10.12 Representing Functions as Power Series

• The center of a power series is also known as the point of expansion
• What is the key difference between a power series and a Taylor series?

  $a =$$0$
• The center of a Taylor series is always zero
• The ratio test is used to find the limit L = \lim_{n \to \infty} \left| \frac{a_{n + 1}}{a_{n}} \right|</latex>, which helps determine the radius of convergence
• Steps to find the radius and interval of convergence:

  1️⃣ Use the Ratio Test to find $L$

  2️⃣ Calculate $R =$$\frac{1}{L}$

  3️⃣ Check convergence at endpoints $x =$$a \pm R$
• What test is used to determine the convergence of a power series?

  Ratio Test
• What is the primary purpose of a power series?

  Represent functions as infinite series
• What is the general form of a Taylor series centered at a = 0</latex>?

  $\sum_{n = 0}^{\infty} f^{(n)}(0)\frac{x^{n}}{n!}$
• The radius of convergence determines the range of $x$ values for which the power series converges.
• What is the general form of a power series?

  $\sum_{n = 0}^{\infty} a_{n}(x - a)^{n}$
• Power series are used to represent functions as infinite series.


  True
• Match the type of series with its form or center:

  Power Series ↔️ $\sum_{n = 0}^{\infty} a_{n}(x - a)^{n}$

  Taylor Series ↔️ $\sum_{n = 0}^{\infty} f^{(n)}(0)\frac{x^{n}}{n!}$
• The radius of convergence determines the range of $x$ values for which a power series converges.


  True
• What does the radius of convergence determine for a power series?

  Range of $x$ values
• The interval of convergence includes the radius of convergence as its endpoints.

  False
• What are the coefficients of a power series denoted by?

  $a_{n}$
• The series converges if $L < 1$, so |x - 2| < 1</latex>, giving a radius of convergence $R =$$1$. The interval of convergence is [1, 3)
• The series converges at $x =$$1$ because it is an alternating harmonic series.


  True
• What is the general form of a power series?

  $\sum_{n = 0}^{\infty} a_{n}(x - a)^{n}$
• In a power series, the $a_{n}$ are called the coefficients
• What is the term 'center' in the context of a power series?

  Point of expansion
• A Taylor series is a special case of a power series centered at $a =$$0$.


  True
• Match the type of series with its center:

  Power Series ↔️ a

  Taylor Series ↔️ 0
• The interval of convergence includes the endpoints of the range where the series converges.

  True
• Steps to find the radius and interval of convergence

  1️⃣ Use the Ratio Test to find the limit $L =$$\lim_{n \to \infty} \left| \frac{a_{n + 1}}{a_{n}} \right|$

  2️⃣ Calculate the radius of convergence: $R =$$\frac{1}{L}$

  3️⃣ Check convergence at endpoints $x =$$a \pm R$
• What happens to the radius of convergence when a power series is differentiated or integrated term-by-term?

  Remains unchanged
• Match the function with its power series representation:

  $e^{x}$ ↔️ $\sum_{n = 0}^{\infty} \frac{x^{n}}{n!}$

  $\sin(x)$ ↔️ $\sum_{n = 0}^{\infty} \frac{( - 1)^{n} x^{2n + 1}}{(2n + 1)!}$
• The radius of convergence for $\sin(x)$ is $\infty$


  True
• A Taylor series is a special case of a power series with a center at $a =$$0$


  True
• The range of $x$ values for which a power series converges is determined by the radius of convergence.
• Steps to find the radius and interval of convergence:

  1️⃣ Use the Ratio Test to find $L =$$\lim_{n \to \infty} \left| \frac{a_{n + 1}}{a_{n}} \right|$

  2️⃣ Calculate $R =$$\frac{1}{L}$

  3️⃣ Check convergence at $x =$$a \pm R$
• Differentiating a power series term-by-term maintains the same radius of convergence
• Integrating a power series term-by-term maintains the same radius of convergence
• What happens to the radius of convergence when differentiating or integrating a power series term-by-term?

  Remains unchanged
• When differentiating a power series, each term is differentiated individually
• The radius of convergence remains infinite when differentiating the power series for $f(x) =$$\sum_{n = 0}^{\infty} \frac{x^{n}}{n!}$


  True
• How is the radius of convergence calculated using the limit $L$ from the Ratio Test?


  $R =$$\frac{1}{L}$
• For the power series \sum_{n = 1}^{\infty} \frac{(x - 2)^{n}}{n}</latex>, the radius of convergence is 1
• Differentiating a power series term-by-term changes its radius of convergence.

  False
• Integrating the power series $\sum_{n = 0}^{\infty} x^{n}$ term-by-term results in \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{n + 1}