探究对称与不对称

Choose the Middle Term!

When looking at arithmetic sequences, why look at them as $a$, $a+d$, $a+2d$, $a+3d$, $a+4d$? Why not look at them as $a-2d$, $a-d$, $a$, $a+d$, $a+2d$? Symmetry makes nice cancellations occur!

Similarly, when looking at geometric sequences why look at them as $a$, $ar$, $ar^2$, $ar^3$, $ar^4$? Why not look at them as $\frac{a}{r^2}$, $\frac{a}r$, $a$, $ar$, $ar^2$?

Problem (ARML 2011): The product of the first five terms of a geometric sequence is $32$. If the fourth term is $17$, compute the second term.

Solution: Let the terms be $\frac{a}{r^2}$, $\frac{a}r$, $a$, $ar$, $ar^2$. Then $\frac{a}{r^2}\frac{a}r(a)(ar)(ar^2)=32$ so $a^5=32$ so $a=2$. Then $17=ar=2r$ so $r=\frac{17}2$. Therefore, the second term is just $\frac{2}{17/2}=\frac4{17}$.

Problem (ARML 2012): The four zeros of $x^4+jx^2+kx+225$ are distinct real numbers in arithmetic progression. Compute the value of $j$.

Solution: This trick doesn't only apply to an odd number of terms! Take $a-\frac32d$, $a-\frac12d$, $a+\frac12d$, $a+\frac14d$. By Vieta's formulas, the roots sum to $0$ so $a=0$! Therefore the roots are just $-\frac32d$, $-\frac12d$, $\frac12d$, $\frac32d$, but their product must be

$$\frac{3^2}{2^4}d^4=225=15^2\implies d=\pm2\sqrt5$$

We can assume $d=2\sqrt5$ by symmetry, so the roots are $-3\sqrt5$, $-\sqrt5$, $\sqrt5$, and $3\sqrt5$. We can easily find $j$ from here.

Problem (104 Number Theory Problems by Titu Andreescu): Give an example of $11$ consecutive positive integers the sum of whose squares is a perfect square.

Solution: Consider $a-5d$, $a-4d$, $\dots$, $a+4d$, $a+5d$:

$$\begin{align*}\sum_{i=-5}^5(a+id)^2&=a^2+\sum_{i=1}^5[(a+id)^2+(a-id)^2]\\&=a^2+\sum_{i=1}^5[2a^2+2i^2d^2]\\&=11a^2+110d^2=11(a^2+10d^2)\end{align*}$$

So $11\mid a^2+10d^2$, therefore $a^2\equiv d^2\pmod{11}$ or $(a-d)(a+d)\equiv0\pmod{11}$. This simplifies the problem vastly.

This technique helps on Olympiads too! Don't believe me?

Problem (Dutch Mathematical Olympiad 1998): Find all real solutions to the following equation:

$$(x+1995)(x+1997)(x+1999)(x+2001)+16=0$$

Solution: Take $u=x+1998$!

$$0=(u-3)(u-1)(u+1)(u+3)+16=(u^2-1)(u^2-9)+16$$

which is a quadratic in $u^2$. We can solve this easily!

Exploiting symmetry in polynomial coefficients

(Note: I believe the title refers to one-variable polynomials, since I clicked on that category to get here. If someone wants to post on multi-variable polynomials I guess they can)

There are numerous kinds of symmetry that sometimes occur in polynomials. But firrst let me go on a slightly irrelevant tangent.

Let $P(x) = x^2 - 4x+3, Q(x) = 3x^2-4x+1$. Then the roots of $P$ are $1,3$ and the roots of $Q$ are $1, \dfrac{1}{3}$

In general, let $P(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_0$ for $a_0 \neq 0$. Let $Q(x) = a_0x^n + a_{1}x^{n-1} + ... + a_n$. We claim the roots of $P,Q$ are recipricols of each other.

This is because, by Vieta's Formulas, if $r_1,r_2,..r_n$ are the roots of $P$, and $s_1, s_2, ..s_n$ are the roots of $Q$, then we have some equations:

$r_1r_2...r_n = (-1)^n \dfrac{a_0}{a_n}$ which is the recipricol of $(-1)^n \dfrac{a_n}{a_0} = s_1s_2...s_n$. We can pair up similar equations to get the desired result.

Now back to the question... Suppose we have a symmetric polynomial- that is, $a_0=a_n,a_1=a_{n-1}$, etc.

Let's take a larger symmetric polynomial for a more instructive example:

$x^6 + 2x^4 + 3x^3 + 2x^2 + 1=0$. Dividing both sides by $x^3$ yields

$\left( x^3 + \dfrac{1}{x^3} \right) + 2 \left( x + \dfrac{1}{x} \right) + 3 = 0$

Then we let $x + \frac{1}{x} = t$, and the equation is equivalent with

$t^3 - 3t + 2t + 3 = 0$ or $t^3-t+3=0$. Although $t$ is not a nice number, this method allows one to reduce the degree of polynomials by a lot.

