我正在写一篇数学论文，在写了一些例子之后，我用来描述某些现象的纯文本变成了斜体。如何让文档使用默认字体书写？
您可以看到斜体字从 0.1.2 预备阶段开始。以下是该论文的代码
\documentclass[]{book} %These tell TeX which packages to use. \usepackage{array,epsfig} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsxtra} \usepackage{amsthm} \usepackage{mathrsfs} \usepackage{color} %Here I define some theorem styles and shortcut commands for symbols I use often \theoremstyle{definition} \newtheorem{defn}{Definition} \newtheorem{thm}{Theorem} \newtheorem{cor}{Corollary} \newtheorem*{rmk}{Remark} \newtheorem{lem}{Lemma} \newtheorem*{joke}{Joke} \newtheorem{ex}{Example} \newtheorem*{soln}{Solution} \newtheorem{prop}{Proposition} \newcommand{\lra}{\longrightarrow} \newcommand{\ra}{\rightarrow} \newcommand{\surj}{\twoheadrightarrow} \newcommand{\graph}{\mathrm{graph}} \newcommand{\bb}[1]{\mathbb{#1}} \newcommand{\Z}{\bb{Z}} \newcommand{\Q}{\bb{Q}} \newcommand{\R}{\bb{R}} \newcommand{\C}{\bb{C}} \newcommand{\N}{\bb{N}} \newcommand{\M}{\mathbf{M}} \newcommand{\m}{\mathbf{m}} \newcommand{\MM}{\mathscr{M}} \newcommand{\HH}{\mathscr{H}} \newcommand{\Om}{\Omega} \newcommand{\Ho}{\in\HH(\Om)} \newcommand{\bd}{\partial} \newcommand{\del}{\partial} \newcommand{\bardel}{\overline\partial} \newcommand{\textdf}[1]{\textbf{\textsf{#1}}\index{#1}} \newcommand{\img}{\mathrm{img}} \newcommand{\ip}[2]{\left\langle{#1},{#2}\right\rangle} \newcommand{\inter}[1]{\mathrm{int}{#1}} \newcommand{\exter}[1]{\mathrm{ext}{#1}} \newcommand{\cl}[1]{\mathrm{cl}{#1}} \newcommand{\ds}{\displaystyle} \newcommand{\vol}{\mathrm{vol}} \newcommand{\cnt}{\mathrm{ct}} \newcommand{\osc}{\mathrm{osc}} \newcommand{\LL}{\mathbf{L}} \newcommand{\UU}{\mathbf{U}} \newcommand{\support}{\mathrm{support}} \newcommand{\AND}{\;\wedge\;} \newcommand{\OR}{\;\vee\;} \newcommand{\Oset}{\varnothing} \newcommand{\st}{\ni} \newcommand{\wh}{\widehat} %Pagination stuff. \setlength{\topmargin}{-.3 in} \setlength{\oddsidemargin}{0in} \setlength{\evensidemargin}{0in} \setlength{\textheight}{9.in} \setlength{\textwidth}{6.5in} \pagestyle{empty} \begin{document} \begin{center} {\Large Hypergeometric Bernoulli Numbers via Umbral Calculus}\\ \textbf{Christopher Ernst}\\ %You should put your name here Begun: 5-5-2017 %You should write the date here. \end{center} \vspace{0.2 cm} \section{Introduction} \subsection{Preliminaries - Action of a Formal Power Series on Polynomials} Let $f(t)=\sum_{j=0}^\infty{a_j\frac{t^j}{j!