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What is the radius of convergence of the power series \(\Sigma c_{k}(x / 2)^{k}\) if the radius of convergence of \(\sum c_{k} x^{k}\) is \(R ?\)

Short Answer

Expert verified
Answer: The radius of convergence for the modified power series is \(R' = \frac{1}{2}R\).

Step by step solution

01

Recall the Ratio Test formula

The Ratio Test states that, if the limit \(\lim_{k\to\infty}\frac{|a_{k+1}|}{|a_{k}|}\) exists and is equal to \(L\), then the series converges when \(L<1\) and diverges when \(L>1\). We will apply this test to the given power series and then to the modified series.
02

Analyzing the given power series \(\sum c_{k}x^{k}\)

The series converges with radius \(R\). By applying the Ratio Test to this power series, we obtain the limit: \(\lim_{k\to\infty} \frac{|c_{k+1}x^{k+1}|}{|c_{k}x^{k}|} = \lim_{k\to\infty} \frac{|c_{k+1}|}{|c_{k}|} |x| = L\) We know that the series converges, so \(L<1\). Since the radius of convergence is \(R\), we must have: \(|x| < R\)
03

Analyzing the modified power series \(\sum c_{k}(x/2)^{k}\)

To analyze the radius of convergence for the modified series, we will also apply the Ratio Test: \(\lim_{k\to\infty} \frac{|c_{k+1}(\frac{x}{2})^{k+1}|}{|c_{k}(\frac{x}{2})^{k}|} = \lim_{k\to\infty} \frac{|c_{k+1}|}{|c_{k}|} |\frac{x}{2}|\) Let's call this limit \(L'\). For the series to converge we need \(L'<1\).
04

Compare the limits

We need to find the relationship between \(L\) and \(L'\) for the convergence of the modified series. From Step 2 and Step 3, we have: \(L = \frac{|c_{k+1}|}{|c_{k}|} |x|\) \(L' = \frac{|c_{k+1}|}{|c_{k}|} |\frac{x}{2}|\) Dividing both equations, we get: \(\frac{L'}{L} = \frac{x/2}{x} = \frac{1}{2}\) So, we have \(L' = \frac{1}{2}L\).
05

Find the radius of convergence for the modified series

We know that for the given series to converge, \(L<1\), and for the modified series to converge \(L'<1\). Since \(L'=\frac{1}{2}L\), we have: \(R' = \frac{1}{2}R\) Therefore, the radius of convergence for the modified power series \(\Sigma c_{k}(x / 2)^{k}\) is \(R'\), which equals to \(\frac{1}{2}R\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Series
A power series is one of the fascinating ways to represent functions using an infinite sum of terms. Think of it as a series of powers, usually involving a variable like x. The general form is given by:
  • \( \sum_{k=0}^{\infty} c_k (x - a)^k \)
Here, each term in the series includes a coefficient \(c_k\) and a power of \(x - a\). The "a" represents the center of the series, or where it is "anchored" on the x-axis.
The series behaves like a polynomial, but the number of terms is infinite. One important thing to note is that power series do not converge everywhere. They are typically valid within a certain interval around the center "a." The extent of this interval is known as the "radius of convergence."
The radius dictates where the series is a good representation of the function. A larger radius means the series converges for a wider range of x values.
Ratio Test
The Ratio Test is a powerful method to determine the convergence of an infinite series. It is particularly useful when dealing with power series, as these series can often include complex expressions that are otherwise hard to manage.
Here's how it works:
  • Consider the series \( \sum a_k \).
  • Calculate the limit \( L = \lim_{k \to \infty} \frac{|a_{k+1}|}{|a_k|} \).
The result of the Ratio Test gives us critical information about the series:
  • If \( L < 1 \), the series converges absolutely.
  • If \( L > 1 \), the series diverges.
  • If \( L = 1 \), the test is inconclusive.
When applied to power series, the Ratio Test helps find the radius of convergence. You calculate the limit considering the expression of the series and use it to identify up to which point (radius) you can expect the series to converge. This process decomposes a potentially complex problem into manageable steps.
Convergence of Series
Understanding convergence is essential when dealing with series. A series converges when the sum of its terms approaches a finite value as more terms are added. The considerations of convergence have deep implications in both theory and application.
The convergence of a series ensures that the infinite sum behaves like a finite quantity. For power series, this translates into a range of x values for which the series provides an accurate representation or solution.
Different tests like the Ratio Test or the Comparison Test help determine if a series converges. In the context of power series, the Radius of Convergence defines the boundary within which the a series converges.
Any x within this interval means that the partial sums of the series converge to a value; outside this interval, convergence isn't guaranteed. This way, mathematicians and scientists can implement power series effectively, knowing where they work accurately, like approximating functions or solving differential equations.

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Most popular questions from this chapter

Here is an alternative way to evaluate higher derivatives of a function \(f\) that may save time. Suppose you can find the Taylor series for \(f\) centered at the point a without evaluating derivatives (for example, from a known series). Explain why \(f^{(k)}(a)=k !\) multiplied by the coefficient of \((x-a)^{k} .\) Use this idea to evaluate \(f^{(3)}(0)\) and \(f^{(4)}(0)\) for the following functions. Use known series and do not evaluate derivatives. $$f(x)=e^{\cos x}$$

Find the function represented by the following series and find the interval of convergence of the series. (Not all these series are power series.) $$\sum_{k=0}^{\infty}(\sqrt{x}-2)^{k}$$

Find the function represented by the following series and find the interval of convergence of the series. (Not all these series are power series.) $$\sum_{k=0}^{\infty}\left(\frac{x^{2}-1}{3}\right)^{k}$$

a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. You do not need to use the definition of the Taylor series coefficients. b. Determine the radius of convergence of the series. $$f(x)=\sqrt{1-x^{2}}$$

a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. You do not need to use the definition of the Taylor series coefficients. b. Determine the radius of convergence of the series. $$f(x)=\left\\{\begin{array}{ll} \frac{\sin x}{x} & \text { if } x \neq 0 \\ 1 & \text { if } x=0 \end{array}\right.$$

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