Beginner LaTeX Guide

LaTeX{\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{\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\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).


Table of Contents

~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 Imgur

Shows up as 1+2+3=61+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 Imgur

Shows up as

1+2+3=61+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 your text here\text{your text here}

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

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


~Basic Operations~

"x+y" gives x+yx+y

"x-y" gives xyx-y

"x=y" gives x=yx=y

"x\times y" gives x×yx\times y

"x\cdot y" gives xyx\cdot y

"x\div y" gives x÷yx\div y

"x\pm y" gives x±yx\pm y

"x\mp y" gives xyx\mp y

x^{y} gives xyx^{y}

x_{y} gives xyx_{y}

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

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

\log_{a}b gives logab\log_{a}b

\ln a gives lna\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 x10x^{10}, then brackets are needed or else it will show up as x10x^10.


~Fractions~

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

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

\dfrac{x}{y} gives a bigger clearer version. However, this takes up more vertical space: xy\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 xx+yy+z2 \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\lim

"\int" gives \int

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

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

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

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

\int_{a}^{b} gives ab\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 i=0n\displaystyle\sum_{i=0}^n

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

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

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


~Modular Arithmetic~

"\equiv" gives \equiv

\mod{a} gives moda\mod{a}

\pmod{a} gives (moda)\pmod{a}

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

"a\mid b" creates aba\mid b, which states that bb is divisible by aa.


~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\sin (as opposed to sinsin)

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

\tan gives tan\tan

\sec gives sec\sec

\csc gives csc\csc

\cot gives cot\cot

\arcsin gives arcsin\arcsin

\arccos gives arccos\arccos

\arctan gives arctan\arctan

Putting a ^{-1} after the trigonometric function designates it as the inverse. For example, \sin^{-1} gives sin1\sin^{-1}.

\sinh gives sinh\sinh

\cosh gives cosh\cosh

\tanh gives tanh\tanh


~Combinatorics~

\binom{x}{y} gives (xy)\binom{x}{y}

\dbinom{x}{y} gives (xy)\dbinom{x}{y}


~Geometry~

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

\angle gives \angle

\Delta gives Δ\Delta, for example ΔABC\Delta ABC

\triangle also does the job: ABC\triangle ABC

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

AB\parallel CD gives ABCDAB\parallel CD

AB\perp CD gives ABCDAB\perp CD

A\cong B gives ABA\cong B

A\sim B gives ABA\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 ddx\dfrac{\text{d}}{\text{d}x}

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

So, \dfrac{\partial}{\partial x} gives x\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)(a+b)

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

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

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

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

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

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


~Fitting Parentheses~

Suppose you want to write (ab)c\left(\dfrac{a}{b}\right)^c. When you try, it gives (ab)c(\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 (ab)c\left(\dfrac{a}{b}\right)^c as desired.

This isn't just for parentheses; you can use them on brackets: {ab}\{\dfrac{a}{b}\} changes into {ab}\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: x3+2x3x2ab\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 3×33\times 3 square with the numbers 191\rightarrow 9, I would put: \ begin{array}{lcr}1 & 2 & 3 \ \ 4 & 5 & 6 \ \ 7& 8 & 9 \ end{array}. This produces: 123456789 \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: 123456789\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

No vote yet
1 vote

  Easy Math Editor

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.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

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

Percy Jackson - 3 weeks ago

Log in to reply

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

Páll Márton - 1 week, 5 days ago

Log in to reply

facepalm

LOL

Percy Jackson - 1 week, 5 days ago

Log in to reply

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

Pepper Mint - 4 months, 2 weeks ago

Log in to reply

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

John Rey Jimenez - 5 months, 2 weeks ago

Log in to reply

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

Vedant Saini - 1 year, 5 months ago

Log in to reply

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

Blan Morrison - 1 year, 5 months ago

Log in to reply

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

X X - 1 year, 5 months ago

Log in to reply

Thanks a lot 😀👍

Vedant Saini - 1 year, 5 months ago

Log in to reply

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

Mohammad Farhat - 1 year, 6 months ago

Log in to reply

What exactly do you mean? Like piecewise functions?

