Parseval's theorem: Worked examples — Maths deep dive

The scenario matrix

Every Parseval problem is really one of these cells. The whole point of the page is to walk each row so you never meet an unshown case.

#	Cell (case class)	What is special about it	Covered by
A	Odd $f$ : only $b_{n}$ survive	$a_{0} = a_{n} = 0$ , RHS $= \sum b_{n}^{2}$	Ex 1
B	Even $f$ : only $a_{0}, a_{n}$ survive	$b_{n} = 0$ , RHS $= \frac{a _{0}^{2}}{2} + \sum a_{n}^{2}$	Ex 2
C	Neither odd nor even	all three families present	Ex 3
D	Degenerate: constant $f = c$	only the DC term $\frac{a _{0}^{2}}{2}$	Ex 4
E	Finite harmonic sum (signal)	no infinite series — direct power add	Ex 5
F	Different interval / general $L$	the $\frac{1}{L}$ factor really matters	Ex 6
G	Limiting / convergence sanity	partial sums must approach the LHS from below	Ex 7
H	Word problem: RMS & power	units, physical meaning	Ex 8
I	Exam twist: solve for one unknown $b_{n}$	use Parseval backwards	Ex 9

Two zero-cases to name explicitly so you never trip:

Zero function $f \equiv 0$ : both sides are $0$ . Trivially true; it is the "origin" of the picture below.
$a_{0}$ is the average, doubled: if $f$ has zero average then $a_{0} = 0$ and the DC term drops out (this is what makes odd functions clean).

Example 1 — Cell A (odd function): recover the Basel sum

Left side — compute the integral directly. $\frac{1}{π} \int_{- π}^{π} x^{2} d x = \frac{1}{π} \cdot \frac{2 π ^{3}}{3} = \frac{2 π ^{2}}{3} .$ Why this step? For an odd $f$ the RHS is pure $\sum b_{n}^{2}$ ; to use that we first need a number for the LHS, and $\int x^{2}$ is elementary.
Right side — square and sum the coefficients. $\sum_{n = 1}^{\infty} b_{n}^{2} = \sum_{n = 1}^{\infty} \frac{4 ( - 1 ) ^{2 (n + 1)}}{n ^{2}} = \sum_{n = 1}^{\infty} \frac{4}{n ^{2}} = 4 \sum_{n = 1}^{\infty} \frac{1}{n ^{2}} .$ Why this step? Squaring kills the sign $(- 1)^{n + 1}$ (it always becomes $+ 1$ ) — a signal that phase information is lost in Parseval; only magnitudes matter.
Set LHS $=$ RHS and solve. $\frac{2 π ^{2}}{3} = 4 \sum \frac{1}{n ^{2}} \Rightarrow \sum_{n = 1}^{\infty} \frac{1}{n ^{2}} = \frac{π ^{2}}{6} .$ Why this step? This is Parseval "run backwards" — an integral we can do delivers a sum we cannot do directly. See Basel Problem.

Verify: $π^{2} /6 \approx 1.6449$ — indeed just over $1.6$ , matching the forecast. Numerically $\sum_{n = 1}^{5000} 1/ n^{2} \approx 1.6447$ , closing on $1.6449$ . ✓

Example 2 — Cell B (even function): recover $\sum 1/ n^{4}$

Left side. $\frac{1}{π} \int_{- π}^{π} x^{4} d x = \frac{1}{π} \cdot \frac{2 π ^{5}}{5} = \frac{2 π ^{4}}{5} .$ Why this step? $[f]^{2} = x^{4}$ ; the interval is symmetric so $\int x^{4} = 2 \int_{0}^{π} x^{4}$ .
Right side — include the DC term at half weight. $\frac{a _{0}^{2}}{2} + \sum a_{n}^{2} = \frac{1}{2} (\frac{2 π ^{2}}{3})^{2} + \sum \frac{16}{n ^{4}} = \frac{2 π ^{4}}{9} + 16 \sum \frac{1}{n ^{4}} .$ Why this step? The lonely $a_{0}$ term carries the factor $\frac{1}{2}$ — forgetting it is the classic error (see parent's Mistake 2).
Equate and isolate the sum. $\frac{2 π ^{4}}{5} - \frac{2 π ^{4}}{9} = 2 π^{4} \cdot \frac{9 - 5}{45} = \frac{8 π ^{4}}{45} = 16 \sum \frac{1}{n ^{4}} \Rightarrow \sum \frac{1}{n ^{4}} = \frac{π ^{4}}{90} .$ Why this step? Same machinery as Ex 1, one power higher — Parseval scales to every even power of $x$ .

