Full Fourier series — coefficients derivation: Worked examples

Everything below uses only these three boxed results from the parent note. We restate them once so no symbol is unearned.

The scenario matrix

Every Fourier problem you can be handed falls into one of these cells. The point of the page is that the seven examples below cover all of them.

Cell	Situation	Symmetry shortcut	What survives	Example
C1	$f$ is odd (e.g. a square wave, or $x$ )	all $a_{n} = 0$	only $b_{m}$ (sines)	Ex 1
C2	$f$ is even (e.g. $∣ x ∣$ )	all $b_{m} = 0$	$a_{0}, a_{m}$ (cosines)	Ex 2
C3	$f$ has no symmetry (e.g. $e^{x}$ , or $x$ shifted)	none — do all three	$a_{0}, a_{m}, b_{m}$	Ex 3
C4	$f$ is a constant (degenerate: zero frequency only)	trivial	only $a_{0}$	Ex 4
C5	$f$ is already one basis wave (limiting / pure tone)	orthogonality	a single coefficient	Ex 5
C6	Real-world word problem + exam twist (interval not centred, must relocate)	reduce to C1–C5	depends	Ex 6
C7	$f$ is discontinuous (a jump / square wave) — Gibbs behaviour	symmetry if present	$b_{m}$ (odd jump)	Ex 7

A visual map of the same seven cells — glance here to place any problem before you start:

Figure — Full Fourier series — coefficients derivation

Example 1 — Cell C1 (odd function): triangle-slope wave $f (x) = x$ on $[- L, L]$

Look at the straight line through the origin and its odd flip:

Worked example Rebuild

$f (x) = x$ on $[- L, L]$ as sines and cosines. Forecast (guess first!): $f (x) = x$ is a straight line through the origin. Flip $x \to - x$ and the whole graph flips upside-down — that is the signature of an odd function. So predict: all cosine coefficients vanish, only sines remain. Do you also expect $a_{0} = 0$ ? (The average of a line symmetric about the origin is zero — yes.)

Step 1 — classify symmetry. $f (- x) = - x = - f (x)$ , so $f$ is odd → Cell C1. Why this step? By Even and odd functions, odd functions have $a_{0} = 0$ and $a_{m} = 0$ automatically. That deletes two of the three integrals before we lift a pen.

Step 2 — compute $b_{m}$ . Because $f sin$ is (odd)(odd)=even, we may fold the interval: $\int_{- L}^{L} = 2 \int_{0}^{L}$ . $b_{m} = \frac{1}{L} \int_{- L}^{L} x sin \frac{mπ x}{L} d x = \frac{2}{L} \int_{0}^{L} x sin \frac{mπ x}{L} d x .$ Why this step? Folding halves the work and uses the symmetry we just found. The extra factor $2$ is a consequence, not the base $\frac{1}{L}$ changing.

Step 3 — integrate by parts. We have polynomial $\times$ trig; differentiating $x$ makes it constant, so pick $u = x, d v = sin \frac{mπ x}{L} d x$ : $\int_{0}^{L} x sin \frac{mπ x}{L} d x = [- \frac{Lx}{mπ} cos \frac{mπ x}{L}]_{0}^{L} + \frac{L}{mπ} \int_{0}^{L} cos \frac{mπ x}{L} d x .$ The remaining integral evaluates to $\frac{L}{mπ} [sin \frac{mπ x}{L}]_{0}^{L} = \frac{L}{mπ} (sin mπ - 0) = 0$ , since $sin mπ = 0$ for every integer $m$ — the zero comes from the boundary values, not from an averaging argument. The bracket term at $x = L$ gives $- \frac{L ^{2}}{mπ} cos mπ = - \frac{L ^{2}}{mπ} (- 1)^{m}$ ; at $x = 0$ it is $0$ . Why this step? Integration by parts is the tool for "power times wiggle" — it trades the $x$ for a constant.

