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Chapter 6: Problem 4

Let \(f_{n}(s)=s e^{-n s}, n \geq 1\). Show that \(f_{n} \rightarrow 0\) uniformly on \([0, \infty)\).

Short Answer

Expert verified
To show that the sequence of functions $f_n(s) = se^{-ns}$ converges uniformly to 0 on $[0, \infty)$, we found the maximum value of $|f_n(s)|$ for all $n \geq 1$ and $s \geq 0$. This maximum value occurs at $s = \frac{1}{n}$, and we found the maximum value function to be $M_n = \left|\frac{1}{n}e^{-1}\right|$. We showed that $M_n \rightarrow 0$ as $n \rightarrow \infty$, proving that $f_n$ converges uniformly to 0 on $[0, \infty)$.

Step by step solution

01

Find the derivative of \(f_n(s)\)

First, let's find the derivative of \(f_n(s)\) with respect to \(s\). Using the product rule, we have: $$\frac{d f_n(s)}{ds} = \frac{d(se^{-ns})}{ds} = e^{-ns} + s(-ne^{-ns}) = e^{-ns}(1 - ns).$$
02

Find the critical points of \(f_n(s)\)

Now let's find the critical points of \(f_n(s)\), which are the points where the first derivative is equal to zero: $$\frac{d f_n(s)}{ds} = e^{-ns}(1 - ns) = 0.$$ Since \(e^{-ns}\) is always positive, this equation is satisfied only when $$1 - ns = 0$$ or $$s = \frac{1}{n}.$$
03

Determine if it's a maximum or minimum

We can now determine if \(s = \frac{1}{n}\) is a maximum or minimum value of \(f_n(s)\). Let's find the second derivative of \(f_n(s)\): $$\frac{d^2 f_n(s)}{ds^2} = \frac{d(e^{-ns}(1 - ns))}{ds} = e^{-ns}(-n)+e^{-ns}(-n)+e^{-ns}(n^2s) = -n^2e^{-ns}(s-1).$$ Since the second derivative is continuous, we know that at \(s = \frac{1}{n}\): $$\frac{d^2 f_n(s)}{ds^2} = -n^2e^{-1}(1-\frac{1}{n}) < 0,$$ which implies that \(s = \frac{1}{n}\) is a maximum point of \(f_n(s)\).
04

Find the maximum value of \(|f_n(s)|\)

Now, we need to find the maximum value of \(|f_n(s)|\). Since \(f_n(0)=0\), \(f_n(s)\) has a maximum absolute value at \(s = \frac{1}{n}\) for all \(n\geq1\). So, the maximum value function will be: $$M_n= |f_n\left(\frac{1}{n}\right)| = \left|\frac{1}{n}e^{-1}\right|. $$
05

Prove that \(f_n \rightarrow 0\) uniformly

Now we have to show that \(M_n \rightarrow 0\) as \(n \rightarrow \infty\). This is straightforward since for \(n\geq1\), we have: $$M_n = \left|\frac{1}{n}e^{-1}\right| \rightarrow 0$$ as \(n \rightarrow \infty\). Therefore, by definition, \(f_n \rightarrow 0\) uniformly on \([0, \infty)\).

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Key Concepts

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

Understanding Sequence of Functions
A sequence of functions is a collection of functions, where each function is associated with a unique number from a sequence, typically integers. For example, if we have the sequence \( f_1, f_2, f_3, ... \), the first function in the sequence is \( f_1 \) and so on. The concept becomes especially interesting when we look at the behavior of these functions as the sequence progresses. This behavior may converge, which means the functions become arbitrarily close to some well-defined function in terms of a certain metric as the sequence moves forward.

Uniform convergence is a strong form of convergence for sequences of functions. It occurs when the functions in the sequence converge to a function uniformly for all points in a certain interval. This means the difference between the functions in the sequence and the limiting function becomes small at the same rate across the entire interval.
Utilizing Critical Point Analysis
Critical point analysis is a calculative approach used to find points on a graph that may correspond to maximum or minimum values. A critical point of a function is where its derivative either is zero or does not exist. By setting the derivative equal to zero, we are effectively searching for points where the function's rate of change switches direction鈥攁 likely indicator of a peak or trough in the function's graph.

