Math Problem Statement

solve step by step

Solve step by step the given mathematical problems, including determining convergence of a series, finding the sum of an infinite series, evaluating a limit using power series, and representing an integral as a power series.

Solution

Let’s address these problems step-by-step.


Problem 1: Convergence of the Power Series

n=1(1)n42n(x2)nn\sum_{n=1}^\infty (-1)^n \frac{4^{2n}(x-2)^n}{\sqrt{n}}

Solution:

  1. Determine absolute convergence:

    • Analyze an=42nx2nn\left|a_n\right| = \frac{4^{2n}|x-2|^n}{\sqrt{n}}.
    • To check convergence, use the Ratio Test: limnan+1an=limn42(n+1)x2n+1/n+142nx2n/n=limn16x2n+1/n\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{4^{2(n+1)}|x-2|^{n+1}/\sqrt{n+1}}{4^{2n}|x-2|^n/\sqrt{n}} = \lim_{n \to \infty} \frac{16|x-2|}{\sqrt{n+1}/\sqrt{n}}
      • As nn \to \infty, the ratio simplifies to 16x216|x-2|.
      • For convergence, 16x2<116|x-2| < 1. Thus, x2<116|x-2| < \frac{1}{16}.
  2. Interval of convergence:

    • Test endpoints:
      • At x=2+116x = 2 + \frac{1}{16} or x=2116x = 2 - \frac{1}{16}, the series becomes an alternating series. Analyze n\sqrt{n} growth to check conditional convergence.
  3. Conclusion:

    • The series converges absolutely for x2<116|x-2| < \frac{1}{16} and conditionally at the boundaries if alternating series criteria are satisfied.

Problem 2: Sum of the Infinite Series

3+92!+273!+814!+3 + \frac{9}{2!} + \frac{27}{3!} + \frac{81}{4!} + \cdots

Solution:

  1. Recognize the pattern:
    The series can be written as: n=03n+1n!\sum_{n=0}^\infty \frac{3^{n+1}}{n!} Factor out the 3: 3n=03nn!3 \sum_{n=0}^\infty \frac{3^n}{n!}

  2. Use the exponential series formula: The series n=0xnn!=ex\sum_{n=0}^\infty \frac{x^n}{n!} = e^x. Thus: n=03nn!=e3\sum_{n=0}^\infty \frac{3^n}{n!} = e^3

  3. Final result: 3e33 \cdot e^3


Problem 3: Power Series to Find Limit

limx0ex21+x25x4\lim_{x \to 0} \frac{e^{-x^2} - 1 + x^2}{5x^4}

Solution:

  1. Expand ex2e^{-x^2} as a power series: ex2=1x2+x42!x63!+e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots

  2. Subtract and simplify: ex21+x2=x42!x63!+e^{-x^2} - 1 + x^2 = \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots

  3. Divide by 5x45x^4: x42!x63!+5x4=110x230+\frac{\frac{x^4}{2!} - \frac{x^6}{3!} + \cdots}{5x^4} = \frac{1}{10} - \frac{x^2}{30} + \cdots

  4. Take the limit as x0x \to 0: limx0ex21+x25x4=110\lim_{x \to 0} \frac{e^{-x^2} - 1 + x^2}{5x^4} = \frac{1}{10}


Problem 4: Indefinite Integral as Power Series

sin(x2)xdx\int \frac{\sin(x^2)}{x} dx

Solution:

  1. Expand sin(x2)\sin(x^2) as a power series: sin(x2)=x2x63!+x105!\sin(x^2) = x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!} - \cdots

  2. Divide by xx: sin(x2)x=xx53!+x95!\frac{\sin(x^2)}{x} = x - \frac{x^5}{3!} + \frac{x^9}{5!} - \cdots

  3. Integrate term by term: sin(x2)xdx=x22x663!+x10105!\int \frac{\sin(x^2)}{x} dx = \frac{x^2}{2} - \frac{x^6}{6 \cdot 3!} + \frac{x^{10}}{10 \cdot 5!} - \cdots

  4. Estimate the integral 00.1sin(x2)xdx\int_0^{0.1} \frac{\sin(x^2)}{x} dx with error no greater than 10910^{-9}:

    • Calculate sufficient terms until the error is 109\leq 10^{-9}.

Would you like detailed clarification on any of these steps?


Follow-up Questions:

  1. How does the Ratio Test help in determining absolute convergence in Problem 1?
  2. What is the role of exponential series in solving Problem 2?
  3. Why is the power series expansion crucial in solving limits as x0x \to 0 in Problem 3?
  4. How does term-by-term integration apply to Problem 4?
  5. What methods ensure error bounds in numerical series approximations?

Tip: Always verify convergence before approximating infinite series or integrals to avoid misleading results.

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Math Problem Analysis

Mathematical Concepts

Series Convergence
Infinite Series Summation
Power Series Expansion
Limit Evaluation
Definite and Indefinite Integration

Formulas

Ratio Test for series: lim(n→∞) |a_(n+1) / a_n| < 1 for convergence
Exponential series: e^x = Σ (x^n / n!)
Power series expansion of sine: sin(x) = Σ (-1)^n * (x^(2n+1) / (2n+1)!)
Term-by-term integration of power series

Theorems

Alternating Series Test
Ratio Test
Taylor Series Theorem

Suitable Grade Level

Undergraduate (Calculus II or III level)