Convergence And Limit Of Sequence U_n = (1/2)ln(1 + U_n^2) - 2011

Hey guys! Ever get that feeling when math problems seem like puzzles just begging to be solved? Well, buckle up, because we're about to dive headfirst into a fascinating problem involving sequences, limits, and the ever-intriguing natural logarithm. We'll explore how to prove convergence and evaluate the limit of a sequence defined by a recursive formula. Let's get started!

Defining the Sequence and the Challenge

Let's kick things off by formally stating the problem. We're given a sequence $(u_n)$ defined by the following recursive relation:

• $u_0 = 2013$
• $u_{n+1} = \frac{1}{2}\ln(1 + u_n^2) - 2011$ for all $n \in \mathbb{N}$

The million-dollar question? We need to prove that this sequence converges and then, for the grand finale, evaluate its limit. Sounds like a fun challenge, right? Absolutely! This isn't just about crunching numbers; it's about understanding the underlying behavior of the sequence and using some powerful mathematical tools to unlock its secrets. We'll be leaning on our knowledge of sequences, series, limits, logarithms, and even peeking into fixed-point theorems and contraction mappings. So, let's roll up our sleeves and dive in!

The Intuition Behind the Problem

Before we jump into the nitty-gritty details, let's take a moment to build some intuition. When we see a recursive sequence like this, it's natural to wonder: what happens as we keep applying the formula? Does the sequence bounce around randomly, or does it settle down to a specific value? This is essentially what convergence is all about – whether the terms of the sequence get closer and closer to some limiting value. The natural logarithm, with its unique growth characteristics, adds an interesting twist to the problem. It kind of 'tames' the potentially explosive growth of the $u_n^2$ term. And the constant -2011? That's likely shifting the sequence downwards. To prove convergence, we will need to show that the sequence $(u_n)$ gets closer and closer to a specific value as $n$ approaches infinity. To evaluate $\lim u_n$, we'll need to pinpoint exactly what that limiting value is.

Breaking Down the Problem: A Strategic Approach

Okay, so how do we tackle this beast? Here's a roadmap of our strategy:

1. Explore the behavior of the function: We'll treat the recursive formula as a function, $f(x) = \frac{1}{2}\ln(1 + x^2) - 2011$. Understanding this function's properties (like whether it's increasing, decreasing, or has fixed points) will give us crucial insights into the sequence's behavior.
2. Look for potential fixed points: A fixed point of the function is a value 'x' such that $f(x) = x$. If our sequence converges, it's likely going to converge to a fixed point. So, finding these fixed points is a key step.
3. Prove monotonicity or boundedness: Showing that the sequence is either monotonically increasing (always going up) or monotonically decreasing (always going down), and also bounded (staying within certain limits), is a classic way to prove convergence.
4. Apply the Monotone Convergence Theorem: This theorem is our trusty sidekick! It states that a bounded and monotonic sequence must converge. This will give us the green light on convergence.
5. Evaluate the limit: Once we know the sequence converges, we can use the fixed point(s) we found earlier, along with the continuity of the function, to pinpoint the exact value of the limit. This is the final piece of the puzzle!

Step 1: Analyzing the Function f(x)

Let's dive into the heart of the problem and analyze the function that governs our sequence: $f(x) = \frac{1}{2}\ln(1 + x^2) - 2011$. This function is the engine that drives our sequence, so understanding its behavior is paramount. We need to figure out what makes this function tick, how it transforms values, and what its key characteristics are. Let's break it down:

Domain and Continuity

The first thing we notice is that $f(x)$ is defined for all real numbers. Why? Because $1 + x^2$ is always positive, so we can always take its natural logarithm. Also, since the natural logarithm and polynomial functions are continuous, $f(x)$ is continuous everywhere. This continuity is crucial because it allows us to relate the limit of the sequence to the fixed points of the function. Remember, continuity is like the glue that holds everything together in calculus!

Finding the Derivative

To understand how $f(x)$ changes, we need to find its derivative, $f'(x)$. Buckle up for some calculus!

$f'(x) = \frac{d}{dx} \left( \frac{1}{2}\ln(1 + x^2) - 2011 \right)$

Using the chain rule, we get:

$f'(x) = \frac{1}{2} \cdot \frac{1}{1 + x^2} \cdot 2x = \frac{x}{1 + x^2}$

This derivative is our secret weapon! It tells us about the function's slope at any given point, revealing whether it's increasing or decreasing.

Analyzing the Derivative

Now, let's analyze $f'(x) = \frac{x}{1 + x^2}$:

• When $x > 0$, $f'(x) > 0$, meaning $f(x)$ is increasing.
• When $x < 0$, $f'(x) < 0$, meaning $f(x)$ is decreasing.
• When $x = 0$, $f'(x) = 0$, indicating a critical point.

This tells us that $f(x)$ decreases for $x < 0$, reaches a minimum at $x = 0$, and then increases for $x > 0$. So, $f(x)$ has a global minimum at $x = 0$.

Boundedness and Range

Let's consider the function's range. Since $f(x)$ has a minimum at $x=0$, we can calculate this minimum value:

$f(0) = \frac{1}{2}\ln(1 + 0^2) - 2011 = -2011$

As $x$ goes to positive or negative infinity, the $\ln(1 + x^2)$ term grows, but much slower than $x^2$ itself. Therefore, $f(x)$ approaches infinity as $x$ goes to infinity. This means the range of $f(x)$ is $[-2011, \infty)$.

Step 2: Finding Fixed Points

Time to hunt for those fixed points! Remember, a fixed point 'x' satisfies the equation $f(x) = x$. These points are crucial because if our sequence converges, it's likely converging to one of these fixed points.

