Post Reply 
 
Thread Rating:
  • 1 Votes - 5 Average
  • 1
  • 2
  • 3
  • 4
  • 5
Rudin 【Principle of Mathematical Analysis】Notes & Solutions
05-07-2012, 12:28 PM
Post: #21
Exercises 1.1~1.2 -- Rudin [Principle of Mathematical Analysis]
Unless the contrary is explicitly stated, all numbers that are mentioned in these exercises are understood to be real.

Ex. 1.1 If $r$ is rational $(r\ne 0)$ and $x$ is irrational, prove that $r+x$ and $rx$ are irrational.
$\quad$ Proof Otherwise $x = (r+x) -r \in\mathbb{Q}$ or $x = (rx) /x \in\mathbb{Q}$ since $\mathbb{Q}$ is a subfield of $\mathbb{R}$

Ex. 1.2 Prove that there is no rational number whose square is 12.
$\quad$ We'll show that $\mathbb{Q}\ni \sqrt[k]{n}=n^{1/k}\; (1< k, n\in\mathbb{N}) \Longleftrightarrow \exists k,m\in\mathbb{N}\;(m < n = m^k) $.
$\quad\quad$Proof Let $p,q\in\mathbb{N}^+\; \gcd(p,q) = 1$ such that $n^{1/k} = p/q$. Then $p^k = n q^k$. If $q > 1$, then there is an prime $d \mid q$ thus $d \mid p$. A contradiction. So $q = 1$ and $n = m^k$ with $m = p$.
$\quad\quad$ Note that for prime $d$, $(d \mid p^k)\implies (d\mid p)$ is a consequence of $(d \mid ab) \implies (d\mid a) \vee (d\mid b) $. To see the later, we 1st show $\gcd(u,v) = \min\{su+tv \mid su+tv > 0\; (s,t\in\mathbb{Z})\}:=L_{u,v}.\quad$ Clearly $\gcd(u,v) \le L_{u,v}$. If $L_{u,v}$ is not a common divisor of $a$ and $b$, say $L_{u,v} \nmid u$, there are some $e,f\in\mathbb{Z}$, $u = eL_{u,v} +f,\; 0< f< L_{u,v}$. But then $f = (1-es)u - etv \in\{su+tv \mid su+tv > 0\; (s,t\in\mathbb{Z})\}$ Contradict to the definition of $L_{u,v}$. Therefore $L_{u,v} \mid \gcd{u,v}\implies L_{u,v}\le \gcd(u,v)$
$\quad\quad$ If prime $d \mid ab$ and $d\nmid a$, then $\gcd(d,a) = 1.\quad$ i.e. $sa +td = 1$ for some $s,t\in\mathbb{Z}$ thus $d\mid s(ab)+(td)b = (sa+td)b=b$. Now we see that $(d \mid p^k)\implies (d\mid p)$
Find all posts by this user
Quote this message in a reply
05-07-2012, 12:28 PM
Post: #22
Exercise 1.3 --Rudin [Principle of Mathematical Analysis] Notes
Ex 1.3 Prove Proposition 1.15.
$\quad\quad$Proof (a) $(x\ne 0)\wedge (xy=xz) \overset{M_5,M_4,M_3}{\implies} y=(\frac{1}{x}x)y = \frac{1}{x}(xy) = \frac{1}{x}(xz)=(\frac{1}{x}x)z = z$
$\quad\quad$ (b) $(x\ne 0)\wedge (xy = x = x\cdot 1) \overset{(a)}\implies y = 1$
$\quad\quad$ (c) $ (xy = 1\overset{x\ne 0, M_5}{=}x\frac{1}{x})\overset{(a)}\implies (y=\frac{1}{x})$
$\quad\quad$ (d) $(1/x)x = 1\overset{(c )}{\implies} x = 1/(1/x)$
Find all posts by this user
Quote this message in a reply
05-07-2012, 12:28 PM
Post: #23
Exercises 1.4~1.5 --Rudin [Principle of Mathematical Analysis] Notes
Ex. 1.4 Let $E \ne \varnothing$ be a subset of an ordered set, $\alpha\;(\beta)$ is a lower (upper) bound of $E$. Prove that $\alpha \le \beta$.
$\quad\quad$ Proof There exists a member $x\in E$, clearly we have $\alpha \le x\le \beta$

