拉馬努金
拉馬努金出生於印度埃羅德，即便拉馬努金的家族屬於印度種姓制度中最高等級，他們家還是相當貧窮。 但拉馬努金憑藉著他超出常人的數學天份，被其他數學家們發現而展露頭角。 人們稱他為從未來穿越而來的天才數學家，而數學家哈代稱他為第二個牛頓。 人們說拉馬努金是天才的原因是，他從小沒有接受正規的訓練，只憑「直覺」寫下幾千個數學公式，由於他慣以不給出證明過程，所以當時沒有人能看懂這些公式。在很多年之後，數學家們才解開了其中的一些公式命題，甚至由此獲得重大數學成就。
Q(1): \(\phi=\small\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt\cdots}}}\)
數字皆為正整數
Ramanujan's Nested Radical\(x=x\)
\(x=\small\sqrt{x^2}\)
\(x=\small\sqrt{1+(x^2-1)}\)
\(x=\small\sqrt{1+(x-1)(x+1)}\)
\(\therefore\left\{\begin{array}{lc}\alpha=\small\sqrt{1+(\alpha-1)(\alpha+1)}&(a)\\(\alpha+1)=\small\sqrt{1+(\alpha)(\alpha+2)}&(b)\\(\alpha+2)=\small\sqrt{1+(\alpha+1)(\alpha+3)}&(c)\\(\alpha+3)=\small\sqrt{1+(\alpha+2)(\alpha+4)}&(d)\\(\alpha+4)=\small\sqrt{1+(\alpha+3)(\alpha+5)}&(e)\\\cdots&\cdots\end{array}\right.\)
\((c)\rightarrow(b)\;\Rightarrow(\alpha+1)=\small\sqrt{1+(\alpha)\small\sqrt{1+(\alpha+1)(\alpha+3)}}\)
\(insert\;(d)\;\Rightarrow(\alpha+1)=\small\sqrt{1+(\alpha)\small\sqrt{1+(\alpha+1)\small\sqrt{1+(\alpha+2)(\alpha+4)}}}\)
\(insert\;(e)\;\Rightarrow(\alpha+1)=\small\sqrt{1+(\alpha)\small\sqrt{1+(\alpha+1)\small\sqrt{1+(\alpha+2)\small\sqrt{1+(\alpha+3)(\alpha+5)}}}}\)
\(\therefore(\alpha+1)=\small\sqrt{1+(\alpha)\small\sqrt{1+(\alpha+1)\small\sqrt{1+(\alpha+2)\small\sqrt{1+(\alpha+3)\small\sqrt{1+\cdots}}}}}\)
\(\therefore3=(2+1)=\small\sqrt{1+2\small\sqrt{1+3\small\sqrt{1+4\small\sqrt\cdots}}}\;\;\;(\alpha)\)
\(\therefore4=(3+1)=\small\sqrt{1+3\small\sqrt{1+4\small\sqrt{1+5\small\sqrt\cdots}}}\;\;\;(\beta)\)
\(\therefore5=(4+1)=\small\sqrt{1+4\small\sqrt{1+5\small\sqrt{1+6\small\sqrt\cdots}}}\;\;\;(\gamma)\)
\((\beta)\rightarrow(\alpha)\;\Rightarrow\;3=(2+1)=\small\sqrt{1+2\cdot4}=\small\sqrt9=3\)
\((\gamma)\rightarrow(\beta)\;\Rightarrow4=(3+1)=\small\sqrt{1+3\cdot5}=\small\sqrt{16}=4\)
