Parseval's theorem: Exercises — Maths deep dive

Level 1 — Recognition

Goal: read a series and pick out the coefficients correctly.

Exercise 1.1

A function on $(- π, π)$ has Fourier series $f (x) = 2 + 3 cos x - cos 2 x + 4 sin x$ . Write down $a_{0}, a_{1}, a_{2}, b_{1}$ , and all others.

Recall Solution

The general series is $f (x) = \frac{a _{0}}{2} + \sum (a_{n} cos n x + b_{n} sin n x)$ (here $L = π$ ). WHAT the constant means: the standalone number is $\frac{a _{0}}{2}$ , NOT $a_{0}$ . So $\frac{a _{0}}{2} = 2 \Rightarrow a_{0} = 4.$ Match term by term:

$cos x$ has coefficient $3 \Rightarrow a_{1} = 3$ .
$cos 2 x$ has coefficient $- 1 \Rightarrow a_{2} = - 1$ .
$sin x$ has coefficient $4 \Rightarrow b_{1} = 4$ .
Every other $a_{n} = b_{n} = 0$ .

WHY the factor-of-2 trap matters: the constant term is written with a built-in $\frac{1}{2}$ so that the coefficient formulas are uniform. Reading $a_{0} = 2$ here would poison every Parseval calculation later.

Exercise 1.2

For the same $f$ , which single quantity does Parseval's left-hand side compute? Name it in words.

Recall Solution

The left side is $\frac{1}{L} \int_{- L}^{L} [f (x)]^{2} d x = \frac{1}{π} \int_{- π}^{π} f^{2} d x$ . In words: twice the mean-square value of $f$ over one period. (Dividing further by 2 gives the mean square; its square root is the RMS value.) It measures the total "energy per unit length" of the signal, before splitting into frequencies.

Level 2 — Application

Goal: plug numbers into Parseval and get a value out.

Exercise 2.1

For $f (x) = 2 + 3 cos x - cos 2 x + 4 sin x$ on $(- π, π)$ , compute $\frac{1}{π} \int_{- π}^{π} f^{2} d x$ using Parseval.

Recall Solution

Coefficients (from Ex 1.1): $a_{0} = 4, a_{1} = 3, a_{2} = - 1, b_{1} = 4$ . $\frac{1}{π} \int_{- π}^{π} f^{2} d x = \frac{a _{0}^{2}}{2} + (a_{1}^{2} + b_{1}^{2}) + (a_{2}^{2} + b_{2}^{2})$ $= \frac{16}{2} + (9 + 16) + (1 + 0) = 8 + 25 + 1 = 34.$ So $\int_{- π}^{π} f^{2} d x = 34 π .$ WHY this is easier than integrating: squaring $f$ directly gives 16 cross terms; orthogonality guarantees all cross terms integrate to zero, so we never touch them.

Exercise 2.2

A signal on $(- π, π)$ has only $a_{0} = 0, a_{3} = 6, b_{5} = 8$ . Find its RMS value.

Recall Solution

Parseval: $\frac{1}{π} \int_{- π}^{π} f^{2} d t = a_{3}^{2} + b_{5}^{2} = 36 + 64 = 100.$ So $\int_{- π}^{π} f^{2} d t = 100 π$ . Mean square = $\frac{1}{2 π} \int f^{2} = \frac{100 π}{2 π} = 50$ . RMS = $50 = 52 \approx 7.07$ . WHY divide by $2 π$ for the mean: the full period length is $2 π$ ; averaging squares over the whole period gives the mean square, whose square root is the root-mean-square.

Exercise 2.3

For $f (x) = x$ on $(- π, π)$ with $b_{n} = \frac{2 ( - 1 ) ^{n + 1}}{n}$ , verify Parseval reproduces $n = 1 \sum \infty \frac{1}{n ^{2}} = \frac{π ^{2}}{6}$ .

