Even and odd functions — graphical and algebraic tests: Visual walkthrough

This page proves the parent result's crown jewel — that any function is secretly the sum of a mirror-symmetric part and a rotation-symmetric part — starting from a blank board. We build every symbol before we use it, and every step has a picture. By the end you will see why $cosh$ and $sinh$ had to exist.

We only assume you can plot points $(x, y)$ and add numbers. Everything else we grow here.

Step 1 — What "a function" and "its mirror twin" even look like

WHAT. A function $f$ is a rule: feed it a number $x$ , it hands back a height $f (x)$ . Plot the point $(x, f (x))$ for every input and you get a curve. We will use a lopsided example on purpose — one with no symmetry — so the magic later is real, not staged: $f (x) = x^{2} + x .$ Here $x$ is the input (how far left/right), $x^{2}$ is a bowl-shaped term (always $\geq 0$ ), and the lone $+ x$ is a tilt that leans the curve.

WHY. Before we can split symmetry out of a function, we must first look at the single most important companion of any curve: its reflected input. Define the mirror value $f (- x)$ — the same rule, but fed the opposite-signed input. Geometrically, reading $f$ at $- x$ means walking to the mirror-image position across the vertical axis.

PICTURE. The blue curve is $f (x)$ . The pink curve is $f (- x)$ : literally the blue curve flipped left-to-right across the $y$ -axis. Notice they meet exactly on the $y$ -axis (where $x = 0$ , so $x$ and $- x$ are the same spot).

Figure — Even and odd functions — graphical and algebraic tests

Step 2 — Two pure symmetries: the fold and the spin

WHAT. There are exactly two "clean" curves we want to manufacture:

An even part $E (x)$ that satisfies $E (- x) = E (x)$ — its mirror twin is itself.
An odd part $O (x)$ that satisfies $O (- x) = - O (x)$ — its mirror twin is itself turned upside-down.

WHY. These are the only two ways a curve can be "self-similar under sign flip." Even $=$ the curve equals its own $y$ -axis reflection (fold the paper on the $y$ -axis, both halves match). Odd $=$ the curve equals its own $18 0^{\circ}$ spin about the origin (turn the paper upside down, it lands on itself). Any other behaviour is a mixture — which is the whole point of what follows.

PICTURE. Left: an even curve; the dashed fold line is the $y$ -axis and the two halves kiss. Right: an odd curve with a marked point $(a, b)$ and its partner $(- a, - b)$ — the same point after a half-turn about the origin (the black dot).

Step 3 — The guess: write $f$ as fold-part plus spin-part

WHAT. We hope our lopsided $f$ can be assembled from one even piece and one odd piece: $f (x) = fold-symmetric E (x) + spin-symmetric O (x) .$

WHY. We don't yet know $E$ or $O$ . But we have one equation and two unknown curves — hopeless alone. The trick that unlocks everything: evaluate this same identity at the mirror input $- x$ . That gives a second, independent equation for free, because we already know how $E$ and $O$ react to a sign flip (Step 2).

PICTURE. The stacked bars show, at a fixed input $x$ , the total height $f (x)$ made of a blue even block sitting on top of a pink odd block. Same idea at $- x$ : the even block is identical, the odd block has flipped below the axis.

Term by term, the equation says:

$f (x)$ — the known total height we can measure.
$E (x)$ — the unknown even contribution.
$O (x)$ — the unknown odd contribution.

Step 4 — Feed it the mirror input to get a second equation

WHAT. Substitute $- x$ everywhere in $f (x) = E (x) + O (x)$ : $f (- x) = E (- x) + O (- x) .$ Now apply Step 2's defining properties $E (- x) = E (x)$ and $O (- x) = - O (x)$ : $f (- x) = E (x) - O (x) .$

WHY. This is the crucial move. Reflecting the input costs $E$ nothing (it's even) but flips $O$ 's sign (it's odd). So the same $E$ and $O$ now appear in a new combination — a subtraction instead of an addition. Two equations, two unknowns: solvable.

PICTURE. Side-by-side "recipe cards." Card 1 (at $x$ ): $E + O$ . Card 2 (at $- x$ ): $E - O$ . The even block is copied unchanged; only the odd block's sign toggles.

Annotating the second equation:

$f (- x)$ — measurable height at the mirror spot.
$E (x)$ — unchanged, because even means "mirror does nothing."
$- O (x)$ — the same odd piece, now subtracted, because odd means "mirror flips the sign."

Step 5 — Add the two equations: the even part falls out

WHAT. Stack the two facts and add them:

f (x) f (- x) = E (x) + O (x) = E (x) - O (x)

Adding column by column, the $+ O$ and $- O$ annihilate: $f (x) + f (- x) = 2 E (x) ⟹ E (x) = \frac{f ( x ) + f ( - x )}{2} .$

WHY. Adding cancels the odd part by design — odd contributions come in equal-and-opposite pairs, so their average is zero. What survives the averaging is exactly the fold-symmetric skeleton. The division by $2$ is because each equation contributed one full copy of $E$ .

PICTURE. The blue curve $f (x)$ and pink curve $f (- x)$ are added; their point-by-point average (yellow) is the even curve $E (x)$ — you can see it is perfectly $y$ -axis-symmetric.

Step 6 — Subtract the two equations: the odd part falls out

WHAT. Same two facts, now subtract the second from the first: $f (x) - f (- x) = 2 O (x) ⟹ O (x) = \frac{f ( x ) - f ( - x )}{2} .$

WHY. Subtracting cancels the even part this time — because $E (x) - E (x) = 0$ . The even skeleton is identical in both equations, so their difference erases it, leaving only the piece that changed under the mirror: the odd tilt. Again a factor $2$ appears because $O$ showed up as $+ O$ and $- O$ .