Blan Morrison - 1 year, 6 months ago

Log in to reply

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

Charn Son - 1 year, 7 months ago

Log in to reply

ppqqpqp \Rightarrow q
111
101
010
001
Is there a way to put lines between the columns?

Johanan Paul - 1 year, 9 months ago

Log in to reply

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

Blan Morrison - 1 year, 9 months ago

Log in to reply

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

Mohammad Farhat - 1 year, 9 months ago

Log in to reply

π=k=0116k[48k+128k+418k+518k+6]=12k=0116k[88k+2+48k+3+48k+418k+7]=116k=01256k[6416k+13216k+41616k+51616k+6+416k+9216k+12116k+13116k+14]=132k=01256k[1281k+2+6416k+3+6416k+41616k+7+816k+10+416k+11+416k+12116k+15]=132k=014096k[25624k+2+19224k+325624k+49624k+69624k+8+1624k+10424k+12324k+15624k+16224k+18124k+20]=164k=014096k[25624k+1+25624k+238424k+325624k+46424k+5+9624k+8+6424k+9+1624k+10+824k+12424k+13+624k+15+624k+16+124k+17+124k+18124k+20124k+20]=196k=014096k[25624k+2+6424k+3+12824k+5+35224k+6+6424k+7+28824k+8+12824k+9+8024k+10+2024k+121624k+14124k+15+624k+16223k+17124k+19+124k+20224k+21]=196k=014096k[25624k+1+32024k+3+25624k+419224k+522424k+66424k+719224k+86424k+96424k+102824k+12424k+13524k+15+324k+17+124k+18+124k+19+124k+21124k+22]=196k=014096k[51224k+125624k+2+6424k+351224k+43224k+6+6424k+7+9624k+8+6424k+9+4824k+101224k+12824k+131624k+14124k+15624k+16224k+18124k+19124k+20124k+21]=14096k=0165536k[1638432k+1819232k+4409632k+5409632k+6+102432k+951232k+1225632k+1325632k+14+6432k+173232k+201632k+211632k+22+432k+25232k+28132k+29132k+30]\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}

Páll Márton - 3 weeks ago

Log in to reply

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 :)

Páll Márton - 3 weeks ago

Log in to reply

@Páll Márton π=k=0116k[48k+128k+418k+518k+6]=12k=0116k[88k+2+48k+3+48k+418k+7]=116k=01256k[6416k+13216k+41616k+51616k+6+416k+9216k+12116k+13116k+14]=132k=01256k[1281k+2+6416k+3+6416k+41616k+7+816k+10+416k+11+416k+12116k+15]=132k=014096k[25624k+2+19224k+325624k+49624k+69624k+8+1624k+10424k+12324k+15624k+16224k+18124k+20]=164k=014096k[25624k+1+25624k+238424k+325624k+46424k+5+9624k+8+6424k+9+1624k+10+824k+12424k+13+624k+15+624k+16+124k+17+124k+18124k+20124k+20]=196k=014096k[25624k+2+6424k+3+12824k+5+35224k+6+6424k+7+28824k+8+12824k+9+8024k+10+2024k+121624k+14124k+15+624k+16223k+17124k+19+124k+20224k+21]=196k=014096k[25624k+1+32024k+3+25624k+419224k+522424k+66424k+719224k+86424k+96424k+102824k+12424k+13524k+15+324k+17+124k+18+124k+19+124k+21124k+22]=196k=014096k[51224k+125624k+2+6424k+351224k+43224k+6+6424k+7+9624k+8+6424k+9+4824k+101224k+12824k+131624k+14124k+15624k+16224k+18124k+19124k+20124k+21]=14096k=0165536k[1638432k+1819232k+4409632k+5409632k+6+102432k+951232k+1225632k+1325632k+14+6432k+173232k+201632k+211632k+22+432k+25232k+28132k+29132k+30]\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}

Páll Márton - 3 weeks ago

Log in to reply

Did you put LaTeX wrapping? @Mohammad Farhat

Yajat Shamji - 3 weeks ago

Log in to reply

@Yajat Shamji - When is BRILLIAthon problem 1 coming?

