Question

A function \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) is given. (a) Give a power series for \(f^{\prime}(x)\) and its interval of convergence. (b) Give a power series for \(\int f(x) d x\) and its interval of convergence. $$\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^{n}$$

Short answer

(a) \(f'(x) = \sum_{n=1}^{\infty} n \left(\frac{1}{2}\right)^{n} x^{n-1}\), interval \((-2, 2)\); (b) \(\int f(x) \, dx = C + \sum_{n=0}^{\infty} \frac{1}{n+1}\left(\frac{1}{2}\right)^{n} x^{n+1}\), interval \((-2, 2)\).

Step-by-step solution

Identify the Given Function

The given power series is a geometric series: \(f(x) = \sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^{n} = \frac{1}{1 - \frac{x}{2}}\) for \(|x| < 2\). This series represents the function \(f(x)\) implicitly.

Differentiate the Series Term by Term

The derivative of a power series \(f(x) = \sum_{n=0}^{\infty} a_{n} x^{n}\) can be found by differentiating each term: \(f'(x) = \sum_{n=1}^{\infty} n a_{n} x^{n-1}\). For the given geometric series, this translates to differentiating: \(f(x) = \sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^{n}\), resulting in \(f'(x) = \sum_{n=1}^{\infty} n \cdot \left(\frac{1}{2}\right)^{n} x^{n-1}\).

Find the Interval of Convergence for the Derivative

Since the differentiation of a power series does not affect its interval of convergence, the interval of convergence for \(f'(x)\) remains the same as for \(f(x)\), which is \(|x| < 2\). The interval of convergence is \((-2, 2)\).

Integrate the Series Term by Term

The integral of a power series \(f(x) = \sum_{n=0}^{\infty} a_{n} x^{n}\) is given by \(\int f(x) \, dx = C + \sum_{n=0}^{\infty} \frac{a_{n}}{n+1} x^{n+1}\). For the given series, integrating each term gives \(\int f(x) \, dx = C + \sum_{n=0}^{\infty} \frac{1}{n+1}\left(\frac{1}{2}\right)^{n} x^{n+1}\).

Find the Interval of Convergence for the Integral

The interval of convergence for integration is typically the same as for the original series, but you must check the endpoints separately. However, due to uniform convergence within the open interval, the interval for \(\int f(x) \, dx\) is also \((-2, 2)\), excluding the endpoints.

Key concepts

Geometric Series

A geometric series is a series of the form \[ a + ar + ar^2 + ar^3 + \ldots = \sum_{n=0}^{\infty} ar^n \]where \(a\) is the first term of the series and \(r\) is the common ratio. In the context of power series, geometric series can represent functions. For example, \[ f(x) = \sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^{n} \]is a geometric series where the first term \(a = 1\) and the common ratio \(r = \frac{x}{2}\). This series converges to \[ \frac{1}{1 - \frac{x}{2}} \]for the values of \(x\) where the absolute value of the ratio is less than 1, i.e., \(|x| < 2\). This gives a clear visual understanding of the series' behavior and its key property: convergence within specific bounds.

• First term (\(a\)): \(1\)
• Common ratio (\(r\)): \(\frac{x}{2}\)
• Convergence condition: \(|x| < 2\)

Understanding these basics helps in efficiently working with power series and determining their functions.

Interval of Convergence

The interval of convergence is crucial for understanding where a power series converges to a particular function. For the geometric series \[ \sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^{n} \], the series converges when the absolute value of the common ratio \(\frac{x}{2}\) is less than 1.
This simplifies to \(|x| < 2\), defining the radius of convergence as 2 and thus establishing the interval of convergence as \((-2, 2)\).Knowing this interval is vital because it tells us where the power series can represent the function exactly.

• Radius of convergence (\(R\)): \(2\)
• Interval of convergence: \((-2, 2)\)

The interval is important when differentiating or integrating the series, as it typically remains unchanged, though endpoints may need additional checking.

Differentiation of Power Series

Differentiating a power series term by term is a straightforward process, preserving the radius of convergence. For a series like \[ f(x) = \sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^{n}, \]the derivative is found by differentiating each term individually:\[ f'(x) = \sum_{n=1}^{\infty} n \cdot \left(\frac{1}{2}\right)^{n} x^{n-1}. \]Here, the coefficient \(n\) arises because of the power rule applied to each term. This operation doesn't change the interval of convergence, remaining \((-2, 2)\).
Differentiation has practical utility, helping us find slopes and rates of change for series-based functions, without altering the convergence properties of the original function.

• Derivative Form: \(f'(x) = \sum_{n=1}^{\infty} n \cdot \left(\frac{1}{2}\right)^{n} x^{n-1}\)
• Interval of Convergence after Differentiation: \((-2, 2)\)

Integration of Power Series

Integrating power series involves integrating term by term, which can lead to new series representations of functions. The integral of a power series like \[ f(x) = \sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^{n} \] is computed by finding the integral of each term:\[ \int f(x) \, dx = C + \sum_{n=0}^{\infty} \frac{1}{n+1} \left(\frac{1}{2}\right)^{n} x^{n+1}, \]where \(C\) is the constant of integration. The process often preserves the original interval of convergence \((-2, 2)\), except boundaries should be individually assessed.
The integration adds flexibility by allowing accumulation rates and the total sum of values over intervals, again without affecting the convergence properties within most of the interval.

• Integrated Form: \(\int f(x) \, dx = C + \sum_{n=0}^{\infty} \frac{1}{n+1} \left(\frac{1}{2}\right)^{n} x^{n+1}\)
• Interval of Convergence after Integration: Typically \((-2, 2)\), endpoints might need checking.