IRRATIONAL NUMBERS ARISING FROM CERTAIN DIFFERENTIAL EQUATIONS

M. RAM MURTY AND V. KUMAR MURTY

Niven [3] gave a simple proof that π is irrational. Koksma [2] modified Niven’s proof to show that e^{r} is irrational for every non-zero rational r. Dixon [1] made a similar modification to show that π is not algebraic of degree 2. In this note, we prove a general theorem which gives Niven’s and Koksma’s results as easy corollaries. A suitable modification in our proof also gives Dixon’s result.

Theorem 1. Let G be non-trivial solution of the equation

L(u) = p_{0}u^{(n)} + p_{1}u^{(n−1)} + ⋯ + p_{n}u = 0

where p_{i} are rational numbers and p_{n} ≠ 0. If b > 0 is such that G(x) ≥ 0 on [0, b] and G^{(i)}(0), G^{(i)}(b) are rational for 0 ≤ i ≤ n − 1, then b is irrational.

Proof. Without any loss of generality, we may suppose that the p_{i} are integers. Suppose b is rational and set b = p/q, (p, q) = 1, p, q ∈ ℤ. Set f_{m}(x) = 1/m!(qx)^{m}(p − qx)^{m}, where m is a natural number. It is easy to see that f_{m}^{(k)}(0) are integers for k ≥ 0 and since f_{m}(x) = f_{m}(b − x), the same is true of f_{m}^{(k)}(b). Now define the sequence {t_{k}} recursively as follows:

t_{0} = 1,
p_{n}t_{1} − p_{n−1}t_{0} = 0,
p_{n}t_{2} − p_{n−1}t_{1} + p_{n−2}t_{0} = 0,
p_{n}t_{n−1} − p_{n−1}t_{n−2} + ⋯ + (−1)^{n−1}p_{1}t_{0} = 0,
p_{n}t_{n+r} − p_{n−1}t_{n+r−1} + ⋯ + (−1)^{n}p_{0}t_{r} = 0 for r ≥ 0.

Clearly, p_{n}^{k}t_{k} is an integer for k ≥ 0. Let

F_{m}(x) = ∑_{r=0}^{2m} t_{r}f_{m}^{(r)}(x).

If L^{∗} is the adjoint of L, we have

L^{∗}(F_{m}(x)) = ∑_{k=0}^{n}(−1)^{k}p_{n−k}F_{m}^{(k)}(x)

= ∑_{k=0}^{n}(−1)^{k}p_{n−k} ∑_{r=0}^{2m} t_{r}f_{m}^{(r+k)}(x)

= ∑_{s=0}^{2m} f_{m}^{(s)}(x) ∑_{r+k=s}(−1)^{k}p_{n−k}t_{r} = p_{n}f_{m}(x).

Received by the editors Sept. 15, 1976.

Letting

P (u, v) = u [p_{n−1}v − \frac{d}{dx}(p_{n−2}v) + ⋯ + (−1)^{n−1} \frac{d^{n−1}}{dx^{n−1}} (p_{0}v)] +

\frac{du}{dx} [p_{n−2}v − \frac{d}{dx}(p_{n−3}v) + ⋯ + (−1)^{n−2} \frac{d^{n−2}}{dx^{n−2}} (p_{0}v)] +

+ ⋯ + \frac{d^{n−1}u}{dx^{n−1}}(p_{0}v),

we have by Lagrange’s identity,

F_{m}(x)L(G) − G(x)L^{∗}(F_{m}(x)) = \frac{d}{dx}P (G, F_{m}),

so that

− ∫_{0}^{b} p_{n}f_{m}(x)G(x) dx = [P (G, F_{m})]_{0}^{b}

since L(G) = 0. As p_{n}^{k}t_{k} is an integer, it follows that p_{n}^{2m}F_{m}^{(w)}(x) is an integer for x = 0 and b, w ≥ 0. Thus, if A denotes the products of the denominators of G^{(i)}(0) and G^{(i)}(b), 0 ≤ i ≤ n − 1 (when expressed in lowest terms), Ap_{n}^{2m} [P (G, F_{m})]_{0}^{b} is an integer for every m. Now

Ap_{n}^{2m} [P (G, F_{m})]_{0}^{b} = −Ap_{n}^{2m+1} ∫_{0}^{b} f_{m}(x)G(x) dx.

