MATH464

Revised 2026-09-17

Lecture 1

Proposition:

Let $θ_{1}, θ_{2} \in R$ . Then $e^{i θ_{1}} \cdot e^{i θ_{2}} = e^{i (θ_{1} + θ_{2})}$ .

Proof:

$e^{i θ_{1}} \cdot e^{i θ_{2}} = (cos θ_{1} + i sin θ_{1}) (cos θ_{2} + i sin θ_{2}) = (cos θ_{1} cos θ_{2} - sin θ_{1} sin θ_{2}) + i (cos θ_{1} sin θ_{2} + sin θ_{1} cos θ_{2})$ $e^{i (θ_{1} + θ_{2})} = cos (θ_{1} + θ_{2}) + i sin (θ_{1} + θ_{2})$

By $sin$ and $cos$ angle sum identities, as desired.

Corollary:

$e^{i θ_{1}} \cdot e^{i θ_{2}} \cdot e^{i θ_{3}} \dots = e^{i (θ_{1} + θ_{2} + θ_{3} + \dots)}$

(Induction on the number of factors: the base case is the Proposition, and the inductive step applies the Proposition to $e^{i (θ_{1} + \dots + θ_{n - 1})} \cdot e^{i θ_{n}}$ .) Taking every $θ_{j} = θ$ , $\Rightarrow (e^{i θ})^{n} = e^{in θ} (De Moivre’s Formula)$

Note:

One can derive the exponential representations of $sin$ and $cos$ : $e^{i θ} = cos θ + i sin θ, e^{- i θ} = e^{i θ} = cos θ - i sin θ$ Adding the two kills the $sin$ term, subtracting them kills the $cos$ term: $\Rightarrow cos θ = \frac{e ^{i θ} + e ^{- i θ}}{2}, sin θ = \frac{e ^{i θ} - e ^{- i θ}}{2 i}$

Lecture 2

Periodic functions.

Definition:

We say a function $f$ is periodic of period $T$ if $f (t + T) = f (t) \forall t \in R$

Fact: If $f$ is periodic of period 1, then $f$ can be “represented” as $n = - \infty \sum \infty c_{n} e^{2 π in t}, c_{n} \in C$

Note that not all infinite sums of exponentiations are convergent, but in most cases that we will encounter in this class, this is an exact representation.

The collection of functions ${e^{2 π in t} : n \in Z}$ can be used to represent any function of period 1.

Now suppose $f$ is of period 1, and we are told $f (t) = n \in Z \sum c_{n} e^{2 π in t}$

How can we compute $c_{n}$ ? $e^{- 2 π ik t} f (t) \Rightarrow \int_{0}^{1} e^{- 2 π ik t} f (t) d t = n \in Z \sum c_{n} e^{- 2 π ik t} e^{2 π in t} = n \in Z \sum c_{n} \int_{0}^{1} e^{2 π i (n - k) t} d t$

Examine $\int_{0}^{1} e^{2 π i (n - k) t} d t$ . Antidifferentiating, we see that this must be split into two cases: $n = k$ and $n \neq = k$ . $\int_{0}^{1} e^{2 π i (n - k) t} d t = ⎩ ⎨ ⎧ \int_{0}^{1} 1 d t = 1 \frac{e ^{2 π i (n - k) t}}{2 π i ( n - k )}_{0}^{1} = \frac{1 - 1}{2 π i ( n - k )} = 0 n = k n \neq = k$ (The numerator vanishes because $e^{2 π i (n - k)} = 1$ for integer $n - k$ .) So every term of the sum dies except $n = k$ : $\Rightarrow \int_{0}^{1} f (t) e^{- 2 π ik t} d t = c_{k}$

Thus, if $f$ is of period 1, then $c_{n} = \int_{0}^{1} f (t) e^{- 2 π in t} d t$

Now suppose $f$ is of period $T$ . Let $g (t) = f (T t)$ . Then $g (t)$ is of period 1. We can observe this fact $g (t + 1) = f (T (t + 1)) = f (T t + T) = f (T t) = g (t)$

We already know how to represent $g (t)$ : $g (t) = n = - \infty \sum \infty c_{n} e^{2 π in t}, c_{n} = \int_{0}^{1} e^{- 2 π in t} g (t) d t$ $\Rightarrow f (T t) = n = - \infty \sum \infty c_{n} e^{2 π in t}$

Let $s = T t \Rightarrow t = \frac{s}{T}$ . We can change variables to obtain: $f (s) = n = - \infty \sum \infty c_{n} e^{\frac{2 π in}{T} s}$

The same substitution in the coefficient formula ( $s = T t$ , so $d s = T d t$ , and $t : 0 \to 1$ becomes $s : 0 \to T$ ): $c_{n} = \int_{0}^{1} g (t) e^{- 2 π in t} d t = \int_{0}^{1} f (T t) e^{- 2 π in t} d t$ $\Rightarrow c_{n} = \frac{1}{T} \int_{0}^{T} f (s) e^{\frac{- 2 π in}{T} s} d s$

Useful result:

If $f \in C^{k}$ ( $k$ -times differentiable), then $c_{n}$ decays like $c_{n} \sim \frac{1}{∣ n ∣ ^{k}}$ .

Why: integrate by parts once. The boundary term cancels because $f$ and $e^{- 2 π in t / T}$ are both $T$ -periodic, so $c_{n} (f^{'}) = \frac{1}{T} (f (t) e^{\frac{- 2 π in t}{T}}_{0}^{T} + \frac{2 π in}{T} \int_{0}^{T} f (t) e^{\frac{- 2 π in t}{T}} d t) = \frac{2 π in}{T} c_{n} (f)$ Iterating $k$ times gives $c_{n} (f) = (\frac{T}{2 π in})^{k} c_{n} (f^{(k)})$ , and $∣ c_{n} (f^{(k)}) ∣ \leq max ∣ f^{(k)} ∣$ is bounded, so $∣ c_{n} ∣ \leq C /∣ n ∣^{k}$ .

Example:

Suppose $f$ is periodic of period $T$ and even.

$c_{n} (Let s := - t) = \frac{1}{T} \int_{0}^{T} e^{\frac{- 2 π in t}{T}} f (t) d t = \frac{1}{T} \int_{0}^{T} e^{\frac{- 2 π in t}{T}} f (- t) d t = \frac{- 1}{T} \int_{0}^{- T} e^{\frac{2 π in s}{T}} f (s) d s = \frac{1}{T} \int_{- T}^{0} e^{\frac{2 π in s}{T}} f (s) d s = \frac{1}{T} \int_{0}^{T} e^{\frac{2 π in s}{T}} f (s) d s = c_{- n}$

The last step shifts the interval $[- T, 0]$ over to $[0, T]$ , which is legal by the Proposition below.

Proposition:

If $g$ is periodic of period $T$ , then $\int_{a}^{a + T} g (t) d t = \int_{0}^{T} g (t) d t$

Proof:

Let $F (x) = \int_{x}^{x + T} g (t) d t$ . $F^{'} (x) = g (x + T) - g (x) = 0 \Rightarrow F is constant.$ In particular $F (a) = F (0)$ , which is the claim.

Example:

Consider the square wave $f$ of period 1 defined on $[0, 1]$ by $f (t) = ⎩ ⎨ ⎧ 1, - 1, 0 \leq t < \frac{1}{2} \frac{1}{2} \leq t < 1$

One can easily find the Fourier coefficients by using casework on the integral between 0 and 1, as this is a piecewise function. Doing it: $c_{0} = \int_{0}^{1/2} 1 d t + \int_{1/2}^{1} (- 1) d t = \frac{1}{2} - \frac{1}{2} = 0$ For $n \neq = 0$ , with antiderivative $\frac{e ^{- 2 π in t}}{- 2 π in}$ , $c_{n} = \int_{0}^{1/2} e^{- 2 π in t} d t - \int_{1/2}^{1} e^{- 2 π in t} d t = \frac{e ^{- π in} - 1}{- 2 π in} - \frac{e ^{- 2 π in} - e ^{- π in}}{- 2 π in} = \frac{[ ( - 1 ) ^{n} - 1 ] - [ 1 - ( - 1 ) ^{n} ]}{- 2 π in} = \frac{2 ( - 1 ) ^{n} - 2}{- 2 π in} = \frac{1 - ( - 1 ) ^{n}}{π in} = ⎩ ⎨ ⎧ 0, \frac{2}{π in}, n even n odd$ using $e^{- π in} = (- 1)^{n}$ and $e^{- 2 π in} = 1$ . This is the series quoted in Lecture 4.

Lecture 3

Suppose $f (t)$ is periodic of period $T$ .

When $f$ is even and real-valued then $c_{n} = c_{- n} = c_{n}$ , or that $c_{n}$ are real.

Example:

The triangle wave is the following. Let $f (t)$ be a periodic function with period 1 $f (t) = ⎩ ⎨ ⎧ \frac{1}{2} + t, \frac{1}{2} - t, - \frac{1}{2} \leq t < 0 0 \leq t < \frac{1}{2}$

Find the Fourier series of $f$ . $c_{n} = \int_{- \frac{1}{2}}^{\frac{1}{2}} e^{- 2 π in t} f (t) d t$ $c_{0} = \int_{- \frac{1}{2}}^{\frac{1}{2}} f (t) d t = 2 \int_{0}^{1/2} (\frac{1}{2} - t) d t = \frac{1}{4}$

For $n \neq = 0$ : split $e^{- 2 π in t} = cos (2 π n t) - i sin (2 π n t)$ . Since $f$ is even, $f (t) sin (2 π n t)$ is odd and integrates to $0$ , while $f (t) cos (2 π n t)$ is even, so the integral is twice the half-range one: $c_{n} = \int_{- \frac{1}{2}}^{0} e^{- 2 π in t} (\frac{1}{2} + t) d t + \int_{0}^{\frac{1}{2}} e^{- 2 π in t} (\frac{1}{2} - t) d t = 2 \int_{0}^{\frac{1}{2}} (\frac{1}{2} - t) cos (2 π n t) d t$ Integrate by parts with $u = \frac{1}{2} - t$ , $d v = cos (2 π n t) d t$ , so $d u = - d t$ and $v = \frac{s i n ( 2 π n t )}{2 π n}$ . The boundary term dies at both ends ( $u = 0$ at $t = \frac{1}{2}$ , $v = 0$ at $t = 0$ ): $c_{n} = 2 ((\frac{1}{2} - t) \frac{sin ( 2 π n t )}{2 π n}_{0}^{1/2} + \frac{1}{2 π n} \int_{0}^{1/2} sin (2 π n t) d t) = \frac{1}{π n} (\frac{- cos ( 2 π n t )}{2 π n}_{0}^{1/2}) = \frac{1 - cos ( π n )}{2 π ^{2} n ^{2}} = \frac{1 - ( - 1 ) ^{n}}{2 π ^{2} n ^{2}} = ⎩ ⎨ ⎧ 0, \frac{1}{π ^{2} n ^{2}}, n even n odd$

Thus, the Fourier series of $f$ is $\frac{1}{4} + n odd, n \in Z \sum \frac{1}{π ^{2} n ^{2}} e^{2 π in t}$

However, since odd numbers come in $\pm$ pairs, we can write $\frac{1}{4} + n odd, n \geq 1 \sum \frac{1}{π ^{2} n ^{2}} cis (2 π n t) + n odd, n \leq - 1 \sum \frac{1}{π ^{2} n ^{2}} cis (2 π n t) = \frac{1}{4} + \frac{2}{π ^{2}} n odd, n \geq 1 \sum \frac{cos ( 2 π n t )}{n ^{2}} = \frac{1}{4} + \frac{2}{π ^{2}} k = 0 \sum \infty \frac{cos ( 2 π ( 2 k + 1 ) t )}{( 2 k + 1 ) ^{2}}$ pairing $n$ with $- n$ and using $e^{2 π in t} + e^{- 2 π in t} = 2 cos (2 π n t)$ , then substituting $n = 2 k + 1$ .

Rayleigh’s identity (Parseval’s Theorem)

Let $f$ be periodic of period $T$ with Fourier series $n = - \infty \sum \infty c_{n} e^{\frac{2 π in t}{T}}$

Assume $f$ has only a finite number of jump discontinuities on the interval $[0, T]$ . Then $n = - \infty \sum \infty ∣ c_{n} ∣^{2} = \frac{1}{T} \int_{0}^{T} ∣ f (t) ∣^{2} d t$

Proof:

$∣ f (t) ∣^{2} = f (t) f (t) = (n = - \infty \sum \infty c_{n} e^{\frac{2 π in t}{T}}) (k = - \infty \sum \infty c_{k} e^{- \frac{2 π ik t}{T}}) = n = - \infty \sum \infty k = - \infty \sum \infty c_{n} c_{k} e^{\frac{2 π i ( n - k ) t}{T}}$ Integrate over one period. By the same two-case computation as in Lecture 2, $\int_{0}^{T} e^{2 π i (n - k) t / T} d t$ equals $T$ when $n = k$ and $0$ otherwise, so only the diagonal survives: $\Rightarrow \int_{0}^{T} ∣ f (t) ∣^{2} d t = n = - \infty \sum \infty k = - \infty \sum \infty c_{n} c_{k} \int_{0}^{T} e^{\frac{2 π i ( n - k ) t}{T}} d t = T n = - \infty \sum \infty c_{n} c_{n}$ Divide by $T$ and use $c_{n} c_{n} = ∣ c_{n} ∣^{2}$ .

Lecture 4

Example:

Refer back to the square wave diagram. We know that the Fourier series is $f (t) \sim n odd \sum \frac{2}{π in} e^{2 π in t}$

Using Parseval’s theorem, and noting $∣ f (t) ∣ = 1$ everywhere so the left side is just 1: $\int_{0}^{1} ∣ f (t) ∣^{2} d t = 1 = n odd \sum \frac{4}{π ^{2} n ^{2}} = \frac{8}{π ^{2}} n odd, n \geq 1 \sum \frac{1}{n ^{2}}$ $= \frac{8}{π ^{2}} k = 0 \sum \infty \frac{1}{( 2 k + 1 ) ^{2}} \Rightarrow k = 0 \sum \infty \frac{1}{( 2 k + 1 ) ^{2}} = \frac{π ^{2}}{8}$ (The 8 is 4 from $\frac{2}{π in}^{2} = \frac{4}{π ^{2} n ^{2}}$ , doubled because $n$ and $- n$ contribute equally.)

Example:

Refer back to the triangle wave. We know that $f (t) = \frac{1}{4} + n odd \sum \frac{1}{π ^{2} n ^{2}} e^{2 π in t}$ The left-hand side of Parseval is, using evenness, $\int_{- \frac{1}{2}}^{\frac{1}{2}} ∣ f (t) ∣^{2} d t = 2 \int_{0}^{1/2} (\frac{1}{2} - t)^{2} d t = 2 \frac{- ( \frac{1}{2} - t ) ^{3}}{3}_{0}^{1/2} = \frac{2}{24} = \frac{1}{12}$ so $\frac{1}{12} = \frac{1}{16} + \frac{1}{π ^{4}} n odd \sum \frac{1}{n ^{4}} \Rightarrow \frac{1}{48} = \frac{2}{π ^{4}} k = 0 \sum \infty \frac{1}{( 2 k + 1 ) ^{4}}$ $\Rightarrow k = 0 \sum \infty \frac{1}{( 2 k + 1 ) ^{4}} = \frac{π ^{4}}{96} .$ Here $\frac{1}{16} = ∣ c_{0} ∣^{2}$ , and the $\pm n$ pairing again supplies the factor of 2.

Theorem: (Pointwise convergence of Fourier series)

Let $f$ be a periodic function of period $T$ . Suppose $f$ and $f^{'}$ are continuous on $[0, T]$ except for a finite number of jump discontinuities. Then the Fourier series converges to $\frac{f ( a + ) + f ( a - )}{2}$ at every number $a$ . That is, $n = - \infty \sum \infty c_{n} e^{\frac{2 π ina}{T}} = \frac{f ( a + ) + f ( a - )}{2}$ $f (a +) = x \to a + lim f (x), f (a -) = x \to a - lim f (x)$

Example:

Return to the square wave series. Pairing $n$ with $- n$ and using $e^{2 π in t} - e^{- 2 π in t} = 2 i sin (2 π n t)$ , $\frac{2}{π in} e^{2 π in t} + \frac{2}{π i ( - n )} e^{- 2 π in t} = \frac{2}{π in} \cdot 2 i sin (2 π n t) = \frac{4}{π n} sin (2 π n t)$ $\Rightarrow f (t) \sim \frac{4}{π} k = 0 \sum \infty \frac{1}{2 k + 1} sin (2 π (2 k + 1) t)$

At $t = 0$ , we have $\frac{f ( 0 + ) + f ( 0 - )}{2} = 0$ . The Fourier series returns the same result—since $sin (0) = 0$ .

