4.7.7 · D5Partial Differential Equations
Question bank — Parseval's theorem
True or false — justify
Every claim below is either subtly right or subtly wrong. The reveal gives the reason, which matters more than the verdict.
Parseval's theorem is just the Pythagoras theorem, but for functions instead of vectors.
True — the "length-squared" of (its energy ) equals the sum of squares of its coordinates (the coefficients) in the orthogonal sine/cosine basis.
Parseval requires to be an even or odd function.
False — it holds for any with a valid Fourier Series; even/odd only decides which coefficients are zero, not whether the theorem applies.
The right-hand side has no in it, so never matters.
False — is hidden in the left side as the averaging factor , and it also sets the frequencies used to compute the coefficients.
If two functions have the exact same set of Fourier coefficients, they have the same energy.
True — the right side depends only on the coefficients, so equal coefficients force equal .
Parseval can be true for a function whose Fourier series does not converge to pointwise.
True — convergence "in the mean" (energy sense) is enough; the series can misbehave at isolated jump points yet still capture all the energy.
The cross terms like () are ignored because they are small.
False — they are not small, they are exactly zero after integration by orthogonality; "small" would only give an approximation, orthogonality gives an identity.
Adding more harmonics to a signal always increases its total energy.
True — each new adds a non-negative slice, so the running sum can only grow or stay equal, never shrink.
For a purely DC signal (constant), the energy lives entirely in the part.
False — a constant has , all other coefficients zero, so all energy sits in the lonely term.
Parseval and Bessel's inequality say the same thing.
False — Bessel gives using any subset of the basis; Parseval upgrades it to precisely because sines and cosines form a complete basis (nothing is left over).
The theorem lets you compute without ever integrating directly.
True — that is its power: once you know the coefficients, the energy is a plain sum of squares, no messy integral of needed.
Spot the error
Each line states a student's step. Find and fix the flaw.
"On I wrote ."
Missing the factor on the left; correct is .
"The DC term contributes to the right side, matching the pattern of the others."
Wrong — the series carries , so squaring gives , and (double the others), combining to , not .
"Since is odd on , its energy is zero."
Confusing with ; is odd but is even and positive, so its energy is genuinely nonzero ().
"I found , so the right side is ."
Parseval sums the squares: , not (which wouldn't even converge nicely nor represent energy).
"The half-range sine series and the full-range series of the same must give the same Parseval sum."
Not necessarily — they are expansions on different intervals representing differently, so both energies and coefficients differ; only within one consistent expansion does Parseval hold.
" only when ."
It is for all (including ), because a cosine times a sine of matching argument is odd across the symmetric interval.
Why questions
State the deep reason, not just the fact.
Why does orthogonality make the cross terms vanish rather than merely cancel?
Because each cross-term integral is individually zero — perpendicular basis functions have zero "overlap", like , so nothing needs to cancel against anything.
Why is Parseval called an energy or power statement in engineering?
measures total energy of a signal, and is the power carried by the -th harmonic, so the theorem literally says total power = sum of harmonic powers (RMS and Power Spectra).
Why does dividing the derivation's final line by make the formula "look nicer"?
It converts into a mean value and rescales every survivor by , so the , terms drop their and the DC term lands exactly on the pattern.
Why can Parseval evaluate impossible-looking infinite sums like ?
It equates a known integral (easy) with a sum built from coefficients (the unknown series), letting you solve for the sum — turning a summation problem into an integration problem (Basel Problem).
Why does the theorem need the basis to be complete, not merely orthogonal?
Orthogonality alone gives Bessel's ; only completeness guarantees no energy escapes into directions outside the basis, upgrading to the exact .
Why is 's Parseval identity "deeper" than 's?
The extra power pushes the sum from to ; higher powers of reach higher zeta values, revealing a whole ladder of results.
Edge cases
The scenarios people never test — but that either break naive formulas or reveal the structure.
What does Parseval give for (the zero function)?
All coefficients are zero and , so both sides are — the trivial but consistent boundary case.
What does Parseval say for a single pure tone ?
with every other coefficient zero — the energy is entirely in that one harmonic, showing each term is a self-contained energy bin.
If has a jump discontinuity, does Parseval still hold?
Yes — energy is an integral (insensitive to isolated points), so a finite number of jumps changes nothing; only the pointwise value at the jump is ambiguous, not the total energy.
What happens to Parseval if you double every coefficient of (i.e. use )?
The right side scales by (each square gets a factor ) and so does the left () — energy is quadratic, so both sides stay balanced.
Can the right-hand sum be finite even though has infinitely many nonzero coefficients?
Yes — as long as converges (which it must for any finite-energy ), the coefficients shrink fast enough that the total is finite.
Does Parseval work for a function that is nonzero only on part of , e.g. a pulse?
Yes — the integral automatically ignores where , and the coefficients encode the pulse; energy of the pulse equals the coefficient-square sum, no special treatment needed.
What is the continuous-signal analogue of Parseval for non-periodic ?
The Plancherel Theorem: — the same energy-conservation idea with a Fourier transform replacing the discrete coefficient sum.
If a signal has , does that mean it carries no energy?
No — only means zero average (no DC offset); energy can still be large, stored entirely in the AC harmonics .
Connections
- Parseval's theorem — the parent statement these traps stress-test.
- Orthogonality of functions — why cross terms are exactly zero.
- Bessel's Inequality — the that Parseval sharpens to .
- Fourier Series — the expansion the whole theorem lives on.
- Basel Problem — the famous payoff sum.
- Plancherel Theorem — the transform-world twin.
- RMS and Power Spectra — the engineering reading of "energy".