If B and C are such that |G(x)| ≤ B, |qx(p − qx)| ≤ C on [0, b] we have

0 < Ap_{n}^{2m+1} |∫_{0}^{b} f_{m}(x)G(x) dx| < \frac{bBAp_{n}^{2m+1}C^{2m}}{m!} .

If m is sufficiently large, the right hand side is < 1, giving a contradiction. Hence b is irrational.

Corollary 1. (1) π^{2} is irrational, (hence so also is π). (2) log r is irrational for every rational r > 0, r ≠ 1. (3) e^{r}, sin r, cos r, cosh r, sinh r are irrational for every non-zero rational r.

Proof. If π^{2} is irrational, consider y′′ + π^{2}y = 0 which has a solution (1/π) sin πx. For b = 1, we get a contradiction. This proves (1). (2) and (3) are proved similarly, using the equation y′ − y = 0 or y′′ ± y = 0.

The following theorem is more arithmetical in nature.

Theorem 2. Let G be a non-trivial solution of y^{(n)}+ty = 0 where t = (u/v,)(u, v) = 1, is a non-zero rational. Suppose G^{(i)}(0) is rational for 0 ≤ i ≤ n − 1 and for some r ≠ 0 with (r, n) = 1, we have G^{(r)}(0) ≠ 0. If β is a non-zero rational, then G^{(n−1)}(β) is irrational.

Proof. Let β = (a/b), (a, b) = 1. Define

f_{p}(x) = \frac{(β − x)^{np}[β^{n} − (β − x)^{n}]^{p−1}b^{np+(n−1)(p−1)}}{(p − 1)!} ,

where p is a prime soon to be specified. If we compute the t_{k} in Theorem 1 for the equation y^{(n)} + ty = 0, we find t_{k} = 0 if k ≢ 0 (mod n) and in case k = sn, t_{sn} = (−1)^{sn−s}t^{s}. If we set

F_{p}(x) = ∑_{k=0}^{M} t_{k}f_{p}^{(k)}(x)

were M = n(2p − 1), we have as in Theorem 1, L^{∗}(F_{p}(x)) = tf_{p}(x), where L^{∗} is the adjoint of L(y) = y^{(n)} + ty. Since f_{p} is a polynomial of degree M, f_{p}^{(k)}(x) ≡ 0 for k > M. If we set β − x = y and g_{p}(y) = y^{np}(β^{n} − y^{n})^{p−1}, then

g_{p}(y) = ∑_{i=0}^{p−1}(−1)^{i}β^{n(p−1−i)} ( \begin{matrix}p − 1 \\ i \end{matrix} ) y^{n(p+1)}

from which it follows at once that f_{p}^{(k)}(β) = 0 for all k ≠ n(p + i), 0 ≤ i ≤ p − 1 and

f_{p}^{(k)}(β) = (−1)^{n(p+i)+i} \frac{b^{M−p+1}}{(p − 1)!} β^{n(p−1−i)} ( \begin{matrix}p − 1 \\ i \end{matrix} ) [n(p + i)]!

for k = n(p + i). Hence v^{2p−1}F_{p}(β) is an integer divisible by p. Since f_{p} has a zero of order p − 1 at x = 0, we have f_{p}^{(k)}(0) = 0 for k < p − 1. Writing f_{p}(x) =