Ex. 1.5 Let $\varnothing\ne A\subset \mathbb{R}$, $A$ is bounded below. and $-A = \{-x: x\in A\}$. Prove that $\inf A = -\sup(-A)$
$\quad\quad$ Proof Let $\alpha = \inf A$, then $(x\in -A)\implies (-x \in A) \implies (\alpha \le -x) \implies (x \le -\alpha)$
$\quad\quad$ So $-\alpha$ is an upper bound of $-A$. Moreover,
$\quad\quad$ $(\beta < -\alpha)\implies (\alpha < -\beta)\implies (\exists y\in A \; (y <-\beta))\implies (\exists z\in -A\; (\beta < z))$
$\quad\quad$ That means $-A$ has no upper bound smaller than $-\alpha$ and so $\sup(-A) = -\alpha = -\inf A. \quad\square$
Find all posts by this user
Quote this message in a reply
05-07-2012, 12:28 PM
Post: #24
Exercises 1.6~.17 --Rudin [Principle of Mathematical Analysis] Notes
Ex. 1.6 Fix b > 1.
$\quad\quad$ (a) $\left(m,n\in\mathbb{Z},\; n,q\in\mathbb{N}^+,\; r = \frac{m}{n} = \frac{p}{q}\right)\implies (b^m)^{1/n}=(b^p)^{1/q}.$ So $b^r = (b^m)^{1/n}$ is well defined.
$\quad\quad$ (b) Prove that $b^{r+s} = b^r b^s$ if $r,s\in\mathbb{Q}$
$\quad\quad$ (c) For $x\in\mathbb{R}$, let $B(x)=\{b^t\mid x\ge t\in\mathbb{Q}\}$. Prove that $b^r = \sup B( r)\;(r\in\mathbb{Q})$.
$\quad\quad\quad\quad$ Hence it makes sense to define $b^x = \sup B(x)$ for every $x\in\mathbb{R}$.
$\quad\quad$ (d) Prove that $b^{x+y} = b^x b^y\; (\forall x,y\in\mathbb{R})$
$\quad\quad$ Proof
$\quad\quad$ (a) By Definition 1.12 ($M_3$), the associative rule, $(b^n)^m = (b^m)^n = b^{mn}$ so
$\quad\quad\quad\quad ((b^m)^{1/n})^{nq} = (((b^m)^{1/n})^n)^q = (b^m)^q =b^{mq} = b^{np}=(((b^p)^{1/q})^q)^n = ((b^p)^{1/q})^{nq}$
$\quad\quad\quad\quad$ is the consequence of the definition of $k$th root. Now by its uniqueness, $(b^m)^{1/n}=(b^p)^{1/q}$
$\quad\quad$ (b) Let $m,p\in\mathbb{Z},\; n,q\in\mathbb{N}^+,\; r = m/n,\; s = p/q$, then by (a) and the Corollary of Theorem 1.21,
$\quad\quad\quad\quad (b^{r+s})^{nq} =(b^{(mq+pn)/(nq)})^{nq} = b^{mq} b^{pn} = ((b^m)^{1/n})^{nq}((b^p)^{1/q})^{qn}=(b^r)^{nq}(b^s)^{nq} = (b^r b^s)^{nq}$
$\quad\quad$ (c) Since $b^{m/n} >1\quad(m,n\in\mathbb{N}^+)$, we have $\forall r,s\in\mathbb{Q}\; (r < s)\implies (b^r < b^r b^{s-r} = b^s)$ by (b).
$\quad\quad\quad\quad$ This in turn implies $b^r = \sup B( r)\quad(\forall r\in\mathbb{Q})\quad$. If $x\in\mathbb{R\setminus Q}$, there are $r,s\in\mathbb{R}$ such that
$\quad\quad\quad\quad x-1 < r < x < s < x+1$. So $\varnothing \ne B(x),\; b^r$ is an upper bound of $B(x), b^x$ is well defined
$\quad\quad$ (d) $\left(\forall (b^r,b^s)\in B(x)\times B(y)\quad (b^r b^s = b^{r+s} \in B(x+y))\Rightarrow (b^r b^s \le b^{x+y})\right ) \Rightarrow (b^x b^y \le b^{x+y})$
$\quad\quad\quad\quad$ We use a fact $(\star)\;b^{1/n} \le 1+(b-1)/n\;(n\in\mathbb{N}^+)$ to exclude the possibility of $b^x b^y < b^{x+y}$:
$\quad\quad\quad\quad$ If $\delta = b^{x+y}-b^x b^y > 0$, take $n\in\mathbb{N},\; r,s\in\mathbb{Q}$ such that \[n\delta > b b^x b^y, \;x-\frac{1}{2n} &#36;\quad\quad\quad\quad&#36; But &#36;\displaystyle{(\beta^n -1 = (\beta-1)\left(\sum_{j=0}^{n-1} \beta^j\right) \ge n(\beta-1),\; \beta = b^{1/n})\Rightarrow (1+(b-1)/n \ge b^{1/n})\quad\quad\square}&#36;

