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Chapter 7: Problem 46

Describe how to differentiate and integrate a power series with a radius of convergence \(R\). Will the series resulting from the operations of differentiation and integration have a different radius of convergence? Explain.

Short Answer

Expert verified
Yes, the operations of differentiation and integration can be performed on a power series term by term. The derived power series will be \(f'(x) = \sum_{n=1}^\infty n.a_n(x-c)^{n-1}\) and the integrated power series will be \(F(x) = C + \sum_{n=0}^\infty \frac{a_n}{n+1}(x-c)^{n+1}\). Regardless of whether we differentiate or integrate, the radius of convergence remains unchanged, i.e. \(R' = R\).

Step by step solution

01

Differentiate the Power Series

Consider a power series \(f(x) = \sum_{n=0}^\infty a_n(x-c)^n\) with a radius of convergence \(R\). The derivative of this power series \(f'(x)\) can be obtained by differentiating term by term. The derived power series is \(f'(x) = \sum_{n=1}^\infty n.a_n(x-c)^{n-1}\).
02

Integrate the Power Series

Now, consider integrating the same power series \(f(x)\). The integral of this power series \(F(x) = \int f(x)dx\) can be obtained by integrating term by term. The integrated power series is \(F(x) = C + \sum_{n=0}^\infty \frac{a_n}{n+1}(x-c)^{n+1}\), where \(C\) is the constant of integration.
03

Determine the Radius of Convergence

Now investigate whether the radius of convergence changes after the operations. According to mathematical principles, the radius of convergence \(R'\) for the derivative or the integral of a power series is the same as the original radius \(R\). This can be proven using the ratio test to determine convergence. Differentiating or integrating does not affect the radius of convergence.

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