\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{Pseudofourier} \[\hat{g}(f) = \frac{1}{{t_2  t_1}}\int_{t_1}^{t_2} g(t)e^{2\pi\,ift}\,dt\] \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 frictionless 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 $\int_{0}^{\infty} t^{x}e^{t} \,dt = x\int_{0}^{\infty} t^{x1}e^{t} \,dt$. So the inverse is $\int_{0}^{\infty} t^{x1}e^{t} \,dt =\frac{\int_{0}^{\infty} t^{x}e^{t} \,dt}{x}$. $x$ 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.\\ \[1 + \sum_{a=1}^{2^n} \left\lfloor \sqrt[n]{\frac{n}{\sum_{b=1}^{a} \left\lfloor \left \cos \left( \frac{(b1)!+1}{b} \pi \right) \right \right\rfloor}} \right\rfloor\] \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}
LaTeX forum ⇒ Math & Science ⇒ package issue
 erikgobrrr
 Posts: 3
 Joined: Sun Sep 17, 2023 11:27 pm
package issue
I want to try displaying a doublestriked 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:
Last edited by erikgobrrr on Mon Sep 18, 2023 2:50 am, edited 2 times in total.

 Posts: 706
 Joined: Tue Mar 25, 2008 5:02 pm
package issue
The package you meant is "amsmath" not "asmmath" (AMS = American Mathematical Society). There is also the bbm package to get "blackboard bold" characters.
 erikgobrrr
 Posts: 3
 Joined: Sun Sep 17, 2023 11:27 pm
package issue
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 doublestrike R (ℝ) like i showed in the main post?

 Posts: 706
 Joined: Tue Mar 25, 2008 5:02 pm
package issue
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 doublestrike 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.
 erikgobrrr
 Posts: 3
 Joined: Sun Sep 17, 2023 11:27 pm
package issue
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 doublestrike 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!
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