At $t = \frac{1}{4}$ , we get $f (t) = 1$ . In the Fourier series: $\frac{4}{π} k = 0 \sum \infty \frac{sin ( \frac{( 2 k + 1 ) π}{2} )}{2 k + 1} = \frac{4}{π} k = 0 \sum \infty \frac{( - 1 ) ^{k}}{2 k + 1}$

Hence, we get a nice sum from an alternating series: $k = 0 \sum \infty \frac{( - 1 ) ^{k}}{2 k + 1} = \frac{π}{4}$

Example:

Let’s return to the triangle wave again. At $t = 0$ , we get $f (t) = \frac{1}{2}$ . Test in the Fourier series: $\frac{1}{4} + \frac{1}{π ^{2}} n odd \sum \frac{1}{n ^{2}} = \frac{1}{4} + \frac{2}{π ^{2}} k = 0 \sum \infty \frac{1}{( 2 k + 1 ) ^{2}}$ $\frac{1}{2} = \frac{1}{4} + \frac{2}{π ^{2}} k = 0 \sum \infty \frac{1}{( 2 k + 1 ) ^{2}} \Rightarrow k = 0 \sum \infty \frac{1}{( 2 k + 1 ) ^{2}} = \frac{π ^{2}}{8}$

Do the same thing at $t = \frac{1}{2}$ . The periodic extension is continuous there with $f (\frac{1}{2}) = 0$ , and $e^{π in} = (- 1)^{n} = - 1$ for odd $n$ , so $0 = \frac{1}{4} + n odd \sum \frac{( - 1 ) ^{n}}{π ^{2} n ^{2}} = \frac{1}{4} - \frac{1}{π ^{2}} n odd \sum \frac{1}{n ^{2}} \Rightarrow n odd \sum \frac{1}{n ^{2}} = \frac{π ^{2}}{4}$

This sum runs over all odd $n \in Z$ , so it is twice the sum over $n \geq 1$ : $\frac{π ^{2}}{4} = 2 k = 0 \sum \infty \frac{1}{( 2 k + 1 ) ^{2}} \Rightarrow k = 0 \sum \infty \frac{1}{( 2 k + 1 ) ^{2}} = \frac{π ^{2}}{8}$ the same result as before.

Toward the trigonometric Fourier series.

Motivation:

Suppose $f$ is periodic of period $T = 2 l$ . Then $f (x) \sim n = - \infty \sum \infty c_{n} e^{\frac{2 π in x}{T}} = n = - \infty \sum \infty c_{n} e^{\frac{π in x}{l}} = n = - \infty \sum \infty c_{n} cos (\frac{nπ x}{l}) + i n = - \infty \sum \infty c_{n} sin (\frac{nπ x}{l})$

Lecture 5

Trigonometric Fourier Series

Let $f (x)$ be periodic of period $T = 2 l$ . For example, we could have $cos (\frac{k π x}{l})$ .

Then $f$ can be represented as $f (x) \sim \frac{a _{0}}{2} + k = 1 \sum \infty a_{k} cos (\frac{k π x}{l}) + k = 1 \sum \infty b_{k} sin (\frac{k π x}{l})$

Now, how do we compute $a_{k}, b_{j}$ , the coefficients?

Recall:

Trig identities: $cos (A + B) = cos A cos B - sin A sin B$ , $sin (A + B) = sin A cos B + sin B cos A$ . From these, we can derive identities for $sin A cos B, sin B cos A, sin A sin B, cos A cos B$ . Writing down the $(A - B)$ versions too and adding/subtracting: $cos A cos B = \frac{1}{2} [cos (A - B) + cos (A + B)], sin A sin B = \frac{1}{2} [cos (A - B) - cos (A + B)]$ $sin A cos B = \frac{1}{2} [sin (A + B) + sin (A - B)]$

Return to $f (x) \sim \frac{a _{0}}{2} + k = 1 \sum \infty a_{k} cos (\frac{k π x}{l}) + k = 1 \sum \infty b_{k} sin (\frac{k π x}{l}) (*)$

Let $m \geq 1$ . Multiply $(*)$ by $cos (\frac{mπ x}{l})$ and integrate over $[- l, l]$ (swapping $\sum$ and $\int$ is fine here, since the series converges uniformly under our assumptions): $\int_{- l}^{l} f (x) cos (\frac{mπ x}{l}) d x = \frac{a _{0}}{2} \int_{- l}^{l} cos (\frac{mπ x}{l}) d x + k = 1 \sum \infty a_{k} \int_{- l}^{l} cos (\frac{k π x}{l}) cos (\frac{mπ x}{l}) d x + k = 1 \sum \infty b_{k} \int_{- l}^{l} sin (\frac{k π x}{l}) cos (\frac{mπ x}{l}) d x$

We know that $\int_{- l}^{l} cos (\frac{mπ x}{l}) d x = \frac{l}{mπ} sin (\frac{mπ x}{l})_{- l}^{l} = \frac{l}{mπ} [sin (mπ) - sin (- mπ)] = 0$ $\int_{- l}^{l} cos (\frac{k π x}{l}) cos (\frac{mπ x}{l}) d x = \frac{1}{2} (\int_{- l}^{l} cos (\frac{( k + m ) π x}{l}) d x + \int_{- l}^{l} cos (\frac{( k - m ) π x}{l}) d x)$ $= \frac{1}{2} (\frac{l}{( m + k ) π} sin (\frac{( m + k ) π x}{l})_{- l}^{l} + \int_{- l}^{l} cos (\frac{( k - m ) π x}{l}) d x)$ The first piece vanishes since $k + m \geq 2$ is a nonzero integer, leaving $= \frac{1}{2} \int_{- l}^{l} cos (\frac{( k - m ) π x}{l}) d x = ⎩ ⎨ ⎧ \frac{l}{2 π ( k - m )} sin (\frac{( k - m ) π x}{l})_{- l}^{l} = 0, \frac{1}{2} \int_{- l}^{l} 1 d x = l, k \neq = m k = m$

Now, return to the sum from before. Only the single term $k = m$ survives: $k = 1 \sum \infty a_{k} \int_{- l}^{l} cos (\frac{k π x}{l}) cos (\frac{mπ x}{l}) d x = l a_{m}$

So we have filled in one nasty sum/integral pair, let’s look at the last one: $\int_{- l}^{l} sin (\frac{k π x}{l}) cos (\frac{mπ x}{l}) d x = \frac{1}{2} (\int_{- l}^{l} sin (\frac{( k + m ) π x}{l}) d x + \int_{- l}^{l} sin (\frac{( k - m ) π x}{l}) d x)$ $= \frac{1}{2} (\frac{- l}{( k + m ) π} cos (\frac{( k + m ) π x}{l})_{- l}^{l} + \int_{- l}^{l} sin (\frac{( k - m ) π x}{l}) d x)$ Since $cos$ is even, the evaluations at $x = l$ and $x = - l$ agree and the first piece is 0: $= \frac{1}{2} \int_{- l}^{l} sin (\frac{( k - m ) π x}{l}) d x = ⎩ ⎨ ⎧ \frac{- l}{2 ( k - m ) π} cos (\frac{( k - m ) π x}{l})_{- l}^{l} = 0, \frac{1}{2} \int_{- l}^{l} 0 d x = 0, k \neq = m k = m$

Shortcut: that whole computation was unnecessary. $sin (\frac{k π x}{l}) cos (\frac{mπ x}{l})$ is odd in $x$ (odd times even), and we integrate over the symmetric interval $[- l, l]$ , so it is 0 immediately.

Now, combine all the sums to obtain $\int_{- l}^{l} f (x) cos (\frac{mπ x}{l}) d x = 0 + l a_{m} + 0 = l a_{m}$ $\Rightarrow a_{m} = \frac{1}{l} \int_{- l}^{l} f (x) cos (\frac{mπ x}{l}) d x$

Now look for $a_{0}$ . Integrate $(*)$ itself over $[- l, l]$ : $\int_{- l}^{l} f (x) d x = \frac{a _{0}}{2} \int_{- l}^{l} d x + k = 1 \sum \infty a_{k} \int_{- l}^{l} cos (\frac{k π x}{l}) d x + k = 1 \sum \infty b_{k} \int_{- l}^{l} sin (\frac{k π x}{l}) d x$ The first integral is $2 l$ ; the $cos$ integrals vanish as computed above; the $sin$ integrals vanish since $sin (\frac{k π x}{l})$ is odd. So $= l a_{0} + 0 + 0 = l a_{0} \Rightarrow a_{0} = \frac{1}{l} \int_{- l}^{l} f (x) d x$ This is exactly the $m = 0$ case of the formula for $a_{m}$ , which is why the constant term of $(*)$ is written $\frac{a _{0}}{2}$ and not $a_{0}$ : it lets one formula cover every $m \geq 0$ .

Using the same methods, we can find $b_{k}$ : multiply $(*)$ by $sin (\frac{mπ x}{l})$ instead. We need one more orthogonality relation, $\int_{- l}^{l} sin (\frac{k π x}{l}) sin (\frac{mπ x}{l}) d x = \frac{1}{2} (\int_{- l}^{l} cos (\frac{( k - m ) π x}{l}) d x - \int_{- l}^{l} cos (\frac{( k + m ) π x}{l}) d x) = {0, l, k \neq = m k = m$ by the identical case analysis. The $\int cos (\frac{k π x}{l}) sin (\frac{mπ x}{l}) d x$ terms vanish by the oddness shortcut, and $\int_{- l}^{l} sin (\frac{mπ x}{l}) d x = 0$ kills the $\frac{a _{0}}{2}$ term, so $\int_{- l}^{l} f (x) sin (\frac{mπ x}{l}) d x = l b_{m} \Rightarrow b_{k} = \frac{1}{l} \int_{- l}^{l} f (x) sin (\frac{k π x}{l}) d x$

Example:

Let $f$ be periodic of period $2 π$ , and $f (x) = {0, x, - π \leq x \leq 0 0 \leq x \leq π$

what is the Fourier series of $f$ ?

Solution: here $2 l = 2 π$ , so $l = π$ and every $\frac{k π x}{l}$ is just $k x$ . $a_{0} = \frac{1}{π} \int_{- π}^{π} f (x) d x = \frac{1}{π} \int_{0}^{π} x d x = \frac{1}{π} \cdot \frac{π ^{2}}{2} = \frac{π}{2}$ For $a_{n}$ , integrate by parts with $u = x$ , $d v = cos (n x) d x$ , $v = \frac{s i n ( n x )}{n}$ : $a_{n} = \frac{1}{π} \int_{0}^{π} x cos (n x) d x = \frac{1}{π} (\frac{x sin ( n x )}{n}_{0}^{π} - \frac{1}{n} \int_{0}^{π} sin (n x) d x) = \frac{1}{π} (0 + \frac{cos ( n x )}{n ^{2}}_{0}^{π}) = \frac{1}{π n ^{2}} [(- 1)^{n} - 1] = ⎩ ⎨ ⎧ \frac{- 2}{π n ^{2}}, 0, n odd n even$ For $b_{n}$ , take $u = x$ , $d v = sin (n x) d x$ , $v = \frac{- c o s ( n x )}{n}$ : $b_{n} = \frac{1}{π} \int_{0}^{π} x sin (n x) d x = \frac{1}{π} (\frac{- x cos ( n x )}{n}_{0}^{π} + \frac{1}{n} \int_{0}^{π} cos (n x) d x) = \frac{1}{π} (\frac{- π ( - 1 ) ^{n}}{n} + \frac{sin ( n x )}{n ^{2}}_{0}^{π}) = \frac{- ( - 1 ) ^{n}}{n} = \frac{( - 1 ) ^{n + 1}}{n}$ Assembling, with constant term $\frac{a _{0}}{2} = \frac{π}{4}$ : $\Rightarrow f (x) \sim \frac{π}{4} - \frac{2}{π} n odd, n \geq 1 \sum \frac{cos ( n x )}{n ^{2}} + n = 1 \sum \infty \frac{( - 1 ) ^{n + 1}}{n} sin (n x)$ $= \frac{π}{4} - \frac{2}{π} k = 0 \sum \infty \frac{cos (( 2 k + 1 ) x )}{( 2 k + 1 ) ^{2}} + n = 1 \sum \infty \frac{( - 1 ) ^{n + 1}}{n} sin (n x) .$

Lecture 6

Relation between complex and trigonometric coefficients

Let $f$ be periodic of period $T = 2 l$ . Since $e^{- \frac{inπ x}{l}} = cos (\frac{nπ x}{l}) - i sin (\frac{nπ x}{l})$ , for $n \geq 1$ $c_{n} = \frac{1}{2 l} \int_{- l}^{l} f (x) e^{- \frac{inπ x}{l}} d x = \frac{a _{n} - i b _{n}}{2}, c_{- n} = \frac{a _{n} + i b _{n}}{2}, c_{0} = \frac{a _{0}}{2}$ $\Rightarrow a_{n} = c_{n} + c_{- n}, b_{n} = i (c_{n} - c_{- n})$

Parseval’s Theorem (trigonometric form):

If $f$ is real-valued, then $∣ c_{n} ∣^{2} = ∣ c_{- n} ∣^{2} = \frac{a _{n}^{2} + b _{n}^{2}}{4}$ , so $\frac{1}{2 l} \int_{- l}^{l} ∣ f (x) ∣^{2} d x = \frac{a _{0}^{2}}{4} + n = 1 \sum \infty \frac{a _{n}^{2} + b _{n}^{2}}{2} \Rightarrow \frac{1}{l} \int_{- l}^{l} ∣ f (x) ∣^{2} d x = \frac{a _{0}^{2}}{2} + n = 1 \sum \infty (a_{n}^{2} + b_{n}^{2})$

Example:

Return to the example from Lecture 5: $f (x) \sim \frac{π}{4} - \frac{2}{π} k = 0 \sum \infty \frac{cos (( 2 k + 1 ) x )}{( 2 k + 1 ) ^{2}} + n = 1 \sum \infty \frac{( - 1 ) ^{n + 1}}{n} sin (n x)$

At $x = 0$ , $f$ is continuous and $f (0) = 0$ : $0 = \frac{π}{4} - \frac{2}{π} k = 0 \sum \infty \frac{1}{( 2 k + 1 ) ^{2}} \Rightarrow k = 0 \sum \infty \frac{1}{( 2 k + 1 ) ^{2}} = \frac{π ^{2}}{8}$

Using Parseval’s theorem, with $\frac{1}{π} \int_{0}^{π} x^{2} d x = \frac{π ^{2}}{3}$ and $\sum a_{n}^{2} = \frac{4}{π ^{2}} \cdot \frac{π ^{4}}{96} = \frac{π ^{2}}{24}$ (Lecture 4): $\frac{π ^{2}}{3} = \frac{π ^{2}}{8} + \frac{π ^{2}}{24} + n = 1 \sum \infty \frac{1}{n ^{2}} \Rightarrow n = 1 \sum \infty \frac{1}{n ^{2}} = \frac{π ^{2}}{6}$

Lecture 7

Theorem:

Any function $f$ can be expressed as the sum of an odd and even function.

Proof:

$f (x) = \frac{f ( x ) + f ( - x )}{2} + \frac{f ( x ) - f ( - x )}{2}$

The first term is even and the second is odd: replacing $x$ by $- x$ leaves the first unchanged and flips the sign of the second.

Theorem:

Suppose $f (x)$ is a periodic function of period $T = 2 l$ with Fourier series $f (x) \sim \frac{a _{0}}{2} + k = 1 \sum \infty a_{k} cos (\frac{k π x}{l}) + k = 1 \sum \infty b_{k} sin (\frac{k π x}{l}) .$ If $f$ is an even function, $b_{k} = 0$ for all $k$ , and if $f$ is an odd function, $a_{k} = 0$ for all $k$ .

Proof:

$a_{k} = \frac{1}{l} \int_{- l}^{l} f (x) cos (\frac{k π x}{l}) d x$ If $f$ is odd, then $f (x) cos (\frac{k π x}{l})$ is odd (odd times even), so its integral over the symmetric interval $[- l, l]$ is 0, and $a_{k} = 0$ for all $k$ (including $k = 0$ ). $b_{k} = \frac{1}{l} \int_{- l}^{l} f (x) sin (\frac{k π x}{l}) d x$ If $f$ is even, then $f (x) sin (\frac{k π x}{l})$ is odd (even times odd), so $b_{k} = 0$ for all $k$ .

Fact:

Let $f (x)$ be defined on $[0, l]$ .

(1) Even periodic extension of $f$ , $f^{e}$ . Extend $f$ as an even function on $[- l, l]$ , and extend this to a periodic function of period $T = 2 l$ . Call this $f^{e} (x)$ . Since $f^{e}$ is even, its Fourier series has only cosine terms: $f^{e} (x) \sim \frac{a _{0}}{2} + k = 1 \sum \infty a_{k} cos (\frac{k π x}{l}), a_{k} = \frac{1}{l} \int_{- l}^{l} f^{e} (x) cos (\frac{k π x}{l}) d x = \frac{2}{l} \int_{0}^{l} f (x) cos (\frac{k π x}{l}) d x$ The last step uses that $f^{e} (x) cos (\frac{k π x}{l})$ is even and $f^{e} = f$ on $[0, l]$ .

This is the Fourier cosine series of $f$ .

(2) Odd periodic extension of $f$ , $f^{o}$ . Extend $f$ as an odd function on $[- l, l]$ , then extend as a periodic function of period $T = 2 l$ . Call this $f^{o} (x)$ . Since $f^{o}$ is odd, its Fourier series has only sine terms: $f^{o} (x) \sim k = 1 \sum \infty b_{k} sin (\frac{k π x}{l}), b_{k} = \frac{1}{l} \int_{- l}^{l} f^{o} (x) sin (\frac{k π x}{l}) d x = \frac{2}{l} \int_{0}^{l} f (x) sin (\frac{k π x}{l}) d x$ since $f^{o} (x) sin (\frac{k π x}{l})$ is even (odd times odd).

This is the Fourier sine series of $f$ .

Example:

Let $f (x) = x$ on $[0, 2]$ . Find the cosine and sine series of $f$ .

