Solutions to MA387 WPRs

# WPR I

Problem 1. Prove that $\sqrt{3}$ is irrational.

See textbook, page 11.

Problem 2. Show that $(|\vec a|-|\vec b|)(|\vec a|+|\vec b|)=\langle \vec a-\vec b,\vec a+\vec b\rangle$.

Recall that $|\vec a|^2=\langle \vec a,\vec a\rangle$. Expanding the left side gives $|\vec a|^2-|\vec b|^2$. Expanding the right side gives $\langle\vec a,\vec a\rangle + \langle\vec a,\vec b\rangle - \langle\vec b,\vec a\rangle - \langle\vec b,\vec b\rangle = \langle\vec a,\vec a\rangle - \langle\vec b,\vec b\rangle = |\vec a|^2-|\vec b|^2$, since inner products are symmetric. $\blacksquare$

Problem 3. For the discontinuous function shown, find an x and an $\epsilon>0$ such that for all $\delta>0$ there exists some x' such that $|x-x'|<\delta$ and $|f(x)-f(x')|>\epsilon$.

The existence of x and $\epsilon$ is guaranteed by the definition of continuity (negated). Let $x=1$ and $\epsilon=1$. Then $f(x)=-1$, and if $\delta=1$ then a value $x'\in(1-\delta,1)$ has a function value greater than +1, i.e., $f(x')>1$. Hence, $|f(x)-f(x')| = 2 > \epsilon$. $\blacksquare$

Problem 4. What is the difference between the "maximum" and the "Least Upper Bound" of a subset of real numbers?

The maximum may not always exist, while the LUB is guaranteed to exist (by completeness of the reals). For example, if $S=(0,1)$, then S has no maximum, whereas $LUB(S)=1$.

Problem 5. Given that $\sqrt{5}$ corresponds to the cut $A|B$, write out definitions of A and B.

The key is to write out the definitions without referencing any nonrationals. Hence, we write $A=\{q\in\mathbb{Q}:q^2<5\}$ and $B=\{q\in\mathbb{Q}:q^2\geq 5\}$.

Problem 6. Let $S\subset\mathbb{R}$ be bounded, nonempty, and let $b=LUB(S)$. Prove that for each $\epsilon>0$ there exists $s\in S$ with $b-\epsilon\leq s\leq b$.

We will prove this by assuming that there is no such s and contradicting the fact that b is the least upper bound of S.

Assume, by way of contradiction, that no such $s\in[b-\epsilon,b]$ exists. Then $b-\epsilon>t$ for all $t\in S$, so that $b-\epsilon$ is also an upper bound. Since $b-\epsilon<b$, the value b cannot be the least upper bound. This contradiction establishes the proof. $\blacksquare$

Problem 7. Given cuts $A|B$ and $C|D$, (i) what does trichotomy imply about A and C, (ii) prove trichotomy by demonstrating what you stated in (i).

Trichotomy implies that either $A\subsetneq C$, $A\supsetneq C$, or $A=C$.

Suppose that $A\neq C$. Then either there exists $a\in A$ which is not in C or there exists $c\in C$ which is not in A. Without loss of generality assume the first. By definition of a cut (see page 11 in the text), that means $a\in D$, and consequently $a>c$ for all $c\in C$. But since if $a>a'$ then $a'\in A$ also, that implies that every $c\in C$ is also contained in A, hence $C\subsetneq A$.
Likewise, if some $c\in C$ is not in A, then $A\subsetneq C$. $\blacksquare$

# WPR II

Problem 1. Explain why, at any given time, there is a highest temperature on the surface of the earth. (List all assumptions.)

Assume that the temperature on the earth is a continuous function, whose domain is the surface of the earth. Assume also that the surface of the earth may be taken to be a subset of $\mathbb{R}^3$. As such, by the Heine-Borel Theorem it is compact, since it is closed and bounded. With these assumptions, the Extreme Value Theorem implies that the temperature function has a maximum, since it is a continuous function on a compact domain.

Problem 2. The middle half Cantor set may be written $C=C_1\cap C_2\cap \cdots$, where $C_1=[0,1]$, $C_2=[0,\frac14]\cup[\frac34,1]$, $C_3=[0,\frac1{16}]\cup[\frac3{16},\frac14]\cup[\frac34,\frac{13}{16}]\cup[\frac{15}{16},1]$, etc. Prove that C is compact.

