Syllabus Outline

In this subchapter, we define what it means for a function to have a limit, what it means to be continuous at a point, and prove the equivalence of the sequence (Heine) and $\epsilon-\delta$ (Cauchy) definitions. Definition of a Limit (Cauchy): We say that $\lim_{x\to x_0} f(x) = L$ if and only if: $\forall \epsilon > 0, \quad \exists \delta > 0 \quad \text{such that} \quad \forall x \in D: \quad 0 < |x - x_0| < \delta \implies |f(x) - L| < \epsilon$ The condition $0 < |x - x_0|$ ensures that $x \neq x_0$. The value of $f(x_0)$ itself is irrelevant to the limit. Definition of Continuity at a Point $x_0$: A function $f: D \to \mathbb{R}$ is continuous at $x_0 \in D$ if and only if $\lim_{x\to x_0} f(x) = f(x_0)$. - Heine (Sequence Continuity): For every sequence $(x_n)_{n\in\mathbb{N}}$ in $D$ that converges to $x_0$ ($\lim_{n\to\infty} x_n = x_0$), the sequence of function values converges to $f(x_0)$ ($\lim_{n\to\infty} f(x_n) = f(x_0)$). - Cauchy ($\epsilon-\delta$ Continuity): $\forall \epsilon > 0, \quad \exists \delta > 0 \quad \text{such that} \quad \forall x \in D: \quad |x - x_0| < \delta \implies |f(x) - f(x_0)| < \epsilon$ Example 1: Equivalence Proof (2023 Exam, Q4c)

Prove that the sequence definition of continuity implies the $\epsilon-\delta$ definition at a point $x_0 \in D$. Proof by Contradiction: Assume the sequence definition holds, but the $\epsilon-\delta$ definition is false. If the $\epsilon-\delta$ condition is false, its logical negation is: $\exists \epsilon > 0 \quad \text{such that} \quad \forall \delta > 0, \quad \exists x \in D: \quad |x - x_0| < \delta \quad \text{and} \quad |f(x) - f(x_0)| \ge \epsilon$ Since this holds for all $\delta > 0$, we can choose a sequence of window sizes $\delta_n = \frac{1}{n}$ for $n \in \mathbb{N}$. For each $\delta_n$, there must exist a point $x_n \in D$ such that: $|x_n - x_0| < \frac{1}{n} \quad \text{and} \quad |f(x_n) - f(x_0)| \ge \epsilon$ Now look at the sequence $(x_n)_{n\in\mathbb{N}}$: - Since $|x_n - x_0| < \frac{1}{n}$ for all $n$, by the Sandwich Theorem, $\lim_{n\to\infty} x_n = x_0$. - By the sequence definition of continuity, this must imply $\lim_{n\to\infty} f(x_n) = f(x_0)$. - However, our sequence satisfies $|f(x_n) - f(x_0)| \ge \epsilon$ for all $n$, which means $f(x_n)$ cannot converge to $f(x_0)$. This is a contradiction. Therefore, the $\epsilon-\delta$ definition must hold. $\square$ Example 2: Rigorous $\epsilon-\delta$ Proof (2023 Exam, Q4d)

Use the $\epsilon-\delta$ definition of continuity to prove that $f(x) = x^2$ is continuous at $x_0 = 1$. Solution: We want to show that for any $\epsilon > 0$, we can find a $\delta > 0$ such that $|x - 1| < \delta \implies |x^2 - 1| < \epsilon$. First, factor the target term: $|x^2 - 1| = |x - 1| \cdot |x + 1|$ To control the size of $|x + 1|$, we assume a preliminary bound $\delta \le 1$. If $|x - 1| < 1$: $-1 < x - 1 < 1 \implies 0 < x < 2 \implies 1 < x + 1 < 3 \implies |x + 1| < 3$ Thus, $|x^2 - 1| < 3 |x - 1|$. We want this to be less than $\epsilon$, so we require $|x - 1| < \frac{\epsilon}{3}$. To satisfy both constraints, we choose: $\delta = \min\left(1, \frac{\epsilon}{3}\right)$ Verification: Let $\epsilon > 0$ be given. If $|x - 1| < \delta$: - Since $\delta \le 1$, we have $|x + 1| < 3$. - Since $\delta \le \frac{\epsilon}{3}$, we have $|x - 1| < \frac{\epsilon}{3}$. - Therefore, $|x^2 - 1| = |x - 1| \cdot |x + 1| < \left(\frac{\epsilon}{3}\right) \cdot 3 = \epsilon$. $\square$ Example 3: Moodle Logic Audit (2025 Exam, CF Q1)