Step 4 — assemble. $b_{m} = \frac{2}{L} \cdot (- \frac{L ^{2}}{mπ} (- 1)^{m}) = \frac{2 L ( - 1 ) ^{m + 1}}{mπ} .$ Why we may now write the series: we have measured how much of every sine wave lives inside $f$ (the $b_{m}$ ), and all cosine amounts are zero. Re-stacking exactly those amounts rebuilds $f$ (see the "matching terms" box above), so $x = \frac{2 L}{π} m = 1 \sum \infty \frac{( - 1 ) ^{m + 1}}{m} sin \frac{mπ x}{L} on (- L, L) .$

Verify: Set $L = π$ : $b_{m} = \frac{2 ( - 1 ) ^{m + 1}}{m}$ , matching Example 2 of the parent note. Numerically $b_{1} = 2, b_{2} = - 1, b_{3} = \frac{2}{3}$ . ✔

The partial sums really do close in on the line (watch the endpoints, discussed later):

Example 2 — Cell C2 (even function): $f (x) = ∣ x ∣$ on $[- π, π]$

The V-shape is its own mirror image across the vertical axis:

Worked example Rebuild

$f (x) = ∣ x ∣$ on $[- π, π]$ (so $L = π$ ). Forecast: $∣ x ∣$ is a V-shape, mirror-symmetric across the vertical axis. That is even. Predict: all sine coefficients vanish, and since the graph sits entirely above the axis, its average $\frac{a _{0}}{2}$ is positive (not zero this time!).

Step 1 — classify. $f (- x) = ∣ - x ∣ = ∣ x ∣ = f (x)$ : even → Cell C2, so all $b_{m} = 0$ . Why this step? Even $\times$ odd sine = odd, integrates to $0$ . Half the labour gone.

Step 2 — $a_{0}$ . Fold ( $f$ even): $a_{0} = \frac{1}{π} \int_{- π}^{π} ∣ x ∣ d x = \frac{2}{π} \int_{0}^{π} x d x = \frac{2}{π} \cdot \frac{π ^{2}}{2} = π .$ So the average value is $\frac{a _{0}}{2} = \frac{π}{2}$ — positive, as forecast. Why this step? The constant term must equal the DC level of the wave.

Step 3 — $a_{m}$ . With $L = π$ the cosine is $cos m x$ ; $f cos$ is even, fold and parts ( $u = x$ ): $a_{m} = \frac{2}{π} \int_{0}^{π} x cos m x d x = \frac{2}{π} ([\frac{x s i n m x}{m}]_{0}^{π} - \frac{1}{m} \int_{0}^{π} sin m x d x) .$ The boundary term vanishes ( $sin mπ = 0$ ). The remaining integral: $\int_{0}^{π} sin m x d x = \frac{1 - c o s mπ}{m} = \frac{1 - ( - 1 ) ^{m}}{m}$ . So $a_{m} = \frac{2}{π} \cdot (- \frac{1}{m}) \cdot \frac{1 - ( - 1 ) ^{m}}{m} = - \frac{2}{π m ^{2}} (1 - (- 1)^{m}) .$ Why this step? Same parts logic as Ex 1; here differentiating $x$ against $cos$ leaves a $sin$ boundary term that dies.

Step 4 — read off cases. $1 - (- 1)^{m} = 0$ for even $m$ , $= 2$ for odd $m$ : $a_{m} = ⎩ ⎨ ⎧ - \frac{4}{π m ^{2}} 0 m odd m even$ Why we may now write the series: we now know the exact amount of every cosine wave ( $a_{0}$ and each $a_{m}$ ) inside $f$ , and all sine amounts are zero. Re-stacking those measured amounts rebuilds the V-shape: $∣ x ∣ = \frac{π}{2} - \frac{4}{π} k = 0 \sum \infty \frac{1}{( 2 k + 1 ) ^{2}} cos ((2 k + 1) x) .$

Verify: Put $x = 0$ : LHS $= 0$ , so $\frac{π}{2} = \frac{4}{π} \sum \frac{1}{( 2 k + 1 ) ^{2}} \Rightarrow \sum_{k \geq 0} \frac{1}{( 2 k + 1 ) ^{2}} = \frac{π ^{2}}{8}$ — a famous identity. ✔ Also $a_{1} = - \frac{4}{π}, a_{2} = 0, a_{3} = - \frac{4}{9 π}$ . ✔

Example 3 — Cell C3 (no symmetry): $f (x) = e^{x}$ on $[- 1, 1]$

Worked example Rebuild

$f (x) = e^{x}$ on $[- 1, 1]$ (so $L = 1$ ). Forecast: $e^{x}$ is neither a mirror image nor a flip of itself, so no shortcut — we must find $a_{0}$ , all $a_{m}$ , and all $b_{m}$ . This is the general case that keeps you honest.

