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Heegner number
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==Other Heegner numbers== For the four largest Heegner numbers, the approximations one obtains<ref>These can be checked by computing <math display=block>\sqrt[3]{e^{\pi\sqrt{d}}-744}</math> on a calculator, and <math display=block>\frac{196\,884}{e^{\pi\sqrt{d}}}</math> for the linear term of the error.</ref> are as follows. <math display=block>\begin{align} e^{\pi \sqrt{19}} &\approx \phantom{000\,0}96^3+744-0.22\\ e^{\pi \sqrt{43}} &\approx \phantom{000\,}960^3+744-0.000\,22\\ e^{\pi \sqrt{67}} &\approx \phantom{00}5\,280^3+744-0.000\,0013\\ e^{\pi \sqrt{163}} &\approx 640\,320^3+744-0.000\,000\,000\,000\,75 \end{align} </math> Alternatively,<ref>{{Cite web|url=http://groups.google.com.ph/group/sci.math.research/browse_thread/thread/3d24137c9a860893?hl=en|title=More on e^(pi*SQRT(163))|access-date=2008-04-19|archive-date=2009-08-11|archive-url=https://web.archive.org/web/20090811214935/http://groups.google.com.ph/group/sci.math.research/browse_thread/thread/3d24137c9a860893?hl=en|url-status=dead}}</ref> <math display=block>\begin{align} e^{\pi \sqrt{19}} &\approx 12^3\left(3^2-1\right)^3\phantom{00}+744-0.22\\ e^{\pi \sqrt{43}} &\approx 12^3\left(9^2-1\right)^3\phantom{00}+744-0.000\,22\\ e^{\pi \sqrt{67}} &\approx 12^3\left(21^2-1\right)^3\phantom{0}+744-0.000\,0013\\ e^{\pi \sqrt{163}} &\approx 12^3\left(231^2-1\right)^3+744-0.000\,000\,000\,000\,75 \end{align} </math> where the reason for the squares is due to certain [[Eisenstein series]]. For Heegner numbers <math>d < 19</math>, one does not obtain an almost integer; even <math>d = 19</math> is not noteworthy.<ref>The absolute deviation of a random real number (picked uniformly from [[unit interval|{{closed-closed|0,1|size=120%}}]], say) is a uniformly distributed variable on {{closed-closed|0,β0.5|size=120%}}, so it has [[absolute average deviation]] and [[median absolute deviation]] of 0.25, and a deviation of 0.22 is not exceptional.</ref> The integer ''j''-invariants are highly factorisable, which follows from the form :<math display=block>12^3\left(n^2-1\right)^3=\left(2^2\cdot 3 \cdot (n-1) \cdot (n+1)\right)^3,</math> and factor as, <math display=block>\begin{align} j\left(\frac{1+\sqrt{-19}}{2}\right) &= \phantom{000\,0}-96^3 = -\left(2^5 \cdot 3\right)^3\\ j\left(\frac{1+\sqrt{-43}}{2}\right) &= \phantom{000\,}-960^3 = -\left(2^6 \cdot 3 \cdot 5\right)^3\\ j\left(\frac{1+\sqrt{-67}}{2}\right) &= \phantom{00}-5\,280^3 = -\left(2^5 \cdot 3 \cdot 5 \cdot 11\right)^3\\ j\left(\frac{1+\sqrt{-163}}{2}\right)&= -640\,320^3 = -\left(2^6 \cdot 3 \cdot 5 \cdot 23 \cdot 29\right)^3. \end{align} </math> These [[transcendental numbers]], in addition to being closely approximated by integers (which are simply [[algebraic numbers]] of degree 1), can be closely approximated by algebraic numbers of degree 3,<ref>{{cite web|url=http://sites.google.com/site/tpiezas/001|title=Pi Formulas}}</ref> <math display=block>\begin{align} e^{\pi \sqrt{19}} &\approx x^{24}-24.000\,31 ; & x^3-2x-2&=0\\ e^{\pi \sqrt{43}} &\approx x^{24}-24.000\,000\,31 ; & x^3-2x^2-2&=0\\ e^{\pi \sqrt{67}} &\approx x^{24}-24.000\,000\,0019 ; & x^3-2x^2-2x-2&=0\\ e^{\pi \sqrt{163}} &\approx x^{24}-24.000\,000\,000\,000\,0011 ; &\quad x^3-6x^2+4x-2&=0 \end{align} </math> The [[root of a function|roots]] of the cubics can be exactly given by quotients of the [[Dedekind eta function]] ''Ξ·''(''Ο''), a modular function involving a 24th root, and which explains the 24 in the approximation. They can also be closely approximated by algebraic numbers of degree 4,<ref>{{cite web|url=http://sites.google.com/site/tpiezas/ramanujan|title=Extending