Math, asked by jhumakardeb, 1 year ago

what are ramanujans infinte roots?

Answers

Answered by Anubhavdeb
2

Answer:

Ramanujan has many infinite expression one of which is$ 3=\sqrt9$\\$3=\sqrt{1+8}$\\ $3=\sqrt{1+2\cdot4}$  \\$3=\sqrt{1+2\cdot\sqrt{16}}$\\$3=\sqrt{1+2\cdot \sqrt{1+15}}$\\$3=\sqrt{1+2\cdot\sqrt{1+3\cdot5} }$\\$3=\sqrt{1+2\cdot\sqrt{1+3\cdot\sqrt{25}} }$\\$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+24}} }$\\$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+4\cdot 6}} }$\\$3=\sqrt{1+2\cdot \sqrt{1+3\cdot \sqrt{1+4\cdot \sqrt{36}}} }$\\so,\\$3=\sqrt{1+2\cdot\sqrt{1+3\cdot\sqrt{1+4\cdot \sqrt{1+ \cdots}}} }$\\\\

Which can be generalised as

$ n=\sqrt(n^2)$  \\$n=\sqrt{1+n^2-1^2}$  \\$n=\sqrt{1+(n-1) \cdot  (n+1)}$\\  $n=\sqrt{1+(n-1)\cdot \sqrt{(n+1)^2}}$  \\$n=\sqrt{1+(n-1)\cdot \sqrt{1+(n+1)^2-1^2}}$ \\$n=\sqrt{1+(n-1)\cdot \sqrt{1+n \cdot (n+2)} }$   \\$n=\sqrt{1+(n-1)\cdot \sqrt{1+n \cdot \sqrt{(n+2)^2}} }$  \\$n=\sqrt{1+(n-1)\cdot \sqrt{1+n \cdot \sqrt{1+(n+2)^2-1^2}} }$\\$n=\sqrt{1+(n-1)\cdot \sqrt{1+n \cdot \sqrt{1+(n+1)\cdot (n+3)}} }$  \\$n=\sqrt{1+(n-1)\cdot \sqrt{1+n \cdot \sqrt{1+(n+1)\cdot \sqrt{(n+3)^2}}} }$ \\

So,

$n=\sqrt{1+(n-1)\cdot \sqrt{1+n \cdot \sqrt{1+(n+1)\cdot \sqrt{1+(n+2)}}} }.....$

Thus using the above expression we can turn any number greater than 1 to this infinte expression.

THANK YOU:)))

Answered by sunitameena45647
2

Answer:

it's too long bhro and sis but I am trying to get the correct answer

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