PICTURE. The pink mirror curve is subtracted from the blue original; the half-difference (yellow) is the odd curve $O (x)$ — verify it has the $18 0^{\circ}$ spin symmetry, passing through the origin.

Applied to our lopsided example $f (x) = x^{2} + x$ :

$f (- x) = (- x)^{2} + (- x) = x^{2} - x$ .
$E (x) = \frac{( x ^{2} + x ) + ( x ^{2} - x )}{2} = x^{2}$ (the even bowl).
$O (x) = \frac{( x ^{2} + x ) - ( x ^{2} - x )}{2} = x$ (the odd tilt).
Check: $x^{2} + x = f (x)$ . ✓ The split is the original.

Step 7 — Prove the two manufactured pieces really are even and odd

WHAT. We called them even and odd — but we must verify the formulas actually obey the Step 2 tests, not just hope. Flip the input in each formula: $E (- x) = \frac{f ( - x ) + f ( - ( - x ))}{2} = \frac{f ( - x ) + f ( x )}{2} = E (x) . ✓ even$ $O (- x) = \frac{f ( - x ) - f ( - ( - x ))}{2} = \frac{f ( - x ) - f ( x )}{2} = - \frac{f ( x ) - f ( - x )}{2} = - O (x) . ✓ odd$

WHY. Because $- (- x) = x$ , feeding $- x$ into $E$ just swaps the two terms in the sum — addition doesn't care about order, so $E$ is unchanged (even). In $O$ , swapping the two terms of a subtraction flips the overall sign (odd). This is the proof the guess in Step 3 was legitimate: the pieces exist and behave.

PICTURE. Two mini-panels: $E (- x)$ landing on top of $E (x)$ (arrows show the term-swap that changes nothing), and $O (- x)$ landing on the reflection of $O (x)$ (the term-swap that introduces a minus).

Step 8 — The degenerate cases: nothing breaks

WHAT. Three edge cases must be checked so the reader never hits a surprise.

$f$ already even. Then $f (- x) = f (x)$ , so $O (x) = \frac{f ( x ) - f ( x )}{2} = 0$ . The odd part is the flat zero line — no tilt to extract. Good.
$f$ already odd. Then $f (- x) = - f (x)$ , so $E (x) = \frac{f ( x ) - f ( x )}{2} = 0$ . The even part vanishes.
Value at $x = 0$ . For an odd part, $O (0) = \frac{f ( 0 ) - f ( 0 )}{2} = 0$ always. So every odd function is forced through the origin. If a curve claiming to be odd misses $(0, 0)$ , it isn't odd.

WHY. These aren't lucky accidents — the formulas guarantee them, which is exactly why the decomposition is trustworthy on any symmetric-domain function.

Step 9 — The payoff: $e^{x}$ births $cosh$ and $sinh$

WHAT. Run the machine on the exponential $f (x) = e^{x}$ (see Polynomial functions-free growth; here from Function transformations). Its mirror is $f (- x) = e^{- x}$ : $E (x) = \frac{e ^{x} + e ^{- x}}{2} \equiv cosh (x), O (x) = \frac{e ^{x} - e ^{- x}}{2} \equiv sinh (x) .$

WHY. The Hyperbolic functions are not arbitrary definitions — they are literally the even and odd halves of $e^{x}$ , forced out by Steps 5–6. That is why $cosh (0) = 1$ (even, peaks at the axis) and $sinh (0) = 0$ (odd, through the origin, per Step 8). And of course $cosh x + sinh x = e^{x}$ rebuilds the original.

PICTURE. The blue $e^{x}$ and its pink mirror $e^{- x}$ ; their average is the yellow $cosh$ bowl (even), their half-difference is the yellow $sinh$ S-curve (odd).

The one-picture summary

Everything above, on one board: the lopsided $f = x^{2} + x$ splitting cleanly into its even bowl $x^{2}$ and its odd tilt $x$ , driven by add-the-mirror and subtract-the-mirror.

Recall Feynman retelling — say it like a story

Take any curve. Draw its mirror image across the vertical line. Now you have two curves: the real one and its flipped twin. Average them and you get a curve that looks the same in a mirror — that's the even part; all the sideways lean has been washed out by the averaging. Take half the difference instead and you get a curve that flips upside-down in a mirror — that's the odd part; the symmetric bowl cancelled out, leaving only the lean. Add these two back together and you rebuild the exact original curve, because average-plus-half-the-difference of two things is just the first thing again. That single trick — evaluate at $x$ , evaluate at $- x$ , then add and subtract — is the whole theorem. It's also why $cosh$ and $sinh$ exist: they're nothing but the mirror-average and mirror-difference of $e^{x}$ .

Recall

Why does adding $f (x) + f (- x)$ give the even part? ::: The odd contributions are equal-and-opposite, so their sum is zero; only the fold-symmetric part survives (halved). Why must every odd function pass through the origin? ::: $O (0) = \frac{f ( 0 ) - f ( 0 )}{2} = 0$ , always. What are $cosh$ and $sinh$ , really? ::: The even and odd parts of $e^{x}$ . When is the split impossible? ::: When the domain is not symmetric, so $f (- x)$ is undefined.

Related uses of this decomposition: Symmetry in calculus (odd integrals over $[- a, a]$ vanish) and Fourier series (even $\to$ cosines only, odd $\to$ sines only), with $cos, sin$ themselves being the even/odd stars from Trigonometric functions.