Percy Jackson - 3 weeks ago

Log in to reply

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

Yajat Shamji - 3 weeks ago

Log in to reply

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

Percy Jackson - 3 weeks ago

Log in to reply

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

Percy Jackson - 3 weeks ago

Log in to reply

\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

Mohammad Farhat - 1 year, 9 months ago

Log in to reply

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

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

Blan Morrison - 1 year, 10 months ago

Log in to reply

\left\lfloor \frac {1000}p \right\rfloor1000p\left\lfloor \frac {1000}p \right\rfloor

X X - 1 year, 10 months ago

Log in to reply

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

Hjalmar Orellana Soto - 1 year, 10 months ago

Log in to reply

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}}}}

Mohammad Farhat - 1 year, 11 months ago

Log in to reply

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

X X - 1 year, 11 months ago

Log in to reply

π=41+122+322+522+ \pi=\cfrac{4}{1+\cfrac{1^2}{2+\cfrac{3^2}{2+\cfrac{5^2}{2+\ddots}}}}

Niranjan Khanderia - 1 year, 9 months ago

Log in to reply

@Niranjan Khanderia Thank you but it is already in my note:π\pi, a beautiful number

Mohammad Farhat - 1 year, 9 months ago

Log in to reply

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))))

Mohammad Farhat - 1 year, 11 months ago

Log in to reply

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

X X - 1 year, 11 months ago

Log in to reply

π2=12n=0(n!)22n+1(2n+1)!=n=0n!(2n=1)!!=1+13+1×23×5+1×2×33×5×7+=1+13(1+25(1+37(1+49(1+))))\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))))

X X - 1 year, 11 months ago

Log in to reply

@X X By the way n=0n!(2n1)!!=4+π2\large \displaystyle \sum_{n=0}^\infty \frac{n!}{(2n-1)}!! = \dfrac{4 + \pi}{2} and not π2\dfrac{\pi}{2}

James Bacon - 1 year, 11 months ago

Log in to reply

@James Bacon It is not - but +

Mohammad Farhat - 1 year, 11 months ago

Log in to reply

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

Mohammad Farhat - 1 year, 11 months ago

Log in to reply

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

X X - 1 year, 11 months ago

Log in to reply

@X X I already edited it in the note

Mohammad Farhat - 1 year, 11 months ago

Log in to reply

oh! Thank you!

Mohammad Farhat - 1 year, 11 months ago

Log in to reply

Can someone help me with this : π=343+24014xx2dx=334+24(11215×25128×27172×29)\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 14\frac{1}{4} is down when it is supposed to be up in the integral

Mohammad Farhat - 1 year, 11 months ago

Log in to reply

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

Daum Equation Editor

Michael Mendrin - 1 year, 11 months ago

Log in to reply

Thanks

Mohammad Farhat - 1 year, 11 months ago

Log in to reply

\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 ^

X X - 1 year, 11 months ago

Log in to reply

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

Mohammad Farhat - 1 year, 11 months ago

Log in to reply

Oh! I missed ^ Thank you!

Mohammad Farhat - 1 year, 11 months ago

Log in to reply

How do we input the Riemann Zeta function

Mohammad Farhat - 1 year, 11 months ago

Log in to reply

ζ(x)\zeta(x) \zeta(x)

X X - 1 year, 11 months ago

Log in to reply

Oh! Thank you very much!

Mohammad Farhat - 1 year, 11 months ago

Log in to reply

How do you write bold AND italicized text.

Mohammad Farhat - 1 year, 11 months ago

Log in to reply

Do you mean this?

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

X X - 1 year, 11 months ago

Log in to reply

Please answer this

X X - 2 years, 1 month ago

Log in to reply

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 1̸47\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!

zico quintina - 2 years, 1 month ago

Log in to reply

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:

Md Zuhair - 2 years, 6 months ago

Log in to reply

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

donglin loo - 2 years, 6 months ago

Log in to reply

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

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

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

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

Munem Shahriar - 2 years, 6 months ago

Log in to reply

Thanks!