Solution: here $l = 2$ . $a_{0} = \frac{2}{2} \int_{0}^{2} x d x = 2$ For $k \geq 1$ , integrate by parts with $u = x$ , $d v = cos (\frac{k π}{2} x) d x$ , $v = \frac{2}{k π} sin (\frac{k π}{2} x)$ : $a_{k} = \frac{2}{2} \int_{0}^{2} x cos (\frac{k π}{2} x) d x = \frac{2 x}{k π} sin (\frac{k π}{2} x)_{0}^{2} - \frac{2}{k π} \int_{0}^{2} sin (\frac{k π}{2} x) d x = 0 + \frac{4}{k ^{2} π ^{2}} cos (\frac{k π}{2} x)_{0}^{2} = \frac{4 [ ( - 1 ) ^{k} - 1 ]}{k ^{2} π ^{2}} = ⎩ ⎨ ⎧ 0, \frac{- 8}{π ^{2} k ^{2}}, k even k odd$ With constant term $\frac{a _{0}}{2} = 1$ , $f (x) \sim 1 - \frac{8}{π ^{2}} n = 0 \sum \infty \frac{cos ( \frac{( 2 n + 1 ) π}{2} x )}{( 2 n + 1 ) ^{2}}$ Similarly, with $u = x$ , $d v = sin (\frac{k π}{2} x) d x$ , $v = - \frac{2}{k π} cos (\frac{k π}{2} x)$ : $b_{k} = \frac{2}{2} \int_{0}^{2} x sin (\frac{k π}{2} x) d x = - \frac{2 x}{k π} cos (\frac{k π}{2} x)_{0}^{2} + \frac{2}{k π} \int_{0}^{2} cos (\frac{k π}{2} x) d x = - \frac{4 ( - 1 ) ^{k}}{k π} + 0 = \frac{4 ( - 1 ) ^{k + 1}}{k π}$ $f (x) \sim \frac{4}{π} k = 1 \sum \infty \frac{( - 1 ) ^{k + 1}}{k} sin (\frac{k π}{2} x)$

Fourier series of derivatives.

Suppose $f$ is a continuous periodic function of period $T$ such that $f^{'}$ is piecewise continuous (finitely many jump discontinuities per period). Write $f (x) \sim n = - \infty \sum \infty c_{n} (f) e^{\frac{2 π in x}{T}}, f^{'} (x) \sim n = - \infty \sum \infty c_{n} (f^{'}) e^{\frac{2 π in x}{T}}$ We want to relate $c_{n} (f^{'})$ to $c_{n} (f)$ . $c_{n} (f^{'}) = \frac{1}{T} \int_{0}^{T} f^{'} (x) e^{- \frac{2 π in}{T} x} d x$ Examine $c_{0} (f^{'})$ : $c_{0} (f^{'}) = \frac{1}{T} \int_{0}^{T} f^{'} (x) d x = \frac{f ( T ) - f ( 0 )}{T} = 0$ by periodicity. Now let $n \neq = 0$ and integrate by parts: $c_{n} (f^{'}) = \frac{1}{T} f (x) e^{- \frac{2 π in}{T} x}_{0}^{T} + \frac{2 π in}{T} \cdot \frac{1}{T} \int_{0}^{T} f (x) e^{- \frac{2 π in x}{T}} d x = \frac{2 π in}{T} c_{n} (f)$ The boundary term vanishes because $f (T) = f (0)$ and $e^{- 2 π in} = 1$ . (If $f^{'}$ has jumps, integrate by parts on each piece; since $f$ itself is continuous, the interior boundary terms cancel.)

Conclude (the case $n = 0$ included): $c_{n} (f^{'}) = \frac{2 π in}{T} c_{n} (f) \forall n$

Notation:

We say $f \in C^{k}$ if $f$ has $k$ continuous derivatives.

Consequence:

Suppose $f \in C^{2}$ (more generally: $f, f^{'}$ continuous and $f^{''}$ piecewise continuous). Applying the formula to $f^{'}$ and then to $f$ , $c_{n} (f^{''}) = \frac{2 π in}{T} c_{n} (f^{'}) = (\frac{2 π in}{T})^{2} c_{n} (f) .$ In general, $c_{n} (f^{(k)}) = (\frac{2 π in}{T})^{k} c_{n} (f)$ .

Lecture 8

Decay of Fourier Coefficients and Differentiability

Suppose $f (x)$ is periodic of period $T = 2 l$ and piecewise continuous, with $f (x) \sim n \in Z \sum c_{n} e^{\frac{2 π in x}{T}} = \frac{a _{0}}{2} + k = 1 \sum \infty a_{k} cos (\frac{k π x}{l}) + k = 1 \sum \infty b_{k} sin (\frac{k π x}{l})$ Parseval’s Theorem says $\frac{1}{T} \int_{0}^{T} ∣ f (x) ∣^{2} d x = n \in Z \sum ∣ c_{n} ∣^{2} < \infty$ and similarly (Lecture 6) $\sum_{k = 0}^{\infty} a_{k}^{2} < \infty$ and $\sum_{k = 1}^{\infty} b_{k}^{2} < \infty$ . The terms of a convergent series tend to 0, so $∣ n ∣ \to \infty lim c_{n} = n \to \infty lim a_{n} = n \to \infty lim b_{n} = 0$ Then an interesting question is: at what rate do the Fourier coefficients decay to 0?

Fact:

The more derivatives $f$ has, the faster the coefficients decay.

Let $C^{k} = {f : f has k continuous derivatives}$

Theorem 1:

Let $f$ be periodic with $f \in C^{1}$ . Then there is $c > 0$ such that $∣ c_{n} (f) ∣ \leq \frac{c}{∣ n ∣}$ , $∣ a_{n} (f) ∣ \leq \frac{c}{n}$ , $∣ b_{n} (f) ∣ \leq \frac{c}{n}$ for all $n \neq = 0$ .

Proof:

$c_{n} (f^{'}) = \frac{1}{T} \int_{0}^{T} e^{\frac{- 2 π in x}{T}} f^{'} (x) d x$ (For the record, $c_{0} (f^{'}) = \frac{1}{T} \int_{0}^{T} f^{'} (x) d x = \frac{1}{T} (f (T) - f (0)) = 0$ .) For any $n \neq = 0$ , Lecture 7 gives $c_{n} (f^{'}) = \frac{2 π in}{T} c_{n} (f)$ On the other hand, since $e^{- 2 π in x / T} = 1$ , $∣ c_{n} (f^{'}) ∣ = \frac{1}{T} \int_{0}^{T} e^{\frac{- 2 π in x}{T}} f^{'} (x) d x \leq \frac{1}{T} \int_{0}^{T} ∣ f^{'} (x) ∣ d x =: M$ Therefore $∣ c_{n} (f) ∣ = \frac{T}{2 π ∣ n ∣} ∣ c_{n} (f^{'}) ∣ \leq \frac{M T}{2 π} \cdot \frac{1}{∣ n ∣}$ Let’s do the same thing for the trigonometric coefficients. Integrating by parts (the boundary term vanishes by periodicity), $b_{n} (f^{'}) = \frac{1}{l} \int_{- l}^{l} f^{'} (x) sin (\frac{nπ x}{l}) d x = \frac{1}{l} f (x) sin (\frac{nπ x}{l})_{- l}^{l} - \frac{π n}{l} \cdot \frac{1}{l} \int_{- l}^{l} f (x) cos (\frac{nπ x}{l}) d x = - \frac{π n}{l} a_{n} (f)$ and in the same way (now the boundary term is $\frac{( - 1 ) ^{n}}{l} [f (l) - f (- l)] = 0$ ) $a_{n} (f^{'}) = \frac{π n}{l} b_{n} (f)$ : $a_{n} (f^{'}) = \frac{π n}{l} b_{n} (f), b_{n} (f^{'}) = - \frac{π n}{l} a_{n} (f) for n \geq 1$ Moreover $∣ b_{n} (f^{'}) ∣ = \frac{1}{l} \int_{- l}^{l} f^{'} (x) sin (\frac{nπ x}{l}) d x \leq \frac{1}{l} \int_{- l}^{l} ∣ f^{'} (x) ∣ d x =: M^{'}$ and likewise $∣ a_{n} (f^{'}) ∣ \leq M^{'}$ . Hence $∣ a_{n} (f) ∣ = \frac{l}{π n} ∣ b_{n} (f^{'}) ∣ \leq \frac{l M ^{'}}{π} \cdot \frac{1}{n}, ∣ b_{n} (f) ∣ = \frac{l}{π n} ∣ a_{n} (f^{'}) ∣ \leq \frac{l M ^{'}}{π} \cdot \frac{1}{n}$ Take $c = max (\frac{M T}{2 π}, \frac{l M ^{'}}{π})$ .

Theorem 2:

Let $f$ be periodic with $f \in C^{k}$ . Then there is $c > 0$ such that $∣ c_{n} (f) ∣ \leq \frac{c}{∣ n ∣ ^{k}}, ∣ a_{n} (f) ∣ \leq \frac{c}{n ^{k}}, ∣ b_{n} (f) ∣ \leq \frac{c}{n ^{k}} (n \neq = 0)$

Proof:

Say $f \in C^{2}$ . Then $f^{'} \in C^{1}$ , and applying Lecture 7 twice, $c_{n} (f^{''}) = \frac{2 π in}{T} c_{n} (f^{'}) = (\frac{2 π in}{T})^{2} c_{n} (f) \Rightarrow c_{n} (f) = (\frac{T}{2 π in})^{2} c_{n} (f^{''})$ Exactly as before, $∣ c_{n} (f^{''}) ∣ = \frac{1}{T} \int_{0}^{T} e^{\frac{- 2 π in x}{T}} f^{''} (x) d x \leq \frac{1}{T} \int_{0}^{T} ∣ f^{''} (x) ∣ d x =: M < \infty$ Therefore, $∣ c_{n} (f) ∣ \leq \frac{M T ^{2}}{4 π ^{2}} \cdot \frac{1}{n ^{2}} = \frac{c}{n ^{2}} .$ For general $k$ the same argument works (formally, induction on $k$ ): $c_{n} (f) = (\frac{T}{2 π in})^{k} c_{n} (f^{(k)})$ and $∣ c_{n} (f^{(k)}) ∣ \leq \frac{1}{T} \int_{0}^{T} ∣ f^{(k)} (x) ∣ d x$ . The bounds on $a_{n} (f)$ and $b_{n} (f)$ follow the same way by applying the boxed relations above $k$ times.

Remark:

The practical upshot: the smoother $f$ is, the fewer Fourier modes are needed to approximate it well. Lecture 9 makes this quantitative: to guarantee accuracy $\frac{1}{100}$ , the $C^{2}$ bound there needs $N = 400$ terms, while the $C^{4}$ bound needs only $N = 6$ .

Lecture 9

Application:

$f (x) = n = - \infty \sum \infty c_{n} e^{\frac{2 π in x}{T}} = n = - N \sum N c_{n} e^{\frac{2 π in x}{T}} + ∣ n ∣ > N \sum c_{n} e^{\frac{2 π in x}{T}} =: S_{N} (x) + E_{N} (x)$ where $S_{N} (x)$ is the $N$ -th partial sum, and $E_{N} (x)$ is the $N$ -th error term.

Example:

Suppose $f$ is a periodic function with $f \in C^{2}$ , and $∣ c_{n} ∣ \leq \frac{2}{n ^{2}}$ . How large must $N$ be to guarantee $∣ f (x) - S_{N} (x) ∣ \leq \frac{1}{100}$ for all $x$ ?

Solution:

Since $e^{2 π in x / T} = 1$ , and the terms $n$ and $- n$ obey the same bound, $∣ f (x) - S_{N} (x) ∣ = ∣ E_{N} (x) ∣ \leq ∣ n ∣ > N \sum ∣ c_{n} ∣ \leq 2 n > N \sum \frac{2}{n ^{2}} = 4 n > N \sum \frac{1}{n ^{2}}$ Compare the sum with an integral: $\frac{1}{x ^{2}}$ is decreasing, so $\frac{1}{n ^{2}} \leq \int_{n - 1}^{n} \frac{d x}{x ^{2}}$ , and $4 n > N \sum \frac{1}{n ^{2}} \leq 4 \int_{N}^{\infty} \frac{1}{x ^{2}} d x = \frac{4}{N}$ $\Rightarrow ∣ f (x) - S_{N} (x) ∣ \leq \frac{4}{N}$ Thus it suffices that $\frac{4}{N} \leq \frac{1}{100}$ , i.e. $N \geq 400$ .

Example:

Now suppose $f$ is periodic with $f \in C^{4}$ , and $∣ c_{n} ∣ \leq \frac{2}{n ^{4}}$ . Now how large must $N$ be to guarantee $∣ f (x) - S_{N} (x) ∣ \leq \frac{1}{100}$ ?

Solution:

The same estimate gives $∣ f (x) - S_{N} (x) ∣ = ∣ E_{N} (x) ∣ \leq 4 n > N \sum \frac{1}{n ^{4}} \leq 4 \int_{N}^{\infty} \frac{1}{x ^{4}} d x = \frac{4}{3 N ^{3}}$ We need $\frac{4}{3 N ^{3}} \leq \frac{1}{100}$ , i.e. $N^{3} \geq \frac{400}{3} \approx 133.3$ . Since $5^{3} = 125 < 133.3 \leq 216 = 6^{3}$ , $N \geq 6$

Note:

What about the converse? That is, suppose $f$ is periodic and the Fourier coefficients satisfy $∣ c_{n} ∣ \leq \frac{c}{∣ n ∣ ^{k}}$ . Is $f \in C^{k}$ ?

The answer is NO (see the example below). But it is at least partially true.

Theorem (partial converse):

Let $f$ be periodic and piecewise continuous. If $∣ c_{n} ∣ \leq \frac{c}{∣ n ∣ ^{k + 1 + δ}}$ for all $n \neq = 0$ , where $δ > 0$ , then $f \in C^{k}$ (after redefining $f$ at its finitely many jump points, if it has any).

Proof sketch:

For $0 \leq j \leq k$ , differentiate the Fourier series term by term $j$ times. The resulting terms satisfy $(\frac{2 π in}{T})^{j} c_{n} e^{\frac{2 π in x}{T}} \leq (\frac{2 π}{T})^{j} \frac{c ∣ n ∣ ^{j}}{∣ n ∣ ^{k + 1 + δ}} \leq (\frac{2 π}{T})^{j} \frac{c}{∣ n ∣ ^{1 + δ}},$ and $\sum_{n \neq = 0} \frac{1}{∣ n ∣ ^{1 + δ}} < \infty$ . By the Weierstrass M-test all of these series converge uniformly, so $g (x) := \sum_{n} c_{n} e^{2 π in x / T}$ is $C^{k}$ (a series whose differentiated series converge uniformly may be differentiated term by term). Integrating term by term, $g$ has the same Fourier coefficients $c_{n}$ as $f$ , so $f - g$ has all coefficients 0, and Parseval gives $\int_{0}^{T} ∣ f - g ∣^{2} d x = 0$ . Hence $f = g$ at every point where $f$ is continuous.

Example:

Recall the triangle wave, with $c_{n} = \frac{1}{π ^{2} n ^{2}}$ for $n$ odd and $c_{n} = 0$ for $n \neq = 0$ even. Then $∣ c_{n} ∣ \leq \frac{c}{n ^{2}}$ , but the triangle wave has corners, so it is not in $C^{1}$ , let alone $C^{2}$ . So the converse fails.

The theorem only gives what is true: $\frac{1}{n ^{2}} = \frac{1}{n ^{0 + 1 + 1}}$ ( $k = 0$ , $δ = 1$ ), so the triangle wave is continuous. To conclude $f \in C^{1}$ we would need $\frac{1}{n ^{2}} = \frac{1}{n ^{1 + 1 + δ}}$ with $δ = 0$ , which is not allowed. So the condition $δ > 0$ is required.

Heat Equation

Consider a uniform rod of length $ℓ$ (with unit cross-sectional area). Let $u (x, t)$ be the temperature of the rod at point $x$ at time $t$ . Assume the rod is insulated along its sides, and that its temperature at time $t = 0$ is $u (x, 0) = f (x)$ . Let $σ$ be the specific heat and $δ$ the density of the rod. Then to raise the temperature of a piece of length $L$ by $τ$ degrees, we need $σ δ Lτ$ units of heat energy.

Take a small piece of the rod, $[x, x + Δ x]$ . Its temperature at time $t$ is approximately $u (x, t)$ , and at a later time $t + Δ t$ approximately $u (x, t + Δ t)$ . So the change in heat energy of the piece from time $t$ to $t + Δ t$ is $\approx [u (x, t + Δ t) - u (x, t)] σ δ Δ x$ This is caused by heat conduction. Experimentally (Fourier’s law), the amount of heat flowing past a point $x$ in the positive direction per unit time (the flux) is proportional to $- \frac{\partial u}{\partial x} (x, t)$ : $Φ (x, t) = - k \frac{\partial u}{\partial x} (x, t)$ The proportionality constant $k > 0$ is called the thermal conductivity of the object.

