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. $$\sum_{n=1}^{\infty} \frac{x^{n}}{n}$$

Short answer

The power series for \(f'(x)\) is \(\sum_{m=0}^{\infty} x^m\) and for \(\int f(x) dx\) is \(\sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)}\), converging for \(|x| < 1\).

Step-by-step solution

Identify the Function Given

The given function is \[ f(x) = \sum_{n=0}^{\infty} a_n x^n = \sum_{n=1}^{\infty} \frac{x^n}{n} \]Here, \(a_n = \frac{1}{n}\) for \(n \geq 1\), and \(f(x)\) is the power series representing the logarithmic function \(-\ln(1-x)\), converging for \(|x| < 1\).

Differentiate the Function

Differentiate \(f(x)\) term by term to find \(f'(x)\):\[ f'(x) = \sum_{n=1}^{\infty} \frac{d}{dx}\left(\frac{x^n}{n}\right) = \sum_{n=1}^{\infty} \frac{nx^{n-1}}{n} = \sum_{n=1}^{\infty} x^{n-1} \]This can be rewritten as:\[ f'(x) = \sum_{m=0}^{\infty} x^m \]This is a geometric series with common ratio \(x\) that converges for \(|x| < 1\).

Provide Power Series for the Integral

The integral of \(f(x)\) is:\[ \int f(x) \, dx = \int \left( \sum_{n=1}^{\infty} \frac{x^n}{n} \right) dx \]Integrating term by term gives:\[ \sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)} + C \]Where \(C\) is the constant of integration. The power series converges for \(|x| < 1\) since it is the integral of a series that converges for \(|x| < 1\).

Conclusion of Integral and Derivative Series

The power series for \(f'(x)\) is \(\sum_{m=0}^{\infty} x^m\) and for \(\int f(x) \), it is \(\sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)}\), with both series having the interval of convergence \(|x| < 1\).

Key concepts

Differentiation of Power Series

Differentiating a power series involves taking the derivative of each term individually. Given a function in the form of a power series, such as \[ f(x) = \sum_{n=0}^{\infty} a_n x^n \]we can derive its derivative by applying the rule of differentiation directly to each term. This means that for the term \(a_n x^n\), its derivative will be \(n a_n x^{n-1}\). In the exercise, the function \[ f(x) = \sum_{n=1}^{\infty} \frac{x^n}{n} \]was provided, and by differentiating term by term, we obtained \[ f'(x) = \sum_{n=1}^{\infty} x^{n-1} \]which simplifies to \[ f'(x) = \sum_{m=0}^{\infty} x^m \]This result forms a geometric series, and it converges in the interval where original series converges. The interval of convergence remains \[ |x| < 1 \]because differentiating a power series does not change the radius or interval of convergence.

Integration of Power Series

Integrating a power series means integrating each term individually with respect to \(x\). For a given power series\[ f(x) = \sum_{n=0}^{\infty} a_n x^n \]the integral will become an added power to each term:\[ \int f(x) \, dx = C + \sum_{n=0}^{\infty} \frac{a_n x^{n+1}}{n+1} \]where \(C\) is the constant of integration. It’s important to note that after integration, every term is increased by a power, changing from \(n\) to \(n+1\), and divided by this new exponent.In this case, integrating\[ \sum_{n=1}^{\infty} \frac{x^n}{n} \]results in\[ \sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)} \]Plus, the constant of integration \(C\) should not be forgotten. The convergence for integrals of power series also remains bound by the same interval as the original series, \[ |x| < 1 \]ensuring consistency in convergence behavior.

Interval of Convergence

The interval of convergence of a power series is essential in defining where the series will produce accurate results. It is defined by the range of \(x\) values for which the series converges to a finite sum.For a general power series\[ \sum_{n=0}^{\infty} a_n x^n \]we determine its interval of convergence using the ratio test, root test, or other forms of analysis. The series given\[ \sum_{n=1}^{\infty} \frac{x^n}{n} \]converges when \[ |x| < 1 \]indicating that for any \(x\) within this interval, the series represents a valid function. This condition does not change upon differentiating or integrating the series, allowing the derivative and integral forms to share the original convergence interval \[ |x| < 1 \] as well. This consistency is key for analyzing the power series within valid bounds.