Statement: "If $f$ and $g$ are continuous at $x = 3$, then $f/g$ is continuous at $x = 3$." Solution: Logically incorrect. If $g(3) = 0$, then $f/g$ is undefined at $x=3$, hence not continuous. To make the statement true, we must add the condition $g(3) \neq 0$. Exercise 1 (Analytical Proof):

Prove that $f(x) = 5x - 7$ is continuous at $x_0 = 3$ using the formal $\epsilon-\delta$ definition. Solution: We want to show that for any $\epsilon > 0$, we can find a $\delta > 0$ such that: $|x - 3| < \delta \implies |(5x - 7) - (5(3) - 7)| < \epsilon$ First, simplify the target term: $|(5x - 7) - 8| = |5x - 15| = 5|x - 3|$ We want this to be less than $\epsilon$: $5|x - 3| < \epsilon \implies |x - 3| < \frac{\epsilon}{5}$ We choose $\delta = \frac{\epsilon}{5}$. Verification: Let $\epsilon > 0$ be given, and choose $\delta = \frac{\epsilon}{5}$. If $|x - 3| < \delta$: $|f(x) - f(3)| = 5|x - 3| < 5\delta = 5\left(\frac{\epsilon}{5}\right) = \epsilon \quad \square$ Exercise 2 (Conceptual Product Logic):

1. Let $f(x) = |x|$ and $g(x) = \text{sgn}(x)$ (where $\text{sgn}(0)=0$, $\text{sgn}(x)=1$ if $x>0$, and $-1$ if $x<0$). Is $h(x) = f(x)g(x)$ continuous at $x_0=0$?

2. Does the continuity of a product function $f \cdot g$ at $x_0$ imply that both $f$ and $g$ must be continuous at $x_0$? Solution:

3. The product is $h(x) = |x|\text{sgn}(x) = x$ for all $x \in \mathbb{R}$. Since $h(x) = x$ is a linear polynomial, it is continuous everywhere, including at $x_0=0$.

4. No. In our example above, $h(x) = f(x)g(x)$ is continuous at $x_0=0$, and $f(x) = |x|$ is continuous at $0$. However, $g(x) = \text{sgn}(x)$ is discontinuous at $x_0=0$ (since $\lim_{x\to0^+} g(x) = 1 \neq \lim_{x\to0^-} g(x) = -1$). Thus, the product of a continuous and a discontinuous function can be continuous. \subsection{Subchapter 1.2: 1D Continuity Theorems (IVT, Bolzano, Bisection, EVT)}

In this subchapter, we define differentiability at a point, study derivative rules (product, quotient, chain, and inverse function rules), and establish a table of basic derivatives for calculator-free computations. 1. What is Differentiability? - "Finding the Slope" - Intuition: The derivative $f'(x_0)$ is the slope of the tangent line to the graph at the point $x_0$. It measures the instantaneous rate of change of the function. - Mathematical Definition: A function $f: D \to \mathbb{R}$ is differentiable at $x_0 \in D$ if and only if the following limit (the difference quotient) exists: $f'(x_0) = \lim_{h\to0} \frac{f(x_0 + h) - f(x_0)}{h} = \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x - x_0}$ - Differentiability implies Continuity: If a function is differentiable at $x_0$, it is guaranteed to be continuous at $x_0$. The converse is not true (e.g., $f(x) = |x|$ is continuous at $0$, but not differentiable there because it has a sharp corner). 2. Derivative Rules - Product Rule: $(f \cdot g)' = f'g + fg'$ - Quotient Rule: $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$ (where $g \neq 0$) - Chain Rule (Composition): $(f(g(x)))' = f'(g(x)) \cdot g'(x)$ - Derivative of an Inverse Function: If $f$ is differentiable, strictly monotonic, and $f'(x) \neq 0$: $\left(f^{-1}\right)'(y_0) = \frac{1}{f'(x_0)} \quad \text{where } y_0 = f(x_0) \text{ and } x_0 = f^{-1}(y_0)$ Intuition: The slope of the inverse function at $y_0$ is the reciprocal of the slope of the original function at $x_0$. 3. Basic Derivatives Table (Everyday Tools) ll|ll@{}} Function $f(x)$ & Derivative $f'(x)$ & Function $f(x)$ & Derivative $f'(x)$

