ИИ умеет с картинки делать LaTeX. Можно ли как-то автоматически преобразовать такой латекс в форумный латекс? Например, картинка из
темы
А вот LaTeX от ИИ, он отличается от форумного формата:
Код:
\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
\section*{Theorem 6.7}
\textbf{Taylor's Theorem with Remainder} \\
Let \( f \) be a function that can be differentiated \( n + 1 \) times on an interval \( I \) containing the real number \( a \). Let \( p_n \) be the \( n \)-th Taylor polynomial of \( f \) at \( a \) and let
\[
R_n(x) = f(x) - p_n(x)
\]
be the \( n \)-th remainder. Then for each \( x \) in the interval \( I \), there exists a real number \( c \) between \( a \) and \( x \) such that
\[
R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x - a)^{n+1}.
\]
If there exists a real number \( M \) such that \( |f^{(n+1)}(x)| \leq M \) for all \( x \in I \), then
\[
|R_n(x)| \leq \frac{M}{(n+1)!} |x - a|^{n+1}
\]
for all \( x \) in \( I \).
\section*{Proof}
Fix a point \( x \in I \) and introduce the function \( g \) such that
\[
g(t) = f(x) - f(t) - f'(t)(x - t) - \frac{f''(t)}{2!}(x - t)^2 - \cdots - \frac{f^{(n)}(t)}{n!}(x - t)^n - R_n(x) \frac{(x - t)^{n+1}}{(x - a)^{n+1}}.
\]
We claim that \( g \) satisfies the criteria of Rolle's theorem. Since \( g \) is a polynomial function (in \( t \)), it is a differentiable function. Also, \( g \) is zero at \( t = a \) and \( t = x \) because
\[
\begin{align*}
g(a) &= f(x) - f(a) - f'(a)(x - a) - \frac{f''(a)}{2!}(x - a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x - a)^n - R_n(x) \\
&= f(x) - p_n(x) - R_n(x) \\
&= 0, \\
g(x) &= f(x) - f(x) - 0 - \cdots - 0 \\
&= 0.
\end{align*}
\]
\end{document}
Бот сделал всё чётко, проверить можно здесь
https://www.overleaf.com/