Reccurence 遞迴數學
若函數 \(f(x)＝x(x+2)\)，其中 \(x\) 是正整數，
則 \(f(x+1)＝(x+1)(x+3)＝{(x+2)}^2-1\)，
因此 \(f(x+1)+1={(x+2)}^2\)，
\(\Rightarrow{(x+2)}^2=1+f(x+1)\)，
得到 \((x+2)＝\small\sqrt{1+f(x+1)}\)，
所以 \(f(x)＝x \small\sqrt{1+f(x+1)}\)。
使用函數 \(f(x) ＝x\small\sqrt{1+f(x+1)}\) 疊代計算可得 \(f(x+1)＝(x+1)\small\sqrt{1+f(x+2)}\) ，
因此 \(f(x)＝x\small\sqrt{1+(x+1)\small\sqrt{1+f(x+2)}}\)。
因為 \(f(x+2)＝(x+2)\small\sqrt{1+f(x+3)}\)，所以
\(f(x)＝x\small\sqrt{1+(x+1)\small\sqrt{1+(x+2)\small\sqrt{1+f(x+3)}}}\)。
因為 \(f(x+3)＝(x+3)\(\small\sqrt{1+f(x+4)}\)，所以
\(f(x)＝x\small\sqrt{1+(x+1)\small\sqrt{1+(x+2)\small\sqrt{1+(x+3)\small\sqrt{1+f(x+4)}}}}\)。
如此疊代計算下去，可得
\(f(x)＝x\small\sqrt{1+(x+1)\small\sqrt{1+(x+2)\small\sqrt{1+(x+3)\small\sqrt{1+(x+4)\small\sqrt{...\small\sqrt{1+(x+n-1)\small\sqrt{1+f(x+n)}}}}}}}\)，其中n是趨向無限大的正整數。
當上列恆等式中的\(x＝1\)時，\(f(1)＝1×3＝3\)
\(3＝f(1)＝1\times{\small\sqrt{1+f(2)}}\)＝\(1\times{\small\sqrt{1+2\times{4}}}\)
\(3＝f(1)＝1\times{\small\sqrt{1+2\small\sqrt{1+f(3)}}}\)＝\(1\times{\small\sqrt{1+2\small\sqrt{1+3\times{5}}}}\)
\(3＝f(1)＝1\times{\small\sqrt{1+2\small\sqrt{1+3\small\sqrt{1+f(4)}}}}\)＝\(1\times{\small\sqrt{1+2\small\sqrt{1+3\small\sqrt{1+4\times{6}}}}}\)
\(3＝f(1)＝1\times{\small\sqrt{1+2\small\sqrt{1+3\small\sqrt{1+4\small\sqrt{1+f(5)}}}}}\)＝\(1\times{\small\sqrt{1+2\small\sqrt{1+3\small\sqrt{1+4\small\sqrt{1+5\times{7}}}}}}\)
In brief,
Let \(f(n,k)=\left\{\begin{array}{lc}\small\sqrt{1+k\cdot f(n,k+1)},&k＜n\\n+1,&k=n\end{array}\right.\)
Then you could interpret that infinitely nested radical as
\(\displaystyle\lim_{n\rightarrow\infty}f(n,2)=\small\sqrt{1+2\cdot\small\sqrt{1+3\cdot\small\sqrt{1+4\cdot\small\sqrt{\dots+n\cdot\small\sqrt1}}}}\)
We can get a brand new general theorem based on this approach.
