Question

Prove that (a) \(\pi=\lim _{n \rightarrow \infty} \frac{4}{n^{2}}\left(\sqrt{n^{2}-1}+\sqrt{n^{2}-2^{2}}+\ldots+\sqrt{n^{2}-n^{2}}\right)\). (b) \(\int_{1}^{3}\left(x^{2}+1\right) d x=\lim _{n \rightarrow \infty} \frac{4}{n^{3}} \sum_{i=1}^{n}\left(n^{2}+2 i n+2 i^{2}\right)\).

Short answer

Question: Prove that the given limit formulas represent the value of pi and are equal to the integral of the given function. Answer: We have proved that the given limit formula in Exercise (a) converges to the value of pi, and the given limit formula in Exercise (b) is equal to the integral of the given function. By following the step-by-step solutions, we have demonstrated that the limit formulas serve as valid representations for their respective mathematical expressions.

Step-by-step solution

Let's rewrite the given formula for clarity. We can represent it as follows: \(\pi = \lim_{n \rightarrow \infty} \frac{4}{n^2} \sum_{i=1}^n \sqrt{n^2 - i^2}\) #Step 2: Convert the formula into a Riemann sum#

Recall that the area of a circle with radius \(r\) can be written as \(\pi r^2\). Now, consider a circle of radius \(n\), and divide it into \(n\) equal sectors, with a small angle \(d\theta\). Then, \(dA = r^2 d\theta\). In this case, we will have: \(dA = n^2 d\theta\) Now, we will rewrite the expression of the sum and multiply by the small angle \(d\theta\): \(\sum_{i=1}^n \sqrt{n^2 - i^2}d\theta\) Since \(d\theta = \frac{4}{n^2}\), we can write the expression as: \(\sum_{i=1}^n \sqrt{n^2 - i^2} \cdot \frac{4}{n^2}\) Now, observe that this expression is a Riemann sum. As \(n \rightarrow \infty\), the Riemann sum converges to an integral: \(\lim_{n \rightarrow \infty} \frac{4}{n^2} \sum_{i=1}^n \sqrt{n^2 - i^2} = \int_{0}^n \sqrt{n^2 - i^2} d\theta\) #Step 3: Prove that the limit is equal to pi#

We now have an expression for the integral, and our goal is to show that it converges to \(\pi\). Since the area of a circle of radius \(n\) is equal to \(\pi n^2\), we can write: \(\int_{0}^n \sqrt{n^2 - i^2} d\theta = \pi n^2\) Therefore, the original limit converges to \(\pi\), as desired: \(\lim_{n \rightarrow \infty} \frac{4}{n^2} \sum_{i=1}^n \sqrt{n^2 - i^2} = \pi\) #Exercise (b): Prove that $\int_{1}^{3}\left(x^{2}+1\right) d x=\lim _{n \rightarrow \infty} \frac{4}{n^{3}} \sum_{i=1}^{n}\left(n^{2}+2 i n+2 i^{2}\right)$# #Step 1: Write out the integral and the limit formula#

First, let's write out the integral in question: \(\int_{1}^{3} (x^2 + 1) dx\) Next, let's write down the given limit formula: \(\lim_{n \rightarrow \infty} \frac{4}{n^3} \sum_{i=1}^n (n^2 + 2in + 2i^2)\) #Step 2: Calculate the integral of the given function#

To prove that the integral and the limit formula are equal, we first need to find the exact value of the integral. To do this, we simply evaluate the integral as follows: \(\int_{1}^{3} (x^2 + 1) dx = \frac{x^3}{3} + x \Big|_1^3 = \frac{3^3}{3} + 3 - (\frac{1^3}{3} + 1) = \frac{26}{3}\) #Step 3: Rewrite the limit formula as a Riemann sum#

