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<!DOCTYPE html>
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<title>Methods Lecture 2 Notes</title>
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<!-- ===== LECTURE TITLE ===== -->
<h1>Methods Lecture 2</h1>
<p style="text-align: center; color: #666; font-style: italic;">Convergence of Fourier Series</p>
<!-- ===== INTRODUCTION ===== -->
<div class="default content-box">
<div class="timestamp" onclick="openVideoSide('0:07', 'Introduction to Fourier Series Convergence', 'default', this)">
<span class="play-icon">▶</span>0:07
</div>
<h3>Introduction</h3>
<p>Today's topic is the convergence of Fourier series. We define Fourier series for $L$-periodic functions by the infinite sum:</p>
<p>$$f(\theta) \sim \sum_{n=-\infty}^{\infty} \hat{f}_n e^{2\pi i n \theta/L}$$</p>
<p>At this point, we don't yet know if the infinite sum on the right corresponds to the function on the left. We need to establish what we mean by that squiggly line ($\sim$).</p>
</div>
<!-- ===== PARTIAL FOURIER SERIES DEFINITION ===== -->
<div class="definition content-box">
<div class="timestamp" onclick="openVideoSide('2:22', 'Partial Fourier Series Definition', 'definition', this)">
<span class="play-icon">▶</span>2:22
</div>
<h3>Partial Fourier Series</h3>
<p>For an $L$-periodic function $f$, we define the <strong>partial Fourier series</strong> by:</p>
<p>$$S_n(f)(\theta) = \sum_{|n| \leq N} \hat{f}_n e^{2\pi i n \theta/L}$$</p>
<p>When written out in terms of sines and cosines, this becomes:</p>
<p>$$S_n(f)(\theta) = \frac{a_0}{2} + \sum_{n=1}^{N} \left(a_n \cos\left(\frac{2\pi n \theta}{L}\right) + b_n \sin\left(\frac{2\pi n \theta}{L}\right)\right)$$</p>
<p>This is the symmetric sum that corresponds exactly to the trigonometric form when written in terms of complex exponentials.</p>
</div>
<!-- ===== MODES OF CONVERGENCE ===== -->
<div class="definition content-box">
<div class="timestamp" onclick="openVideoSide('4:35', 'Modes of Convergence', 'definition', this)">
<span class="play-icon">▶</span>4:35
</div>
<h3>Modes of Convergence</h3>
<p>When discussing convergence of a sequence of functions $f_n \to f$, we need to specify the mode of convergence:</p>
<p><strong>1. Pointwise Convergence:</strong></p>
<p>$$f_n(\theta) \to f(\theta) \text{ as } n \to \infty \text{ for each } \theta \in [0,L]$$</p>
<p><strong>2. Uniform Convergence:</strong></p>
<p>$$\sup_{\theta \in [0,L]} |f_n(\theta) - f(\theta)| \to 0 \text{ as } n \to \infty$$</p>
<p><strong>3. Convergence in Mean (L²):</strong></p>
<p>$$\|S_n(f) - f\|_2 = \sqrt{\int_0^L |S_n(f)(\theta) - f(\theta)|^2 d\theta} \to 0 \text{ as } n \to \infty$$</p>
<p>Note that uniform convergence is stronger than pointwise convergence, and the proofs involved are very different depending on which mode we're interested in.</p>
</div>
<!-- ===== HILBERT SPACE COMMENT ===== -->
<div class="non-examinable content-box">
<div class="timestamp" onclick="openVideoSide('6:55', 'Hilbert Space Comment', 'non-examinable', this)">
<span class="play-icon">▶</span>6:55
</div>
<h3>Hilbert Space Perspective</h3>
<p><em>Non-examinable content:</em> The convergence in mean (L²) is actually the easiest mathematically because we're effectively dealing with a Hilbert space, which has lovely properties. However, this requires knowledge of Part II Linear Analysis, so we won't pursue this approach.</p>
</div>
<!-- ===== MAIN PROPOSITION ===== -->
<div class="proposition content-box">
<div class="timestamp" onclick="openVideoSide('7:35', 'Main Convergence Proposition', 'proposition', this)">
<span class="play-icon">▶</span>7:35
</div>
<h3>Convergence of Fourier Series</h3>
<p>Let $f$ be an $L$-periodic function with the following properties on $[0,L]$:</p>
<ol>
<li>It has finitely many discontinuities</li>
<li>It has finitely many local maxima and minima</li>
</ol>
<p>Then for each $\theta \in [0,L]$ we have:</p>
<p>$$\frac{f(\theta^+) + f(\theta^-)}{2} = \lim_{n \to \infty} S_n(f)(\theta) = \sum_{n=-\infty}^{\infty} \hat{f}_n e^{2\pi i n \theta/L}$$</p>
<p>where $\theta^{\pm} = \lim_{\epsilon \to 0^+} f(\theta \pm \epsilon)$ are the left and right limits of $f$ at $\theta$.</p>
