20th April 2019

# Homework 7

## Rudin Chapter 3 Q6

1. The partial sum of first $$N$$ terms evaluate to $$\sqrt{N+1} - 1$$ which obviously diverges.

2. For each $$n\in\mathbb{N}$$ we have \begin{align*} a_n &= \frac{\sqrt{n+1}-\sqrt{n}}n \\ &- \frac{1}{n(\sqrt{n+1}+\sqrt{n})} \end{align*} and since $$n(\sqrt{n+1}+\sqrt{n}) > n\sqrt{n}$$, $$a_n < 1/n^{3/2}$$. Because $$\sum \frac{1}{n^{3/2}}$$ converges by comparison test $$\sum a_n$$ converges.

3. $$\sqrt[n]{\left\lvert a_n \right\rvert} = \sqrt[n]{n} - 1$$. From Theorem 3.20 we have $$\lim_{n\to\infty} \sqrt[n]{\left\lvert a_n \right\rvert} = 1 - 1 = 0$$ so by root test $$\sum a_n$$ converges.

4. Whenever $$\left\lvert z \right\rvert > 1$$, \begin{align*} \lim_{n\to\infty}\left\lvert \frac{a_{n+1}}{a_n} \right\rvert &= \lim_{n\to\infty}\left\lvert \frac{1 + z^n}{1 + z^{n+1}} \right\rvert \\ &= \lim_{n\to\infty}\frac{\left\lvert \frac{1}{z^n} + 1 \right\rvert}{\left\lvert \frac 1{z^n} + z \right\rvert} \\ &< \frac1{\left\lvert z \right\rvert} < 1 \end{align*} so $$\sum a_n$$ converges by ratio test.

In the other case when $$\left\lvert z \right\rvert\geq 1$$ then $\left\lvert 1 + z^n \right\rvert \leq 1 + \left\lvert z \right\rvert^n \leq 2$ so for all $$n\in\mathbb{N}$$, $$\left\lvert a_n \right\rvert \geq \frac12$$ and $$\sum a_n$$ diverges.

## Rudin Chapter 3 Q9

1. $$\sum n^3 z^n$$ \begin{align*} \lim_{n\to\infty} \left\lvert \frac{(n+1)^3z^{n+1}}{n^3z^3} \right\rvert &= \left\lvert z \right\rvert \lim_{n\to\infty} \left\lvert \frac{n^3 + O(n^2)}{n^3} \right\rvert \\ &= \left\lvert z \right\rvert \end{align*} so $$R=1$$.

2. $$\sum \frac{2^n}{n!}z^n$$ \begin{align*} \lim_{n\to\infty} \left\lvert \cfrac{\frac{2^{n+1}}{(n+1)!}}{\frac{2^n}{n!}} \right\rvert &= \lim_{n\to\infty} \left\lvert \frac{2}{n+1} \right\rvert = 0 \end{align*} so $$R = \infty$$, this series converges everywhere in $$\mathbb{C}$$.

3. $$\sum \frac{2^n}{n^2}z^n$$ \begin{align*} \lim_{n\to\infty} \left\lvert \cfrac{\frac{2^{n+1}}{(n+1)^2}}{\frac{2^n}{n^2}} \right\rvert &= \lim_{n\to\infty} \left\lvert \frac{2n^2}{(n+1)^2} \right\rvert = 2 \end{align*} so $$R = \frac12$$

4. $$\sum \frac{n^3}{3^n}z^n$$ \begin{align*} \lim_{n\to\infty} \left\lvert \cfrac{\frac{3^{n+1}}{(n+1)^3}}{\frac{3^n}{n^3}} \right\rvert &= \lim_{n\to\infty} \left\lvert \frac{3n^3}{(n+1)^3} \right\rvert = 3 \end{align*} so $$R = \frac13$$

## Rudin Chapter 3 Q10

Suppose for a contradiction the power series of almost all non-zero integer coefficients $$\sum a_n z^n$$ converges but for some $$z$$ with $$\left\lvert z \right\rvert>1$$. By convergence it is necessary that $$\lim_{n\to\infty} \left\lvert a_n z^n \right\rvert = 0$$, but by hypothesis for each $$N\in\mathbb{N}$$ there exists $$n>N$$ with integer coefficient $$a_n \ne 0$$, and because $$\left\lvert z \right\rvert>1$$, $$\left\lvert a_nz^n \right\rvert >1$$, which contradicts the claim that limit goes to $$0$$.