The net amount of heat flowing into the piece from time $t$ to $t + Δ t$ is $[Φ (x, t) - Φ (x + Δ x, t)] Δ t = [\frac{\partial u}{\partial x} (x + Δ x, t) - \frac{\partial u}{\partial x} (x, t)] k Δ t$ Equating the two, $[u (x, t + Δ t) - u (x, t)] σ δ Δ x \approx [\frac{\partial u}{\partial x} (x + Δ x, t) - \frac{\partial u}{\partial x} (x, t)] k Δ t$ Divide by $σ δ Δ x Δ t$ and let $Δ x, Δ t \to 0$ : the left side becomes $\frac{\partial u}{\partial t}$ and the right side becomes $\frac{k}{σ δ} \frac{\partial ^{2} u}{\partial x ^{2}}$ , so $\frac{\partial u}{\partial t} (x, t) = \frac{k}{σ δ} \frac{\partial ^{2} u}{\partial x ^{2}} (x, t)$

Since $\frac{k}{σ δ} > 0$ , write $α^{2} := \frac{k}{σ δ}$ , called the thermal diffusivity. This equation is also called the Diffusion Equation.

It remains to solve the PDE to obtain $u (x, t)$ , the temperature of the rod: $\frac{\partial u}{\partial t} (x, t) = α^{2} \frac{\partial ^{2} u}{\partial x ^{2}} (x, t), 0 < x < ℓ, t > 0.$

Here is what we need:

The initial temperature of the rod: $u (x, 0) = f (x)$ , $0 < x < ℓ$ .
Assume the endpoints of the rod $x = 0$ and $x = ℓ$ are maintained at temperature 0, that is, $u (0, t) = 0$ and $u (ℓ, t) = 0$ .

So we must solve the initial boundary value problem $\frac{\partial u}{\partial t} (x, t) = α^{2} \frac{\partial ^{2} u}{\partial x ^{2}} (x, t) (PDE)$ $u (0, t) = 0, u (ℓ, t) = 0, t > 0 (boundary conditions)$ $u (x, 0) = f (x), 0 < x < ℓ (initial condition)$

Find solutions of the PDE of the form $u (x, t) = X (x) T (t)$ (method of separation of variables).

If $X (x) T (t)$ is a solution, then $X (x) T^{'} (t) = α^{2} X^{''} (x) T (t) \Rightarrow \frac{X ^{''} ( x )}{X ( x )} = \frac{T ^{'} ( t )}{α ^{2} T ( t )}$ Observe that the left side does not depend on $t$ , and the right side does not depend on $x$ . Therefore both sides equal some constant $λ$ : $X^{''} (x) - λ X (x) = 0, T^{'} (t) - λ α^{2} T (t) = 0$ Now, we have two ODEs! Their solutions are:

Case 1: $λ = 0$ . Then $X (x) = d_{1} + d_{2} x$ , $T (t) = d_{3}$ .

Case 2: $λ \neq = 0$ . Then $X (x) = d_{1} e^{λ x} + d_{2} e^{- λ x}$ , $T (t) = d_{3} e^{λ α^{2} t}$ . Here $λ$ may be complex, and we allow complex constants $d_{i}$ .

Since the PDE is linear, any linear combination $\sum_{i} a_{i} X_{i} (x) T_{i} (t)$ of such solutions is again a solution.

However, we need to satisfy the boundary conditions, i.e. we need $\sum_{i} a_{i} X_{i} (0) T_{i} (t) = 0$ and also $\sum_{i} a_{i} X_{i} (ℓ) T_{i} (t) = 0$ . We can satisfy these by requiring $X_{i} (0) = 0$ and $X_{i} (ℓ) = 0$ .

Case 1: $λ = 0$ . $X (0) = d_{1} = 0$ , and then $X (ℓ) = d_{2} ℓ = 0$ forces $d_{2} = 0$ . So $X \equiv 0$ , and we only get the trivial solution $X (x) T (t) = 0$ , which is useless.

Case 2: $λ \neq = 0$ . $X (0) = 0 \Rightarrow d_{1} + d_{2} = 0 \Rightarrow X (x) = d_{1} (e^{λ x} - e^{- λ x})$ Now use the other boundary condition. For a nonzero solution we need $d_{1} \neq = 0$ , so $X (ℓ) = 0 \Rightarrow e^{λ ℓ} = e^{- λ ℓ} \Rightarrow e^{2 λ ℓ} = 1 \Rightarrow 2 λ ℓ = 2 π ik, k \in Z ∖ {0}$ so $λ = \frac{k π i}{ℓ}$ and $λ = - \frac{k ^{2} π ^{2}}{ℓ ^{2}}$ . In particular $λ$ must be negative. Since $k$ and $- k$ give the same $λ$ , take $k \geq 1$ .

Write it out: $X (x) = d_{1} (e^{\frac{k π}{ℓ} i x} - e^{- \frac{k π}{ℓ} i x}) = 2 i d_{1} sin (\frac{k π}{ℓ} x)$ Now that we have an expression for $λ$ , we obtain $T (t) = d_{3} e^{λ α^{2} t} = d_{3} e^{- \frac{k ^{2} π ^{2}}{ℓ ^{2}} α^{2} t}$ So for each $k \geq 1$ , a solution of the PDE and the boundary conditions is $sin (\frac{k π}{ℓ} x) e^{- \frac{k ^{2} π ^{2}}{ℓ ^{2}} α^{2} t}$ Take any linear combination: $k = 1 \sum \infty b_{k} sin (\frac{k π}{ℓ} x) e^{- \frac{k ^{2} π ^{2} α ^{2}}{ℓ ^{2}} t}$ We want to choose $b_{k}$ such that when $t = 0$ , $k = 1 \sum \infty b_{k} sin (\frac{k π}{ℓ} x) = f (x), 0 < x < ℓ$ This is exactly the Fourier sine series of $f$ from Lecture 7: any piecewise continuous $f$ on $(0, ℓ)$ can be extended to an odd periodic function of period $2 ℓ$ , which expands in sines. So take $b_{k} = \frac{2}{ℓ} \int_{0}^{ℓ} f (x) sin (\frac{k π}{ℓ} x) d x$ Then the solution of the problem is $u (x, t) = k = 1 \sum \infty b_{k} sin (\frac{k π}{ℓ} x) e^{- \frac{π ^{2} α ^{2} k ^{2}}{ℓ ^{2}} t}$ where the $b_{k}$ are determined by the initial condition: $b_{k} = \frac{2}{ℓ} \int_{0}^{ℓ} f (x) sin (\frac{k π}{ℓ} x) d x .$

Equilibrium Temperature of the Rod

Since the two ends of the rod are maintained at temperature 0, physically we expect the rod’s equilibrium state to be $t \to \infty lim u (x, t) = 0$ The formula confirms this. Each $∣ b_{k} ∣ \leq \frac{2}{ℓ} \int_{0}^{ℓ} ∣ f (x) ∣ d x =: B$ , and $k^{2} \geq k$ , so with $r = e^{- π^{2} α^{2} t / ℓ^{2}} < 1$ , $∣ u (x, t) ∣ \leq B k = 1 \sum \infty e^{- \frac{π ^{2} α ^{2} k ^{2}}{ℓ ^{2}} t} \leq B k = 1 \sum \infty r^{k} = \frac{B r}{1 - r} t \to \infty 0.$ Every mode decays, and the slowest one ( $k = 1$ ) decays like $e^{- π^{2} α^{2} t / ℓ^{2}}$ .

Lecture 10

Example:

Solve: $⎩ ⎨ ⎧ u_{t} = 9 u_{xx}, u (0, t) = 0, u (2, t) = 8, u (x, 0) = 2 x^{2}, 0 < x < 2, t > 0 t > 0 0 < x < 2$

We convert this to a problem where the boundary conditions are 0. So what is the equilibrium temperature $g (x)$ ? It does not depend on $t$ , so $g_{t} = 0$ and the PDE becomes $g^{''} = 0$ : $g$ is linear. Matching the boundary conditions $g (0) = 0$ , $g (2) = 8$ gives $g (x) = 4 x$ , which satisfies both the PDE and the boundary conditions.

Now suppose $u (x, t)$ is the solution, and let $w (x, t) = u (x, t) - g (x)$ which satisfies $⎩ ⎨ ⎧ w_{t} = 9 w_{xx} w (0, t) = u (0, t) - g (0) = 0, w (2, t) = u (2, t) - g (2) = 0 w (x, 0) = u (x, 0) - g (x) = 2 x^{2} - 4 x$ This is a problem with zero boundary conditions, which we know how to solve ( $α^{2} = 9$ , $ℓ = 2$ ): $w (x, t) = k = 1 \sum \infty b_{k} sin (\frac{k π}{2} x) e^{- \frac{9}{4} π^{2} k^{2} t}$ $b_{k} = \frac{2}{2} \int_{0}^{2} (2 x^{2} - 4 x) sin (\frac{k π}{2} x) d x$ With $a = \frac{k π}{2}$ (so $2 a = k π$ and $cos (2 a) = (- 1)^{k}$ , $sin (2 a) = 0$ ), integrating by parts once and twice gives $\int_{0}^{2} x sin (a x) d x = - \frac{2 ( - 1 ) ^{k}}{a}, \int_{0}^{2} x^{2} sin (a x) d x = - \frac{4 ( - 1 ) ^{k}}{a} + \frac{2 [ ( - 1 ) ^{k} - 1 ]}{a ^{3}}$ so $b_{k} = 2 (- \frac{4 ( - 1 ) ^{k}}{a} + \frac{2 [ ( - 1 ) ^{k} - 1 ]}{a ^{3}}) - 4 (- \frac{2 ( - 1 ) ^{k}}{a}) = \frac{4 [ ( - 1 ) ^{k} - 1 ]}{a ^{3}} = \frac{32 [ ( - 1 ) ^{k} - 1 ]}{π ^{3} k ^{3}} = ⎩ ⎨ ⎧ \frac{- 64}{π ^{3} k ^{3}}, 0, k odd k even$

Finally, conclude $u (x, t) = w (x, t) + g (x) = k = 1 \sum \infty b_{k} sin (\frac{k π}{2} x) e^{- \frac{9}{4} π^{2} k^{2} t} + 4 x$ As $t \to \infty$ , $u (x, t) \to 4 x$ , the equilibrium.

Neumann Boundary Conditions

Now, suppose we know that both ends of the rod are perfectly insulated (no heat flux through the ends): $⎩ ⎨ ⎧ u_{t} = α^{2} u_{xx}, u_{x} (0, t) = 0, u_{x} (ℓ, t) = 0, u (x, 0) = f (x), 0 < x < ℓ, t > 0 t > 0 0 < x < ℓ$

Let’s use the same process of separation of variables.

Lecture 11

Neumann Boundary Conditions (Continued)

Again seek $u (x, t) = X (x) T (t)$ , which leads to $X^{''} (x) - λ X (x) = 0, T^{'} (t) - λ α^{2} T (t) = 0$ with solutions

Case 1: $λ = 0$ . $X (x) = d_{1} + d_{2} x$ , $T (t) = d_{3}$ .

Case 2: $λ \neq = 0$ . $X (x) = d_{1} e^{λ x} + d_{2} e^{- λ x}$ , $T (t) = d_{3} e^{λ α^{2} t}$ .

Check the boundary conditions: $u (x, t) = \sum_{i} a_{i} X_{i} (x) T_{i} (t)$ should satisfy $u_{x} (0, t) = 0$ and $u_{x} (ℓ, t) = 0$ . Requiring $X_{i}^{'} (0) = 0$ and $X_{i}^{'} (ℓ) = 0$ works as the simplest choice.

Case 1: $λ = 0$ . $X (x) = d_{1} + d_{2} x, X^{'} (x) = d_{2}, X^{'} (0) = 0 \Rightarrow d_{2} = 0$ $\Rightarrow X (x) = d_{1} \Rightarrow u = X (x) T (t) = d_{1} d_{3} (constant)$ Unlike the Dirichlet problem, $λ = 0$ now gives a nontrivial solution.

Case 2: $λ \neq = 0$ . $X^{'} (x) = λ (d_{1} e^{λ x} - d_{2} e^{- λ x})$ $X^{'} (0) = 0 \Rightarrow λ (d_{1} - d_{2}) = 0 \Rightarrow d_{1} = d_{2}$ $\Rightarrow X (x) = d_{1} (e^{λ x} + e^{- λ x})$ $X^{'} (ℓ) = 0 \Rightarrow λ d_{1} (e^{λ ℓ} - e^{- λ ℓ}) = 0$ For a nonzero solution $d_{1} \neq = 0$ , so, exactly as for the Dirichlet problem, $e^{λ ℓ} - e^{- λ ℓ} = 0 \Rightarrow e^{2 λ ℓ} = 1 \Rightarrow λ = \frac{k π}{ℓ} i \Rightarrow λ = - \frac{π ^{2}}{ℓ ^{2}} k^{2}, k \geq 1$ $\Rightarrow X (x) = d_{1} (e^{\frac{k π}{ℓ} x i} + e^{- \frac{k π}{ℓ} x i}) = 2 d_{1} cos (\frac{k π}{ℓ} x)$ Also, $T_{k} (t) = e^{- \frac{π ^{2}}{ℓ ^{2}} k^{2} α^{2} t}$ . Together with the constant solution from Case 1, superposition gives $u (x, t) = \frac{a _{0}}{2} + k = 1 \sum \infty a_{k} cos (\frac{k π}{ℓ} x) e^{- \frac{π ^{2}}{ℓ ^{2}} α^{2} k^{2} t} .$ What remains is the initial condition. To get $u (x, 0) = f (x)$ on $(0, ℓ)$ , we need to choose the $a_{k}$ such that $\frac{a _{0}}{2} + k = 1 \sum \infty a_{k} cos (\frac{k π}{ℓ} x) = f (x),$ which is the Fourier cosine series of $f$ (Lecture 7). So choose $a_{k} = \frac{2}{ℓ} \int_{0}^{ℓ} f (x) cos (\frac{k π}{ℓ} x) d x, k \geq 0,$ and the solution is $u (x, t) = \frac{a _{0}}{2} + k = 1 \sum \infty a_{k} cos (\frac{k π}{ℓ} x) e^{- \frac{π ^{2}}{ℓ ^{2}} α^{2} k^{2} t}$

Since the rod is completely insulated, no heat escapes, and we expect the equilibrium temperature of the rod to be the constant average of the initial temperature. Indeed, every $k \geq 1$ term decays, leaving $t \to \infty lim u (x, t) = \frac{a _{0}}{2} = \frac{1}{ℓ} \int_{0}^{ℓ} f (x) d x .$

The Wave Equation

We would like to solve a system governed by $⎩ ⎨ ⎧ u_{tt} = c^{2} u_{xx}, u (0, t) = 0, u (ℓ, t) = 0, u (x, 0) = f (x), u_{t} (x, 0) = g (x), 0 < x < ℓ, t > 0 t > 0 0 < x < ℓ$ Now we need two initial conditions (initial position and initial velocity), since the equation is second order in time.

Once again, use the method of separation of variables: seek solutions of the PDE and BC of the form $u (x, t) = X (x) T (t)$ . Then $X T^{''} = c^{2} X^{''} T$ , so $\frac{X ^{''} ( x )}{X ( x )} = \frac{T ^{''} ( t )}{c ^{2} T ( t )} = λ$ $\Rightarrow X^{''} (x) - λ X (x) = 0, T^{''} (t) - λ c^{2} T (t) = 0$ Consider $u (x, t) = \sum_{i} a_{i} X_{i} (x) T_{i} (t)$ . Then $u (0, t) = 0$ and $u (ℓ, t) = 0$ if $X_{i} (0) = 0$ and $X_{i} (ℓ) = 0$ . This is exactly the boundary value problem for $X$ we solved for the heat equation, so $λ = - \frac{π ^{2} k ^{2}}{ℓ ^{2}}, X_{k} (x) = sin (\frac{k π}{ℓ} x), k \geq 1.$ Then $T^{''} + (\frac{π k c}{ℓ})^{2} T = 0$ , so $T_{k} (t) = a_{k} cos (\frac{π k c}{ℓ} t) + b_{k} sin (\frac{π k c}{ℓ} t)$ , and $u (x, t) = k = 1 \sum \infty (a_{k} cos (\frac{π c k}{ℓ} t) + b_{k} sin (\frac{π c k}{ℓ} t)) sin (\frac{π k}{ℓ} x)$ We want $f (x) = u (x, 0) = \sum_{k = 1}^{\infty} a_{k} sin (\frac{π k}{ℓ} x)$ . This is the sine series of $f$ , so $a_{k} = \frac{2}{ℓ} \int_{0}^{ℓ} f (x) sin (\frac{π k}{ℓ} x) d x$ Also, differentiating term by term in $t$ , $g (x) = u_{t} (x, 0) = k = 1 \sum \infty \frac{π c k}{ℓ} b_{k} sin (\frac{π k}{ℓ} x)$ so $\frac{π c k}{ℓ} b_{k}$ is the $k$ -th sine coefficient of $g$ : $\frac{π c k}{ℓ} b_{k} = \frac{2}{ℓ} \int_{0}^{ℓ} g (x) sin (\frac{π k}{ℓ} x) d x \Rightarrow b_{k} = \frac{2}{π c k} \int_{0}^{ℓ} g (x) sin (\frac{π k}{ℓ} x) d x$ Thus both families of constants are determined by the initial conditions, and the solution is complete.

Lecture 12

The Fourier Transform

Motivation:

Suppose $f : R \to C$ is nonperiodic. We approximate $f$ by periodic functions as follows.

