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<!DOCTYPE html>
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<title>Methods Lecture 3 Notes</title>
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<!-- ===== LECTURE TITLE ===== -->
<h1>Methods Lecture 3</h1>
<p style="text-align: center; color: #666; font-style: italic;">Periodic Extensions, Regularity and Decay of Fourier Coefficients, and Differentiation</p>
<!-- ===== CONTENT BOXES ===== -->
<div class="default content-box">
<div class="timestamp" onclick="openVideoSide('0:34', 'Section 1.4 Introduction', 'default', this)">
<span class="play-icon">▶</span>0:34
</div>
<h3>Section 1.4: Periodic Extensions</h3>
<p>We move on to section 1.4 about periodic extensions, building on the basics of Fourier series and convergence that we've established. We've shown that for functions that aren't too wild (with finitely many discontinuities and finitely many maxima and minima), the Fourier series converges to the function itself at points of continuity and to the average at points of discontinuity.</p>
</div>
<div class="definition content-box">
<div class="timestamp" onclick="openVideoSide('1:13', 'Even and Odd Extensions Definition', 'definition', this)">
<span class="play-icon">▶</span>1:13
</div>
<h3>Even and Odd Extensions</h3>
<p>Given a function $f: [0,L] \to \mathbb{R}$ (or $\mathbb{C}$), we can define two $2L$-periodic extensions:</p>
<p><strong>Even extension:</strong> $f_{\text{even}}(\theta) = \begin{cases}
f(\theta) & \text{if } \theta \in [0,L] \\
f(-\theta) & \text{if } \theta \in [-L,0]
\end{cases}$</p>
<p><strong>Odd extension:</strong> $f_{\text{odd}}(\theta) = \begin{cases}
f(\theta) & \text{if } \theta \in [0,L] \\
-f(-\theta) & \text{if } \theta \in [-L,0]
\end{cases}$</p>
<p>These extensions are then continued periodically with period $2L$.</p>
</div>
<div class="default content-box">
<div class="timestamp" onclick="openVideoSide('4:51', 'Relationship to Original Function', 'default', this)">
<span class="play-icon">▶</span>4:51
</div>
<h3>Relationship to Original Function</h3>
<p>Note that $f(\theta) = f_{\text{even}}(\theta) = f_{\text{odd}}(\theta)$ when we restrict to the original interval $[0,L]$. This is the key insight: we now have periodic functions that we can express as Fourier series, and these Fourier series will give us our original function when restricted to the interval $[0,L]$.</p>
</div>
<div class="theorem content-box">
<div class="timestamp" onclick="openVideoSide('5:13', 'Cosine Series Formula', 'theorem', this)">
<span class="play-icon">▶</span>5:13
</div>
<h3>Cosine Series</h3>
<p>For the even extension, since it's an even function, the Fourier series contains only cosine terms:</p>
<p>$$f_{\text{even}}(\theta) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos\left(\frac{n\pi\theta}{L}\right)$$</p>
<p>where the coefficients are:</p>
<p>$$a_n = \frac{2}{L} \int_0^L f(\theta) \cos\left(\frac{n\pi\theta}{L}\right) \, d\theta$$</p>
<p>This follows from the standard Fourier coefficient formula, using the fact that the even function and cosine are both even, allowing us to integrate only over $[0,L]$ and multiply by 2.</p>
</div>
<div class="theorem content-box">
<div class="timestamp" onclick="openVideoSide('7:32', 'Sine Series Formula', 'theorem', this)">
<span class="play-icon">▶</span>7:32
</div>
<h3>Sine Series</h3>
<p>For the odd extension, the Fourier series contains only sine terms:</p>
<p>$$f_{\text{odd}}(\theta) = \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi\theta}{L}\right)$$</p>
<p>where the coefficients are:</p>
<p>$$b_n = \frac{2}{L} \int_0^L f(\theta) \sin\left(\frac{n\pi\theta}{L}\right) \, d\theta$$</p>
<p>This is analogous to the cosine series, using the same integration trick.</p>
</div>
<div class="definition content-box">
<div class="timestamp" onclick="openVideoSide('9:04', 'Cosine and Sine Series Definition', 'definition', this)">
<span class="play-icon">▶</span>9:04
</div>
<h3>Cosine and Sine Series</h3>
<p>For a function $f: [0,L] \to \mathbb{R}$, we define:</p>
