Math Problem Statement

ans

The value of lim as n approaches infinity of the sum from k=1 to n of ((n-k)/n^2) * cos(4k/n) equals:

Solution

To solve the problem, let’s analyze the given expression:

limnk=1nnkn2cos(4kn)\lim_{n \to \infty} \sum_{k=1}^{n} \frac{n - k}{n^2} \cos\left(\frac{4k}{n}\right)

This expression resembles a Riemann sum. To convert it into an integral, we need to interpret the terms in the summation accordingly.

Step-by-Step Solution:

  1. Rewrite the Sum as a Riemann Sum:

    • As nn \to \infty, the sum can be interpreted as: \approx \int_{0}^{1} (1 - x) \cos(4x) \, dx$$
    • Here, we let x=knx = \frac{k}{n} so that Δx=1n\Delta x = \frac{1}{n}.
    • Then, nkn=1x\frac{n - k}{n} = 1 - x and 4kn=4x\frac{4k}{n} = 4x.
  2. Set Up the Integral: 01(1x)cos(4x)dx\int_{0}^{1} (1 - x) \cos(4x) \, dx

  3. Integrate by Parts: To solve this integral, use integration by parts where:

    • Let u=1xu = 1 - x and dv=cos(4x)dxdv = \cos(4x) \, dx.
    • Then du=dxdu = -dx and v=sin(4x)4v = \frac{\sin(4x)}{4}.

    Using the integration by parts formula udv=uvvdu\int u \, dv = uv - \int v \, du: 01(1x)cos(4x)dx=[(1x)sin(4x)4]01+01sin(4x)4dx\int_{0}^{1} (1 - x) \cos(4x) \, dx = \left[ (1 - x) \frac{\sin(4x)}{4} \right]_{0}^{1} + \int_{0}^{1} \frac{\sin(4x)}{4} \, dx

  4. Evaluate Each Part:

    • For the first term: [(1x)sin(4x)4]01=(0)(1sin(0)4)=0\left[ (1 - x) \frac{\sin(4x)}{4} \right]_{0}^{1} = (0) - \left(\frac{1 \cdot \sin(0)}{4}\right) = 0
    • For the second term, integrate sin(4x)4\frac{\sin(4x)}{4} from 00 to 11: 01sin(4x)4dx=1401sin(4x)dx=14[cos(4x)4]01\int_{0}^{1} \frac{\sin(4x)}{4} \, dx = \frac{1}{4} \int_{0}^{1} \sin(4x) \, dx = \frac{1}{4} \left[ -\frac{\cos(4x)}{4} \right]_{0}^{1} =14(cos(4)4+14)=116(1cos(4))= \frac{1}{4} \cdot \left( -\frac{\cos(4)}{4} + \frac{1}{4} \right) = \frac{1}{16} (1 - \cos(4))

Final Answer:

Thus, the value of the limit is:

116(1cos(4))\frac{1}{16} (1 - \cos(4))

Correct Option:

The answer is (d) 116(1cos4)\frac{1}{16} (1 - \cos 4).

Would you like further explanation or have any questions?

Related Questions:

  1. How does converting a summation to a Riemann integral help in finding limits?
  2. Can integration by parts always be used with trigonometric functions?
  3. What are some typical forms that indicate a problem can be solved using a Riemann sum?
  4. How do we know which substitution to make when turning a discrete sum into an integral?
  5. What is the significance of the cos(4)\cos(4) term in this particular limit?

Tip:

When working with Riemann sums, identifying the correct form of f(x)f(x) and dxdx in the context of the integral is key to solving the problem correctly.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Limits
Summation
Riemann Sum
Integration
Trigonometric Functions

Formulas

Riemann Sum approximation: \( \sum f(x_i) \Delta x \approx \int f(x) dx \)
Integration by Parts: \( \int u \, dv = uv - \int v \, du \)

Theorems

Riemann Sum to Integral Conversion
Integration by Parts

Suitable Grade Level

Undergraduate Calculus