# Beginner LaTeX Guide

${\LaTeX}$ is an extremely useful typesetting language to learn, especially in a math environment like this. However, the quick instructions Brilliant.org gives just aren't good enough to use for most situations.

This is why I've decided to create a beginner's ${\LaTeX}$ guide. There is a table of contents for easy symbol or format finding. I hope you can refer to this guide later, when writing solutions, problems, or notes.

Note: You can also view Latex codes by hovering over the equation. Read Seeing actual $\LaTeX$ for more details!

To quickly navigate to the part you want via the Table of Contents, press CTRL+F, and type in the section you want (including the tilde's ~ before and after the section).

~Using LaTeX~

~Text~

~Basic Operations~

~Fractions~

~Sums, Products, Limits, and Integrals~

~Modular Arithmetic~

~Trigonometry~

~Combinatorics~

~Geometry~

~Calculus~

~Parentheses~

~Fitting Parentheses~

~Tables and Arrays~

~Other~

~Using LaTeX~

To use LaTeX, put a backslash and a left parenthesis before the math you want to LaTeXify, and put a backslash and a right parenthesis after the math you want to LaTeXify. For example:

Imgur

Shows up as $1+2+3=6$

However, if you want your math to be more conspicuous and centered, you can use a backslash then a left bracket, then your math, then a backslash then a right bracket. For example:

Imgur

Shows up as

$1+2+3=6$

This second option is the display text. A lot of other math operations will look better in this text. To force the first option to also use display text, you can add a \displaystyle at the beginning.

~Text~

To write text in LaTeX use \text{your text here}. This gives $\text{your text here}$

To use bolded text, use \textbf{your text here}. This gives $\textbf{your text here}$

Italicized text is similar: \textit{your text here}. This gives $\textit{your text here}$

~Basic Operations~

"x+y" gives $x+y$

"x-y" gives $x-y$

"x=y" gives $x=y$

"x\times y" gives $x\times y$

"x\cdot y" gives $x\cdot y$

"x\div y" gives $x\div y$

"x\pm y" gives $x\pm y$

"x\mp y" gives $x\mp y$

x^{y} gives $x^{y}$

x_{y} gives $x_{y}$

\sqrt{x} gives $\sqrt{x}$

\sqrt[y]{x} gives $\sqrt[y]{x}$

\log_{a}b gives $\log_{a}b$

\ln a gives $\ln a$ (that's a lowercase "l" in the beginning, not an uppercase "i")

Note that many of you use "*" or "." for multiplying. This shows up as $*$ and $.$ which don't look good. Use $\times$ or $\cdot$ instead.

Also, the brackets in x^{y} or x_{y} may be omitted if the index is a single character. However, if it is more than one character like $x^{10}$, then brackets are needed or else it will show up as $x^10$.

~Fractions~

Many people simply put a slash between the numerator and denominator to represent a fraction: $x/y$. However, there are neater ways in LaTeX.

\frac{x}{y} is the standard way to write fractions: $\frac{x}{y}$

\dfrac{x}{y} gives a bigger clearer version. However, this takes up more vertical space: $\dfrac{x}{y}$ the "d" stands for "display text".

EXTRA

\cfrac{x}{y} is a special type of fraction formatting. This is for continued fractions, hence the "c". typing \cfrac{x}{x+\cfrac{y}{y+\cfrac{z}{2}}} gives $\cfrac{x}{x+\cfrac{y}{y+\cfrac{z}{2}}}$

~Sums, Products, Limits, and Integrals~

These four are in the same group because they format differently than other symbols.

"\sum" gives $\sum$

"\prod" gives $\prod$

"\lim" gives $\lim$

"\int" gives $\int$

We can add the other elements of each thing by using _ and ^:

\sum_{i=0}^n gives $\sum_{i=0}^n$

\prod_{i=0}^n gives $\prod_{i=0}^n$

\lim_{x\rightarrow n} gives $\lim_{x\rightarrow n}$

\int_{a}^{b} gives $\int_a^b$

However, these don't look very good. However, once putting it on display text, either using the brackets or using \displaystyle as said in the beginning of the guide, we can make them look normal.

\displaystyle\sum_{i=0}^n gives $\displaystyle\sum_{i=0}^n$

\displaystyle\prod_{i=0}^n gives $\displaystyle\prod_{i=0}^n$

\displaystyle\lim_{x\rightarrow n} gives $\displaystyle\lim_{x\rightarrow n}$

\displaystyle\int_{a}^{b} gives $\displaystyle\int_{a}^{b}$

~Modular Arithmetic~

"\equiv" gives $\equiv$

\mod{a} gives $\mod{a}$

\pmod{a} gives $\pmod{a}$

\bmod{a} is \mod{a} without the space before it: $a\bmod{b}$ versus $a\mod{b}$

"a\mid b" creates $a\mid b$, which states that $b$ is divisible by $a$.

~Trigonometry~

Many of you simply put "sin" and "cos" and be done with it; however, adding a backslash before those two make it look much better.

