Download Irrationality of ratios of solutions to tanx = x and related matter

Irrationality of ratios of solutions to tan x = x and
related matter
K. Bruce Erickson
April 4, 2008
1
A simple result
Consider the equation
(1)
tan x = x
One easily verifies that this equation has countably many isolated positive solutions.
Theorem 1. Each solution to (1) is transcendental and the ratio of any two distinct
positive solutions is irrational.
Before proving the theorem, we need some
2
Algebraic preliminaries
Suppose x and y both solve (1). In order to be able to rule out a relation of the form
nx = my we will need certain identities for tan nx and tan my.
Define two sequences of polynomials in the variable t as follows:
P0 = 0, Q0 = 1 and for n ≥ 1 :
(2a)
Pn = Pn−1 + t Qn−1 ,
(2b)
Qn = − t Pn−1 + Qn−1
1
Thus P1 (t) = t, Q1(t) = 1, P2 (t) = 2t, Q2(t) = −t2 + 1, P3 (t) = −t3 + 3t, Q3(t) =
−3t2 + 1, P4 (t) = −4t3 + 4t, Q4 (t) = t4 − 6t2 + 1, etc.
For real t a simple induction argument shows that:
(3a)
(3b)
i
(1 + it)n − (1 − it)n = =(1 + it)n
2
1
Qn (t) = (1 + it)n + (1 − it)n = <(1 + it)n
2
Pn (t) = −
and the following facts are quickly established from the definition and formulas (3):
(i) Pn (0) = 0 and Qn (0) = 1. The number 0 is a simple zero of Pn (t), n ≥ 1.
(ii) These polynomials have real integer coefficients.
(iii) Qn (t) has only even powers of t and Pn (t) has only odd powers of t.
(iv) For k ≥ 1 deg(Q2k+1) = deg(Q2k ) = 2k and deg(P2k ) = deg(P2k−1 ) = 2k − 1.
(v) Pn (t) and Qn (t), have no zeros in common.
Pn (tan z)
for all complex z
Qn (tan z)
(including z for which either | tan z| = ∞ or | tan nz| = ∞, or Q(tan z) = 0, provided
one interprets the identity as a limit at such points.)
Lemma 1. Let Pn and Qn be as above. Then tan nz =
This is easily proved using induction and the formula
i tan nz =
einz − e−inz
.
einz + e−inz
Corollary 1. For n ≥ 1, the zeros of Pn (t) are tan(jπ/n), j = 0, 1, . . . , n−1, with the
exception that j 6= n/2 when n is even. The zeros of Qn (t) are tan((2j+1)π/2n), j =
0, 1, . . . , n − 1 with the exception that j 6= (n − 1)/2 in the case that n is odd.
Lemma 2. Let n 6= m. Then for any number r 6= 0, the rational function
(4)
t 7→
Pn (t)
Pm (rt)
−
Qn (t) Qm (rt)
is non-constant and has algebraic (rational) coefficients if r is algebraic (rational).
2
We may assume 1 ≤ m < n and that r 6= 0. Suppose, to the contrary, that the
function on the right side of (4) equals a constant for all t (excluding the zeros of the
denominators.) Then that constant must be 0 because the P s vanish at t = 0. It
follows that
(5)
Qm (rt)Pn (t) = Pm (rt)Qn (t)
for all complex t
Case 1. m ≤ n − 2. Then deg(Qm (rt)) ≤ deg(Qn (t)) − 1 and deg(Pm (rt)) ≤
deg(Pn (t)) − 1. Therefore the polynomials Pn (t) and Qn (t) must have a common
non-constant polynomial factor. (See [3], §5.8.) But this is ruled out by item (v).
Case 2. m = n − 1. From (5) we have
(6)
Qn−1 (rt)Pn (t) = Pn−1 (rt)Qn (t) for all t
Now either deg(Qn−1 (rt)) = n−2, deg(Qn (t)) = n, deg(Pn (t)) = n−1 = deg(Pn−1 (rt))
(n even), or else the same degree formulas are true but with the P s and Qs interchanged (n odd). In either case it follows that one side of (6) is a polynomial of
degree 2n − 3 in t and the other is a polynomial of degree 2n − 1 in t. This is a
contradiction and implies that (5) is once again impossible.
3
Proof of Theorem 1
The transcendence of the non-zero solutions is a consequence of Lindemann’s theorem.
Let x and y be distinct positive solutions to (1). Suppose that nx = my where
m and n are relatively prime positive integers. By Lemma 1
Pm (y)
Pn (x)
= tan nx = tan my =
Qn (x)
Qm (y)
Replacing y by nx/m it follows that
(7)
Pn (x)
Pm (nx/m)
−
=0
Qn (x) Qm (nx/m)
for this particular x. However, the rational function on the left has rational coefficients
and is not identically constant. On the other hand x is transcendental so it cannot
3
be a zero of a non-constant rational function with rational coefficients. Therefore (7)
is impossible. The resolution of this contradiction is the conclusion that y/x must be
irrational.N
Corollary. Let x, y solve (1) with 0 < x < y. Then at least one of the numbers y/x
or y − x is transcendental.
Put b = y − x, c = y/x, so b = (c − 1)x. By Theorem 1, c must be irrational. If
c is an algebraic number, then b must be transcendental because x is transcendental.
N
4
An odd result
Proposition. Suppose x, y, z are distinct positive solutions to (1) such that
(8)
kz + nx + my = 0
for some non-zero integers k, n, and m; then the ratios of any two of x, y, z must be
transcendental. More generally, a relationship of the form y = f(x), y = g(z), x =
u(y), etc. where f, g, u, etc., is a polynomial with algebraic coefficients, is impossible.
