Problem 64 - Thomas Calculus

Question

Proof of Theorem 21 Assume that \(a=0\) in Theorem 21 and that \(f(x)=\sum_{n=0}^{\infty} c_{n} x^{n}\) converges for \(- R < x < R .\) Let \(g(x)=\sum_{n=1}^{\infty} n c_{n} x^{n-1} .\) This exercise will prove that \(f^{\prime}(x)=g(x)\) that is, \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=g(x)\) $$ \begin{array}{c}{\text { a. Use the Ratio Test to show that } g(x) \text { converges for }} \\ {-R < x < R .} \\ {\text { b. Use the Mean Value Theorem to show that }} \\ {\frac{(x+h)^{n}-x^{n}}{h}=n c_{n}^{n-1}} \\ {\text { for some } c_{n} \text { between } x \text { and } x+h \text { for } n=1,2,3, \ldots}\\\\{\text { c. Show that }} \\ {\quad g(x)-\frac{f(x+h)-f(x)}{h}|=| \sum_{n=2}^{\infty} n a_{n}\left(x^{n-1}-c_{n}^{n-1}\right) |} \\ {\text { d. Use the Mean Value Theorem to show that }} \\\ {\frac{x^{n-1}-c_{n}^{n-1}}{x-c_{n}}=(n-1) d_{n-1}^{n-2}} \\ {\text { for some } d_{n-1} \text { between } x \text { and } c_{n} \text { for } n=2,3,4, \ldots}\\\\{\text { e. Explain why }\left|x-c_{n}\right| < h \text { and why }} \\ {\left|d_{n-1}\right| \leq \alpha=\max \\{|x|,|x+h|\\}} \\ {\text { f. Show that }} \\ {\left|g(x)-\frac{f(x+h)-f(x)}{h}\right| \leq|h| \sum_{n=2}^{\infty}\left|n(n-1) a_{n} \alpha^{n-2}\right|}\\\\{\text { g. Show that } \sum_{n=2}^{\infty} n(n-1) \alpha^{n-2} \text { converges for }- R < x < R \text { . }} \\ {\text { h. Let } h \rightarrow 0 \text { in part (f) to conclude that }} \\ {\quad \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=g(x)}\end{array} $$

Step-by-Step Solution

Answer

The proof shows that \(f'(x) = g(x)\) by demonstrating convergence and bounding derivative limits using series expansion and the Mean Value Theorem.

Step 1: Verify Convergence of g(x) Using Ratio Test

First, apply the Ratio Test to the series \(g(x) = \sum_{n=1}^{\infty} n c_{n} x^{n-1}\). Calculate the ratio of terms: \(\frac{a_{n+1}}{a_n} = \left|\frac{(n+1) c_{n+1} x^n}{n c_n x^{n-1}}\right| = \left|\frac{(n+1) c_{n+1} x}{n c_n}\right|\). Take the limit as \(n\) approaches infinity. If the limit is \(|x|/R < 1\), \(g(x)\) converges for \(-R < x < R\).

Step 2: Use Mean Value Theorem for Polynomial Differences

Use the Mean Value Theorem on \( (x+h)^n - x^n \) implying \((x+h)^n - x^n = n c_n^{n-1} h\) for some \(c_n\) between \(x\) and \(x+h\), as it concerns the derivative of \(x^n\).

Step 3: Show Expression for g(x) and f'(x) Using Series

Demonstrate that \(g(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\) by showing how summation differences lead to the same limit: \(\left| g(x) - \frac{f(x+h) - f(x)}{h} \right| = \left| \sum_{n=2}^{\infty} n c_n (x^{n-1} - c_n^{n-1}) \right| \).

Step 4: Mean Value Application on Series Terms

Use the Mean Value Theorem in \(\frac{x^{n-1} - c_n^{n-1}}{x - c_n} = (n-1) d_{n-1}^{n-2}\) where \(d_{n-1}\) is between \(x\) and \(c_n\); apply to each term in the series.