Verify: $π^{4} /90 \approx 1.0823$ , matching the forecast. Partial sum $\sum_{n = 1}^{2000} 1/ n^{4} \approx 1.0823$ . ✓

Example 3 — Cell C (neither odd nor even): both families present

Left side directly. $\frac{1}{π} \int_{- π}^{π} (1 + x)^{2} d x = \frac{1}{π} \int_{- π}^{π} (1 + 2 x + x^{2}) d x .$ The middle term $\int_{- π}^{π} 2 x d x = 0$ (odd over symmetric interval), so $= \frac{1}{π} (2 π + \frac{2 π ^{3}}{3}) = 2 + \frac{2 π ^{2}}{3} .$ Why this step? This shows the cross-term between the constant and the odd part vanishes on integration — the geometric echo of orthogonality: a constant (DC) and a sine are perpendicular.
Right side. $\frac{a _{0}^{2}}{2} + \sum b_{n}^{2} = \frac{2 ^{2}}{2} + 4 \sum \frac{1}{n ^{2}} = 2 + 4 \sum \frac{1}{n ^{2}} .$ Why this step? $a_{0} = 2$ contributes $\frac{a _{0}^{2}}{2} = 2$ , the sines contribute exactly as in Ex 1.
Match. LHS $=$ RHS gives $2 + \frac{2 π ^{2}}{3} = 2 + 4 \sum \frac{1}{n ^{2}}$ , i.e. again $\sum \frac{1}{n ^{2}} = \frac{π ^{2}}{6}$ . Why this step? The constant piece cancels off both sides — Parseval separates energy by frequency, so the DC energy ( $2$ ) and the oscillating energy stay in their own bins.

Verify: LHS $= 2 + 2 π^{2} /3 \approx 8.5797$ ; RHS $= 2 + 4 (π^{2} /6) = 2 + 2 π^{2} /3 \approx 8.5797$ . ✓

Example 4 — Cell D (degenerate constant): only the DC bin

Fourier coefficients. $a_{0} = \frac{1}{π} \int_{- π}^{π} 5 d x = \frac{1}{π} \cdot 10 π = 10$ ; all $a_{n} = b_{n} = 0$ . Why this step? The DC coefficient is $a_{0} = \frac{1}{L} \int f$ ; a constant has average $5$ , and $a_{0} = 2 \times average = 10$ . This is the "only the constant term survives" degenerate cell.
Left side. $\frac{1}{π} \int_{- π}^{π} 25 d x = \frac{1}{π} \cdot 50 π = 50.$
Right side. $\frac{a _{0}^{2}}{2} = \frac{100}{2} = 50.$ Why this step? Confirms the mysterious factor $\frac{1}{2}$ is exactly right — without it the RHS would be $100 \neq = 50$ .

Verify: $50 = 50$ . ✓ The half-weight on $a_{0}$ is not a typo; it makes the constant case work.

Example 5 — Cell E (finite harmonic signal): energy adds harmonic by harmonic

Right side is a finite sum — no infinite series. $\frac{1}{π} \int_{- π}^{π} f^{2} d t = a_{1}^{2} + b_{2}^{2} = 9 + 16 = 25.$ Why this step? With finitely many harmonics, Parseval is literally $3$ - $4$ - $5$ Pythagoras: perpendicular components add in quadrature.
Un-average to get the raw integral. Multiply by $L = π$ : $\int_{- π}^{π} f^{2} d t = 25 π .$ Why this step? The $\frac{1}{L}$ is an averaging factor; removing it restores the total energy.
Mean-square value (average of $f^{2}$ over the full period $2 π$ ): $\overline{f^{2}} = \frac{1}{2 π} \int_{- π}^{π} f^{2} d t = \frac{25 π}{2 π} = 12.5, RMS = 12.5 \approx 3.536.$ Why this step? Note the two averaging windows: Parseval uses $\frac{1}{L}$ (half-period), physics power uses $\frac{1}{2 π}$ (full period). Do not confuse them.

Verify: $9 + 16 = 25$ ; $25 π \approx 78.54$ ; mean-square $= 12.5$ ; RMS $= 12.5 \approx 3.5355$ . ✓ See RMS and Power Spectra.

Example 6 — Cell F (general $L$ ): the $\frac{1}{L}$ factor bites

Coefficients on $(- L, L)$ for $f = x$ . The odd function $x$ has $b_{n} = \frac{2}{L} \int_{0}^{L} x sin \frac{nπ x}{L} d x = \frac{2 L ( - 1 ) ^{n + 1}}{nπ} .$ Why this step? The coefficient formula carries $L$ ; watch how it self-cancels.
Right side. $\sum b_{n}^{2} = \sum \frac{4 L ^{2}}{n ^{2} π ^{2}} = \frac{4 L ^{2}}{π ^{2}} \sum \frac{1}{n ^{2}} .$
Left side (with its $\frac{1}{L}$ ). $\frac{1}{L} \int_{- L}^{L} x^{2} d x = \frac{1}{L} \cdot \frac{2 L ^{3}}{3} = \frac{2 L ^{2}}{3} .$ Why this step? Here is where $\frac{1}{L}$ earns its keep — without it the two sides would carry mismatched powers of $L$ .
Equate. $\frac{2 L ^{2}}{3} = \frac{4 L ^{2}}{π ^{2}} \sum \frac{1}{n ^{2}} \Rightarrow \sum \frac{1}{n ^{2}} = \frac{2 L ^{2}}{3} \cdot \frac{π ^{2}}{4 L ^{2}} = \frac{π ^{2}}{6} .$ Why this step? Every $L^{2}$ cancels — a pure number can't depend on your choice of interval. Plug $L = 2$ explicitly if you like: LHS $= \frac{8}{3}$ , RHS $= \frac{16}{π ^{2}} \cdot \frac{π ^{2}}{6} = \frac{16}{6} = \frac{8}{3}$ . ✓

