Problem 4 - The Calculus with Analytic Geometry

Question

A function \(f\) is defined by a power series. In each exercise do the following: (a) Find the radius of convergence of the given power series and the domain of \(f\); (b) write the power series which defines the function \(f^{\prime}\) and find its radius of convergence by using methods of Sec. \(16.7\) (thus verifying Theorem 16.8.1); (c) find the domain of \(f^{\prime}\).\(f(x)=\sum_{n=2}^{+\infty} \frac{(x-2)^{n}}{\sqrt{n-1}}\)

Step-by-Step Solution

Answer

The radius of convergence is. The domains of and are both.

Step 1: Identify the Power Series

The given function is represented by the power series:

Step 2: Analyze the General Term

Identify the general term of the series:

Step 3: Apply the Ratio Test

Utilize the ratio test to determine the radius of convergence. , we get .

Step 4: Solve the Limit of the Ratio Test

Compute the limit: Therefore, we have .

Step 5: Find the Radius of Convergence

The solution for the limit gives us the radius of convergence: .

Step 6: Determine the Domain of

The domain of is given by:

Step 7: Differentiate term-wise

Differentiate the power series term-by-term to find .

Step 8: Identify the New Series

Apply differentiation to find the new series for :

Step 9: Apply the Ratio Test for

Utilize the ratio test again for the radius of convergence of:

Step 10: Simplify the limit for

The ratio test computation yields the same radius of convergence as the original series.

Step 11: Verify with Theorem

The radius of convergence is verified by the theorem applied.

Step 12: Determine the Domain of

Thus, the domain of is similar to that of.

Key Concepts

Radius of Convergence

To understand the radius of convergence, we first need to look at what a power series is. A power series is an infinite series of the form \(\sum_{n=0}^{\infty} a_n (x - c)^n\), where \(a_n\) are coefficients and \(c\) is the center of the series. The radius of convergence is the distance from the center \(c\) within which the series converges.

In mathematical terms, a series converges if the terms approach a specific number as \(n\) becomes large. We can use the ratio test to find the radius of convergence. If \(\lim_{{n \to \infty}} \left| \frac{{a_{n+1}(x-c)^{n+1}}}{{a_n (x-c)^n}} \right| < 1\), then the series converges. For our given power series, we apply the ratio test to identify the radius within which the function is well-defined and convergent.

Ratio Test

The ratio test is a powerful tool to determine the convergence of a series. It involves taking the limit of the absolute value of the ratio of successive terms \(a_{n+1}\) and \(a_n\) as \(n\) approaches infinity. Mathematically, the ratio test is expressed as:

\[\lim_{{n \to \infty}} \left| \frac{{a_{n+1}}}{{a_n}} \right| = L.\]

If \(L < 1\), the series converges. If \(L > 1\), the series diverges. If \(L = 1\), the test is inconclusive.

In step-by-step solutions, we apply this test to determine how far from the center (2 in this case) the series converges. For the given series \(\sum_{n=2}^{+\infty} \frac{(x-2)^n}{\sqrt{n-1}}\), we utilize the ratio test on the general term \(a_n = \frac{(x-2)^n}{\sqrt{n-1}}\) and find the radius of convergence to be 1.

Term-by-term Differentiation

Term-by-term differentiation is an approach where we differentiate each term of a power series individually. This method can sometimes be simpler than differentiating the entire series at once.

For example, if we have \(f(x) = \sum_{n=2}^{+\infty} \frac{(x-2)^n}{\sqrt{n-1}}\), we can differentiate it term by term:

\[f'(x) = \sum_{n=2}^{+\infty} \frac{d}{dx} \left( \frac{(x-2)^n}{\sqrt{n-1}} \right).\]

By differentiating each term individually, you get:

\[f'(x) = \sum_{n=2}^{+\infty} \frac{n (x-2)^{n-1}}{\sqrt{n-1}}.\]

This new power series represents the derivative of the original function. Next, we apply the ratio test to this new series to determine its radius of convergence.

Analytic Functions

An analytic function is a function that is locally given by a convergent power series. This means around any point within the radius of convergence, the function can be expressed as a power series. Such functions are infinitely differentiable and can be very effectively approximated by their Taylor series.

In the context of our problem, given \(f(x) = \sum_{n=2}^{+\infty} \frac{(x-2)^n}{\sqrt{n-1}}\), we say function \(f\) is analytic within its radius of convergence, which is determined to be 1.

This implies \(f\) is smoothly defined and differentiable within this range. Also, any derivatives like \(f'(x)\) also share similar properties, by being represented by another power series within the same radius of convergence.