\({(n+Cm)}^2=n^2+C^2m^2+2Cnm=m^2+(C^2-1)m^2+2Cnm+n^2\)
\(=m^2+\lbrack n^2+2Cnm+(C^2-1)m^2\rbrack\)
\(=m^2+\lbrack n^2+2Cnm+(C-1)(C+1)m^2\rbrack\)
\(=m^2+\lbrack n+(C-1)m\rbrack\lbrack n+(C+1)m\rbrack\)
\(\therefore\forall n\forall m,\;n+Cm=\sqrt{m^2+\lbrack n+(C-1)m\rbrack\lbrack n+(C+1)m\rbrack},\;C\in\mathcal N\;正整數\)
apply to
\({(n+m)}^2=m^2+n(n+2m)\)
\(n+m=\sqrt{m^2+n(n+2m)}\)
\(\Rightarrow n+m=\sqrt{m^2+n\sqrt{m^2+(n+m)(n+3m)}}\)
\(\Rightarrow n+m=\sqrt{m^2+n\sqrt{m^2+(n+m)\sqrt{m^2+(n+2m)(n+4m)}}}\)
\(\Rightarrow n+m=\sqrt{m^2+n\sqrt{m^2+(n+m)\sqrt{m^2+(n+2m)\sqrt{m^2+(n+3m)(n+5m)}}}}\)
Q(2): \(\phi=\frac{1+\sqrt5}2=1+\frac1{1+\frac1{1+\frac1{\ddots+\frac n{1+\ddots}}}}\)
\(x^2-x-1=0\)
\(x^2=x+1\)
\(\Rightarrow x=1+\frac1x\)
\(as\;\;x=1+\frac1x\Rightarrow\;x=1+\frac1{1+\frac1x}\Rightarrow x=1+\frac1{1+\frac1{1+\frac1x}}\;and\;so\;on\)
\(as\;we\;know\;x\neq0\)
\(multiplied\;x\Rightarrow x^2=x+1\)
\(on\;the\;other\;hand\;\frac{x^2}x=\frac{x+1}x\Rightarrow x=1+\frac1x\)
\(x^2-x-1=0,\;x=\frac{1\pm\sqrt{1+4}}2\)
\(x≮0\;\therefore x=\frac{1+\sqrt5}2\)
Q(3): \(\phi=3+\frac1{3+\frac1{3+\frac1{\ddots+\frac n{3+\ddots}}}}\)
\(x=3+\frac1{3+\frac1{\ddots+{\textstyle\frac n{3+\ddots}}}}\)
\(x=3+\frac1{3+\frac1{\ddots+{\textstyle\frac n{3+\ddots}}}}\;implies\;x=3+\frac1x\;and\;x\neq0\)
\(x^2=3x+1\;\Rightarrow\;x^2-3x-1=0\)
\(x=\frac{3\pm\sqrt{3^2+4\cdot1}}2=\frac{3\pm\sqrt{13}}2\)
\(x≮0\;\therefore x=\frac{3+\sqrt{13}}2\)
\(x^2-3x-1=0\;\Rightarrow x^2=3x+1\)
\(x=3+\frac1x\)
\(=3+\frac1{3+\frac1x}\)
Q(4): \(\sqrt[3]{\sqrt[3]2-1}=\frac{1-\sqrt[3]2+\sqrt[3]4}x\)
\(x=\frac{1-\sqrt[3]2+\sqrt[3]4}{\sqrt[3]{\sqrt[3]2-1}}\)
\(let\;y=\sqrt[3]2\)
\(x=\frac{1-y+y^2}{\sqrt[3]{y-1}}\)
\(x^3=\frac{{(y^2-y+1)}^3}{y-1}\)
\(=\frac{{(y^2-y+1)}^3(y+1)}{(y-1)(y+1)}\)
\(=\frac{{(y^2-y+1)}^2\lbrack y^3-y^2+y+y^2-y+1\rbrack}{y^2-1}\)
\(=\frac{{(y^2-y+1)}^2(y^3+1)}{y^2-1}\)
\(=\frac{{(y^2-y+1)}^2({(\sqrt[3]2)}^3+1)}{y^2-1}\)
\(=3\cdot\frac{y^4-y^3+y^2-y^3+y^2-y+y^2-y+1}{y^2-1}\)
\(=3\cdot\frac{2y-2+y^2-2+y^2-y+y^2-y+1}{y^2-1}\)
\(=3\cdot\frac{3y^2-3}{y^2-1}=9\)
\(\Rightarrow x^3=9\)
\(x=\sqrt[3]{3^2}\)