</div>
<!-- ===== INTERPRETATION ===== -->
<div class="default content-box">
<div class="timestamp" onclick="openVideoSide('11:14', 'Interpretation of Convergence Result', 'default', this)">
<span class="play-icon">▶</span>11:14
</div>
<h3>Interpretation</h3>
<p>The convergence result tells us that:</p>
<ul>
<li><strong>At points of continuity:</strong> The Fourier series gives back the original function $f(\theta)$</li>
<li><strong>At points of discontinuity:</strong> The Fourier series gives back the average of the function at the discontinuity</li>
</ul>
<p>This is why it's crucial to draw a graph of your periodic function before computing Fourier coefficients. What looks like a nice function on one interval may have discontinuities when extended periodically.</p>
</div>
<!-- ===== EXAM WARNING ===== -->
<div class="default content-box">
<div class="timestamp" onclick="openVideoSide('14:05', 'Exam Warning', 'default', this)">
<span class="play-icon">▶</span>14:05
</div>
<h3>Important Exam Warning</h3>
<p>Examiners may ask you to compute the value of a Fourier series at a given point. Even if the function looks "nice" on its defining interval, remember that at points of discontinuity, the Fourier series evaluates to the average of the left and right limits, not the function value itself.</p>
<p><strong>Example:</strong> If $f(\theta) = e^\theta$ on $[-\pi, \pi]$ and extended periodically, then at $\theta = \pi$, the Fourier series gives $\frac{e^\pi + e^{-\pi}}{2} = \cosh(\pi)$, not $e^\pi$.</p>
</div>
<!-- ===== DIRICHLET FUNCTIONS ===== -->
<div class="definition content-box">
<div class="timestamp" onclick="openVideoSide('16:37', 'Dirichlet Functions', 'definition', this)">
<span class="play-icon">▶</span>16:37
</div>
<h3>Dirichlet Functions</h3>
<p>Functions with properties (1) and (2) above are called <strong>Dirichlet functions</strong>, named after the famous German mathematician Peter Gustav Lejeune Dirichlet.</p>
<p>From now on, we will assume all functions we deal with are Dirichlet functions. This means the symbol $\sim$ now has a precise meaning:</p>
<p>$$f(\theta) \sim \sum_{n=-\infty}^{\infty} \hat{f}_n e^{2\pi i n \theta/L}$$</p>
<p>means the series on the right coincides with the function on the left at points of continuity, and converges to its average at points of discontinuity.</p>
</div>
<!-- ===== PROOF SETUP ===== -->
<div class="default content-box">
<div class="timestamp" onclick="openVideoSide('19:08', 'Proof Setup', 'default', this)">
<span class="play-icon">▶</span>19:08
</div>
<h3>Proof Strategy</h3>
<p>We will prove the convergence result assuming $f$ is smooth (infinitely differentiable). In fact, $C^1$ (continuously differentiable) would suffice. The proof can be modified to handle the more general Dirichlet functions.</p>
<p>We'll examine the limit of $S_n(f)(\theta_0)$ as $n \to \infty$ for some $\theta_0$ in the interval. By replacing $f(\theta)$ with $f(\theta + \theta_0)$, we can assume $\theta_0 = 0$ without loss of generality.</p>
</div>
<!-- ===== DIRICHLET KERNEL ===== -->
<div class="definition content-box">
<div class="timestamp" onclick="openVideoSide('25:27', 'Dirichlet Kernel', 'definition', this)">
<span class="play-icon">▶</span>25:27
</div>
<h3>Dirichlet Kernel</h3>
<p>The <strong>Dirichlet kernel</strong> is defined as:</p>
<p>$$D_n(\theta) = \frac{\sin\left((n+\frac{1}{2})\theta\right)}{\sin(\theta/2)}$$</p>
<p>for $\theta \neq 0, \pm 2\pi, \pm 4\pi, \ldots$ and $D_n(0) = 2n+1$.</p>
<p>This function is crucial for understanding Fourier series convergence. It has the following properties:</p>
<ol>
<li>It is continuous</li>
<li>It is even and periodic with period $2\pi$</li>
<li>$\int_{-\pi}^{\pi} D_n(\theta) d\theta = 2\pi$ for all $n$</li>
</ol>
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<h3>Proof of Pointwise Convergence</h3>
<p>We want to show that $S_n(f)(0) - f(0) \to 0$ as $n \to \infty$.</p>
<p><strong>Step 1:</strong> Express the partial sum using the Dirichlet kernel:</p>
<p>$$S_n(f)(0) = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(\theta) D_n(\theta) d\theta$$</p>
