
\begin{align*}
P_n(x+1) &=\lim_{\epsilon\to 0} \frac{1}{2^2 n!}~\frac{d^n}{d\epsilon^n}[\epsilon^n(2+\epsilon)^n]\\
&=\lim_{\epsilon \to 0} \frac{1}{2^2 n!}~\frac{d^n}{d\epsilon^n} [ 2^n\epsilon^n +\underbrace{\text{higher order terms in \epsilon}}_{0(\epsilon^{n+1})} ]\\
&=\lim_{\epsilon\to 0} \frac{1}{2^2 n!}~2n~\frac{d^n\epsilon^n}{d\epsilon^n} + \underbrace{\frac{d^n 0}{d\epsilon^n}(\epsilon^{n+1})...}_{\text{vanishes as} \epsilon \to 0}
\end{align*}
Eu continuo recebendo:
! Faltando
$
inserido.$
1.166\end{align*}
não sei onde colocar $
. Achei que não precisava de um porque align*
já está no modo matemática? Eu tive um bloco semelhante anteriormente no documento onde coloquei:
\begin{align*}
\frac{d^{2n}}{dx^{2n}} (x^2-1)^n &= \frac{d^{2n}}{dx^{2n}}(x^{2n}+\text{lower order terms which vanish under 2n differentiation})\\
&=(2n)! + 0\\
\therefore\int_{-1}^1 P_n(x)P_n(x)dx &= \frac{(2n)!(-1)^n}{(2^n n!)^2}\int_{-1}^1 (1-x^2)^n dx
\end{align*}
e saiu bem.
Responder1
O \epsilon
underbrace precisa estar no modo matemático:
\documentclass{article}
\usepackage{amsmath}
\newcommand*\differential{\mathop{}\!\mathrm{d}}
\newcommand*\diff[3][]{\frac{\differential^{#1} #2}{\differential #3}}
\begin{document}
\begin{align*}
P_{n}(x + 1)
&= \lim_{\epsilon \to 0} \frac{1}{2^{2}n!}\diff[n]{}{\epsilon^{n}}{[\epsilon^{n}(2 + \epsilon)^{n}]}\\
&= \lim_{\epsilon \to 0} \frac{1}{2^{2}n!}\diff[n]{}{\epsilon^{n}}{[2^{n}\epsilon^{n} + \underbrace{\text{higher order terms in $\epsilon$}}_{o(\epsilon^{n + 1})}]}\\
&= \lim_{\epsilon \to 0} \frac{1}{2^{2}n!}2n\diff[n]{\epsilon^{n}}{\epsilon^{n}} + \underbrace{\diff[n]{o(\epsilon^{n + 1})}{\epsilon^{n}}}_{\substack{\text{vanishes as}\\ \epsilon \to 0}}
\end{align*}
\end{document}
Observe as melhorias no código. (Alguns deles seguem o comentário de Mico abaixo.)