Step 1 — classify. $e^{- x} \neq = e^{x}$ and $\neq = - e^{x}$ : neither even nor odd → Cell C3. Why this step? Confirming "no symmetry" tells us not to expect any coefficient to vanish for free.

Step 2 — $a_{0}$ . With $L = 1$ : $a_{0} = \int_{- 1}^{1} e^{x} d x = e - e^{- 1} = 2 sinh 1$ (using the definition above). Why this step? Straight integral; sets the average level $\frac{a _{0}}{2} = sinh 1$ .

Step 3 — $a_{m}$ by integrating by parts twice (not a black box). We want $I_{c} = \int e^{x} cos (mπ x) d x$ . Apply parts once with $u = e^{x}, d v = cos (mπ x) d x$ : $I_{c} = \frac{e ^{x} s i n ( mπ x )}{mπ} - \frac{1}{mπ} \int e^{x} sin (mπ x) d x .$ The leftover integral $I_{s} = \int e^{x} sin (mπ x) d x$ is the same type, so apply parts again ( $u = e^{x}, d v = sin (mπ x) d x$ ): $I_{s} = - \frac{e ^{x} c o s ( mπ x )}{mπ} + \frac{1}{mπ} \int e^{x} cos (mπ x) d x = - \frac{e ^{x} c o s ( mπ x )}{mπ} + \frac{1}{mπ} I_{c} .$ Why this step? Each time we integrate the trig and differentiate $e^{x}$ , we come back to the same shape — so the integral feeds on itself, and we can solve for it algebraically instead of forever. Substitute $I_{s}$ into $I_{c}$ : $I_{c} = \frac{e ^{x} s i n ( mπ x )}{mπ} - \frac{1}{mπ} (- \frac{e ^{x} c o s ( mπ x )}{mπ} + \frac{1}{mπ} I_{c}) \Rightarrow I_{c} (1 + \frac{1}{m ^{2} π ^{2}}) = \frac{e ^{x} ( c o s mπ x + mπ s i n mπ x )}{m ^{2} π ^{2}} .$ Clearing the bracket gives the closed form $I_{c} = \frac{e ^{x} ( c o s mπ x + mπ s i n mπ x )}{1 + m ^{2} π ^{2}} .$ Now evaluate on $[- 1, 1]$ : at $x = \pm 1$ , $sin (\pm mπ) = 0$ and $cos (\pm mπ) = (- 1)^{m}$ , so $a_{m} = [I_{c}]_{- 1}^{1} = \frac{( - 1 ) ^{m}}{1 + m ^{2} π ^{2}} (e - e^{- 1}) = \frac{2 ( - 1 ) ^{m} s i n h 1}{1 + m ^{2} π ^{2}} .$

Step 4 — $b_{m}$ from the same $I_{s}$ . Solving the pair for $I_{s}$ likewise gives $I_{s} = \frac{e ^{x} ( s i n mπ x - mπ c o s mπ x )}{1 + m ^{2} π ^{2}} .$ The $sin$ boundary terms vanish, leaving $b_{m} = [I_{s}]_{- 1}^{1} = \frac{- mπ ( - 1 ) ^{m}}{1 + m ^{2} π ^{2}} (e - e^{- 1}) = \frac{- 2 mπ ( - 1 ) ^{m} s i n h 1}{1 + m ^{2} π ^{2}} .$ Why this step? Same self-feeding pair; note $b_{m}$ carries the extra $mπ$ , so the sine amplitudes decay slower here.

Verify: Numerically $sinh 1 \approx 1.1752$ . Then $a_{0} \approx 2.3504$ , $a_{1} = \frac{- 2 s i n h 1}{1 + π ^{2}} \approx - 0.2160$ , $b_{1} = \frac{2 π s i n h 1}{1 + π ^{2}} \approx 0.6785$ . All three families are nonzero, as the no-symmetry cell demands. ✔

Example 4 — Cell C4 (degenerate constant): $f (x) = 5$ on $[- L, L]$

Worked example Rebuild the constant

$f (x) = 5$ . Forecast: A flat line has no wiggles at all. So every genuine harmonic ( $m \geq 1$ ) should be zero, and only the "zero-frequency" term $a_{0}$ can be nonzero. This is the degenerate edge of the matrix.

