Srinivasa Ramanujan

Ramanujan todisti monia alkeellisia mutta kiehtovia tuloksia:

${\displaystyle (3x^{2}+5xy-5y^{2})^{3}+(4x^{2}-4xy+6y^{2})^{3}+(5x^{2}-5xy-3y^{2})^{3}=(6x^{2}-4xy+4y^{2})^{3}}$

${\displaystyle {\sqrt {(n+a)^{2}+x\,a+x\,{\sqrt {(n+a)^{2}+(x+n)\,a+(x+n)\,{\sqrt {(n+a)^{2}+(x+2n)\,a+(x+2n)\,{\sqrt {\dots }}}}}}}}}$
${\displaystyle =x\,+\,n\,+\,a.}$

Hän kehitti monia approksimaatioita piille:

${\displaystyle {\frac {9}{5}}+{\sqrt {\frac {9}{5}}}=3.1416^{+}}$

${\displaystyle {\sqrt[{4}]{3^{4}+2^{4}+{\frac {1}{2+({\frac {2}{3}})^{2}}}}}={\sqrt[{4}]{\frac {2143}{22}}}=3.14159\ 2652^{+}.}$

Identiteettejä juurille:

${\displaystyle {\sqrt[{4}]{\frac {3+2{\sqrt[{4}]{5}}}{3-2{\sqrt[{4}]{5}}}}}={\frac {{\sqrt[{4}]{5}}+1}{{\sqrt[{4}]{5}}-1}}={\tfrac {1}{2}}\left(3+{\sqrt[{4}]{5}}+{\sqrt {5}}+{\sqrt[{4}]{125}}\right),}$

${\displaystyle {\sqrt {{\sqrt[{3}]{28}}-{\sqrt[{3}]{27}}}}={\tfrac {1}{3}}\left({\sqrt[{3}]{98}}-{\sqrt[{3}]{28}}-1\right),}$

${\displaystyle {\sqrt[{3}]{{\sqrt[{5}]{\frac {32}{5}}}-{\sqrt[{5}]{\frac {27}{5}}}}}={\sqrt[{5}]{\frac {1}{25}}}+{\sqrt[{5}]{\frac {3}{25}}}-{\sqrt[{5}]{\frac {9}{25}}},}$

${\displaystyle {\sqrt[{3}]{\ {\sqrt[{3}]{2}}\ -1}}={\sqrt[{3}]{\frac {1}{9}}}-{\sqrt[{3}]{\frac {2}{9}}}+{\sqrt[{3}]{\frac {4}{9}}}.}$

Ramanujan tutki Bernoullin lukuja ja löysi monia kiehtovia ominaisuuksia:

${\displaystyle {{m+3} \choose {m}}B_{m}={\begin{cases}{{m+3} \over 3}-\sum \limits _{j=1}^{m/6}{m+3 \choose {m-6j}}B_{m-6j},&{\mbox{jos}}\ m\equiv 0{\pmod {6}};\\{{m+3} \over 3}-\sum \limits _{j=1}^{(m-2)/6}{m+3 \choose {m-6j}}B_{m-6j},&{\mbox{jos}}\ m\equiv 2{\pmod {6}};\\-{{m+3} \over 6}-\sum \limits _{j=1}^{(m-4)/6}{m+3 \choose {m-6j}}B_{m-6j},&{\mbox{jos}}\ m\equiv 4{\pmod {6}}.\end{cases}}}$

Ramanujan löysi monia äärettömiä sarjoja piille:

${\displaystyle {\frac {1}{\pi }}={\frac {2{\sqrt {2}}}{9801}}\sum _{k=0}^{\infty }{\frac {(4k)!(1103+26390k)}{(k!)^{4}396^{4k}}}.}$

Toinen hänen tuloksistaan äärettömille sarjoille on

${\displaystyle \left[1+2\sum _{n=1}^{\infty }{\frac {\cos(n\theta )}{\cosh(n\pi )}}\right]^{-2}+\left[1+2\sum _{n=1}^{\infty }{\frac {\cosh(n\theta )}{\cosh(n\pi )}}\right]^{-2}={\frac {2\Gamma ^{4}\left({\frac {3}{4}}\right)}{\pi }}}$

kaikille ${\displaystyle \theta }$, missä ${\displaystyle \Gamma (z)}$ on gammafunktio. Hän todisti myös monia tuloksia hypergeometrisille sarjoille, kuten:

${\displaystyle 1-5\left({\frac {1}{2}}\right)^{3}+9\left({\frac {1\times 3}{2\times 4}}\right)^{3}-13\left({\frac {1\times 3\times 5}{2\times 4\times 6}}\right)^{3}+\cdots ={\frac {2}{\pi }}}$

${\displaystyle 1+9\left({\frac {1}{4}}\right)^{4}+17\left({\frac {1\times 5}{4\times 8}}\right)^{4}+25\left({\frac {1\times 5\times 9}{4\times 8\times 12}}\right)^{4}+\cdots ={\frac {2^{\frac {3}{2}}}{\pi ^{\frac {1}{2}}\Gamma ^{2}\left({\frac {3}{4}}\right)}}.}$

Ensimmäinen tulos oli jo tunnettu, mutta toinen oli todennäköisesti uusi.

