Intuition What this page is forThe parent topic note told you the rules . This page is a firing range : we shoot every kind of function at those rules — smooth points, jump points, period-wrap points, functions that pass, functions that fail — so that when an exam throws one at you, you've already seen its twin.
The single question we answer at every point x
S ( x ) = 2 f ( x + ) + f ( x − ) . Here f ( x + ) f towards x right " and f ( x − ) left ". Read that formula as: the series lands at the average of the two heights the function has on either side of x .
Throughout this page we say "condition 1/2/3". Here they are, spelled out before any example uses them. They are the Dirichlet conditions applied over one period of a periodic f
Definition The Dirichlet checklist
Condition 1 — absolute integrability. The total area under ∣ f ∣ ∫ − L L ∣ f ( x ) ∣ d x < ∞ (Bans infinite blow-ups; makes the coefficient integrals exist.) Condition 2 — finitely many finite jumps. Every discontinuity is a jump where both one-sided limits f ( x − ) , f ( x + ) (Bans infinite discontinuities and points with no one-sided limit.) Condition 3 — finitely many extrema. f (Bans infinite oscillation.)
If all three hold, the series converges at every x S ( x ) = 2 f ( x + ) + f ( x − ) sufficient , not necessary. Mnemonic: I-J-W = I ntegrable, finite J umps, finite W iggles.
Before working anything, list every distinct situation a Fourier convergence problem can present. Each worked example below is tagged with the one cell it covers.
#
Case class
What makes it different
Example
A
Continuous interior point f ( x + ) = f ( x − ) = f ( x ) Ex 1
B
Finite jump inside the interval f ( x + ) = f ( x − ) Ex 2
C
Period-wrap jump at the endpoint jump created by periodicity, not by f
Ex 3
D
Symmetric jump landing on zero left and right cancel to 0
Ex 3
E
Asymmetric jump (non-zero midpoint) midpoint is some non-zero number
Ex 4
F
Removable / "hole" point (defined weirdly at one spot)f Ex 5
G
FAILS: infinite wiggles condition 3 broken
Ex 6
H
FAILS: infinite blow-up (both sides) condition 1 broken (not absolutely integrable)
Ex 7
K
One-sided infinite limit one side finite, the other blows up
Ex 8
L
Endpoint continuity (wrap agrees) f ( π − ) = f ( − π + ) S ( π ) = f ( π ) Ex 9
I
Word problem / real signal a physical square pulse (with Gibbs overshoot)
Ex 10
J
Exam twist: series value used to sum a number series plug a clever x
Ex 11
We now hit every cell.
Figure 1 — a black function with a jump at x 0 f ( x 0 − ) f ( x 0 + ) red dot sits exactly midway between them. That red dot is S ( x 0 )
Intuition Why halfway, in one sentenceThe series is built from sines and cosines, which are perfectly symmetric left-right; at a jump they have no reason to prefer one side, so they split the difference — the Dirichlet kernel samples both sides with equal weight.
Worked example Example 1 —f ( x ) = x ( − π , π ) x = 2 π Forecast: guess the series value before reading on. Is x = π /2
Check the point is interior and continuous. Why this step? The midpoint rule only differs from f f − π π f ( x ) = x π /2 Since continuous, f ( x + ) = f ( x − ) = f ( π /2 ) Why? Walking towards π /2 Apply the formula:
S ( 2 π ) = 2 f ( 2 π + ) + f ( 2 π − ) = 2 2 π + 2 π = 2 π .
Verify: at a point of continuity the average of a number with itself is that number, so S = f ( π /2 ) = π /2 ≈ 1.5708 π /2 ( − π , π )
Worked example Example 2 — Square wave, jump atx = 0 f ( x ) = { − 1 + 1 − π < x < 0 0 < x < π 2 π S ( 0 )
Forecast: the function is never 0 0
Identify the limits at x = 0 Why? 0 + 1 − 1 f ( 0 + ) = + 1 , f ( 0 − ) = − 1. Both limits are finite (condition 2 satisfied) , so the midpoint rule applies. Why this matters: if a side blew up to infinity we could not average — see Ex 8.Average them:
S ( 0 ) = 2 ( + 1 ) + ( − 1 ) = 0.
Verify: the square wave is an odd function, and every term of its series is sin ( n x ) 0 x = 0 0 x = 0 N 0
Figure 2 — square wave. Open circles at f ( 0 − ) = − 1 f ( 0 + ) = + 1 red dot at height 0 S ( 0 )
Worked example Example 3 —f ( x ) = x ( − π , π ) x = π Forecast: the formula f ( x ) = x
Remember the function is periodic with period 2 π Why this step? The Fourier series only knows one period and then repeats it. So just to the right of x = π − π Compute the two limits at x = π
f ( π − ) = π ( left piece reaches π ) , f ( π + ) = − π ( wrap restarts at − π ) . Why − π Periodicity: f ( π + ) = f (( − π ) + ) = − π Average:
S ( π ) = 2 π + ( − π ) = 0.