}}\in \mathcal{F}$ be a formal power series where $a_j\in \mathbb{F}$ where $char(\mathbb{F})=0$ and let $p(x)=\sum_{k=0}^n{c_kx^k}\in P$, the algebra of polynomials of a single variable over $\mathbb{C}$. We know then that $P^*$ is the vector space of all linear functionals on $P$. The notation $$ \langle L \mid p(x) \rangle $$ is used to denote the action of a linear functional $L$ on a polynomial $p(x)$. As $P^*$ is a vector space, we have by linearity that for linear functionals $L$ and $M$ and field elements $a,b\in \mathbb{C}$, $$ \langle aL+bM \mid p(x) \rangle = a\langle L \mid p(x) \rangle + b\langle M \mid p(x) \rangle $$ It is important then to note that since linear functionals are uniquely determined by the action on a basis, we know that $L$ is uniquely determined by the sequence of constants $ \langle L \mid x^n \rangle $. The formal power series $f$ defined above defines a linear functional on $P$ if we define the following; \begin{equation} \langle f(t) \mid x^n \rangle = a_n \end{equation} for all nonnegative integers. Thus, the formal power series $t^k$ gives $$\langle t^j \mid x^n \rangle=n!\delta_{n,j}$$ where $\delta_{n,j}$ is the Kronecker delta. Something to take note of then is that any linear functional $L\in P^*$ has the form of $f(t)$. If then \begin{equation} f_L(t)=\sum_{j=0}^\infty {\langle L \mid x^j \rangle}\frac{t^j}{j!} \end{equation} we have that $f_L(t)=L$ as linear functionals. Defining a map $L\rightarrow f_L(t)$, it can be shown this map is a vector space isomorphism and we can now think of linear functionals as formal power series. Let $$h(t)=f(t)+g(t)=\sum_{j=0}^\infty{a_j\frac{t^j}{j!}}+\sum_{j=0}^\infty{b_j\frac{t^j}{j!}}=\sum_{j=0}^\infty{(a_j+b_j)\frac{t^j}{j!}}$$ and $$ j(t)=f(t)g(t)=\left(\sum_{j=0}^\infty{a_j\frac{t^j}{j!}}\right)\left(\sum_{j=0}^\infty{b_j\frac{t^j}{j!}}\right)= \sum_{j=0}^\infty\left[\sum_{i=0}^j\binom{j}{i}a_ib_{j-i}\right]\frac{t^j}{j!}$$ Then \begin{equation} \langle h(t) \mid x^n \rangle = a_n+b_n \end{equation} and \begin{equation} \langle j(t) \mid x^n \rangle = \sum_{i=0}^j\binom{j}{i}a_ib_{j-i} \end{equation} Noticing then the relationship between the formal power series have with the linear functionals, we can show then that \begin{equation} \langle f+g \mid x^n \rangle = \langle f \mid x^n \rangle + \langle g \mid x^n \rangle \end{equation} \begin{equation} \langle fg \mid x^n \rangle = \sum_{k=0}^n\binom{n}{k}\langle f \mid x^k \rangle \langle g \mid x^{n-k} \rangle \end{equation} and it is these operations that allow us to extend our vector space idea and can now label $\mathcal{F}$ as an algebra structure. In fact, this structure is known as the \it{umbral algebra}. \begin{ex} $f(t)=e^{yt}$ $$\langle e^{yt}\mid x^n \rangle =\left\langle \sum_{j=0}^\infty \frac{(yt)^j}{j!} \middle| x^n\right\rangle=\left\langle \sum_{j=0}^\infty y^j\frac{t^j}{j!} \middle| x^n\right\rangle=y^n$$ and $$\langle e^{yt} \mid p(x) \rangle = \left\langle e^{yt} \middle| \sum_{k=0}^nc_kx^k \right\rangle=\sum_{k=0}^n{c_k}\langle e^{yt} \mid x^k \rangle=\sum_{k=0}^n{c_k}y^k=p(y)$$ \end{ex} \begin{ex} $g(t)=f'(t)$ $$\langle g(t) \mid x^n \rangle =\langle f'(t) \mid x^n \rangle =\left\langle \frac{d}{dx}\sum_{j=0}^\infty a_j\frac{t^j}{j!