Problem 5
Question
\(\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \sin ^{2 k} \frac{r \pi}{2 n}\) is equal to (A) \(\frac{2 k !}{2^{2 k}(k !)^{2}}\) (B) \(\frac{2 k !}{2^{k}(k !)}\) (C) \(\frac{2 k !}{2^{k}(k !)^{2}}\) (D) None of these
Step-by-Step Solution
Verified Answer
The limit evaluates to option (A): \( \frac{2k!}{2^{2k}(k!)^2} \).
1Step 1: Understanding the Limit and Sum
Given the expression is \( \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \sin^{2k} \frac{r\pi}{2n} \), our goal is to find this limit. Notice that the fraction serves as an approximation to an integral for a large \( n \).
2Step 2: Approximate the Sum with an Integral
Recognize that as \( n \to \infty \), the Riemann sum \( \frac{1}{n} \sum_{r=1}^{n} \sin^{2k} \frac{r\pi}{2n} \) approaches the integral \( \int_{0}^{1} \sin^{2k} \left( \frac{\pi x}{2} \right) \, dx \). This is because \( \frac{r\pi}{2n} \) is a partition of the interval \([0, \frac{\pi}{2}]\).
3Step 3: Change of Variables
To solve \( \int_{0}^{1} \sin^{2k} \left( \frac{\pi x}{2} \right) \, dx \), perform a substitution: let \( y = \frac{\pi x}{2} \), then \( dx = \frac{2}{\pi} dy \). The limits of integration change from \( 0 \) to \( \pi/2 \).
4Step 4: Evaluate the Integral
The integral becomes \( \frac{2}{\pi} \int_{0}^{\pi/2} \sin^{2k} y \, dy \). This is a known integral related to the Wallis's integrals. Evaluate it using the formula: \( \int_{0}^{\pi/2} \sin^{2k} x \, dx = \frac{(2k)!}{2^{2k}(k!)^2} \).
5Step 5: Substitute and Simplify the Result
Substitute back the evaluated integral into the approximation. Therefore, \( \frac{2}{\pi} \cdot \frac{(2k)!}{2^{2k} (k!)^2} \) simplifies to \( \frac{2k!}{2^{2k} (k!)^2} \).
6Step 6: Conclusion Comparison
Compare this result with the multiple choice options given. It matches option (A): \( \frac{2k!}{2^{2k}(k!)^2} \).
Key Concepts
Riemann sumIntegral approximationWallis's integrals
Riemann sum
A Riemann sum is a method for approximating the total area under a curve, also known as an integral. This method breaks the area into small rectangles, where each rectangle's width is a small change in the x-value, and the height is the function's value at that x-value.
As the number of rectangles increases (n approaches infinity), the approximation becomes more accurate. In this context, the limit of Riemann sums leads to the concept of an integral. This is why it is significant in the original exercise. It turns a complicated sum into a simpler continuous integral.
As the number of rectangles increases (n approaches infinity), the approximation becomes more accurate. In this context, the limit of Riemann sums leads to the concept of an integral. This is why it is significant in the original exercise. It turns a complicated sum into a simpler continuous integral.
- Definition: Approach for approximating integration using sums of rectangular areas.
- Purpose: Provides a bridge to understanding integrals through discrete sums.
Integral approximation
Integral approximation is a way to find approximate values for definite integrals. It helps us understand what these integrals represent and compute them more easily. In simple terms, we're finding the area under the curve.
For example, in the exercise given, the goal is to approximate the integral of function involving sine. A change of variables is used to simplify and solve the integral. Thus the sum is transformed from a discrete set of terms into a smooth area calculation.
For example, in the exercise given, the goal is to approximate the integral of function involving sine. A change of variables is used to simplify and solve the integral. Thus the sum is transformed from a discrete set of terms into a smooth area calculation.
- Definition: Finding approximate solutions to integrals.
- Applications: Useful when exact solutions are difficult or impossible to get.
Wallis's integrals
Wallis's integrals refer to a specific class of integrals involving sine and cosine functions raised to a power. These integrals are notably related to trigonometric identities and play a crucial role in various mathematical fields.
In the context of the original exercise, Wallis's formula simplifies the integral from the Riemann sum, giving us an exact value. It involves specific formulas, such as the one for the sine function used in this exercise: \[\int_{0}^{\pi/2} \sin^{2k} x \, dx = \frac{(2k)!}{2^{2k}(k!)^2}\] Leading to the precise result needed.
In the context of the original exercise, Wallis's formula simplifies the integral from the Riemann sum, giving us an exact value. It involves specific formulas, such as the one for the sine function used in this exercise: \[\int_{0}^{\pi/2} \sin^{2k} x \, dx = \frac{(2k)!}{2^{2k}(k!)^2}\] Leading to the precise result needed.
- Definition: Integral formulas for powered sine/cosine functions.
- Significance: They allow for precise resolution of complex trigonometric integrals.
Other exercises in this chapter
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