Math Hell

Power Series Representation

  • We begin with an expression ( x43\frac{x^4}{3} ) alongside a constant 6 that can be factored out.

  • After factoring, we express the equation as:

    • 3x26×11+x43\frac{3x^2}{6} \times \frac{1}{1 + \frac{x^4}{3}}

  • This simplifies on the left to:

    • x22×11−(−x43)\frac{x^2}{2} \times \frac{1}{1 - \left(-\frac{x^4}{3}\right)}

  • Importance of noticing that we're looking at a geometric series.

  • The correct format is 1 over 1 minus a common ratio, where the common ratio ( r=−x43r = -\frac{x^4}{3} ).

Converting to Power Series

  • The expression can be transformed into a geometric series by using the formula a1−r=∑n=0∞arn\frac{a}{1-r} = \sum_{n=0}^{\infty} ar^n:

    • x22∑n=0∞(−x43)n\frac{x^2}{2} \sum_{n=0}^{\infty} \left(-\frac{x^4}{3}\right)^n

    • Which expands further to: x22∑n=0∞(−1)n(x43)n\frac{x^2}{2} \sum_{n=0}^{\infty} (-1)^n \left(\frac{x^4}{3}\right)^n

  • This represents an unsimplified but valid power series.

  • We can bring constant factors in and out of the summation:

    • Important to recognize integration simplifies similarly, but summation is flexible in handling constants.

  • The radius of convergence (ROC) for this geometric series is determined by |r| < 1:

    • |-\frac{x^4}{3}| < 1

    • \frac{|x^4|}{3} < 1

    • |x^4| < 3

    • |x| < \sqrt[4]{3}

  • Thus, the radius of convergence is R=34R = \sqrt[4]{3}. The interval of convergence must be checked at endpoints, but for basic geometric series, it is (−34,34)(-\sqrt[4]{3}, \sqrt[4]{3}).

Rule of Exponents

  • Since the term involves exponent rules, we apply them to simplify the general term:

    • Convert to: (−1)n(x4)n⋅x23n=(−1)nx4n⋅x23n(-1)^n \frac{(x^4)^n \cdot x^2}{3^n} = (-1)^n \frac{x^{4n} \cdot x^2}{3^n}

    • Which simplifies to: (−1)nx4n+23n(-1)^n \frac{x^{4n + 2}}{3^n}

  • Resulting in the final warranted representation after including the 1/21/2 factor from the front:

    • ∑n=0∞(−1)nx4n+22⋅3n\sum_{n=0}^{\infty} (-1)^n \frac{x^{4n + 2}}{2 \cdot 3^n}

Integrating Geometric Functions

  • Integrating a power series can be done term-by-term within its radius of convergence. If f(x)=∑<em>n=0∞c</em>n(x−a)nf(x) = \sum<em>{n=0}^{\infty} c</em>n (x-a)^n, then the integral is:

    • ∫f(x)dx=C+∑<em>n=0∞∫c</em>n(x−a)ndx=C+∑<em>n=0∞c</em>n(x−a)n+1n+1\int f(x) dx = C + \sum<em>{n=0}^{\infty} \int c</em>n (x-a)^n dx = C + \sum<em>{n=0}^{\infty} c</em>n \frac{(x-a)^{n+1}}{n+1}

  • The radius of convergence of the integrated series remains the same as the original series.

  • However, the interval of convergence may change at the endpoints; specific tests (like the Ratio Test or Endpoint Tests such as the Alternating Series Test or p-series test) are needed to determine convergence at the endpoints after integration.

  • For example, if we were to integrate the original function x2211+x43\frac{x^2}{2} \frac{1}{1 + \frac{x^4}{3}} (which is x2233+x4=3x22(3+x4)\frac{x^2}{2} \frac{3}{3+x^4} = \frac{3x^2}{2(3+x^4)}), we would get a new power series whose terms correspond to the integrals of (−1)nx4n+22⋅3n\frac{(-1)^n x^{4n + 2}}{2 \cdot 3^n}:

    • ∫(∑<em>n=0∞(−1)nx4n+22⋅3n)dx=C+∑</em>n=0∞(−1)nx4n+32⋅3n(4n+3)\int \left( \sum<em>{n=0}^{\infty} (-1)^n \frac{x^{4n + 2}}{2 \cdot 3^n} \right) dx = C + \sum</em>{n=0}^{\infty} (-1)^n \frac{x^{4n + 3}}{2 \cdot 3^n (4n + 3)}