Verify: the sawtooth's series is 2 ∑ n ≥ 1 n ( − 1 ) n + 1 sin ( n x ) x = π sin ( nπ ) = 0 S ( π ) = 0 cell C (wrap jump) and , because the two heights are ± π cell D (symmetric ⇒
Figure 3 — sawtooth f ( x ) = x x = π π − π red dot at 0 S ( π )
Worked example Example 4 — a step that doesnot average to zero
f ( x ) = { 0 4 − π < x < 0 0 < x < π 2 π S ( 0 ) S ( π )
Forecast: which jump gives a non-zero answer, and which gives zero?
At x = 0 the two pieces meet with f ( 0 − ) = 0 f ( 0 + ) = 4 Why this step? We read each side's height right at the meeting point.
S ( 0 ) = 2 4 + 0 = 2. Why non-zero? The two sides are not symmetric about zero, so their average is a genuine non-zero value — cell E .At x = π just left is the top piece f ( π − ) = 4 f ( π + ) = f (( − π ) + ) = 0 S ( π ) = 2 4 + 0 = 2. Why also 2 Same two heights (4 0
Verify: Trick — split f = 2 + 2 g g ± 1 2 and every jump (constants have no jump). At x = 0 g ( 0 ) = 0 S ( 0 ) = 2 + 2 ⋅ 0 = 2 x = π g = 0 S ( π ) = 2
First, the tool we lean on. A convergent Fourier series is reconstructed from coefficients built by orthogonality :
Worked example Example 5 — the value at a single point cannot be seen by the seriesTake f ( x ) = x ( − π , π ) artificially set f ( 0 ) = 100 S ( 0 )
Forecast: does the series "see" that we spiked the value to 100
The coefficients a n , b n f Why this step? Changing f single point changes the integral by an area of width zero, i.e. by nothing. So a n , b n f ( x ) = x Compute the one-sided limits, ignoring the spike. Why ignore it? Limits look at what happens approaching 0 at 0 f ( 0 + ) = 0 , f ( 0 − ) = 0. Average:
S ( 0 ) = 2 0 + 0 = 0.
Verify: f ( x ) = x 0 0 100 limits , not the assigned point value). So S ( 0 ) = 0 = 100 piecewise-smooth
Worked example Example 6 —f ( x ) = sin ( 1/ x ) 0 < ∣ x ∣ < π 2 π Define the function precisely first. Why? You can only take a Fourier series of a fully specified periodic function. Set f ( x ) = sin ( 1/ x ) 0 < ∣ x ∣ < π f ( 0 ) = 0 2 π Forecast: will Dirichlet's theorem guarantee convergence here?
Count the maxima/minima as x → 0 Why this step? Condition 3 caps the number of wiggles per period. sin ( 1/ x ) 1/ x = 2 π + 2 π k x = π /2 + 2 π k 1 infinitely many peaks bunching up at 0 Also check the limit at x = 0 Why this second check? Because sin ( 1/ x ) − 1 + 1 x → 0 f ( 0 + ) f ( 0 − ) do not exist — so condition 2 (a finite jump needs both one-sided limits to exist) fails too.Two conditions fail , so Dirichlet's theorem says nothing . Why this matters: the theorem is a sufficient checklist; unticked boxes withdraw the guarantee — they do not by themselves prove divergence, they simply leave us without a promise.
Verify: the peaks sit at x k = 1/ ( π /2 + 2 π k ) k = 0 , 1 , 2 x 0 = π 2 ≈ 0.6366 x 1 ≈ 0.1273 x 2 ≈ 0.0688 + 1 − 1 0
Figure 4 — sin ( 1/ x ) 0 x = 0 0
Worked example Example 7 —f ( x ) = x 2 1 ( − π , π ) Forecast: the graph shoots to + ∞ both sides of 0
Test absolute integrability ∫ − π π ∣ f ∣ d x Why this step? Condition 1 needs a finite total area under ∣ f ∣ a n , b n ∫ 0 π x 2 1 d x = [ − x 1 ] 0 π = − π 1 − ( − ∞ ) = + ∞. The area is infinite, so condition 1 fails. Why this step? The whole midpoint machinery only kicks in after the three conditions are met; here the very first one (finite area) is broken, so we must stop and report "no guarantee" rather than pretend to average. This infinite discontinuity — both one-sided limits = + ∞
Verify: ∫ ε π x 2 1 d x = ε 1 − π 1 → + ∞ ε → 0 + finite jump (Ex 2), which is completely fine.
