## Wednesday, November 21, 2007

### What Complex proof?

You may or may not recall the proof which I posted about some time ago. It was for a theorem from Complex Analysis, but naturally it holds in real analysis as well. I had been trying to prove the theorem by contradiction (i.e. me being perverse) but this had led me to a problem, of which I may have a solution (but need to write it up).

However, as I don't like to leave shoe laces untied (NOTE: If you know me then please don't say that I walk around with mine untied!) OK fine, I will correct myself and say that on this occasion I feel it neccessary to tie the laces! (In my defense it can annoying having to bend down and tie your shoe laces every other minute...)

So now I will provide a conventional proof for the theorem. This theorem was presented to us in real analysis and a rather long and dodgy proof was given for it. I say dodgy because we had to choose epsilon as one and then play around. I had asked DC whether or not the nice and short proof that I will write, was also correct. It was and hence in yesterdays lecture he gave that to us too.

Alas, once again due to my perverse nature (I am liking that word more and more!) I will not be posting the proof in the normal way in this post. That is I am going to let you do the LaTeXing!

For those without TeXnic centre, winedt or any other editor, just copy past everything in between the dashed lines into this link and then click submit. (Trust me, you wouldn't NOT want to do this! There might be a mistake lurking...) Having done that a document should appear with the nice proof.

Unfortunately due to my not using LaTeX for quite some time (darn these coursework tests), I seem to have made a mess of a few things. I don't really have that much of an understanding of my preamble anymore, but I can claim to once having a slight idea. Leave nothing unquestioned (apart from the black square at the end)!

------------------------------------------------------------------------------------

\documentclass[titlepage, a4paper]{article}
\oddsidemargin=0in
\usepackage{fancyhdr}
\usepackage{amssymb}
\usepackage{upquote}
\usepackage{amsmath}
\usepackage[top=3.5cm,bottom=2.7cm,text={6.5in,9in},centering]{geometry} %can change lengths to suit own needs
\usepackage[pdftex]{graphicx}
\pagestyle{fancy}
\cfoot{\thepage} % page number?
\parindent=0in %when we press enter for a new paragraph we don't have to write \ noindent.
\date{}
\begin{document}

\textbf{Theorem}\\
If $f$ is differentiable at $a\in \mathbb{R}$ then $f$is continuous at $a$.\\

\textbf{Proof}\\
Consider $$f(x) - f(a) = \frac{f(x)-f(a)}{x-a} \times (x-a) \quad a \neq 0.$$

Let $x\to a$ so,

$$\begin{array}{cc}\displaystyle \lim_{x \to a} \frac{f(x)-f(a)}{x-a} = f'(a) & \text{ which exists by our assumption and}\\*[2ex] \lim_{x \to a} (x-a) =0.& \end{array}$$\\

Since both limits exist we can use the product rule for limits to say that,

$$\begin{array}{ccc}\displaystyle \lim_{x \to a} \left ( \frac{f(x) - f(a)}{x-a} (x-a) \right ) & = & \displaystyle \lim_{x \to a}\frac{f(x)-f(a)}{x-a} \times \lim_{x \to a} (x-a) \\*[2ex] &=& f'(a) \times 0\\*[2ex] &=& 0\\ \end{array}.\\$$

Hence $\displaystyle \lim_{x \to a} (f(x) - f(a)) =0$ i.e. $\displaystyle \lim_{x \to a} f(x) = f(a)$.\\

Thus $f$ is continuous at $x=a$ as required. \\
\flushright{$\blacksquare$ }
\end{document}
-------------------------------------------------------------------------------------------------

Wow - that took ages.

Now I must retire to my misery, for silly Mac has made me feel this way. And the premiership is meant to be one of the best leagues in the world... I need to start writing my speech as well, but maybe another day!

PS: I will be posting the proof with LaTeX on my blog if you haven't managed to see it (or couldn't be bothered). And what say you if I was England's manager?

Jake said...