\sin gives $\sin$ (as opposed to $sin$)

\cos gives $\cos$ (as opposed to $cos$)

\tan gives $\tan$

\sec gives $\sec$

\csc gives $\csc$

\cot gives $\cot$

\arcsin gives $\arcsin$

\arccos gives $\arccos$

\arctan gives $\arctan$

Putting a ^{-1} after the trigonometric function designates it as the inverse. For example, \sin^{-1} gives $\sin^{-1}$.

\sinh gives $\sinh$

\cosh gives $\cosh$

\tanh gives $\tanh$

~Combinatorics~

\binom{x}{y} gives $\binom{x}{y}$

\dbinom{x}{y} gives $\dbinom{x}{y}$

~Geometry~

x^{\circ} gives $x^{\circ}$ the degree symbol

\angle gives $\angle$

\Delta gives $\Delta$, for example $\Delta ABC$

\triangle also does the job: $\triangle ABC$

\odot gives $\odot$, for example $\odot O$

AB\parallel CD gives $AB\parallel CD$

AB\perp CD gives $AB\perp CD$

A\cong B gives $A\cong B$

A\sim B gives $A\sim B$

~Calculus~

We've already learned to use $\int$. However, there is much more to calculus than integrals!

There is no command for the total derivative, so you have to use \text{d} to get around it.

For example, \dfrac{\text{d}}{\text{d}x} gives $\dfrac{\text{d}}{\text{d}x}$

Fortunately, there is a symbol for partial derivatives: \partial gives $\partial$.

So, \dfrac{\partial}{\partial x} gives $\dfrac{\partial}{\partial x}$

Double or even triple integrals can be condensed into \iint and \iiint, respectively. This gives $\displaystyle\iint$ and $\displaystyle\iiint$ (I am using display text).

EXTRA

Line integrals can be written as \oint: $\displaystyle \oint$.

~Parentheses~

( and ) are standard for parentheses: $(a+b)$

[ and ] are used for brackets: $[a+b]$

{ and } are used for curly brackets: $\{a+b\}$

\lfloor and \rfloor are used for the floor function: $\lfloor a+b\rfloor$

\lceil and \rceil are used for the ceiling function: $\lceil a+b\rceil$

\langle and \rangle are used for vectors: $\langle a,b\rangle$

The vertical line symbol | (not a capital "i" or a lowercase "l"!) is used for absolute value: $|a+b|$

~Fitting Parentheses~

Suppose you want to write $\left(\dfrac{a}{b}\right)^c$. When you try, it gives $(\dfrac{a}{b})^c$. How did I stretch the parentheses to fit?

To stretch the parentheses, use \left before the left parenthesis and \right before the right one, like this: \left( and \right). When put back into the expression, this yields $\left(\dfrac{a}{b}\right)^c$ as desired.

This isn't just for parentheses; you can use them on brackets: $\{\dfrac{a}{b}\}$ changes into $\left\{\dfrac{a}{b}\right\}$

You can also use this technique on things that use only one parenthesis/bracket/etc. However, just putting \left or \right will yield an error. This is because \left and \right come in pairs. In orer to sidestep this, you can put a period after the one that you do not need (i.e \left. or \right.). This way it will not produce an error, and it will stretch the parenthesis to size. For example, this: \left. \dfrac{x^3+2x}{3x^2}\right|_0^3 gives this: $\left. \dfrac{x^3+2x}{3x^2}\right|_a^b$

~Tables and Arrays~~

To make tables and arrays, use \ begin{array}{[modifiers]} ... \ end{array}. (A space is put before "begin" and before "end" to prevent the LaTeX from prematurely rendering. Even though there are no brackets around to make it render, it does so anyways, I don't know why.)

In the modifiers section, you put either l for left, c for center, or r for right, per column. For example, to make an array with 3 columns, all formatted to align along the right edge, you put "rrr" inside the modifier. It would look like this: \ begin{array}{rrr} ... \ end{array}.

To add a vertical line between two columns, put the vertical line symbol | between two modifiers: for example, if you wanted a horizontal line between the first two columns in the previous example, then you would put \ begin{array}{r|rr} ... \ end{array}.

For actual inputting in the array, there are two rules: put a "&" sign to notify to switch to the next column, and put a "\ \" divider (again a space is added in between to prevent it from rendering) to notify to switch to the next row. When building the table, always fill in row by row: in the first row, fill in all the corresponding columns, and then switch to the next row; then continue in this manner. For example, if I wanted to make a $3\times 3$ square with the numbers $1\rightarrow 9$, I would put: \ begin{array}{lcr}1 & 2 & 3 \ \ 4 & 5 & 6 \ \ 7& 8 & 9 \ end{array}. This produces: $\begin{array}{lcr}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7& 8 & 9 \end{array}$.

To insert horizontal lines between any two rows, put \hline after the divider that separates the two rows. For example, if I wanted to add horizontal lines and vertical lines in the previous example to look like a tic tac toe board, this would be my code: \ begin{array}{l|c|r}1 & 2 & 3 \ \ \hline 4 & 5 & 6 \ \ \hline 7& 8 & 9 \ end{array} and it will produce: $\begin{array}{l|c|r}1 & 2 & 3 \\ \hline 4 & 5 & 6 \\ \hline 7& 8 & 9 \end{array}$

~Other~

To negate any symbol, put \not before the symbol. For example, "\not =" gives $\not =$

Look here for a big list of symbols.