&#36;\quad\quad&#36; <span style="font-weight: bold;">Remark Ex. 6 There are some rather lengthy but correct proof and somewhat wrong proof for (d) in pdf seen from the web. Finding a proof without too many lemmas and additional notions (W. Rudin's style) is an interesting and rewarding job. Strongly encourage finding better and pretty proofs. One of a wrong proof says basically that since $b^x b^y \ge b^r b^s = b^{r+s}\in B(x+y)\; (b^r\in B(x),\; b^s\in B(y))$, we must have $b^x b^y \ge b^{x+y}$, but from the former, we can only say the contrary: $b^x b^y \le b^{x+y}$, as I did. Welcome to more detailed discussion about this exercise.

Ex. 1.7 Fix $b>1,\; y > 0$, prove that $\exists ! x\in\mathbb{R}\; (b^x = y)$ by completing the following outline.
$\quad\; $ (This $x$ is called the logarithm of $y$ to the base $b$.)
$\quad\quad$ (a) $b^n -1 \ge n(b-1).\quad (n\in\mathbb{N})$
$\quad\quad$ (b) Hence $b-1 \ge n(b^{1/n}-1)$
$\quad\quad$ (c) If $t > 1$ and $n > (b-1)/(t-1)$, then $b^{1/n} < t$
$\quad\quad$ (d) If $b^w < y$, then $b^{w+(1/n)} < y$ for sufficiently large $n$; apply (c) with $t = y\cdot b^{-w}$ to see this.
$\quad\quad$ (e) If $b^w > y$, then $b^{w-(1/n)} > y$ for sufficiently large $n$.
$\quad\quad$ (f) Let $A$ be the set of all $w$ such that $b^w < y$, and show that $x=\sup A$ satisfies $b^x = y$
$\quad\quad$ (g) Prove that this x is unique.
$\quad$ Proof
$\quad\quad$ (a) $b^n -1 = b^n - 1^n = (b-1)\sum_{j=0}^{n-1} b^j \ge n(b-1)$
$\quad\quad$ (b) is the result of replacing $b$ by $b^{1/n}$ in (a)
$\quad\quad$ (c) By (b) we then have $b^{1/n} \le 1+(b-1)/n < 1+n(t-1)/n = t$
$\quad\quad$ (d) Since $b^w < y$, for $t = b^{-w} y > 1$ (c) gives $b^{w+1/n} $\quad\quad$ (e) This time $t = b^w / y > 1$ and (c) gives $y < b^{w-1/n}$ when $n > (b-1)/(t-1)$
$\quad\quad$ (f) Since $b >1,\; y > 0$, (a) shows that $b^n > y > b^{-n}$ when $n$ is big enough and so $\varnothing \ne A$
$\quad\quad\quad\quad$ and $A$ is bounded above. Thus $x = \sup A$ always exists.
$\quad\quad\quad\quad$ If $b^x < y$ then (d) shows that $x$ is not an upper bound of $A$;
$\quad\quad\quad\quad$ If $b^x > y$ then (e) shows that $x$ is not the least upper bound of $A$. So $b^x = y$
$\quad\quad$ (g) We only need to show that $\forall w>0\; (b^w > 1 )$. But this is clear for $w\in\mathbb{Q}$, and also true by
$\quad\quad\quad\quad$ general definition of $b^x$ in Ex. 6. $\quad\quad\square$
Find all posts by this user
Quote this message in a reply
05-12-2012, 10:00 AM
Post: #25
Exercises 1.8~1.15 --Rudin [Principle of Mathematical Analysis]
Ex. 1.8 Prove that no order can be defined in the complex field that turns it into an ordered field.
$\quad\quad$ Hint: $-1$ is a square.
Proof If $\mathbb{C}$ became an ordered field somehow, by Proposition 1.18(d), $-1 = i^2 > 0,\; 1 = 1^2 > 0$
$\quad\quad$ thus by Definition1.12 ($A_5$) and Definition 1.17, $0 = -1+1 = i^2+1>0$. A contradiction!