$x^r$ & $r x^{r-1}$ & $\sin(x)$ & $\cos(x)$

$e^x$ & $e^x$ & $\cos(x)$ & $-\sin(x)$

$\ln(x)$ & $1/x$ & $\tan(x)$ & $1 + \tan^2(x) = \frac{1}{\cos^2(x)}$

$\arctan(x)$ & $\frac{1}{1 + x^2}$ & $\arcsin(x)$ & $\frac{1}{\sqrt{1 - x^2}}$

Example 1: Difference Quotient Proof (2023 Exam, Q2b)

Let $n \in \mathbb{N}$. Prove that the derivative of $f(x) = x^n$ is $f'(x) = n x^{n-1}$ using the definition of the difference quotient. Proof: By the definition of the derivative: $f'(x) = \lim_{h\to0} \frac{(x + h)^n - x^n}{h}$ Using the Binomial Theorem to expand $(x+h)^n$: $(x+h)^n = x^n + n x^{n-1} h + \binom{n}{2} x^{n-2} h^2 + \dots + h^n$ Substitute this back: $\frac{(x+h)^n - x^n}{h} = \frac{n x^{n-1} h + \binom{n}{2} x^{n-2} h^2 + \dots + h^n}{h} = n x^{n-1} + \binom{n}{2} x^{n-2} h + \dots + h^{n-1}$ Now, take the limit as $h \to 0$. Every term containing $h$ vanishes, leaving: $f'(x) = n x^{n-1} \quad \square$ Example 2: Inverse Function Derivative (Classic Exam Template)

Let $f(x) = x^3 + x$. Find the value of $(f^{-1})'(2)$ without a calculator. Solution: - We want to find $(f^{-1})'(2)$. Here $y_0 = 2$. - Find $x_0$ such that $f(x_0) = 2 \implies x_0^3 + x_0 = 2$. - By inspection (guessing small integers), we see that $x_0 = 1$ works: $1^3 + 1 = 2$. Since $f'(x) = 3x^2 + 1 > 0$, the function is strictly increasing, meaning $x_0 = 1$ is the unique solution. - Compute the derivative of $f(x)$: $f'(x) = 3x^2 + 1$. - Evaluate the derivative at $x_0 = 1$: $f'(1) = 3(1)^2 + 1 = 4$. - Use the inverse derivative formula: $\left(f^{-1}\right)'(2) = \frac{1}{f'(1)} = \frac{1}{4} \quad \square$ Example 3: Differentiability of Piecewise Functions

Let $f(x) = x^2$ if $x \ge 0$ and $x$ if $x < 0$. Determine if $f$ is differentiable at $x_0 = 0$. Solution: - Check Continuity: $\lim_{x\to0^+} f(x) = 0^2 = 0$, and $\lim_{x\to0^-} f(x) = 0$. Since $f(0) = 0$, the function is continuous. - Check Derivatives from both sides: - Right-hand derivative: $\lim_{h\to0^+} \frac{f(h) - f(0)}{h} = \lim_{h\to0^+} \frac{h^2 - 0}{h} = \lim_{h\to0^+} h = 0$. - Left-hand derivative: $\lim_{h\to0^-} \frac{f(h) - f(0)}{h} = \lim_{h\to0^-} \frac{h - 0}{h} = \lim_{h\to0^-} 1 = 1$. - Since the left-hand derivative ($1$) and the right-hand derivative ($0$) are not equal, $f$ is not differentiable at $x_0 = 0$ (the graph has a sharp kink at the origin). $\square$ Exercise 1 (Inverse Derivative):

Let $f(x) = x^3 + 2x$. Find $(f^{-1})'(3)$ without a calculator. Solution: - Let $y_0 = 3$. Find $x_0$ such that $f(x_0) = 3 \implies x_0^3 + 2x_0 = 3$. - By inspection, $x_0 = 1$ is the unique solution (since $1^3 + 2(1) = 3$). - Compute the derivative: $f'(x) = 3x^2 + 2$. - Evaluate at $x_0 = 1$: $f'(1) = 3(1)^2 + 2 = 5$. - Use the formula: $\left(f^{-1}\right)'(3) = \frac{1}{f'(1)} = \frac{1}{5}$ Exercise 2 (Piecewise Differentiability):