If you don't know how to do something or see something missing in this guide, please do comment below so I can add it! Together, we can make a great LaTeX guide!

Note by Daniel Liu
6 years, 7 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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Note: you must add a full line of space before and after lists for them to show up correctly
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Sort by:

@Daniel Liu - I'm inspired by your guide, so I'm making my own, hope you don't mind :)

- 3 weeks ago

I'm inspired by your guide, so I'm making my own, hope you don't mind :) LOL

- 1 week, 5 days ago

facepalm

LOL

- 1 week, 5 days ago

How can I use packages such as dsfont for writing \mathds{R}?

- 4 months, 2 weeks ago

Hi! How will I align series of equivalent equations by their equal signs? Thank you.

- 5 months, 2 weeks ago

Hey Daniel, it is a great guide but I was curious to know how do you put a line over a number abcd to show that a, b, c and d are digits.

Thanks

- 1 year, 5 months ago

Using two vertical lines (\parallel) is also common for concatenation: $a\parallel b$

- 1 year, 5 months ago

\overline{abcd} shows $\overline{abcd}$

- 1 year, 5 months ago

Thanks a lot 😀👍

- 1 year, 5 months ago

I forgot how to align brackets. Can someone help me?

- 1 year, 6 months ago

What exactly do you mean? Like piecewise functions?

- 1 year, 6 months ago

Wow LaTeX is so hard... "x+y=5" that is the only thing I can remember..

- 1 year, 7 months ago

 $p$ $q$ $p \Rightarrow q$ 1 1 1 1 0 1 0 1 0 0 0 1
Is there a way to put lines between the columns?

- 1 year, 9 months ago

I feel that this list needs a logic category; for instance, things like and, or, & not.

- 1 year, 9 months ago

Help:

\begin{aligned} \pi & = \sum_{k=0}^\infty \frac{1}{16^k} \left[\frac{4}{8k+1} - \frac{2}{8k+4} - \frac{1}{8k+5} - \frac{1}{8k+6}\right] \\ &= \frac{1}{2} \sum_{k=0}^\infty \frac{1}{16^k} \left[\frac{8}{8k+2} + \frac{4}{8k+3} + \frac{4}{8k+4} - \frac{1}{8k+7} \right] \\ &= \frac{1}{16} \sum_{k=0}^\infty \frac{1}{256^k} \left[\frac{64}{16k+1} - \frac{32}{16k+4} - \frac{16}{16k+5} - \frac{16}{16k+6} + \frac{4}{16k+9} - \frac{2}{16k+12} - \frac{1}{16k+13} - \frac{1}{16k+14} \right] \\ &= \frac{1}{32} \sum_{k=0}^\infty \frac{1}{256^k} \left[\frac{128}{1k+2} + \frac{64}{16k+3}+\frac{64}{16k+4}-\frac{16}{16k+7} + \frac{8}{16k+10}+\frac{4}{16k+11}+\frac{4}{16k+12}-\frac{1}{16k+15}\right] \\ &= \frac{1}{32} \sum_{k=0}^\infty \frac{1}{4096^k} \left[\frac{256}{24k+2}+\frac{192}{24k+3}-\frac{256}{24k+4}-\frac{96}{24k+6}-\frac{96}{24k+8}+\frac{16}{24k+10}-\frac{4}{24k+12}-\frac{3}{24k+15}-\frac{6}{24k+16}-\frac{2}{24k+18}-\frac{1}{24k+20}\right] \\ &= \frac{1}{64} \sum_{k=0}^\infty \frac{1}{4096^k} \left[\frac{256}{24k+1}+\frac{256}{24k+2}-\frac{384}{24k+3}-\frac{256}{24k+4}-\frac{64}{24k+5}+\frac{96}{24k+8}+\frac{64}{24k+9}+\frac{16}{24k+10}+\frac{8}{24k+12}-\frac{4}{24k+13}+\frac{6}{24k+15}+\frac{6}{24k+16}+\frac{1}{24k+17}+\frac{1}{24k+18}-\frac{1}{24k+20}-\frac{1}{24k+20}\right] \\ &= \frac{1}{96} \sum_{k=0}^\infty \frac{1}{4096^k}\left[\frac{256}{24k+2}+\frac{64}{24k+3}+\frac{128}{24k+5}+\frac{352}{24k+6}+\frac{64}{24k+7}+\frac{288}{24k+8}+\frac{128}{24k+9}+\frac{80}{24k+10}+\frac{20}{24k+12}-\frac{16}{24k+14}-\frac{1}{24k+15}+\frac{6}{24k+16}-\frac{2}{23k+17}-\frac{1}{24k+19}+\frac{1}{24k+20}-\frac{2}{24k+21}\right] \\ &= \frac{1}{96} \sum_{k=0}^\infty \frac{1}{4096^k} \left[\frac{256}{24k+1} + \frac{320}{24k+3} + \frac{256}{24k+4} - \frac{192}{24k+5}-\frac{224}{24k+6}-\frac{64}{24k+7}-\frac{192}{24k+8}-\frac{64}{24k+9}-\frac{64}{24k+10}-\frac{28}{24k+12}-\frac{4}{24k+13}-\frac{5}{24k+15}+\frac{3}{24k+17}+\frac{1}{24k+18}+\frac{1}{24k+19}+\frac{1}{24k+21}-\frac{1}{24k+22}\right] \\ & = \frac{1}{96} \sum_{k=0}^\infty \frac{1}{4096^k} \left[\frac{512}{24k+1}-\frac{256}{24k+2}+\frac{64}{24k+3}-\frac{512}{24k+4}-\frac{32}{24k+6}+\frac{64}{24k+7}+\frac{96}{24k+8}+\frac{64}{24k+9}+\frac{48}{24k+10}-\frac{12}{24k+12}-\frac{8}{24k+13}-\frac{16}{24k+14}-\frac{1}{24k+15}-\frac{6}{24k+16}-\frac{2}{24k+18}-\frac{1}{24k+19}-\frac{1}{24k+20}-\frac{1}{24k+21}\right] \\ &=\frac{1}{4096} \sum_{k=0}^\infty \frac{1}{65536^k} \left[\frac{16384}{32k+1}-\frac{8192}{32k+4}-\frac{4096}{32k+5}-\frac{4096}{32k+6}+\frac{1024}{32k+9}-\frac{512}{32k+12}-\frac{256}{32k+13}-\frac{256}{32k+14}+\frac{64}{32k+17}-\frac{32}{32k+20}-\frac{16}{32k+21}-\frac{16}{32k+22}+\frac{4}{32k+25}-\frac{2}{32k+28}-\frac{1}{32k+29}-\frac{32k+30}\right] \end{aligned}

