Every function this course decomposes is periodic. Every decomposition is built out of sines and cosines. So before any of that machinery starts, two facts about a function's shape need to be automatic: how often it repeats, and whether it's symmetric enough that half its Fourier coefficients vanish for free. Neither idea is new. Both are about to matter in a much more concrete way than they did the first time around.
Fold a graph in half along the y-axis. An even function's two halves land exactly on top of each other. The left side mirrors the right, the way a butterfly's two wings mirror each other across its body. Now spin the graph 180° around the origin instead, like giving a two-bladed pinwheel a half-turn. An odd function lands back on its own picture after that spin. An even one doesn't, since a mirror image isn't the same thing as a half-turn. Neither name has anything to do with odd or even numbers showing up in the formula. They're names for which physical symmetry the picture has, mirror or pinwheel, and a graph can easily have neither.
The trap is trusting a formula's look instead of doing the fold or the spin. f(x) = x^2 + x has an x^2 in it, which reads as "even-shaped" at a glance. But the lone x term breaks the mirror. f(-2) = 2, while f(2) = 6 and -f(2) = -6. Neither test passes, so f is neither. The power-counting shortcut only ever applies to a sum of pure powers. The instant a function is built some other way, the fold-it-and-look test is the only one that still works.
A function f is periodic with period P if f(x+P) = f(x) for every x in its domain. If a smallest positive such P exists, it's called the fundamental period. Periods stack: if P works, so does 2P, 3P, and every other positive multiple. That's exactly why "the period" always means the fundamental one unless stated otherwise.
f is even if f(-x) = f(x) for every x in its domain, and odd if f(-x) = -f(x). These are properties of the whole graph, not of any one term in a formula. Checking f(-x) directly is the only test that always works. For a function built entirely from a sum of pure powers of x (a polynomial), the power pattern happens to predict the answer. Every term an even power, including a constant, makes f even. Every term an odd power makes f odd. Mixing the two makes it neither. That shortcut is a consequence of the definition, for polynomials specifically. It says nothing about \sin x, |x|, or a function defined piecewise, where f(-x) has to be computed directly.
Symmetry earns its place here because of what it does to an integral over a symmetric interval [-a, a]: \int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx if f is even, and \int_{-a}^{a} f(x)\,dx = 0 if f is odd. The negative side either doubles or exactly cancels the positive side. Chapter 3's Fourier coefficient formulas are integrals over a symmetric interval by construction. This single fact is what makes an odd function's cosine coefficients vanish outright, and an even function's sine coefficients vanish outright, before a single integral is computed.
How the period responds to a frequency multiplier
Drag b. The curve still oscillates between -1 and 1, but how many times does it repeat in the same window?
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Flipping the sign is a half-turn through the origin
Slide a from positive to negative. Watch the curve rotate 180° about the origin rather than mirror across an axis.
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Recipe Classify a function as even, odd, or neither Use when: you're given a formula for f and need to know whether f(-x) equals f(x), equals -f(x), or matches neither Goal: a settled even, odd, or neither label, backed by an identity that holds for every x, not a check at one value
Quick recipe
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1
Look at whether f is built entirely as a sum of pure powers of x, meaning a polynomial with nothing else mixed in.
Example
f(x) = x^4+2x^2: both powers (4 and 2) are even, so the shortcut predicts even. h(x)=x^3+x^2: the powers 3 and 2 are mixed, so the shortcut already says neither. g(x)=x\cos x: \cos x isn't a pure power of x, so the shortcut doesn't apply, and Step 2 is needed.
- every term has an even power, including a bare constant — the shortcut predicts even; go to Step 3 to confirm.
- every term has an odd power, with no constant term — the shortcut predicts odd; go to Step 3 to confirm.
- the powers are mixed, some even and some odd — the shortcut already says neither; go to Step 3 to confirm.
- f isn't a sum of pure powers of x at all, such as \sin x, |x|, a product, or a piecewise rule — the shortcut doesn't apply; continue to Step 2.
Why: the power pattern is a genuine shortcut only for a sum of pure powers of x; it says nothing about a product or a transcendental function like cosine.
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2
Replace every x in the formula with -x, then simplify with known identities, such as cosine being even, until nothing further reduces.
Example
g(-x) = (-x)\cos(-x) = (-x)(\cos x) = -x\cos x.
Why: f(-x) is the one quantity that actually decides even or odd; no shortcut replaces computing it once f isn't a sum of pure powers.