(x^{p−1}/(p−1)!)h_{p}(x) we see from (p−1)!f_{p}^{(k)}(x) = ∑_{s=0}^{k} ( \begin{matrix}k \\ s \end{matrix} ) [x^{p−1}]^{(s)}[h_{p}(x)]^{(k−s)}

that f_{p}^{(k)}(0) = ( \begin{matrix}k \\ p − 1 \end{matrix} ) h_{p}^{(k−p+1)}(0) for k ≥ p − 1. Clearly f_{p}^{(k)}(0) is an integer

as h_{p}^{(k−p+1)}(0) is an integer. Also ( \begin{matrix}k \\ p − 1 \end{matrix} ) is divisible by p if k ≥ p, and k ≢ −1

(mod p). If k ≥ p and k − p + 1 ≡ 0 (mod p), then h_{p}^{(k−p+1)}(0) is divisible by p. If k = p − 1, f_{p}^{(p−1)}(0) = n^{p−1}a^{np+(n−1)(p−1)}. Hence, f_{p}^{(k)}(0) is divisible by p unless k = p − 1. As (r, n) = 1, let p be a prime > na, congruent to −r (mod n). As G^{(r)}(0) ≠ 0, and p − 1 ≡ n − r − 1 (mod n), the term f_{p}^{(p−1)}(0) occurs once and only once in ∑_{k=0}^{n−1}(−1)^{k}G^{(k)}(0)F_{p}^{(n−k−1)} and that is in the expression for F_{p}^{(n−r−1)}(0). Let N be the product of all the denominators of the rationals G(0), G′(0), …, G^{(n−1)}(0), G^{(n−1)}(β). (Here we are supposing G^{(n−1)}(β) is rational and will arrive at a contradiction). Thus, if p > max (na, NG^{(r)}(0), uv), all terms in

Nv^{2p−1}{G^{(n−1)}(β)F_{p}(β) − ∑_{k=0}^{n−1}(−1)^{k}G^{(k)}(0)F_{p}^{(n−k−1)}(0)}

are divisible by p except one term (the one involving G^{(r)}(0) ≠ 0). Now, as in the proof of Theorem 1,

−uNv^{2p} ∫_{0}^{β} G(x)f_{p}(x) dx = Nv^{2p−1}{G^{(n−1)}(β)
×F_{p}(β) − ∑_{k=0}^{n−1}(−1)^{k}G^{(k)}(0)F_{p}^{(n−k−1)}(0)}.

Thus, it follows uNv^{2p} ∫_{0}^{β} G(x)f_{p}(x) dx ≠ 0 for an infinity of primes p, using Dirichlet’s theorem. On the other hand, we know uNv^{2p} ∫_{0}^{β} G(x)f_{p}(x) dx is an

4 M. RAM MURTY AND V. KUMAR MURTY

integer. This is a contradiction since

lim_{p→∞} |uNv^{2p} ∫_{0}^{β} G(x)f_{p}(x) dx| = 0.

This proves the theorem. Corollary 2. Let p be an odd prime and G a non-trivial solution of y^{(p)} +ty = 0, t a non-zero rational. If G(0), …, G^{(p−1)}(0) are rational and at least two of them are non-zero, then G(β), G′(β), …, G^{(p−1)}(β) are irrational for any non-zero rational β. Remark 1. The case p = 2 has been covered by a corollary of Theorem 1. Proof. As at least two of G(0), G′(0), …, G^{(p−1)}(0) are non-zero, there is an r such that G^{(r)}(0) ≠ 0 and (r, p) = 1. The conditions of the theorem are satisfied and so G^{(p−1)}(β) is irrational. As G^{(i)}(x) also satisfies the conditions of the theorem for 0 < i ≤ p − 1 the result follows.

Acknowledgement. We would like to thank Professor P. R. Beesack for his helpful suggestions.

References

[1] J. D. Dixon, π is not algebraic of degree one or two, Amer. Math. Monthly, 69 (1962), 632.

[2] J. F. Koksma, On Niven’s proof that π is irrational, Nieuw Archief voor Wiskunde, (2) 23 (1949), 39.

[3] I. Niven, A simple proof that π is irrational, Bull. Amer. Math. Soc., 53 (1947), 509. Department of Mathematics, Carleton University, Ottawa, Ontario