Infinite Product Series

Question

I was solving my school book and got bored so I made my own question for solving myself and to entertain myself.

I made something like this: Find solution of $$ \log_{e}x= \sqrt{x\sqrt{x\sqrt{x\sqrt{x...\infty}}}} $$

What I did in solution was this(sorry I was not able to add the solution in coded form for better representation as I am using stack exchange from my Android):-

https://bit.ly/3cLFCgv

So, I got x=1 as my solution BUT if we put x=1 in the former statement it will not hold true as it will come as 0=1 which is actually not true!!

Please let me know where have I mistaken!!

Thank you in Advance!!

Answer 1

The expression is equivalent to $\ln x=\sqrt{x\ln x}$ which can only be true if $\ln x\ge 0$ or $x\ge 1$. Now for that set of values of $x$, we can square both sides of the equality to get $x=\ln x$ which is again equivalent to $x=e^x$ for which you can clearly see that no solutions exist.

Answer 2

Congratulations! You just proved (by contradiction) that no solution exists. Your proof is valid under the assumption that $\ln x = \sqrt{x\sqrt{x\cdots}}$ exists. To formalize this, you need only add at the beginning of your proof: "Assume that some $x$ satisfies..." and proceed from here.

A common mistake, particularly when dealing with infinite sums and infinite products, is to assume that a solution exists when it does not. When we start with the assumption "Suppose a solution exists," then either 1) we find out by contradiction that we were wrong, or 2) we may find out some interesting properties that any solution satisfies which can lead us to a solution.