Ramanujan laski monia mielenkiintoisia integraaleja, kuten

${\displaystyle \int _{0}^{\infty }{\cfrac {1+{x}^{2}/({b+1})^{2}}{1+{x}^{2}/({a})^{2}}}\times {\cfrac {1+{x}^{2}/({b+2})^{2}}{1+{x}^{2}/({a+1})^{2}}}\times \cdots \;\;dx={\frac {\sqrt {\pi }}{2}}\times {\frac {\Gamma (a+{\frac {1}{2}})\Gamma (b+1)\Gamma (b-a+{\frac {1}{2}})}{\Gamma (a)\Gamma (b+{\frac {1}{2}})\Gamma (b-a+1)}}.}$

Ramanujan löysi suuren määrän kaavoja ketjumurtoluvuille:

${\displaystyle 1+{\frac {1}{1\cdot 3}}+{\frac {1}{1\cdot 3\cdot 5}}+{\frac {1}{1\cdot 3\cdot 5\cdot 7}}+{\frac {1}{1\cdot 3\cdot 5\cdot 7\cdot 9}}+\cdots +{1 \over 1+{1 \over 1+{2 \over 1+{3 \over 1+{4 \over 1+{5 \over 1+\cdots }}}}}}={\sqrt {\frac {e\cdot \pi }{2}}}}$

${\displaystyle {\sqrt {\varphi +2}}-\varphi ={\cfrac {e^{-2\pi /5}}{1+{\cfrac {e^{-2\pi }}{1+{\cfrac {e^{-4\pi }}{1+{\cfrac {e^{-6\pi }}{1+\,\cdots }}}}}}}}}$

missä ${\displaystyle \varphi ={\frac {1+{\sqrt {5}}}{2}}}$ on kultainen leikkaus;

${\displaystyle {\frac {1}{\displaystyle 1+{\frac {e^{-2\pi {\sqrt {5}}}}{\displaystyle 1+{\frac {e^{-4\pi {\sqrt {5}}}}{\displaystyle 1+{\frac {e^{-6\pi {\sqrt {5}}}}{\ddots }}}}}}}}={\Biggl (}{\frac {\sqrt {5}}{1+^{5}{\sqrt {5^{\frac {3}{4}}({\frac {{\sqrt {5}}-1}{2}})^{\frac {5}{2}}-1}}}}-{\frac {{\sqrt {5}}+1}{2}}{\Biggr )}\cdot e^{2\pi /{\sqrt {5}}}.}$

${\displaystyle \;_{1}\psi _{1}\left[{\begin{matrix}a\\b\end{matrix}};q,z\right]=\sum _{n=-\infty }^{\infty }{\frac {(a;q)_{n}}{(b;q)_{n}}}z^{n}={\frac {(b/a,q,q/az,az;q)_{\infty }}{(b,b/az,q/a,z;q)_{\infty }}}}$

jos |q| < 1 ja |b/a| < |z| < 1.

Rogersin–Ramanujanin identiteetit:

${\displaystyle G(q)=\sum _{n=0}^{\infty }{\frac {q^{n^{2}}}{(q;q)_{n}}}={\frac {1}{(q;q^{5})_{\infty }(q^{4};q^{5})_{\infty }}}=1+q+q^{2}+q^{3}+2q^{4}+2q^{5}+3q^{6}+\cdots \,}$

ja

${\displaystyle H(q)=\sum _{n=0}^{\infty }{\frac {q^{n^{2}+n}}{(q;q)_{n}}}={\frac {1}{(q^{2};q^{5})_{\infty }(q^{3};q^{5})_{\infty }}}=1+q^{2}+q^{3}+q^{4}+q^{5}+2q^{6}+\cdots \,}$ .

Muita q-sarjoja:

${\displaystyle \sum _{k=0}^{\infty }p(5k+4)q^{k}=5{\frac {(q^{5})_{\infty }^{5}}{(q)_{\infty }^{6}}}}$

${\displaystyle \sum _{k=0}^{\infty }p(7k+5)q^{k}=7{\frac {(q^{7})_{\infty }^{3}}{(q)_{\infty }^{4}}}+49q{\frac {(q^{7})_{\infty }^{7}}{(q)_{\infty }^{8}}}}$

missä p(n) on partitiofunktio. Tästä saadaan korollaareina kongruenssit

${\displaystyle p(5k+4)\equiv 0{\pmod {5}}}$

${\displaystyle p(7k+5)\equiv 0{\pmod {7}}.}$

Ramanujan löysi myös kolmannen kongruenssin:

${\displaystyle p(11k+6)\equiv 0{\pmod {11}}.}$

Määritellään

${\displaystyle L(q)=1-24\sum _{n=1}^{\infty }{\frac {nq^{n}}{1-q^{n}}}=E_{2}(\tau )}$

${\displaystyle M(q)=1+240\sum _{n=1}^{\infty }{\frac {n^{3}q^{n}}{1-q^{n}}}=E_{4}(\tau )}$

${\displaystyle N(q)=1-504\sum _{n=1}^{\infty }{\frac {n^{5}q^{n}}{1-q^{n}}}=E_{6}(\tau ),}$

silloin on

${\displaystyle q{\frac {dL}{dq}}={\frac {L^{2}-M}{12}}}$

${\displaystyle q{\frac {dM}{dq}}={\frac {LM-N}{3}}}$

${\displaystyle q{\frac {dN}{dq}}={\frac {LN-M^{2}}{2}}.}$

Nämä kaavat johtavat korollaareihin aritmeettisille funktioille.