Q(5): \(\phi=\small\sqrt{2+\small\sqrt{2-\small\sqrt{2+\small\sqrt{2-\cdots}}}}\)
\(x=\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\cdots}}}}\)
\(x^2=2+\sqrt{2-\sqrt{2+\sqrt{2-\cdots}}}\)
\(x^2-2=\sqrt{2-\sqrt{2+\sqrt{2-\cdots}}}\)
set \(y=\sqrt{2-\sqrt{2+\sqrt{2-\cdots}}}\)
\(y^2-2=-\sqrt{2+\sqrt{2-\cdots}}\)
\(x^2-2=y\)
\(y^2-2=-x\)
\(x^2-y^2=x+y\)
\((x+y)(x-y)=x+y\)
\(x-y=1\)
\(x=y+1\)
\(x^2-2=y=x-1\)
\(x^2-x-1=0\)
\(x=\frac{1\pm\sqrt5}2\Rightarrow\frac{1+\sqrt5}2\)
\(x^2-x-1=0\)
\(x^2=x+1\)
\(x=\sqrt{x+1}\)
\(x=x^2-1\)
\(x=\sqrt{2+x-1}\)
\(=\sqrt{2+\sqrt{{(x-1)}^2}}\)
\(=\sqrt{2+\sqrt{x^2-2x+1}}\)
\(=\sqrt{2+\sqrt{(x^2-1)-2x+2}}\)
\(=\sqrt{2+\sqrt{x-2x+2}}\)
\(=\sqrt{2+\sqrt{2-x}}\)
\(=\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}}\)
Q(6): \(\phi=\small\sqrt{2\small\sqrt{4\small\sqrt{8\cdots}}}\)
\(\sqrt{2\sqrt{4\sqrt{8\cdots}}}=2^\frac12\cdot4^{\frac12\cdot\frac12}\cdot8^{\frac12\cdot\frac12\cdot\frac12}\cdot16^{\frac12\cdot\frac12\cdot\frac12\cdot\frac12}\cdot\cdots\)
\(=2^\frac12\cdot2^\frac24\cdot2^\frac38\cdot2^\frac4{16}\cdot\cdots\)
\(=2\cdot2^\frac38\cdot2^\frac4{16}\cdot\cdots\)
\(\sum_{n=1}^k\frac n{2^n}=2^{-k}(-k+2^{k+1}-2)\)
Let \(S=\sum_{n=1}^\infty\frac n{2^n}\)
\(S=\frac1{2^1}+\frac2{2^2}+\frac3{2^3}+\frac4{2^4}+\frac5{2^5}+\cdots\)
\(2S=\frac1{2^0}+\frac2{2^1}+\frac3{2^2}+\frac4{2^3}+\frac5{2^4}+\cdots\)
\(2S-S=\frac1{2^0}+(\frac2{2^1}-\frac1{2^1})+(\frac3{2^2}-\frac2{2^2})+(\frac4{2^3}-\frac3{2^3})+\cdots\)
\(S=1+\frac1{2^1}+\frac1{2^2}+\frac1{2^3}+\frac1{2^4}+\cdots\)
\(S=\frac1{1-{\frac12}}=2\)
\(\therefore2^\frac12\cdot2^\frac24\cdot2^\frac38\cdot2^\frac4{16}\cdot\cdots\)
\(=2^{(\frac1{2^1}+\frac2{2^2}+\frac3{2^3}+\frac4{2^4}+\frac5{2^5}+\cdots)}\)
\(=2^2=4\) ◼
Q(11): \(x^3+{(x^2-3x+4)}^2+2\)
Q(12): \(\frac1\pi=\frac{2\sqrt2}{9801}{\displaystyle\small\sum_{n=0}^\infty}\frac{{(4n)}!\cdot{(1103+26390n)}}{{(n!)}^4{\cdot}396^{4n}}.\)
Q(13): \(\sqrt{\phi+2}-\phi=\frac{e^{-\frac{2\pi}5}}{1+{\frac{e^{-2\pi}}{1+{\frac{e^{-4\pi}}{1+{\frac{e^{-6\pi}}{1+\cdots}}}}}}}=0.2840...\;\Rightarrow\phi=\frac{1+\sqrt5}2\)
Q(14): \(e=3+\frac{-1}{4+{\frac{-2}{5+{\frac{-3}{6+{\frac{-4}{7+\cdots}}}}}}}\)
