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(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 \(S=\sum_{n=1}^\infty\frac n{2^n}\)

\(S=\frac1{2^1}+\frac2{2^2}+\frac3{2^3}+\frac4{2^4}+\frac5{2^5}+\cdots\)

\(2S=\frac1{2^0}+\frac2{2^1}+\frac3{2^2}+\frac4{2^3}+\frac5{2^4}+\cdots\)

\(2S-S=\frac1{2^0}+(\frac2{2^1}-\frac1{2^1})+(\frac3{2^2}-\frac2{2^2})+(\frac4{2^3}-\frac3{2^3})+\cdots\)

\(S=1+\frac1{2^1}+\frac1{2^2}+\frac1{2^3}+\frac1{2^4}+\cdots\)

\(S=\frac1{1-{\frac12}}=2\)

\(\therefore2^\frac12\cdot2^\frac24\cdot2^\frac38\cdot2^\frac4{16}\cdot\cdots\)

\(=2^{(\frac1{2^1}+\frac2{2^2}+\frac3{2^3}+\frac4{2^4}+\frac5{2^5}+\cdots)}\)

\(=2^2=4\) ◼

Elliptic integral

\( \lambda = \left(\frac{a - b}{a + b}\right)^2 \)

\( s = \pi(a + b)\frac{1 + 3\lambda}{10 + \sqrt{4 - 3\lambda}} \)

The system of equations describes Ramanujan's Second Approximation (published in 1914) for the perimeter (circumference) of an ellipse. Its perimeter requires evaluating a complex infinite series. Ramanujan designed this highly accurate empirical fraction to perfectly match that series.
The Equations:
\( \lambda = \left(\frac{a - b}{a + b}\right)^2 \)
\( s = \pi(a + b)\frac{1 + 3\lambda}{10 + \sqrt{4 - 3\lambda}} \)

1. The True Infinite Series

The exact perimeter (\(P\)) of an ellipse with semi-major axis \(a\) and semi-minor axis \(b\) is calculated using a complete elliptic integral of the second kind. When expanded as an infinite series using the parameter \(\lambda\), the exact formula yields:
\( P = \pi(a+b) \left[ 1 + \frac{1}{4}\lambda + \frac{1}{64}\lambda^2 + \frac{1}{256}\lambda^3 + \frac{25}{16384}\lambda^4 + \dots \right] \)

2. Ramanujan's Approximation Method

Ramanujan did not leave behind a traditional line-by-line derivation for how he structured the fraction \(\frac{1 + 3\lambda}{10 + \sqrt{4-3\lambda}}\). Instead, he derived it using a technique called asymptotic matching (or a truncated continued fraction expansion) to ensure the terms aligned seamlessly with the calculus-based expansion.

To see why his formula works so perfectly, we can expand his approximation as a Taylor series around \(\lambda = 0\) (which corresponds to a perfect circle where \(a=b\) and \(\lambda=0\)):

1. First, isolate the square root term in the denominator and apply a binomial expansion:
   \( \sqrt{4 - 3\lambda} = 2\left(1 - \frac{3}{4}\lambda\right)^{1/2} = 2 - \frac{3}{4}\lambda - \frac{9}{64}\lambda^2 - \frac{27}{512}\lambda^3 - \dots \)
2. Substitute this back into the full denominator:
   \( 10 + \sqrt{4 - 3\lambda} = 12 - \frac{3}{4}\lambda - \frac{9}{64}\lambda^2 - \frac{27}{512}\lambda^3 - \dots \)
3. Expand the entire rational fraction expression \(\frac{1 + 3\lambda}{10 + \sqrt{4-3\lambda}}\) into a power series: \( \frac{1 + 3\lambda}{10 + \sqrt{4-3\lambda}} = 1 + \frac{1}{4}\lambda + \frac{1}{64}\lambda^2 + \frac{1}{256}\lambda^3 + \frac{25}{16384}\lambda^4 + \dots \)

3. The Brilliant Result

When you multiply this expansion by the leading term \(\pi(a+b)\), you get:
\( s \approx \pi(a+b) \left[ 1 + \frac{1}{4}\lambda + \frac{1}{64}\lambda^2 + \frac{1}{256}\lambda^3 + \frac{25}{16384}\lambda^4 + \dots \right] \)

Ramanujan's brilliantly engineered algebraic expression perfectly matches the exact calculus-based infinite series up to the \(\lambda^3\) term. Even at the \(\lambda^4\) term, the error is so remarkably small that for most real-world ellipses, the formula is accurate to within a tiny fraction of a percent.