Consider the piecewise function: $g(x) =$\begin{cases} x^2 + x & \text{if } x \ge 1 3x - 1 & \text{if } x < 1 \end{cases}$$ Determine if $g(x)$ is differentiable at $x_0 = 1$ without a calculator. Solution: - Check Continuity at $x_0 = 1$: - $\lim_{x\to1^+} g(x) = 1^2 + 1 = 2$. - $\lim_{x\to1^-} g(x) = 3(1) - 1 = 2$. - $g(1) = 2$. Since the limits match the function value, $g$ is continuous at $x_0 = 1$. - Check Derivatives: - For $x > 1$, $g'(x) = (x^2+x)' = 2x + 1$. Thus, the right-hand derivative is $g'_+(1) = 2(1) + 1 = 3$. - For $x < 1$, $g'(x) = (3x-1)' = 3$. Thus, the left-hand derivative is $g'_-(1) = 3$. - Since the left-hand and right-hand derivatives are equal ($g'_+(1) = g'_-(1) = 3$), the function $g(x)$ is differentiable at $x_0 = 1$, and its derivative is $g'(1) = 3$. $\square$ Exercise 3 (Kink proof):

Use the difference quotient definition to prove that $f(x) = |x|$ is not differentiable at $x_0 = 0$. Solution: The difference quotient limit at $x_0 = 0$ is: $\lim_{h\to0} \frac{|0 + h| - |0|}{h} = \lim_{h\to0} \frac{|h|}{h}$ Let's evaluate the limit from both sides: - From the right ($h > 0$), $|h| = h$, so: $\lim_{h\to0^+} \frac{h}{h} = 1$. - From the left ($h < 0$), $|h| = -h$, so: $\lim_{h\to0^-} \frac{-h}{h} = -1$. Since the limits from the left and right do not match ($1 \neq -1$), the limit of the difference quotient does not exist. Thus, $f(x) = |x|$ is not differentiable at $x_0 = 0$. $\square$ \subsection{Subchapter 2.2: 1D Curve Sketching (Monotonicity, Extrema, Convexity)}

In this subchapter, we explore how the definite integral is constructed as a limit of Riemann sums (using upper and lower sums), the algebraic properties of integrals, and the Fundamental Theorem of Calculus (FTC) which links differentiation and integration. 1. Antiderivatives (Indefinite Integrals) - Definition: Let $f(x)$ be defined on an open interval $I$. A function $F(x)$ is called an antiderivative or primitive of $f(x)$ on $I$ if: $F'(x) = f(x) \quad \text{for all } x \in I$ - Uniqueness: If $F(x)$ is an antiderivative of $f(x)$, then $F(x) + C$ (where $C \in \mathbb{R}$ is any constant) is also an antiderivative, because the derivative of any constant is zero: $\frac{d}{dx}[F(x) + C] = F'(x) + 0 = f(x)$. - Notation: We write the indefinite integral as: $\int f(x) \, dx = F(x) + C$ 2. The Definite Integral (Riemann Integral) - Intuition: The definite integral $\int_a^b f(x) \, dx$ represents the signed area between the graph of $f(x)$ and the $x$-axis from $x=a$ to $x=b$. - Partitioning: To find this area formally, we divide the interval $[a,b]$ into $n$ equal subintervals of width $\Delta x = \frac{b-a}{n}$. The boundaries are $x_k = a + k \Delta x$ for $k = 0, \dots, n$. - Lower & Upper Sums: - The lower sum $L_n(f)$ approximates the area using the minimum height of the function in each subinterval: $L_n(f) = \sum_{k=0}^{n-1} \left(\min_{x \in [x_k, x_{k+1}]} f(x)\right) \Delta x$ - The upper sum $U_n(f)$ approximates the area using the maximum height of the function in each subinterval: $U_n(f) = \sum_{k=0}^{n-1} \left(\max_{x \in [x_k, x_{k+1}]} f(x)\right) \Delta x$ - Riemann Integrability: If $f(x)$ is continuous on $[a,b]$, then as $n \to \infty$, the lower and upper sums converge to the exact same value. This limit is the definite integral: $\int_a^b f(x) \, dx = \lim_{n\to\infty} L_n(f) = \lim_{n\to\infty} U_n(f)$ - Gauß Summation Formula: To compute limits of Riemann sums by hand, we use: $\sum_{k=0}^{n-1} k = \frac{n(n-1)}{2}$ 3. Key Properties of Definite Integrals - Linearity: $\int_a^b (\alpha f(x) + \beta g(x)) \, dx = \alpha \int_a^b f(x) \, dx + \beta \int_a^b g(x) \, dx$ - Reversing Limits: $\int_a^b f(x) \, dx = -\int_b^a f(x) \, dx \quad$ (and $\int_a^a f(x) \, dx = 0$) - Splitting the Domain: For any $c \in [a,b]$, $\int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx$ - Triangle Inequality: $\left| \int_a^b f(x) \, dx \right| \le \int_a^b |f(x)| \, dx$ 4. Mean Value Theorem (MVT) for Integrals - If $f(x)$ is continuous on $[a,b]$, there exists some value $\xi \in [a,b]$ such that: $\int_a^b f(x) \, dx = f(\xi)(b-a)$ Here, $f(\xi) = \frac{1}{b-a}\int_a^b f(x) \, dx$ is the average value of the function on the interval. 5. The Fundamental Theorem of Calculus (FTC) - Part I (Area Function Derivative): Let $f(x)$ be continuous on $[a,b]$. The function $F(x) = \int_a^x f(y) \, dy$ is differentiable, and its derivative is: $F'(x) = \frac{d}{dx} \left[ \int_a^x f(y) \, dy \right] = f(x)$ - Leibniz Rule (Chain Rule for Variable Limits): If the upper limit is a function $h(x)$, then: $\frac{d}{dx} \left[ \int_a^{h(x)} f(y) \, dy \right] = f(h(x)) \cdot h'(x)$ - Part II (Integral Evaluation): If $F(x)$ is any antiderivative of $f(x)$, then: $\int_a^b f(x) \, dx = F(b) - F(a) = [F(x)]_a^b$ Example 1: Definite Integral from definition (Riemann Sum)