Recall Solution

Left: $\frac{1}{π} \int_{- π}^{π} x^{2} d x = \frac{1}{π} \cdot \frac{2 π ^{3}}{3} = \frac{2 π ^{2}}{3} .$ Right: $\sum b_{n}^{2} = \sum \frac{4}{n ^{2}} = 4 \sum \frac{1}{n ^{2}}$ (the sign $(- 1)^{n + 1}$ squares to $1$ ). Equate: $\frac{2 π ^{2}}{3} = 4 \sum \frac{1}{n ^{2}} \Rightarrow \sum \frac{1}{n ^{2}} = \frac{π ^{2}}{6} .$ This is the Basel Problem.

Level 3 — Analysis

Goal: reason about which terms survive and why.

Exercise 3.1

Without computing any integral, explain why $\int_{- π}^{π} (3 cos x) (4 sin x) d x = 0$ , and why this fact is what makes Parseval possible.

Recall Solution

$cos x$ and $sin x$ are orthogonal over a full period: $\int_{- π}^{π} cos \frac{mπ x}{L} sin \frac{nπ x}{L} d x = 0 (always) .$ WHAT this means geometrically: think of each basis function as a perpendicular axis in an infinite-dimensional space (see figure). A product of two different axes integrates to zero, exactly like a dot product of perpendicular vectors: $\overset{x}{^} \cdot \overset{y}{^} = 0$ . WHY Parseval needs it: expanding $(f)^{2}$ produces every pairwise product; orthogonality zeroes all the cross products, leaving only the diagonal squared terms. Parseval is literally Pythagoras in this space.

Exercise 3.2

When we integrate $\int_{- π}^{π} cos^{2} n x d x$ we get $π$ , but $\int_{- π}^{π} 1 d x = 2 π$ . Explain how these two different values produce the asymmetry $\frac{a _{0}^{2}}{2}$ vs $a_{n}^{2}$ .

Recall Solution

The DC piece of the series is $\frac{a _{0}}{2}$ . Its squared integral: $\int_{- π}^{π} (\frac{a _{0}}{2})^{2} d x = \frac{a _{0}^{2}}{4} \cdot 2 π = \frac{a _{0}^{2} π}{2} .$ A cosine piece: $\int_{- π}^{π} (a_{n} cos n x)^{2} d x = a_{n}^{2} \cdot π .$ Dividing everything by $L = π$ : $DC \to \frac{a _{0}^{2}}{2}, harmonic \to a_{n}^{2} .$ WHY the DC term is different: the constant "1" has period-length norm $2 π$ (double), but the $\frac{1}{2}$ in $\frac{a _{0}}{2}$ squares to $\frac{1}{4}$ (quarter). $2 π \times \frac{1}{4} = \frac{π}{2}$ per unit $a_{0}^{2}$ , versus $π \times 1$ for harmonics — leaving the lonely factor $\frac{1}{2}$ .

Exercise 3.3

A student claims: "If two functions have the same Fourier coefficients magnitudes but different signs, they have different energy." True or false — justify.

Recall Solution

False. Parseval depends only on $a_{n}^{2}$ and $b_{n}^{2}$ . Since $(+ c)^{2} = (- c)^{2}$ , flipping any sign leaves every squared term unchanged, so the total energy $\frac{1}{L} \int f^{2}$ is identical. WHAT changes and what doesn't: the shape (waveform) changes because relative phases shift, but the energy is phase-blind. This is exactly why $b_{n} = \frac{2 ( - 1 ) ^{n + 1}}{n}$ in Ex 2.3 gave $\sum 4/ n^{2}$ with the alternating sign gone.

Level 4 — Synthesis

Goal: combine coefficient-finding with Parseval to derive a new result.

Exercise 4.1

Use $f (x) = x^{2}$ on $(- π, π)$ (with $a_{0} = \frac{2 π ^{2}}{3}$ , $a_{n} = \frac{4 ( - 1 ) ^{n}}{n ^{2}}$ , $b_{n} = 0$ ) to prove $n = 1 \sum \infty \frac{1}{n ^{4}} = \frac{π ^{4}}{90}$ .