Q(15): \(\frac e{e-2}=4-\frac1{5-{\frac2{6-{\frac3{7-{\frac4{8-\cdots}}}}}}}\)
Q(16): \(\frac4{3\pi-8}=3-\frac{1\cdot1}{6-{\frac{2\cdot3}{9-{\frac{3\cdot5}{12-{\frac{4\cdot7}{15-\cdots}}}}}}}\)
Q(17): \(\frac8{\pi^2}=1-\frac{2\cdot1^4-3\cdot1^3}{7-{\frac{2\cdot2^4-3\cdot2^3}{19-{\frac{2\cdot3^4-3\cdot3^3}{37-{\frac{2\cdot4^4-3\cdot4^3}{61-\ddots}}}}}}}\)
Q(18): \(\frac{e^{\frac{-2\pi}5}}{1+{\frac{e^{-2\pi}}{1+{\frac{e^{-4\pi}}{1+\ddots}}}}}=\sqrt{\frac{5+\sqrt{5}}2}-\frac{\sqrt{5}+1}2\)
Q(19): \(\frac{e^{\frac{-\pi}5}}{1+{\frac{e^{-\pi}}{1+{\frac{e^{-2\pi}}{1+\ddots}}}}}=\sqrt{\frac{5-\sqrt{5}}2}-\frac{\sqrt{5}-1}2\)
Q(20): \(\frac1\pi=\frac{2\sqrt2}{9801}{\displaystyle\small\sum_{k=0}^\infty}\frac{(4k)!(1103+26390k)}{(k!)^4396^{4k}}\) 1914
Q(21): \(\frac4\pi=\frac1{882}{\displaystyle\small\sum_{k=0}^\infty}\frac{{(-1)}^k(4k)!(1123+21460k)}{(4^kk!)^4882^{2k}}\) 1914
Q(22): \(\frac1\pi=12{\displaystyle\small\sum_{k=0}^\infty}\frac{{(-1)}^k(6k)!(13591409+545140134k)}{(3k)!(k!)^3{(640320^3)}^{k+{\small\frac12}}}\) Chudonovsky 1987
Q(23): \(\phi=\small\sqrt{\frac12(5+\sqrt5)}-\frac{e^{-\frac{2\pi}5}}{1+\frac{e^{-2\pi}}{1+\frac{e^{-4\pi}}{1+\frac{e^{-6\pi}}{1+\ddots}}}}\)
Q(24): \(\frac{e^{-\frac{2\pi}5}}{1+\frac{e^{-2\pi}}{1+\frac{e^{-4\pi}}{1+\frac{e^{-6\pi}}{1+\ddots}}}}=\sqrt[4]5\sqrt\phi-\phi\)
Q(25): \(\sqrt{\frac{\pi e}2}=1+\frac1{1⋅3}+\frac1{3⋅5}+\cdots+\frac1{(2n-1)!!}+\cdots+\frac{1}{1+\frac1{1+{\frac2{1+{\frac3{\ddots+{\frac n{1+\ddots}}}}}}}}\)
https://www.youtube.com/watch?v=6iTdNmDHfV0&t=1268s
Q(26): \(\frac{e^\pi-1}{e^\pi+1}=\frac\pi{2+{\frac{\pi^2}{6+{\frac{\pi^2}{10+{\frac{\pi^2}{14+\ddots}}}}}}}\)
Nested Radical
Q(B1): \(\sqrt[n]{a\cdot\sqrt[n]{a\cdot\sqrt[n]{a\cdots}}}=\sqrt[n-1]a\)
\(x=\sqrt[n]{a\cdot x}\)
\(x=\sqrt[n]{a\cdot\sqrt[n]{a\cdot x}}\)
\(x^n=a\cdot x\)
\(x^{n-1}=a\)
\(\therefore x=\sqrt[n-1]a\)
Q(B2): \(\sqrt[n]{\frac {a}{\sqrt[n]{\frac {a}{\sqrt[n]{\frac {a}\ddots}}}}}=\sqrt[n+1]a\)
\(x=\sqrt[n]{\frac ax}\)
\(x^n=\frac ax\)
\(x^{n+1}=a\)
\(x=\sqrt[n+1]a\)
Q(B3): \(\sqrt[n]{a\cdot\sqrt[m]{a\cdot\sqrt[n]{a\cdot\sqrt[m]{a\cdots}}}}=\sqrt[m\cdot n-1]{a^{(m+1)}}\)
