Polynomial Graphs and Symmetry
Geoff Goehle
Mitsuo Kobayashi
April 8, 2012
When is 7 even?
When it’s a function, of course.
As is well-known, a function f which is symmetric with respect to the
y-axis, is called even and satisfies f (x) = f (−x). Similarly, an odd function
is rotationally symmetric about the origin and satisfies f (x) = −f (−x). For
polynomials, there is an easy way to tell the difference: even polynomials
only have even power terms; odd polynomials have odd power terms. A
polynomial with a mix of terms, as is typical, is neither even nor odd. This
is not the whole story, however.
The polynomial on the left in Figure 1 looks even, but isn’t—because its
axis of symmetry is shifted away from the y-axis. Similarly, the polynomial
on the right looks odd, but it is also in the wrong place. The issue is that
the definition of even and odd, quite natural from an algebraic point of
view, geometrically only detects graphs having symmetry with respect to
1
5
1
4
3
-3
-2
-1
1
2
-1
1
-2
-1
2
3
-3
1
-1
-4
Figure 1: Polynomials with existing symmetry.
the origin. How can we tell when a polynomial has symmetry even when it’s
not centered correctly? If we build a polynomial from scratch, say
f (x) = 5x5 + 4x4 + 3x3 + 2x2 + x,
how can we tell if its graph has symmetry?
Call a polynomial f even at s if it has reflection symmetry about the line
x = s, and odd at s if it has rotation symmetry about the point (s, f (s)). In
either case we will call the x-coordinate s of the line or point the symmetric
center. Knowing where this line or point of symmetry is, we can translate
the function to the origin and apply the usual tests.
We show that there is a way to read off a polynomial’s symmetric center
from its coefficients. Surprisingly, only the first two coefficients are needed!
As soon as you see 5x5 + 4x4 + · · · , you can stop reading and say “If it’s
symmetric, it must be odd at...”
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Low Degree
A polynomial of degree zero is a constant and, as we noted at the outset, a
constant is even—even, in fact, at any value of x. Moving on to degree one,
such graphs are lines with non-zero slope. These have an odd symmetric
center at any x-value. What about quadratics? It is well known that every
quadratic f (x) = ax2 + bx + c is symmetric about its axis of symmetry; its
symmetric center is s = −b/(2a).
It is not as well known, but true, that every cubic is rotationally symmetric [1]. Once this is pointed out, it is easy to see where the symmetric
center must lie. A cubic must be rotationally symmetric about its point of
inflection, where the concavity changes.
We can see graphically that all cubics are odd with respect to their
inflection points. First note that the inflection point of a cubic f (x) =
ax3 + bx2 + cx + d occurs at s = −b/(3a). Its derivative is the quadratic
f 0 (x) = 3ax2 + 2bx + c, which is symmetric at exactly the same point. By
the fundamental theorem of calculus, we recover f by integrating f 0 in either direction starting from s. As we see in Figure 2, that f 0 (x) has even
symmetry about x = s implies that integrating in opposite directions from
s yields equal values, opposite in sign, so f is odd at s. To summarize, the
even symmetry of f 0 at s produces odd symmetry in f at the same value.
Returning to the quadratic case, we retroactively consider its derivative,
the line, as odd about its point of intersection with the x-axis. This fact
3
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
1
2
3
4
Figure 2: The graph of a potential f 0 (x). The area between s + t and s is
equal to the area between s and s − t.
gives the parabola its symmetry.
It is tempting to hope that we can continue this process to show that
all polynomials have symmetry; however, a quick look at some graphs of
quartics shows that this is far from true. Many quartic polynomials do not
have symmetry. What goes wrong?
Consider f (x) = x4 /4 + x, which has f 0 (x) = x3 + 1 as its derivative.
We know that f 0 is odd with respect to its inflection point (0, 1). For f to
inherit the symmetry of f 0 and so be even with respect to x = 0 we need
f (x) = f (−x). That is, we need to get the same value when we integrate
f 0 (x) from x = 0 in either direction. However, it is clear from the graph of
f 0 (x) in Figure 3a that this is not the case.