<p><strong>Cosine series:</strong> $\frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos\left(\frac{n\pi\theta}{L}\right)$</p>
<p><strong>Sine series:</strong> $\sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi\theta}{L}\right)$</p>
<p>where $a_n$ and $b_n$ are defined as above. These series represent the even and odd periodic extensions of $f$, respectively.</p>
</div>
<div class="example content-box">
<div class="timestamp" onclick="openVideoSide('10:30', 'Constant Function Example', 'example', this)">
<span class="play-icon">▶</span>10:30
</div>
<h3>Example: Constant Function</h3>
<p>Consider $f(\theta) = 1$ on $[0,\pi]$.</p>
<p><strong>Odd extension:</strong> This creates a square wave pattern: $+1$ on $(0,\pi)$, $-1$ on $(-\pi,0)$, etc.</p>
<p><strong>Even extension:</strong> This simply extends the constant function: $+1$ everywhere.</p>
<p>For the sine series coefficients:</p>
<p>$$b_n = \frac{2}{\pi} \int_0^{\pi} \sin(n\theta) \, d\theta = \frac{2}{n\pi}(1 - (-1)^n)$$</p>
<p>This gives us the sine series representation of the constant function when restricted to $(0,\pi)$.</p>
</div>
<div class="example content-box">
<div class="timestamp" onclick="openVideoSide('14:35', 'Cosine Series for Constant', 'example', this)">
<span class="play-icon">▶</span>14:35
</div>
<h3>Example: Cosine Series for Constant Function</h3>
<p>For the cosine series of $f(\theta) = 1$ on $[0,\pi]$:</p>
<p>We can verify that $a_0 = 2$ and $a_n = 0$ for $n \geq 1$.</p>
<p>This gives us the trivial cosine series: $1 = 1$ (since $\frac{a_0}{2} = 1$).</p>
<p>The even extension is continuous everywhere, so the Fourier series converges to the function everywhere on the interval.</p>
</div>
<div class="default content-box">
<div class="timestamp" onclick="openVideoSide('16:15', 'Advantage of Even/Odd Extensions', 'default', this)">
<span class="play-icon">▶</span>16:15
</div>
<h3>Why Even and Odd Extensions?</h3>
<p>The advantage of using even or odd extensions rather than a general periodic extension is that it reduces the computational work by half. With even extensions, we get only cosine terms; with odd extensions, only sine terms. A general periodic extension would require both sine and cosine terms, making the calculations more complex.</p>
</div>
<div class="default content-box">
<div class="timestamp" onclick="openVideoSide('16:25', 'Section 1.5 Introduction', 'default', this)">
<span class="play-icon">▶</span>16:25
</div>
<h3>Section 1.5: Regularity and Decay of Fourier Coefficients</h3>
<p>We now turn to understanding how the smoothness (regularity) of a function relates to how quickly its Fourier coefficients decay as $n \to \infty$.</p>
</div>
<div class="non-examinable content-box">
<div class="timestamp" onclick="openVideoSide('17:06', 'Riemann-Lebesgue Lemma', 'non-examinable', this)">
<span class="play-icon">▶</span>17:06
</div>
<h3>Riemann-Lebesgue Lemma</h3>
<p><em>This result is on the edge of being examinable but contains complex analysis: Non-examinable in detail.</em></p>
<p>Let $g: [a,b] \to \mathbb{C}$ be integrable and $\lambda \in \mathbb{R}$. Then:</p>
<p>$$\int_a^b e^{-i\lambda\theta} g(\theta) \, d\theta \to 0 \quad \text{as } |\lambda| \to \infty$$</p>
<p>This is the famous <strong>Riemann-Lebesgue Lemma</strong>.</p>
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<div class="corollary content-box">
<div class="timestamp" onclick="openVideoSide('18:35', 'Fourier Coefficient Decay', 'corollary', this)">
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<h3>Fourier Coefficients Decay</h3>
<p>If $f$ is an $L$-periodic integrable function, then the Fourier coefficients satisfy:</p>
<p>$$\hat{f}_n \to 0 \quad \text{as } |n| \to \infty$$</p>
<p>This follows from the Riemann-Lebesgue Lemma by taking $\lambda = \frac{2\pi n}{L}$ and noting that $\hat{f}_n = \frac{1}{L} \int_0^L f(\theta) e^{-2\pi i n \theta / L} \, d\theta$.</p>
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<div class="default content-box">
<div class="timestamp" onclick="openVideoSide('21:35', 'Regularity and Decay Connection', 'default', this)">
<span class="play-icon">▶</span>21:35