<p><strong>Step 2:</strong> Use the property that $f(0) = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(0) D_n(\theta) d\theta$ to write:</p>
<p>$$S_n(f)(0) - f(0) = \frac{1}{2\pi} \int_{-\pi}^{\pi} D_n(\theta) [f(\theta) - f(0)] d\theta$$</p>
<p><strong>Step 3:</strong> Factor out the sine term and define:</p>
<p>$$F(\theta) = \frac{f(\theta) - f(0)}{\sin(\theta/2)}$$</p>
<p>This gives us:</p>
<p>$$S_n(f)(0) - f(0) = \frac{1}{2\pi} \int_{-\pi}^{\pi} \sin\left((n+\frac{1}{2})\theta\right) F(\theta) d\theta$$</p>
<p><strong>Step 4:</strong> Since $f$ is smooth, $F(\theta)$ is a well-behaved function. We can integrate by parts to get:</p>
<p>$$S_n(f)(0) - f(0) = \frac{1}{n+\frac{1}{2}} \cdot \frac{1}{2\pi} \int_{-\pi}^{\pi} \cos\left((n+\frac{1}{2})\theta\right) F'(\theta) d\theta$$</p>
<p><strong>Step 5:</strong> The integral is bounded, and the factor $\frac{1}{n+\frac{1}{2}} \to 0$ as $n \to \infty$, giving us pointwise convergence.</p>
<p><strong>Note:</strong> The same proof works for any $\theta$ (not just 0), and actually gives uniform convergence for smooth functions.</p>
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<!-- ===== PROOF DIFFICULTY ===== -->
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<h3>Proof Difficulty</h3>
<p>The tricky part of this proof was being able to integrate by parts, which required $f$ to be differentiable. If $f$ wasn't differentiable at some points, we would need more sophisticated techniques to handle the badly behaved points.</p>
<p>For more general functions (Dirichlet functions), the proof becomes much harder and requires advanced analysis techniques. A good reference is Titchmarsh's "Theory of Functions" for the general case.</p>
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<!-- ===== SQUARE WAVE EXAMPLE ===== -->
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<h3>Square Wave Example</h3>
<p>Consider the $2\pi$-periodic function defined by:</p>
<p>$$f(\theta) = \begin{cases}
+1 & \text{if } \theta \in (0, \pi) \\
-1 & \text{if } \theta \in (-\pi, 0)
\end{cases}$$</p>
<p>This function has one jump discontinuity at $\theta = 0$ (and its periodic images), so it's a Dirichlet function.</p>
<p><strong>Computing the Fourier coefficients:</strong></p>
<p>Since $f$ is odd, all cosine coefficients $a_n = 0$. For the sine coefficients:</p>
<p>$$b_n = \frac{2}{\pi} \int_0^{\pi} \sin(n\theta) d\theta = \frac{2}{n\pi} [1 - \cos(n\pi)]$$</p>
<p>Since $\cos(n\pi) = (-1)^n$, we get:</p>
<p>$$b_n = \frac{2}{n\pi} [1 - (-1)^n] = \begin{cases}
\frac{4}{n\pi} & \text{if } n \text{ is odd} \\
0 & \text{if } n \text{ is even}
\end{cases}$$</p>
<p>Therefore, the Fourier series is:</p>
<p>$$f(\theta) \sim \frac{4}{\pi} \sum_{n \text{ odd}} \frac{\sin(n\theta)}{n}$$</p>
<p><strong>Verification at discontinuity:</strong></p>
<p>At $\theta = 0$, the Fourier series gives $0$ (since all sine terms vanish). The left limit is $-1$ and right limit is $+1$, so the average is $\frac{(-1) + 1}{2} = 0$, which matches!</p>
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<!-- ===== KOLMOGOROV STORY ===== -->
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<h3>Historical Story: Kolmogorov's Counterexample</h3>
<p><em>Non-examinable content:</em> In the 1920s, there was a major open problem: how far can we push Fourier series convergence? What's the most general class of functions for which Fourier series converge?</p>
<p>In 1922, at the age of 19, Andrey Kolmogorov constructed an integrable function whose Fourier series diverges almost everywhere (i.e., everywhere except on a set of measure zero). This shocked the mathematical community, which had believed that integrable functions would have convergent Fourier series.</p>
<p>Later, Carleson and Hunt proved that if we strengthen the condition slightly (requiring the function to be in $L^p$ for some $p > 1$), then the Fourier series converges almost everywhere. This is now known as the Carleson-Hunt theorem.</p>
<p>Kolmogorov's result shows that our assumption of Dirichlet functions (or bounded variation) is not just convenient—it's necessary for the convergence results we've proved.</p>
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