Ex. 1.9 Suppose $z=a+bi,\; w=c+di$. Define $z < w$ if $(a $\quad\quad$ Prove that this turns the set of all complex numbers into an ordered set. (The order is called
$\quad\quad$ a dictionary order, or lexicographic order, for obvious reasons.)
$\quad$ Does this ordered set have the least-upper-bound property?
Proof $(z\ne w)\implies ((a\ne c)\vee ((a=c)\wedge(b\ne d))) \implies (zw)$
$\quad\; (z < w)\implies (aw))$
$\quad\; (z = w)\implies (a=c)\wedge (b=d)\implies \lnot ((zw))$
$\quad\, \begin{align} (z< w )\wedge (w & \Rightarrow (a < e)\vee ((a=e)\wedge(b $\quad$ So the relation $<$ turns the set of all complex numbers into an ordered set.
$\quad\quad$ Let $E = \{z \mid \operatorname{Re}(z) < 0\}$ then $\varnothing\ne E$ and $E$ has upper bound $0$. If $z = a+bi$ is an upper bound
$\quad\quad$ of $E$, so is $w := 0+(b-1)i < z$. Thus the ordered set has no least-upper-bound property. $\;\square$

Ex. 1.10 Suppose $z = a+bi,\; w = u+iv$, and \[a = \left(\frac{|w|+u}{2} \right )^{1/2},\quad b=\left(\frac{|w|-u}{2} \right )^{1/2}.\] $\quad$ Prove that $z^2 = w\;(v\ge 0)$ and that $\overline{z}^2 = w\; (v\le 0)$. Conclude that every
$\quad$ complex number (with one exception!) has two complex square roots.
Proof $\displaystyle{z^2 = a^2-b^2 + 2ab i = u + |v|i}$. Thus $\pm z\;(v>0)$ or $\pm \overline{z}\;(v<0)$ are two square roots of $w$.
$\quad\quad$ but $0$ has only one distinct square root: $0$.
$\quad\quad$ If $(x+yi)^2 = u+iv\; (x,y,u,v\in\mathbb{R})$, then $x^2 - y^2 = u,\; 2xy = v$ thus $(x^2+y^2)^2 = u^2+v^2$
$\quad\quad$ So $\displaystyle{|x| = \left(\frac{|w|+u}{2}\right)^{1/2},\; |y| = \left(\frac{|w|-u}{2}\right)^{1/2}}$ hence $u+iv$ has no more than 2 square roots.

Ex. 1.11 If $z$ is a complex number, prove that there exists an $r\ge 0$ and a complex number $w$ with
$\quad\; |w|=1$ such that $z = rw$. Are $w$ and $r$ always uniquely determined by $z$?
Proof If $z \ne 0$, let $r = |z|,\; w = z/r$, then $z = rw,\; r > 0,\; |w|=1$. If $r > 0,\; |w|=1,\; z=rw$,
$\quad\quad$ then $r = r|w| = |rw|=|z|,\; w = z/r = z/|z|$ are uniquely determined by $z$.
$\quad\quad$ Notice that $ 0 = 0 w$ for any $w\in\mathbb{C}$ with $|w|=1$. When $z=0$, $w$ is not uniquely determined.

Ex. 1.12 Prove that $\quad |z_1+z_2+\cdots + z_n|\le |z_1|+|z_2|+\cdots +|z_n|.\quad (z_1,\cdots,z_n\in\mathbb{C})$
Proof $|z_1+z_2|^2 = (z_1+z_2)(\overline{z}_1+\overline{z}_2) = |z_1|^2+(z_1 \overline{z}_2 + \overline{z}_1 z_2) + |z_2|^2 \le (|z_1|+|z_2|)^2$.
$\quad\quad$ General case is obtained by induction.

