Putnam 2000 A1: A Deep Dive Into A Classic Math Problem

Let's break down the Putnam 2000 A1 problem. This problem is a classic example of a Putnam exam question, known for its challenging and thought-provoking nature. It often requires a blend of creativity, problem-solving skills, and a solid foundation in mathematical principles. So, buckle up, guys, as we explore this intriguing problem and dissect its solution.

Understanding the Putnam Competition

Before diving into the specifics of Putnam 2000 A1, it's essential to understand the Putnam Competition itself. The William Lowell Putnam Mathematical Competition is a prestigious mathematics competition for undergraduate students in the United States and Canada. It's renowned for its difficulty, with problems that demand a deep understanding of mathematical concepts and creative problem-solving techniques. The exam consists of two sessions, each containing six problems, and students are given three hours to solve each set. The problems cover a wide range of mathematical topics, including algebra, analysis, combinatorics, number theory, and geometry. The Putnam Competition is not just about finding the correct answer; it's about demonstrating a rigorous and elegant solution that showcases mathematical insight and ingenuity.

The problems are designed to be challenging, often requiring students to think outside the box and apply their knowledge in novel ways. The Putnam Competition is a great way for students to test their mathematical abilities and compete with some of the brightest minds in North America. It also provides a valuable experience for those interested in pursuing careers in mathematics or related fields. Participating in the Putnam Competition can enhance problem-solving skills, deepen mathematical understanding, and build confidence in tackling complex challenges. The competition is a celebration of mathematical talent and a testament to the power of human intellect.

The Problem Statement

Alright, let's get to the heart of the matter! The Putnam 2000 A1 problem states:

Evaluate

$$\int_0^1 \int_0^1 \frac{x^2 - y^2}{(x^2 + y^2)^2} \, dy \, dx.$$

That's it! Looks simple, right? Well, don't be fooled. This integral requires careful consideration and a strategic approach. Many students find themselves tripped up by the apparent symmetry and the potential for singularities. The key is to evaluate the integral in the correct order and to be mindful of the behavior of the integrand.

Initial Observations and Potential Pitfalls

Before we jump into the calculations, let's make some initial observations. First, notice that the integrand, $\frac{x^2 - y^2}{(x2 + y²⁾2}$, is an odd function with respect to $y$. That is, if we replace $y$ with $-y$, the integrand changes sign. This suggests that there might be some cancellation involved when we integrate with respect to $y$. However, we need to be cautious because the limits of integration are from 0 to 1, not symmetric around 0. Second, the integrand has a potential singularity at $(x, y) = (0, 0)$. Although this point is not in the interior of the region of integration, we need to be mindful of its potential impact on the integral.

Another potential pitfall is to assume that we can simply switch the order of integration. While Fubini's theorem allows us to do this under certain conditions, we need to verify that those conditions are met. In particular, we need to ensure that the integral is absolutely convergent. If the integral is not absolutely convergent, then switching the order of integration might lead to a different result. Therefore, we need to proceed with caution and carefully justify each step in our solution.

Solving the Integral

Let's evaluate the inner integral first:

$$I(x) = \int_0^1 \frac{x^2 - y^2}{(x^2 + y^2)^2} \, dy$$

To solve this integral, we can use the substitution $y = x \tan(\theta)$. Then, $dy = x \sec^2(\theta) \, d\theta$. When $y = 0$, $\theta = 0$, and when $y = 1$, $\theta = \arctan(1/x)$. Substituting these into the integral, we get:

$$I(x) = \int_0^{\arctan(1/x)} \frac{x^2 - x^2 \tan^2(\theta)}{(x^2 + x^2 \tan^2(\theta))^2} x \sec^2(\theta) \, d\theta$$

Simplifying, we have:

$$I(x) = \int_0^{\arctan(1/x)} \frac{x^2(1 - \tan^2(\theta))}{x^4(1 + \tan^2(\theta))^2} x \sec^2(\theta) \, d\theta$$

