Question

Suppose \(\left\\{\alpha_{j}\right\\}_{1}^{\infty} \subset(0,1)\). a. \(\prod_{1}^{\infty}\left(1-\alpha_{j}\right)>0\) iff \(\sum_{1}^{\infty} \alpha_{j}<\infty\). (Compare \(\sum_{1}^{\infty} \log \left(1-\alpha_{j}\right)\) to \(\left.\sum \alpha_{j} .\right)\) b. Given \(\beta \in(0,1)\), exhibit a sequence \(\left\\{\alpha_{j}\right\\}\) such that \(\prod_{1}^{\infty}\left(1-\alpha_{j}\right)=\beta\).

Short answer

The product \(\prod_{j=1}^{\infty} (1-\alpha_j) > 0\) iff \(\sum_{j=1}^{\infty} \alpha_j < \infty\). For \(\beta\), choose \(1-\alpha_j = 1-\exp(-\frac{\log(\beta)}{2^j})\).

Step-by-step solution

Understand the problem a

The problem is asking for a characterization of the infinite product \(\prod_{j=1}^{\infty}(1-\alpha_j)\). Specifically, we need to determine when this product is positive and relate it to the sum \(\sum_{j=1}^{\infty} \alpha_j\). For this, we consider the logarithm of the product and compare it to the sum conditions.

Relate Infinite Product and Series

We know from analysis that an infinite product \(\prod_{j=1}^{\infty} (1-\alpha_j)\) converges to a positive number if and only if the corresponding sum of logarithms \(\sum_{j=1}^{\infty} \log(1-\alpha_j)\) converges to a finite value. Using the approximation \(\log(1-x) \approx -x\) for small \(x\), we have \(\sum_{j=1}^{\infty} \log(1-\alpha_j) \approx \sum_{j=1}^{\infty} (-\alpha_j)\). This implies the convergence of \(\sum_{j=1}^{\infty} \alpha_j\) is necessary for a positive product.

Analyze Convergence Conditions

If \(\sum_{j=1}^{\infty} \alpha_j < \infty\), then each \(\alpha_j\) must be sufficiently small for large \(j\), ensuring that \(\log(1-\alpha_j)\) is also small, leading to a finite negative sum. If \(\sum_{j=1}^{\infty} \alpha_j = \infty\), the logarithm series diverges negatively, causing the product to approach zero, thus proving part a.

Approach Problem b

Now, given \(\beta\in(0,1)\), we want a sequence \(\{\alpha_j\}\) such that \(\prod_{j=1}^{\infty} (1-\alpha_j) = \beta\). Since \(\log \prod_{j=1}^{\infty} (1-\alpha_j) = \sum_{j=1}^{\infty} \log(1-\alpha_j)\), set this equal to \(\log(\beta)\) and find suitable \(\alpha_j\) values.

Construct the Sequence for Problem b

To construct \(\alpha_j\), we use \(1-\alpha_j = 1 - \exp\left(-\frac{\log(\beta)}{2^j}\right)\). This choice ensures that the cumulative product \(\prod_{j=1}^{n} (1-\alpha_j)\) approximates \(\beta\) as \(n\to\infty\), since the exponential function ensures quick convergence.

Key concepts

Convergence of Series

Convergence is a crucial concept in mathematics, especially when dealing with infinite series and products. In this problem, we are analyzing the conditions under which an infinite product converges to a positive number. To do this, we relate it to the convergence of a series, specifically the sum of the series of terms \( \alpha_j \).

This relationship is established by considering the logarithmic transformation of the product. Series \( \sum_{j=1}^{\infty} \alpha_j \) must converge for the product \( \prod_{j=1}^{\infty}(1-\alpha_j) \) to converge to a positive number. If the series \( \sum_{j=1}^{\infty} \alpha_j \) diverges, the cumulative effect makes the product approach zero, thus failing to be positive.

Understanding convergence can sometimes be tricky. Here are some key points:

• A series converges when the sum of its terms approaches a finite limit as more terms are added.
• If the series of logarithms \( \sum_{j=1}^{\infty} \log(1-\alpha_j) \) converges, it indicates that the original product converges to a value greater than zero.
• Otherwise, if the sum diverges, the product tends to zero.

Recognizing this connection between series and product convergence helps in comprehending the behavior of infinite products.

Logarithmic Approximation

The logarithmic approximation plays a vital role in simplifying complex expressions, especially when dealing with small quantities. In this problem, we're using the approximation \( \log(1-x) \approx -x \) for small \( x \). This approximation helps in analyzing the convergence of the logarithmic series \( \sum_{j=1}^{\infty} \log(1-\alpha_j) \).

The key insights from this approximation include:

• For small \( \alpha_j \), \( \log(1-\alpha_j) \) can be approximated as \(-\alpha_j\), simplifying the behavior of the series.
• This approximation holds as long as \( \alpha_j \) is close to zero, which is a typical assumption for convergence issues.

Through this approximation, we transform the product problem into a sum problem, enabling easier determination of convergence behavior.

By handling complex logarithms in this manner, students can readily analyze the problem without diving into complicated calculations. It's a fine example of how approximations support clearer insights in mathematical analysis.

Sequence Construction

Constructing a sequence that satisfies certain conditions involves a blend of creativity and mathematical procedures. In this exercise, given a target product value \( \beta \) between 0 and 1, our task is to find a sequence \( \{\alpha_j\} \) such that \( \prod_{j=1}^{\infty}(1-\alpha_j) = \beta \).

Here's a practical approach:

• Understand that to achieve the desired product, we use the relation \( \log(\beta) = \sum_{j=1}^{\infty} \log(1-\alpha_j) \).
• By breaking \( \log(\beta) \) into smaller steps, each \( \alpha_j \) is incrementally determined to ensure the collective product approaches \( \beta \).
• A suitable sequence can be constructed using \( 1-\alpha_j = 1 - \exp\left(-\frac{\log(\beta)}{2^j}\right) \), which ensures rapid convergence with manageable calculations.

This method guarantees that the sequence produces a product converging to \( \beta \), demonstrating the interplay of exponential and logarithmic functions in crafting custom solutions.