} \middle| x^n\right\rangle=\left\langle \sum_{j=0}^\infty a_{j+1}\frac{t^j}{j!} \middle| x^n\right\rangle=a_{n+1}$$ \end{ex} \begin{prop} $$\langle f^{(k)}(x) \mid x^n \rangle=a_{n+k}$$ \end{prop} \begin{ex} $f(t)=\frac{e^x-1}{x}$ Note that $$\frac{e^x-1}{x}=\frac{1}{x}\sum_{j=1}^\infty\frac{x^j}{j!}=\sum_{j=1}^\infty\frac{x^{j-1}}{j!}=\sum_{j=0}^\infty\frac{x^j}{(j+1)!}=\sum_{j=0}^\infty{\frac{1}{(j+1)}}\frac{x^j}{j!}$$ Therefore, $$\left\langle \frac{e^x-1}{x} \middle| x^n \right\rangle = \frac{1}{n+1}$$ \end{ex} \begin{ex} $f(t)=\frac{e^x-1-x}{x^2}$ Note that $$\frac{e^x-1-x}{x^2}=\frac{1}{x^2}\sum_{j=2}^\infty\frac{x^j}{j!}=\sum_{j=2}^\infty\frac{x^{j-2}}{j!}=\sum_{j=0}^\infty\frac{x^j}{(j+2)!}=\sum_{j=0}^\infty{\frac{1}{(j+2)(j+1)}}\frac{x^j}{j!}$$ Therefore, $$\left\langle \frac{e^x-1-x}{x^2} \middle| x^n \right\rangle = \frac{1}{(n+2)(n+1)}$$ \end{ex} \begin{prop} It makes sense then that if $T_n(x)$ is the Taylor polynomial $$T_n(x)=\sum_{j=0}^n\frac{x^j}{j!}$$ then $$\left\langle \frac{e^x-T_{k-1}(x)}{x^k} \middle| x^n \right\rangle = \frac{1}{(n+k)_k}$$ where $(n+k)_k=(n+k)(n+k-1)...(n+2)(n+1)$ \end{prop} \begin{ex} Replacing $x^n$ with an arbitrary polynomials $p(x)=c_0+c_1x+...+c_nx^n$ \begin{eqnarray*} \left\langle \frac{e^x-1}{x} \middle| p(x) \right\rangle &=& >\left\langle \frac{e^x-1}{x} \middle| c_0+c_1x+...+c_nx^n \right\rangle \\ >&=&c_0\left\langle \frac{e^x-1}{x} \middle| 1 \right\rangle+c_1\left\langle \frac{e^x-1}{x} \middle| x \right\rangle+...+c_n\left\langle \frac{e^x-1}{x} \middle| x^n \right\rangle \\ &=&\frac{c_0}{1}+\frac{c_1}{2}+...+\frac{c_n}{n+1} \\ &=&\int_0^1[c_0+c_1x+...+c_nx^n]dx \\ &=&\int_0^1p(x)dx \end{eqnarray*} \end{ex} \begin{ex} \begin{eqnarray*} \left\langle \frac{e^x-1-x}{x^2} \middle| p(x) \right\rangle &=& \left\langle \frac{e^x-1-x}{x^2} \middle| c_0+c_1x+...+c_nx^n \right\rangle \\ &=&c_0\left\langle \frac{e^x-1-x}{x^2} \middle| 1 \right\rangle+c_1\left\langle \frac{e^x-1-x}{x^2} \middle| x \right\rangle+...+c_n\left\langle \frac{e^x-1-x}{x^2} \middle| x^n \right\rangle \\ &=&\frac{c_0}{1\cdot 2} +\frac{c_1}{2\cdot 3}+...+\frac{c_n}{(n+1)(n+2)} \\ >&=&\int_0^1\left[\int_0^1(c_0+c_1y+...+c_ny^n)dy\right]dx \\ &=&\int_0^1\int_0^1p(y)dydx \end{eqnarray*} \end{ex} \begin{prop} $$\left\langle \frac{e^x-T_{k-1}(x)}{x^k} \middle| p(x) \right\rangle =\underbrace{\int_0^1 \int_0^1 \int_0^1}_{k\text{ times}}p(x_1)dx_1dx_2...dx$$ \end{prop} \subsection{Preliminaries - Formal Power Series as Linear Operators} Let \begin{equation} t^kx^n=(n)_kx^{n-k} \end{equation} It is clear that $t^k$ then acts as the $k$-th derivative on the monomial $x^n$. Now if we consider a formal power series acting on $x^n$, we see that \end{document}
编辑：我知道这可能会被人反对，但我还是把整个代码贴出来了。我对 TeX 还比较陌生，而且我自学的，从未正式教过，所以我很难根据我看到的回复找到问题所在。如果我这样做违反了任何规则，我很抱歉。