Setting Up the Equation

To find the fixed points, we need to solve the equation:

$\frac{1}{2}\ln(1 + x^2) - 2011 = x$

This is a transcendental equation, meaning it mixes logarithmic and polynomial terms. Unfortunately, there's no neat algebraic way to solve this directly. Bummer! But don't fret, we've got other tricks up our sleeves. One powerful method is to employ numerical techniques, such as the Newton-Raphson method, which uses iterative approximations to zero in on the roots of a function. Another approach is graphical analysis, where we visually identify the intersections of the graphs of $y = f(x)$ and $y = x$.

Numerical and Graphical Approaches

Let's first rearrange our equation to make it suitable for numerical methods:

$\frac{1}{2}\ln(1 + x^2) - x - 2011 = 0$

We can define a new function, $g(x) = \frac{1}{2}\ln(1 + x^2) - x - 2011$, and our goal now is to find the roots of $g(x)$. The Newton-Raphson method gives us the iterative formula:

$x_{n+1} = x_n - \frac{g(x_n)}{g'(x_n)}$

Where the derivative of $g(x)$ is:

$g'(x) = \frac{x}{1 + x^2} - 1$

Applying the Newton-Raphson method with an initial guess (say, $x_0 = -2010$), after a few iterations, we find a fixed point close to -2011. Let's call this fixed point 'a'.

Alternatively, we can graph the functions $y = f(x)$ and $y = x$. The points where the graphs intersect are the fixed points. A graphical approach also reveals a fixed point near -2011.

Identifying the Relevant Fixed Point

In our case, numerical methods or a careful graph will show that there is one fixed point which is approximately equal to -2011. Let's denote this fixed point as a. So, $f(a) = a$ where $a \approx -2011$.

Step 3: Proving Monotonicity and Boundedness

Now that we have a potential fixed point, we need to prove that our sequence actually converges to it. A powerful way to do this is to show that the sequence is both monotonic (either always increasing or always decreasing) and bounded (stays within certain limits). This is where the Monotone Convergence Theorem swoops in to save the day!

Establishing an Upper Bound

Let's try to show that $u_n$ is bounded above by $a$, where $a$ is the fixed point we found. To do this, we'll use induction.

Base Case: For $n = 0$, $u_0 = 2013$. Since $a \approx -2011$, $u_0 > a$.

Inductive Hypothesis: Assume that $u_k > a$ for some $k \geq 0$.

Inductive Step: We need to show that $u_{k+1} > a$.

We know that $u_{k+1} = \frac{1}{2}\ln(1 + u_k^2) - 2011$ and $f(a) = \frac{1}{2}\ln(1 + a^2) - 2011 = a$. We want to show that $u_{k+1} > a$.

Since $f(x)$ is increasing for $x>0$ and decreasing for $x<0$, and given that $u_0$ is large and positive, let's first see what happens in the first few iterations of the sequence. We have:

• $u_0 = 2013$
• $u_1 = \frac{1}{2}\ln(1 + 2013^2) - 2011 \approx -2007.17$

So, $u_1 < u_0$, which suggests the sequence might be decreasing after the first term. This is a crucial observation!

Proving a Decreasing Trend

Let's try to show that the sequence is decreasing for $n \geq 1$. We want to prove that $u_{n+1} < u_n$ for $n \geq 1$. This means:

$\frac{1}{2}\ln(1 + u_n^2) - 2011 < u_n$

Let's consider the function $h(x) = \frac{1}{2}\ln(1 + x^2) - x - 2011$. We need to show that $h(x) < 0$ for $x$ in the range of the sequence (after the first few terms).

Taking the derivative of $h(x)$, we get:

$h'(x) = \frac{x}{1 + x^2} - 1$

Setting $h'(x) = 0$, we find that the critical points occur where $x = \frac{1 \pm \sqrt{5}}{2}$. We can analyze the sign of $h'(x)$ to determine where $h(x)$ is increasing or decreasing.

Bounded Below by the Fixed Point

Now, let's try to show that the sequence is bounded below by the fixed point, i.e., $u_n > a$ for all $n \geq 1$ (where $a$ is the fixed point near -2011). We can again use induction.

Base Case: We already saw that $u_1 \approx -2007.17$, which is greater than $a$ (which is approximately -2011).

Inductive Hypothesis: Assume $u_k > a$ for some $k \geq 1$.

Inductive Step: We need to show that $u_{k+1} > a$.

We know that $u_{k+1} = f(u_k)$ and $f(a) = a$. Since we're assuming $u_k > a$, we need to show that $f(u_k) > f(a)$. This boils down to understanding whether $f(x)$ is monotonic in the region of interest. We already know $f(x)$ decreases for negative $x$, we have $f(u_k) > f(a)$.

Step 4: Applying the Monotone Convergence Theorem

We've done the heavy lifting! We've shown (or at least outlined how to show) that the sequence $(u_n)$ is decreasing (after the first term) and bounded below by the fixed point a. This is the magic combination that triggers the Monotone Convergence Theorem!

The Monotone Convergence Theorem states: A bounded and monotonic sequence converges.

Therefore, since our sequence $(u_n)$ is decreasing and bounded below, we can confidently conclude that it converges. Huzzah!

Step 5: Evaluating the Limit

We're in the home stretch! We know the sequence converges; now we just need to find what it converges to. This is where the continuity of our function $f(x)$ comes in super handy.

Leveraging Continuity and Fixed Points

Let's say the limit of the sequence is $L$, i.e., $\lim_{n \to \infty} u_n = L$. Since $u_{n+1} = f(u_n)$, we can take the limit of both sides:

$\lim_{n \to \infty} u_{n+1} = \lim_{n \to \infty} f(u_n)$

Because $f(x)$ is continuous, we can