Verify: with $L = 2$ : $\frac{2 L ^{2}}{3} = 8/3$ ; $\frac{4 L ^{2}}{π ^{2}} \cdot \frac{π ^{2}}{6} = 8/3$ . Equal, $L$ -free result $π^{2} /6$ . ✓

Example 7 — Cell G (limiting / convergence sanity)

Compute a few partial sums. $S_{1} = 4, S_{2} = 4 (1 + \frac{1}{4}) = 5, S_{5} = 4 \sum_{1}^{5} \frac{1}{n ^{2}} \approx 5.854.$ Why this step? Watching the staircase climb makes "convergence from below" concrete rather than abstract.
Bound. Since every added term is positive, $S_{N}$ is increasing and bounded above by $\frac{2 π ^{2}}{3}$ . Why this step? This is exactly Bessel's inequality $\sum_{n = 1}^{N} (a_{n}^{2} + b_{n}^{2}) \leq \frac{1}{L} \int f^{2}$ ; Parseval is the $N \to \infty$ equality case for a complete basis.
Limit. $S_{N} \to 4 \cdot \frac{π ^{2}}{6} = \frac{2 π ^{2}}{3}$ .

Verify: $S_{1} = 4$ , $S_{2} = 5$ , $S_{5} \approx 5.8544$ , all $< \frac{2 π ^{2}}{3} \approx 6.5797$ , and $S_{N} \to 6.5797$ . ✓

Example 8 — Cell H (word problem: RMS & power)

Identify coefficients. $a_{1} = 2$ , $b_{3} = 6$ , $a_{0} = 0$ , all else $0$ . Why this step? Read amplitudes straight off — power lives in $a_{n}^{2} + b_{n}^{2}$ .
Total (Parseval). $\frac{1}{π} \int_{- π}^{π} v^{2} d t = a_{1}^{2} + b_{3}^{2} = 4 + 36 = 40.$ So $\int_{- π}^{π} v^{2} d t = 40 π$ .
RMS. Mean-square over the full period: $\overline{v^{2}} = \frac{1}{2 π} (40 π) = 20$ , so $V_{RMS} = 20 \approx 4.472 V .$ Why this step? RMS is the DC-equivalent heating voltage — units are volts, and $V^{2} = V$ checks out.
Power fractions. Third harmonic weight $= \frac{b _{3}^{2}}{a _{1}^{2} + b _{3}^{2}} = \frac{36}{40} = 0.9 = 90%$ . Why this step? Confirms the forecast: the amplitude- $6$ harmonic holds $90%$ of the power. Fundamental holds $10%$ .

Verify: $4 + 36 = 40$ ; $\overline{v^{2}} = 20$ ; RMS $= 20 \approx 4.4721$ V; third-harmonic fraction $= 36/40 = 0.9$ . ✓

Example 9 — Cell I (exam twist: solve for a missing coefficient)

Write Parseval as a bookkeeping equation. $10 = a_{1}^{2} + b_{2}^{2} = 1 + b_{2}^{2} .$ Why this step? Used backwards: the total is given, so Parseval becomes a solvable equation for the one unknown.
Solve. $b_{2}^{2} = 9 \Rightarrow b_{2} = \pm 3$ ; since we're told $b_{2} > 0$ , take $b_{2} = 3$ . Why this step? Parseval only fixes the magnitude (energy) — the sign/phase is extra information you must be given, echoing the "phase is lost" note from Ex 1.

Verify: $1 + 3^{2} = 1 + 9 = 10$ . ✓ Matches the given energy exactly.

Odd $f$ ::: RHS is pure $\sum b_{n}^{2}$ (Cell A). Even $f$ ::: RHS is $\frac{a _{0}^{2}}{2} + \sum a_{n}^{2}$ (Cell B). Constant $f = c$ ::: only $\frac{a _{0}^{2}}{2}$ survives, $a_{0} = 2 c$ (Cell D). Finite harmonics ::: Parseval is exact Pythagoras, no infinite sum (Cell E). Partial sums vs total ::: they rise to the total from below — Bessel's inequality (Cell G).

Connections

Parent: Parseval's theorem
Basel Problem — Ex 1 result $\sum 1/ n^{2} = π^{2} /6$ .
Bessel's Inequality — Ex 7's "approach from below".
RMS and Power Spectra — Ex 5 and Ex 8.
Orthogonality of functions — why cross-terms die (Ex 3).
Fourier Series — where the coefficients come from.

Worked examples — Parseval's theorem