I've previously discussed an interesting math problem posed by Srinivasa Ramanujan way back in 1911.
In this third posting on this topic, I'll present a proof that there are an infinite number of valid solutions to his puzzle.
Ramanujan said the solution was \( \{x=3\} \). I'm claiming the proper solution is \( \{x=3+n : n \ge 0\} \). Lets see how that works out.
We start by unwrapping the first few radicals and simplifying a bit.
If we work out the full radical at each step, we rapidly end up with very high-order polynomials on the right side of the equation. However, the 0th and 1st order terms will always only depend on the 0th and 1st order term of the polynomial for the previous radical. Since we're only working with values of n greater than zero, all of the higher-order terms [in red] will have positive values.
This allows us to simply discard them, converting the equations into inequalities. This cleans up our work a lot.
- the-biologist-is-in.blogspot.ca/2014/12/mathematical-recreations-ramanujans.html
- the-biologist-is-in.blogspot.com/2016/08/mathematical-recreations-ramanujans-2.html
In this third posting on this topic, I'll present a proof that there are an infinite number of valid solutions to his puzzle.
\( x = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\cdots}}}} \)
Ramanujan said the solution was \( \{x=3\} \). I'm claiming the proper solution is \( \{x=3+n : n \ge 0\} \). Lets see how that works out.
\( 3+n = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\cdots}}}} \)
We start by unwrapping the first few radicals and simplifying a bit.
\( \sqrt{1+3\sqrt{1+4\sqrt{\cdots}}} = \frac{(3+n)^2 -1}{2} = \frac{9+6n+n^2-1}{2} = 4+3n+\frac{1}{2}n^2 \)
\( \sqrt{1+4\sqrt{1+5\sqrt{\cdots}}} = \frac{(4+3n+\frac{1}{2}n^2)^2 -1}{3} = \frac{16+24n+13n^2+3n^3+\frac{1}{4}n^4-1}{3} = 5+8n+\frac{13}{3}n^2+n^3+\frac{1}{12}n^4 \)
If we work out the full radical at each step, we rapidly end up with very high-order polynomials on the right side of the equation. However, the 0th and 1st order terms will always only depend on the 0th and 1st order term of the polynomial for the previous radical. Since we're only working with values of n greater than zero, all of the higher-order terms [in red] will have positive values.
\( \sqrt{1+3\sqrt{1+4\sqrt{\cdots}}} = 4+3n \color{red}{+ \frac{1}{2}n^2} \)
\( \sqrt{1+4\sqrt{1+5\sqrt{\cdots}}} = 5+8n \color{red}{+ \frac{13}{3}n^2+n^3+\frac{1}{12}n^4} \)
This allows us to simply discard them, converting the equations into inequalities. This cleans up our work a lot.
\( \sqrt{1+3\sqrt{1+4\sqrt{\cdots}}} \ge 4+3n \)
\( \sqrt{1+4\sqrt{1+5\sqrt{\cdots}}} \ge 5+8n \)
Unwrap another radical. I'm using a bit of an unusual notation in the right half of the inequality, where I'm using equalities inside of the large parentheses for keeping track of a few steps of simplification.
Unwrap another radical. I'm using a bit of an unusual notation in the right half of the inequality, where I'm using equalities inside of the large parentheses for keeping track of a few steps of simplification.
\( \sqrt{1+5\sqrt{1+6\sqrt{\cdots}}} \ge \left(\frac{(5+8n)^2
-1}{4} = \frac{25+80n+64n^2-1}{4} =
6+20n+16n^2\right) \)
\( \sqrt{1+5\sqrt{1+6\sqrt{\cdots}}} \ge 6+20n\color{red}{+16n^2} \)
Since we already have an inequality, we can go ahead and discard this higher-order term.
\( \sqrt{1+5\sqrt{1+6\sqrt{\cdots}}} \ge 6+20n \)
The first order terms of the polynomials increase with each unwrapping of a radical. These terms form a series that has very simple behavior.
Because the trimmed polynomials representing each successive radical are <= the true polynomials, and they increase towards infinity, the true polynomials also increase towards infinity.
Unwrapping successive radicals from Ramanujan's solution results in a simple series that races upwards to infinity.
Unwrapping successive radicals from my solution results in a more complicated series that also races upwards to infinity.
As long as the value of 'n' is greater than zero, no contradictions (such as negative values for a radical) arise, thus Ramanujan's solution to his puzzle is incomplete.
A more complete solution is:
Ramanujan's puzzle remains of interest to many and seems to inspire ongoing conversations in various online forums. I have come across a few discussions where people mention calculating subsequent radicals for different starting values, the method at the root of my proof, but I've never come across anyone discussing an actual proof. Ramanujan's solution to his puzzle held for 115 years, but I've now proven his solution to be incomplete. I wonder how long it will take for my proof to start appearing in some of those forum discussions.
Unfortunately, my solution remains incomplete. As of yet, I do not have a proof to back up my intuition that all values less than three are invalid solutions. The method of proof I used here is not simply applied to show values less than 3 are invalid solutions, but I am pondering on methods of doing so.
Stay tuned for further developments.
References
\( \left[ k_0 = 1; k_m = 2k_{m-1}+2^{m-1} \right] \)
\( \lim\limits_{m \to \infty} k_m = \infty \)
Because the trimmed polynomials representing each successive radical are <= the true polynomials, and they increase towards infinity, the true polynomials also increase towards infinity.
Unwrapping successive radicals from Ramanujan's solution results in a simple series that races upwards to infinity.
\( [3, 4, 5, 6, 7, \cdots, \infty ] \)
Unwrapping successive radicals from my solution results in a more complicated series that also races upwards to infinity.
\( [3+1n, 4+3n\color{red}{[+\cdots]}, 5+8n\color{red}{[+\cdots]}, 6+20n\color{red}{[+\cdots]}, 7+48n\color{red}{[+\cdots]}, \cdots, \infty ] \)
As long as the value of 'n' is greater than zero, no contradictions (such as negative values for a radical) arise, thus Ramanujan's solution to his puzzle is incomplete.
 |
| Profiles from valid solutions (>= 3) are in green. Profiles from invalid solutions (< 3) are in red. Note the change in scale above and below zero. |
\( \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\cdots}}}} \ge 3 \)
Ramanujan's puzzle remains of interest to many and seems to inspire ongoing conversations in various online forums. I have come across a few discussions where people mention calculating subsequent radicals for different starting values, the method at the root of my proof, but I've never come across anyone discussing an actual proof. Ramanujan's solution to his puzzle held for 115 years, but I've now proven his solution to be incomplete. I wonder how long it will take for my proof to start appearing in some of those forum discussions.
Unfortunately, my solution remains incomplete. As of yet, I do not have a proof to back up my intuition that all values less than three are invalid solutions. The method of proof I used here is not simply applied to show values less than 3 are invalid solutions, but I am pondering on methods of doing so.
Stay tuned for further developments.
References
I found a general solution for a diferent case. See academia.edu, George Plousos, paper "Nested radicals and trigonometry" (pdf).
ReplyDelete