Let $T > 0$ , and let $f_{T} (x)$ be the periodic function of period $T$ that agrees with $f$ on $[- \frac{T}{2}, \frac{T}{2}]$ . Then $lim_{T \to \infty} f_{T} (x) = f (x)$ for each $x$ , and $f_{T} (x) = n = - \infty \sum \infty c_{n} e^{\frac{2 π in}{T} x} = n = - \infty \sum \infty \frac{1}{T} (\int_{- \frac{T}{2}}^{\frac{T}{2}} f (t) e^{\frac{- 2 π in t}{T}} d t) e^{\frac{2 π in x}{T}}$ Let $x_{n} = \frac{n}{T}$ , and $Δ x_{n} = \frac{1}{T} = Δ x$ . Then $f_{T} (x) = n = - \infty \sum \infty (\int_{- \frac{T}{2}}^{\frac{T}{2}} f (t) e^{- 2 π i x_{n} t} d t) e^{2 π i x_{n} x} Δ x_{n}$ Heuristically, as $T \to \infty$ the inner integral becomes an integral over all of $R$ : $f (x) = T \to \infty lim f_{T} (x) \approx n = - \infty \sum \infty (\int_{- \infty}^{\infty} f (t) e^{- 2 π i x_{n} t} d t) e^{2 π i x_{n} x} Δ x_{n}$ This looks like a Riemann sum, $n = - \infty \sum \infty h (x_{n}) Δ x_{n} where h (y) = (\int_{- \infty}^{\infty} f (t) e^{- 2 π i y t} d t) e^{2 π i y x},$ with mesh $Δ x \to 0$ , so it should converge to $\int_{- \infty}^{\infty} h (y) d y = \int_{- \infty}^{\infty} (\int_{- \infty}^{\infty} f (t) e^{- 2 π i y t} d t) e^{2 π i y x} d y$ Call the inner integral $f^(y)$ . Then the heuristic says $f (x) = \int_{- \infty}^{\infty} f^(y) e^{2 π i y x} d y$ , which is the inversion formula (Lecture 13, proved in Lecture 15).

Definition:

The Fourier transform of $f (x)$ is $F f (ζ) = f^(ζ) = \int_{- \infty}^{\infty} f (x) e^{- 2 π i ζ x} d x$ if the integral is defined.

Definition:

Let $f : R \to C$ . We say $f \in L^{1} (R)$ ( $f$ is “absolutely integrable”) if $R \to \infty lim \int_{- R}^{R} ∣ f (x) ∣ d x exists (and is finite).$ If the limit exists, then we define $\int_{- \infty}^{\infty} ∣ f (x) ∣ d x = R \to \infty lim \int_{- R}^{R} ∣ f (x) ∣ d x$

Remark:

Suppose $f \in L^{1} (R)$ . Then $\int_{- \infty}^{\infty} ∣ f (x) ∣ d x = \int_{- R}^{R} ∣ f (x) ∣ d x + \int_{{x : ∣ x ∣ \geq R}} ∣ f (x) ∣ d x$ $\Rightarrow \int_{{x : ∣ x ∣ \geq R}} ∣ f (x) ∣ d x = \int_{- \infty}^{\infty} ∣ f (x) ∣ d x - \int_{- R}^{R} ∣ f (x) ∣ d x$ Since $f \in L^{1}$ , $lim_{R \to \infty} \int_{{x : ∣ x ∣ \geq R}} ∣ f (x) ∣ d x = 0$ : the tails of an $L^{1}$ function are small.

Example:

Let $f$ be bounded, say $∣ f ∣ \leq B$ , and assume there are $R_{0}, δ, C > 0$ such that $∣ f (x) ∣ \leq \frac{C}{∣ x ∣ ^{1 + δ}}$ for $∣ x ∣ \geq R_{0}$ . Then $f \in L^{1}$ .

Proof:

On the middle piece, $\int_{- R_{0}}^{R_{0}} ∣ f (x) ∣ d x \leq 2 R_{0} B$ . On the tails, $\int_{R_{0}}^{\infty} \frac{C}{x ^{1 + δ}} d x + \int_{- \infty}^{- R_{0}} \frac{C}{( - x ) ^{1 + δ}} d x = - \frac{C}{δ x ^{δ}}_{R_{0}}^{\infty} + \frac{C}{δ ( - x ) ^{δ}}_{- \infty}^{- R_{0}} = \frac{C}{δ R _{0}^{δ}} + \frac{C}{δ R _{0}^{δ}}$ So $\int_{- R}^{R} ∣ f (x) ∣ d x \leq 2 R_{0} B + \frac{2 C}{δ R _{0}^{δ}}$ for every $R$ . This is increasing in $R$ and bounded above by a finite value, so the limit exists.

Example:

Let $f (x)$ be bounded (e.g. continuous) on $[a, b]$ , say $∣ f ∣ \leq B$ there, and $f (x) = 0$ when $x \in / [a, b]$ . Then $f \in L^{1}$ , since $\int_{- R}^{R} ∣ f (x) ∣ d x \leq (b - a) B$ for every $R$ .

Example:

Recall the Taylor series of $e^{x}$ . In particular, for $x \geq 0$ , $e^{x} \geq \frac{x ^{2}}{2 !}$ , so $e^{- x} \leq \frac{2}{x ^{2}}$ for $x > 0$ . Let $a > 0$ and $f (x) = {e^{- a x}, 0, x \geq 0 x < 0$ Then $∣ f ∣ \leq 1$ and $∣ f (x) ∣ \leq \frac{2}{a ^{2} x ^{2}}$ for $x > 0$ , so $f \in L^{1}$ by the first example (with $δ = 1$ ). (Directly: $\int_{0}^{\infty} e^{- a x} d x = \frac{1}{a}$ .)

Example:

$g (x) = e^{- x^{2}}$ . By the same technique ( $e^{x^{2}} \geq \frac{x ^{4}}{2}$ , so $e^{- x^{2}} \leq \frac{2}{x ^{4}}$ for $x \neq = 0$ , and $e^{- x^{2}} \leq 1$ ), $g \in L^{1}$ .

Fourier Transform Calculations:

Suppose $f \in L^{1}$ . Then $∣ f (x) e^{- 2 π i ζ x} ∣ = ∣ f (x) ∣$ is integrable on $R$ , so $f^(ζ)$ is well-defined for every $ζ$ .

Lecture 13

Note: If $f \in L^{1}$ , then $∣ f^(ζ) ∣ \leq \int_{- \infty}^{\infty} ∣ f (x) ∣ d x$ for all $ζ$ , since $e^{- 2 π i x ζ} = 1$ .

Example (the rectangle function):

Let $R (x) = ⎩ ⎨ ⎧ 1, 0, ∣ x ∣ < \frac{1}{2} ∣ x ∣ > \frac{1}{2}$ (also written $Π$ ). Then for $ζ \neq = 0$ , $R^(ζ) = \int_{- \frac{1}{2}}^{\frac{1}{2}} e^{- 2 π i x ζ} d x = \frac{e ^{- 2 π i x ζ}}{- 2 π i ζ}_{- \frac{1}{2}}^{\frac{1}{2}} = \frac{e ^{π i ζ} - e ^{- π i ζ}}{2 π i ζ} = \frac{sin ( π ζ )}{π ζ},$ and $R^(0) = \int R = 1$ .

Example:

Let $a > 0$ and $f (x) = e^{- a x}$ for $x \geq 0$ , $f (x) = 0$ for $x < 0$ . Then $f^(ζ) = \int_{0}^{\infty} e^{- (a + 2 π i ζ) x} d x = \frac{e ^{- (a + 2 π i ζ) x}}{- ( a + 2 π i ζ )}_{0}^{\infty} = \frac{1}{a + 2 π i ζ}$ (the upper limit contributes 0 since $e^{- (a + 2 π i ζ) x} = e^{- a x} \to 0$ ).

Example:

Let $a > 0$ and $f (x) = e^{- a ∣ x ∣}$ . Splitting the integral at 0, $f^(ζ) = \int_{- \infty}^{0} e^{(a - 2 π i ζ) x} d x + \int_{0}^{\infty} e^{- (a + 2 π i ζ) x} d x = \frac{1}{a - 2 π i ζ} + \frac{1}{a + 2 π i ζ} = \frac{2 a}{a ^{2} + 4 π ^{2} ζ ^{2}}$

Theorem (Fourier inversion; proved in Lecture 15):

If $f, f^\in L^{1}$ , then at every $x$ where $f$ is continuous, $f (x) = \int_{- \infty}^{\infty} f^(ζ) e^{2 π i ζ x} d ζ .$

Lecture 14

Example:

Let $f (x) = e^{- π x^{2}}$ . $f^(ζ) = \int_{- \infty}^{\infty} e^{- 2 π i x ζ} e^{- π x^{2}} d x$ Differentiate under the integral sign (justified in Lecture 17, since $x e^{- π x^{2}} \in L^{1}$ ): $\frac{d f ^}{d ζ} = - 2 π i \int_{- \infty}^{\infty} e^{- 2 π i x ζ} x e^{- π x^{2}} d x$ Since $x e^{- π x^{2}} = - \frac{1}{2 π} \frac{d}{d x} (e^{- π x^{2}})$ , and then integrating by parts (the boundary terms vanish because $e^{- π x^{2}} \to 0$ ), $= i \int_{- \infty}^{\infty} e^{- 2 π i x ζ} \frac{d}{d x} (e^{- π x^{2}}) d x = - i \int_{- \infty}^{\infty} \frac{d}{d x} (e^{- 2 π i x ζ}) e^{- π x^{2}} d x$ $= - i (- 2 π i ζ) \int_{- \infty}^{\infty} e^{- 2 π i x ζ} e^{- π x^{2}} d x = - 2 π ζ f^(ζ)$ This is the ODE $f^^{'} = - 2 π ζ f^$ , so $\Rightarrow f^(ζ) = c e^{- π ζ^{2}}, c = f^(0)$ But $c = f^(0) = \int_{- \infty}^{\infty} e^{- π x^{2}} d x = 1$ . (Polar coordinates: $(\int e^{- π x^{2}} d x)^{2} = \iint e^{- π (x^{2} + y^{2})} d x d y = \int_{0}^{2 π} \int_{0}^{\infty} e^{- π r^{2}} r d r d θ = 2 π \cdot \frac{1}{2 π} = 1$ .) Hence $\Rightarrow f^(ζ) = e^{- π ζ^{2}} = f (ζ)$ So the Fourier transform of this Gaussian is the same Gaussian. Among the Gaussians $e^{- δ x^{2}}$ , this is the only one whose Fourier transform is equal to the function itself (by the formula for $e^{- δ x^{2}}$ below, this forces $δ = π$ ).

Note:

Let $a \neq = 0$ . Given a function $f (x)$ , how does the graph of $f_{a} (x) = f (a x)$ compare with that of $f (x)$ ? It’s a horizontal contraction ( $∣ a ∣ > 1$ ) or stretch ( $∣ a ∣ < 1$ ), reflected if $a < 0$ .

Theorem: (The Stretch Theorem)

Let $f \in L^{1}$ and $a \neq = 0$ . Let $f_{a} (x) = f (a x)$ . Then $f^_{a} (ζ) = \frac{1}{∣ a ∣} f^(\frac{ζ}{a})$ .

Proof:

$f^_{a} (ζ) = \int_{- \infty}^{\infty} e^{- 2 π i x ζ} f_{a} (x) d x = \int_{- \infty}^{\infty} e^{- 2 π i x ζ} f (a x) d x$ Substitute $y = a x$ , $d y = a d x$ . If $a > 0$ the limits stay $- \infty \to \infty$ and we get a factor $\frac{1}{a}$ . If $a < 0$ the limits flip, and flipping them back gives a factor $- \frac{1}{a}$ . Either way the factor is $\frac{1}{∣ a ∣}$ : $= \frac{1}{∣ a ∣} \int_{- \infty}^{\infty} e^{- 2 π i y \frac{ζ}{a}} f (y) d y = \frac{1}{∣ a ∣} f^(\frac{ζ}{a})$

Note:

The more stretched out a function, the more squeezed its Fourier transform (and vice versa).

Example:

Let $δ > 0$ and $g (x) = e^{- δ x^{2}} \in L^{1}$ . Find $g^(ζ)$ .

Solution:

Let $f (x) = e^{- π x^{2}}$ . Recall that $f^(ζ) = e^{- π ζ^{2}}$ .

Express $g$ as $f_{a}$ for some $a$ . We want $f_{a} (x) = g (x)$ : $f (a x) = g (x) \Rightarrow e^{- π a^{2} x^{2}} = e^{- δ x^{2}} \Rightarrow a = \frac{δ}{π}$ Hence, $g (x) = f_{δ / π} (x)$ , and $g^(ζ) = f^_{δ / π} (ζ) = \frac{π}{δ} f^(\frac{π}{δ} ζ) = \frac{π}{δ} e^{- \frac{π ^{2} ζ ^{2}}{δ}}$

Theorem: (The Shift Theorem)

For any $a \in R$ , define the shift operator $(S_{a} f) (x) = f (x + a)$ . Then $S_{a} f (ζ) = e^{2 π ia ζ} f^(ζ)$

Proof:

$S_{a} f (ζ) = \int_{- \infty}^{\infty} e^{- 2 π i x ζ} (S_{a} f) (x) d x = \int_{- \infty}^{\infty} e^{- 2 π i x ζ} f (x + a) d x$ Let $y = x + a$ , $d y = d x$ : $= \int_{- \infty}^{\infty} e^{- 2 π i (y - a) ζ} f (y) d y = e^{2 π ia ζ} f^(ζ) .$

Example:

Let $f (x) = e^{- x^{2} + 4 x} \in L^{1}$ . Find $f^(ζ)$ .

Solution:

Complete the square in the exponent: $f (x) = e^{- (x - 2)^{2} + 4} = e^{4} e^{- (x - 2)^{2}}$ Let $g (x) = e^{- x^{2}}$ . Then $f (x) = e^{4} g (x - 2) = e^{4} (S_{- 2} g) (x)$ By the previous example with $δ = 1$ , $g^(ζ) = π e^{- π^{2} ζ^{2}}$ . By the Shift Theorem, $f^(ζ) = e^{4} S_{- 2} g (ζ) = e^{4} e^{2 π i (- 2) ζ} g^(ζ)$ $\Rightarrow f^(ζ) = π e^{4} e^{- 4 π i ζ} e^{- π^{2} ζ^{2}}$

Lecture 15

Recall: if $a \neq = 0$ and $f_{a} (x) = f (a x)$ , then $f^_{a} (ζ) = \frac{1}{∣ a ∣} f^(\frac{ζ}{a})$ .

If $(S_{b} f) (x) = f (x + b)$ , then $S_{b} f (ζ) = e^{2 π ib ζ} f^(ζ)$ .

Example:

Let $h (x) = f (a x + b)$ with $a \neq = 0$ . What is $h^(ζ)$ ?

Solution:

Write $h (x) = f (a (x + \frac{b}{a}))$ . Let $g (x) = f (a x) = f_{a} (x)$ . Then $h (x) = g (x + \frac{b}{a}) = (S_{b / a} g) (x),$ so first the Shift Theorem and then the Stretch Theorem give $h^(ζ) = e^{2 π i \frac{b}{a} ζ} g^(ζ) = \frac{e ^{2 π i \frac{b}{a} ζ}}{∣ a ∣} f^(\frac{ζ}{a}) .$

Theorem: (Inversion)

Assume $f, f^\in L^{1}$ . Then at every $x$ where $f$ is continuous, $f (x) = \int_{- \infty}^{\infty} e^{2 π i x ζ} f^(ζ) d ζ$

Proof (assuming also that $f$ is bounded, which covers every example in this course):

Insert a Gaussian damping factor $e^{- δ ζ^{2}}$ . Since $f^\in L^{1}$ and $e^{- δ ζ^{2}} \to 1$ as $δ \to 0$ , $\int_{- \infty}^{\infty} e^{2 π i x ζ} f^(ζ) d ζ = δ \to 0 lim \int_{- \infty}^{\infty} e^{2 π i x ζ} e^{- δ ζ^{2}} f^(ζ) d ζ$ (The difference is at most $\int (1 - e^{- δ ζ^{2}}) ∣ f^(ζ) ∣ d ζ$ . Split it at $∣ ζ ∣ = A$ : the tail is small by the Remark in Lecture 12, and on $∣ ζ ∣ \leq A$ the factor $1 - e^{- δ ζ^{2}} \leq 1 - e^{- δ A^{2}} \to 0$ .) Now write out $f^$ : $= δ \to 0 lim \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} e^{2 π i x ζ} e^{- δ ζ^{2}} e^{- 2 π i y ζ} f (y) d y d ζ$ This double integral converges absolutely ( $f \in L^{1}$ and $e^{- δ ζ^{2}} \in L^{1}$ ), so we may swap the order of integration: $= δ \to 0 lim \int_{- \infty}^{\infty} (\int_{- \infty}^{\infty} e^{- 2 π i (y - x) ζ} e^{- δ ζ^{2}} d ζ) f (y) d y$ The inner integral is the Fourier transform of $e^{- δ ζ^{2}}$ evaluated at $y - x$ , which by Lecture 14 is $\frac{π}{δ} e^{- \frac{π ^{2} ( y - x ) ^{2}}{δ}}$ : $= δ \to 0 lim \frac{π}{δ} \int_{- \infty}^{\infty} e^{- \frac{π ^{2} ( y - x ) ^{2}}{δ}} f (y) d y$ Substitute $y = x + δ t$ , $d y = δ d t$ : $= δ \to 0 lim \frac{π}{δ} \int_{- \infty}^{\infty} e^{- π^{2} t^{2}} f (x + δ t) δ d t = π δ \to 0 lim \int_{- \infty}^{\infty} e^{- π^{2} t^{2}} f (x + δ t) d t$ As $δ \to 0$ , $f (x + δ t) \to f (x)$ by continuity of $f$ at $x$ . Moving the limit inside the integral is justified by the Theorem at the end of Lecture 19 (with $g (t) = π e^{- π^{2} t^{2}}$ , which has $\int g = 1$ ). So the expression equals $f (x) \cdot π \int_{- \infty}^{\infty} e^{- π^{2} t^{2}} d t = f (x) \cdot π \cdot \frac{1}{π} = f (x),$ using $\int_{- \infty}^{\infty} e^{- a t^{2}} d t = \frac{π}{a}$ with $a = π^{2}$ .