\(x=\sqrt[n]{a\cdot\sqrt[m]{a\cdot x}}\)
\(x=\sqrt[n]{a\cdot\sqrt[m]{a\cdot\sqrt[n]{a\cdot\sqrt[m]{a\cdot x}}}}\\x=\sqrt[n]{a\cdot\sqrt[m]{a\cdot x}}\)
\(x^n=a\cdot\sqrt[m]{a\cdot x}\)
\(\frac1ax^n=\sqrt[m]{a\cdot x}\)
\(\frac1{a^{m+1}}x^{n\cdot m-1}=1\)
\(x^{n\cdot m-1}=a^{m+1}\)
\(x=\sqrt[n\cdot m-1]{a^{m+1}}\)
Q(B4): \(\sqrt[n]{a\cdot\sqrt[m]{b\cdot\sqrt[n]{a\cdot\sqrt[m]{b\cdots}}}}=\sqrt[m\cdot n-1]{a^m\cdot b}\)
\(x=\sqrt[n]{a\cdot\sqrt[m]{b\cdot x}}\)
\(x=\sqrt[n]{a\cdot\sqrt[m]{b\cdot\sqrt[n]{a\cdot\sqrt[m]{b\cdot x}}}}\)
\(x^n=a\cdot\sqrt[m]{b\cdot x}\)
\(\frac1ax^n=\sqrt[m]{b\cdot x}\)
\(\frac1{a^m}x^{n\cdot m}=b\cdot x\)
\(\frac1{a^m\cdot b}x^{n\cdot m-1}=1\)
\(x^{n\cdot m-1}=a^m\cdot b\)
\(x=\sqrt[n\cdot m-1]{a^m\cdot b}\)
Q(B5): \(\sqrt{a+\sqrt{a+\sqrt{a+\cdots}}}=\frac{1+\sqrt{4a+1}}2\)
\(x=\sqrt{a+x}\)
\(x^2=a+x\)
\(x^2-x-a=0\)
\(x=\frac{1\pm\sqrt{4a+1}}2\)
\(x＞0\;\therefore x=\frac{1+\sqrt{4a+1}}2\)
Q(B6): \(\sqrt{a-\sqrt{a-\sqrt{a-\cdots}}}=\frac{-1+\sqrt{4a+1}}2\)
\(x=\sqrt{a-x}\)
\(x^2=a-x\)
\(x^2+x-a=0\)
\(x=\frac{-1\pm\sqrt{4a+1}}2\)
\(x＞0\;\therefore x=\frac{-1+\sqrt{4a+1}}2\)
Egyptian fraction
An Egyptian fraction is a finite sum of distinct unit fractions, each fraction in the expression has a numerator equal to 1 and a denominator that is a positive integer, and all the denominators differ from each other.
\(\frac27=\frac14+\frac1{28}=\frac15+\frac1{20}+\frac1{28}\)
Greedy algorithm
\(\frac{19}{20}\Rightarrow\frac{20}{19}=1,\;\frac{19}{20}-\frac1{1+1}=\frac9{20}\)
\(\frac9{20}\Rightarrow\frac{20}9=2,\;\frac9{20}-\frac1{1+2}=\frac7{60}\)
\(\frac7{60}\Rightarrow\frac{60}7=8,\;\frac7{60}-\frac1{1+8}=\frac1{180}\)
\(\frac{19}{20}=\frac12+\frac13+\frac19+\frac1{180}\)
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Euler’s Series of Pi
https://www.youtube.com/watch?v=apODop1vnBg
https://www.youtube.com/watch?v=wlbKFvKTMNw
https://en.wikipedia.org/wiki/Madhava_of_Sangamagrama
https://mathworld.wolfram.com/NestedRadical.html
Ramanujan-Nagell equation
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∮ ∯ ∰ ∫∬ ∭⨌ ＞ ＜ ⋉ % § ↯
½, ↉, ⅓, ⅔, ¼, ¾, ⅕, ⅖, ⅗, ⅘, ⅙, ⅚, ⅐, ⅛, ⅜, ⅝, ⅞, ⅑, ⅒