Worked example Example 8 — one side finite, the other blows upf ( x ) = ⎩ ⎨ ⎧ x 1 0 0 < x < π − π < x < 0 2 π x = 0
Forecast: the left side is a flat 0 + ∞
Read each one-sided limit. Why this step? The midpoint rule needs both limits to be finite numbers before we may average them.
f ( 0 − ) = 0 ( finite ) , f ( 0 + ) = lim x → 0 + x 1 = + ∞ ( does not exist as a finite number ) . One side is infinite , so this is a one-sided infinite discontinuity . Why it matters: condition 1 already fails — ∫ 0 π x 1 d x = + ∞ no guarantee .Contrast with cell B. Why include this? A finite jump (both sides finite, Ex 2) is allowed; the moment one side is infinite, the checklist breaks — the finiteness of the other side cannot rescue it.
Verify: ∫ ε π x 1 d x = ln π − ln ε → + ∞ ε → 0 + ( 0 , π ) f ( 0 + ) f ( 0 − ) = 0
Worked example Example 9 —f ( x ) = x 2 ( − π , π ) x = π Forecast: at the endpoint of f ( x ) = x every endpoint jump? Guess before reading.
Compute the left limit at x = π Why this step? We must know the height the last piece reaches: f ( π − ) = π 2 Compute the wrap (right) limit. Why? Just past x = π x = − π f ( π + ) = f (( − π ) + ) = ( − π ) 2 = π 2 The key observation: because x 2 even , the two ends have the same height — the wrap agrees, so there is no jump at the endpoint.Since both sides equal π 2
S ( π ) = 2 π 2 + π 2 = π 2 .
Verify: f ( − π + ) = π 2 = f ( π − ) x 2 π 2 ≈ 9.8696 f ( π ) = π 2 no midpoint discount. ✓ This completes the endpoint picture: a wrap creates a jump only when the two ends differ (Ex 3); when they agree, the endpoint behaves like any interior continuous point.
Figure 5 — periodic f ( x ) = x 2 x = π π 2 π 2 red dot ) — no gap, S ( π ) = π 2
Worked example Example 10 — a rectangular voltage pulseA digital clock line holds 0 5 V ( t ) = { 0 5 0 < t < T /2 T /2 < t < T , period T . exactly at the switching instant t = T /2
Forecast: hardware says it's a clean step from 0 5 series draw at that instant?
Read the one-sided limits at the switch t = T /2 Why this step? t = T /2 V ( T / 2 − ) = 0 , V ( T / 2 + ) = 5. Both finite ⇒
S ( T /2 ) = 2 0 + 5 = 2 5 = 2.5 volts .
Verify: units are volts throughout; the midpoint 2.5 V 0 5 V
The Gibbs overshoot (see Figure 6): near the switch the finite partial sum S N not rise smoothly — it overshoots to about 9% Why does this happen? Each finite partial sum is a smooth wave that, forced to climb a vertical cliff of height 5 V 0.0895 × 5 ≈ 0.45 V 5.45 V − 0.45 V does not shrink as N → ∞ at the switch stays exactly 2.5 V Gibbs phenomenon in the flesh — the overshoot decorates the jump, but the limit value is still the midpoint.
Figure 6 — the 0 5 V N red overshoot horns near the switch never vanish (Gibbs), but the crossover passes exactly through the midpoint 2.5 V
Worked example Example 11 — evaluate1 − 3 1 + 5 1 − 7 1 + ⋯ f ( x ) = π 4 ∑ k = 0 ∞ 2 k + 1 s i n ( ( 2 k + 1 ) x ) . x = 2 π
Forecast: what value does the left side take at x = π /2
Evaluate f x = π /2 Why here? π /2 ( 0 , π ) f = + 1 continuity (cell A), so S ( π /2 ) = f ( π /2 ) = 1 Evaluate the sines: sin ( ( 2 k + 1 ) 2 π ) + 1 , − 1 , + 1 , − 1 , … k = 0 , 1 , 2 , 3 Why? sin ( odd ⋅ 2 π ) = ( − 1 ) k Equate both sides:
1 = π 4 ( 1 − 3 1 + 5 1 − 7 1 + ⋯ ) ⟹ 1 − 3 1 + 5 1 − 7 1 + ⋯ = 4 π .
Verify: the partial sum 1 − 3 1 + 5 1 − 7 1 + 9 1 − 11 1 ≈ 0.7440 π /4 ≈ 0.7854 Leibniz series — an exam favourite born from a continuity point.
Recall Which cell is this? (hide and answer)A function jumps from − 3 + 3 2 3 + ( − 3 ) = 0 f ( x ) = x 2 f f ( x ) = x 2 x = π S ( π ) = π 2 f 1/ x 2 x = 0 0 = 1/ x → ∞ 2.5 0 5
Mnemonic The one question that solves every cellAt any point ask: "What are the two heights on my left and right?" Average them. Continuity (interior or an agreeing wrap) ⇒ ⇒ f ( x ) ⇒ ⇒ ⇒
same height cells A and L
two finite heights cells B C D E I
both sides infinite cell H
infinitely many peaks cell G
Two heights left and right?
Meet in the middle equals midpoint
Cell J plug clever x to sum a series
