Question

Prove that if \(f(x)=\sum_{k=0}^{\infty} c_{k} x^{k}\) converges on the interval \(I,\) then the power series for \(x^{m} f(x)\) also converges on \(I\) for positive integers \(m\)

Short answer

Question: Show that if the power series of a function \(f(x)\) converges on an interval \(I\), then the power series for \(x^m f(x)\) also converges on the same interval \(I\). Answer: By using the Ratio Test, we showed that if the power series of a function \(f(x)\) converges in an interval \(I\), the same happens for the power series of \(x^m f(x)\) since the limit of their ratio is less than 1.

Step-by-step solution

Re-write the series for \(x^m f(x)\)

Recall that the original power series for \(f(x)\) is given by \(f(x) = \sum_{k=0}^{\infty} c_{k} x^{k}\). Let's now multiply this series by \(x^m\). We obtain: \((x^m f(x)) = x^m \sum_{k=0}^{\infty} c_{k} x^{k} = \sum_{k=0}^{\infty} c_{k} x^{k+m}\).

Analyze the convergence of the new series

To prove the new power series converges on \(I\), we will use the Ratio Test. The Ratio Test states that a power series converges absolutely if the limit: \(\lim_{k\to\infty} \left| \frac{c_{k+1} x^{(k+1)+m}}{c_{k} x^{k+m}} \right| = L < 1\) for all \(x\) in the interval \(I\).

Simplify the limit

We can now simplify the limit in the Ratio Test: \(\lim_{k\to\infty} \left| \frac{c_{k+1} x^{(k+1)+m}}{c_{k} x^{k+m}} \right| = \lim_{k\to\infty} \left| \frac{c_{k+1} x^{k+m+1}}{c_{k} x^{k+m}} \right| = \lim_{k\to\infty} \left| \frac{c_{k+1}}{c_{k}} \right| \cdot \left| x \right|\).

Utilize the original convergence of \(f(x)\)

Since the original series \(f(x)\) converges in the interval \(I\), this implies that the ratio of the coefficients: \(\lim_{k\to\infty} \left| \frac{c_{k+1}}{c_{k}} \right| = \frac{1}{\rho}\) for some constant \(\rho > 0\).

Find the limit of the new series

Now, we plug this limit ratio between the coefficients of \(f(x)\) back into the limit for the expanded series: \(\lim_{k\to\infty} \left| \frac{c_{k+1}}{c_{k}} \right| \cdot \left| x \right| = \frac{1}{\rho} \left| x \right|\).

Determine the convergence of the new series

To guarantee the convergence of the new series for all \(x\in I\), the limit calculated previously must be less than 1. Therefore, we have the following inequality: \(\frac{1}{\rho} \left| x \right| < 1\). Since both \(x\) and \(\frac{1}{\rho}\) are positive, the inequality holds for any \(x \in I\). This means that the series for \(x^m f(x)\) converges in the same interval \(I\) on which the original series \(f(x)\) converges.

Key concepts

Ratio Test

The Ratio Test is a powerful tool used to determine the convergence of an infinite series. It involves examining the limit of the absolute value of the ratio of consecutive terms in a series. Specifically, if we consider a series \(\sum_{k=0}^{\infty} a_{k}\), the Ratio Test focuses on the following limit:

• \( \lim_{{k \to \infty}} \left| \frac{a_{k+1}}{a_{k}} \right| \).

If this limit \( L < 1 \), then the series converges absolutely. Absolute convergence means the series converges even when all terms are replaced with their absolute values. If \( L > 1 \) or the limit is infinite, the series diverges. If \( L = 1 \), the test is inconclusive, and one must use other methods to determine convergence.
This test is very helpful when dealing with power series because it can confirm the series' behavior within its interval of convergence.

Interval of Convergence

The interval of convergence is a critical concept related to power series. A power series around center \( a \) is typically represented as \( \sum_{k=0}^{\infty} c_{k} (x-a)^k \). The interval of convergence is the set of all values of \( x \) for which the series converges.

• For series convergence, testing points in its derived interval helps confirm where the series is valid.
• The interval can be finite or infinite, depending on the nature of the series.
• The endpoints of the interval should be tested separately since the radius of convergence provided by ratio test doesn’t always suffice to determine behavior at the extremes.

Finding the interval involves using tests like the Ratio Test to ensure the convergence of the power series within a radius defined from center \( a \). This radius, called the radius of convergence, is crucial because it tells how far the variable \( x \) can be taken from \( a \) while keeping the series convergent.

Absolute Convergence

Absolute convergence of a series \( \sum_{k=0}^{\infty} a_{k} \) means that the series \( \sum_{k=0}^{\infty} |a_{k}| \) also converges. It is a stronger form of convergence than ordinary convergence because if the series converges absolutely, it also converges normally.

• To examine absolute convergence, it involves checking if the series remains convergent when all terms are replaced by their absolute values.
• If a series converges absolutely, then rearranging the terms of the series does not affect its sum.

Absolute convergence implies robustness in the series behavior, which is why it’s frequently verified in instances where the position of terms might vary, such as when the series is multiplied by different factors or manipulated.

Multiplication of Series

Multiplying power series involves essentially multiplying each term of one series with every term of another. This operation results in a new power series, whose convergence depends on the convergence properties of the original series themselves.

• A well-behaved power series multiplied by a polynomial (like \( x^m \)) simply shifts its terms but doesn't change its convergence unless it's multiplied by a problematic series.
• Multiplication can modify the convergence pattern if the multiplier affects term ratios significantly, which is usually not the case with polynomials.
• Conceptually, each coefficient in the resulting series comes from the summation of products of coefficients from the multiplied series.

Thus, when multiplying power series, especially by simple expressions like \( x^m \), the main concern is whether convergence within the original interval holds. For controlled multiplications like this, convergence generally persists.