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How can I convert the old tex file below into a pdf? I would very much like to know as I have a few other files that I would like to convert. Thanks.

PS: I selected the shortest example I had to hand.

\hrule
\vskip 2cm
\centerline{\magtwotenrm Some reciprocal summation identities with 
applications to the}
\centerline{\magtwotenrm Fibonacci and Lucas numbers}

\vskip 1cm

\centerline{\magonetenit Derek Jennings}
\centerline{\magonetenrm Faculty of Mathematics, University of Southampton,}
\centerline{\magonetenrm Southampton, Hampshire, SO17 1BJ, England.}

\vskip 2cm

\centerline{\magtwotenrm 1: Introduction} 
\vskip 1cm 

\noindent In this arcticle we use theta functions and simple telescoping
of series to produce some reciprocal summation results for the Fibonacci
and Lucas numbers. The two results that we prove are the following:
\gap
{\noindent\bf Theorem 1}
$${\left( \sum_{n=1}^\infty {1 \over F_{2n-1}} \right)}^2 -
\sum_{n=1}^\infty {1 \over F_{2n-1}^2} = 
2\sum_{n=1}^\infty {1 \over F_{4n-2}^2}$$
\gap
{\noindent\bf Theorem 2} 
$$\sum_{n=1}^\infty {\left( {2^n \over L_{2^n} } \right)}^2 = {4 \over 5}$$
\gap
\noindent $F_n$ and $L_n$ are the Fibonacci and Lucas numbers respectively,
satisfying the usual recurrence $U_{n+1}=U_n+U_{n-1}$ where $F_0=0$, $F_1=1$,
$L_0=2$ and $L_1=1$.
\vskip 2cm

\centerline{\magtwotenrm 2: Proof of Theorem 1} 
\vskip 1cm
\noindent By simple series rearrangement we have 

$$\sum_{n=1}^\infty {q^{2n-1} \over (1+q^{2n-1})^2} = 
\sum_{n=1}^\infty (-1)^{n-1}{ nq^n \over 1-q^{2n}}.$$

\noindent Replacing $q$ by $q^2$ in the above then splitting the summation 
over the odd and even numbers we obtain

$$\eqalign{
\sum_{n=1}^\infty {q^{4n-2} \over (1+q^{4n-2})^2} &= 
\sum_{n=1}^\infty (-1)^{n-1}{ nq^{2n} \over 1-q^{4n}} \cr
&= \sum_{n=1}^\infty (2n-1){q^{4n-2} \over 1-q^{8n-4}} -
2\sum_{n=1}^\infty {nq^{4n} \over 1-q^{8n}}. \cr }\eqno(1)$$

\noindent Now a theorem originally due to Guass gives

$$\sum_{n=1}^\infty (2n-1){q^{8n-4} \over 1-q^{16n-8}} =
{ \left\{ {1 \over 2} \theta_2(q^4) \right\} }^4$$

\noindent where 

$$\theta_2(q) = \sum_{-\infty}^\infty q^{(n+1/2)^2} =
2q^{1/4} \prod_{n=1}^\infty(1-q^{2n})(1+q^{2n})^2.$$

\noindent Therefore 

$$\sum_{n=1}^\infty (2n-1){q^{4n-2} \over 1-q^{8n-4}} = 
{\left( \sum_{n=1}^\infty {q^{2n-1} \over 1+q^{4n-2} } \right)}^2,\eqno(2)$$

\noindent using 

$$\theta_2(q^2)^2 = 4 \sum_{n=1}^\infty {q^{2n-1} \over 1+q^{4n-2} }.$$

\noindent The last equality follows directly from equating the coefficient
of $x$ in (2) of {\bf [1]}.

\noindent Also, by simple series rearrangement we have 

$$\sum_{n=1}^\infty { nq^n \over 1-q^{2n}} =
\sum_{n=1}^\infty {q^{2n-1} \over (1-q^{2n-1})^2},$$

\noindent which on replacing $q$ by $q^4$ gives

$$\sum_{n=1}^\infty { nq^{4n} \over 1-q^{8n}} =
\sum_{n=1}^\infty {q^{8n-4} \over (1-q^{8n-4})^2}.\eqno(3)$$

\noindent Using (2) and (3) to substitute for the summations in the RHS of
(1) gives

$$\sum_{n=1}^\infty {q^{4n-2} \over (1+q^{4n-2})^2}=
{\left( \sum_{n=1}^\infty {q^{2n-1} \over 1+q^{4n-2} } \right)}^2 -2
\sum_{n=1}^\infty {q^{8n-4} \over (1-q^{8n-4})^2}.\eqno(4)$$