How can we fix this? To make the areas equal, the derivative must have
odd symmetry somewhere on the x-axis, as in Figure 3b. Any quartic polynomial with the latter graph as its derivative will be even with respect to
their shared centers.
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2.5
1.0
2.0
0.5
1.5
1.0
- 1.5
- 1.0
- 0.5
0.5
1.0
1.5
0.5
- 0.5
- 1.5
- 1.0
- 0.5
0.5
1.0
1.5
- 1.0
- 0.5
(a)
(b)
Figure 3
A General Polynomial
For any polynomial, if f is even at s, then the slopes of the graph of f have
opposite sign about s, hence f 0 is odd at s. Likewise, if f is odd at s, f 0 is
even there. Reversing the process by integrating, however, requires care: an
f even at s does have an integral odd at s, but if f is odd at s, its integral
is even at s as long as f (s) = 0.
One of the more interesting consequences of the above relationship is that
for any given polynomial there is only one possible point of symmetry. Let
f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 with n > 1 be symmetric with center
s. Observe that if n is even (as a number, not a function!) then n − 1 is odd
and vice versa. In either case the previous paragraph implies f (n−1) (s) = 0,
whence
0 = f (n−1) (s) = n!an s + (n − 1)!an−1 .
We conclude that s = −an−1 /(nan ) and the symmetric center of f depends
only on the first two coefficients, as we claimed.
In [2], the symmetric center s is called the center of gravity of a poly5
nomial. This terminology derives from the observation that the symmetric
center is in fact the mean value of the roots (real and complex) of f . If the
roots on the complex plane are interpreted as point masses, multiple roots
weighted accordingly, the symmetric center is where a pivot must be placed
to balance the plane.
Now that we know where the center of symmetry must be, we can shift
the polynomial to the origin and use the usual polynomial tests for symmetry.
In lieu of an actual shift, we can Taylor-expand about s. If we find
f (x) = bn (x − s)n + bn−1 (x − s)n−1 + · · · + b0 ,
then f (x + s) is even exactly when bi = 0 for odd i. Since bi = f (i) (s)/i!, we
need only check that f (i) (s) = 0 for all odd i.
The case when f is odd is similar. We check that f (i) (s) = 0 for even i,
except that we do not require b0 to be zero (since our generalized symmetry
allows for vertical shifts). Thus, we have the following theorem.
Theorem. If f is a polynomial with even degree n > 1 then f will be even
at s if and only if f (i) (s) = 0 for all odd i. Similarly, if f is a polynomial
with odd degree n > 1 then f will be odd at s if and only if f (i) (s) = 0 for all
even i ≥ 2.
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The Ancillary Polynomial
In view of the theorem, if a degree n polynomial is not symmetric, we can
blame certain coefficients of its Taylor expansion about s for not being zero,
namely the (n − j)th coefficients for 1 ≤ j ≤ n − 1, j odd. This amounts to
saying that the polynomial
n−1
X
f (n−j) (s)
(x − s)n−j
p(x) =
(n − j)!
j=1
j odd
is nonzero. We call p the ancillary polynomial of f . Not only does subtracting
off the ancillary polynomial p make f symmetric, but the degree of p measures
the asymmetry of f . We define the degree of asymmetry of f to be the degree
of its ancillary polynomial, or in terms of derivatives, the largest m such that
f (m) (x) has odd degree and is nonzero at the symmetric center of f .
What is the highest possible degree of asymmetry for a polynomial of
degree n? Surprisingly, it is not n − 1 because, as we have seen, the center s
of f is the x-intercept of the linear (n−1)-st derivative. This implies that the
largest index for which a derivative could have odd degree and be nonzero
at s is the cubic (n − 3)-rd derivative. Thus the degree of asymmetry of a
polynomial f of degree n can be no larger than n − 3.
To justify our use of the ancillary polynomial p to measure the asymmetry
of f we show that p is minimal in the sense that p is the polynomial with
the smallest degree one can subtract from the original polynomial in order
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to make it symmetric. Formally we show that if f is a polynomial of degree
n > 3, q is a polynomial such that f − q is symmetric, and p is the ancillary
polynomial of f , then the degree of q is greater than or equal to the degree
of p.