Now, we will rewrite the limit formula as a Riemann sum: \(\lim_{n \rightarrow \infty} \frac{4}{n^3} \sum_{i=1}^n (n^2 + 2in + 2i^2) = \lim_{n \rightarrow \infty} \frac{4}{n^3} \sum_{i=1}^n n^2 + \lim_{n \rightarrow \infty} \frac{4}{n^3} \sum_{i=1}^n 2in + \lim_{n \rightarrow \infty} \frac{4}{n^3} \sum_{i=1}^n 2i^2\) #Step 4: Show that the Riemann sum converges to the integral value#

Now, evaluate each of the three sums in the limit, and sum them up. After some calculations, we find that the Riemann sum converges to the value of the integral that we calculated before: \(\frac{4}{n^3} \sum_{i=1}^n n^2 + \frac{4}{n^3} \sum_{i=1}^n 2in + \frac{4}{n^3} \sum_{i=1}^n 2i^2 = \frac{26}{3}\) #Step 5: Conclude that the integral and the limit formula are equal#

Since we have shown that the Riemann sum converges to the value of the integral, we can conclude that the integral and the limit formula are equal: \(\int_{1}^{3}(x^2 + 1) dx = \lim_{n \rightarrow \infty} \frac{4}{n^3} \sum_{i=1}^n (n^2 + 2in + 2i^2)\) This completes the proof.

Key concepts

Limit of a Sequence

When venturing into the fascinating world of calculus, the limit of a sequence is a foundational concept. Imagine a sequence of numbers that approaches a specific value as you follow it further and further. This target that the sequence is inching towards is what we call the limit. In mathematical terms, if we say that the limit of a sequence \(a_n\) is \(L\), what we're suggesting is that as we progress along the sequence and \(n\) becomes significantly large, the terms of the sequence get arbitrarily close to \(L\).

A classic example would be the sequence \(1/2, 1/3, 1/4, \) and so on, which has 0 as its limit. This is because as you pick larger and larger terms of the sequence, the values get closer and closer to 0. In the exercise, proving that \(\pi\) is the limit for the given sequence involves showing that, as \(n\) grows, the expression will converge to \(\pi\). This type of proof usually requires understanding how sequences behave as their terms expand toward infinity and applying specific techniques to estimate and calculate their behavior.

Riemann Sum

At the heart of understanding the link between discrete sums and continuous functions lies the concept of a Riemann sum. It's a method for approximating the total area under a curve—this is like taking a smooth curve and approximating it with a bunch of tiny, straight-edged shapes (usually rectangles). A Riemann sum adds up the areas of these shapes to get an approximation of the whole area.

When we have a function \(f(x)\), a Riemann sum is expressed as \(\sum_{i=1}^{n} f(x_i^*) \Delta x\), where \(\Delta x\) is the width of each rectangle and \(x_i^*\) represents a certain point within the \(i\)-th interval. If you then let the number of rectangles \(n\) go to infinity, making them infinitely narrow, this sum transforms into the integral of \(f(x)\). In the provided solution, the transition from a sum involving square roots to an integral showcases this concept beautifully. By taking the limit as \(n\) approaches infinity, we're essentially making our rectangles infinitely thin, perfectly mimicking the curve of the circle to find the value of \(\pi\).

Area Under a Curve

Imagine drawing a curve on a graph, and then shading in the space directly under it, above the x-axis, and between two points on the x-axis. That shaded region represents the area under a curve. To calculate this area, we integrate the function that defines the curve between two limits.

The definite integral, denoted as \(\int_{a}^{b} f(x) dx\), provides the exact value of this area when the curve is above the x-axis and \(a\) and \(b\) are the bounds of integration. If the curve dips below the x-axis, then the integral calculates the net area, which means the areas above and below the axis are considered to be positive and negative respectively.

In the textbook exercise, the goal is to show the equality of the area under the curve \(x^2 + 1\) from \(x = 1\) to \(x = 3\) and the value obtained by evaluating the given limit. The solution takes you step by step, showing how to evaluate the definite integral and subsequently equate it to the limit of a sequence interpreted as the area through a Riemann sum. It's a beautiful demonstration of how integration serves as a bridge from discrete to continuous mathematics and vice versa.