\noindent Setting $q=(1-\sqrt 5)/2$ in (4), and using the Binet form 
$F_n = (\alpha^n - \beta^n)/(\alpha -\beta)$ of the Fibonacci numbers
($\alpha = (1+\sqrt 5)/2$, $\beta = (1-\sqrt 5)/2$) we obtain Theorem 1.
\vskip 2cm

\centerline{\magtwotenrm 3: Proof of Theorem 2} 
\vskip 1cm
\noindent Our starting point is the identity 

$${q \over (1+q)^2} + {4q^2 \over (1-q^2)^2 } = {q \over (1-q)^2}.$$

\noindent This telescopes to

$${q \over (1+q)^2} + {4q^2 \over (1+q^2)^2 } + {16q^4 \over (1-q^4)^2}
= {q \over (1-q)^2}$$

\noindent and continuing the expansion process we arrive at

$$\sum_{n=0}^\infty {2^{2n} q^{2^n} \over (1+q^{2^n})^2} = 
{q \over (1-q)^2}.$$

\noindent Now we put $q=(1-\sqrt 5)/2$ in the above identity and use the
Binet form of $L_n = \alpha^n + \beta^n$ of the Lucas numbers ($\alpha$,
$\beta$ as in section 2) to obtain Theorem 2.
\vskip 2cm

\centerline{\magonetenrm References}
\vskip .75cm
\item{[1]} D.Jennings {\sl An identity for the Fibonacci and Lucas numbers}
Glasgow Math. J. {\bf 35} (1993) 381--384.
\vfill\eject
share|improve this question
    
Could whoever voted to close this please remember that it's important to leave a comment explaining why. (In this case, I don't see a reason to close.) –  Loop Space Aug 14 '10 at 17:54
    
I voted to close, because this question is very specialized, asking to reconstruct one person's forgotten set of macros. It is very unlikely to be useful to anybody except the asker. I didn't comment because I thought the "too localized" tag, "...an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet" didn't need any further clarification. –  Lev Bishop Aug 14 '10 at 19:37
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2 Answers 2

I would built a small LaTeX frame document for this purpose and use \input to include such tex code. Do know the definition of commands like \magtwotenrm, \magonetenit etc.? If you don't have it, define them by \newcommand. These look like font declarations \rm, \it.

The plain bundle contains files to use plain TeX code in LaTeX documents. So, together with Carls lines and plain.tex I could create a pdf document with this LaTeX code:

\documentclass{article}
\input{plain}
\font\magtwotenrm=cmr10 scaled\magstep2
\let\magonetenrm\rm
\let\magonetenit\it
\def\gap{\vskip16pt}
\begin{document}
\input{file1}
\end{document}

Of course you could use pdftex instead if you just want to get a pdf. But if you would like to modify page dimensions or the content, you could go further with LaTeX.

To convince that it works, heres the pdf output of that file: main.pdf. I still had to confirm some errors but the pdf has been produced. Now the formatting could be enhanced.

share|improve this answer
1  
I no longer have the exact definitions of those commands, which is part of my problem. I believe they define certain fonts along with their size. I'll play around with it a bit more, in the light of your advice, and see if I have any more success. Thanks. –  Derek Jennings Aug 14 '10 at 12:52
    
Derek, I've added example code and output. –  Stefan Kottwitz Aug 14 '10 at 13:34
    
Stefan, I'm convinced. Amazing! Thanks. –  Derek Jennings Aug 14 '10 at 13:42
    
Stefan, I would like to vote up your answer but I don't have the reputation points. –  Derek Jennings Aug 14 '10 at 13:45
    
Don't worry, I'm glad that it helped. –  Stefan Kottwitz Aug 14 '10 at 13:47
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Here's a quick and dirty way. Put the following at the top of the file:

\font\magtwotenrm=cmr10 scaled\magstep2
\let\magonetenrm\rm
\let\magonetenit\it
\def\gap{\vskip16pt}

Then put "\bye" at the bottom of the file, and run it through pdftex.

The "mag" commands were intended to change the size of the text, and you can redefine them to scale differently if you want.

share|improve this answer
    
Looks like a quick way to get started. –  Joseph Wright Aug 14 '10 at 13:16
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