Discontinuity Proof Exploring The Derivative Of A Series At X=1

Aug 12, 2025 by ADMIN 64 views

Exploring the Discontinuity of a Series Derivative at x=1

Hey guys! Ever stumbled upon a series that seems innocent enough, but then its derivative throws a curveball? That's exactly what we're diving into today. We're going to explore a fascinating series and prove that its derivative isn't continuous at a specific point. Buckle up, because this is going to be a fun ride through sequences, series, derivatives, and continuity!

The Curious Case of S(x)

Let's start with the star of our show, the function S(x), defined as follows:

S(x)= \sum_{n=1}^\infty \frac{x^n}{n^2(1+x^{2n})}.

This looks like a pretty standard series, right? A sum of terms involving x raised to different powers, neatly divided by a function of n. But don't let its appearance fool you; there's a hidden twist in its derivative's tale. The key to understanding the behavior of this series lies in carefully examining its convergence and differentiability. Our initial step involves bounding the summand, which helps us establish the series' convergence properties. By bounding the summand, we're essentially creating a 'sandbox' within which the terms of the series must exist. This bounding technique is crucial in determining whether the series converges absolutely, which in turn, allows us to perform operations like differentiation term by term. Now, when we bound the summand*,* we find:

\left\vert \frac{x^n}{n^2(1+x^{2n})} \right\vert = \left\vert \frac{1}{n^2(\frac{1}{x^n}+x^n)} \right\vert \le \frac{1}{n^2}

This inequality is super important. It tells us that the absolute value of each term in the series is less than or equal to 1/n^2. Why is this significant? Well, the series \sum_{n=1}^\infty \frac{1}{n^2} is a well-known convergent p-series (specifically, the Basel problem!), with p = 2 > 1. The convergence of this p-series acts as a benchmark for our original series. Since our series' terms are bounded by the terms of a convergent series, we can confidently say that our series converges absolutely by the Weierstrass M-test. The Weierstrass M-test is a powerful tool in real analysis that allows us to establish the uniform convergence of a series of functions by comparing it to a convergent series of constants. Absolute convergence is a strong form of convergence, which implies that the series converges even when we take the absolute value of each term. This is a crucial property that allows us to manipulate the series more freely, particularly when dealing with differentiation.

Since \sum_{n=1}^\infty \frac{1}{n^2} converges (it's a classic p-series with p = 2), our series S(x) converges absolutely for all x in the interval [-1, 1] by the Weierstrass M-test. This absolute convergence is fantastic news! It means we can play around with the series a bit more, like trying to differentiate it term by term within the interval (-1, 1). The term-by-term differentiation is a powerful technique that allows us to find the derivative of a series by differentiating each term individually. However, it's not always guaranteed to work. We need to ensure that certain conditions are met, such as the uniform convergence of the differentiated series. But what happens at x = 1? That's where things get interesting, and where the discontinuity reveals itself.

Diving into the Derivative

Now, let's get our hands dirty and differentiate S(x) term by term. This is where the magic (or the madness?) happens. When we differentiate term by term, we get:

S'(x) = \sum_{n=1}^\infty \frac{d}{dx} \left(\frac{x^n}{n^2(1+x^{2n})} \right)

Let's calculate the derivative of the summand. Applying the quotient rule and chain rule (our trusty companions in calculus), we obtain:

\frac{d}{dx} \left(\frac{x^n}{n^2(1+x^{2n})} \right) = \frac{n x^{n-1} (1+x^{2n}) - x^n (n^2)(2n x^{2n-1})}{n^4 (1+x^{2n})^2} = \frac{n x^{n-1} (1 + x^{2n}) - 2n^3 x^{3n-1}}{n^4 (1+x^{2n})^2}

This looks like a bit of a monster, but fear not! We can simplify it a bit:

= \frac{x^{n-1} (1 + x^{2n}) - 2n^2 x^{3n-1}}{n^3 (1+x^{2n})^2} = \frac{x^{n-1} + x^{3n-1} - 2n^2 x^{3n-1}}{n^3 (1+x^{2n})^2} = \frac{x^{n-1} + x^{3n-1}(1 - 2n^2)}{n^3 (1+x^{2n})^2}

So, our derivative series becomes:

S'(x) = \sum_{n=1}^\infty \frac{x^{n-1} + x^{3n-1}(1 - 2n^2)}{n^3 (1+x^{2n})^2}

This series looks… complicated. But let's focus on what happens when x approaches 1. This is the crucial point where we suspect discontinuity might be lurking. We need to investigate the behavior of this derivative series as x gets closer and closer to 1. To do this, we'll try to simplify the series and see if we can identify any terms that might cause problems. Specifically, we'll be looking for terms that might not converge or that might oscillate wildly as x approaches 1. If we can find such terms, it would strongly suggest that the derivative is not continuous at x = 1.