## Rudin Chapter 3 Q13

To simplify the question we can assume terms from both series are $$\geq 0$$, so given two absolutely convergent series of nonnegative terms, denote $A_n = \sum_{i=1}^n a_i \to A,\quad B_n = \sum_{j=1}^n b_j \to B$ we just have to show $$C_n := \sum_{i=0}^n \sum_{j=0}^i a_j b_{i-j}$$ is a bounded sequence of partial sums.

For each $$i\in\mathbb{N}$$ denote \begin{align*} c_i &= \sum_{j=0}^i a_j b_{i-j} \\ &= \sum_{j=0}^{i-1} A_j(b_{i-j} - b_{i-(j+1)}) + A_ib_0 - A_{-1}b_i\\ &= \sum_{j=0}^i A_j (b_{i-j} - b_{i-j-1}) \end{align*} where we set $$A_{-1} = b_{-1} = 0$$ to simplify our term. Now for each $$n\in\mathbb{N}$$ \begin{align*} C_n &= \sum_{i=0}^n c_i \\ &= \sum_{i=0}^n \sum_{j=0}^i A_j (b_{i-j} - b_{i-j-1}) \\ &= \sum_{j=0}^n \sum_{i=j}^n A_j (b_{i-j} - b_{i-j-1}) \\ &= \sum_{j=0}^n A_j \sum_{i=j}^n (b_{i-j} - b_{i-j-1}) \\ &= \sum_{j=0}^n A_j b_{n-j} \\ &\leq \sum_{j=0}^n A b_{n-j} \\ &= A \sum_{k=0}^n b_{k} = A B_n \leq AB \\ \end{align*} so the partial sums are bounded.

# Homework 8

## Rudin Chapter 3 Q18

Choose $$x_1>\sqrt[p]{\alpha}$$, the sequence decreases monotonically with limit $$\sqrt[p]{\alpha}$$.

Suppose $$x_n \geq \sqrt[p]{\alpha}$$ we show $$x_{n+1} \geq \sqrt[p]{\alpha}$$ too. By AM-GM inequality \begin{align*} x_{n+1} &= \frac{p-1}{p}x_n + \frac{\alpha}{p} x_n^{-p+1} \\ &= \frac{\overbrace{x_n + x_n + \dots + x_n}^{p-1\text{ copies}} + \alpha\cdot x_n^{-p+1}}p \\ &\geq \sqrt[p]{x_n^{p-1}\cdot\alpha\cdot x_n^{-p+1}} \\ &= \sqrt[p]{\alpha} \end{align*} so by induction the sequence is bounded below by $$\sqrt[p]{\alpha}$$.

Suppose $$x_n\geq \sqrt[p]{\alpha}$$ we show $$x_{n+1} \leq x_n$$. \begin{align*} x_{n+1} &= \frac{p-1}{p}x_n + \frac{\alpha}{p} x_n^{-p+1} \\ &= x_n \frac1p \left(p-1 + \frac{\alpha}{x_n^p} \right) \end{align*} Since $$x_n\geq \sqrt[p]{\alpha}$$ by assumption we observe that \begin{align*} x_n^p &\geq \alpha \\ \frac{\alpha}{x_n^p} &\leq 1\\ p - 1 + \frac{\alpha}{x_n^p} &\leq p \\ \frac1p\left(p - 1 + \frac{\alpha}{x_n^p}\right) &\leq 1 \end{align*} where equality holds if and only if it holds in our assumption. This means $$x_{n+1} \leq x_n$$ where equality holds iff $$x_n = \sqrt[p]{\alpha}$$.

## Q1

Suppose $$\sum_{n=1}^\infty a_n$$ converges where each $$a_n\geq 0$$. Determine whether $$\sum_n \sqrt{a_n a_{n+1}}$$ also converges.

As $$\sum_{n=1}^\infty a_n$$ converges, we can drop the first term so $$\sum_{n=1}^\infty a_{n+1}$$ still converges. Then we see that $\sum_{n=1}^\infty \frac{a_n + a_{n+1}}2$ converges. Then by AM-GM inequality $\sqrt{a_na_{n+1}} \leq \frac{a_n + a_{n+1}}2$ and by comparison test $$\sum_n \sqrt{a_n a_{n+1}}$$ converges.

