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erikgobrrr
Posts: 3
Joined: Sun Sep 17, 2023 11:27 pm

package issue

Postby erikgobrrr » Sun Sep 17, 2023 11:34 pm

I want to try displaying a double-striked R (ℝ) but nothing works. I can't paste the letter in, the packages fail to get added, I don't know what to do. Here's my code:
\documentclass{article}
\usepackage{graphicx} % Required for inserting images
\usepackage{hyperref}
\usepackage{asmmath} % Problem happens here
\title{The equations that I know}
\author{Made by Erik Andrade Lins}
\date{July 2023}

\begin{document}

\maketitle
\begin{center}
    Version 1
\end{center}
\section{Introduction}
In this file made with this \href{https://www.overleaf.com}{\LaTeX\,editor}, I'll be showing you some equations that I know.

\section{The equations}
\subsection{Fourier transforms}
Credits to \href{https://www.youtube.com/watch?v=spUNpyF58BY}{3Blue1Brown} for the fourier transform formulas.
\subsubsection{Pseudo-fourier}
\[\hat{g}(f) = \frac{1}{{t_2 - t_1}}\int_{t_1}^{t_2} g(t)e^{-2\pi\,ift}\,dt\]
\subsubsection{Actual}
This one is exactly the same as the one on section 2.1.1 except that the fraction $\frac{1}{t_2 - t_1}$ is not multiplied to the integral. The equation is:\\
\[\hat{g}(f) = \int_{t_1}^{t_2} g(t)e^{-2\pi\,ift}\,dt\]
\subsection{Exponentiating $e$}
\subsubsection{Taylor's expansion of $e^{x}$}
I credit \href{https://youtu.be/B1J6Ou4q8vE?t=454}{Alan} for this one. The equation is this:\\
\[e^{x} = \sum\limits_{n=0}^{\infty} \frac{x^{n}}{n!}\]
\subsubsection{Euler's formula for $e^{ix}$}
Also crediting \href{https://youtu.be/B1J6Ou4q8vE?t=256}{Alan}. The equation is $e^{ix} = \cos\,x + i\sin\,x$. Did you know that $e^{i\pi} = -1$?
\subsection{Newton's law of force}
$F$ is the force you require to move something of mass $m$ on a friction-less surface where the acceleration you want is $a$. The equation is $F = ma$, though you could write it as $F = am$.
\subsection{Exchange for mass and energy}
Everyone knows this equation. It was developed by Einstein and it's $E = mc^{2}$.
\subsection{Disproofs of commutation on division, exponentiation and subtraction}
Please note that the disproofs are no longer true if the equation $x = y$ is satisfied, though for division you need to be careful because if you define $x$ and/or $y$ to be $0$, you'll be handling with a division by $0$, which is undefined (It's exactly why $\int_{0}^{\infty} t^{x}e^{-t} \,dt$ has asymptotes. You have to divide by $0$ every time $x$ is a negative number to satisfy $\int_{0}^{\infty} t^{x}e^{-t} \,dt = x\int_{0}^{\infty} t^{x-1}e^{-t} \,dt$. So the inverse is $\int_{0}^{\infty} t^{x-1}e^{-t} \,dt =\frac{\int_{0}^{\infty} t^{x}e^{-t} \,dt}{x}$. $x$ in this case is $0$ so, it means to get $\int_{0}^{\infty} t^{-1}e^{-t} \,dt$, we have to divide by zero. If you wanted to know the exact fraction, here it is: $\frac{1}{0} = \int_{0}^{\infty} t^{-1}e^{-t} \,dt$ ($\int_{0}^{\infty} t^{0}e^{-t} \,dt$ is $1$).).
\subsubsection{Division}
$\frac{x}{y} \neq \frac{y}{x}$. We can prove this using proof by contradiction. Let's suppose that $\frac{x}{y} = \frac{y}{x}$, $x = 1$ and that $y = 2$. $\frac{1}{2}$ is $0.5$, but $\frac{2}{1}$ is $2$, implying that $\frac{1}{2} \neq \frac{2}{1}$ and that $\frac{1}{2} - \frac{2}{1} \neq 0$, disproving $\frac{x}{y} = \frac{y}{x}$.
\subsubsection{Exponentiation}
$x^{y} \neq y^{x}$. Again, we can prove this using proof by contradiction. Assume again that $x^{y} = y^{x}$, $x = 1$ and $y = 2$. $1^{2} = 1$, but again, $2^{1} = 2$, suggesting again that $1^{2} \neq 2^{1}$ and that $1^{2} - 2^{1} \neq 0$, again disproving $x^{y} = y^{x}$.
\subsubsection{Subtraction}
I think you know where this is going. $x - y \neq y - x$, because, again, using proof by contradiction and assuming that $x - y = y - x$, $x = 1$ and that $y = 2$, $1 - 2$ is  $-1$ but $2 - 1$ is $1$, indicating one last time that $x - y \neq y - x$ and that $(x - y) - (y - x) \neq 0$, contradicting $x - y = y - x$ again.
\subsection{The golden ratio}
$\phi$, which is commonly used to represent the golden ratio, can be calculated using $\frac{1 + \sqrt{5}}{2}$ or by solving for $x$ in $x^{2} = x + 1$ or in $\frac{1}{x} = x - 1$. Or you can also approximate it by making a fraction where a Fibonnachi number is the numerator and the one before it the denominator. The higher the number, the more precision.
\subsection{Prime computer}
This formula is for calculating the $n$th prime ($n$ must be a real positive integer that is not $\infty$). $n$ is determined by the person.\\
\[1 + \sum_{a=1}^{2^n} \left\lfloor \sqrt[n]{\frac{n}{\sum_{b=1}^{a} \left\lfloor \left| \cos \left( \frac{(b-1)!+1}{b} \pi \right) \right| \right\rfloor}} \right\rfloor\]
\subsection{Expanding $x!$ to the \mathbb{R}eals} % This is unfinished because of the problem
\section{End}
Thanks so much for viewing my document. Now go! This document is not copyrighted. Share it with others.
\end{document}
Last edited by erikgobrrr on Mon Sep 18, 2023 2:50 am, edited 2 times in total.