Note:

The inversion formula says $f (x) = F (F f) (- x)$ , i.e. $F (F f) (x) = f (- x) .$ In particular, if $f$ is even, then $F (F f) = f$ .

Applications:

Let $f (x) = \frac{s i n ( π x )}{π x} = sinc (x)$ . What is $f^(ζ)$ ?

Recall that $\frac{s i n ( π x )}{π x} = R^(x)$ (the rectangle function). Therefore, $f = F R$ , so $F f (ζ) = F (F R) (ζ) = R (- ζ) = R (ζ)$ since $R$ is even. So $f^(ζ) = R (ζ)$ .

Caveat: $sinc \in / L^{1}$ (it decays only like $\frac{1}{∣ x ∣}$ ), so here $f^(ζ)$ must be understood as $lim_{A \to \infty} \int_{- A}^{A}$ . In that sense the formula is correct, with value $\frac{1}{2}$ (the average) at the jumps $ζ = \pm \frac{1}{2}$ .

The triangle function:

Let $T (x) = {1 - ∣ x ∣, 0, ∣ x ∣ \leq 1 ∣ x ∣ > 1$ $T$ is even, so the sine part of $e^{- 2 π i x ζ}$ integrates to 0. For $ζ \neq = 0$ , integrate by parts with $u = 1 - x$ , $v = \frac{s i n ( 2 π ζ x )}{2 π ζ}$ ; the boundary term vanishes at both ends: $T^(ζ) = 2 \int_{0}^{1} (1 - x) cos (2 π ζ x) d x = 2 \int_{0}^{1} \frac{sin ( 2 π ζ x )}{2 π ζ} d x = \frac{1 - cos ( 2 π ζ )}{2 π ^{2} ζ ^{2}} = \frac{2 sin ^{2} ( π ζ )}{2 π ^{2} ζ ^{2}} = \frac{sin ^{2} ( π ζ )}{π ^{2} ζ ^{2}}$ and $T^(0) = \int T = 1$ . (This is the same computation as the triangle wave in Lecture 3. Lecture 18 gives a second derivation via $T = R * R$ .)

Note:

We can also compute integrals. For example, compute $\int_{- \infty}^{\infty} \frac{sin ^{2} ( π x )}{π ^{2} x ^{2}} d x .$ This is simply the integral of the Fourier transform of the triangle function. Since $T, T^\in L^{1}$ and $T$ is continuous, the inversion theorem at $x = 0$ gives $\int_{- \infty}^{\infty} T^(ζ) d ζ = \int_{- \infty}^{\infty} T^(ζ) e^{2 π i \cdot 0 \cdot ζ} d ζ = T (0) = 1.$

Note:

Recall, if $a > 0$ and $f^{a} (t) = {0, e^{- a t}, t \leq 0 t > 0$ then $f^{a} (ζ) = \frac{1}{a + 2 π i ζ}$ .

Example:

Find $g$ so that $g^(ζ) = \frac{1}{ζ ^{2} - 5 i ζ - 6}$ .

Solution:

The roots of $ζ^{2} - 5 i ζ - 6$ are $ζ = \frac{5 i \pm - 25 + 24}{2} = \frac{5 i \pm i}{2}$ , i.e. $3 i$ and $2 i$ . Partial fractions: $g^(ζ) = \frac{1}{( ζ - 3 i ) ( ζ - 2 i )} = \frac{i}{ζ - 2 i} - \frac{i}{ζ - 3 i}$ To match the form $\frac{1}{a + 2 π i ζ}$ , multiply the top and bottom of each term by $2 π i$ : $\frac{i}{ζ - 2 i} = \frac{2 π i ^{2}}{2 π i ζ + 4 π} = \frac{- 2 π}{4 π + 2 π i ζ}, - \frac{i}{ζ - 3 i} = \frac{- 2 π i ^{2}}{2 π i ζ + 6 π} = \frac{2 π}{6 π + 2 π i ζ}$ $\Rightarrow g^(ζ) = - 2 π f^{4 π} (ζ) + 2 π f^{6 π} (ζ) = F (- 2 π f^{4 π} + 2 π f^{6 π}) (ζ)$ $\Rightarrow g (x) = - 2 π f^{4 π} (x) + 2 π f^{6 π} (x) = {2 π (e^{- 6 π x} - e^{- 4 π x}), 0, x > 0 x \leq 0$ (Check: $g^(0) = \int g = 2 π (\frac{1}{6 π} - \frac{1}{4 π}) = - \frac{1}{6}$ , which agrees with $\frac{1}{0 - 0 - 6}$ .)

Example:

Find $h$ where $h^(ζ) = \frac{e ^{6 i ζ}}{7 + 2 π i ζ}$ .

Solution:

$\frac{1}{7 + 2 π i ζ} = f^{7} (ζ)$ , and $e^{6 i ζ} = e^{2 π i \cdot \frac{3}{π} \cdot ζ}$ is exactly the factor produced by the Shift Theorem with $b = \frac{3}{π}$ : $S_{3/ π} f^{7} (ζ) = e^{2 π i \frac{3}{π} ζ} f^{7} (ζ) = \frac{e ^{6 i ζ}}{7 + 2 π i ζ}$ Hence $h (x) = f^{7} (x + \frac{3}{π}) = ⎩ ⎨ ⎧ e^{- 7 (x + \frac{3}{π})} = e^{- 7 x - \frac{21}{π}}, 0, x > - \frac{3}{π} x \leq - \frac{3}{π}$

Example:

Let $h (x) = \frac{1}{1 + x ^{2}}$ . Find $h^(ζ)$ .

Solution:

Recall that if $a > 0$ , then $e^{- a ∣ x ∣} (ζ) = \frac{2 a}{a ^{2} + 4 π ^{2} ζ ^{2}}$ . Then $h (x) = \frac{4 π ^{2}}{4 π ^{2} + 4 π ^{2} x ^{2}} = π (\frac{4 π}{( 2 π ) ^{2} + 4 π ^{2} x ^{2}}) = π F (e^{- 2 π ∣ \cdot ∣}) (x)$ $\Rightarrow h^(ζ) = π F (F (e^{- 2 π ∣ \cdot ∣})) (ζ) = π e^{- 2 π ∣ - ζ ∣} = π e^{- 2 π ∣ ζ ∣} .$

Example:

Find $\int_{- \infty}^{\infty} \frac{1}{( 1 + x ^{2} ) ( 6 + x ^{2} )} d x$ .

(Partial fractions plus the inversion theorem; this is worked out at the start of Lecture 16. The answer is $\frac{π}{5} (1 - \frac{1}{6})$ .)

Lecture 16

Example:

Let $h (x) = \frac{1}{( 1 + x ^{2} ) ( 6 + x ^{2} )}$ . What is $h^(ζ)$ ?

Solution:

$h (x) = \frac{1}{5} (\frac{1}{1 + x ^{2}} - \frac{1}{6 + x ^{2}})$ . For $a > 0$ , let $g^{a} (x) = e^{- a ∣ x ∣}$ , so that $g^{a} (ζ) = \frac{2 a}{a ^{2} + 4 π ^{2} ζ ^{2}}$ .

Consider $\frac{1}{1 + x ^{2}} = \frac{4 π ^{2}}{4 π ^{2} + 4 π ^{2} x ^{2}} = π g^{2 π} (x) = π F (g^{2 π}) (x)$ . Therefore, $F (\frac{1}{1 + x ^{2}}) (ζ) = π g^{2 π} (- ζ) = π e^{- 2 π ∣ ζ ∣}$ .

Consider $\frac{1}{6 + x ^{2}} = \frac{4 π ^{2}}{24 π ^{2} + 4 π ^{2} x ^{2}}$ . With $a = 2 π 6$ we have $a^{2} = 24 π^{2}$ and $2 a = 4 π 6$ , so $\frac{1}{6 + x ^{2}} = \frac{4 π ^{2}}{4 π 6} \cdot \frac{4 π 6}{24 π ^{2} + 4 π ^{2} x ^{2}} = \frac{π}{6} g^{2 π 6} (x) \Rightarrow F (\frac{1}{6 + x ^{2}}) (ζ) = \frac{π}{6} e^{- 2 π 6 ∣ ζ ∣}$ Therefore $h^(ζ) = \frac{π}{5} (e^{- 2 π ∣ ζ ∣} - \frac{1}{6} e^{- 2 π 6 ∣ ζ ∣})$ (In general, the same computation gives $F (\frac{1}{a ^{2} + x ^{2}}) (ζ) = \frac{π}{a} e^{- 2 π a ∣ ζ ∣}$ for $a > 0$ .)

Application:

We want to evaluate the integral $\int_{- \infty}^{\infty} \frac{1}{( 1 + x ^{2} ) ( 6 + x ^{2} )} d x = F (\frac{1}{( 1 + x ^{2} ) ( 6 + x ^{2} )}) (0) = h^(0) = \frac{π}{5} (1 - \frac{1}{6})$ (Check with arctangents: $\frac{1}{5} (π - \frac{π}{6})$ , the same.)

Derivation:

Let’s find the Fourier transforms of $f (x) cos (a x)$ and $f (x) sin (a x)$ in terms of the Fourier transform of $f$ , with $a \in R$ . $f (x) cos (a x) = \frac{1}{2} [f (x) e^{ai x} + f (x) e^{- ai x}]$ Consider $f (x) e^{ai x}$ : $f (x) e^{ai x} (ζ) = \int_{- \infty}^{\infty} e^{- 2 π i ζ x} f (x) e^{ai x} d x = \int_{- \infty}^{\infty} e^{(ai - 2 π i ζ) x} f (x) d x$ $= \int_{- \infty}^{\infty} e^{- 2 π i (ζ - \frac{a}{2 π}) x} f (x) d x = f^(ζ - \frac{a}{2 π})$ Thus, $f (x) e^{ai x} (ζ) = f^(ζ - \frac{a}{2 π})$ , and likewise $f (x) e^{- ai x} (ζ) = f^(ζ + \frac{a}{2 π})$ . Hence $f (x) cos (a x) (ζ) = \frac{1}{2} [f^(ζ - \frac{a}{2 π}) + f^(ζ + \frac{a}{2 π})] .$ The same for $sin (a x) = \frac{1}{2 i} [e^{ai x} - e^{- ai x}]$ : $f (x) sin (a x) (ζ) = \frac{1}{2 i} [f^(ζ - \frac{a}{2 π}) - f^(ζ + \frac{a}{2 π})] .$

Example:

Compute $\int_{0}^{\infty} \frac{c o s ( 2 x )}{1 + x ^{2}} d x$ .

Solution:

The integrand is an even function, so $\int_{0}^{\infty} \frac{cos ( 2 x )}{1 + x ^{2}} d x = \frac{1}{2} \int_{- \infty}^{\infty} \frac{cos ( 2 x )}{1 + x ^{2}} d x = \frac{1}{2} F (\frac{cos ( 2 x )}{1 + x ^{2}}) (0)$ Using $f (x) = \frac{1}{1 + x ^{2}}$ , $f^(ζ) = π e^{- 2 π ∣ ζ ∣}$ , and $a = 2$ (so $\frac{a}{2 π} = \frac{1}{π}$ ), $F (\frac{cos ( 2 x )}{1 + x ^{2}}) (0) = \frac{1}{2} [f^(- \frac{1}{π}) + f^(\frac{1}{π})]$ We know $f^(\frac{1}{π}) = f^(- \frac{1}{π}) = π e^{- 2}$ so $F (\frac{c o s ( 2 x )}{1 + x ^{2}}) (0) = π e^{- 2}$ . Thus, the integral is equal to $\int_{0}^{\infty} \frac{cos ( 2 x )}{1 + x ^{2}} d x = \frac{π}{2 e ^{2}}$

Plancherel’s (Parseval’s) Identity

Let $f, g$ be continuous with $f, g, f^, g^\in L^{1}$ . Then $\int_{- \infty}^{\infty} f (x) g (x) d x = \int_{- \infty}^{\infty} f^(ζ) g^(ζ) d ζ$

Corollary:

Take $f = g$ . Then $\int_{- \infty}^{\infty} ∣ f (x) ∣^{2} d x = \int_{- \infty}^{\infty} ∣ f^(ζ) ∣^{2} d ζ$ (Analog of Parseval’s (Rayleigh’s) Identity for Fourier series, Lecture 3.)

Proof:

Conjugating $g^(ζ) = \int g (x) e^{- 2 π i x ζ} d x$ gives $g^(ζ) = \int g (x) e^{2 π i x ζ} d x$ . So $\int_{- \infty}^{\infty} f^(ζ) g^(ζ) d ζ = \int_{- \infty}^{\infty} f^(ζ) \int_{- \infty}^{\infty} g (x) e^{2 π i x ζ} d x d ζ$ The double integral converges absolutely, since $\iint ∣ f^(ζ) ∣∣ g (x) ∣ d x d ζ = \int ∣ f^∣ \cdot \int ∣ g ∣ < \infty$ . So swap the order of integration: $= \int_{- \infty}^{\infty} g (x) (\int_{- \infty}^{\infty} f^(ζ) e^{2 π i x ζ} d ζ) d x = \int_{- \infty}^{\infty} g (x) f (x) d x$ by the inversion theorem (Lecture 15).

Example:

$\int_{- \infty}^{\infty} \frac{1}{( 1 + x ^{2} ) ^{2}} d x$

Solution:

Let $f (x) = \frac{1}{1 + x ^{2}}$ , so $f^(ζ) = π e^{- 2 π ∣ ζ ∣}$ (both in $L^{1}$ ). The integral is $\int ∣ f ∣^{2}$ , so by Plancherel it equals $\int_{- \infty}^{\infty} ∣ f^(ζ) ∣^{2} d ζ = π^{2} \int_{- \infty}^{\infty} e^{- 4 π ∣ ζ ∣} d ζ = 2 π^{2} \int_{0}^{\infty} e^{- 4 π ζ} d ζ = 2 π^{2} \cdot \frac{1}{4 π} = \frac{π}{2}$

Example:

$\int_{- \infty}^{\infty} \frac{sin ^{2} ( π x )}{x ^{2}} d x$

Method 1:

Use the fact that $\frac{s i n ^{2} ( π x )}{x ^{2}}$ is related to the transform of the triangle function: $\frac{sin ^{2} ( π x )}{x ^{2}} = π^{2} \frac{sin ^{2} ( π x )}{π ^{2} x ^{2}} = π^{2} T^(x) .$ By inversion at 0 (Lecture 15), $\int_{- \infty}^{\infty} \frac{sin ^{2} ( π x )}{x ^{2}} d x = π^{2} \int_{- \infty}^{\infty} T^(x) d x = π^{2} T (0) = π^{2}$

Method 2:

Use Parseval, since $\frac{s i n ( π x )}{x} = π \frac{s i n ( π x )}{π x} = π R^(x)$ is related to the rectangle function: $\int_{- \infty}^{\infty} \frac{sin ^{2} ( π x )}{x ^{2}} d x = π^{2} \int_{- \infty}^{\infty} ∣ R^(x) ∣^{2} d x = π^{2} \int_{- \infty}^{\infty} ∣ R (x) ∣^{2} d x = π^{2} \cdot 1 = π^{2}$ (Strictly, $R^\in / L^{1}$ , so this uses the extension of Plancherel to all $f$ with $\int ∣ f ∣^{2} < \infty$ . Method 1 avoids the issue.)

Note:

We should study the link between the decay of $f^(ζ)$ and the differentiability of $f$ .

Lecture 17

(These notes had no entry for Lecture 17. The material below is a reconstruction of what Lecture 18 builds on: how the Fourier transform interacts with derivatives, and the resulting link between smoothness of $f$ and decay of $f^$ . Check it against the lecture.)

Theorem (Transform of a derivative):

Suppose $f$ is continuous and piecewise $C^{1}$ , with $f, f^{'} \in L^{1}$ . Then $f^{'} (ζ) = 2 π i ζ f^(ζ)$

Proof:

First, $f (x) \to 0$ as $x \to \pm \infty$ . Indeed, $f (x) = f (0) + \int_{0}^{x} f^{'} (s) d s$ , and since $f^{'} \in L^{1}$ the integral has a limit as $x \to \infty$ . So $lim_{x \to \infty} f (x)$ exists, and it must be 0, or else $∣ f ∣$ would not be integrable. The same holds as $x \to - \infty$ .

Now integrate by parts on $[- R, R]$ (on each piece where $f$ is $C^{1}$ ; the interior boundary terms cancel because $f$ is continuous): $\int_{- R}^{R} f^{'} (x) e^{- 2 π i x ζ} d x = f (x) e^{- 2 π i x ζ}_{- R}^{R} + 2 π i ζ \int_{- R}^{R} f (x) e^{- 2 π i x ζ} d x$ Let $R \to \infty$ . The boundary term tends to 0 since $∣ f (\pm R) ∣ \to 0$ , which leaves $f^{'} (ζ) = 2 π i ζ f^(ζ)$ .

