Problem 53 - Thomas Calculus

Question

For what values of \(x\) does the series $$1-\frac{1}{2}(x-3)+\frac{1}{4}(x-3)^{2}+\cdots+\left(-\frac{1}{2}\right)^{n}(x-3)^{n}+\cdots$$ converge? What is its sum? What series do you get if you differentiate the given series term by term? For what values of \(x\) does the new series converge? What is its sum?

Step-by-Step Solution

Answer

The original series converges for \(1 < x < 5\) with sum \(\frac{2}{5-x}\). The differentiated series converges for \(1 < x < 5\) with sum \(\frac{-1}{5-x}\).

Step 1 - Identify the Series Type

The given series is \(1 - \frac{1}{2}(x-3) + \frac{1}{4}(x-3)^2 + \cdots + \left(-\frac{1}{2}\right)^n(x-3)^n + \cdots\). This is a geometric series with the first term \(a = 1\) and the common ratio \(r = -\frac{1}{2}(x-3)\).

Step 2 - Determine Convergence Criteria

A geometric series \(\sum_{n=0}^{\infty} ar^n\) converges if and only if the common ratio \(|r| < 1\). In this series, the common ratio is \(|r| = \left| -\frac{1}{2}(x-3) \right| < 1\).

Step 3 - Solve the Inequality

Solve \(\left| -\frac{1}{2}(x-3) \right| < 1\) to find \(x\):1. \(\left| x-3 \right| < 2\), to remove the negative sign and constant.2. Solve absolute inequality: \(-2 < x-3 < 2\).3. Add 3 to all sides: \(1 < x < 5\).Thus, the series converges for \(1 < x < 5\).

Step 4 - Calculate the Sum of the Series

The sum \(S\) of a convergent geometric series \(a + ar + ar^2 + \cdots\) is given by \(S = \frac{a}{1-r}\), where \(|r| < 1\). Here \(a = 1\) and \(r = -\frac{1}{2}(x-3)\), thus:\[ S = \frac{1}{1 + \frac{1}{2}(x-3)} = \frac{2}{5-x} \] for \(1 < x < 5\).

Step 5 - Differentiate the Series Term by Term

If you differentiate the original series term by term, you get:\[ 0 - \frac{1}{2} + \frac{1}{2}(x-3) - \frac{3}{4}(x-3)^2 + \cdots \] This results from differentiating each term \(\left(-\frac{1}{2}\right)^n(x-3)^n\) as \(n\left(-\frac{1}{2}\right)^n(x-3)^{n-1}\).

Step 6 - Identify the New Series Type

The differentiated series is \(-\frac{1}{2} + \frac{1}{2}(x-3) - \frac{3}{4}(x-3)^2 + \cdots\), which is again a geometric series with first term \(a = -\frac{1}{2}\) and common ratio \(r = -\frac{1}{2}(x-3)\).

Step 7 - Determine Convergence Criteria for New Series

As it is the derivative of the original series, the same convergence interval \(1 < x < 5\) applies as this condition ensures \(|r| < 1\) in the differentiated series as well.

Step 8 - Calculate the Sum of the New Series

The sum of the differentiated series can be found by summing the geometric series with first term \(a = -\frac{1}{2}\) and the same ratio \(r = -\frac{1}{2}(x-3)\):\[ S' = \frac{-\frac{1}{2}}{1 + \frac{1}{2}(x-3)} = \frac{-1}{5-x} \] for \(1 < x < 5\).

Key Concepts

Series Convergence

When you hear about series convergence, it's like asking whether the series has a finite limit as we keep adding more terms. In a geometric series like this one, a series converges when its common ratio, often denoted by \(r\), satisfies the condition \(|r| < 1\). This condition ensures that the series doesn't wildly grow out to infinity but rather settles down to a specific value.

For our series, the common ratio is \(-\frac{1}{2}(x-3)\). To check convergence, we find when \(|-\frac{1}{2}(x-3)| < 1\). Solving this tells us that the series converges when \(1 < x < 5\). That means for any \(x\) within this range, the infinite sum will tend towards a fixed number.

This understanding of convergence is crucial in calculus and many real-life applications like physics and finance, where knowing the behavior of sums over time matters.

Sum of Series

Once we know a series converges, we can actually find this resulting finite limit or sum. For geometric series, there's a handy formula: \(S = \frac{a}{1-r}\), where \(a\) is the first term of the series, and \(r\) is the common ratio. This formula gives us a quick way to compute the sum without manually adding each term, which is practically impossible for infinite series!

In our example, plugging in \(a = 1\) and \(r = -\frac{1}{2}(x-3)\), we get:

• \(S = \frac{2}{5-x}\)

for \(1 < x < 5\). This gives a neat and concise expression for the sum, making it easy to evaluate for any given \(x\) in the convergence range. It demonstrates how powerful mathematical tools can simplify what would otherwise be an unending task.

Term Differentiation

Differentiating a series term by term might sound complex, but it's quite a straightforward process if done correctly. The idea is, we take each term of our series and apply the rules of differentiation, just like we do with polynomials or functions.

So, if we differentiate each term \((-\frac{1}{2})^n(x-3)^n\), the term becomes \(n(-\frac{1}{2})^n(x-3)^{n-1}\). It's important to note that the first term (which was a constant, 1) vanishes upon differentiation, leading to a new series:

\(-\frac{1}{2} + \frac{1}{2}(x-3) - \frac{3}{4}(x-3)^2 + \cdots\).

The cool thing about differentiating a power series is that it mirrors our earlier rules of convergence. That means this new series also converges over the same interval \(1 < x < 5\).

The sum of this new series can be determined in a similar way, yielding \(S' = \frac{-1}{5-x}\). Differentiation turns series expansion into an exercise of understanding changes, making it immensely useful in contexts like physics, where changes in motion or other parameters are analyzed.