation*} f'(1) = \frac{y - f(1)}{(x-1)}. \end{equation*}\]

Now isolate \(y\).

Answer

\(y = 1 + 2(x-1)\)

Question b#

A function \(f : \mathbb{R}^2 \to \mathbb{R}\) is given by:

\[\begin{equation*} f(x,y) = x^2 + y^2. \end{equation*}\]

Find the equation of the tangent plane at the point \((1,1)\).

Hint

Calculate the gradient \(\nabla f(1,1)\), and use the formula

\[\begin{equation*} z = f(1,1) + \frac{\partial f}{\partial x}(1,1)(x-1) + \frac{\partial f}{\partial y}(1,1)(y-1). \end{equation*}\]

Answer

\(z = 2 + 2(x-1) + 2(y-1)\)

Intuitive explanation

A tangent line is the line that “touches” a curve at a single point without crossing it. Similarly, a tangent plane is a surface that “touches” a 3D surface at a single point. The tangent line and tangent plane are described by first-degree Taylor polynomials.

II: Taylor Polynomials in One Variable#

Question a#

Find the first-degree Taylor polynomial of

\[\begin{equation*} f(x) = \sin(x) \end{equation*}\]

from the expansion point \(x_0 = 0\). This polynomial is also called the approximating polynomial of degree one.

Hint

Use \(f(0)\) and \(f'(0)\).

Answer

\(P_1(x) = 0 + 1\cdot x = x\)

Question b#

Find the second-degree Taylor polynomial of

\[\begin{equation*} f(x) = \cos(x) \end{equation*}\]

from the expansion point \(x_0 = 0\). This polynomial is also called the approximating polynomial of degree two.

Hint

Use \(f(0)\), \(f'(0)\) and \(f''(0)\).

Answer

\(P_2(x) = 1 + 0\cdot x + \frac{-1}{2}x^2 = 1 - \frac{1}{2}x^2\)

Intuitive explanation

Taylor polynomials provide a way to approximate a “complicated” function by a sum of simple polynomial terms. The first-degree polynomial gives a linear approximation, while the second-degree polynomial includes a quadratic term for better accuracy.

III: “Big O” Notation and Growth Rate#

Note

Here, we go slightly beyond the curriculum: “Big O” notation is used to describe how fast a function approximately grows. For example, if a given function is \(O(n^2)\) then its growth rate is at most proportional to \(n^2\) as \(n\) approaches infinity.

For a given functional expression, it is usually sufficient to focus on the term that grows the fastest, as lower-order terms and constants do not affect asymptotic growth. For example, a function \(f : \mathbb{N} \to \mathbb{N}\) given by

\[\begin{equation*} f(n) = 2^n + 100 n^5 \end{equation*}\]

is \(O(2^n)\), because exponential growth indicated by \(2^n\) causes a much faster increase for growing \(n\) than polynomial growth with the term \(n^5\). (As \(n\) grows to very large values, the term \(2^n\) will eventually dominate with \(100n^5\) becoming comparatively insignificant - when considering asymptotic growth, only the fastest-growing term is relevant.)

Let

\[\begin{equation*} f(n) = 4n^2 + 10n + 6. \end{equation*}\]

Show that \(f(n)\) is \(O(n^2)\). That is, find constants \(C > 0\) and \(n_0 \in \mathbb{N}\) such that for all \(n \geq n_0\), the following holds:

\[\begin{equation*} |4n^2 + 10n + 6| \leq C \cdot n^2. \end{equation*}\]

Hint

For \(n \geq 1\) note that:

\[\begin{equation*} 10n \leq 10n^2 \quad \text{and} \quad 6 \leq 6n^2. \end{equation*}\]

Use these inequalities to show that:

\[\begin{equation*} 4n^2 + 10n + 6 \leq 4n^2 + 10n^2 + 6n^2 = 20n^2. \end{equation*}\]

Answer

With the help in the hint, we find that we, for example, can choose \(C = 20\) and \(n_0 = 1\).

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