Recall Solution

Left: $\frac{1}{π} \int_{- π}^{π} x^{4} d x = \frac{1}{π} \cdot \frac{2 π ^{5}}{5} = \frac{2 π ^{4}}{5} .$ Right: $\frac{a _{0}^{2}}{2} + \sum a_{n}^{2} = \frac{1}{2} (\frac{2 π ^{2}}{3})^{2} + \sum \frac{16}{n ^{4}} = \frac{2 π ^{4}}{9} + 16 \sum \frac{1}{n ^{4}} .$ Equate and isolate: $\frac{2 π ^{4}}{5} - \frac{2 π ^{4}}{9} = 2 π^{4} \cdot \frac{9 - 5}{45} = \frac{8 π ^{4}}{45} = 16 \sum \frac{1}{n ^{4}} .$ $\sum \frac{1}{n ^{4}} = \frac{8 π ^{4}}{45 \cdot 16} = \frac{π ^{4}}{90} . ■$ WHY higher power → higher zeta: each power of $x$ pushes the coefficients' denominators up (here $1/ n^{2}$ ), and squaring in Parseval sends $1/ n^{2} \to 1/ n^{4}$ . This is $ζ (4)$ .

Exercise 4.2

Take $f (x) = x$ on $(- π, π)$ but now ask for the sum $n = 1 \sum \infty \frac{1}{( 2 n - 1 ) ^{2}}$ (odd terms only). Derive its value from $\sum \frac{1}{n ^{2}} = \frac{π ^{2}}{6}$ .

Recall Solution

Split the full sum into odd and even indices: $\sum_{n = 1}^{\infty} \frac{1}{n ^{2}} = S_{odd} odd \sum \frac{1}{n ^{2}} + S_{even} even \sum \frac{1}{n ^{2}} .$ Even terms: $n = 2 k$ , so $S_{even} = k = 1 \sum \infty \frac{1}{( 2 k ) ^{2}} = \frac{1}{4} \sum \frac{1}{k ^{2}} = \frac{1}{4} \cdot \frac{π ^{2}}{6} = \frac{π ^{2}}{24} .$ Therefore $S_{odd} = \frac{π ^{2}}{6} - \frac{π ^{2}}{24} = \frac{4 π ^{2} - π ^{2}}{24} = \frac{3 π ^{2}}{24} = \frac{π ^{2}}{8} .$ WHY this trick works: the even-indexed terms are a scaled copy of the whole sum (factor $\frac{1}{4}$ ), so subtracting recovers the odd part without a new integral.

Exercise 4.3

Design a signal on $(- π, π)$ whose total Parseval energy $\frac{1}{π} \int f^{2}$ equals exactly $10$ , using only $a_{0}$ and $b_{1}$ . Give one valid choice.

Recall Solution

Constraint: $\frac{a _{0}^{2}}{2} + b_{1}^{2} = 10.$ One clean choice: let $a_{0} = 2 \Rightarrow \frac{4}{2} = 2$ , then need $b_{1}^{2} = 8 \Rightarrow b_{1} = 22 .$ Check: $\frac{2 ^{2}}{2} + (22)^{2} = 2 + 8 = 10.$ ✓ So $f (x) = 1 + 22 sin x$ works (recall the constant is $\frac{a _{0}}{2} = 1$ ). WHY infinitely many answers exist: we have two free knobs and one equation — a circle of solutions in $(a_{0}, b_{1})$ space. Any point on it is valid; energy is a level curve.

Level 5 — Mastery

Goal: push into limits, degenerate cases, and cross-theorem links.

Exercise 5.1

Show that for any $f$ with a valid Fourier series, the partial-energy sum $E_{N} = \frac{a _{0}^{2}}{2} + \sum_{n = 1}^{N} (a_{n}^{2} + b_{n}^{2})$ satisfies $E_{N} \leq \frac{1}{L} \int_{- L}^{L} f^{2} d x$ for every finite $N$ , and equals it only as $N \to \infty$ . Name the two theorems involved.

