3 TeXed

# Why are powers of exp(pi*sqrt(163))$\exp(\pi\sqrt{163})$ almost integers?

I've been prodded to ask a question expanding this one on Ramanujan's constantexp(π√163). $R=\exp(\pi\sqrt{163})$.

Recall that R = exp(π √163) $R$ is very close to an integer; specifically it's 262537412640768744 $R=262537412640768744 - ε \epsilon$ where ε $\epsilon$ is about $0.75 × 10-12\times 10^{-12}$. Call the integer here N, $N$, so $R = N - ε.\epsilon$.

So R2$R^2 = N2N^2 - 2Nε 2N\epsilon + ε2. \epsilon^2$. It turns out that Nε $N\epsilon$ is itself nearly an integer, namely 196884, $196884$, and so R2$R^2$ is again an almost-integer. More precisely,

j(τ)

$$j(\tau) = 1/q + 744 + 196884q + 21493760q221493760q^2 + O(q3)O(q^3)$$

where $q = exp(2πiτ). \exp(2\pi i\tau)$. For τ $\tau = (1+√(-163))/2, 1+\sqrt{-163})/2$, and hence $q = exp(-π√163), \exp(-\pi\sqrt{163})$, it's known that the left-hand side is an integer. Squaring both sides,

j(τ)2

$$j(\tau)^2 = 1/q21/q^2 + 1488/q + 974304 + 335950912q + O(q2).O(q^2).$$

To show that 1/q2$1/q^2$ is nearly an integer, we can rearrange a bit to get

j(τ)2

$$j(\tau)^2 - 1/q21/q^2 - 974304 = 1488/q + 335950912q + O(q2)O(q^2)$$

and we want the left-hand side to be nearly zero. 1488/q $1488/q$ is nearly an integer since 1/q $1/q$ is nearly an integer; since q is small the higher-order terms on the right-hand side are small.

As noted by Mark Thomas in this question, R5$R^5$ is also very close to an integer -- but as I pointed out, that integer is not N5. $N^5$. This isn't special to fifth powers. R, R2, R3, R4, R5, R6$R$, $R^2$, $R^3$, $R^4$, $R^5$, $R^6$, respectively differ from the nearest integer by less than 10-12, 10-9, 10-8, 10-6, 10-5, 10-4, $10^{-12}$, $10^{-9}$, $10^{-8}$, $10^{-6}$, $10^{-5}$, $10^{-4}$, and 10-2 . $10^{-2}$. But the method of proof outlined above doesn't work for higher powers, since the coefficients of the q-expansion $q$-expansion of j(τ)5$j(\tau)^5$ (for example) grow too quickly. Is there some explanation for the fact that these higher powers are almost integers?

2 retag
1

# Why are powers of exp(pi*sqrt(163)) almost integers?

I've been prodded to ask a question expanding this one on Ramanujan's constant exp(π√163).

Recall that R = exp(π √163) is very close to an integer; specifically it's 262537412640768744 - ε where ε is about 0.75 × 10-12. Call the integer here N, so R = N - ε.

So R2 = N2 - 2Nε + ε2. It turns out that Nε is itself nearly an integer, namely 196884, and so R2 is again an almost-integer. More precisely,

j(τ) = 1/q + 744 + 196884q + 21493760q2 + O(q3)

where q = exp(2πiτ). For τ = (1+√(-163))/2, and hence q = exp(-π√163), it's known that the left-hand side is an integer. Squaring both sides,

j(τ)2 = 1/q2 + 1488/q + 974304 + 335950912q + O(q2).

To show that 1/q2 is nearly an integer, we can rearrange a bit to get

j(τ)2 - 1/q2 - 974304 = 1488/q + 335950912q + O(q2)

and we want the left-hand side to be nearly zero. 1488/q is nearly an integer since 1/q is nearly an integer; since q is small the higher-order terms on the right-hand side are small.

As noted by Mark Thomas in this question, R5 is also very close to an integer -- but as I pointed out, that integer is not N5. This isn't special to fifth powers. R, R2, R3, R4, R5, R6 respectively differ from the nearest integer by less than 10-12, 10-9, 10-8, 10-6, 10-5, 10-4, and 10-2 . But the method of proof outlined above doesn't work for higher powers, since the coefficients of the q-expansion of j(τ)5 (for example) grow too quickly. Is there some explanation for the fact that these higher powers are almost integers?