Recall that (i) the finite union of closed sets is closed, and (ii) the infinite intersection of closed sets is closed. By the first, each $C_i$ is closed. By the second, the intersection of the $C_i$ is closed. Hence, C is closed, and it is clearly bounded, proving that it is compact (by the Heine-Borel Theorem). $\blacksquare$

Problem 3. Prove, directly from the definition of an open set, that the union of two open sets is open.

Let A and B be open, and let $p\in A\cup B$. We want to show that there exists an open neighborhood of p which is contained in $A\cup B$. Note that either $p\in A$ or $p\in B$ (or both). Without loss of generality, suppose that $p\in A$. Then, since A is open, p has an open neighborhood N contained in A, that is $p\in N\subset A$. But then $N\subset A\cup B$ as well, verifying that p has an open neighborhood contained in the union $A\cup B$. Since p was chosen arbitrarily, this completes the proof. $\blacksquare$

Problem 4. Let S consist of concentric circles of radius $1+\frac1n$ for natural numbers $n\geq 1$, as shown. (i) Describe the limit points of S which are not contained in S. (ii) Is S open, closed, or neither? (iii) Is S compact? (iv) What are the interior, closure, and boundary of S?

(i) The limit points not contained in S are precisely the unit circle.
(ii) S is not open, since no point contains a neighborhood within S (consequently, every point of S is a boundary point). It is not closed, since it does not contain all of its limit points. Hence it is neither.
(iii) Since S is not closed, it is not compact.
(iv) The interior of S is empty; the boundary of S is all of S; the closure of S is S together with the standard unit circle.

Problem 5. Let $f(x)=\cos(x)$, and let $(x_n)$ be the sequence defined by setting $x_n=f(n)$ for $n\geq1$. Also, let $(y_n)$ be the sequence $y_n=f(2\pi n)$. (i) Which sequence converges? (ii) Do both sequences have convergent subsequences? Why?

(i) The sequence $(y_n)$ converges, since $\cos(2\pi n)=1$ for all $n\in\mathbb{N}$. The sequence $(x_n)$ does not converge, since the terms $\cos(1),\cos(2),\ldots$ do not converge.
(ii) Both sequences have convergent subsequences, since they are both bounded. Indeed, every bounded sequence in $\mathbb{R}$ has a convergent subsequence… this is guaranteed by compactness.

Problem 6. Define $x_n=\frac{(-1)^n}{3^n}$. Prove that $(x_n)$ converges, using the $\epsilon$ definition of convergence.

Note first that the limit, provided it exists, would be 0.

Let $\epsilon>0$. We need to find some $N\in\mathbb{N}$ such that for $n>N$, $|x_n-0|<\epsilon$. We can find the appropriate N by finding the value at which $|x_p|=3^{-p}=\epsilon$. Taking the log of both sides of this equation, we have $\log(3^{-p})=\log(\epsilon) \Rightarrow -p \log(3) = \log(\epsilon) \Rightarrow p = - \frac{log(\epsilon)}{\log(3)}$. Note that if $\epsilon>0$ is small, this value will be positive. Since we're looking for $N\in\mathbb{N}$, we choose N to be the smallest positive integer greater than $-\frac{\log(\epsilon)}{\log(3)}$. Then if $n> N$, $|x_n|=3^{-n} < 3^{-N} < 3^{-p} = \epsilon$. This establishes convergence to 0. $\blacksquare$

A function f defined on a domain D is said to be uniformly continuous if for every $\epsilon>0$ there exists $\delta>0$ such that for all $x\in D$, if $d(x,y)<\delta$ then $d(f(x),f(y))<\epsilon$.

Problem 7. Here is the definition of uniform continuity:

Prove that the function $f(x)=x^2$ is uniformly continuous on the domain $[0,1]$. (Hint: $|x^2-y^2|=|x+y|\cdot|x-y|\leq 2|x-y|$ when $x,y\in[0,1]$).

Let $\epsilon>0$, and let $\delta=\frac\epsilon2$. Let $x,y\in D$ be chosen such that $d(x,y)<\delta$. Then, $d(f(x),f(y))=|x^2-y^2|\leq 2|x-y|=2 d(x,y) \leq 2\delta = \epsilon$. $\blacksquare$

# WPR III

Problem 1. Using the definition of the derivative, show that the function $f(x)=\begin{cases}x^2\sin\frac{1}{x} & x\neq 0\\ 0 & x=0\end{cases}$ is differentiable at $x=0$, and find the derivative $f'(0)$.