Error is in the last characters but I could not find it

- 1 year, 9 months ago

$\begin{array}{ll} \pi &= \sum_{k=0}^\infty \frac{1}{16^k} \left[\frac{4}{8k+1} - \frac{2}{8k+4} - \frac{1}{8k+5} - \frac{1}{8k+6}\right] \\ &= \frac{1}{2} \sum_{k=0}^\infty \frac{1}{16^k} \left[\frac{8}{8k+2} + \frac{4}{8k+3} + \frac{4}{8k+4} - \frac{1}{8k+7} \right] \\ &= \frac{1}{16} \sum_{k=0}^\infty \frac{1}{256^k} \left[\frac{64}{16k+1} - \frac{32}{16k+4} - \frac{16}{16k+5} - \frac{16}{16k+6} + \frac{4}{16k+9} - \frac{2}{16k+12} - \frac{1}{16k+13} - \frac{1}{16k+14} \right] \\ &= \frac{1}{32} \sum_{k=0}^\infty \frac{1}{256^k} \left[\frac{128}{1k+2} + \frac{64}{16k+3}+\frac{64}{16k+4}-\frac{16}{16k+7} + \frac{8}{16k+10}+\frac{4}{16k+11}+\frac{4}{16k+12}-\frac{1}{16k+15}\right] \\ &= \frac{1}{32} \sum_{k=0}^\infty \frac{1}{4096^k} \left[\frac{256}{24k+2}+\frac{192}{24k+3}-\frac{256}{24k+4}-\frac{96}{24k+6}-\frac{96}{24k+8}+\frac{16}{24k+10}-\frac{4}{24k+12}-\frac{3}{24k+15}-\frac{6}{24k+16}-\frac{2}{24k+18}-\frac{1}{24k+20}\right] \\ &= \frac{1}{64} \sum_{k=0}^\infty \frac{1}{4096^k} \left[\frac{256}{24k+1}+\frac{256}{24k+2}-\frac{384}{24k+3}-\frac{256}{24k+4}-\frac{64}{24k+5}+\frac{96}{24k+8}+\frac{64}{24k+9}+\frac{16}{24k+10}+\frac{8}{24k+12}-\frac{4}{24k+13}+\frac{6}{24k+15}+\frac{6}{24k+16}+\frac{1}{24k+17}+\frac{1}{24k+18}-\frac{1}{24k+20}-\frac{1}{24k+20}\right] \\ &= \frac{1}{96} \sum_{k=0}^\infty \frac{1}{4096^k}\left[\frac{256}{24k+2}+\frac{64}{24k+3}+\frac{128}{24k+5}+\frac{352}{24k+6}+\frac{64}{24k+7}+\frac{288}{24k+8}+\frac{128}{24k+9}+\frac{80}{24k+10}+\frac{20}{24k+12}-\frac{16}{24k+14}-\frac{1}{24k+15}+\frac{6}{24k+16}-\frac{2}{23k+17}-\frac{1}{24k+19}+\frac{1}{24k+20}-\frac{2}{24k+21}\right] \\ &= \frac{1}{96} \sum_{k=0}^\infty \frac{1}{4096^k} \left[\frac{256}{24k+1} + \frac{320}{24k+3} + \frac{256}{24k+4} - \frac{192}{24k+5}-\frac{224}{24k+6}-\frac{64}{24k+7}-\frac{192}{24k+8}-\frac{64}{24k+9}-\frac{64}{24k+10}-\frac{28}{24k+12}-\frac{4}{24k+13}-\frac{5}{24k+15}+\frac{3}{24k+17}+\frac{1}{24k+18}+\frac{1}{24k+19}+\frac{1}{24k+21}-\frac{1}{24k+22}\right] \\ &=\frac{1}{96} \sum_{k=0}^\infty \frac{1}{4096^k} \left[\frac{512}{24k+1}-\frac{256}{24k+2}+\frac{64}{24k+3}-\frac{512}{24k+4}-\frac{32}{24k+6}+\frac{64}{24k+7}+\frac{96}{24k+8}+\frac{64}{24k+9}+\frac{48}{24k+10}-\frac{12}{24k+12}-\frac{8}{24k+13}-\frac{16}{24k+14}-\frac{1}{24k+15}-\frac{6}{24k+16}-\frac{2}{24k+18}-\frac{1}{24k+19}-\frac{1}{24k+20}-\frac{1}{24k+21}\right] \\ &=\frac{1}{4096} \sum_{k=0}^\infty \frac{1}{65536^k} \left[\frac{16384}{32k+1}-\frac{8192}{32k+4}-\frac{4096}{32k+5}-\frac{4096}{32k+6}+\frac{1024}{32k+9}-\frac{512}{32k+12}-\frac{256}{32k+13}-\frac{256}{32k+14}+\frac{64}{32k+17}-\frac{32}{32k+20}-\frac{16}{32k+21}-\frac{16}{32k+22}+\frac{4}{32k+25}-\frac{2}{32k+28}-\frac{1}{32k+29}-\frac{1}{32k+30}\right] \end{array}$

