Question

Suppose that \(f(x)=\sum a_{n}\left(x-x_{0}\right)^{n}\) has radius of convergence \(R\) and \(0

Short answer

To show that the following inequality holds for all x with \(|x - x_0| \leq r\) and k ≥ k: $$ \left| f(x) - \sum_{n = 0}^k a_n (x - x_0)^n \right| \le \left( \frac{r}{R_1} \right)^{k+1} \frac{R_1}{R_1 - r} $$ One must follow the following steps: 1. Define the partial sums of the series as \(S_k(x) = \sum_{n=0}^k a_n (x-x_0)^n\). 2. Rewrite the inequality in terms of the remainder of the series \(R_k(x) = \sum_{n=k+1}^\infty a_n (x - x_0)^n\). 3. Use properties of radius of convergence to find that \(\tilde{R} = \frac{R_1}{r}R\). 4. Establish bounds on the remainder using the new radius of convergence. 5. Bound the expression using the geometric series formula.

Step-by-step solution

Define partial sums of the series

Let's define the partial sums of the series as \(S_k(x) = \sum_{n=0}^k a_n (x-x_0)^n.\) Therefore, the given inequality can be written as: $$ \left|f(x) - S_k(x)\right| \le \left(\frac{r}{R_1}\right)^{k+1} \frac{R_1}{R_1 - r}, $$ which needs to be proved for all \(x\) with \(|x - x_0| \le r\) and \(k \ge k\).

Use the remainder of the series

Since the series converges to \(f(x)\) within its radius of convergence, we can write the function as \(f(x) = S_k(x) + R_k(x)\), where \(R_k(x) = \sum_{n=k+1}^\infty a_n (x - x_0)^n\) is the remainder after the \(k\)th term. Hence, the inequality can be rewritten as: $$ \left|R_k(x)\right| \le \left(\frac{r}{R_1}\right)^{k+1} \frac{R_1}{R_1 - r} $$

Use properties of radius of convergence

For power series \(\sum a_n (x - x_0)^n\) converging to \(f(x)\) within its radius of convergence \(R\), we know that \(R^{-1} = \limsup_{n \to \infty} |a_n|^{1/n}\). Likewise, for power series in the form \(\sum a_n (x - x_0)^n \left(\frac{r}{R_1}\right)^n\), its radius of convergence is \(\tilde{R} = \frac{R_1}{r}R\), which comes from \(\tilde{R}^{-1} = \limsup_{n \to \infty} \left|a_n \left(\frac{r}{R_1}\right)^n\right|^{1/n}\).

Establish bounds on the remainder using the new radius of convergence

Since \(0 < r < R_1 < R\), we have \(\tilde{R} > R\). This means the power series with radius of convergence \(\tilde{R}\) converges absolutely when \(x\) is in the range \(|x - x_0| \le r\). Thus, for all \(x\) within this range, we can apply the Cauchy Hadamard theorem to bound the remainder: $$ |R_k(x)| \le \left| \sum_{n=k+1}^\infty a_n (x - x_0)^n \left(\frac{r}{R_1}\right)^n \right| \le \sum_{n=k+1}^\infty \left| a_n (x - x_0)^n \left(\frac{r}{R_1}\right)^n \right|. $$ Now, since \(\left| a_n (x - x_0)^n \left(\frac{r}{R_1}\right)^n \right| \le |a_n| \cdot r^n \cdot \left( \frac{r}{R_1} \right)^n\): $$ |R_k(x)| \le \sum_{n=k+1}^\infty |a_n| r^n \left(\frac{r}{R_1}\right)^n = \sum_{n=k+1}^\infty |a_n| \left(\frac{r}{R_1}\right)^{n+1} R_1^n, $$ where the last term follows by the fact that \(\frac{r}{R_1} < 1\).

Bound using geometric series

The expression in Step 4 is a geometric series with the ratio \(\frac{r}{R_1}\). Thus, we can bound it using the sum of geometric series formula \(S = \frac{a}{1-q}\): $$ |R_k(x)| \le \frac{|a_{k+1}| \left(\frac{r}{R_1}\right)^{k+1} R_1^{k+1}}{1 - \frac{r}{R_1}} = \left(\frac{r}{R_1}\right)^{k+1} \frac{R_1}{R_1 - r}, $$ which is what was required to prove.

Key concepts

Radius of Convergence

A power series, such as \(f(x) = \sum a_n (x-x_0)^n\), only converges for certain values of \(x\). The range of \(x\) for which the series converges is determined by the radius of convergence \(R\). This radius marks a boundary within which the series is guaranteed to converge.

To find the radius of convergence, we use the formula: \[ R^{-1} = \limsup_{n \to \infty} |a_n|^{1/n} \].

This means that we examine the sequence of coefficients \(a_n\) as \(n\) becomes very large. The radius of convergence tells us how far out from \(x_0\) we can go before the series stops converging.

In simpler terms:

• The smaller the coefficients \(|a_n|\), typically, the larger the radius of convergence.
• If you exceed the radius, the series may diverge, meaning it doesn't settle down to a specific value.

In our problem, this concept helps us understand limits on \(x\) for which our original series \(\sum a_n (x-x_0)^n\) will converge so that our manipulations remain valid.

Partial Sums

Partial sums are when we take only a finite number of terms from a series to form an approximation. For a power series \(f(x) = \sum a_n (x-x_0)^n\), a partial sum through the \(k\)th term is denoted as \(S_k(x) = \sum_{n=0}^k a_n (x-x_0)^n\).

The sum \(S_k(x)\) represents an approximation of \(f(x)\) using just the first \(k+1\) terms.

Why use one? Because calculating an infinite sum is practically impossible, partial sums give us a tangible way to estimate the value of the series. The more terms you include, the closer \(S_k(x)\) gets to \(f(x)\) inside the radius of convergence.

In the exercise, partial sums allow us to see how well the series approximates \(f(x)\) and how the error—called the remainder \(R_k(x)\)—behaves.

• \(R_k(x)\), the remainder, is the difference between \(f(x)\) and \(S_k(x)\).
• The smaller the remainder, the closer the partial sum is to the actual value of the series at a given \(x\).

Partial sums thus provide a practical method to work with infinite series like power series by only focusing on a manageable part at a time.

Cauchy-Hadamard Theorem

The Cauchy-Hadamard Theorem is a powerful tool in understanding the behavior of a power series. It directly relates to finding the radius of convergence \(R\) of a series \(\sum a_n (x-x_0)^n\).

This theorem states: \(R^{-1} = \limsup_{n \to \infty} |a_n|^{1/n}\), but in our instance, we consider a modified series like \(\sum a_n (x-x_0)^n \left(\frac{r}{R_1}\right)^n\). The modified version helps in analyzing convergence within new parameters by introducing this factor of \(\left(\frac{r}{R_1}\right)^n\).

Here's what it implies practically:

• It provides a method to compute or estimate \(R\), using the "lim sup" concept, which determines the least upper bound of the sequence elements as \(n\) grows.
• This theorem allows us to evaluate convergence for more complex scenarios, especially when dealing with scaled versions of the series.

By applying this theorem, we confirm if certain conditions (like \(|x-x_0| \le r\) and \(R_1
In layman's terms, it gives us a mathematical magnifying glass to verify if and how the series converges under specified changes or scales.