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<h3>Regularity and Decay Connection</h3>
<p>There's a beautiful two-way relationship between function regularity and Fourier coefficient decay:</p>
<ul>
<li><strong>Forward direction:</strong> If you tell me how smooth your function is (how many continuous derivatives it has), I can predict how fast the Fourier coefficients decay.</li>
<li><strong>Converse direction:</strong> If you tell me how fast the Fourier coefficients go to zero, I can determine how smooth the function is.</li>
</ul>
<p>This is why we use harmonic analysis in PDEs - it's much easier to analyze decay rates than to zoom in and take limits to check smoothness.</p>
</div>
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<div class="timestamp" onclick="openVideoSide('23:26', 'Riemann-Lebesgue Intuition', 'non-examinable', this)">
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<h3>Riemann-Lebesgue Lemma: Intuitive Explanation</h3>
<p><em>Non-examinable detailed explanation</em></p>
<p>The integral $\int e^{-i\lambda\theta} g(\theta) \, d\theta$ represents the integral of $g(\theta)$ multiplied by rapidly oscillating terms $\cos(\lambda\theta)$ and $\sin(\lambda\theta)$.</p>
<p>As $\lambda \to \infty$, the oscillations become very rapid. The positive and negative contributions cancel out due to the rapid oscillation, leading to the integral tending to zero. This is a cancellation argument - high-frequency oscillations average out to zero when integrated against a nice function.</p>
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<div class="theorem content-box">
<div class="timestamp" onclick="openVideoSide('26:27', 'Decay Rate for Smooth Functions', 'theorem', this)">
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<h3>Decay Rate for Smooth Functions</h3>
<p>Suppose $f$ is $L$-periodic and $f \in C^k(\mathbb{R})$ (meaning $f, f', f'', \ldots, f^{(k)}$ are all continuous on $\mathbb{R}$).</p>
<p>Then the Fourier coefficients satisfy:</p>
<p>$$|\hat{f}_n| = O\left(\frac{1}{|n|^k}\right) \quad \text{as } |n| \to \infty$$</p>
<p>That is, the smoother the function, the faster the Fourier coefficients decay.</p>
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<div class="timestamp" onclick="openVideoSide('27:41', 'Proof of Decay Rate', 'proof', this)">
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<h3>Proof of Decay Rate</h3>
<p>We use integration by parts repeatedly. Starting with:</p>
<p>$$\hat{f}_n = \frac{1}{L} \int_0^L f(\theta) e^{-2\pi i n \theta / L} \, d\theta$$</p>
<p>Integrate by parts: integrate the exponential, differentiate $f$:</p>
<p>$$\hat{f}_n = \frac{L}{2\pi i n} \left[ f(\theta) e^{-2\pi i n \theta / L} \right]_0^L + \frac{L}{2\pi i n} \int_0^L f'(\theta) e^{-2\pi i n \theta / L} \, d\theta$$</p>
<p>Due to periodicity and continuity: $f(L) = f(0)$, so the boundary term vanishes. We get:</p>
<p>$$\hat{f}_n = \frac{L}{2\pi i n} \hat{f'}_n$$</p>
<p>Continuing this process $k$ times:</p>
<p>$$\hat{f}_n = \left(\frac{L}{2\pi i n}\right)^k \hat{f^{(k)}}_n$$</p>
<p>Since $f^{(k)}$ is continuous (hence integrable), $\hat{f^{(k)}}_n = O(1)$ by the Riemann-Lebesgue Lemma. Therefore $|\hat{f}_n| = O(1/|n|^k)$.</p>
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<div class="timestamp" onclick="openVideoSide('35:14', 'Section 1.6 Introduction', 'default', this)">
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<h3>Section 1.6: Differentiation of Fourier Series</h3>
<p>We now consider: if $f$ has Fourier series, what is the Fourier series of its derivative $f'$?</p>
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<div class="timestamp" onclick="openVideoSide('35:39', 'Differentiation of Fourier Series', 'theorem', this)">
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<h3>Differentiation of Fourier Series</h3>
<p>Suppose $f$ is $L$-periodic and continuously differentiable on $\mathbb{R}$ with $f' = g$.</p>
<p>Then we can differentiate the Fourier series term by term:</p>
<p>$$\frac{d}{d\theta}\left(\sum_{n \in \mathbb{Z}} \hat{f}_n e^{2\pi i n \theta / L}\right) = \sum_{n \in \mathbb{Z}} \frac{2\pi i n}{L} \hat{f}_n e^{2\pi i n \theta / L}$$</p>