- 3 weeks ago

So the problem was that your last contains a fraction without numerator. The second thing is that your code is too long, so I used a left aligned array. And the fractions are too small, so I will make a better one :)

- 3 weeks ago

$\begin{array}{ll} \pi &= \sum_{k=0}^\infty \cfrac{1}{16^k} \left[\cfrac{4}{8k+1} - \cfrac{2}{8k+4} - \cfrac{1}{8k+5} - \cfrac{1}{8k+6}\right] \\ &= \cfrac{1}{2} \sum_{k=0}^\infty \cfrac{1}{16^k} \left[\cfrac{8}{8k+2} + \cfrac{4}{8k+3} + \cfrac{4}{8k+4} - \cfrac{1}{8k+7} \right] \\ &= \cfrac{1}{16} \sum_{k=0}^\infty \cfrac{1}{256^k} \left[\cfrac{64}{16k+1} - \cfrac{32}{16k+4} - \cfrac{16}{16k+5} - \cfrac{16}{16k+6} + \cfrac{4}{16k+9} - \cfrac{2}{16k+12} - \cfrac{1}{16k+13} - \cfrac{1}{16k+14} \right] \\ &= \cfrac{1}{32} \sum_{k=0}^\infty \cfrac{1}{256^k} \left[\cfrac{128}{1k+2} + \cfrac{64}{16k+3}+\cfrac{64}{16k+4}-\cfrac{16}{16k+7} + \cfrac{8}{16k+10}+\cfrac{4}{16k+11}+\cfrac{4}{16k+12}-\cfrac{1}{16k+15}\right] \\ &= \cfrac{1}{32} \sum_{k=0}^\infty \cfrac{1}{4096^k} \left[\cfrac{256}{24k+2}+\cfrac{192}{24k+3}-\cfrac{256}{24k+4}-\cfrac{96}{24k+6}-\cfrac{96}{24k+8}+\cfrac{16}{24k+10}-\cfrac{4}{24k+12}-\cfrac{3}{24k+15}-\cfrac{6}{24k+16}-\cfrac{2}{24k+18}-\cfrac{1}{24k+20}\right] \\ &= \cfrac{1}{64} \sum_{k=0}^\infty \cfrac{1}{4096^k} \left[\cfrac{256}{24k+1}+\cfrac{256}{24k+2}-\cfrac{384}{24k+3}-\cfrac{256}{24k+4}-\cfrac{64}{24k+5}+\cfrac{96}{24k+8}+\cfrac{64}{24k+9}+\cfrac{16}{24k+10}+\cfrac{8}{24k+12}-\cfrac{4}{24k+13}+\cfrac{6}{24k+15}+\cfrac{6}{24k+16}+\cfrac{1}{24k+17}+\cfrac{1}{24k+18}-\cfrac{1}{24k+20}-\cfrac{1}{24k+20}\right] \\ &= \cfrac{1}{96} \sum_{k=0}^\infty \cfrac{1}{4096^k}\left[\cfrac{256}{24k+2}+\cfrac{64}{24k+3}+\cfrac{128}{24k+5}+\cfrac{352}{24k+6}+\cfrac{64}{24k+7}+\cfrac{288}{24k+8}+\cfrac{128}{24k+9}+\cfrac{80}{24k+10}+\cfrac{20}{24k+12}-\cfrac{16}{24k+14}-\cfrac{1}{24k+15}+\cfrac{6}{24k+16}-\cfrac{2}{23k+17}-\cfrac{1}{24k+19}+\cfrac{1}{24k+20}-\cfrac{2}{24k+21}\right] \\ &= \cfrac{1}{96} \sum_{k=0}^\infty \cfrac{1}{4096^k} \left[\cfrac{256}{24k+1} + \cfrac{320}{24k+3} + \cfrac{256}{24k+4} - \cfrac{192}{24k+5}-\cfrac{224}{24k+6}-\cfrac{64}{24k+7}-\cfrac{192}{24k+8}-\cfrac{64}{24k+9}-\cfrac{64}{24k+10}-\cfrac{28}{24k+12}-\cfrac{4}{24k+13}-\cfrac{5}{24k+15}+\cfrac{3}{24k+17}+\cfrac{1}{24k+18}+\cfrac{1}{24k+19}+\cfrac{1}{24k+21}-\cfrac{1}{24k+22}\right] \\ &=\cfrac{1}{96} \sum_{k=0}^\infty \cfrac{1}{4096^k} \left[\cfrac{512}{24k+1}-\cfrac{256}{24k+2}+\cfrac{64}{24k+3}-\cfrac{512}{24k+4}-\cfrac{32}{24k+6}+\cfrac{64}{24k+7}+\cfrac{96}{24k+8}+\cfrac{64}{24k+9}+\cfrac{48}{24k+10}-\cfrac{12}{24k+12}-\cfrac{8}{24k+13}-\cfrac{16}{24k+14}-\cfrac{1}{24k+15}-\cfrac{6}{24k+16}-\cfrac{2}{24k+18}-\cfrac{1}{24k+19}-\cfrac{1}{24k+20}-\cfrac{1}{24k+21}\right] \\ &=\cfrac{1}{4096} \sum_{k=0}^\infty \cfrac{1}{65536^k} \left[\cfrac{16384}{32k+1}-\cfrac{8192}{32k+4}-\cfrac{4096}{32k+5}-\cfrac{4096}{32k+6}+\cfrac{1024}{32k+9}-\cfrac{512}{32k+12}-\cfrac{256}{32k+13}-\cfrac{256}{32k+14}+\cfrac{64}{32k+17}-\cfrac{32}{32k+20}-\cfrac{16}{32k+21}-\cfrac{16}{32k+22}+\cfrac{4}{32k+25}-\cfrac{2}{32k+28}-\cfrac{1}{32k+29}-\cfrac{1}{32k+30}\right] \end{array}$