Step 1 — $a_{0}$ . $a_{0} = \frac{1}{L} \int_{- L}^{L} 5 d x = \frac{1}{L} \cdot 5 \cdot 2 L = 10$ . Why this step? The constant term $\frac{a _{0}}{2} = 5$ must equal the function itself. ✔

Step 2 — $a_{m}, b_{m}$ for $m \geq 1$ . $a_{m} = \frac{5}{L} \int_{- L}^{L} cos \frac{mπ x}{L} d x = 0$ and $b_{m} = \frac{5}{L} \int_{- L}^{L} sin \frac{mπ x}{L} d x = 0$ , because each integral evaluates to a difference of sines/cosines at the endpoints that cancels. Why this step? This is the orthogonality of each wave with the constant $1$ — the DC part is "perpendicular" to every wiggle.

Verify: Series $= \frac{a _{0}}{2} = 5 = f (x)$ exactly, with all other terms zero. ✔ Degenerate case confirmed: a constant is its own Fourier series.

Example 5 — Cell C5 (limiting: already a basis wave) $f (x) = sin \frac{2 π x}{L} + 3 cos \frac{5 π x}{L}$

Worked example Rebuild a function that is

already a sum of two basis waves. Forecast: By Orthogonality of functions, the machine should just hand back the coefficients you can read off the formula: $b_{2} = 1$ , $a_{5} = 3$ , everything else $0$ . If it did anything else, orthogonality would be broken.

Step 1 — read off the target coefficients. Comparing with $\sum [a_{n} cos \frac{nπ x}{L} + b_{n} sin \frac{nπ x}{L}]$ , the only nonzero pieces sit at $n = 2$ (a sine, amplitude $1$ ) and $n = 5$ (a cosine, amplitude $3$ ). Why this step? If the answer is obvious, use it to test the integral machinery rather than trusting it blindly.

Step 2 — confirm $b_{2}$ by integrating. $b_{2} = \frac{1}{L} \int_{- L}^{L} (sin \frac{2 π x}{L} + 3 cos \frac{5 π x}{L}) sin \frac{2 π x}{L} d x .$ The cross term $\int cos \frac{5 π x}{L} sin \frac{2 π x}{L} d x = 0$ (sine–cosine orthogonality, always). The self term $\int sin^{2} \frac{2 π x}{L} d x = L$ . So $b_{2} = \frac{1}{L} \cdot L = 1$ . ✔ Why this step? This is orthogonality in action: multiplying by $sin \frac{2 π x}{L}$ isolates the $n = 2$ sine and silences the rest.

Step 3 — confirm $a_{5}$ . Multiply by $cos \frac{5 π x}{L}$ : the sine term dies (orthogonal), the $3 cos^{2} \frac{5 π x}{L}$ gives $\frac{1}{L} \cdot 3 L = 3$ . So $a_{5} = 3$ . ✔

Step 4 — everything else. For any other $m$ , both self-terms are at $n \neq = m$ , so orthogonality gives $0$ .

Verify: Nonzero coefficients: $b_{2} = 1, a_{5} = 3$ , all others $0$ . The series reproduces $f$ exactly — a basis function is its own series. ✔

Example 6 — Cell C6 (word problem + exam twist): a heater on an off-centre interval

A full period is a full period no matter where the window starts — the orange and teal windows below capture identical coefficients:

Worked example A metal rod of length

$L$ starts at temperature $u_{0} (x) = x$ for $0 < x < L$ . To solve the heat equation by separation of variables we need this as a sine series on $[0, L]$ . But suppose the exam instead gives $g (x) = x$ on the shifted interval $[a, a + 2 L]$ with $a \neq = - L$ and asks for the full series with period $2 L$ . Show these are compatible.

Forecast: Fourier coefficients depend only on one full period, not where you start integrating. So sliding the window from $[- L, L]$ to $[a, a + 2 L]$ must give the same $a_{m}, b_{m}$ — provided you extend $g$ periodically. The "off-centre" twist is a trap: it changes nothing.

Step 1 — the shift-invariance principle. For any $2 L$ -periodic integrand $h$ , $\int_{a}^{a + 2 L} h (x) d x = \int_{- L}^{L} h (x) d x .$ Why this step? Periodicity means the piece you drop off the right end reappears at the left end — the total over any full period is fixed. So you may always re-centre to $[- L, L]$ .

Step 2 — reduce to Cell C1. Once re-centred, $g (x) = x$ on $[- L, L]$ is exactly Example 1, an odd function → all $a_{m} = 0$ and $b_{m} = \frac{2 L ( - 1 ) ^{m + 1}}{mπ}$ . Why this step? The scary shifted interval collapses onto a case we already solved.