Recall Solution

Let $S_{N}$ be the $N$ -term partial sum of the series and $r_{N} = f - S_{N}$ the remainder. Because the basis is orthogonal, $\int f^{2} = \int S_{N}^{2} + \int r_{N}^{2}$ (the cross term vanishes — $r_{N}$ is orthogonal to every kept mode). Dividing by $L$ : $\frac{1}{L} \int f^{2} = E_{N} + \geq 0 \frac{1}{L} \int r_{N}^{2} .$ Since $\frac{1}{L} \int r_{N}^{2} \geq 0$ always, $E_{N} \leq \frac{1}{L} \int_{- L}^{L} f^{2} d x .$ This inequality is Bessel's Inequality. As $N \to \infty$ the remainder energy $\to 0$ (completeness of the sine–cosine basis), giving equality — that equality is Parseval's theorem itself. WHAT it looks like: each new harmonic fills in a slice of the total energy bar; the bar can only rise toward the true total, never overshoot (see figure).

Exercise 5.2 (Degenerate case)

What does Parseval say for the zero function $f \equiv 0$ , and for a pure constant $f \equiv c$ ? Confirm both are consistent.

Recall Solution

Zero function: all coefficients $0$ . Left $= \frac{1}{L} \int 0 d x = 0$ ; right $= 0$ . Consistent — no energy in, none out. Constant $f = c$ : only the DC term, $\frac{a _{0}}{2} = c \Rightarrow a_{0} = 2 c$ ; all $a_{n}, b_{n} = 0$ . Left: $\frac{1}{L} \int_{- L}^{L} c^{2} d x = \frac{1}{L} \cdot c^{2} \cdot 2 L = 2 c^{2} .$ Right: $\frac{a _{0}^{2}}{2} = \frac{( 2 c ) ^{2}}{2} = \frac{4 c ^{2}}{2} = 2 c^{2} .$ ✓ WHY test degenerates: they are the cheapest sanity check — if the $a_{0} /2$ bookkeeping were wrong, the constant case would immediately disagree.

Exercise 5.3 (Synthesis with power spectra)

A current $i (t)$ over $(- π, π)$ has harmonics $a_{1} = 5, b_{1} = 0, a_{3} = 0, b_{3} = 12$ . Find (a) the RMS current and (b) the fraction of total power carried by the 3rd harmonic.

Recall Solution

Total: $\frac{1}{π} \int i^{2} d t = a_{1}^{2} + b_{3}^{2} = 25 + 144 = 169.$ So $\int i^{2} d t = 169 π .$ (a) RMS: mean square $= \frac{1}{2 π} \int i^{2} = \frac{169 π}{2 π} = 84.5$ ; $RMS = 84.5 \approx 9.192.$ (b) Power fractions: power in $n$ -th harmonic $\propto a_{n}^{2} + b_{n}^{2}$ . $3rd harmonic fraction = \frac{b _{3}^{2}}{a _{1}^{2} + b _{3}^{2}} = \frac{144}{169} \approx 0.852 (85.2%) .$ WHY this is the whole point of RMS and Power Spectra: Parseval lets an engineer read total power as a sum over frequency bins, and instantly see which harmonic dominates — here the 3rd carries $85%$ of the power.

Recall Ladder

Recall Which level tested what? (click)

L1 ::: reading coefficients, spotting $\frac{a _{0}}{2}$ .
L2 ::: plugging into Parseval, RMS values.
L3 ::: why cross terms die, the $\frac{1}{2}$ asymmetry, phase-blindness.
L4 ::: deriving $ζ (4) = \frac{π ^{4}}{90}$ , splitting sums, designing energy.
L5 ::: Bessel→Parseval limit, degenerate cases, power-spectrum fractions.

Connections

Parseval's theorem — the parent this drills.
Fourier Series · Orthogonality of functions — the machinery.
Basel Problem · Bessel's Inequality · Plancherel Theorem · RMS and Power Spectra — where these results live next.