This is the Fourier transform analog of $c_{n} (f^{'}) = \frac{2 π in}{T} c_{n} (f)$ from Lecture 7.

Corollary:

If $f \in C^{n}$ and $f, f^{'}, \dots, f^{(n)} \in L^{1}$ , then applying the theorem $n$ times, $f^{(n)} (ζ) = (2 π i ζ)^{n} f^(ζ) .$ Combining with $∣ f^{(n)} (ζ) ∣ \leq \int ∣ f^{(n)} ∣$ (Lecture 13), $∣ f^(ζ) ∣ \leq \frac{1}{( 2 π ∣ ζ ∣ ) ^{n}} \int_{- \infty}^{\infty} ∣ f^{(n)} (x) ∣ d x = \frac{C}{∣ ζ ∣ ^{n}} (ζ \neq = 0) .$ This is the Fourier transform version of Lecture 8: the more derivatives $f$ has, the faster $f^$ decays.

Example:

The triangle function $T$ is continuous and piecewise $C^{1}$ , with $T^{'} (x) = ⎩ ⎨ ⎧ 1, - 1, 0, - 1 < x < 0 0 < x < 1 ∣ x ∣ > 1 = R (x + \frac{1}{2}) - R (x - \frac{1}{2}) .$ By the Shift Theorem, $T^{'} (ζ) = (e^{π i ζ} - e^{- π i ζ}) R^(ζ) = 2 i sin (π ζ) \cdot \frac{sin ( π ζ )}{π ζ} = \frac{2 i sin ^{2} ( π ζ )}{π ζ}$ and indeed, as the theorem predicts, $2 π i ζ T^(ζ) = 2 π i ζ \cdot \frac{sin ^{2} ( π ζ )}{π ^{2} ζ ^{2}} = \frac{2 i sin ^{2} ( π ζ )}{π ζ} .$ This matches the decay picture: $R$ (discontinuous) has $∣ R^∣ \leq \frac{C}{∣ ζ ∣}$ , while $T$ (continuous, with piecewise continuous derivative) has $∣ T^∣ \leq \frac{C}{∣ ζ ∣ ^{2}}$ .

Theorem (Derivative of a transform):

Suppose $f \in L^{1}$ and $x f (x) \in L^{1}$ . Then $f^$ is differentiable and $\frac{d}{d ζ} f^(ζ) = (- 2 π i x) f (ζ) = - 2 π i \int_{- \infty}^{\infty} x f (x) e^{- 2 π i x ζ} d x$

Proof:

Differentiate $f^(ζ) = \int f (x) e^{- 2 π i x ζ} d x$ under the integral sign. This is legitimate because the $ζ$ -derivative of the integrand, $- 2 π i x f (x) e^{- 2 π i x ζ}$ , is bounded in absolute value by $2 π ∣ x f (x) ∣$ , which is integrable and does not depend on $ζ$ .

This justifies the first step of the Gaussian computation in Lecture 14. Iterating: if $x^{k} f (x) \in L^{1}$ for all $k \leq n$ , then $f^\in C^{n}$ . So the two theorems are mirror images of each other: $smoothness of f ⟷ decay of f^, decay of f ⟷ smoothness of f^.$

Lecture 18

Fact (Lecture 17):

If $f \in C^{n}$ with $f, f^{'}, \dots, f^{(n)} \in L^{1}$ , then $∣ f^(ζ) ∣ \leq \frac{C}{∣ ζ ∣ ^{n}}$ for $ζ \neq = 0$ .

Converse:

If $∣ f^(ζ) ∣ \leq \frac{C}{∣ ζ ∣ ^{n}}$ for all $ζ \neq = 0$ , is $f \in C^{n}$ ?

Answer:

No. Take the rectangle function ( $n = 1$ ). We have $∣ R^(ζ) ∣ = \frac{sin ( π ζ )}{π ζ} \leq \frac{1/ π}{∣ ζ ∣}$ However, $R \in / C^{1}$ . It’s not even continuous!

Let’s try the triangle function then, which is one order higher ( $n = 2$ ): $∣ T^(ζ) ∣ = \frac{sin ^{2} ( π ζ )}{π ^{2} ζ ^{2}} \leq \frac{1/ π ^{2}}{∣ ζ ∣ ^{2}}$ But $T \in / C^{2}$ ; in fact $T \in / C^{1}$ , because of its corners.

Theorem:

Let $f \in L^{1}$ be continuous. If $∣ f^(ζ) ∣ \leq \frac{C}{∣ ζ ∣ ^{n + 1 + δ}}$ for some $δ > 0$ and all $ζ \neq = 0$ , then $f \in C^{n}$ .

Proof:

Take $n = 1$ first, so $∣ f^(ζ) ∣ \leq \frac{C}{∣ ζ ∣ ^{2 + δ}}$ . In particular $f^\in L^{1}$ ( $f^$ is bounded by $\int ∣ f ∣$ near 0 and decays like $∣ ζ ∣^{- 2 - δ}$ at infinity), so by inversion $f (x) = \int_{- \infty}^{\infty} e^{2 π i x ζ} f^(ζ) d ζ$ Differentiating under the integral sign, $f^{'} (x) = 2 π i \int_{- \infty}^{\infty} e^{2 π i x ζ} ζ f^(ζ) d ζ$ This is legitimate because the $x$ -derivative of the integrand is bounded by $2 π ∣ ζ f^(ζ) ∣$ , independent of $x$ , and $∣ ζ f^(ζ) ∣ \leq \frac{C}{∣ ζ ∣ ^{1 + δ}}$ (and is bounded near 0), so $ζ f^(ζ) \in L^{1}$ . The same domination shows that this integral is continuous in $x$ . Therefore, $f \in C^{1}$ .

For general $n$ , repeat: $∣ ζ ∣^{k} ∣ f^(ζ) ∣ \leq \frac{C}{∣ ζ ∣ ^{1 + δ}}$ for $k \leq n$ , so we may differentiate $n$ times, getting $f^{(k)} (x) = \int (2 π i ζ)^{k} e^{2 π i x ζ} f^(ζ) d ζ$ , continuous in $x$ .

(Compare Lecture 9. Applied to $T$ : $∣ T^∣ \leq \frac{C}{∣ ζ ∣ ^{0 + 1 + 1}}$ , so $T \in C^{0}$ , and that is all one can conclude.)

Definition:

The convolution of $f$ and $g$ , denoted by $f * g$ , is defined by $f * g (x) = \int_{- \infty}^{\infty} f (x - y) g (y) d y$ (Substituting $y \to x - y$ shows $f * g = g * f$ .)

Theorem: (Convolution Theorem)

If $f, g \in L^{1}$ , then $f * g (ζ) = f^(ζ) g^(ζ)$

Proof:

$f * g (ζ) = \int_{- \infty}^{\infty} e^{- 2 π i x ζ} \int_{- \infty}^{\infty} f (x - y) g (y) d y d x$ This double integral converges absolutely, since $\iint ∣ f (x - y) ∣∣ g (y) ∣ d x d y = \int ∣ f ∣ \cdot \int ∣ g ∣ < \infty$ . So we may swap the order of integration: $= \int_{- \infty}^{\infty} g (y) (\int_{- \infty}^{\infty} f (x - y) e^{- 2 π i x ζ} d x) d y$ In the inner integral, substitute $s = x - y$ (this is the Shift Theorem): $\int_{- \infty}^{\infty} f (s) e^{- 2 π i (s + y) ζ} d s = e^{- 2 π i y ζ} f^(ζ)$ Hence $f * g (ζ) = f^(ζ) \int_{- \infty}^{\infty} g (y) e^{- 2 π i y ζ} d y = f^(ζ) g^(ζ) .$

The same computation shows $f * g \in L^{1}$ , with $\int ∣ f * g ∣ \leq \int ∣ f ∣ \cdot \int ∣ g ∣$ .

Example:

$R * R = T$ . Indeed, $(R * R) (x) = \int R (x - y) R (y) d y$ is the length of the overlap of $(- \frac{1}{2}, \frac{1}{2})$ and $(x - \frac{1}{2}, x + \frac{1}{2})$ , which is $1 - ∣ x ∣$ for $∣ x ∣ \leq 1$ and 0 otherwise. So $T^= R^^{2} = \frac{s i n ^{2} ( π ζ )}{π ^{2} ζ ^{2}}$ , recovering the computation from Lecture 15.

Example:

Evaluate $\int_{- \infty}^{\infty} \frac{sin ( π ζ )}{ζ} \frac{sin ( ζ )}{ζ} e^{π i ζ} d ζ$

Solution:

Recall $R^(ζ) = \frac{s i n ( π ζ )}{π ζ}$ , so $\frac{s i n ( π ζ )}{ζ} = π R^(ζ)$ . For the second factor, use the Stretch Theorem with $R_{π} (x) = R (π x)$ , which is 1 for $∣ x ∣ < \frac{1}{2 π}$ and 0 for $∣ x ∣ > \frac{1}{2 π}$ : $R^_{π} (ζ) = \frac{1}{π} R^(\frac{ζ}{π}) = \frac{1}{π} \cdot \frac{sin ζ}{ζ} \Rightarrow \frac{sin ζ}{ζ} = π R^_{π} (ζ)$ The factor $e^{π i ζ} = e^{2 π i \cdot \frac{1}{2} \cdot ζ}$ is a shift: by the Shift Theorem, $g (x) := π R_{π} (x + \frac{1}{2})$ has $g^(ζ) = π e^{π i ζ} R^_{π} (ζ) = \frac{sin ζ}{ζ} e^{π i ζ} .$ So, by the Convolution Theorem, the integrand is $π R^(ζ) g^(ζ) = π R * g (ζ)$ Both $π R * g$ and its transform are in $L^{1}$ (the transform decays like $\frac{1}{ζ ^{2}}$ ), and $π R * g$ is continuous, so by the inversion theorem at $x = 0$ we only need $\int_{- \infty}^{\infty} π R * g (ζ) d ζ = (π R * g) (0) = π^{2} \int_{- \infty}^{\infty} R (0 - y) R_{π} (y + \frac{1}{2}) d y$ Now $R (- y) = 1$ exactly for $- \frac{1}{2} < y < \frac{1}{2}$ , and $R_{π} (y + \frac{1}{2}) = 1$ exactly for $- \frac{1}{2} - \frac{1}{2 π} < y < - \frac{1}{2} + \frac{1}{2 π}$ . These intervals overlap in $(- \frac{1}{2}, - \frac{1}{2} + \frac{1}{2 π})$ , of length $\frac{1}{2 π}$ . Therefore $\int_{- \infty}^{\infty} \frac{sin ( π ζ )}{ζ} \frac{sin ( ζ )}{ζ} e^{π i ζ} d ζ = π^{2} \cdot \frac{1}{2 π} = \frac{π}{2}$ (Sanity check: the imaginary part is the integral of an odd function, so the answer had to be real.)

Applications: ODE and PDE.

Example:

Solve $u^{''} (x) - u (x) = f (x)$ on $R$ (looking for a solution with $u, u^{'}, u^{''} \in L^{1}$ ).

Take the Fourier transform, using $u^{''} (ζ) = (2 π i ζ)^{2} u^(ζ)$ (Lecture 17): $u^{''} (ζ) - u^(ζ) = f^(ζ) \Rightarrow (2 π i)^{2} ζ^{2} u^(ζ) - u^(ζ) = f^(ζ)$ $\Rightarrow u^(ζ) = - \frac{f ^ ( ζ )}{1 + 4 π ^{2} ζ ^{2}} = - f^(ζ) \frac{1}{1 + 4 π ^{2} ζ ^{2}} = - \frac{1}{2} f^(ζ) e^{- ∣ x ∣} (ζ)$ since $e^{- ∣ x ∣} (ζ) = \frac{2}{1 + 4 π ^{2} ζ ^{2}}$ (Lecture 13 with $a = 1$ ). With $g (x) = e^{- ∣ x ∣}$ , the Convolution Theorem gives $u^= - \frac{1}{2} g * f$ , so $\Rightarrow u (x) = - \frac{1}{2} (g * f) (x) = - \frac{1}{2} \int_{- \infty}^{\infty} g (x - y) f (y) d y = - \frac{1}{2} \int_{- \infty}^{\infty} e^{- ∣ x - y ∣} f (y) d y$ The function $G (x) = - \frac{1}{2} e^{- ∣ x ∣}$ , for which $u = G * f$ , is called the Green’s function of the ODE.

Lecture 19

Example (heat equation on the line):

Solve $u_{t} (x, t) = u_{xx} (x, t), x \in R, t > 0, u (x, 0) = f (x) .$

Solution:

Fix $t$ and take the Fourier transform in $x$ . By Lecture 17, $u_{xx} (ζ, t) = (2 π i ζ)^{2} u^(ζ, t)$ , so for each fixed $ζ$ , $\frac{d}{d t} u^(ζ, t) = (2 π i ζ)^{2} u^(ζ, t) = - 4 π^{2} ζ^{2} u^(ζ, t), u^(ζ, 0) = f^(ζ)$ This is a first-order linear ODE in $t$ , so $\Rightarrow u^(ζ, t) = e^{- 4 π^{2} ζ^{2} t} f^(ζ) \forall t > 0$ To invert, we want a function whose transform is $e^{- 4 π^{2} t ζ^{2}}$ . Recall that for $δ > 0$ , if $h_{δ} (x) = e^{- δ x^{2}}$ , then $h^_{δ} (ζ) = \frac{π}{δ} e^{- \frac{π ^{2}}{δ} ζ^{2}}$ . Taking $δ = \frac{1}{4 t}$ , $h_{\frac{1}{4 t}} (ζ) = e^{- \frac{x ^{2}}{4 t}} (ζ) = 4 π t e^{- 4 π^{2} t ζ^{2}}$ Thus, if $k_{t} (x) = \frac{1}{4 π t} e^{- \frac{x ^{2}}{4 t}}$ , then $k^_{t} (ζ) = e^{- 4 π^{2} t ζ^{2}}$ , and $u^(ζ, t) = k^_{t} (ζ) f^(ζ) = k_{t} * f (ζ) \Rightarrow u (x, t) = (k_{t} * f) (x)$ Hence $u (x, t) = \int_{- \infty}^{\infty} k_{t} (x - y) f (y) d y = \frac{1}{4 π t} \int_{- \infty}^{\infty} e^{- \frac{( x - y ) ^{2}}{4 t}} f (y) d y$

Definition:

$k_{t} (x) = \frac{1}{4 π t} e^{- \frac{x ^{2}}{4 t}} is the heat kernel .$ Note that $\int k_{t} (x) d x = k^_{t} (0) = 1$ for every $t > 0$ .

Note:

The formula doesn’t make sense at $t = 0$ ! So we should check that $u (x, t) \to f (x)$ as $t ↓ 0$ .

For $t > 0$ , $u (x, t) = \frac{1}{4 π t} \int_{- \infty}^{\infty} e^{- \frac{( x - y ) ^{2}}{4 t}} f (y) d y$ Let $s = \frac{x - y}{2 t}$ , so $y = x - 2 t s$ and $d y = - 2 t d s$ (the minus sign is absorbed by flipping the limits back to $- \infty \to \infty$ ): $u (x, t) = \frac{1}{4 π t} \int_{- \infty}^{\infty} e^{- s^{2}} f (x - 2 t s) 2 t d s = \frac{1}{π} \int_{- \infty}^{\infty} e^{- s^{2}} f (x - 2 t s) d s$ As $t ↓ 0$ , $f (x - 2 t s) \to f (x)$ if $f$ is continuous at $x$ . So, for $f$ bounded and continuous, $t ↓ 0 lim u (x, t) = f (x) \cdot \frac{1}{π} \int_{- \infty}^{\infty} e^{- s^{2}} d s = f (x) \cdot \frac{π}{π} = f (x) .$ Moving the limit inside the integral is justified by the Theorem at the end of this lecture, since $k_{t} = g_{2 t}$ for $g (s) = \frac{1}{π} e^{- s^{2}}$ .

Example:

Solve $u_{xx} (x, y) + u_{y y} (x, y) = 0$ for $x \in R$ , $y > 0$ , with $u (x, 0) = f (x)$ and $u$ bounded.

Take the Fourier transform in $x$ , to obtain $\frac{d ^{2}}{d y ^{2}} u^(ζ, y) - 4 π^{2} ζ^{2} u^(ζ, y) = 0, u^(ζ, 0) = f^(ζ)$ For each fixed $ζ \neq = 0$ this is a constant-coefficient ODE in $y$ , with general solution $u^(ζ, y) = c_{1} (ζ) e^{- 2 π ∣ ζ ∣ y} + c_{2} (ζ) e^{2 π ∣ ζ ∣ y}$ The second term blows up as $y \to \infty$ ; we discard it to keep the solution bounded, so $c_{2} = 0$ . Then $f^(ζ) = u^(ζ, 0) = c_{1} (ζ)$ $\Rightarrow u^(ζ, y) = e^{- 2 π ∣ ζ ∣ y} f^(ζ)$ To invert, use Lecture 13 with $a = 2 π y$ : $e^{- 2 π y ∣ x ∣} (ζ) = \frac{4 π y}{4 π ^{2} y ^{2} + 4 π ^{2} ζ ^{2}} = \frac{1}{π} \cdot \frac{y}{y ^{2} + ζ ^{2}}$ So for $y > 0$ let $P_{y} (x) = \frac{1}{π} \cdot \frac{y}{y ^{2} + x ^{2}} .$ Then $P_{y} = F (e^{- 2 π y ∣ \cdot ∣})$ , and since $e^{- 2 π y ∣ x ∣}$ is even, inversion gives $P^_{y} (ζ) = e^{- 2 π y ∣ ζ ∣}$ Thus $u^(ζ, y) = P^_{y} (ζ) f^(ζ) = P_{y} * f (ζ)$ , and $u (x, y) = P_{y} * f (x)$ : $u (x, y) = \int_{- \infty}^{\infty} P_{y} (x - t) f (t) d t = \frac{1}{π} \int_{- \infty}^{\infty} \frac{y}{y ^{2} + ( x - t ) ^{2}} f (t) d t, y > 0$

Definition:

$P_{y} (x) = \frac{1}{π} \cdot \frac{y}{y ^{2} + x ^{2}}$ is called the Poisson kernel for the Laplace operator ( $Δ = \partial_{x}^{2} + \partial_{y}^{2}$ ) in the upper half plane.

