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Following was proposed by Ramanujan:

$ \sqrt{11-2\sqrt{11+2\sqrt{11-2\sqrt{11+\cdots}}}}=1+4\sin(10^o)$

Working on this I got the radical on the left equal to $(1+2\sqrt{2})$ implying that $\sin(10^o)=1/\sqrt{2}$

How is this possible? What is wrong here?

Mrigank Shekhar Pathak
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  • $\sin(\color{red}{45}^o)=\frac{1}{\sqrt{2}}$ ? – Donald Splutterwit Jun 10 '17 at 13:50
  • I think you could show how you obtained $1+2\sqrt{2}$ for the left hand side since that seems to be the issue – user12345 Jun 10 '17 at 13:55
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    I got $2 \sqrt{2} - 1$ for the LHS. – Dando18 Jun 10 '17 at 13:58
  • @Dando18 Even W|A says that. – Jaideep Khare Jun 10 '17 at 14:03
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    It is a more long for comment but I write it: $$2\sqrt(2)-1=\sqrt{(2\sqrt(2)-1)^2}=\sqrt{9-4\sqrt(2)}=\sqrt {11-2(2\sqrt(2)+1)}$$ $$=\sqrt {11-2\sqrt {(2\sqrt(2)+1)^2}}=\sqrt {11-2\sqrt {9+4\sqrt(2)}}=\sqrt {11-2\sqrt {11+2(2\sqrt(2)-1)}}$$ $$ =\sqrt {11-2\sqrt {11+2(\sqrt{(2\sqrt(2)-1)^2}}}=\sqrt {11-2\sqrt {11+2\sqrt {11-2\sqrt{\cdots}}}} $$ Maybe useful. – Amin235 Jun 10 '17 at 15:04

3 Answers3

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Actually you have written the identity incorrectly. The original, by Ramanujan, stated:

$$ 1 + 4\sin 10^\circ = \sqrt{11 - 2\sqrt{11 + 2\sqrt{11 - 2(1 + 4\sin 10^\circ) \dots }}} $$

Edit: note the period of signs is $-, +, -, -, +, -, -, +, -, \dots$

or letting $ x = 1 + 4\sin 10^\circ $

$$ x = \sqrt{11 - 2\sqrt{11 + 2\sqrt{11 - 2x \dots }}} $$

According to WA, we have $x \approx 1.695$ and again $1 + 4 \sin 10^\circ \approx 1.695$

Mr Pie
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Dando18
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  • Try this ... https://www.wolframalpha.com/input/?i=sqrt%5B11+-+2sqrt%5B11%2B+2x%5D%5D+%3D+x – Donald Splutterwit Jun 10 '17 at 14:16
  • @DonaldSplutterwit won't let me follow that link – Dando18 Jun 10 '17 at 14:24
  • @Dando18 you could visit : https://www.google.co.in/url?url=https://pdfs.semanticscholar.org/ae75/da0be9fb455e2c55daa5fca46ae63e6a60bd.pdf&rct=j&sa=U&ved=0ahUKEwjizq-CvbPUAhXMRo8KHS-jB2kQFggpMAE&sig2=xJKzJdEKawNUHnFw83UhAA&q=problems+published+by+ramanujan&usg=AFQjCNHBMQq-GgLprHX8m7qQuztDwbjwPg and see page 4 – Mrigank Shekhar Pathak Jun 10 '17 at 14:25
  • sqrt[11 - 2sqrt[11+ 2x]] = x ... <- input this into Wolfie @Dando18 – Donald Splutterwit Jun 10 '17 at 14:25
  • @DonaldSplutterwit I'm not sure what you're implying. The expression in my post is different in signs $-, +, -, -, +, ... $ – Dando18 Jun 10 '17 at 14:40
  • @MrigankShekharPathak see the top link I added. and also below number b in your link, it says the period of signs is $-, +, -, -, +, -, -, +, +, \dots$ – Dando18 Jun 10 '17 at 14:42
  • @Dando18 ... The OP confused me into thinking the $+$ & $-$ alternated ... You are right, we need to look more carefully at the original result. It repeats every third time ... $(-,+,-)$ ! – Donald Splutterwit Jun 10 '17 at 14:48
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Let $x=\sqrt{11-2\sqrt{11+2\sqrt{11-2\sqrt{11+\cdots}}}}$ and $ y=\sqrt{11+2\sqrt{11-2\sqrt{11+2\sqrt{11-\cdots}}}}$ then \begin{eqnarray*} x=\sqrt{11-2y} \\ y=\sqrt{11+2x} \end{eqnarray*} so $x^2=11-2y$ & $y^2=11+2x$ ... after a little algebra ... \begin{eqnarray*} x^4-22x^2-8x+77=0 \\ (x^2+2x-7)(x^2+2x-11)=0 \end{eqnarray*} So we have the possible solutions $x=-1 \pm 2 \sqrt{2}$ & $x=1 \pm 2 \sqrt{3}$. One can verify that \begin{eqnarray*} \sqrt{11-2\sqrt{11+2\sqrt{11-2\sqrt{11+\cdots}}}}=\color{red}{2 \sqrt{2}-1}. \end{eqnarray*} If the $+$ and $-$'s alternate then the above value is correct ... Ramanujan actually does the $(-,+,-)$ repeating every $3$ times ... then the result is \begin{eqnarray*} \sqrt{11-2\sqrt{11+2\sqrt{11-2\sqrt{11\color{red}{-}\cdots}}}}=1+ \sin(10^o). \end{eqnarray*}

