Ramanujan's Formula for Nested Radical

Ramanujan discovered in 1911 that:

$x+n+a = \sqrt{ax+(n+a)^2+x\sqrt{a(x+n)+(n+a)^2+(x+n)\sqrt{a(x+2n)+(n+a)^2+(x+2n)\sqrt \cdots}}}$

Note: I hope someone can provide a proof for the formula.

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Comments

let $f(x) = x+n+a$ and note that $f(x+n) = x+2n+a$

$f(x) = \sqrt{(x+n+a)^2} = \sqrt{x^2+a^2+n^2+2ax+2an+2nx} = \sqrt{ax+(n^2+2an+a^2)+x(x+2n+a)} = \sqrt{ax+(n+a)^2+xf(x+n)}$

$= \sqrt{ax+(n+a)^2+x\sqrt{ax+(n+a)^2+xf(x+2n)}} = \sqrt{ax+(n+a)^2+x\sqrt{ax+(n+a)^2+\sqrt{ax+(n+a)^2+(x+2n)\sqrt{\cdots}}}}$

Tommy Li - 3 years, 5 months ago

Thanks.

Chew-Seong Cheong - 3 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$