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3
Match what Step 2 produced against the original f(x) and against its negative, -f(x); whichever one it equals settles the label.
Example
g(-x) = -x\cos x = -g(x) exactly, so g is odd. Checking f(x)=x^4+2x^2 directly confirms Step 1's shortcut: f(-x) = (-x)^4+2(-x)^2 = x^4+2x^2 = f(x).
- the result matches f(x) exactly — f is even.
- the result matches -f(x) exactly — f is odd.
- the result matches neither — f is neither even nor odd.
Why: even and odd are properties of the whole function, so only a match holding for every x counts, not a coincidence at one value.
Common mistakes
- Trusting the power-pattern shortcut on a function that isn't a pure sum of powers of x, such as \sin x, |x|, or a piecewise rule.
- Letting one matching term stand in for the whole check, when a single mismatched term breaks the symmetry outright.
- Checking f(-x) against only f(x) or only -f(x), and stopping before ruling out the other.
- Concluding a pattern from one numeric value of x instead of confirming the identity holds in general.
Recipe Find the fundamental period of a sum of periodic terms Use when: you have a sum of two or more periodic terms, each with its own known period, and need the period of the whole combination Goal: the smallest positive number that is a common period of every term, found without graphing
Quick recipe
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1
For a term like \cos(bx) or \sin(bx), the period is \frac{2\pi}{|b|}; for any other periodic piece, use its own known period directly.
Example
\sin(4x) has b=4, so its period is \frac{2\pi}{4}=\frac{\pi}{2}. \cos(6x) has b=6, so its period is \frac{2\pi}{6}=\frac{\pi}{3}.
Why: the sum's own period can only be built from the periods already present in each individual term.
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2
Express each period as a number of units of \pi, so the periods become plain numbers ready to compare.
Example
\frac{\pi}{2} = \frac12\cdot\pi and \frac{\pi}{3}=\frac13\cdot\pi, so the numbers to compare, in units of \pi, are \frac12 and \frac13.
Why: comparing how many times one period fits into a candidate only makes sense once every period is measured in the same unit.
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3
Find the smallest positive number that is a whole-number multiple of every individual period at once.
Example
\frac12's multiples are \frac12,1,\frac32,2,\dots and \frac13's multiples are \frac13,\frac23,1,\frac43,\dots; the smallest number on both lists is 1, so the period is 1\cdot\pi=\pi.
- every period is a rational multiple of the common unit — a smallest common multiple exists, and it's the fundamental period.
- the ratio between two periods is irrational — no common period exists, and the sum isn't periodic at all.
Why: a shared period has to complete a whole number of cycles of every single term, or some term would be left mid-cycle.
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4
Check that nothing smaller than the candidate from Step 3 is also a whole-number multiple of every individual period.
Example
No number smaller than 1 (in units of \pi) is a whole multiple of both \frac12 and \frac13, so \pi is genuinely the smallest, not just a common one.
Why: a common period always has infinitely many multiples that also work, but the fundamental period is specifically the smallest positive one, by definition.
Common mistakes
- Adding the two periods together instead of finding their smallest common multiple.
- Assuming the sum's period always equals the smaller of the individual periods.
- Finding a common multiple but skipping the check for a smaller one that also works.
- Assuming every sum of periodic functions is itself periodic, when an irrational ratio of periods rules that out.
Worked example
Classifying by the power pattern
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1
Classify f(x) = x^4 - 3x^2 as even, odd, or neither. This runs Steps 1 and 3 of the recipe above, the power-pattern shortcut followed by its confirmation, on a fresh function.
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2
Every term's power of x is even (4 and 2), so the power-pattern shortcut predicts even. Confirm directly: f(-x) = (-x)^4 - 3(-x)^2 = x^4 - 3x^2 = f(x).
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3
\boxed{f \text{ is even}}. Its graph mirrors across the y-axis.
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Worked example
A product of two symmetries
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1
Classify g(x) = \sin(x)\cos(x). \sin is odd and \cos is even, and neither is a pure power of x. So the power-pattern shortcut doesn't apply at all here, the last case in Step 1 of the recipe above. The product has to be checked directly, Steps 2 and 3.
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2g(-x) = \sin(-x)\cos(-x) = (-\sin x)(\cos x) = -\sin(x)\cos(x) = -g(x).
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3
\boxed{g \text{ is odd}}. An odd function times an even function is always odd, the same way a negative times a positive is always negative.