Ex. 1.13 Prove that $\quad ||x|-|y||\le |x-y|\quad(\forall x,y\in\mathbb{C})$
Proof Assume $|x|\ge|y|$, then $||x|-|y|| = |x|-|y| = |(x-y)+y|-|y|\le |x-y|+|y|-|y|=|x-y|$
$\quad\quad$ But exchange $x$ and $y$ will not affect anything.$\quad\quad\square$

Ex. 1.14 If $z\in\mathbb{C}$ with $|z|=1=z\overline{z}$, compute $|1+z|^2+|1-z|^2.$
Solution $|1+z|^2+|1-z|^2 = (1+z)(1+\overline{z})+(1-z)(1-\overline{z}) = 2+2\operatorname{Re}(z) + 2-\operatorname{Re}(z) = 4.$

Ex. 1.15 Under what conditions does equality hold in the Schwarz inequality?
Solution If $(\star)\quad \exists C, B\in\mathbb{C}$ with $(B\ne 0)\vee (C\ne 0))$ such that $Ba_j - Cb_j = 0\; (j=\overline{1,n})$
$\quad\quad$ Then $\left|\sum a_j\overline{b}_j \right|= |k|\sum |a_j|^2,\; \sum|a_j|^2\sum|b_j|^2 = |k|^2\left(\sum |a_j|^2 \right )^2$ and the equality holds.
$\quad\quad$ Conversely, if the equality holds, $B,\;C$ as in the proof of Theorem 1.35, clearly $(\star)$ hold.
Find all posts by this user
Quote this message in a reply
05-15-2012, 12:28 AM
Post: #26
Exercises 1.16~1.19 --Rudin [Principle of Mathematical Analysis]
1.16 Suppose $k\ge 3,\; \mathbf{x,y}\in\mathbb{R}^k,\; |\mathbf{x}-\mathbf{y}| = d>0$, and $r>0$. Prove:
$\quad\quad$ (a) If $2r > d$, there are infinitely many $\mathbf{z}\in\mathbb{R}^k$ such that \[|\mathbf{z}-\mathbf{x}|=|\mathbf{z}-\mathbf{y}|=r.\]$\quad\quad$ (b) If $2r=d$, there is exactly one such $\mathbf{z}$.
$\quad\quad$ (c) If $2r < d$, there is no such $\mathbf{z}$.
$\quad\quad$ How must these statements be modified if $k$ is $2$ or $1$?
Proof Since $r=|\mathbf{z-x}|,\;d=|\mathbf{y-x}|$ and $\displaystyle{(|\mathbf{z-x}|^2 = |\mathbf{z-y}|^2)\Leftrightarrow \bigg(\mathbf{z}-\frac{\mathbf{x+y}}{2}\bigg)\cdot (\mathbf{y-x})=0}$
$\quad\quad$ we have $\displaystyle{\mathbf{z-x}=\bigg(\mathbf{z}-\frac{\mathbf{x+y}}{2}\bigg)+\frac{\mathbf{y-x}}{2},\; r^2 = |\mathbf{z-x}|^2 = \left|\mathbf{z}-\frac{\mathbf{x+y}}{2}\right|^2 + \frac{d^2}{4}}\quad (\blacktriangle )$
$\quad\quad$ Thus $(\bigstar)\; \displaystyle{\mathbf{z} = \frac{\mathbf{x+y}}{2} + \mathbf{w}}\;$ with $\; \mathbf{w}\cdot (\mathbf{y-x}) = 0,\; |\mathbf{w}|^2 = r^2 - d^2/4$

$\quad\quad$ Assume $k\ge 3$, let $\mathbf{e}_j =\underset{j\text{th}}{(0,\cdots,0,1,0,\cdots,0)}\in\mathbb{R}^k$ be the $j$th unit coordinate vector,
$\quad\quad\;\mathbf{u} = \mathbf{x-y}$, then $\mathbf{u} = \sum_1^k (\mathbf{u\cdot e}_j)\mathbf{e}_j$. There are $l,m,n\in\{1,\cdots,k\}$, pairwise distinct, such
$\quad\quad$ that $a = \mathbf{u}\cdot \mathbf{e}_l \ne 0,\; b = \mathbf{u}\cdot \mathbf{e}_m,\; c=\mathbf{u}\cdot \mathbf{e}_m$. With these,
$\quad\quad$ define $\mathbf{w}_1 = b\mathbf{e}_l - a\mathbf{e}_m,\; \mathbf{w}_2 = c\mathbf{e}_l - a\mathbf{e}_n$, then $\mathbf{u}\cdot \mathbf{w}_1 = ba-ab = 0 = ca - ac = \mathbf{u}\cdot \mathbf{w}_2$