Compute the definite integral $\int_0^1 x \, dx$ using the limit of Riemann lower sums. Solution: - Set up the partition: We divide $[0,1]$ into $n$ intervals. Width $\Delta x = \frac{1-0}{n} = \frac{1}{n}$. The boundaries are $x_k = \frac{k}{n}$ for $k=0, \dots, n$. - Formulate the lower sum $L_n(f)$: Since $f(x)=x$ is strictly increasing, the minimum value in each subinterval $[x_k, x_{k+1}]$ is at the left endpoint $x_k$: $L_n(f) = \sum_{k=0}^{n-1} f(x_k) \Delta x = \sum_{k=0}^{n-1} \left( \frac{k}{n} \right) \frac{1}{n} = \frac{1}{n^2} \sum_{k=0}^{n-1} k$ - Apply Gauß's Summation Formula: $L_n(f) = \frac{1}{n^2} \left[ \frac{n(n-1)}{2} \right] = \frac{n^2 - n}{2n^2} = \frac{1}{2} - \frac{1}{2n}$ - Take the limit as $n \to \infty$: $\int_0^1 x \, dx = \lim_{n\to\infty} \left( \frac{1}{2} - \frac{1}{2n} \right) = \frac{1}{2} \quad \square$ Example 2: FTC Part I and the Chain Rule

Find the derivative $g'(x)$ of the function $g(x) = \int_0^{x^2} \cos(t) \, dt$ for all $x \in \mathbb{R}$. Solution: Let $h(x) = x^2$. By the Leibniz Rule: $g'(x) = \cos(h(x)) \cdot h'(x) = \cos(x^2) \cdot (2x) = 2x \cos(x^2) \quad \square$ Example 3: Antiderivative Initial Value Problem (2023 Exam style)

Find the antiderivative $F(x)$ of $f(x) = x^2 + 2x + 3$ which satisfies the condition $F(0) = 2$. Solution: - Find the general antiderivative: $F(x) = \int (x^2 + 2x + 3) \, dx = \frac{x^3}{3} + x^2 + 3x + C$ - Apply the initial condition $F(0) = 2$: $F(0) = \frac{0^3}{3} + 0^2 + 3(0) + C = 2 \implies C = 2$ - Write down the specific antiderivative: $F(x) = \frac{x^3}{3} + x^2 + 3x + 2 \quad \square$ Exercise 1 (Riemann Definition):