## Q2

Fix $$\alpha \in (0,1)$$, show that $$\sum_{n=1}^\infty \frac{\cos nx}{n^\alpha}$$ converges for all $$x$$ except multiples of $$2\pi$$.

In the case that $$x$$ is an integer multiple of $$2\pi$$, the series becomes $$\sum_{n=1}^\infty \frac1{n^\alpha}$$ which diverges because $$\alpha \leq 1$$.

In the other case assume $$x \ne 2\pi k$$ for all $$k\in\mathbb{Z}$$. We show that $\sum_{n=1}^\infty \frac{e^{inx}}{n^\alpha}$ converges.

First see that $$\frac1{n^\alpha}$$ is a decreasing sequence of real numbers with limit $$0$$. Next we set $$b_n := e^{inx}$$ and $$b_N := \sum_{n=1}^N b_n$$, Now we note that for each $$n\in\mathbb{N}$$, as $$x$$ is not a multiple of $$2\pi$$, we have the finite geometric sum $\left\lvert B_n \right\rvert = \left\lvert e^{ix}\left(\frac{1-e^{inx}}{1-e^{ix}}\right) \right\rvert \leq \frac{2}{\left\lvert 1-e^{ix} \right\rvert}$ note that $$x$$ is a fixed constant so the partial sums are bounded. Applying Dirichlet’s test (3.42) the series converges.

## Q3

### Part (a)

We can re-express our terms for readability as $a_n = \begin{cases} -\frac2{n^2} &\text{if $$n$$ is odd}\\ \frac2{n} &\text{if $$n$$ is even} \end{cases}$ to show $$\sum_{n=1}^\infty a_n$$ diverges consider the series of every 2 terms grouped together, $\sum_{n=1}^\infty b_n \text{ where } b_n := -\frac2{n^2} + \frac2{(n+1)}$ for all $$n\geq 3$$ it holds that \begin{align*} n^2 &> 2(n+1) \\ \frac1{n^2} &< \frac{1}{2(n+1)} \\ \frac1{n+1} - \frac1{n^2} &> \frac{1}{2(n+1)} b_n > \frac1{n+1} \end{align*} then by comparison test with harmonic series $$\sum b_n$$ diverges.

### Part (b)

Given $$a_n = \frac1{(\log n)^{\log n}}$$ we want to find convergence of $$\sum_{n=2}^\infty a_n$$. By theorem of Cauchy it converges iff this series converges $\sum_{k=1}^\infty 2^ka_k = \sum_{n=1}^\infty 2^k \frac1{(\log 2^k)^{\log 2^k}}.$ Define each summand in RHS as $$c_n$$ and simplify \begin{align*} c_n &= \frac{2^n}{(n\log 2)^{n\log 2}} \\ &= \frac1{n^{n\log 2}}\cdot\left( \frac{2}{(\log 2)^{\log 2}} \right)^n \end{align*} To see that $$\sum_{n=1}^\infty c_n$$ converges we first note that $$r :=\frac{2}{(\log 2)^{\log 2}} < 1$$ so $$\sum_{n=1}^\infty r^n$$ is a convergent geometric series. Next we see that $$\sum_{i=1}^\infty \frac1{n^{n\log 2}}$$ also converges by ratio test, as \begin{align*} \lim_{n\to\infty} \left\lvert \frac{n^{n\log 2}}{(n+1)^{(n+1)\log 2}} \right\rvert &= \lim_{n\to\infty} \left\lvert \frac{n^{n}}{(n+1)^{(n+1)}} \right\rvert^{\log 2} \\ &= \lim_{n\to\infty} \left\lvert \left(\frac{n}{n+1}\right)^n \frac 1{n+1} \right\rvert^{\log 2} \\ &= \lim_{n\to\infty} \left\lvert \frac{1}{\left(1+\frac 1n\right)^n} \frac 1{n+1} \right\rvert^{\log 2} \\ &= 0 \end{align*} Therefore the original series was convergent.

