Problem 64 - Calculus Single Variable

Question

Let \(x_{j}=1+j / N\) for \(0 \leq j \leq N .\) Let \(\mathcal{S}_{N}=\left\\{s_{j}=\sqrt{x_{j-1} \cdot x_{j}}\right.\) : \(1 \leq j \leq N\\}\) a. Verify that \(\mathcal{S}_{N}\) is a valid choice of points for the order \(N\) uniform partition of interval [1,2]. b. Let \(f(x)=1 / x^{2}\). Show that $$ \mathcal{R}\left(f, \mathcal{S}_{N}\right)=N \sum_{j=1}^{N}\left(\frac{1}{N+j-1}-\frac{1}{N+j}\right) $$ c. For the same \(f\), prove that \(\mathcal{R}\left(f, \mathcal{S}_{N}\right)=1 / 2\) for every \(N\).

Step-by-Step Solution

Answer

The geometric means \( \mathcal{S}_N \) are valid partition points; they ensure \( \mathcal{R}(f, \mathcal{S}_N) = \frac{1}{2}\) due to term cancellation.

Step 1: Analyze the Given Function for Partition

We define \(x_j = 1 + \frac{j}{N}\) where \(0 \leq j \leq N\). Each \(x_j\) represents a partition point from 1 to 2. Notice that \(x_0 = 1\) and \(x_N = 2\). So it divides the interval [1,2] into \(N\) subintervals.

Step 2: Understand the Elements of \( \mathcal{S}_N \)

For the sequence \( \mathcal{S}_N \), each \( s_j = \sqrt{x_{j-1} \cdot x_{j}}\). These are the geometric means of consecutive \(x_j\) points, creating an intermediate point in each subinterval.

Step 3: Verify \( \mathcal{S}_N \) as Part of Uniform Partition

To show \( \mathcal{S}_N \) is a valid choice, we check if all \(s_j\) are in [1, 2] and ordered. Since \( s_j = \sqrt{(1 + \frac{j-1}{N})(1 + \frac{j}{N})} \) and both factors are in [1, 2], their product and hence square root, \(s_j\), will also be in [1, 2]. The ordering follows from the non-decreasing nature of \(x_j\).

Step 4: Reframe \( \mathcal{R}(f, \mathcal{S}_N) \) for the Given Function

Given \(f(x) = \frac{1}{x^2}\), we explore the sum: \( \mathcal{R}(f, \mathcal{S}_N) = N \sum_{j=1}^{N} \left(\frac{1}{N+j-1}-\frac{1}{N+j}\right) \). This structure suggests a telescoping series where many intermediate terms cancel.

Step 5: Evaluate the Expression to Simplify

Expanding the sum, we observe many terms cancel out due to the form \( \frac{1}{N+j-1} - \frac{1}{N+j} \). The series reduces to \(1 - \frac{1}{N+N} = 1 - \frac{1}{2} = \frac{1}{2}\).

Step 6: Prove Constant Result for All \(N\)

Since only the first and last fractions remain after cancellation and their difference is always \( \frac{1}{2} \) regardless of \(N\), the Reimann sum is constant. Hence, \( \mathcal{R}(f, \mathcal{S}_N) = \frac{1}{2}\).

Key Concepts

Partition of Interval

When we talk about a partition of an interval, we are essentially dividing that interval into smaller, non-overlapping subintervals. In mathematical terms, if we have an interval \[[a, b]\] and want to partition it into \(N\) subintervals, we create \(N+1\) points from \(x_0\) to \(x_N\), where \(x_0 = a\) and \(x_N = b\).

• The key is to ensure that each subinterval is part of the original interval.
• The points \(x_j\) must be in a non-decreasing order.
• The selection of these points affects the accuracy of approximating areas under functions.

In our exercise, we see this in action with the interval [1,2], which is partitioned by the points \(x_j = 1 + \frac{j}{N}\). This divides [1,2] into \(N\) equal parts, helping to methodically approach finding the Riemann sum.

Geometric Mean

The geometric mean is an important mathematical concept especially relevant when determining mean values in different scenarios. For two numbers, \(a\) and \(b\), the geometric mean is given by \(\sqrt{a \cdot b}\). This gives a value that is typically somewhere between \(a\) and \(b\).

• It's often used in sequences where the data is exponentially varying.
• In our problem, \(s_j = \sqrt{x_{j-1} \cdot x_{j}}\), acts as the geometric mean of two consecutive partition points \(x_{j-1}\) and \(x_j\).

Using geometric means helps in segmenting the interval points, ensuring an even spread of values and maintaining relevance within the context, such as ensuring each value falls within the evaluated interval.

Telescoping Series

A telescoping series is a special type of series in mathematics where the sum of its terms results in a large number of terms canceling out, leaving only a few terms that contribute to the overall sum.

• Each term in a telescoping series 'cancels out' with a subsequent term.
• This results in a simplified series where only the first and last terms remain after cancellation.

In the given exercise, the Riemann sum involves a difference of fractions that set up a telescoping format: \[\sum_{j=1}^{N}\left(\frac{1}{N+j-1}-\frac{1}{N+j}\right)\]With careful expansion, most terms cancel each other, and one is left with only the first few terms contributing to the sum.

Uniform Partition

A uniform partition is a specific type of partition where each subinterval is of equal length. This is especially helpful when approximating calculations such as integrals or Riemann sums as it simplifies computations.

• The length of each subinterval is calculated as \(\frac{b-a}{N}\), where \([a,b]\) is the interval and \(N\) is the number of subintervals.
• Uniform partitions ensure a regular division making mathematical analyses like the Riemann sum straightforward.

In our problem, because the partition is uniform, it simplifies calculating the Riemann sum: each segment contributes equally to the sum, allowing a consistent approach regardless of \(N\), the number of partitions. This uniform structure is crucial in demonstrating that the complex expression simplifies to \(\frac{1}{2}\) for all \(N\).