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. $$\sum_{n=0}^{\infty} n x^{n}$$

Short answer

Differentiate and integrate term-by-term, the intervals match the original function's interval.

Step-by-step solution

Differentiate the Power Series

Given the function as a power series: \( f(x) = \sum_{n=0}^{\infty} a_n x^n \), it's needed to find the derivative. Differentiate term by term: \[ f'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1}. \] The constant term \(a_0\) disappears upon differentiation because \(0 \cdot a_0 \cdot x^{-1} = 0\). Hence, the power series for \(f'(x)\) is started from \(n=1\) with each term becoming \(n\cdot a_n \cdot x^{n-1}\).

Determine the Interval of Convergence for the Derivative

The interval of convergence for the derivative of a power series is the same as the original function's interval of convergence, minus any possible endpoints where differentiation might cause divergence. However, since the power series is shifted by reducing the power of \(x\), the endpoints often need special attention unless known generally to hold.

Integrate the Power Series

To find the integral of the function expressed as a series, integrate term by term: \[ \int f(x) \, dx = \sum_{n=0}^{\infty} \int a_n x^n \, dx = \sum_{n=0}^{\infty} \frac{a_n}{n+1} x^{n+1} + C. \] Here, \(C\) is the constant of integration and each term \(a_n x^n\) becomes \(\frac{a_n}{n+1} x^{n+1}\).

Determine the Interval of Convergence for the Integral

Integrating a power series generally increases its convergence radius or maintains it, not reducing it. Thus, the interval of convergence for the integral will largely coincide with the function’s interval, except both ends can be verified if integral conditions don't diverge!

Apply to Specific Function

Applying to the specific given function \(\sum_{n=0}^{\infty} n x^n\):- Derivative: Using \(f'(x) = \sum_{n=1}^{\infty} n n x^{n-1} = \sum_{n=1}^{\infty} n^2 x^{n-1}\).- Integral: \(\int f(x) \, dx = \sum_{n=0}^{\infty} \frac{n}{n+1} x^{n+1} + C\)Its special convergence properties depend exactly on the same intervals.

Key concepts

Derivative of a Power Series

To differentiate a power series, you follow a simple term-by-term process. Given a function as a power series, like \( f(x) = \sum_{n=0}^{\infty} a_n x^n \), the derivative \( f'(x) \) is obtained by performing standard differentiation to each term. This involves bringing down the exponent and reducing the power of \( x \) by one.
As we carry out this operation, our series transforms to:

• Start from \( n=1 \) since the constant term disappears after differentiation.
• Each term turns into \( n \cdot a_n \cdot x^{n-1} \).
• The resulting series becomes \( f'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} \).

This differentiation process is incredibly useful and maintains intuitive results in terms of polynomial constants and structures, allowing for efficient methods in calculus regarding function approximation. The actual change is straightforward and highlights the elegance of calculus within infinite series.

Integral of a Power Series

When integrating a power series, the process again follows a term-by-term approach. Here, given a power series \( f(x) = \sum_{n=0}^{\infty} a_n x^n \), integration is applied to each term:

• Add one to the exponent of \( x \).
• Divide by the new exponent, which is \( n+1 \).
• Include a constant of integration, \( C \), as part of the solution.

This term-by-term approach results in a new series: \[ \int f(x) \, dx = \sum_{n=0}^{\infty} \frac{a_n}{n+1} x^{n+1} + C. \]Calculating integrals this way not only accumulates functions but also simplifies numerical approximations, sustaining accuracy over continuous spans of \( x \).
The resulting integral reveals insights similar to its parent function while introducing the constant \( C \), which accounts for differences in initial value across any domain.

Interval of Convergence

The interval of convergence is crucial when dealing with derivatives and integrals of power series. It defines the set of \( x \)-values for which the series converges to a function. For the original power series \( \sum_{n=0}^{\infty} a_n x^n \), you first identify this interval.
For the derivative, generally, the interval of convergence remains the same, although the behavior at the endpoints needs careful examination, as differentiation might lead to divergence or maintain convergence.
Consequently, whenever examining an integral of a power series \( \int f(x) \, dx \), typically, the interval maintains or expands, since integration tends to "smooth out" functions making them converge more easily.

• Same start and end for most intervals.
• Verify endpoints separately, as behaviors might differ.

Understanding convergence intervals helps ensure the validity of estimates or transformations back to functions, maintaining consistency across different types of calculus applications. It's the anchor for the practical utility that power series bring to mathematical modeling and computations.