Question

Scaling power series If the power series \(f(x)=\sum c_{k} x^{k}\) has an interval of convergence of \(|x|

Short answer

Answer: The interval of convergence for the power series of \(f(ax)\) is \(|x|<\frac{R}{|a|}\).

Step-by-step solution

Substitute ax for x in the power series

We are given the power series \(f(x) = \sum_{k} c_{k} x^{k}\), and we want to find the power series for \(f(ax)\). We substitute \(ax\) for \(x\) in the power series to obtain: \(f(ax) = \sum_{k} c_{k} (ax)^{k}\)

Simplify the power series

Now, we can simplify the power series by using the property of the exponent, \((ab)^n = a^n * b^n\): \(f(ax) = \sum_{k} c_{k} (a^{k}x^{k})\) Now, we have the power series for \(f(ax) = \sum_{k} (c_{k} a^{k})x^{k}\).

Apply the Ratio Test

To find the interval of convergence of our new power series, we will use the Ratio Test. The Ratio Test states that the interval of convergence is determined by the limit: \(\lim_{k \to \infty} \left|\frac{x^{k+1}(c_{k+1}a^{k+1})}{x^k(c_ka^k)}\right| < 1\) Solve for the limit: \(\lim_{k \to \infty} \left|\frac{a^{k+1}}{a^k}\frac{c_{k+1}}{c_k}\right| < 1\) \(\lim_{k \to \infty} \left|a\frac{c_{k+1}}{c_k}\right| < 1\) Now, we have the inequality given by the Ratio Test. In order to find the interval of convergence, we need to solve this inequality for \(x\).

Find the interval of convergence

From the original power series, we are given that the interval of convergence is \(|x|<R\). Now, we have the following inequality from the Ratio Test in Step 3: \(\left|a\frac{c_{k+1}}{c_k}\right| < 1\) To solve this for \(x\), we can rewrite the inequality as follows: \(|x|<\frac{1}{|a|}|a\frac{c_{k+1}}{c_k}|\) Thus, the new interval of convergence for the power series of \(f(ax)\) is: \(|x|<\frac{R}{|a|}\)

Key concepts

Interval of Convergence

The interval of convergence for a power series is the range of values within which the series converges, or adds up to a finite number. Mathematically, consider a power series of the form \[f(x) = \sum_{k} c_{k} x^{k}.\]This series converges for values of \(x\) within an interval that can be expressed as \[|x| < R,\]where \(R\) represents the radius of convergence. Within this interval, the infinite sum converges; outside it, the series diverges.

Sometimes, the interval of convergence includes the endpoints, depending on the specific series. Determining whether these endpoints converge requires testing them individually.

For example, if the original series converges on the interval \(-R < x < R\),the endpoints \(x = -R\)and \(x = R\)need to be checked separately to know for certain if they belong to the interval of convergence.

Ratio Test

The Ratio Test is a method used to determine the convergence or divergence of an infinite series. This test applies particularly well to power series. To apply the Ratio Test, we consider the limit of the absolute value of the ratio between consecutive terms in the series:

• If this ratio is less than 1 as the number of terms goes to infinity, the series converges absolutely.
• If the ratio is greater than 1, or if it becomes infinite, the series diverges.
• If the ratio equals 1, the test is inconclusive.

Let's apply it to a series \[\sum_{k} c_{k} x^{k}.\]We calculate:
\[\lim_{k \to \infty} \left| \frac{c_{k+1}x^{k+1}}{c_kx^k} \right| = |x| \cdot \lim_{k \to \infty} \left| \frac{c_{k+1}}{c_k} \right|.\]This allows us to find the value of \(x\) such that the series converges, thereby establishing the interval of convergence. This step-wise method helps determine how far you can "stretch" or "contract" \(x\) while still ensuring the series remains convergent.

Scaling Power Series

Scaling a power series involves changing the variable by a certain factor, often denoted as \(a\) in the expression \(f(ax)\). When scaling, it is crucial to understand how this transformation affects the interval of convergence. For a series \[f(x) = \sum_{k} c_{k} x^{k},\]which converges within \(|x| < R,\)applying a scaling factor \(a\) creates the new series
\[f(ax) = \sum_{k} c_{k} (a^k x^k) = \sum_{k} (c_{k} a^k) x^k.\]The crucial change lies in the interval of convergence. By substituting \(ax\) for \(x\), you essentially "compress" or "expand" the interval depending on \(a's\) magnitude.

To find the suitable values for \(x\) in the new scaled series, note that the new convergence criterion will be \(|x| < \frac{R}{|a|}.\)So, if \(a\) is greater than 1, the interval becomes narrower, and if \(0 < a < 1\), the interval widens. This adjustment allows you to determine where the series remains convergent based on the scaling applied. This concept illustrates the powerful flexibility of power series in adapting to various functions.