We need to check to see whether $\lim_{h\to0}\frac{f(h)-f(0)}{h}$ exists. This limit is $\lim_{h\to0}\frac{h^2\sin(\frac1h)-0}{h}=\lim_{h\to 0}h\sin(\frac1h)=0$. The last step follows since $\sin$ is a bounded function. Hence, $f'(0)=0$. $\blacksquare$

Problem 2. For what values of p does the series $\sum_{n=1}^\infty \frac{2008 n}{5 n^p+5}$ converge? (Justify your answer.)

Note that $\frac{2008 n}{5 n^p+5} = \frac{2008}{5} \left(\frac{n}{n^p+5}\right) = \frac{2008}{5}\left(\frac{1}{n^{p-1}+5n^{-1}}\right) < \frac{2008}{5}\left(\frac{1}{n^{p-1}}\right).$ We know by the power test that $\sum_{n=1}^\infty\frac{1}{n^{p-1}}$ converges for $p>2$, hence $\frac{2008}{5}\sum_{n=1}^\infty\frac{1}{n^{p-1}}$. Therefore, by the comparison test, the original series converges for $p>2$ as well. Similarly, it diverges for $p<2$, and we can say nothing about $p=2$.

Problem 3. Prove that the function $\sin(x)$ is uniformly continuous. (Hint: use the Mean Value Theorem to obtain the necessary inequality.)

To solve this problem, we need the assumption that $|\cos(x)|\leq 1$. Then the mean value theorem (with absolute values) states that, given any $x,y\in\mathbb{R}$, there exists some value $c\in(x,y)$ such that $|\sin(x)-\sin(y)|=|f'(c)|\cdot|x-y|=|\cos(c)|\cdot|x-y|\leq|x-y|$.
We now give the proof. Let $\epsilon>0$, and choose $\delta=\epsilon$. Then if $x,y\in\mathbb{R}$ are chosen such that $|x-y|<\delta$, then by the above inequality $|f(x)-f(y)|\leq|x-y|<\delta=\epsilon$, verifying uniform continuity. $\blacksquare$

Problem 4. Suppose a function $f(x)$ has a local maximum near $x=0$. (i) What does this imply about $f(t)-f(0)$ for t near 0? (ii) Show that the sign of $\frac{f(t)-f(0)}{t-0}$ (for t near 0) depends entirely on whether t is positive or negative. (iii) Why does this prove that $f'(0)=0$?

(i) Since $f(0)$ is the maximum, we must have $f(t)<f(0)$ for t near 0, meaning $f(t)-f(0)<0$.
(ii) Since $f(t)-f(0)<0$ for all t near 0, whereas the sign of the denominator $t-0$ changes depending on t, we see that $\frac{f(t)-f(0)}{t-0}<0$ if $t>0$, and $\frac{f(t)-f(0)}{t-0}>0$ if $t<0$.
(iii) If the derivative exists and $f'(0)=0$, then $\lim_{t\to 0}\frac{f(t)-f(0)}{t-0}$ must exist and be equal to the same value, no matter how t approaches zero. We have already seen that if $t\to 0$ with positive t, the values in the limit are negative, hence $\lim_{t\to 0}\frac{f(t)-f(0)}{t-0}\leq 0$. Likewise, if t approaches zero from the negative side, the values are positive, so that $\lim_{t\to 0}\frac{f(t)-f(0)}{t-0}\geq 0$. Hence, the only possible value for the limit is 0. $\blacksquare$

Problem 5. The ratio mean value theorem states that if f and g are continuous real functions on $[a,b]$ and differentiable on $(a,b)$, then there is a point $\theta\in(a,b)$ such that $\left(\frac{f(b)-f(a)}{b-a}\right)g'(\theta)=\left(\frac{g(b)-g(a)}{b-a}\right)f'(\theta).$ (i) Provide an interpretation of this theorem in terms of secant and tangent lines. (ii) Verify the theorem for $f(x)=x^2$, $g(x)=x^3$ on the interval $[0,1]$.

(i) Rearranging terms a bit, the theorem says that there is a point at which the ratio of the tangent slopes of f and g is equal to the ratio of secant slopes of f and g between a and b. The statement needs to be altered slightly if a denominator slope is 0, but the flavor is still the same.
(ii) Since $f(0)=g(0)=0$ and $f(1)=g(1)=1$, both secant slopes are equal to 1. Hence, the objective is to find a value of $\theta$ for which $\frac{f'(\theta)}{g'(\theta)}=1$ or more simply $f'(\theta)=g'(\theta)$. We have $f'(x)=2x$ and $g'(x)=3x^2$, so that we want to solve the equation $2\theta=3\theta^2 \Rightarrow \frac23=\theta$.