Step 3 — the half-range connection, carried all the way through. The rod problem wants a sine-only series on $[0, L]$ . The half-range sine expansion is defined by a recipe: take $u_{0}$ given only on $[0, L]$ , reflect it as an odd function onto $[- L, L]$ , i.e. set $u_{0}^{odd} (x) = u_{0} (x)$ for $x > 0$ and $u_{0}^{odd} (- x) = - u_{0} (x)$ , then run the full-series machinery. Because the extension is odd, all $a_{m} = 0$ , and the surviving sine coefficients are $b_{m} = \frac{1}{L} \int_{- L}^{L} u_{0}^{odd} (x) sin \frac{mπ x}{L} d x .$ The integrand $u_{0}^{odd} sin$ is (odd)(odd)=even, so we may fold: $b_{m} = \frac{2}{L} \int_{0}^{L} u_{0}^{odd} (x) sin \frac{mπ x}{L} d x = \frac{2}{L} \int_{0}^{L} u_{0} (x) sin \frac{mπ x}{L} d x,$ and on $[0, L]$ the extension equals $u_{0}$ itself, so the odd reflection has vanished from the final formula. That last expression is the half-range sine formula. Now specialise to $u_{0} (x) = x$ : $b_{m} = \frac{2}{L} \int_{0}^{L} x sin \frac{mπ x}{L} d x = \frac{2 L ( - 1 ) ^{m + 1}}{mπ},$ which is exactly the coefficient from Step 2 (and from Example 1). So the off-centre full series, the re-centred full series, and the half-range sine series all deliver the same $b_{m}$ . Why this step? It closes the loop: the full-series derivation and the half-range recipe are not two rules but one — an odd function's full series is pure sine, and the $\frac{2}{L} \int_{0}^{L}$ is just the folded $\frac{1}{L} \int_{- L}^{L}$ .

Conclusion. The exam twist (shifted interval) and the physics goal (half-range sine on $[0, L]$ ) both reduce to Cell C1. The temperature profile expands as $u_{0} (x) = x = \frac{2 L}{π} \sum_{m = 1}^{\infty} \frac{( - 1 ) ^{m + 1}}{m} sin \frac{mπ x}{L} on (0, L) .$

Verify: With $L = π$ : $b_{m} = \frac{2 ( - 1 ) ^{m + 1}}{m}$ , so $b_{1} = 2, b_{2} = - 1, b_{3} = \frac{2}{3}$ — the same numbers as Example 1's check, obtained regardless of the starting endpoint $a$ . ✔

Example 7 — Cell C7 (discontinuous jump): square wave, endpoints and Gibbs

The square wave leaps from $- 1$ to $+ 1$ ; note where the series lands at the jump, and the stubborn overshoot beside it:

Worked example Rebuild the square wave

$f (x) = {- 1 + 1 - L < x < 0 0 < x < L$ and describe what happens at the jump and at the endpoints $x = \pm L$ . Forecast: The graph leaps from $- 1$ to $+ 1$ at $x = 0$ — a genuine discontinuity. It is also odd (flip $x \to - x$ and the value flips sign), so predict only sines. But a new question appears: what does the series do exactly at the jump and at the ends, and does it overshoot near them?

Step 1 — classify. $f (- x) = - f (x)$ : odd → Cell C7 with odd symmetry, so all $a_{m} = 0$ . Why this step? Odd kills the cosines; a jump does not change that.

Step 2 — compute $b_{m}$ . Since $f sin$ is even, fold: $b_{m} = \frac{2}{L} \int_{0}^{L} (1) sin \frac{mπ x}{L} d x = \frac{2}{L} \cdot \frac{L}{mπ} [- cos \frac{mπ x}{L}]_{0}^{L} = \frac{2}{mπ} (1 - cos mπ) .$ With $cos mπ = (- 1)^{m}$ : $b_{m} = \frac{2}{mπ} (1 - (- 1)^{m}) = {\frac{4}{mπ} 0 m odd m even$ Why this step? Same folding + endpoint evaluation as Ex 1; the odd terms carry all the weight. So $f (x) = \frac{4}{π} k = 0 \sum \infty \frac{1}{2 k + 1} sin \frac{( 2 k + 1 ) π x}{L}$ .

