# Winter 2014 MATH 226: Limits and infinite seriesBranko Ćurgus

Friday, March 15, 2014

Wednesday, February 26, 2014

• In the note On two common sequences I prove that the sequences \begin{alignat*}{2} S_n &= \frac{1}{0!}+\frac{1}{1!} +\frac{1}{2!}+\cdots +\frac{1}{n!} = \sum_{k=0}^{n} \frac{1}{k!}, \ \ \ & & n \in {\mathbb N}, \\ P_n &= \left( 1 + \frac{1}{n}\right)^{\!n}, \ \ \ & & n \in {\mathbb N}, %\label{eqdss} \end{alignat*} converge to the same limit which is, by definition, the famous number $e$. As usual ${\mathbb N}$ denotes the set of positive integers.

Tuesday, February 25, 2014

• Updated list of the topics for the final exam. The topics covered on the coming exam are emphasized.

Monday, February 10, 2014

• Today in class I proved that the function $\ln$ is continuous on its domain. Here is a proof.
1. First we recall the inequality $1 - \frac{1}{v} \leq \ln v \leq v - 1 \qquad \text{is valid for all} \qquad v \gt 0,$ which we proved using the integral definition of $\ln$.
2. An inequality for $|\ln v|$ will be useful in the proof of the continuity below. Such an inequality can be obtained from the inequality in item 1 by considering two cases: \begin{align*} |\ln v| & \leq \left\{ \begin{array}{lll} v-1 & \text{if} & 1 \leq v \\ -\left( 1 - \frac{1}{v} \right) & \text{if} & 0 \lt v \lt 1 \end{array} \right\} \\ & = \left\{ \begin{array}{lll} v - 1 & \text{if} & 1 \leq v \6pt] -\frac{v - 1}{v} & \text{if} & 0 \lt v \lt 1 \end{array} \right\} \\ & = \left\{ \begin{array}{lll} |v - 1| & \text{if} & 1 \leq v \\[6pt] \frac{|v - 1|}{v} & \text{if} & 0 \lt v \lt 1 \end{array} \right\} \end{align*} Next we will restrict v to the interval (1/2,3/2). That is we assume v \in (1/2,3/2). Then we have that |v-1|/v \leq 2|v-1|. Since always |v-1| \leq 2|v-1|, we have that \[ |\ln v| \leq 2 |v-1| \qquad \text{is valid for all} \qquad v \in (1/2,3/2).
3. Let $a \gt 0$ be arbitrary. Let $x \in (a/2, 3a/2)$. Then $x/a \in (1/2, 3/2)$ and we can simplify the expression $| \ln x - \ln a |$ which appears in the definition of continuity. In the next sequence of inequalities we first use a property of logarithm, then the inequality in item 2 and simple algebra to get: \begin{align*} | \ln x - \ln a | & = | \ln \frac{x}{a} | \\ & \leq 2 \left| \frac{x}{a} - 1 \right| \\ & = 2 \left| \frac{x-a}{a} \right| \\ & = 2 \frac{|x-a|}{a} \\ & = \frac{2}{a} \, |x-a|. \end{align*} Thus, we proved that $\boxed{ | \ln x - \ln a | \leq \frac{2}{a} \, |x-a| \qquad \text{is valid for all} \qquad x \in (a/2,3a/2).}$
4. To finish the proof of continuity let $\epsilon \gt 0$ be arbitrary and set $\delta(\epsilon) = \min \left\{\frac{a\epsilon}{2}, \frac{a}{2} \right\}.$ Clearly $\delta(\epsilon) \gt 0$. Next we will prove the implication $|x-a| \lt \min \left\{\frac{a\epsilon}{2}, \frac{a}{2} \right\} \qquad \Rightarrow \qquad |\ln x - \ln a| \lt \epsilon.$ Assume $|x-a| \lt \min \left\{\frac{a\epsilon}{2}, \frac{a}{2} \right\}$. Then $|x-a| \lt \frac{a\epsilon}{2}$ and $|x-a| \lt \frac{a}{2}$. Since $|x-a| \lt \frac{a}{2}$, we have $x \in (a/2,3a/2)$ and therefore, by item 3, we have $| \ln x - \ln a | \leq \frac{2}{a} \, |x-a|.$ Since $|x-a| \lt \frac{a\epsilon}{2}$ we have $\frac{2}{a} \, |x-a| \lt \epsilon.$ The last two displayed inequalities yield $| \ln x - \ln a | \lt \epsilon.$