### Part (c)

We can see that whenever $$\left\lvert \alpha \right\rvert \geq 1$$, $$n\mapsto \sqrt[n]{\left\lvert \alpha \right\rvert}$$ is decreasing, so $$\left\lvert \frac{\sqrt[n]{\left\lvert \alpha \right\rvert}}n \right\rvert$$ is decreasing so the conditions of 3.43 is satisfied and the series converges.

When $$0 < \left\lvert \alpha \right\rvert < 1$$, we find $$N$$ such that for any $$n>N$$, $$\sqrt[n]{\alpha} > \frac12$$. Now if we examine the tail of the series each term is greater than $$b_n := \frac{1}{2(N+n)}$$. But $$\sum b_n$$ is the tail sum of (half of) harmonic series, which diverges. So by comparison test the original series diverges.

Convergence is obvious for $$\alpha=0$$.

# Homework 9

## Rudin Chapter 4 Q2

Let $$y\in f(\overline{E})$$, want to show $$y\in\overline{f(E)}$$. Find $$x\in\overline{E}$$ such that $$f(x) = y$$, then there exists a sequence $$x_1,x_2,\dots \in E$$ such that $$x_n \to x$$. Note that for each $$x_n\in E$$, we have $$f(x_n) \in f(E)$$, then because $$f$$ is continuous $y = f(x) = f\left(\lim_{n\to\infty} x_n\right) = \lim_{n\to\infty} f(x_n)$ which shows $$y\in\overline{f(E)}$$.

## Rudin Chapter 4 Q3

Note that $$Z(f) := {\left\{ x\in X:f(x)=0 \right\}} = f^{-1}({\left\{ 0 \right\}})$$ the pre-image of $${\left\{ 0 \right\}}$$. Since $${\left\{ 0 \right\}}$$ is closed in $$\mathbb{R}$$ and $$f$$ is continuous it follows that $$Z(f)$$ is closed.

## Rudin Chapter 4 Q16

For each integer $$n\in\mathbb{Z}$$, $$[x]$$ and $$(x)$$ are continuous on the interval $$(n,n+1)$$, which is easier to see by sketching the graph.

To see that $$(x)$$ is discontinuous at all integer points, consider $$\varepsilon=\frac12$$, then for any $$\delta>0$$ there is an $$x \in (n-\delta,n+\delta)$$ such that $$(x) \geq \frac12$$ (concretely maybe choose $$x=\max(n-\frac1m, n-\frac12)$$ where $$\frac1m < \delta$$).

To see that case of $$[x]$$ is similar we see that the difference between $$[x]$$ and $$(x)$$ is a continuous function since it was defined that $$(x) = x - [x]$$.

## Rudin Chapter 4 Q18

For any $$x\in\mathbb{R}$$, we show that $$\lim_{y\to x} f(y) = 0$$.

Given $$\varepsilon>0$$ we find the smallest $$N\in\mathbb{N}$$ such that $\frac1{N-1}\geq \varepsilon> \frac1N.$

Next to each $$n=1,2,\dots,N$$, let $$z_n$$ such that $\frac{z_n}n \leq x < \frac{z_n+1}n.$ Then define $\delta_n := \begin{cases} \min(x-\frac{z_n}n,\frac{z_n+1}n - x) &\text{if } x\ne \frac{z_n}n; \\ \frac1n &\text{otherwise}. \end{cases}$

Let $$\delta=\min(\delta_1,\dots,\delta_n$$ and for any $$y\in (x-\delta,x+\delta)\smallsetminus{\left\{ x \right\}}$$, if $$y$$ is irrational $$f(y)=0<\varepsilon$$, if $$y$$ is rational then $$y=\frac{a}b$$ for some $$a,b\in\mathbb{Z}$$ where $$b\ne 0,\gcd(a,b)=1$$. By our previous definitions it cannot be the case that $$b<N$$ so $$0<f(y)<\frac1N < \varepsilon$$.

The claim $$\lim_{y\to x} f(y) = 0$$ shows both claims.