- 3 weeks ago

Did you put LaTeX wrapping? @Mohammad Farhat

- 3 weeks ago

@Yajat Shamji - When is BRILLIAthon problem 1 coming?

- 3 weeks ago

Give me time to roam around the Community like I do every morning then I'll post it.

- 3 weeks ago

Oh, its morning for you......... Its early morning as in 3 am here.......LOL

- 3 weeks ago

WTF is this!?!?!?! Finding an error in this will take aeons! @Mohammad Farhat

- 3 weeks ago

\displaystyle \pi = 4 \sum{k=0}^\infty \frac{(-1)^k}{2k+1} = 3 \sum{k=0}^\infty (-1)^k \left[\frac{1}{6k+1} + \frac{1}{6k+5} \right] = 4 \sum{k=0}^\infty (-1)^k \left[\frac{1}{10k+1} - \frac{1}{10k+3} + \frac{1}{10k+5} - \frac{1}{10k+7} + \frac{1}{10k+9}\right] = \sum{k=0}^\infty (-1)^k \left[\frac{3}{14k+1} - \frac{3}{14k+3} + \frac{3}{14k+5} - \frac{3}{14k+7} + \frac{3}{14k+9} - \frac{3}{14k+11} + \frac{3}{14k+13}\right]

How do I make the equal signs aligned

- 1 year, 9 months ago

How do I fit the floor/ceiling function around fractions?

$\lfloor\frac{1000}{p}\rfloor$

- 1 year, 10 months ago

\left\lfloor \frac {1000}p \right\rfloor$\left\lfloor \frac {1000}p \right\rfloor$

- 1 year, 10 months ago

You forgot to tell about matrices and piecewise functions, but still good job. $\displaystyle \oint _{S} \vec{E} \cdot d\vec{s} =\frac{Q_{enclosed}} {\epsilon _{\circ}}$

- 1 year, 10 months ago

why is the continued fraction not working:

\pi=\cfrac{4}{1+\cfrac{1^2}{2+\cfrac{2+\cfrac{3^2}{2+\cfrac{5^2}{2+\ddots}}}}

- 1 year, 11 months ago

\pi=\cfrac{4}{1+\cfrac{1^2}{2+\cfrac{3^2}{2+\cfrac{5^2}{2+\ddots}}}}

- 1 year, 11 months ago

$\pi=\cfrac{4}{1+\cfrac{1^2}{2+\cfrac{3^2}{2+\cfrac{5^2}{2+\ddots}}}}$

- 1 year, 9 months ago

Thank you but it is already in my note:$\pi$, a beautiful number

- 1 year, 9 months ago

What's wrong in this:\displaystyle \frac{\pi}{2} = \frac{1]{2} \sum_{n=0}^\infty \frac{(n!)^2 2^{n+1}}{(2n+1)!} = \sum_{n=0}^\infty \frac{n!}{(2n=1)!! = 1 + \frac{1}{3} + \frac{1 \times 2}{3 \times 5} + \frac{1 \times 2 \times 3}{3 \times 5 \times 7} + \cdots = 1 + \frac{1}{3}(1 + \frac{2}{5}(1 + \frac{3}{7}(1 + \frac{4}{9}(1 + \cdots))))