$\quad\quad$ (i)when $2r > d$, $\displaystyle{\mathbf{w}_t =\sqrt{r^2-d^2/4}\frac{\mathbf{w}_1 +t\mathbf{w}_2}{|\mathbf{w}_1 +t\mathbf{w}_2|}}\; (t\in\mathbb{R})$ satisfies $(\bigstar)$ and $\mathbf{w}_{\alpha} \ne \mathbf{w}_{\beta}\;(\alpha \ne \beta)$

$\quad\quad$ (ii) If $2r = d$, by ($\blacktriangle$), $\mathbf{z} = (\mathbf{x+y})/2$ is the only solution.

$\quad\quad$ (iii) If $2r < d$ then ($\blacktriangle$) can never be satisfied thus no solution $\mathbf{z}$ exist.

$\quad\quad$ If $k = 2$, then the above construction can only lead to $\mathbf{w}=\pm\mathbf{w}_1$ when $2r > d$; while clearly
$\quad\quad\;2r = d$ determines the only solution $\mathbf{z} = (\mathbf{x+y})/2$ and no solutions when $2r < d$.

$\quad\quad$ If $k = 1$, then $(|z - x|^2 = |z - y|^2)\Leftrightarrow (z = (x+y)/2)\Leftrightarrow (r = |z-x|=|z-y|=d/2)$
$\quad\quad$ Thus if and only if $2r = d$, solution $z$ exists and it's $z = (x+y)/2$

1.17 Prove that $\forall \mathbf{x,\; y \in\mathbb{R}^k:\; (|\mathbf{x+y}|^2+|\mathbf{x-y}|^2=2|\mathbf{x}|^2+2|\mathbf{y}|^2)}$
$\quad\quad$ Interpret this geometrically, as a statement about parallelograms.
Proof $LHS= (|\mathbf{x}|^2+|\mathbf{x}|^2+2\mathbf{x}\cdot \mathbf{y})+(|\mathbf{x}|^2+|\mathbf{x}|^2-2\mathbf{x}\cdot\mathbf{y}) =RHS$. Treat $\mathbf{x,\; y}$ as two sides
$\quad\quad$ of a parallelogram, then the given equation reads The sum of the squares of the diagonals of
$\quad\quad$ a parallelogram equals the sum of the squares of its four sides
.

1.18 If $k\ge 2$ and $\mathbf{x}\in\mathbb{R}^k$, prove that $\exists \mathbf{y}\in\mathbb{R}^k:\; (\mathbf{y}\ne\mathbf{0})\wedge (\mathbf{x}\cdot\mathbf{y} = 0).\;$ Is it also true if $k=1$?
Proof Let $\mathbf{e}_j,\; \text{Proj}_{\mathbf{v}}(\mathbf{u})$ be as in the solution of Ex. 1.16. If $\mathbf{e}_1 = \text{Proj}_{\mathbf{x}}(\mathbf{e}_1)$ for some $j$, we are done
$\quad\quad$ (pick $\mathbf{y}=\mathbf{e}_2$). Otherwise let $\mathbf{y} = \mathbf{e}_1-\text{Proj}_{\mathbf{x}}(\mathbf{e}_1)\ne \mathbf{0}$ we have $\mathbf{x}\cdot\mathbf{y}=\mathbf{x}\cdot\mathbf{e}_1 - \mathbf{x}\cdot\mathbf{e}_1 = 0$

$\quad\quad$ If $k=1$, this is in general not true unless $x = 0$.