Just a small typo (s/this/thus) and a couple of small LaTeX errors (black square needs $s, a $$missing etc.) ...this should work fine: \documentclass[titlepage, a4paper]{article} \oddsidemargin=0in \usepackage{fancyhdr} \usepackage{amssymb} \usepackage{upquote} \usepackage{amsmath} \usepackage[top=3.5cm,bottom=2.7cm,text={6.5in,9in},centering]{geometry} %can change lengths to suit own needs \usepackage[pdftex]{graphicx} \pagestyle{fancy} \rhead{Your name} \lhead{document details} \cfoot{\thepage} % page number? \parindent=0in %when we press enter for a new paragraph we don't have to write \ noindent. \date{} \begin{document} \textbf{Theorem}\\ If f is differentiable at a\in \mathbb{R} then f is continuous at a.\\ \textbf{Proof}\\ Consider$$f(x) - f(a) = \frac{f(x)-f(a)}{x-a} \times (x-a) \quad a \neq 0.$$Let x\to a so,$$ \begin{array}{cc}\displaystyle \lim_{x \to a} \frac{f(x)-f(a)}{x-a} = f'(a) & \text{ which exists by our assumption and}\\*[2ex] \lim_{x \to a} (x-a) =0.& \end{array}$$\\ Since both limits exist we can use the product rule for limits to say that,$$ \begin{array}{ccc}\displaystyle \lim_{x \to a} \left ( \frac{f(x) - f(a)}{x-a} (x-a) \right ) & = & \displaystyle \lim_{x \to a}\frac{f(x)-f(a)}{x-a} \times \lim_{x \to a} (x-a) \\*[2ex] &=& f'(a) \times 0\\*[2ex] &=& 0\\ \end{array}.$$\\ Hence \displaystyle \lim_{x \to a} (f(x) - f(a)) =0 i.e. \displaystyle \lim_{x \to a} f(x) = f(a).\\ Thus f is continuous at x=a as required. \\ \flushright{\blacksquare} \end{document} Беанс said... Thanks, I have edited it. I thought you might have changed the black square to a white one for a second... ;) steve said... I hope it helps but I have updated your code as there were a number of things that could do with improving. There now should be no warnings or bad boxes. Just a few of the changes: 1.$$...$$is no longer used and has been replaced by $...$ (see Why use $...$ in place of$$ ...$? 2. Use amsthm to automate your theorems and proofs. For example it automatically italicises the statement of the theorem and auto adds a qed symbol. 3. Don't use array to lay out your equations. That's exactly what align (or in this case align*) is designed for. It knows how to vertically space the equations, lines up the equals signs and puts everything in display mode without requiring $...$ or \displaystyle. I hope you agree aligned equations look so much better. 4. LaTeX is not like other programming languages. If you want to set a length you need to use \setlength; you can't just write \variable=0. 5. The geometry package complained about the text settings as you had overspecified them, so they are left out. So here's the new code: \documentclass[titlepage, a4paper]{article} \usepackage{fancyhdr} \usepackage{amsmath,amssymb,amsthm} %amsthm is ued for theorems and proofs \usepackage{upquote} \usepackage[top=3.5cm,bottom=2.7cm,centering]{geometry} %can change lengths to suit own needs \usepackage[pdftex]{graphicx} \pagestyle{fancy} \rhead{Your name} \lhead{document details} \cfoot{\thepage} % page number? \setlength{\parindent}{0in} %when we press enter for a new paragraph we don't have to write \ noindent. \date{} \begin{document} \newtheorem*{theorem}{Theorem} %set up theorem environment * means don't want theorem number \renewcommand{\qedsymbol}{\ensuremath{\blacksquare}} %change automated qed symbol to a black square \begin{theorem} Iff$is differentiable at$a\in \mathbb{R}$then$f$is continuous at$a$. \end{theorem} \begin{proof} Consider $f(x) - f(a) = \frac{f(x)-f(a)}{x-a} \times (x-a) \quad a \neq 0.$ Let$x\to aso, \begin{align*} \lim_{x \to a} \frac{f(x)-f(a)}{x-a} &= f'(a) \text{ which exists by our assumption and}\\ \lim_{x \to a} (x-a) &=0. \end{align*} Since both limits exist we can use the product rule for limits to say that, \begin{align*} \lim_{x \to a} \left ( \frac{f(x) - f(a)}{x-a} (x-a) \right ) & = \lim_{x \to a}\frac{f(x)-f(a)}{x-a} \times \lim_{x \to a} (x-a) \\ &=f'(a) \times 0\\ &= 0 \end{align*} Hence \displaystyle \lim_{x \to a} (f(x) - f(a)) =0$i.e.$ \displaystyle \lim_{x \to a} f(x) = f(a)$. \vspace{2ex} Thus$f$is continuous at$x=a\$ as required.
\end{proof}
\end{document}

Беанс said...

Hi Steve,

Thanks a lot for all the updates. For the first time ever, I think I produced a document with no warnings and bad boxes!

I remember having a discussion with you about the use of array in the AoPS forum. I seem to keep on remembering array rather than align (and had thought that array was the nicer one to use!)

Is the statement of the theorem italicised by default? What would I do if I didn't want it like that? (And would I just write \textbf{\begin{Proof}} if I wanted that in bold? )

The updated code looks much better too, and it's easier to follow as well. (No messing around with \displaystyle!) I seem to want a coursework now, just so I can use LaTeX.

steve said...

Is the statement of the theorem italicised by default? What would I do if I didn't want it like that? (And would I just write \textbf{\begin{Proof}} if I wanted that in bold? )

If you want the theorem style not italicised then before \newtheorem* add the line

\theoremstyle{definition}

(there are 3 predefined theoremstyles: plain, definition and remark to try out).

For proof you add what you want to see using []. For upright bold text use

\begin{proof}[\textup{\textbf{Proof}}]

or, you can play and use something silly like

\begin{proof}[\huge{Isn't this an amazing Proof!}]

to see how easy it is to change things.

Jake said...

I thought you might have changed the black square to a white one for a second... ;)

LOL. I just don't like drawing black squares; I don't mind typed ones.

Thanks for the amended code Steve, looks pretty pro so I will use it as a template for typing up notes and the like.

Беанс said...

Hi Steve,

Thanks a lot!

However, I don't see what was so silly. ;) Although I would write:

\begin{proof}[\textup{\textbf{\huge{Isn't this an amazing Proof!}}}]

Беанс said...

Hi Jake,
I don't know how people can draw the empty ones and then NOT colour them in! They are asking to be coloured in when drawn.