Step 5: Analyze Bounds for h and Component Magnitudes

Explain the inequality \(\left|x - c_n\right| < h\) and establish that \(|d_{n-1}| \leq \alpha\), where \(\alpha = \max\{|x|, |x+h|\}\). This helps to bound the terms in the series.

Step 6: Establish Bound for the Difference Using Series

Show that \(\left|g(x) - \frac{f(x+h)-f(x)}{h}\right| \leq |h| \sum_{n=2}^{\infty} |n(n-1) c_n \alpha^{n-2}|\). This states the difference's growth is limited by a series expression.

Step 7: Prove Series Convergence for Bounds

Confirm that the series \(\sum_{n=2}^{\infty} n(n-1) \alpha^{n-2}\) converges within \(-R < x < R\) using properties of geometric-like series combined with the behavior of \(\alpha\).

Step 8: Conclude with Inferential Limit and Final Equality

As \(h \rightarrow 0\), the bound tends to 0, and \(\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} = g(x)\), formally proving \(f'(x) = g(x)\).

Key Concepts

Ratio Test

The Ratio Test is a powerful tool used to determine the convergence of infinite series. It's especially useful when dealing with power series, like those involving terms of the form \( g(x) =\sum_{n=1}^{\infty} n c_{n} x^{n-1} \). The test involves finding the limit of the absolute value of the ratio between successive terms in the series.
For the series \( g(x) \), we calculate the ratio as follows:

• Consider the terms \( a_n = n c_n x^{n-1} \) and \( a_{n+1} = (n+1) c_{n+1} x^n \).
• The ratio \( \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(n+1)c_{n+1}x}{nc_n} \right| \).

By taking the limit as \( n \) approaches infinity, \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \), if this limit is less than 1, the series \( g(x) \) converges for that value of \( x \). In our context, it leads to \( \frac{|x|}{R} < 1 \), where \( R \) is the radius of convergence for the series.

Mean Value Theorem

The Mean Value Theorem (MVT) is a fundamental concept in calculus that relates the derivative of a function to its values at two points. For polynomials, it implies that the instantaneous rate of change (the derivative) is represented by the average rate of change across an interval.
In this exercise, we apply MVT to show:

• \( \frac{(x+h)^n - x^n}{h} = n c_n^{n-1} \) for some \( c_n \) between \( x \) and \( x+h \).

This equation stems from the power rule for differentiation applied to \( x^n \), as evaluated using the MVT:

• Focus on the polynomial \( x^n \), ensuring differentiation aligns with MVT outcomes.
• This is significant in approximations, showing changes in \( f(x) \) can be captured by differences in \( h \).

The MVT helps establish the core relationship between \( f(x) \) and its derivative \( g(x) \), thus bridging the gap between theoretical and computational methods of differentiation.

Uniform Convergence

Uniform convergence is a type of convergence for sequences of functions where the speed of convergence must be consistent across the domain. It ensures that functions converge not just pointwise but do so uniformly, maintaining bounds.
This is particularly helpful when dealing with series of functions like power series:

• It influences how assuredly we can interchange limits and other operations like differentiation or integration.
• In the current exercise, uniform convergence helps show that the inequality \( \left| g(x) - \frac{f(x+h) - f(x)}{h} \right| \) tends towards zero uniformly as \( h \to 0 \).

The concept is crucial when we want seamless behavior of function sequences, ensuring analyses hold true across entire intervals rather than merely at arbitrary points.

Series Convergence

Series convergence is concerned with whether an infinite sum of terms leads to a finite result or not. To find if a series converges involves examining how the partial sums behave as they include more terms.
In the exercise, establishing series convergence is key in confirming the behavior of \( g(x) \) in relation to its differentiability:

• The convergence test \( \sum_{n=2}^{\infty} n(n-1)\alpha^{n-2} \) validates the convergent behavior within \(-R < x < R\).
• Tools like the Ratio Test and understanding of power series help clarify where these series have finite sums.

A properly converging series implies that the theoretical justifications align with the expected computations. This fundamentally helps in verifying our derivations, such as \( g(x) = f'(x) \).