1.19 Suppose $\mathbf{a,\; b}\in\mathbb{R}^k$. Find $\mathbf{c}\in\mathbb{R}^k$ and $r > 0$ such that \[|\mathbf{x}-\mathbf{a}|=2|\mathbf{x}-\mathbf{b}|\]$\quad\quad$ if and only if $|\mathbf{x}-\mathbf{c}| = r.\quad\quad$ (Solution: $3\mathbf{c}=4\mathbf{b}-\mathbf{a},\; 3r = 2|\mathbf{b}-\mathbf{a}|$)
Sol. $(|\mathbf{x-a}|=2|\mathbf{x-b}|)\Leftrightarrow (|\mathbf{x-a}|^2=4|\mathbf{x-b}|^2)\Leftrightarrow (3|\mathbf{x}|^2 = 2\mathbf{x}\cdot (4\mathbf{b}-\mathbf{a}) + |\mathbf{a}|^2-4|\mathbf{b}|^2)$
$\displaystyle{\quad\quad \Leftrightarrow (3\mathbf{x\cdot x} -2(4\mathbf{b}-\mathbf{a})\cdot \mathbf{x} + 4\mathbf{b\cdot b}-\mathbf{a\cdot a} = 0)\Leftrightarrow \left| \mathbf{x}-\frac{4\mathbf{b}-\mathbf{a}}{3} \right |^2 = \frac{4}{9}\left|\mathbf{b}-\mathbf{a}\right|^2} \quad\square$.
Find all posts by this user
Quote this message in a reply
05-16-2012, 04:01 PM
Post: #27
RE: Rudin [Principle of Mathematical Analysis] Notes
1.20 With reference to the Appendix, suppose that property (III) were omitted from the definition of a cut. Keep the same definitions of order and addition. Show that the resulting ordered set has the least-upper-bound property, that addition satisfies axioms ($A_1$) to ($A_4$) (with a slightly different zero-element!) but that ($A_5$) fails.
Find all posts by this user
Quote this message in a reply
05-16-2012, 07:18 PM
Post: #28
Remark on Some Chap.1 Exercises --Rudin [Principle of Mathematical Analysis] Notes
Ex. 1.16 It is beneficial to obtain the exact solution set for the cases $k\ge 3,\; 2r > d$.
Solution Let $\mathbf{e}_j =\underset{j\text{th}}{(0,\cdots,0,1,0,\cdots,0)}\in\mathbb{R}^k$ be the $j$th unit coordinate vector, $\;\mathbf{u} = \mathbf{x-y}$, then
$\quad\quad\; \mathbf{u} = \sum_1^k (\mathbf{u\cdot e}_j)\mathbf{e}_j,\; \mathbf{u\cdot u}=\sum_1^k (\mathbf{u\cdot e}_j)^2$. Without lose of generality, assume $(\mathbf{u\cdot e}_1)^2 > 0$
$\quad\quad$ Let $\mathbf{u}_1 = \mathbf{u},\; \mathbf{u}_p = \mathbf{e}_p\;(1< p\le k)$. Clearly $\displaystyle{\forall \mathbf{x}\in\mathbb{R}^k\;\exists ! (\lambda_1,\cdots,\lambda_k)\in\mathbb{R}:\;\bigg(\mathbf{x}=\sum_{j=1}^k \lambda_j\mathbf{u}_j \bigg)}$
$\quad\quad$ $\{\mathbf{u}_1,\cdots,\mathbf{u}_k\}$ is called a basis of $\mathbb{R}^k$ due to the unique representation property above.
$\quad\quad$ Call $\displaystyle{ \text{Proj}_{\mathbf{v}}(\mathbf{u}) = \frac{\mathbf{u\cdot v}}{\mathbf{v\cdot v}}\mathbf{v}}=\bigg(\mathbf{u}\cdot \frac{\mathbf{v}}{|\mathbf{v}|} \bigg)\frac{\mathbf{v}}{|\mathbf{v}|}$ the projection of $\mathbf{u}$ to $\mathbf{v}$ with $\mathbf{v}\ne \mathbf{0}$.
$\quad\quad$ Let $\displaystyle{\mathbf{v}_1 = \mathbf{u}_1,\quad \mathbf{v}_{p+1} = \mathbf{u}_{p+1} - \sum_{j=1}^p \text{Proj}_{\mathbf{v}_j}(\mathbf{u}_{p+1})}\quad (1\le p < k).\quad$ Notice that $\mathbf{v}_j \ne \mathbf{0}$:
$\quad\quad$ otherwise $\displaystyle{\mathbf{u}_p = \sum_{j=1}^{p-1}\text{Proj}_{\mathbf{v}_j}(\mathbf{u}_p)}$ for some $p>1$. But the right hand side can be written as
$\quad\quad$ a linear combination of $\mathbf{u}_1,\cdots,\mathbf{u}_{p-1}$ by the definition of projection, which means $\mathbf{u}_p$ has more
$\quad\quad$ than $1$ way to be represented by basis $\{\mathbf{u}_1,\cdots,\mathbf{u}_k\}$, impossible! So $\{\mathbf{v}_1,\cdots,\mathbf{v}_k\}$ is well defined.
$\quad\quad$ Its not hard to see that it spans $\mathbb{R}^k$. We now show $\mathbf{v}_i\cdot\mathbf{v}_j = 0\; (i\ne j)$
$\quad\quad\; \displaystyle{\mathbf{v}_1\cdot\mathbf{v}_2 = \mathbf{u}_1\cdot\mathbf{u}_2-\mathbf{u}_1\cdot \bigg(\frac{\mathbf{u}_1\cdot\mathbf{u}_2}{|\mathbf{u}_1|^2}\mathbf{u}_1\bigg)=0​}.\quad$. Suppose $\mathbf{v}_i\cdot\mathbf{v}_j = 0\; (1\le i $\quad\quad$ then for $j \le p$ we have $\mathbf{v}_j\cdot\mathbf{v}_{p+1} = \mathbf{v}_j\cdot\mathbf{u}_{p+1}-\mathbf{v}_j\cdot\text{Proj}_{\mathbf{v}_j}(\mathbf{u}_{p+1}) = 0$
$\quad\quad$ We've constructed an orthonormal basis $\{\mathbb{w}_1,\cdots,\mathbf{w}_k\}$ of $\mathbb{R}^k$ where $\displaystyle{\mathbf{w}_j = \frac{\mathbf{v}_j}{|\mathbf{v}_j|}}\quad(j=\overline{1,k})$
$\quad\quad$ with an important property $\mathbf{w}\cdot (\mathbf{y-x}) \Leftrightarrow \mathbf{w}\in \text{span}\{\mathbf{w}_2,\cdots,\mathbf{w}_k\}\left(=(\mathbf{y-x})^{\perp}\right)$
$\quad\quad$ The rest work should be simple enough....
Find all posts by this user
Quote this message in a reply
05-16-2012, 07:18 PM
Post: #29
RE: Rudin [Principle of Mathematical Analysis] Notes
   