Given kz = −nx − my it follows that
−P (x) −P (y) n
m
+
Pk (−nx − my)
Pn (x)Qm (y) + Pm (y)Qn(x)
Qn (x)
Qm(y)
=
=
−Pn (x) −Pm (y)
Qk (−nx − my)
Pn (x)Pm (y) − Qn (x)Qm (y)
1−
Qn (x)
Qm (y)
Which implies
H(x, y) = 0
for some polynomial in x, y with integer coefficients. Suppose that y = f(x) for some
polynomial f with algebraic coefficients. Then
H(x, f(x)) = 0
In other words, x is a zero of a polynomial with algebraic coefficents. But this is
impossible because x is transcendental.N
4
5
A generalization
One way to generalize Theorem 1 is to put a function on the righthand side of (1)
thus:
(9)
tan x = h(x)
Theorem 2. Suppose that h(x) is a non-constant rational function with algebraic
coefficients such that h(x) is real valued (exculding poles) for real x, then equation
(9) has countably infinitely many positive isolated solutions. Each non-zero solution
is transcendental. If x and y are any two distinct positive solutions, then y/x is
irrational.
Here is a rapid proof of only the last assertion. If nx = my for some relatively
prime integers n and m, then
Pm (h(y))
Pn (h(x))
= tan my = tan nx =
Qm (h(y))
Qn (h(x))
so that
(10)
Pn (h(x))Qm (h(nx/m)) − Pm (h(nx/m))Qn (h(x)) = 0
The left hand side of this equation is a non-constant rational function of x with
algebraic coefficients. However x is a transcendental number. It follows that the
lefthand side of the last equation cannot vanish. This contradiction is again resolved
by the conclusion that y/x must be irrational.
6
Some Conjectures
.
Conjecture 1. The last part of the conclusion of Theorem 1 can be strengthened
to the statement that y/x is transcendental.
Conjecture 2. Let x1, x2, . . . , x` be any collection of distinct positive roots of the
equation (1). Then r1 x1 + r2 x2 + · · · + r` x` 6= 0 for any rational numbers r1 , r2 , . . . r` .
5
7
Other results
7.1
Asymptotic form of the roots
Let x = xn , n ≥ 1, be the nth root of tan x = x. Then nπ < x < (2n + 1)π/2. Write
x=
1
− z,
w
w=
2
(2n + 1)π
then 0 < z < π and tan(w−1 − z) = w−1 − z ↑ ∞ and this implies z ↓ 0. But
cos z
= cot z = tan(w−1 − z) = w−1 − z
sin z
Write
z
z(cos z + z sin z)
w=
,
f(z) =
f(z)
sin z
Now the function f has a removable singularity at 0 and is therefore analytic in
|z| < π. Also f(0) = limz→0 f(z) = 1. An application of the Lagrange inversion
formula shows that
∞
X
1 dk−1 f(z)k
k
z=
ck w ,
ck =
k!
dz k−1
z=0
k=1
(See [1], p.22.) Because f is an even function f(z)k is also even, which implies odd
derivatives at z = 0 vanish, hence ck = 0 for k even. Therefore
X
1
2
xn = −
ck w k ,
w=
w
(2n + 1)π
...
k=1,3,5, ...
Here are some of the coefficients (calculated with Mathematica 10):
c1 = 1
c3 =
2
3
c5 =
13
15
c7 =
146
105
c9 =
781
315
c11 =
16328
3465
These coefficients appear to be positive rational numbers which steadily increase.1
Futher numerical investigtion shows that for n running through the odd integers
−1/n
1, 3, . . . , 801, the sequence cn
steadily decreases and these numbers are bounded
below by .64389 which merely suggests that the radius of convergence of the series
for z in terms of w does not exceed .64389.
1
That the coefficients are rational numbers is provable, but even so, that assertion doesn’t seem
to cast any light on the conjectures of the previous section, so I omit the proof.
6
7.2
Asymptotic form of the solutions to tan x = xp .
Assume, for convenience, that p is a positive odd integer. The solutions to
tan x = xp
again have the form
xn = w−1 − z,
w=
2
(2n + 1)π
and z → 0 as w → 0 ( n → ∞). Hence, as before,
w=
(sin z)1/p
z(sin z)1/p + (cos z)1/p
To get to the Lagrange form, we make a change of variables:
z = tp
Now
w=
t
,
f(t)
f(t) = tp+1 + t (cot tp)1/p
def.
For p an odd integer, f(z) is an even analytic function of t and f(0) ≡ f(0±) = 1.
Applying Lagrange inversion formulas, this gives
t = c1 w + c2 w 2 + · · ·
where
1
ck =
k!
Hence
d
dt
k−1
f(t) k
t=0
(= 0 for even k),
1
− (w + c3 w3 + c5 w5 + · · · )1/p
w
1
c3
= − w1/p − w2+1/p + O w4+1/p
w
p
2+1/p
(2n + 1)π c3
2
1
=
−
+ O 4+1/p ,
2
p (2n + 1)π
n
xn =
7
n→∞
Example: tan x = x3
In this case
c1 = 1,
c2 = c3 = c4 = ceven = 0,
c5 = 1,
1
c7 = − ,
9
c9 = 4
so that
1/3
1 7
1
5
9
xn = − w + w − w + O w
w
9
1
1
1
= − w1/3 − w13/3 + w19/3 + O w25/3 ,
w
3
27
n→∞ w=
2
(2n + 1)π
References
[1] de Bruijn, N.G. (1981). Asymptotic Methods in Analysis. Dover, New York.
[2] Niven, I. (1956). Irrational Numbers. The Carus Mathematical Monographs 11.
Math. Assoc. of America.
[3] van der Waerden, B. L. (1970) Algebra Volume 1, 7th ed. Translated by Fred
Blum and John R. Schulenberger. Frederic Ungar. New York.
8