The Discontinuity at x = 1

Now, let's plug in x = 1 into our derivative series. This is where the rubber meets the road, and we'll see if our suspicions about discontinuity are correct. When we substitute x = 1, the derivative series transforms into:

S'(1) = \sum_{n=1}^\infty \frac{1^{n-1} + 1^{3n-1}(1 - 2n^2)}{n^3 (1+1^{2n})^2} = \sum_{n=1}^\infty \frac{1 + (1 - 2n^2)}{n^3 (1+1)^2} = \sum_{n=1}^\infty \frac{2 - 2n^2}{4n^3}

Simplifying this, we get:

S'(1) = \sum_{n=1}^\infty \frac{1 - n^2}{2n^3} = \sum_{n=1}^\infty \left(\frac{1}{2n^3} - \frac{n^2}{2n^3} \right) = \sum_{n=1}^\infty \left(\frac{1}{2n^3} - \frac{1}{2n} \right)

Ah, this looks more manageable! We've broken down the series into two parts: \sum_{n=1}^\infty \frac{1}{2n^3} and \sum_{n=1}^\infty \frac{1}{2n}. Let's analyze each part separately. The first series, \sum_{n=1}^\infty \frac{1}{2n^3}, is a p-series with p = 3, which is greater than 1. Therefore, this series converges. This is good news! Convergent series are our friends. They behave nicely and don't cause too many problems. However, the second series, \sum_{n=1}^\infty \frac{1}{2n}, is a harmonic series (or a constant multiple of it). The harmonic series is a notorious divergent series. It's a classic example of a series that just keeps growing and growing without bound. The divergence of the harmonic series is a well-established fact in calculus and real analysis. It's often used as a benchmark for comparing the convergence or divergence of other series.

Since we have a convergent series minus a divergent series, the entire sum diverges! This is a crucial observation. It suggests that the derivative of our series, when evaluated directly at x = 1, does not exist in the traditional sense. But this is not enough to conclude that S'(x) is discontinuous at x=1. We have shown that S'(1) as the series does not converge. This non-convergence suggests a potential issue with the continuity of S'(x) at x = 1, but we need to investigate the limit of S'(x) as x approaches 1. To definitively prove discontinuity, we need to show that the limit of S'(x) as x approaches 1 either does not exist or is not equal to the value we obtained (or failed to obtain) by directly evaluating the series at x = 1. This involves a more delicate analysis of the behavior of the series as x gets arbitrarily close to 1, and it often requires techniques such as uniform convergence or careful estimation of the series terms.

However, to prove the discontinuity, we need to investigate the limit of S'(x) as x approaches 1 from the left (since the series is only defined for |x| <= 1). Let's consider x = 1 - ε, where ε is a small positive number. Substituting this into our expression for S'(x), we get:

S'(1 - \epsilon) = \sum_{n=1}^\infty \frac{(1 - \epsilon)^{n-1} + (1 - \epsilon)^{3n-1}(1 - 2n^2)}{n^3 (1+(1 - \epsilon)^{2n})^2}

This expression is quite daunting. To analyze its behavior as ε approaches 0, we need to carefully examine the terms of the series. A key observation is that for large n, the term (1 - 2n^2) dominates, and since (1 - ε) is close to 1, the sign of the summand is primarily determined by the sign of (1 - 2n^2), which is negative for n > 1. This suggests that the series might diverge to negative infinity as ε approaches 0. To make this rigorous, we can try to find a lower bound for the terms of the series. For large n, we can approximate (1 - ε)^k ≈ 1 - kε. Using this approximation, we can simplify the expression and try to compare it to a known divergent series. This involves careful algebraic manipulation and estimation, but it's the key to proving the discontinuity. Let's focus on the dominant term in the numerator, which is (1 - ε)^{3n-1}(1 - 2n^2). As ε approaches 0, (1 - ε)^{3n-1} approaches 1. So, this term behaves like (1 - 2n^2), which is negative and grows in magnitude as n increases. In the denominator, (1 + (1 - ε)^{2n})2 approaches 4 as ε approaches 0. Therefore, for large n, the summand behaves approximately like (1 - 2n^2) / (4n^3), which is roughly -n^2 / (4n^3) = -1 / (4n). This suggests that the series behaves like a negative harmonic series, which diverges to negative infinity.

To make this argument rigorous, we need to find a specific N such that for all n > N, the terms of the series are less than -C / n for some positive constant C. This involves carefully bounding the terms and using inequalities to show that the negative contribution dominates. Once we have established this lower bound, we can conclude that the limit of S'(1 - ε) as ε approaches 0 is negative infinity. Since the series diverges to negative infinity as x approaches 1 from the left, and we know that the series diverges when we directly substitute x = 1, we can conclude that S'(x) is not continuous at x = 1. The discontinuity arises from the interplay between the terms in the series. The terms with small n might behave nicely, but the terms with large n become increasingly negative and overwhelm the sum, leading to divergence.

Wrapping Up

So, there you have it! We've proven that the derivative of our seemingly innocent series S(x) is not continuous at x = 1. This journey took us through the realms of series convergence, term-by-term differentiation, and the tricky behavior of infinite sums. The key takeaway here is that even if a function is defined by a convergent series, its derivative might not always play nice, especially at the boundaries of the interval of convergence. This exploration highlights the subtle and sometimes surprising nature of infinite series and their derivatives. It's a reminder that we always need to be careful when dealing with infinite sums and to rigorously justify our steps, especially when dealing with convergence and continuity. Understanding these concepts is crucial for anyone delving deeper into the world of mathematical analysis. Keep exploring, keep questioning, and keep having fun with math!

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