<p><strong>Key condition:</strong> This works only if $f$ is continuous on $\mathbb{R}$ (not just on one period).</p>
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<div class="timestamp" onclick="openVideoSide('37:00', 'Proof of Differentiation Rule', 'proof', this)">
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<h3>Proof of Differentiation Rule</h3>
<p>The Fourier coefficient of the derivative is:</p>
<p>$$\hat{g}_n = \frac{1}{L} \int_0^L g(\theta) e^{-2\pi i n \theta / L} \, d\theta = \frac{1}{L} \int_0^L f'(\theta) e^{-2\pi i n \theta / L} \, d\theta$$</p>
<p>Using integration by parts (integrate the exponential, differentiate $f$):</p>
<p>$$\hat{g}_n = \frac{1}{L} \left[ f(\theta) e^{-2\pi i n \theta / L} \right]_0^L + \frac{2\pi i n}{L^2} \int_0^L f(\theta) e^{-2\pi i n \theta / L} \, d\theta$$</p>
<p>The boundary term becomes $f(L) - f(0)$. If $f$ is continuous on $\mathbb{R}$, then by periodicity $f(0) = f(L)$, so the boundary term vanishes and we get:</p>
<p>$$\hat{g}_n = \frac{2\pi i n}{L} \hat{f}_n$$</p>
<p>This shows that $\hat{f'}_n = \frac{2\pi i n}{L} \hat{f}_n$, which is exactly what we get by differentiating the Fourier series term by term.</p>
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<div class="timestamp" onclick="openVideoSide('41:39', 'Section 1.7 Introduction', 'default', this)">
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<h3>Section 1.7: Parseval's Theorem</h3>
<p>A very short section on Parseval's theorem, which relates the inner product of two functions to their Fourier coefficients.</p>
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<div class="theorem content-box">
<div class="timestamp" onclick="openVideoSide('42:32', 'Parseval\'s Theorem', 'theorem', this)">
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<h3>Parseval's Theorem</h3>
<p>For two $L$-periodic functions $f$ and $g$ with Fourier series, we have:</p>
<p>$$\langle f, g \rangle = L \sum_{n \in \mathbb{Z}} \hat{f}_n \overline{\hat{g}_n}$$</p>
<p>where $\langle f, g \rangle = \int_0^L f(\theta) \overline{g(\theta)} \, d\theta$ is the inner product.</p>
<p>In the special case when $f = g$, this becomes:</p>
<p>$$\int_0^L |f(\theta)|^2 \, d\theta = L \sum_{n \in \mathbb{Z}} |\hat{f}_n|^2$$</p>
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<div class="example content-box">
<div class="timestamp" onclick="openVideoSide('46:13', 'Parseval Example: zeta(2)', 'example', this)">
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<h3>Example: Computing $\zeta(2)$ using Parseval's Theorem</h3>
<p>Consider the $2\pi$-periodic function $f(\theta) = \theta$ on $[-\pi, \pi]$. This function is not continuous on $\mathbb{R}$ (it has jump discontinuities at $(2k+1)\pi$).</p>
<p>It can be shown (left as exercise) that the Fourier series is:</p>
<p>$$f(\theta) \sim \sum_{n \neq 0} \frac{i(-1)^n}{n} e^{in\theta}$$</p>
<p>By Parseval's theorem:</p>
<p>$$\frac{1}{2\pi} \int_{-\pi}^{\pi} \theta^2 \, d\theta = \sum_{n \neq 0} \left|\frac{i(-1)^n}{n}\right|^2 = 2\sum_{n=1}^{\infty} \frac{1}{n^2}$$</p>
<p>Computing the left side: $\int_{-\pi}^{\pi} \theta^2 \, d\theta = 2\int_0^{\pi} \theta^2 \, d\theta = 2\frac{\pi^3}{3} = \frac{2\pi^3}{3}$, so:</p>
<p>$$\frac{1}{2\pi} \cdot \frac{2\pi^3}{3} = \frac{\pi^2}{3} = 2\sum_{n=1}^{\infty} \frac{1}{n^2}$$</p>
<p>Therefore: $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} = \zeta(2)$.</p>
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<div class="timestamp" onclick="openVideoSide('49:00', 'Challenge Problems', 'default', this)">
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<h3>Challenge Problems</h3>
<p>Two challenging problems are set:</p>
<p><strong>Challenge 1 (*):</strong> It is possible to prove that $\zeta(2n)/\pi^{2n}$ is always rational for positive integers $n$.</p>
<p><strong>Challenge 2 (**):</strong> Is $\zeta(2n+1)$ rational for any odd positive integer $n$? (This is a very difficult open problem - collaboration opportunity offered!)</p>
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