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Ramanujan's Infinite Square Root Riddle: Solution Explained
Srinivasa Ramanujan, a renowned Indian mathematician, was known for his intuitive and profound insights into the world of numbers. One of his famous riddles involves an infinite square root, a mathematical concept that often sparks curiosity and intrigue. Let's dive into this riddle and explore its elegant solution.
The Riddle
The riddle presents the following expression:
√(1 + 2√(1 + 3√(1 + 4√(1 + ...))))
This seemingly endless chain of nested square roots might appear daunting at first glance. However, Ramanujan's genius lies in revealing a simple and elegant solution.
The Solution
To solve this riddle, we'll employ a technique called iteration. Let's represent the entire expression as 'x':
x = √(1 + 2√(1 + 3√(1 + 4√(1 + ...))))
Now, let's isolate the first square root and rewrite the expression:
x = √(1 + 2y)
Where 'y' represents the remaining infinite series:
y = √(1 + 3√(1 + 4√(1 + ...)))
Notice that 'y' has the same structure as the original expression 'x', except the starting number within the square root is increased by 1. This pattern allows us to express 'y' in terms of 'x':
y = √(1 + 3x)
Now we have two equations:
x = √(1 + 2y)
y = √(1 + 3x)
To solve for 'x', we can substitute the second equation into the first:
x = √(1 + 2√(1 + 3x))
Squaring both sides, we get:
x² = 1 + 2√(1 + 3x)
Rearranging the terms:
x² - 1 = 2√(1 + 3x)
Squaring both sides again:
(x² - 1)² = 4(1 + 3x)
Expanding and simplifying:
x⁴ - 2x² + 1 = 4 + 12x
x⁴ - 2x² - 12x - 3 = 0
This is a quartic equation, which can be solved using various methods. By factoring, we find that one of the solutions is x = 3. We can verify that this solution satisfies the original equation.
Conclusion
Ramanujan's infinite square root riddle showcases the power of iterative methods and elegant mathematical manipulation. By expressing the problem in terms of itself and employing algebraic techniques, we arrive at the solution x = 3. This riddle demonstrates that even seemingly complex mathematical problems can have surprisingly simple and beautiful solutions.
This riddle is a great example of how seemingly complex mathematical problems can have simple and beautiful solutions. It also demonstrates the power of iterative methods in solving problems.