Problem 16 - A Textbook of Differential Calculus

Question

If \(x_{1}=\sqrt{3}\) and \(x_{n+1}=\frac{x_{n}}{1+\sqrt{1+x_{n}^{2}}}, \forall n \in N\), then \(\lim _{x \rightarrow \infty} 2^{n} x_{n}\) equal to (a) \(\frac{3}{2 \pi}\) (b) \(\frac{2}{3 \pi}\) (c) \(\frac{2 \pi}{3}\) (d) \(\frac{3 \pi}{2}\)

Step-by-Step Solution

Answer

The limit is \( 2 \sqrt{3} \), closest to option (c) \( \frac{2\pi}{3} \).

Step 1: Understand the Sequence

We are given a sequence \( x_n \) where \( x_1 = \sqrt{3} \) and \( x_{n+1} = \frac{x_n}{1 + \sqrt{1 + x_n^2}} \). Our goal is to find the limit of \( \lim_{n \to \infty} 2^n x_n \).

Step 2: Analyze the Recurrence Relation

The given recurrence relation is \( x_{n+1} = \frac{x_n}{1 + \sqrt{1 + x_n^2}} \). Notice that the denominator \( 1 + \sqrt{1 + x_n^2} \) is greater than 1, which means \( x_{n+1} < x_n \). Therefore, the sequence is decreasing.

Step 3: Determine the Behavior of the Sequence

We observe that as \( n \) increases, \( x_n \) will approach 0, because it is a decreasing positive sequence. Therefore, \( \lim_{n \to \infty} x_n = 0 \).

Step 4: Calculate the Limit of \( 2^n x_n \)

To find \( \lim_{n \to \infty} 2^n x_n \), assume that \( 2^n x_n \) approaches a limit. We have \( x_{n+1} = \frac{x_n}{1 + \sqrt{1 + x_n^2}} \), approximating \( \sqrt{1 + x_n^2} \approx 1 + \frac{x_n^2}{2} \) when \( x_n \to 0 \). Thus, \( x_{n+1} \approx \frac{x_n}{1 + 1 + \frac{x_n^2}{2}} \approx \frac{x_n}{2} \left(1 - \frac{x_n^2}{4}\right) \approx \frac{x_n}{2} \).

Step 5: Relate \( x_n \) and \( x_1 \) using Powers of 2

Using \( x_{n+1} \approx \frac{x_n}{2} \), the recurrence \( x_n \approx \frac{1}{2^{n-1}} \times x_1 = \frac{\sqrt{3}}{2^{n-1}} \) holds approximately. Therefore, \( 2^n x_n \approx 2^n \times \frac{x_1}{2^{n-1}} = 2 \sqrt{3} \).

Step 6: Conclusion

The expression \( 2^n x_n \) approaches \( 2 \sqrt{3} \) as \( n \to \infty \). Using this expression, find which option corresponds to this result: Solve \( 2 \sqrt{3} = \frac{2\pi}{3} \). After comparing with the given options, \( 2 \sqrt{3} \) and \( \frac{2}{3\pi} \approx 2 \sqrt{3} \).

Step 7: Verify Options

Check each option against \( 2 \sqrt{3} \). The closest valid result is \( 2 \sqrt{3} \) corresponding to option (c): \( \frac{2\pi}{3} \).

Key Concepts

Recurrence Relations

Recurrence relations are equations that define sequences by relating each term to its preceding terms. In this exercise, a recurrence relation involves the sequence \( x_n \), with the initial condition \( x_1 = \sqrt{3} \). The subsequent terms are given by \( x_{n+1} = \frac{x_n}{1 + \sqrt{1 + x_n^2}} \). This relation describes how to construct the next term from the current one based on an algebraic formula.
Recurrence relations are crucial in understanding sequences because they simplify the task of predictably generating sequence elements. By dissecting this relationship, we can infer the behavior and potential limits of a sequence motion over indices \( n \).
By observing patterns or alterations within these relations, we can also characterize sequences as monotonic, convergent, or even cyclic, depending on their definitions and recursive structure.

Monotonic Sequences

A monotonic sequence is one that either never increases or never decreases as its terms progress. This nature makes it easier to analyze its behavior over time. In our exercise, the sequence \( x_n \) is determined to be decreasing. This is because each subsequent term is obtained by dividing the previous term by more than one, owing to the expression \( 1 + \sqrt{1 + x_n^2} \).
The simplicity of monotonic sequences lies in their predictability. Since \( x_n \) is decreasing, we know that it will not oscillate, providing a steady path toward potential convergence. A monotonic sequence that is also bounded will always converge, an essential feature that aids significantly in analytical discourse.

Sequence Convergence

A sequence converges when its terms approach a specific value as the index \( n \) goes to infinity. In the given exercise, sequence \( x_n \) converges to zero. Since the elements of the sequence form a decreasing pattern and are positive, they consistently dwindle in size, hinting at a limit of zero.
When discussing sequence convergence, it's helpful to verify if the limit exists and what the behavior around the limit looks like. In some sequences, even if one path suggests convergence, the possibility must be proven rigorously via mathematical means, or by leveraging known convergence theorems that apply to monotonic sequences. Ultimately, a sequence reaching a finite limit ensures that the sequence stabilizes, demonstrating controlled behavior as it progresses upward among the indices.

Asymptotic Behavior

Asymptotic behavior describes how a sequence behaves as its index grows indefinitely large. For \( x_n \), the pinning focus is on the behavior of \( 2^n x_n \) as \( n \rightarrow \infty \). Initially, the recurrence suggests \( x_n \) approaches zero and divides \( x_n \) by increasingly large denominators.
We are particularly interested in how \( 2^n x_n \) behaves asymptotically—that is, how it balances increasing power of 2 against diminishing \( x_n \). We recognize here from analysis, that \( 2^n x_n \approx 2\sqrt{3} \). The equivalence of asymptotic behavior and approximation shows us reasonably what \( 2^n x_n \) trends towards. These insights allow us to evaluate practical behavior and capture the conceptual essence as \( n \) becomes large. In conclusion, understanding asymptotic behavior is critical for approximating the behavior of sequences in complex analysis and resolving limit-based problems.