On Question by Professor Scholze

May 10th, 2019 by Aziz Lokhandwala

Question: $$-4 = {\sqrt{1 - 3* \sqrt{1-4...} }}$$

Solution: let's assume, $${x(n) = (n-3)*(n-5)}$$ This can be also written as, $${(n-3)* \sqrt{1 + (n-4)*(n-6)}}$$ $${(n-3)* \sqrt{1 + x(n-1)}}$$ $${(n-3)* \sqrt{1 + (n-4)* \sqrt{1 + x(n-2)}}}$$ $${(n-3)* \sqrt{1 + (n-4)* \sqrt{1 + (n-5)* \sqrt{1 + x(n-3)}}}}$$ $${(n-3)* \sqrt{1 + (n-4)* \sqrt{1 + (n-5)* \sqrt{1 + (n-6)* \sqrt{1 + ...}}}}}$$

Now, let's put n = 1, $$8 = -2*{\sqrt{1 - 3* \sqrt{1-4...} }}$$ $$-4 = {\sqrt{1 - 3* \sqrt{1-4...} }}$$

Hence Proved