Ramanujan's Dedekind Eta Quotients}}</ref> <math display=block>\begin{align} e^{\pi \sqrt{19}} &\approx 3^5 \left(3-\sqrt{2\left(1- \tfrac{96}{24}+1\sqrt{3\cdot19}\right)} \right)^{-2}-12.000\,06\dots\\ e^{\pi \sqrt{43}} &\approx 3^5 \left(9-\sqrt{2\left(1- \tfrac{960}{24}+7\sqrt{3\cdot43}\right)} \right)^{-2}-12.000\,000\,061\dots\\ e^{\pi \sqrt{67}} &\approx 3^5 \left(21-\sqrt{2\left(1- \tfrac{5\,280}{24} +31\sqrt{3\cdot67}\right)} \right)^{-2}-12.000\,000\,000\,36\dots\\ e^{\pi \sqrt{163}} &\approx 3^5 \left(231-\sqrt{2\left(1- \tfrac{640\,320}{24}+2\,413\sqrt{3\cdot163}\right)} \right)^{-2}-12.000\,000\,000\,000\,000\,21\dots \end{align} </math> If <math>x</math> denotes the expression within the parenthesis (e.g. <math>x=3-\sqrt{2\left(1- \tfrac{96}{24}+1\sqrt{3\cdot19}\right)}</math>), it satisfies respectively the [[quartic equation]]s <math display=block>\begin{align} x^4 -\phantom{00} 4\cdot 3 x^3 + \phantom{000\,0}\tfrac23( 96 +3) x^2 - \phantom{000\,000}\tfrac23\cdot3(96-6)x - 3&=0\\ x^4 -\phantom{00} 4\cdot 9x^3 + \phantom{000\,}\tfrac23( 960 +3) x^2 - \phantom{000\,00}\tfrac23\cdot9(960-6)x - 3&=0\\ x^4 -\phantom{0} 4\cdot 21x^3 + \phantom{00}\tfrac23( 5\,280 +3) x^2 - \phantom{000}\tfrac23\cdot21(5\,280-6)x - 3&=0\\ x^4 - 4\cdot 231x^3 + \tfrac23( 640\,320 +3) x^2 - \tfrac23\cdot231(640\,320-6)x - 3&=0\\ \end{align} </math> Note the reappearance of the integers <math>n = 3, 9, 21, 231</math> as well as the fact that <math display=block>\begin{align} 2^6 \cdot 3\left(-\left(1- \tfrac{96}{24}\right)^2+ 1^2 \cdot3\cdot 19 \right) &= 96^2\\ 2^6 \cdot 3\left(-\left(1- \tfrac{960}{24}\right)^2+ 7^2\cdot3 \cdot 43 \right) &= 960^2\\ 2^6 \cdot 3\left(-\left(1- \tfrac{5\,280}{24}\right)^2+ 31^2 \cdot 3\cdot67 \right) &= 5\,280^2\\ 2^6 \cdot 3\left(-\left(1- \tfrac{640\,320}{24}\right)^2+ 2413^2\cdot 3 \cdot163 \right) &= 640\,320^2 \end{align} </math> which, with the appropriate fractional power, are precisely the ''j''-invariants. Similarly for algebraic numbers of degree 6, <math display=block>\begin{align} e^{\pi \sqrt{19}} &\approx \left(5x\right)^3-6.000\,010\dots\\ e^{\pi \sqrt{43}} &\approx \left(5x\right)^3-6.000\,000\,010\dots\\ e^{\pi \sqrt{67}} &\approx \left(5x\right)^3-6.000\,000\,000\,061\dots\\ e^{\pi \sqrt{163}} &\approx \left(5x\right)^3-6.000\,000\,000\,000\,000\,034\dots \end{align} </math> where the ''x''s are given respectively by the appropriate root of the [[sextic equation]]s, <math display=block>\begin{align} 5x^6-\phantom{000\,0}96x^5-10x^3+1&=0\\ 5x^6-\phantom{000\,}960x^5-10x^3+1&=0\\ 5x^6-\phantom{00}5\,280x^5-10x^3+1&=0\\ 5x^6-640\,320x^5-10x^3+1&=0 \end{align} </math> with the ''j''-invariants appearing again. These sextics are not only algebraic, they are also [[Solvable group|solvable]] in [[Nth root|radicals]] as they factor into two [[Cubic equation|cubics]] over the extension <math>\Q\sqrt{5}</math> (with the first factoring further into two [[Quadratic equation|quadratics]]). These algebraic approximations can be ''exactly'' expressed in terms of Dedekind eta quotients. As an example, let <math>\textstyle \tau = \frac{1+\sqrt{-163}}{2}</math>, then, <math display=block>\begin{align} e^{\pi \sqrt{163}} &= \left( \frac{e^\frac{\pi i}{24} \eta(\tau)}{\eta(2\tau)} \right)^{24}-24.000\,000\,000\,000\,001\,05\dots\\ e^{\pi \sqrt{163}} &= \left( \frac{e^\frac{\pi i}{12} \eta(\tau)}{\eta(3\tau)} \right)^{12}-12.000\,000\,000\,000\,000\,21\dots\\ e^{\pi \sqrt{163}} &= \left( \frac{e^\frac{\pi i}{6} \eta(\tau)}{\eta(5\tau)} \right)^{6}-6.000\,000\,000\,000\,000\,034\dots \end{align} </math> where the eta quotients are the algebraic numbers given above.
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