This completes the proof of the continuity of the logarithm function $\ln$.
• Today in class I also proved that the function $\exp$ is continuous on its domain. Here is a proof.
1. We first substitute $v = \exp u = e^u$ in the inequality in item 1 above: $1 - \frac{1}{e^u} \leq \ln e^u \leq e^u - 1 \qquad \text{is valid for all} \qquad u \in \mathbb R.$ Simplifying we get $1 - \frac{1}{e^u} \leq u \leq e^u - 1.$ We need a squeeze for $e^u$. Above we already have one side of the squeeze. That is $u+1 \leq e^u$. To get the other side we transform $1 - \frac{1}{e^u} \leq u$ to $1 - u \leq \frac{1}{e^u}.$ To get a useful inequality we need to take the reciprocals in the last inequality. For that we need $1 - u \gt 0$, that is we need to assume that $u \lt 1$. Assuming that $u \lt 1$ we have $e^u \leq \frac{1}{1 - u} .$ Together with $u+1 \leq e^u$, we have that $u+1 \leq e^u \leq \frac{1}{1 - u} \qquad \text{is valid for all} \qquad u \lt 1.$
2. An inequality for $|e^u - 1|$ will be useful in the proof of the continuity below. The last inequality in item I yields that $u \leq e^u - 1 \leq \frac{u}{1 - u} \qquad \text{is valid for all} \qquad u \lt 1.$ To get an inequality for $|e^u - 1|$ we consider two cases: \begin{align*} |e^u - 1| & \leq \left\{ \begin{array}{lll} \frac{u}{1 - u} & \text{if} & 0 \leq u \lt 1 \\ -u & \text{if} & u \lt 0 \end{array} \right\} \\ & = \left\{ \begin{array}{lll} \frac{|u|}{1 - u} & \text{if} & 0 \leq u \lt 1 \\ |u| & \text{if} & u \lt 0 \end{array} \right\} \end{align*} Next we will restrict $u$ to the interval $(-1/2,1/2)$. That is we assume $u \in (-1/2,1/2)$. Then we have that $|u|/(1-u) \leq 2|u|$. Since always $|u| \leq 2|u|$, we have that $|e^u - 1| \leq 2 |u| \qquad \text{is valid for all} \qquad u \in (-1/2,1/2).$
3. Let $a \gt 0$ be arbitrary. Let $x \in (a-1/2, a+1/2)$. Then $x-a \in (-1/2, 1/2)$ and we can simplify the expression $| e^x - e^a |$ which appears in the definition of continuity. Next we use a property of the exponential function and the last inequality in item II to get: $| e^x - e^a | = e^a | e^{(x-a)} - 1 | \leq 2 e^a |x-a| .$ Thus, we proved that $\boxed{ | e^x - e^a | \leq 2 e^a |x-a| \qquad \text{is valid for all} \qquad x \in (a-1/2,a+1/2). }$
4. To finish the proof of the continuity let $\epsilon \gt 0$ be arbitrary and set $\delta(\epsilon) = \min \left\{\frac{\epsilon}{2e^a}, \frac{1}{2} \right\}.$ Clearly $\delta(\epsilon) \gt 0$. Next we will prove the implication $|x-a| \lt \min \left\{\frac{\epsilon}{2e^a}, \frac{1}{2} \right\} \qquad \Rightarrow \qquad |e^x - e^a| \lt \epsilon.$ Assume $|x-a| \lt \min \left\{\frac{\epsilon}{2e^a}, \frac{1}{2} \right\}$. Then $|x-a| \lt \frac{\epsilon}{2e^a}$ and $|x-a| \lt \frac{1}{2}$. Since $|x-a| \lt \frac{1}{2}$, we have $x \in (a-1/2,a+1/2)$ and therefore, by item III, we have $| e^x - e^a | \leq 2 e^a |x-a|.$ Since $|x-a| \lt \frac{\epsilon}{2e^a}$ we have $2 e^a |x-a| \lt \epsilon.$ The last two displayed inequalities yield $| e^x - e^a | \lt \epsilon.$
This completes the proof of the continuity of of the exponential function $\exp$.

