Editing Heegner number (section)

===Detail===
In what follows, j(z) denotes the [[j-invariant]] of the [[complex number]] z. Briefly, <math>\textstyle j\left(\frac{1+\sqrt{-d}}{2}\right)</math> is an integer for&nbsp;''d'' a Heegner number, and
<math display=block>e^{\pi \sqrt{d}} \approx -j\left(\frac{1+\sqrt{-d}}{2}\right) + 744</math>
via the ''q''-expansion.

If <math>\tau</math> is a quadratic irrational, then its ''j''-invariant <math>j(\tau)</math> is an [[algebraic integer]] of degree <math>\left|\mathrm{Cl}\bigl(\mathbf{Q}(\tau)\bigr)\right|</math>, the [[Class number (number theory)|class number]] of <math>\mathbf{Q}(\tau)</math> and the minimal (monic integral) polynomial it satisfies is called the 'Hilbert class polynomial'. Thus if the imaginary quadratic extension <math>\mathbf{Q}(\tau)</math> has class number 1 (so ''d'' is a Heegner number), the ''j''-invariant is an integer.

The [[Q-expansion|''q''-expansion]] of ''j'', with its [[Fourier series]] expansion written as a [[Laurent series]] in terms of <math>q=e^{2 \pi i \tau}</math>, begins as:
<math display=block>j(\tau) = \frac{1}{q} + 744 + 196\,884 q + \cdots.</math>

The coefficients <math>c_n</math> asymptotically grow as
<math display=block>\ln(c_n) \sim 4\pi \sqrt{n} + O\bigl(\ln(n)\bigr),</math>
and the low order coefficients grow more slowly than <math>200\,000^n</math>, so for <math>\textstyle q \ll \frac{1}{200\,000}</math>, ''j'' is very well approximated by its first two terms.  Setting <math>\textstyle\tau = \frac{1+\sqrt{-163}}{2}</math> yields
<math display=block> q=-e^{-\pi \sqrt{163}} \quad\therefore\quad \frac{1}{q}=-e^{\pi \sqrt{163}}. </math>
Now
<math display=block>j\left(\frac{1+\sqrt{-163}}{2}\right)=\left(-640\,320\right)^3,</math>
so,
<math display=block>\left(-640\,320\right)^3=-e^{\pi \sqrt{163}}+744+O\left(e^{-\pi \sqrt{163}}\right).</math>
Or,
<math display=block>e^{\pi \sqrt{163}}=640\,320^3+744+O\left(e^{-\pi \sqrt{163}}\right)</math>
where the linear term of the error is,
<math display=block>\frac{-196\,884}{e^{\pi \sqrt{163}}} \approx \frac{-196\,884}{640\,320^3+744}
\approx -0.000\,000\,000\,000\,75</math>
explaining why <math>e^{\pi \sqrt{163}}</math> is within approximately the above of being an integer.