Visualization of Ex1.19 in $\mathbb{R}^2$. If $D$ is the middle point of point of $\overline{AX}$, then $|\overline{DX}|=|\overline{BX}|$
where $X_1=\frac{1}{3}A+\frac{2}{3}B,\;X_2 =-\frac{1}{3}A+\frac{4}{3}B,\; C = \frac{1}{2}(X_1+X_2)$ and $X \in \bigodot {C}$

Pure algebra gives the following general result:

$\begin{align}\text{If }0<\lambda \ne 1 \text{ then }(|\mathbf{x-a}| = \lambda |\mathbf{x-b}|) & \Longleftrightarrow \left((\lambda^2-1)|\mathbf{x}|^2 - 2(\lambda^2 \mathbf{b}-\mathbf{a})\cdot \mathbf{x}+\lambda^2|\mathbf{b}|^2-|\mathbf{a}|^2 \right ) \\
& \Longleftrightarrow \bigg(\, \left|\mathbf{x}-\frac{\lambda^2\mathbf{b}-\mathbf{a}}{\lambda^2-1}\right|^2 = \frac{\lambda^2|\mathbf{a-b}|^2}{(\lambda^2-1)^2} \bigg)\end{align}$
Find all posts by this user
Quote this message in a reply
05-16-2012, 07:18 PM
Post: #30
Forum Notes about {Rudin [Principle of Mathematical Analysis] Notes}
It is nice to group notes/exercise solutions according to Rudin's chapters/sections as we tried so. But this will produce very long math posts and make the MathJax rendering of the page(maximal 10 posts per page) very slow.

So we have to split long post into smaller ones while keep the material well organized.
Find all posts by this user
Quote this message in a reply
Post Reply 


Forum Jump:


Contact Us | Software Frontier | Return to Top | Return to Content | Lite (Archive) Mode | RSS Syndication