Learn with Fun: Solve a Mysterious Mathematical Puzzle to Learn Stability Analysis of Infinite Exponentiation
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Learn with Fun: Solve a Mysterious Mathematical Puzzle to Learn Stability Analysis of Infinite Exponentiation

Puzzles have perplexed human mind for centuries. In this article I am going to talk about one such baffling puzzle on infinite exponentiation which will lead to a simple non-linear equation. During the course of solving we will learn how to analyze stability of a solution of an algebraic equation. This will reveal one application of non-linear dynamics on simple mathematical problems.

Mathematics is said to be the queen of all sciences and one aspect of its beauty is revealed by mathematical puzzles. Puzzles have perplexed human mind for centuries and in this article I am going to talk about one such baffling puzzle. During the course of solving we will also learn how to analyze stability of a solution of an algebraic equation.

The puzzle:

Let us first consider the problem (a). The solution can be obtained by observing that exponent of the bottom x is same as the whole expression, i.e. 2. This is because if we remove the bottom x, the expression still remains same due to infinite exponentiation. Thus, we obtain, x^2=2 and the solution becomes sqrt(2). Therefore x=sqrt(2) should satisfy the given equation.

Following the same procedure in problem (b) we obtain x^4=4 and the solution is again x=sqrt(2).  Now the problem occurs: if we exponentiate sqrt(2) infinitely, which number should it converge to, 2 or 4?


Clearly, the arithmetic shows that both are correct but if you take a calculator and try doing the operation you will always get 2 and never 4. This is because the given equation is a non-linear equation which has solution for a certain range of x. Given a value of x within that region you will obtain, in general, two solutions; out of which one is stable and another is unstable solution. So, you will always converge to the stable solution which is found to be 2 in this case and never reach to 4.

The explanation requires stability analysis of the solutions, which is the most interesting part of the problem.

Explanation by stability analysis:

Let us first consider a general equation obtained by replacing 2 by a variable, say “y”, and look for solutions.

The region for which solution exists is shown in a plot below.

We observe the following properties of the graph.

  1. It has a maximum at  y = e = 2.71828 where value of  x = e^(1/e) = 1.44467. Clearly when value of x is greater than this maximum we have no solution for y and the operation of exponentiation diverges, leading to infinity.
  2. For large values of y the plot is almost parallel to y-axis and the asymptotic value of x is 1. Therefore when x is less than 1, we have only one solution and the operation always converges to that value of y.
  3. The interesting region is when 1< x <1.44467, where two solutions exist. Corresponding to a value of x, the value of y which is on the left side of the vertical pink colored line is stable solution, whereas the point on the right side is unstable.

To ensure stability we perform the following analysis. Let us denote the result of operation at n-th state by “S_n” (suffix). Let “y” be the actual solution and “delta y_n” be the difference of “S_n” and “y”. These are shown below.

In our puzzle  x=sqrt(2) so that  ln(x)=(ln 2)/2.

Case (i): When y=2, we have R= ln (2) = 0.693 < 1. Therefore, difference of S_n from the actual value y decreases further and further as we increase the value of n, i.e. we exponentiate x more times. Thus, y=2 is the stable solution.

Case (ii): When y=4, we have R= 2 ln (2) = 1.386 > 1. Therefore, unlike the above case, here the difference increases further as we exponentiate x more times. This proves that y=4 is the unstable solution.

Consequently, we will always reach to the stable solution, which in our case is 2, in spite of the fact that mathematically both are correct solutions.

Thus we see that, though the puzzle looks bizarre at the first sight, it comes out to be completely fine in light of stability analysis and this unpuzzles our puzzle.

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