Thursday, January 30, 2014

• Today in class I demonstrated how to use Mathematica to illustrate what we proved formally. I used Mathematica version 5.2. Here is the small notebook that I created 20140130.nb. The file is called 20140130.nb. Right-click on the underlined phrase "20140130.nb"; in the pop-up menu that appears, your browser will offer you to save the file in your directory. Make sure that you save it with the exactly same name. After saving the file you can open it with Mathematica version 5.2. You need to find a campus computer with Mathematica v 5.2 installed on it (for example BH 209, BH 215 and many more campus computers). You will find Mathematica 5.2 as follows (this sequence might differ on different campus computers)
Start -> Programs -> Mathematica 5.2 -> Mathematica.
Open Mathematica first, then open the specific file that you want to read in Mathematica. You can execute the entire file by the following manu sequence (in Mathematica):
Kernel -> Evaluation -> Evaluate Notebook.
• More information on how to use Mathematica you can find on my Mathematica page.
• If you have problems running files that I posted please let me know. If you spend some time learning how to use Mathematica you will enhance your understanding of math that you are studying.
• We also have Mathematica 8. It is only available in BH 215. Since Mathematica 5.2 is more widely available I decided to use it in this class. These two versions are not compatible. However, the command structure is very similar. Version 8 will usually recognize the differences and correct them.
• Updated list of the topics for the final exam.

Friday, January 24, 2014

Thursday, January 16, 2014

• In this pdf file I summarize steps involved in limit proofs for $\displaystyle \lim_{x\to +\infty} f(x) = L$.

Monday, January 13, 2014

• Today we discussed the concept of a constant function. We gave two definitions. In the items below $A$ and $B$ are nonempty sets.
• Here is the first definition.
• A function $f: A \to B$ is a constant function if there exists $c \in B$ such that $f(x) = c$ for all $x \in A.$
• This definition written using logical symbols: $\exists \, c \in B \ \ \forall \, x \in A \ \ f(x) = c$.
• It is a good practice to formulate a negation of a mathematical statement. The negation of the definition of a constant function in English is: A function $f: A \to B$ is not a constant function if for every $c \in B$ there exists $x \in A$ such that $f(x) \neq c$.
• The negation written using logical symbols: $\forall \, c \in B \ \ \exists \, x \in A \ \ f(x) \neq c$.
• It is interesting to point out that the definition of a constant function on Wikipedia is different. Here is Wikipedia's definition.
• A function $f: A \to B$ is a constant function if for all $x, y \in A$ we have $f(x) = f(y).$
• This definition written using logical symbols: $\forall \, x \in A \ \ \forall \, y \in A \ \ f(x) = f(y)$.
• The negation of the Wikipedia definition is: A function $f: A \to B$ is not a constant function if there exist $x$ and $y$ in $A$ such that $f(x)\neq f(y)$.
• The negation written using logical symbols: $\exists \, x \in A \ \ \exists \, y \in A \ \ f(x) \neq f(y)$.
• One can prove that two definitions are equivalent.
• As one more exercise in using logical symbols I will write the definition of the range of a function $f: A \to B$ using logical symbols. The range of $f$ is the following set $\bigl\{ y \in B \, : \ \exists \, x \in A \ \ y = f(x) \bigr\}.$
• I point out that in class I gave a detailed proof of the following equivalence. Let $a \in \mathbb R$ and $r \gt 0$. Then $x \in (a-r,a+r) \quad \Leftrightarrow \quad |x-a| \lt r.$

Friday, January 10, 2014

• I started the list of the topics for the final exam. I will update this list as we make progress through the material. So far we have covered the first item.

Tuesday, January 7, 2014