Using Symmetry

What makes an expression symmetric

An expression in two variables is symmetric when swapping the two variables leaves it unchanged. Trade every $x$ for a $y$ and every $y$ for an $x$; if what you get is identical to what you started with, the expression is symmetric.

The sum $x + y$ passes the test, since swapping gives $y + x$, the same number. The product $xy$ passes, since swapping gives $yx$. So does $x^2 + y^2$, and so does $\frac{1}{x} + \frac{1}{y}$ (for nonzero $x$ and $y$), which swaps into $\frac{1}{y} + \frac{1}{x}$. Now test two that fail. The difference $x - y$ swaps into $y - x$, which is the opposite sign, not the same expression, so $x - y$ is not symmetric. Neither is $x^2 y$, which swaps into $y^2 x$; those two agree only for special values of $x$ and $y$, never as expressions.

Here is the leverage hidden in that plain definition. If an expression truly cannot tell $x$ from $y$, then it cannot depend on them one at a time. Such an expression must instead be expressible through quantities that are themselves unmoved by the swap. For two variables there are two such basic building blocks, and every symmetric expression is assembled from them: the sum and the product.

Every symmetric expression is built from the sum and product

Fix two abbreviations once and for all: let $s = x + y$ be the sum and $p = xy$ the product. Both are symmetric, since swapping $x$ and $y$ leaves $x + y$ and $xy$ untouched. The claim of the last section was that every symmetric expression is built from these two, and the way to see it is simply to compute.

Before computing anything, it is worth watching the swap fail to do damage. A rectangle carries both $s$ and $p$ on its face: its area is the product of the two sides, and its perimeter is twice their sum. The figure below reports both for whatever pair of sides you set. A swap of the two sides is therefore a swap of $x$ and $y$ that you can perform and then look at.

Build $3$ wide and $4$ tall. The readout says perimeter $14$, area $12$, which is $2s$ and $p$ for the pair $\{3, 4\}$. Now swap them, to $4$ wide and $3$ tall. The first half of the sentence changes, from ”$3$ units wide and $4$ units tall” to ”$4$ units wide and $3$ units tall”. The picture turns on its side, but the perimeter and the area do not move at all. That is this lesson’s definition happening in front of you. The width on its own is not symmetric, and the swap catches it out immediately. The sum and the product are symmetric, and the swap has no grip on them. Try any other pair and its mirror, $2$ and $7$, or $5$ and $6$. The two readouts hold still under every such swap, while the description of the shape does not.

The same trick reapplies one level up. Setting $a = x^2$ and $b = y^2$ in $a^2 + b^2 = (a + b)^2 - 2ab$ gives $x^4 + y^4 = (x^2 + y^2)^2 - 2(xy)^2$, which becomes $(s^2 - 2p)^2 - 2p^2$ once you substitute the reduction above. Higher powers keep falling this way, each one leaning on the reductions already found.

Turning a symmetric system into one quadratic

The sum and product are useful for evaluating, but their real power shows up when you have to solve. In the sums and products of roots lesson you learned that a monic quadratic $t^2 - st + p = 0$ has two roots that add to $s$ and multiply to $p$. Read that fact backward. If you are handed the sum $s$ and the product $p$ of two unknown numbers, then those two numbers are exactly the roots of $t^2 - st + p = 0$. A symmetric system in $x$ and $y$ has become a single quadratic in $t$.

Look carefully at what that answer does and does not say. Solving for $s$ and $p$ hands back the unordered pair $\{2, 3\}$, and never a label announcing which number is $x$ and which is $y$. That is not a defect of the method. The system you started with, $x + y = 5$ together with $xy = 6$, is itself symmetric: it reads exactly the same after you swap $x$ and $y$. A problem that cannot tell the two apart cannot possibly have an answer that does. Its solutions are therefore forced to come in swapped pairs, here $(2, 3)$ and $(3, 2)$. The method returns precisely what the problem determines, no more and no less.

The same pair $s$ and $p$ also settles, at a glance, whether $x$ and $y$ are real at all. Since $(x - y)^2 = s^2 - 4p$ and the square of a real number is never negative, a real pair exists exactly when $s^2 \ge 4p$. When $s^2 < 4p$ the two numbers still exist, but they form a complex conjugate pair of the kind from the complex numbers chapter. Every worked example and exercise in this lesson keeps $s^2 \ge 4p$, so the pairs stay real.

Often a system is symmetric without handing you $s$ and $p$ directly. The move then is to reduce whatever you are given to $s$ and $p$ first, using the identities from the previous section, and only then build the quadratic.

Pairing a number with its reciprocal

A close cousin of the two-variable story appears with a single variable, whenever a problem pairs $x$ with its reciprocal $\frac{1}{x}$. For any nonzero $x$, set

This is the sum-and-product idea with the product already fixed, because $x \cdot \frac{1}{x} = 1$, so here $p = 1$ automatically. Every power sum in $x$ and $\frac{1}{x}$ therefore reduces to $k$ alone, exactly as the two-variable sums reduced to $s$ and $p$.

Square $k$ and watch the middle term collapse, since $x$ times $\frac{1}{x}$ is $1$:

so subtracting $2$ from each side gives

Cube $k$ and the same cancellation groups the cross terms into another copy of $k$:

which rearranges to

These are the two-variable identities $s^2 - 2p$ and $s^3 - 3ps$ with $p = 1$, so nothing new had to be memorized.

Palindromic equations

The reciprocal pairing is precisely what cracks a whole family of high-degree equations that would otherwise demand machinery you do not have. Look at the coefficients of

Read left to right they are $2, -9, 14, -9, 2$, and read right to left they are the very same list. An equation whose coefficients form a palindrome is called a reciprocal or palindromic equation, and that mirror in the coefficients is a symmetry you can trade on.

The axis of symmetry

Symmetry has been a tool for solving so far. It is also the explanation behind a formula you have leaned on since you first met the quadratic formula. The two roots of $ax^2 + bx + c = 0$ are

The only difference between them is the sign in front of the square root. That means one root sits as far below $\frac{-b}{2a}$ as the other sits above it, so the two roots are mirror images across the vertical line $x = -\frac{b}{2a}$. Their midpoint is the average of the two, and adding them makes the $\pm\sqrt{\cdot}$ cancel outright:

So the line $x = -\frac{b}{2a}$ is the axis of symmetry of the parabola, lying exactly halfway between the roots, and the vertex rides on that line. This is why the vertex sits at $x = -\frac{b}{2a}$: it is not a separate fact to memorize but the midpoint of the two mirror-image roots. The square root measures how far each root strays from that midpoint.

You have used this reflect-and-pair trick once before without naming it. In the arithmetic series lesson, Gauss added $1 + 2 + \cdots + n$ by writing the sum forwards and backwards and pairing each term with its mirror. That pairing works because the first term plus the last, the second plus the second-to-last, and so on all share the same total $n + 1$. Pairing all $n$ terms that way gives $n(n + 1)$ for twice the sum, so the sum is $\frac{n(n + 1)}{2}$. The pairing worked for the same reason the quartic did: the arrangement reads the same forwards and backwards. A reflection that leaves the whole thing unchanged is exactly a symmetry to exploit.

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