我怎样才能将下面的旧 tex 文件转换为 pdf？我非常想知道，因为我还有其他几个文件需要转换。谢谢。
附言：我选择了我能找到的最短的例子。
\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
答案1
我将为此目的建立一个小型 LaTeX 框架文档，并用它来包含此类 tex 代码。您知道诸如、等\input命令的定义吗？如果您不知道，请通过 定义它们。这些看起来像字体声明、。\magtwotenrm\magonetenit\newcommand\rm\it
这普通捆包含在 LaTeX 文档中使用纯 TeX 代码的文件。因此，结合 Carls 行和 plain.tex，我可以使用以下 LaTeX 代码创建 pdf 文档：
\documentclass{article}
\input{plain}
\font\magtwotenrm=cmr10 scaled\magstep2
\let\magonetenrm\rm
\let\magonetenit\it
\def\gap{\vskip16pt}
\begin{document}
\input{file1}
\end{document}
当然，如果您只想获取 pdf，也可以使用 pdftex。但如果您想修改页面尺寸或内容，可以进一步使用 LaTeX。
为了证明其有效，以下是该文件的 pdf 输出：主页.pdf。我仍需确认一些错误，但 pdf 已生成。现在可以增强格式。
答案2
这是一个快速而简单的方法。将以下内容放在文件顶部：
\font\magtwotenrm=cmr10 scaled\magstep2
\let\magonetenrm\rm
\let\magonetenit\it
\def\gap{\vskip16pt}
然后将“\bye”放在文件底部，并通过 pdftex 运行它。
“mag”命令用于改变文本的大小，如果您愿意，可以重新定义它们以进行不同的缩放。