Compute the definite integral $\int_0^1 (2x + 1) \, dx$ using the limit of Riemann lower sums. Solution: - Set up the partition: Width $\Delta x = \frac{1}{n}$. Boundaries $x_k = \frac{k}{n}$ for $k=0, \dots, n$. - Formulate the lower sum $L_n(f)$: Since $f(x)=2x+1$ is increasing, the minimum on $[x_k, x_{k+1}]$ is at $x_k$: $L_n(f) = \sum_{k=0}^{n-1} f(x_k) \Delta x = \sum_{k=0}^{n-1} \left( 2\frac{k}{n} + 1 \right) \frac{1}{n} = \frac{2}{n^2} \sum_{k=0}^{n-1} k + \frac{1}{n} \sum_{k=0}^{n-1} 1$ Using $\sum_{k=0}^{n-1} 1 = n$ and Gauß's formula: $L_n(f) = \frac{2}{n^2} \left[ \frac{n(n-1)}{2} \right] + \frac{1}{n} (n) = \frac{n-1}{n} + 1 = 1 - \frac{1}{n} + 1 = 2 - \frac{1}{n}$ - Take the limit as $n \to \infty$: $\int_0^1 (2x+1) \, dx = \lim_{n\to\infty} \left( 2 - \frac{1}{n} \right) = 2 \quad \square$ Exercise 2 (FTC Part I with Variable Limits):

Find the derivative $h'(x)$ of the function $h(x) = \int_2^{e^x} \ln(t) \, dt$ for all $x > 0$. Solution: Apply Leibniz Rule with upper limit $u(x) = e^x$: $h'(x) = \ln(u(x)) \cdot u'(x) = \ln(e^x) \cdot e^x = x \cdot e^x = x e^x \quad \square$ Exercise 3 (Antiderivative with Initial Condition):

Find the function $y(x)$ satisfying $\frac{dy}{dx} = \cos(x) - \sin(x)$ with initial condition $y(0) = 3$. Solution: - Integrate to find the general solution: $y(x) = \int (\cos(x) - \sin(x)) \, dx = \sin(x) - (-\cos(x)) + C = \sin(x) + \cos(x) + C$ - Impose $y(0) = 3$: $y(0) = \sin(0) + \cos(0) + C = 0 + 1 + C = 3 \implies C = 2$ - Write the final function: $y(x) = \sin(x) + \cos(x) + 2 \quad \square$ \subsection{Subchapter 3.2: Advanced Integration Techniques (Substitution & Parts)}

In this subchapter, we study periodic functions, how to extend a function defined on a finite interval to the entire real line periodically, and the continuity conditions at the boundaries. 1. Periodicity - Definition: A function $f: \mathbb{R} \to \mathbb{C}$ is called periodic with period $T > 0$ if: $f(x + T) = f(x) \quad \text{for all } x \in \mathbb{R}$ - Normalization ($T=1$): In this course, we normalize the period to $T = 1$. A function is 1-periodic if $f(x+1) = f(x)$. If we know the function values on $[0,1)$, we know them everywhere on $\mathbb{R}$. 2. Periodic Extension - Definition: If a function $f(x)$ is defined only on the half-open interval $[0,1)$, we extend it to a 1-periodic function on the entire real line $\mathbb{R}$ using the fractional part of $x$, denoted $\{x\}$: $f(x) := f(\{x\}) \quad \text{for all } x \in \mathbb{R}, \quad \text{where } \{x\} = x - \lfloor x \rfloor \in [0,1)$ 3. Continuity of Periodic Extensions - A periodic extension $f(x) = f(\{x\})$ is continuous on $\mathbb{R}$ if and only if: - $f(x)$ is continuous on the open interval $(0,1)$. - The limits at the boundaries match: $\lim_{x\to0^+} f(x) = \lim_{x\to1^-} f(x)$. - If the boundary limits do not match, the periodic extension will have a jump discontinuity of size $\left| \lim_{x\to1^-} f(x) - \lim_{x\to0^+} f(x) \right|$ at every integer $x \in \mathbb{Z}$. 4. Even and Odd Symmetries - Even Functions: Symmetric about the $y$-axis: $f(-x) = f(x)$. $\text{Integral property: } \int_{-L}^L f(x) \, dx = 2 \int_0^L f(x) \, dx$ - Odd Functions: Symmetric about the origin: $f(-x) = -f(x)$. $\text{Integral property: } \int_{-L}^L f(x) \, dx = 0$ Example 1: Continuity Audit of Sawtooth vs. Triangle Waves