Another sidenote: We can prove that $x^n + \dfrac{1}{x^n}$ is a polynomial in $x + \dfrac{1}{x}$.

Note that if $b_n = x^n + \dfrac{1}{x^n}$ then we have a recursion:

$b_n = b_{n-1}b_1 - b_{n-2}$.

Thus, we may use induction. Our base cases $n=0,1$ are easy. Now for $n \ge 2$, since $b_{n-1}, b_1, b_{n-2}$ are each polynomials in $b_1$, any product or sum of them will be too, yielding that $b_n$ is also a polynomial, which gives us the desired.

Now, I will do the exact reverse process and see what hapens...

Let's take the equation $x^3 - 3x +1 = 0$ and do the reverse: Let $x = w + \dfrac{1}{w}$ for a possibly complex value of $w$. Then our equation becomes

$w^6+w^3+1=0$

This is very easy to solve! It becomes $\dfrac{x^9-1}{x^3-1} = 0$, which I'll leave up to the reader to solve.

Why is this equation easy to solve? It's because saying $x = w + \dfrac{1}{w}$ is really saying $x = 2 \cos \theta$, given that $|w|=1$.

And guess what happens when we plug in $x = 2 \cos \theta$ into the equation? We get

$2(4 \cos^3 \theta - 3 \theta) = -1$, or, by the triple-angle formula, $2 \cos 3\theta = -1$, which is very easy to solve! Thus, whenever you see an equation that looks like a trig identity, the substitution described may help a lot.

Elementary Symmetric Sums

The elementary symmetric sums of a set of numbers, $S=\{a, b, c, d, \cdots\}$, is defined as follows: $$S_1=a+b+c+d+\cdots,$$ $$S_2 = ab+ac+ad+bc+bd+cd+\cdots,$$ $$S_3 = abc+abd+bcd+acd+\cdots,$$ $$\cdots$$ where the $S_n$ denotes the $n$-th symmetric sum and is the sum of the product of the elements of $S$ when taken $n$ at a time. The $n$-th symmetric sum of a function $P(x)$ is represented by $\sum_{\text{sym}}^nP(x)$

Here is an example of all the elementary symmetric sums of the set $A= \{1,2,3\}$:

$1^{\text{st}}$ Symmetric Sum: $A_1=1+2+3=6$

$2^{\text{nd}}$ Symmetric Sum: $A_2=1\cdot2+1\cdot3+2\cdot3=11$

$3^{\text{rd}}$ Symmetric Sum: $A_3=1\cdot2\cdot3=6$

We took the sum of the elements to get $A_1$, added all the possible combinations of the set when two of the three elements were multiplied, and took the product all of the elements to get $A_3$.

Now moving on...

Application in Vieta's Formulas

Given a polynomial $G(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+ax+a_0$ with roots $r_1, r_2, r_3, \cdots, r_n$, Vieta's Formulas tells us the following information:

$$-\dfrac{a_{n-1}}{a_n}=r_1+r_2+r_3+\cdots+r_n,$$ $$\dfrac{a_{n-2}}{a_n}=r_1r_2+r_1r_3+r_2r_3+\cdots+r_{n-1}r_n,$$ $$-\dfrac{a_{n-3}}{a_n}=r_1r_2r_3+r_1r_2r_4+\cdots+r_{n-2}r_{n-1}r_n,$$

etc. The RHS of each equation is an elementary symmetric sum! Therefore, $G_1=-\dfrac{a_{n-1}}{a_n}$, $G_2=\dfrac{a_{n-2}}{a_n}$, ..., $G_k=\dfrac{a_{n-k}}{a_n}$. Hence, if $a_n=1$ (monic polynomial), then the coefficient of $x^{n-j}$ is $G_j$.

Some Uses

A symmetric sum (different from elementary symmetric sums) is a sum where all the permutations of the numbers/variables does not change the final value. Some examples include $x+y+z$ and $x^2+y^2+z^2$.

Every symmetric sum can be written as a function of elementary symmetric sums. Example: $x^2+y^2+z^2=(x+y+z)^2-2(xy+xz+yz)=S_1^2-2S_2$.

Elementary symmetric sums can also evaluate power sums like $x^3+y^3+z^3$ with Vieta's formulas, and can be tied to Newton's Sums.

Symmetry is often an extremely useful tool. For example, suppose you have a rectangle, and you draw a diagonal splitting the rectangle into two triangles. The two triangles are congruent, so their areas are equal. That means both of the triangles have half the area of the rectangle! (This is basically how you find the area of a triangle in terms of base and height.)