Step 3 — value at the interior jump $x = 0$ . Every $sin 0 = 0$ , so the series sums to $0$ there. That is exactly the midpoint of the jump, $\frac{( - 1 ) + ( + 1 )}{2} = 0$ . By Dirichlet's theorem, a Fourier series always converges to the average of the left and right limits at a jump — never to either side. Why this step? It answers the edge-case question "what is the series at the break?" that smooth examples never raise.

Step 4 — value at the endpoints $x = \pm L$ (the other jump). The $2 L$ -periodic extension of this square wave jumps again at $x = \pm L$ , where the right value ( $+ 1$ , just left of $L$ ) meets the wrap-around value ( $- 1$ , just right of $L$ ). At $x = L$ every $sin \frac{( 2 k + 1 ) π L}{L} = sin ((2 k + 1) π) = 0$ , so the series equals $0$ — again the midpoint of $+ 1$ and $- 1$ . So the endpoints obey the same Dirichlet midpoint rule as any interior jump. Why this step? For a "full" series the endpoints $\pm L$ are glued together by periodicity; if $f (- L^{+}) \neq = f (L^{-})$ they form a hidden discontinuity, and the series lands on their average — a case students routinely forget.

Step 5 — the Gibbs phenomenon. Near each jump the partial sums overshoot by a fixed fraction (about $9%$ of the jump height) no matter how many terms you add — the ripple narrows but never shrinks. This is the Gibbs phenomenon, the signature of approximating a discontinuity with smooth waves. (In Example 1's figure the same overshoot appears at $x = \pm L$ , where the sawtooth line $x$ secretly jumps from $+ L$ back to $- L$ .) Why this step? It warns you not to expect uniform closeness at jumps — only convergence to the midpoint at the jump and to $f$ at points away from it.

Verify: $b_{1} = \frac{4}{π}, b_{2} = 0, b_{3} = \frac{4}{3 π}$ ; the series value at $x = 0$ and at $x = L$ is $0$ (midpoint of the jump). Also, setting $L = 1, x = \frac{1}{2}$ gives $\frac{4}{π} \sum_{k} \frac{( - 1 ) ^{k}}{2 k + 1} = \frac{4}{π} \cdot \frac{π}{4} = 1$ , matching $f (\frac{1}{2}) = + 1$ . ✔

Endpoint & discontinuity convergence — the edge case for a "full" series

Intuition What value does the series actually take at a jump or at

$x = \pm L$ ? A full Fourier series treats $x = - L$ and $x = + L$ as the same point (period $2 L$ wraps them together). So there are two edge cases the smooth examples hide:

At any jump inside $(- L, L)$ : by Dirichlet's theorem, the series converges to $\frac{1}{2} (f (x^{-}) + f (x^{+}))$ , the midpoint of the two one-sided limits — see Ex 7, $x = 0$ .
At the endpoints $x = \pm L$ : the series converges to $\frac{1}{2} (f (- L^{+}) + f (L^{-}))$ . If those two values disagree (as for the square wave, and secretly for $f (x) = x$ where $+ L$ meets $- L$ ), the endpoints are themselves a jump and the series lands on their average, not on either end value.
Where $f$ is continuous: the series equals $f (x)$ exactly, but near any jump it always overshoots by ~9% (Gibbs) — an overshoot that never disappears, only narrows.

Recall Case-map self-test

Odd function: which coefficients survive? ::: only the $b_{m}$ (sines); $a_{0} = a_{m} = 0$ . Even function: which survive? ::: only $a_{0}$ and the $a_{m}$ (cosines); $b_{m} = 0$ . A constant $c$ : full series? ::: $a_{0} = 2 c$ , everything else $0$ ; the series is just $c$ . Does the starting endpoint of a full-period integral matter? ::: No — any interval of length $2 L$ gives the same coefficients. At a jump, what value does the series converge to? ::: the midpoint (average of the left and right limits). At $x = \pm L$ , what value? ::: the average of $f (- L^{+})$ and $f (L^{-})$ — a jump if they differ. What is the Gibbs phenomenon? ::: a fixed ~9% overshoot of the partial sums near a discontinuity that never disappears. $\sum_{k \geq 0} \frac{1}{( 2 k + 1 ) ^{2}} = ?$ (from Ex 2 at $x = 0$ ) ::: $\frac{π ^{2}}{8}$ .

Related: Convergence of Fourier series (Dirichlet conditions) (does the series actually equal $f$ at jumps?), Complex (exponential) Fourier series (same coefficients repackaged), Parseval's theorem (energy check on the coefficients above).