- 1 year, 11 months ago

displaystyle \frac{\pi}{2} = \frac{1}{2}..., change ] to }

- 1 year, 11 months ago

$\displaystyle \frac{\pi}{2} = \frac{1}{2} \sum_{n=0}^\infty \frac{(n!)^2 2^{n+1}}{(2n+1)!} = \sum_{n=0}^\infty \frac{n!}{(2n=1)!!} = 1 + \frac{1}{3} + \frac{1 \times 2}{3 \times 5} + \frac{1 \times 2 \times 3}{3 \times 5 \times 7} + \cdots = 1 + \frac{1}{3}(1 + \frac{2}{5}(1 + \frac{3}{7}(1 + \frac{4}{9}(1 + \cdots))))$

- 1 year, 11 months ago

@X X By the way $\large \displaystyle \sum_{n=0}^\infty \frac{n!}{(2n-1)}!! = \dfrac{4 + \pi}{2}$ and not $\dfrac{\pi}{2}$

- 1 year, 11 months ago

It is not - but +

- 1 year, 11 months ago

@X X I already know! Thank you for helping me even if I bothered you or wasted your time!

- 1 year, 11 months ago

No, because there is one more mistake. What is (2n=1)!!

- 1 year, 11 months ago

@X X I already edited it in the note

- 1 year, 11 months ago

oh! Thank you!

- 1 year, 11 months ago

Can someone help me with this : $\displaystyle \pi = \frac{3}{4} \sqrt{3} + 24 \int_{0}{\frac{1}{4}} \sqrt{x-x^2} \text{dx} = \frac{3 \sqrt{3}}{4} + 24 \left(\frac{1}{12} - \frac{1}{5 \times 2^5} - \frac{1}{28 \times 2^7} - \frac{1}{72 \times 2^9} \cdots \right)$ i do not understand why the $\frac{1}{4}$ is down when it is supposed to be up in the integral

- 1 year, 11 months ago

Here's freeware you can use to make your Latex life easier:

- 1 year, 11 months ago

Thanks

- 1 year, 11 months ago

\displaystyle \pi = \frac{3}{4} \sqrt{3} + 24 \int_{0}^{\frac{1}{4}} \sqrt{x-x^2} \text{dx} = \frac{3 \sqrt{3}}{4} + 24 \left(\frac{1}{12} - \frac{1}{5 \times 2^5} - \frac{1}{28 \times 2^7} - \frac{1}{72 \times 2^9} \cdots \right

Insert the ^

- 1 year, 11 months ago

Did you see my note: $\pi$ is a beautiful number?

- 1 year, 11 months ago

Oh! I missed ^ Thank you!

- 1 year, 11 months ago

How do we input the Riemann Zeta function

- 1 year, 11 months ago

$\zeta(x)$ \zeta(x)

- 1 year, 11 months ago

Oh! Thank you very much!

- 1 year, 11 months ago

How do you write bold AND italicized text.

- 1 year, 11 months ago

Do you mean this?

You can type it like $^{***}text^{***}$,but don't use LaTeX.

- 1 year, 11 months ago

- 2 years, 1 month ago

Is there a 'strikethrough' feature in LaTeX? I'm trying to show a a product of fractions with several factors in numerators and denominators cancelling each other out. So far all I've found is the \not feature, but this works very poorly; e.g. when I try \not{147}, I get $\not{147}$, with only the 1 crossed out.

I would like to be able to cross out entire numbers; ideally, I'd also like to show either the strikethrough or the number (preferably not both) in a variety of colors.

Do you know whether this can be done? Thanks!

- 2 years, 1 month ago

Hey how to do this?

Like writing the catalyst of reaction above the arrowhead like in this https://www.google.co.in/search?q=wurtz+reaction&dcr=0&source=lnms&tbm=isch&sa=X&ved=0ahUKEwisMDt-evYAhVBQo8KHcgzAaoQAUICigB&biw=1366&bih=637#imgrc=F_pnGGhHIpA7NM:

- 2 years, 6 months ago

how do we type greater,lesser,greater or equal to and lesser or equal to symbol by LaTeX?

- 2 years, 6 months ago

• $\backslash(>\backslash)$ appears as $>$

• $\backslash(<\backslash)$ appear as $<$

• $\backslash(\text{\ge}\backslash)$ appear as $\ge$

• $\backslash(\text{\le} \backslash)$ appear as $\le$

- 2 years, 6 months ago

Thanks!

- 2 years, 6 months ago

One way of quickly figuring out what the code for a special character is, is this tool.