Since $1 + \tan^2(\theta) = \sec^2(\theta)$, we can further simplify to:

$$I(x) = \frac{1}{x} \int_0^{\arctan(1/x)} \frac{1 - \tan^2(\theta)}{\sec^2(\theta)} \, d\theta$$

$$I(x) = \frac{1}{x} \int_0^{\arctan(1/x)} (\cos^2(\theta) - \sin^2(\theta)) \, d\theta$$

$$I(x) = \frac{1}{x} \int_0^{\arctan(1/x)} \cos(2\theta) \, d\theta$$

Now, we can integrate:

$$I(x) = \frac{1}{x} \left[ \frac{1}{2} \sin(2\theta) \right]_0^{\arctan(1/x)}$$

$$I(x) = \frac{1}{2x} \sin(2 \arctan(1/x))$$

Using the double angle formula, $\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$, we have:

$$I(x) = \frac{1}{x} \sin(\arctan(1/x)) \cos(\arctan(1/x))$$

Let $\alpha = \arctan(1/x)$. Then, $\tan(\alpha) = 1/x$. Consider a right triangle with opposite side 1 and adjacent side $x$. The hypotenuse is $\sqrt{x^2 + 1}$. Therefore,

$$\sin(\alpha) = \frac{1}{\sqrt{x^2 + 1}} \quad \text{and} \quad \cos(\alpha) = \frac{x}{\sqrt{x^2 + 1}}$$

Substituting these into the expression for $I(x)$, we get:

$$I(x) = \frac{1}{x} \cdot \frac{1}{\sqrt{x^2 + 1}} \cdot \frac{x}{\sqrt{x^2 + 1}} = \frac{1}{x^2 + 1}$$

Now, we need to evaluate the outer integral:

$$\int_0^1 I(x) \, dx = \int_0^1 \frac{1}{x^2 + 1} \, dx$$

This is a standard integral:

$$\int_0^1 \frac{1}{x^2 + 1} \, dx = \left[ \arctan(x) \right]_0^1 = \arctan(1) - \arctan(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$$

Therefore, the value of the double integral is $\frac{\pi}{4}$.

Switching the Order of Integration

Now, let's consider what happens if we switch the order of integration:

$$\int_0^1 \int_0^1 \frac{x^2 - y^2}{(x^2 + y^2)^2} \, dx \, dy$$

We can rewrite the integrand as:

$$\frac{x^2 - y^2}{(x^2 + y^2)^2} = - \frac{y^2 - x^2}{(x^2 + y^2)^2}$$

Notice that the new inner integral is the negative of the original inner integral with $x$ and $y$ swapped. Therefore, if we evaluate the integral with the reversed order, we will get:

$$\int_0^1 \int_0^1 \frac{x^2 - y^2}{(x^2 + y^2)^2} \, dx \, dy = - \frac{\pi}{4}$$

This demonstrates that the value of the integral changes when we switch the order of integration. This is because the integral is not absolutely convergent. In other words, the integral of the absolute value of the integrand is not finite. This highlights the importance of verifying the conditions of Fubini's theorem before switching the order of integration.

Key Takeaways

So, what did we learn from this Putnam problem? Here are some key takeaways:

1. Careful Evaluation: Always be meticulous when evaluating integrals, especially when dealing with potential singularities or complex integrands.
2. Trigonometric Substitution: Trigonometric substitutions can be powerful tools for simplifying integrals involving expressions of the form $x^2 + y^2$.
3. Fubini's Theorem: Be aware of the conditions required for applying Fubini's theorem to switch the order of integration. Always check for absolute convergence.
4. Problem-Solving Strategies: Putnam problems often require a combination of mathematical knowledge and creative problem-solving strategies. Don't be afraid to try different approaches and think outside the box.

Conclusion

The Putnam 2000 A1 problem is a great example of the type of challenging and rewarding problems that appear on the Putnam exam. By carefully evaluating the integral, using appropriate substitutions, and being mindful of the conditions for switching the order of integration, we were able to arrive at the correct solution. This problem serves as a reminder that success in mathematics requires not only a solid foundation in fundamental concepts but also the ability to apply those concepts in creative and innovative ways. Keep practicing, keep exploring, and keep challenging yourselves with these types of problems! You've got this, guys!