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Worked example
The power pattern misleads
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1
Classify h(x) = x^2 + x.
Common trap. x^2 alone would be even, and it's tempting to stop there. But h has a second term, x, with an odd power (the mixed-power case from Step 1 of the recipe above).
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2h(-x) = (-x)^2 + (-x) = x^2 - x.
This matches neither h(x) = x^2+x nor -h(x) = -x^2-x.
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3
\boxed{h \text{ is neither even nor odd}}. One mismatched term is enough to break both symmetries at once.
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Worked example
Finding a fundamental period from two frequencies
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1
Find the fundamental period of k(x) = \cos(3x) + \sin(2x), following the period-finding recipe above.
Recognition. The two terms have different fundamental periods on their own. \cos(3x) repeats every \frac{2\pi}{3}, and \sin(2x) repeats every \frac{2\pi}{2}=\pi. So the sum's period is the smallest number both of those divide evenly, not either one alone.
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2
\frac{2\pi}{3} and \pi both divide 2\pi evenly (3\times\frac{2\pi}{3} and 2\times\pi), and no smaller positive number works for both.
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3
\boxed{P = 2\pi}. This is exactly the situation a Fourier series is built from: a whole family of frequencies, each a whole-number multiple of one fundamental frequency, whose sum repeats at that fundamental's own period.
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Sort by symmetry, not by formula shape
Classify all six graphs as even, odd, or neither, by eye, from the picture alone.
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Where this goes next
Fourier coefficient formulas, chapter 3
Every Fourier coefficient formula in this course is an integral over a symmetric interval like [-\pi,\pi]. The moment a function being expanded is even, every sine coefficient is guaranteed to be an integral of an odd integrand over a symmetric interval. That integral is zero, with no computation needed. An odd function kills every cosine coefficient the same way. Recognizing symmetry before integrating is routinely the difference between computing half as many integrals and computing all of them.
Check your understanding (counts toward your progress)
Classify f(x) = x^3 - 4x.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Every term has an ODD power of x (3 and 1) — that pattern predicts odd, not even.
Both terms share the same (odd) power pattern, so nothing mismatches here — check f(-x) directly and it matches -f(x) exactly.
The power pattern alone settles this for a polynomial, with no graph needed.
Why
f(-x) = (-x)^3 - 4(-x) = -x^3+4x = -(x^3-4x) = -f(x), so f is odd.
Find the fundamental period of h(x) = \cos(4x).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
A general \cos(bx) has fundamental period \frac{2\pi}{b}; here b=4, giving \frac{2\pi}{4}=\frac{\pi}{2}.
Correct answer
Also accepted:
Why
Compare \cos(4x) to the base form \cos(bx) and identify b first.
Classify g(x) = \cos(x) + 1.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
\cos is even, and adding a constant doesn't introduce any odd-symmetry behavior — check g(-x) directly.
\cos(x) is even and the +1 is a constant (an even term, power 0) — nothing here breaks the mirror symmetry.
Why
g(-x) = \cos(-x)+1 = \cos(x)+1 = g(x), since cosine itself is even.
Practice: Classify by symmetry, not by appearance (4 problems · doesn’t affect your progress)
Classify p(x) = x^6 + 2x^2 - 1.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Every power of x present (6, 2, and the constant's power 0) is even, so p(-x)=p(x).
Even
Classify q(x) = x^3 - x^2.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
The powers present are 3 (odd) and 2 (even) — a mix, so neither symmetry test passes: q(-x)=-x^3-x^2 matches neither q(x) nor -q(x).
Neither
Classify r(x) = x\sin(x).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
x is odd and \sin x is odd — odd times odd is EVEN, not odd. Check r(-x) directly.
Both factors are pure odd-symmetry pieces, so their product doesn't land in the 'neither' case.
Why
r(-x) = (-x)\sin(-x) = (-x)(-\sin x) = x\sin x = r(x), so r is even — odd times odd is always even, the same way a negative times a negative is positive.
Even
Find the fundamental period of k(x) = \sin(4x) + \cos(6x).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
\sin(4x) has period \frac{2\pi}{4}=\frac{\pi}{2} and \cos(6x) has period \frac{2\pi}{6}=\frac{\pi}{3}. In units of \pi, that's \frac12 and \frac13; their smallest common multiple is 1, so the fundamental period is \pi.
Correct answer
Also accepted:
Why
Write each term's period as a fraction of \pi, then find the smallest number both fractions divide evenly.