Math Problem Statement

The function f(x) = 3x ln(1 + 2x) is represented as a power series. Find the following coefficients in the power series: c0, c1, c2, c3, c4. Also, find the radius of convergence R of the series.

Solution

To solve this problem, we need to expand the function f(x)=3xln(1+2x)f(x) = 3x \ln(1 + 2x) as a power series and find specific coefficients c0,c1,c2,c3,c4c_0, c_1, c_2, c_3, c_4, as well as the radius of convergence RR of the series.

Step 1: Expand ln(1+2x)\ln(1 + 2x) as a Power Series

We can start by using the Taylor series expansion for ln(1+u)\ln(1 + u): ln(1+u)=uu22+u33u44+\ln(1 + u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \frac{u^4}{4} + \cdots Setting u=2xu = 2x, we get: ln(1+2x)=2x(2x)22+(2x)33(2x)44+\ln(1 + 2x) = 2x - \frac{(2x)^2}{2} + \frac{(2x)^3}{3} - \frac{(2x)^4}{4} + \cdots Simplifying each term, we find: ln(1+2x)=2x2x2+8x332x4+\ln(1 + 2x) = 2x - 2x^2 + \frac{8x^3}{3} - 2x^4 + \cdots

Step 2: Multiply by 3x3x to Find f(x)f(x)

Now, multiply this series by 3x3x to obtain f(x)=3xln(1+2x)f(x) = 3x \ln(1 + 2x): f(x)=3x(2x2x2+8x332x4+)f(x) = 3x \left(2x - 2x^2 + \frac{8x^3}{3} - 2x^4 + \cdots \right) Expanding each term, we get: f(x)=6x26x3+8x46x5+f(x) = 6x^2 - 6x^3 + 8x^4 - 6x^5 + \cdots

Step 3: Identify Coefficients c0,c1,c2,c3,c4c_0, c_1, c_2, c_3, c_4

Now we can identify the coefficients of f(x)f(x) in the power series form f(x)=n=0cnxnf(x) = \sum_{n=0}^{\infty} c_n x^n:

  • c0=0c_0 = 0 (no constant term)
  • c1=0c_1 = 0 (no xx term)
  • c2=6c_2 = 6
  • c3=6c_3 = -6
  • c4=8c_4 = 8

Step 4: Find the Radius of Convergence RR

The radius of convergence RR of the series cnxn\sum c_n x^n can be found using the formula: 1R=limncn+1cn\frac{1}{R} = \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| Since the function f(x)=3xln(1+2x)f(x) = 3x \ln(1 + 2x) involves ln(1+2x)\ln(1 + 2x), the radius of convergence for this series is determined by the interval of convergence for ln(1+2x)\ln(1 + 2x), which converges for 2x<1|2x| < 1, or x<12|x| < \frac{1}{2}.

Thus, R=12R = \frac{1}{2}.

Final Answers

  • c0=0c_0 = 0
  • c1=0c_1 = 0
  • c2=6c_2 = 6
  • c3=6c_3 = -6
  • c4=8c_4 = 8
  • Radius of convergence R=12R = \frac{1}{2}

Would you like further clarification on any part of this solution?

Related Questions

  1. How do you derive the Taylor series for ln(1+u)\ln(1 + u)?
  2. What is the general formula for finding the radius of convergence of a power series?
  3. Why do certain functions, like logarithmic functions, have restricted radii of convergence?
  4. How would the solution change if the function were 3x2ln(1+2x)3x^2 \ln(1 + 2x)?
  5. What is the significance of each coefficient in the power series expansion?

Tip

When dealing with logarithmic functions in series expansions, remember that they often converge within a limited interval due to their inherent nature.

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Math Problem Analysis

Mathematical Concepts

Power Series
Taylor Series Expansion
Logarithmic Functions

Formulas

ln(1 + u) = u - u^2/2 + u^3/3 - u^4/4 + ...
Radius of convergence R: 1/R = lim (n -> ∞) |c_(n+1)/c_n|

Theorems

Taylor Series
Radius of Convergence Formula

Suitable Grade Level

Undergraduate Mathematics