Donald Splutterwit
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  • From your solution it reveals that $\sin 10^o = \frac {\sqrt {2} -1} {2} \approx 0.20711$ where as the actual value of $\sin 10^o$ is approximately equal to $0.1736$.Why is this happening? – Arnab Chattopadhyay. Jun 10 '17 at 14:35
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Inspired by Ramanujan's above formula, I added some similar results here:

$ $

$ 4 \cos \left(\frac{4 \pi }{9}\right)+1=\sqrt{11-2 \sqrt{11+2 \sqrt{11-2 \left(4 \cos \left(\frac{4 \pi }{9}\right)+1\right)}}} $

$ 4 \cos \left(\frac{2 \pi }{9}\right)+1=\sqrt{11+2 \sqrt{11-2 \sqrt{11-2 \left(4 \cos \left(\frac{2 \pi }{9}\right)+1\right)}}} $

$ 4 \cos \left(\frac{\pi }{9}\right)-1=\sqrt{11-2 \sqrt{11-2 \sqrt{11+2 \left(4 \cos \left(\frac{\pi }{9}\right)-1\right)}}} $

$ $

$ 4 \cos \left(\frac{\pi }{4}\right)+1=\sqrt{11+2 \sqrt{11-2 \left(4 \cos \left(\frac{\pi }{4}\right)+1\right)}} $

$ 4 \cos \left(\frac{\pi }{4}\right)-1=\sqrt{11-2 \sqrt{11+2 \left(4 \cos \left(\frac{\pi }{4}\right)-1\right)}} $

$ $

$ 4 \cos \left(\frac{\pi }{6}\right)+1=\sqrt{11+2 \left(4 \cos \left(\frac{\pi }{6}\right)+1\right)} $

$ 4 \cos \left(\frac{\pi }{6}\right)-1=\sqrt{11-2 \left(4 \cos \left(\frac{\pi }{6}\right)-1\right)} $

$ $

$ 4 \cos \left(\frac{\pi }{4}\right)+1=\sqrt{7+2 \left(4 \cos \left(\frac{\pi }{4}\right)+1\right)} $

$ 4 \cos \left(\frac{\pi }{4}\right)-1=\sqrt{7-2 \left(4 \cos \left(\frac{\pi }{4}\right)-1\right)} $

$ $

Other similar results can be found here:

https://math.stackexchange.com/a/4233146/954936

https://math.stackexchange.com/a/4232525/954936

http://eslpower.org

http://eslpower.org/Notebook.htm

Chen Shuwen
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