Note:

Once again we should check that the solution tends to $f (x)$ as $y ↓ 0$ . This follows from the next theorem, since $P_{y} = g_{y}$ for $g (x) = \frac{1}{π} \cdot \frac{1}{1 + x ^{2}}$ , which has $\int g = \frac{1}{π} \cdot π = 1$ .

Theorem:

Let $g \in L^{1}$ and $\int_{- \infty}^{\infty} g (x) d x = 1$ . For $δ > 0$ , let $g_{δ} (x) = \frac{1}{δ} g (\frac{x}{δ})$ . If $f$ is bounded and continuous, then for every $x$ $δ ↓ 0 lim (g_{δ} * f) (x) = f (x)$

Proof:

Substituting $y = δ t$ , $(g_{δ} * f) (x) = \int_{- \infty}^{\infty} g_{δ} (y) f (x - y) d y = \frac{1}{δ} \int_{- \infty}^{\infty} g (\frac{y}{δ}) f (x - y) d y = \int_{- \infty}^{\infty} g (t) f (x - δ t) d t$ Since $\int g = 1$ , we can write $f (x) = \int g (t) f (x) d t$ , so $(g_{δ} * f) (x) - f (x) = \int_{- \infty}^{\infty} g (t) [f (x - δ t) - f (x)] d t$ Let $∣ f ∣ \leq M$ and fix $ε > 0$ . By the Remark in Lecture 12, choose $A$ with $\int_{∣ t ∣ > A} ∣ g (t) ∣ d t < ε$ . Splitting the integral at $∣ t ∣ = A$ , $∣ (g_{δ} * f) (x) - f (x) ∣ \leq ∣ s ∣ \leq δ A sup ∣ f (x - s) - f (x) ∣ \int_{- \infty}^{\infty} ∣ g (t) ∣ d t + 2 M ε$ As $δ ↓ 0$ the first term tends to 0 by continuity of $f$ at $x$ . So $lim sup_{δ ↓ 0} ∣ (g_{δ} * f) (x) - f (x) ∣ \leq 2 M ε$ for every $ε > 0$ , which proves the claim.

Remark:

$\int g_{δ} = \int g = 1$ for every $δ$ , while $g_{δ}$ concentrates near 0 as $δ ↓ 0$ (an “approximate identity”). Both kernels from this lecture are of this form: $k_{t} = g_{2 t} with g (s) = \frac{1}{π} e^{- s^{2}}, P_{y} = g_{y} with g (x) = \frac{1}{π} \cdot \frac{1}{1 + x ^{2}},$ so the theorem shows $u (x, t) \to f (x)$ as $t ↓ 0$ for the heat equation, and $u (x, y) \to f (x)$ as $y ↓ 0$ for the Laplace equation.

Lecture 20

Applications:

Recall the Poisson kernel, $P_{y} (x) = \frac{1}{π} (\frac{y}{y ^{2} + x ^{2}}), y > 0$

Let $g (x) = \frac{1}{π} (\frac{1}{1 + x ^{2}})$ . $g \in L^{1}$ since $∣ g (x) \leq \frac{c}{x ^{2}}$ for $∣ x ∣ \geq 1$ . Also, $\int_{- \infty}^{\infty} g (x) d x = 1$ . For $g > 0$ , $P_{y} (x) = \frac{1}{π} (\frac{y}{y ^{2} ( 1 + ( \frac{x}{y} ) ^{2} )})$ $= \frac{1}{y} \frac{1}{π} (\frac{1}{1 + ( \frac{x}{y} ) ^{2}}) = \frac{1}{y} g (\frac{x}{y}) = g_{y} (x)$

By the Theorem, $g_{y} * f (x) = P_{y} * f (x) \to f (x)$ as $y ↓ 0$ .

Applications: $k_{t} (x) = \frac{1}{2 π t} e^{\frac{- x ^{2}}{2 t}}$

Let $g (x) = \frac{1}{2 π} e^{- \frac{x ^{2}}{2}}$ . $\int_{- \infty}^{\infty} g (x) d x = 1$ . For $t > 0$ , $g_{t} (x) = \frac{1}{t} g (\frac{x}{t}) = \frac{1}{t} \frac{1}{2 π} e^{- \frac{x ^{2}}{2 t ^{2}}}$ $\Rightarrow g_{t} (x) = \frac{1}{2 π t} e^{- \frac{x ^{2}}{t}} = k_{t} (x)$ $\Rightarrow k_{t} * f (x) = g_{t} * f (x) \to f (x) as t \to 0^{+} by the theorem.$

Theorem:

Let $f \in C^{k} (R), g \in C^{m} (R)$ . Assume that $D^{j} f$ for $0 \leq j \leq k$ and $D^{ℓ} y$ for $0 \leq ℓ \leq m$ are in $L^{1}$ . Then $f * g \in C^{m + k} (R)$

Proof:

$g (x) = \int_{- \infty}^{\infty} f (x - y) g (y) d y$ $D (f * g) (x) = \int_{- \infty}^{\infty} D f (x - y) g (y) d y$ $D^{k} (f * g) (x) = \int_{- \infty}^{\infty} D^{k} f (x - y) g (y) d y = \int_{- \infty}^{\infty} D^{k} f (y) g (x - y) d y$

Since $g$ is $m$ times differentiable, the $k$ -th derivative of the convolution is $m$ times differentiable, therefore the convolution must be $m + k$ times differentiable.

Theorem: (Poisson Summation Formula)

Suppose $∣ f (x) ∣ \leq \frac{C}{∣ x ∣ ^{2}}$ , and $∣ f^(ζ) ∣ \leq \frac{C}{∣ ζ ∣ ^{2}}$ . Then $n = - \infty \sum \infty f (x + n) = n = - \infty \sum \infty f^(n) e^{2 π in x}$

In particular, when $x = 0$ we get $n = - \infty \sum \infty f (n) = n = - \infty \sum \infty f^(n)$

Proof:

Define $F (x) = \sum_{n = - \infty}^{\infty} f (x + n)$ . $F$ is periodic of period 1. $\Rightarrow F (x) = n = - \infty \sum \infty c_{n} e^{2 π in x}, c_{n} = \int_{0}^{1} e^{- 2 π in t} F (t) d t$ $\Rightarrow c_{n} = \int_{0}^{1} e^{- 2 π in t} (k = - \infty \sum \infty f (t + k)) d t = k = - \infty \sum \infty \int_{0}^{1} e^{- 2 π in t} f (t + k) d t$

Let $y = t + k$ . Then $= k = - \infty \sum \infty \int_{k}^{k + 1} e^{2 π ink} e^{- 2 π in y} f (y) d y$ $\Rightarrow k = - \infty \sum \infty \int_{k}^{k + 1} e^{- 2 π in y} f (y) d y = \int_{- \infty}^{\infty} e^{- 2 π in y} f (y) d y = f^(n)$ $\Rightarrow n = - \infty \sum \infty f (x + n) = F (x) = n = - \infty \sum \infty c_{n} e^{2 π in x} = n = - \infty \sum \infty f^(n) e^{2 π in x}$

Applications:

Let $f (x) = \frac{s i n ^{2} ( π x )}{π ^{2} x ^{2}}$ . We know that $f (x) = F T (x)$ , where $T$ denotes the triangle function. In addition, $F (F T) (x) = F f (x)$ $\Rightarrow T (- x) = F f (x) = f^(x) \Rightarrow f^(x) = T (x)$

By the Poisson Summation Formula, we can obtain a quite beautiful result: $\frac{1}{π ^{2}} - \infty \sum \infty \frac{sin ^{2} ( π ( x + n ))}{( x + n ) ^{2}} = n = - \infty \sum \infty T (x + n) e^{2 π in x}$ $\Rightarrow \frac{1}{π ^{2}} n = - \infty \sum \infty \frac{sin ^{2} ( π x )}{( x + π ) ^{2}} = 1$ $\Rightarrow if x \neq \in Z, n = - \infty \sum \infty \frac{1}{( x + n ) ^{2}} = \frac{π ^{2}}{sin ^{2} ( π x )}$

Example:

For $a > 0$ , let $f_{a} (x) = {e^{- a x}, 0, x > 0 x \leq 0$ . Recall that $f^_{a} (ζ) =$ (finish this later, Poisson summation formula is used to get $h_{a} (n)$ sum over the integers with $h_{a} (x) = x f_{a} (x)$ and find formula for sum of $\frac{1}{( a + 2 π in ) ^{2}}$ )

Lecture 21

Example:

For $a > 0$ , let $f_{a} (x) = {x e^{- a x}, 0, x > 0 x \leq 0$ . Then $f^_{a} (ζ) = \frac{1}{- 2 π i} \frac{d}{d ζ} (\frac{1}{a + 2 π i ζ}) = \frac{1}{( a + 2 π i ζ ) ^{2}}$ $\Rightarrow n \in Z \sum f_{a} (n) = n \in Z \sum \frac{1}{( a + 2 π in ) ^{2}}$ $\Rightarrow n = 1 \sum \infty n e^{- an} = n = - \infty \sum \infty \frac{1}{( a + 2 π in ) ^{2}}$

We claim that LHS is easy to evaluate. We can compute the sum of $\sum_{n = 1}^{\infty} e^{- an} = \frac{1}{e ^{a} - 1}$ easily as it is geometric. Then, we simply differentiate with respect to $a$ to get a formula: $n = 1 \sum \infty n e^{- an} = \frac{e ^{a}}{( e ^{a} - 1 ) ^{2}}$

Since this is real, then the RHS sum must have only real part as well in the summand. $n = - \infty \sum \infty \frac{( a - 2 π in ) ^{2}}{a ^{2} + 4 π ^{2} n ^{2}} = n = - \infty \sum \infty (\frac{a ^{2} - n ^{2}}{a ^{2} + 4 π ^{2} n ^{2}}) - 4 π i n = - \infty \sum \infty \frac{n}{a ^{2} + n ^{2}} = \frac{e ^{a}}{( e ^{a} - 1 ) ^{2}}$

So another interesting result is that the sum on the factor $4 π i$ is equal to zero.

Fourier sine and cosine transforms:

Let $f : [0, \infty) \to C$ be an $L^{1}$ function, so $\int_{0}^{\infty} ∣ f (x) ∣ d x < \infty$ . Let $f^{e} (x), f^{o} (x) : R \to C$ be defined by $f^{e} (x) = {f (x), f (- x), x \geq 0 x < 0$ $f^{o} (x) = {f (x), - f (- x), x > 0 x < 0$

Then take the Fourier transforms: $f^^{e} (ζ) = \int_{- \infty}^{\infty} e^{- 2 π i x ζ} f^{e} (x) d x = \int_{- \infty}^{\infty} cos (2 π x ζ) f^{e} (x) d x - i \int_{- \infty}^{\infty} sin (2 π x ζ) f^{e} (x) d x$ $= 2 \int_{0}^{\infty} cos (2 π x ζ) f^{3} (x) d x = 2 \int_{0}^{\infty} cos (2 π x ζ) f (x) d x$ $f^^{o} (ζ) = \int_{- \infty}^{\infty} cos (2 π x ζ) f^{o} (x) d x - i \int_{- \infty}^{\infty} sin (2 π x ζ) f^{o} (x) d x$ $= 2 i \int_{0}^{\infty} sin (2 π x ζ) f (x) d x$

Definition:

The Fourier cosine transform of $f$ is $f^_{c} (ζ) = \int_{0}^{\infty} cos (2 π x ζ) f (x) d x$

and the Fourier sine transform of $f$ is $f^_{s} (ζ) = \int_{0}^{\infty} sin (2 π x ζ) f (x) d x$

Inversion:

We know that the Fourier transform of an even function is even, and the Fourier transform of an odd function is odd. So $f^{e} (x) = \int_{- \infty}^{\infty} e^{2 π i x ζ} f^^{e} (ζ) d ζ = 2 \int_{0}^{\infty} cos (2 π x ζ) f^^{e} (ζ) d ζ = 4 \int_{0}^{\infty} cos (2 π x ζ) f^_{c} (ζ) d ζ$

Thus, when $x > 0$ , $f (x) = f^{e} (x) = 4 \int_{0}^{\infty} cos (2 π x ζ) f^_{c} (ζ) d ζ$

Likewise, the same holds for sine transform. $f (x) = 4 \int_{0}^{\infty} sin (2 π x ζ) f^_{s} (ζ) d ζ$

Example:

Let $f (x) = e^{- x}$ for $x > 0$ . Compute $f^_{s} (ζ)$ . $f^_{s} (ζ) = \int_{0}^{\infty} e^{- x} sin (2 π x ζ) d x = \frac{2 π ζ}{1 + 4 π ^{2} ζ ^{2}}$

By the inversion formula, for $x > 0$ , $e^{- x} = 4 \int_{0}^{\infty} \frac{2 π ζ}{1 + 4 π ^{2} ζ ^{2}} sin (2 π x ζ) d ζ$

Let $y = 2 π ζ \Rightarrow \frac{d y}{2 π} = d ζ$ $\Rightarrow e^{- x} = \frac{2}{π} \int_{0}^{\infty} \frac{y sin ( x y )}{1 + y ^{2}} d y$

Theorem:

Suppose $f, f^{'} \in L^{1} ([0, \infty))$ and $lim_{x \to \infty} f (x) = 0$ . Then $(D f)_{s} (ζ) = - 2 π ζ f^_{c} (ζ), (D f)_{c} (ζ) = - f (0) + 2 π ζ f^_{s} (ζ)$

Proof:

(please finish this)

Lecture 22

Application:

Solve the heat equation: $⎩ ⎨ ⎧ u_{t} (x, t) = k u_{xx} (x, t), u (x, 0) = f (0), u (0, t) = g (t) x > 0, t > 0 x > 0$

Take the sine transform of the PDE. $⎩ ⎨ ⎧ \frac{d}{d t} u^_{s} (ζ t) = 2 k π ζ u (0, t) - 4 π^{2} k ζ^{2} u^_{s} (ζ, t) u^_{s} (ζ, 0) = f^_{s} (ζ)$

Let $y (t) = u^_{s} (ζ, t)$ , then simplify: $\frac{d}{d t} (e^{4 π^{2} k ζ^{2} t} y (t)) = e^{4 π^{2} k ζ^{2} t} 2 k π ζ g (t)$ $\Rightarrow y (t) = 2 k π ζ \int_{0}^{t} e^{4 π^{2} k ζ^{2} (r - t)} g (r) d r + y (0) e^{- 4 π^{2} k ζ^{2} t}$ $\Rightarrow u^_{s} (ζ, t) = 2 k π ζ \int_{0}^{t} e^{4 π^{2} k ζ^{2} (r - t)} g (r) d r + f^_{s} (ζ) e^{- 4 π^{2} k ζ^{2} t}$

By the inversion formula, $u (x, t) = 4 \int_{0}^{\infty} sin (2 π x ζ) u^_{s} (ζ, t) d ζ$

Can also try using the Cosine transform. $⎩ ⎨ ⎧ u_{t} (x, t) = k u_{xx} (x, t), u (x, 0) = f (x), u_{x} (0, t) = g (t) x > 0, t > 0 x > 0$

Then $⎩ ⎨ ⎧ \frac{d}{d t} u^_{c} (ζ, t) = k (- 4 π^{2} ζ^{2} u^_{c} (ζ, t) - u_{x} (0, t)) u^_{s} (ζ, 0) = f^_{s} (ζ)$

The Laplace Transform

Definition:

Given $f : [0, \infty) \to C$ , define the Laplace Transform of $f$ , denoted $L f (s)$ or $F (s)$ by $L f (s) = F (s) = \int_{0}^{\infty} e^{- s t} f (t) d t .$

Suppose $∣ f (t) ∣ \leq A e^{B t}$ . Then $L f$ makes sense. $L f (s) = \int_{0}^{\infty} e^{- s t} f (t) d t \leq A \int_{0}^{\infty} e^{- s t} e^{B t} d t = A \int_{0}^{\infty} e^{(- s + B) t} d t < \infty if s > - B$

Properties:

Linearity: $L (a f + b g) = a L f + b L g$

Scaling: Let $a > 0$ , then $L (f (a t)) s = \frac{1}{a} L f (\frac{1}{a})$

Shifting: $L (e^{a t} f (t)) (s) = L f (s - a)$

Derivative: $L f^{'} (s) = s L f (s) - f (0)$

j / k	move cursor over links (counts work: 5j)
gg / G	first / last link
Enter / l	open link under cursor
h	back
/	filter entries
Esc	clear filter and cursor, close
j / k	move cursor over links
Enter / l	open link under cursor
d / u	scroll half page
gg / G	first / last link
] / [	next / previous heading
h	back
t	toggle dark mode
?	this help