Donglin Loo - 2 years, 6 months ago

Log in to reply

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

Stefan van der Waal - 2 years, 7 months ago

Log in to reply

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

(p1)p1(modp2){(p-1)}^p \equiv -1 \pmod {p^2}

p21(modp2)\equiv {p^2-1} \pmod {p^2}

(p1)(p+1)(modp2)\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?

zico quintina - 2 years, 8 months ago

Log in to reply

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

Percy Jackson - 2 weeks, 5 days ago

Log in to reply

If you have other LaTeX\LaTeX{} doubts, ask the community by posting it on my LaTeX\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

Percy Jackson - 2 weeks, 5 days ago

Log in to reply

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

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

(p1)p1(modp)2p21(modp)2(p+1)(p1)(modp)2\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\LaTeX{} command that lets us create an equation array, and this helps us align the == or \equiv symbols. The picture below is the LaTeX\LaTeX{} command that I used.

The Latex for the equation that I wrote :) 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!

Percy Jackson - 2 weeks, 5 days ago

Log in to reply

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.

Niranjan Khanderia - 3 years ago

Log in to reply

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

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

:)

Swagat Panda - 3 years, 1 month ago

Log in to reply

How to write Right arrow symbol on LaTex?

Munem Shahriar - 3 years, 1 month ago

Log in to reply

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

Kishore S. Shenoy - 3 years, 1 month ago

Log in to reply

Thanks. I just want the second one \Rightarrow

Munem Shahriar - 3 years, 1 month ago

Log in to reply

@Munem Shahriar You may also want \implies :     \implies

Kishore S. Shenoy - 3 years, 1 month ago

Log in to reply

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

Aaryan Maheshwari - 3 years, 1 month ago

Log in to reply

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} ....

Ankit Kumar Jain - 3 years, 4 months ago

Log in to reply

What is the issue with this?

I wanted to color only 1022102^2 but this did not happen

1002+1012+1022++9982+9992+10002100^2 + 101^2 + \color{#3D99F6}{102^2} + \cdots + 998^2 + 999^2 + 1000^2

Ankit Kumar Jain - 3 years, 4 months ago

Log in to reply

1002+1012+1003+1042+....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).

Niranjan Khanderia - 1 year, 11 months ago

Log in to reply

Use ...\color{blue} {102^2}\color{black}... to unflag it. I don't know the reason though...​ 1002+1012+1022++9982+9992+10002100^2 + 101^2 + \color{#3D99F6}{102^2} \color{#333333} + \cdots + 998^2 + 999^2 + 1000^2

Kishore S. Shenoy - 3 years, 4 months ago

Log in to reply

I got the error...See this

1002+1012+1022++9982+9992+10002100^2 + 101^2 + {\color{#3D99F6}{102^2}} + \cdots + 998^2 + 999^2 + 1000^2

Ankit Kumar Jain - 3 years, 4 months ago

Log in to reply

code for less than and greater than plzz

Nivedit Jain - 3 years, 4 months ago

Log in to reply

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

Kishore S. Shenoy - 3 years, 4 months ago

Log in to reply

i dont need quality

Nivedit Jain - 3 years, 4 months ago

Log in to reply

@Nivedit Jain OK. Use

  • < for <<
  • > for >>

Kishore S. Shenoy - 3 years, 4 months ago

Log in to reply

Whats wrong with this?

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

Ankit Kumar Jain - 3 years, 4 months ago

Log in to reply

Use \color{Blue} {S = x^2 - 8\lfloor x \rfloor + 10} to get S=x28x+10\color{#3D99F6} {S = x^2 - 8\lfloor x \rfloor + 10}

Kishore S. Shenoy - 3 years, 4 months ago

Log in to reply

Thanks!!

Ankit Kumar Jain - 3 years, 4 months ago

Log in to reply

S=x28x+10\color{#3D99F6}{S = x^2 - 8\lfloor x \rfloor + 10}

Ankit Kumar Jain - 3 years, 4 months ago

Log in to reply

Who to write infinity??

Nivedit Jain - 3 years, 4 months ago

Log in to reply

Who to write infinity??

Nivedit Jain - 3 years, 4 months ago

Log in to reply

\infty

Mohammad Farhat - 1 year, 11 months ago

Log in to reply

\infty

James Bacon - 2 years, 8 months ago

Log in to reply

\infty will give you \infty, infinity symbol.

Kishore S. Shenoy - 3 years, 4 months ago
<