Let n be the degree of f and let s be the symmetric center of f − q. Then
s is the (unique) zero of the (n − 1)st derivative of f − q. However, if q has
degree less than n − 3 then
0 = (f − q)(n−1) (s) = f (n−1) (s) − 0 = f (n−1) (s).
Hence, s is also the symmetric center of f . Let d be the degree of asymmetry
of f . Since, by definition, the dth derivative of f has odd degree, it follows
from our theorem that (f − q)(d) (s) = 0. Consequently
q (d) (s) = f (d) (s) = p(d) (s).
Now d is the degree of p so p(d) (s) 6= 0 which implies the degree of q is at
least the degree of p.
Our discussion suggests that asymmetric quartics are minimally so. They
have the mildest possible asymmetry, namely such that can be repaired by
subtracting a line.
8
Symmetry and Arbitrary Functions
Our original motivation was that the usual notion of even and odd did not
take into account the natural symmetries often present in polynomials. Of
course, many other functions possess some sort of symmetry and there is
nothing preventing us from defining symmetries at a point for them. We
only really run into a problem when we try to pin down the symmetric
center. Even the class of analytic functions is too varied to have a consistent
symmetric center. For instance, trigonometric functions like sine and cosine
have infinitely many points of symmetry. On the other hand, every symmetric
center of an analytic function f has to be a root of either the first or second
derivative of f . Since a non-zero analytic function can have at most countably
many roots we may infer that any non-linear analytic function has at most
countably many points of symmetry.
Now, given a specific s we can extend our theorem to include power series.
We start with the usual decomposition of a function f into the sum of an
even and an odd function: f = fe + fo where
fe (x) =
f (x) + f (2s − x)
− f (s)
2
fo (x) =
f (x) − f (2s − x)
+ f (s).
2
and
Basic computations show that fe is even with respect to s, fo is odd with
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respect to s, and fo (s) = f (s). Furthermore, it is not hard to show that
this decomposition is unique. Because of this uniqueness a function f will
be even with respect to s if and only if fo (x) = f (s) for all x and f will be
odd with respect to s if and only if fe = 0.
Combined with the machinery of Taylor series these facts allow us to
extend our theorem. Let s be a real number, f an analytic function, and I
the open interval of convergence for the Taylor series of f at s. Since f is
analytic,
f (x) =
∞
X
f (n) (s)
n=0
n!
(x − s)n ,
for all x ∈ I. Using the decomposition f (x) = fe (x) + fo (x) we know that f
is even if and only if fo (x) = f (s). A relatively painless computation shows
that
fo (x) =
∞
X
f (2j+1) (s)
j=0
(2j + 1)!
(x − s)2j+1 + f (s)
and clearly fo (x) = f (s) if and only if f (2j+1) (s) = 0 for all j ≥ 0. Thus f
is even with respect to s on I if and only if f (i) (s) = 0 for all odd i ≥ 1.
Similar arguments show that f is odd with respect to s on I if and only if
f (i) (s) = 0 for all even i ≥ 2.
This result allows us to use the Taylor expansion of a function to search
for symmetry, or measure the lack thereof. For example, we know that the
exponential function is not symmetric about any point. How asymmetric
P
es
n
is it? Recall that the Taylor expansion of ex about s is ∞
n=0 n! (x − s) .
Regardless of the values of s and n, none of the coefficients es /n! are zero.
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Thus, its “ancillary polynomial” has infinite degree! In this sense ex is as
asymmetric as possible for an analytic function.
Put together, our tools allow us to analyze the even and odd symmetries
in any analytic function—even 7.
Summary. Most quadratic functions are not even, but every parabola
has symmetry with respect to some vertical line. Similarly, every cubic has
rotational symmetry with respect to some point, though most cubics are not
odd. We show that every polynomial has at most one point of symmetry
and give conditions under which the polynomial has rotational or horizontal
symmetry with respect to that point.
References
[1] Sheldon P. Gordon, On symmetries of polynomials, PRIMUS 9 (1999),
no. 1, 13–20.
[2] Frank Irwin and H. N. Wright, Some properties of polynomial curves,
Annals of Mathematics 19 (1917), no. 2, 152–158.
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