- 2 years, 7 months ago

Thank you for this guide; I'm just learning how to use LaTex and this is all extremely helpful.

One question: is there a way to push lines slightly to the right (like the Tab feature in word processors) without pushing them all the way to the middle? For example, in the lines

${(p-1)}^p \equiv -1 \pmod {p^2}$

$\equiv {p^2-1} \pmod {p^2}$

$\equiv {(p-1)(p+1)} \pmod {p^2}$

I would really like to have the equivalence symbol in the second and third lines line up with the one in the first line. Is there a way to do that?

- 2 years, 8 months ago

I just realized that your comment is $\huge Freaking \ 2 \ years \ and \ 7 \ months \ old$ @zico quintina

- 2 weeks, 5 days ago

If you have other $\LaTeX{}$ doubts, ask the community by posting it on my $\LaTeX{}$ Discussion Note, which you will find in my feed(after looking around a little). I shall answer the doubt quickly, or ask few other BRILLIANT users that I know of to tell you the solution, and either way, your question will be answered quickly. @zico quintina

- 2 weeks, 5 days ago

@zico quintina - I know the perfect command that does wonders like this -

$\underline{Centered \ and \ Aligned \ Equation}$

\begin{aligned} (p-1)^{p} & \equiv -1 \pmod p^{2} \\ &\equiv p^{2}-1 \pmod p^{2} \\ &\equiv (p+1)(p-1) \pmod p^{2}\end{aligned}

I used the $\LaTeX{}$ command that lets us create an equation array, and this helps us align the $=$ or $\equiv$ symbols. The picture below is the $\LaTeX{}$ command that I used.

The Latex for the equation that I wrote :)

Whenever you want to break the line as in go to the next line, give \ and don't forget to add & behind the thing you want to align, as in the above code, I have typed & right behind all three \equiv as I wanted to align them. Hope you understood!

- 2 weeks, 5 days ago

With my mac \left, \right does not work. Big work as under.
big ( your text big ). Same for left and right.
Four sizes: big,....bigg,......Big,......Bigg.

- 3 years ago

We can also use \mathrm{d} for the total derivative command you are talking about.

Like $\dfrac{\mathrm{d}}{\mathrm{d}x}.e^{x}=e^x$

:)

- 3 years, 1 month ago

How to write Right arrow symbol on LaTex?

- 3 years, 1 month ago

• Use \rightarrow for $\rightarrow$
• Use \Rightarrow for $\Rightarrow$

- 3 years, 1 month ago

Thanks. I just want the second one $\Rightarrow$

- 3 years, 1 month ago

You may also want \implies : $\implies$

- 3 years, 1 month ago

I think you should make a column for the arrows...

- 3 years, 1 month ago

Like when we conclude two things from a statement , we use \begin{cases} and \end{cases}.

What do we use when we conclude one thing from two things??Like the opposite kind of braces I want as in \begin{cases} ....

- 3 years, 4 months ago

What is the issue with this?

I wanted to color only $102^2$ but this did not happen

$100^2 + 101^2 + \color{#3D99F6}{102^2} + \cdots + 998^2 + 999^2 + 1000^2$

- 3 years, 4 months ago

$100^2+101^2+{ \color{#3D99F6}{100^3} }+104^2+....$
code:....100^2+101^2+ { \color{blue}{100^3} } +104^2+....
100^2+101^2+$\color{#D61F06} { \{}$ \color{blue}{100^3} $\color{#D61F06}{\}}$ +104^2+....
See the extra pair of { } around "" \color{blue}{100^3} "" which I have shown in red, but { } are in normal color( here, in black).

- 1 year, 11 months ago

Use ...\color{blue} {102^2}\color{black}... to unflag it. I don't know the reason though...​ $100^2 + 101^2 + \color{#3D99F6}{102^2} \color{#333333} + \cdots + 998^2 + 999^2 + 1000^2$

- 3 years, 4 months ago

I got the error...See this

$100^2 + 101^2 + {\color{#3D99F6}{102^2}} + \cdots + 998^2 + 999^2 + 1000^2$

- 3 years, 4 months ago

code for less than and greater than plzz

- 3 years, 4 months ago

• Use \le for $\le$
• Use \ge for $\ge$

- 3 years, 4 months ago

i dont need quality

- 3 years, 4 months ago

OK. Use

• < for $<$
• > for $>$

- 3 years, 4 months ago

Whats wrong with this?

\color{#3D99F6}\text{S = x^2 - 8\lfloor x \rfloor + 10}

- 3 years, 4 months ago

Use \color{Blue} {S = x^2 - 8\lfloor x \rfloor + 10} to get $\color{#3D99F6} {S = x^2 - 8\lfloor x \rfloor + 10}$

- 3 years, 4 months ago

Thanks!!

- 3 years, 4 months ago

$\color{#3D99F6}{S = x^2 - 8\lfloor x \rfloor + 10}$

- 3 years, 4 months ago

Who to write infinity??

- 3 years, 4 months ago

Who to write infinity??

- 3 years, 4 months ago

\infty

- 1 year, 11 months ago

\infty will give you $\infty$, infinity symbol.