# Homework 10

## Rudin Chapter 4 Q9

Fix metric spaces $$X,Y$$ and function $$f: X\to Y$$. We begin by writing out definitions. $$f$$ is uniformly continuous means $\forall \varepsilon>0 \exists \delta>0 \forall p,q\in X \left[ d_X(p,q)<\delta \implies d_Y(f(p),f(q)) < \varepsilon \right]$

The question wants to rephrase it as $\forall \varepsilon>0 \exists \delta>0 \forall E\subset X\left[ \operatorname{diam}E < \delta \implies \operatorname{diam}f(E) < \varepsilon \right]$ where $$\operatorname{diam}A := \sup_{x,y\in A} d(x,y)$$. Looking at the similarity of definitions we can just show logical equivalence of $\forall p,q\in X \left[ d_X(p,q)<\delta \implies d_Y(f(p),f(q)) < \varepsilon \right] \tag{1}$ and $\forall E\subset X\left[ \operatorname{diam}E < \delta \implies \operatorname{diam}f(E) < \varepsilon \right] \tag{2}$ fixing $$\varepsilon$$ and $$\delta$$.

$$(2)\implies (1)$$
Assume $$(2)$$, consider $${\left\{ p,q \right\}}\subset X$$.
$$(1)\implies (2)$$
Assume $$(1)$$, let $$E\subset X$$ with $$\operatorname{diam}E = \sup_{x,y\in E} d_X(x,y) < \delta$$. Want to show $$\sup_{x,y\in E} d_Y(f(x),f(y)) < \varepsilon$$. This holds because from assumption $$\operatorname{diam}E < \delta$$, so for all $$x,y\in E$$, $$d_X(x,y) < \delta$$ and by $$(1)$$, $$d_Y(f(x),f(y)) < \varepsilon$$. Since the bound holds for all $$x,y\in E$$ the bound is above the sup.

## Rudin Chapter 4 Q12

More precisely the statement says given $$f:X\to Y$$ and $$g:Y\to Z$$ uniformly continuous $$g\circ f: X\to Z$$ is also uniformly continuous.

Given $$\varepsilon>0$$ by uniform continuity of $$g$$ $\exists \varepsilon_1>0.~ \forall a,b\in Y.~ d_Y(a,b)<\varepsilon_1 \implies d_Z(g(a),g(b)) <\varepsilon.$

Now as $$f$$ is uniformly continuous, $\exists \delta>0.~ \forall c,d\in X.~ d_X(c,d)<\delta \implies d_Y(f(c),f(d))<\varepsilon_1$ and chaining the implications we get $$d_Z(g(f(c)),g(f(d))) < \varepsilon$$, so $$g\circ f$$ is uniformly continuous.

## Rudin Chapter 4 Q14

Define $$g(x) := f(x) - x$$ which is continuous. If $$g(0) = 0$$ or $$g(1) = 0$$ we are done. In the other case $$g(0) = f(0) > 0$$ and $$g(1) = f(1)-1 < 1-1 = 0$$, so by IVT there is $$c\in[0,1]$$ with $$g(c) = 0 \implies f(c) = c$$.

## Q (squeeze)

Given $$\varepsilon>0$$ there exist $$\delta_1,\delta_2$$ such that $0<\left\lvert x-a \right\rvert<\delta_1\implies \left\lvert f(x)-L \right\rvert<\varepsilon$ $0<\left\lvert x-a \right\rvert<\delta_2\implies \left\lvert h(x)-L \right\rvert<\varepsilon$ so choose $$\Delta = \min(\delta,\delta_1,\delta_2)$$, then for all $$x\in (x-\Delta,x+\Delta)\smallsetminus{\left\{ a \right\}}$$, we have $f(x) - L \leq g(x) - L \leq h(x) - L$ which means that $\left\lvert g(x)-L \right\rvert \leq \max\left( \left\lvert f(x)-L \right\rvert,\left\lvert h(x)-L \right\rvert \right) < \varepsilon$

To show that $$f(x)$$ is continuous at $$0$$ we first amend its definition $f(x) := \begin{cases} 0 &\text{if } x = 0 \\ x\sin \frac1x &\text{otherwise} \end{cases}$ To use previous result observe that for all $$x$$ near $$0$$ we have $-\left\lvert x \right\rvert \leq f(x) \leq \left\lvert x \right\rvert$ and because $$\lim_{x\to 0} \left\lvert x \right\rvert = \lim_{x\to 0} -\left\lvert x \right\rvert = 0$$ by previous result $$\lim_{x\to 0} f(x) = 0$$. From definition of $$f$$ we see that $$f$$ is continuous everywhere else hence $$f$$ is continuous at every point on $$\mathbb{R}$$.