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kaiserkarl13
Posts: 707
Joined: Tue Mar 25, 2008 5:02 pm

package issue

Postby kaiserkarl13 » Mon Sep 18, 2023 1:27 am

The package you meant is "amsmath" not "asmmath" (AMS = American Mathematical Society). There is also the bbm package to get "blackboard bold" characters.

User avatar
erikgobrrr
Posts: 3
Joined: Sun Sep 17, 2023 11:27 pm

package issue

Postby erikgobrrr » Mon Sep 18, 2023 2:39 am

kaiserkarl13 wrote:The package you meant is "amsmath" not "asmmath" (AMS = American Mathematical Society). There is also the bbm package to get "blackboard bold" characters.

can you also give the command to put a double-strike R (ℝ) like i showed in the main post?

kaiserkarl13
Posts: 707
Joined: Tue Mar 25, 2008 5:02 pm

package issue

Postby kaiserkarl13 » Mon Sep 18, 2023 2:47 am

erikgobrrr wrote:
kaiserkarl13 wrote:The package you meant is "amsmath" not "asmmath" (AMS = American Mathematical Society). There is also the bbm package to get "blackboard bold" characters.

can you also give the command to put a double-strike R (ℝ) like i showed in the main post?


It's right there on the first page of the link to the bbm package documentation.

User avatar
erikgobrrr
Posts: 3
Joined: Sun Sep 17, 2023 11:27 pm

package issue

Postby erikgobrrr » Mon Sep 18, 2023 2:49 am

kaiserkarl13 wrote:
erikgobrrr wrote:
kaiserkarl13 wrote:The package you meant is "amsmath" not "asmmath" (AMS = American Mathematical Society). There is also the bbm package to get "blackboard bold" characters.

can you also give the command to put a double-strike R (ℝ) like i showed in the main post?


It's right there on the first page of the link to the bbm package documentation.

thanks!

bouamaied
Posts: 1
Joined: Tue May 28, 2024 6:40 pm

package issue

Postby bouamaied » Tue May 28, 2024 6:44 pm

hi, i receive this message in compilation:Package inputenc Error: Keyboard character used is undefined(inputenc) in inputencoding `latin1'. \newblock On the inversion of liapunovâ€; can you help me please; i think the problem in the package between latin and "utf8"


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