Analyze the continuity of the periodic extensions of: - $f(x) = x$ on $[0,1)$ (Sawtooth wave). - $g(x) = \frac{1}{2} - \left|x - \frac{1}{2}\right|$ on $[0,1)$ (Triangle wave). Solution: - Sawtooth wave $f(x) = x$: Check the boundary limits: $\lim_{x\to0^+} f(x) = 0 \quad \text{and} \quad \lim_{x\to1^-} f(x) = 1$ Since the limits $0 \neq 1$ do not match, the periodic extension of $f(x) = x$ is not continuous on $\mathbb{R}$. It has jump discontinuities at $x \in \mathbb{Z}$. - **Triangle wave g(x) = \frac{1**{2} - \left|x - \frac{1}{2}\right|:} Check the boundary limits: $\lim_{x\to0^+} g(x) = \frac{1}{2} - \left|0 - \frac{1}{2}\right| = 0$ $\lim_{x\to1^-} g(x) = \frac{1}{2} - \left|1 - \frac{1}{2}\right| = 0$ Since $0 = 0$, the boundary limits match. Thus, the periodic extension of $g(x)$ is continuous on $\mathbb{R}$. $\square$ Example 2: Symmetries & Definite Integrals

Show that the function $f(x) = \sin(2\pi x)$ is odd and evaluate its integral over a symmetric period $[-1/2, 1/2]$. Solution: - Check odd symmetry: $f(-x) = \sin(2\pi(-x)) = \sin(-2\pi x) = -\sin(2\pi x) = -f(x) \implies f(x) \text{ is odd.}$ - Evaluate integral: Since $f(x)$ is odd and the interval $[-1/2, 1/2]$ is symmetric around $0$: $\int_{-1/2}^{1/2} \sin(2\pi x) \, dx = 0 \quad \square$ Exercise 1 (Periodic Extension Audit):

Let $f(x) = x^2$ on $[0,1)$. - Compute the periodic extension values $f(2.3)$ and $f(-0.7)$. - Determine if the periodic extension of $f(x)$ is continuous on $\mathbb{R}$. Solution: - **Use the fractional part $\{x\**$:} - For $x = 2.3$: $\{2.3\} = 2.3 - 2 = 0.3$. So $f(2.3) = f(0.3) = (0.3)^2 = 0.09$. - For $x = -0.7$: $\{-0.7\} = -0.7 - \lfloor -0.7 \rfloor = -0.7 - (-1) = 0.3$. So $f(-0.7) = f(0.3) = (0.3)^2 = 0.09$. - Check boundary limits: $\lim_{x\to0^+} f(x) = 0^2 = 0 \quad \text{and} \quad \lim_{x\to1^-} f(x) = 1^2 = 1$ Since $0 \neq 1$, the periodic extension of $f(x) = x^2$ is not continuous on $\mathbb{R}$. It has jump discontinuities at all integers. $\square$ Exercise 2 (Symmetry Check & Integration):

Let $g(x) = \left(x - \frac{1}{2}\right)^2$ on $[0,1)$. - Show that $g(x)$ is symmetric around $x = 1/2$. - Compute the integral $\int_0^1 \left(x - \frac{1}{2}\right)^2 \, dx$. Solution: - Check symmetry around $x = 1/2$: We shift $g(x)$ by setting $t = x - 1/2$, which yields $h(t) = t^2$ on $[-1/2, 1/2)$. Since $h(-t) = (-t)^2 = t^2 = h(t)$, the shifted function is even, meaning $g(x)$ is symmetric around $x=1/2$. - Compute the integral: Substitute $t = x - 1/2 \implies dt = dx$. Limits: $x=0 \implies t=-1/2$, $x=1 \implies t=1/2$. $\int_0^1 \left(x - \frac{1}{2}\right)^2 \, dx = \int_{-1/2}^{1/2} t^2 \, dt$ Since $t^2$ is even, we can write: $= 2 \int_0^{1/2} t^2 \, dt = 2 \left[ \frac{t^3}{3} \right]_0^{1/2} = 2 \left( \frac{1}{24} - 0 \right) = \frac{1}{12} \quad \square$ \subsection{Subchapter 4.2: Trigonometric Polynomials & Fourier Series}