For example, when analyzing the critical points of the function \( f_n(s) = se^{-ns} \), we set its derivative to zero, leading us to find that \( s = 1/n \) is a critical point. By further examining the second derivative at this point, we can determine if it's a local maximum or minimum, which provides valuable insights into the behavior of the function.
Determining the Maximum Value of a Function
Finding the maximum value of a function is a vital application in calculus, particularly when identifying optimal solutions. After identifying critical points, we want to know whether these points correspond to the highest value the function achieves. This involves evaluating the function at critical points, endpoints, or looking at the behavior as variables approach infinity.

In our example with the function \( f_n(s) \), after finding the critical point of \( s = 1/n \) and establishing it as a local maximum, we evaluated \( f_n(s) \) at this point, \( M_n = |f_n(1/n)| \) to find the maximum value of the function. For functions that are defined over an interval, finding maximum values is crucial, as it helps in understanding the limits of the function and assessing convergence.
Applying the Product Rule for Differentiation
The product rule is an essential differentiation technique used when taking the derivative of the product of two functions. It states that \( (fg)' = f'g + fg' \), where \( f \) and \( g \) are functions of the same variable. This rule is pivotal in finding derivatives efficiently without expanding the product, which can be cumbersome or impractical for certain functions.

When we applied the product rule to the function \( f_n(s) \) in our example, it was the key step in finding its derivative: \( \frac{d f_n(s)}{ds} = e^{-ns}(1 - ns) \). This derivative then led us to analyze critical points and ultimately assess the function鈥檚 uniform convergence, demonstrating the interconnectedness of these fundamental calculus concepts in solving complex mathematical problems.

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Most popular questions from this chapter

This question investigates the Binomial series. (i) Show that \(\sum_{n=0}^{\infty}\left(\begin{array}{c}\alpha \\\ n\end{array}\right) s^{n}\) converges for \(|s|<1 .\) Recall that \(\alpha \in \mathbb{R}\) and $$ \left(\begin{array}{c} \alpha \\ n \end{array}\right)=\frac{\alpha(\alpha-1) \ldots(\alpha-n+1)}{n !} $$ (ii) If \(f(s)=\sum_{n=0}^{\infty}\left(\begin{array}{c}\alpha \\\ n\end{array}\right) s^{n}\), show that \((1+s) f^{\prime}(s)=\alpha f(s)\) where \(|s|<1\) Show that any function satisfying (ii) must be of the form \(f(s)=\) \(c(1+s)^{\alpha}\) for some constant \(c \in \mathbb{R} .\) Use this fact to establish that $$ (1+s)^{\alpha}=\sum_{n=0}^{\infty}\left(\begin{array}{c} \alpha \\ n \end{array}\right) s^{n} \quad \text { for }|s|<1 $$ Here the equality means that the series on the right is uniformly convergent to the function on the left of the equality.

Let \(A\) be a compact subset of a normed linear space \(X\). Let \(x\) be a point in \(X\) and \(\lambda\) a real number. Show that the sets \(A_{1}=\\{x+a: a \in A\\}\) and \(A_{2}=\\{\lambda a: a \in A\\}\) are both compact.

Suppose the power series \(\sum_{n=0}^{\infty} a_{n} s^{n}\) is convergent for \(|s|

Let \(f: X \rightarrow Y\) be such that \(f^{-1}(\\{y\\})\) is compact for each \(y \in Y\). Let \(A_{n}\) be a decreasing sequence of closed sets in \(X\) with \(f\left(A_{n}\right)=Y\). Show that \(f\left(\cap_{1}^{\infty} A_{n}\right)=Y\). [Hint: Choose \(y \in Y\) and consider the sets \(A_{n} \cap f^{-1}(\\{y\\})\).]

Are the following possible? (i) A continuous mapping of \([0,1]\) onto \(\mathbb{R}\). (ii) A uniformly continuous mapping of \((0,1)\) onto \(\mathbb{R}\).
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