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This post is to help me with LaTeX so i have set it up with Melody.

Anyone can help.

 

 

But please Do Not Interupt With Unnecessary Things.

 

 

Thanks.

 May 2, 2015

Best Answer 

 #86
avatar+4691 
+15

Hello Melody,

 

I was wondering if there was anything else i could learn / practise on LaTeX.

 

Thanks.

 

 

 

EDIT: I've been trying to find an equation similar to that Quadratic formula...But I couldn't find anything.I want a challenge like that even if it doesn't make sense :D

 A challenge is always fun!

 May 6, 2015
 #1
avatar+4691 
+10

$$\frac{1}{2}$$

 

All I know so far is fractions.

 

 

$$5\frac{1}{2}$$

.
 May 2, 2015
 #2
avatar+11885 
+5

as much as ive used latex.....i use $$$$....this sign to give gaps!you can use it to...try?

 May 2, 2015
 #3
avatar+4691 
+10

Hi Rosala,

 

Gap?

 

so like.

$$Hello$$$$I$$$$am$$$$MathsGod1$$

 

Or just one  $$$Hello$I$Am$MathsGod1$$

.
 May 2, 2015
 #4
avatar+11885 
+10

mathsGod...honestly i take chances in this!

 

$$$Hi! Wass Up?$$$

 

so if you want the line to be with spaces....write like this!

 May 2, 2015
 #5
avatar+4691 
+5

$$\frac{1}{2}$$\frac{2}{5}$$

 

How many dollar signs are needed

 

 

 

Edit

 

 

 

Can it be used on fractions?

 

$$$\frac{1}{2}and\frac{2}{5}$$$

.
 May 2, 2015
 #6
avatar+4691 
+5

Oh ok.

 

 

So only 3

 

$$$I'm doing good!$$$

.
 May 2, 2015
 #7
avatar+11885 
0

$$$$$$$ Any much! $$$$$$

 

that was to show you ...you can put as many as you want and leaving a gap and writing or not leaving a gap and writing makes no difference!

 

 

yes it can be use din fractions!

 May 2, 2015
 #8
avatar+4691 
+10

$$$$$\frac{1}{2} \frac{2}{5}$$$$$

 

Nice.

 

$$$\frac{1}{2} \frac{2}{5}$$$

 

spacing matters?

 May 2, 2015
 #9
avatar+11885 
+5

do u want to say you cnat add in a sign between them

 

$$$$\frac{2}{3}$$,$$\frac{4}{5}$$$$

 

is that what you were asking?

 May 2, 2015
 #10
avatar+105894 
+5

There are many different ways of doing things in LaTex.

I have never used the $ like rosala just showed you.  Thanks Rosala.

I also do not understand why yours jumped to different lines but Rosala's stayed on the same line.

If I want a break I usually use \;  or  \:  or  \quad    or   \qquad

\; and \: give little breaks, \quad gives a slightly bigger break (maybe 4 units) and \qquad gives a bigger gap again (maybe 8 units)

 

1 \: 2 \; 3 \quad 4 \qquad5       See how the gaps get bigger :)

$$1\:2\;3\quad 4\qquad5$$

 

spaces in Latex, using the space bar, are usually meaningless :)

 

An important thing to know is that the symbol for a new line is \\

If you want to skip a line you would uses \\\\

 May 2, 2015
 #11
avatar+11885 
+5

not much it does!

 May 2, 2015
 #12
avatar+105894 
+10

I see what you are doing MG, you are using the enter button to move to a new line.

Yes that will work,  however if you use fractions afterwards, they stuff up.

So it is better to use  \\   or   \\\\  or  \\\\\\  to go to a new line or to skip lines :)      [corrected]

 May 2, 2015
 #13
avatar+11885 
+5

$$$ I didnt $$$
know that$$$
Melody!$$

 

 

is that correct?why is there such a big gap between lines?

 May 2, 2015
 #14
avatar+4691 
+5

$$Cool\:I\:didn't\:know\:that!$$

 

$$Cool\quad I\quad didn't\quad know\quad that$$

 

$$Cool\;I\;didn't\;know\;that$$

 

$$Cool\qquad I\qquad didn't\qquad know\qquad that\qquad$$

.
 May 2, 2015
 #15
avatar+4691 
+10

$$Ok\:I'll\:try\:that

\frac{1}{2} /// \frac{2}{5}$$

.
 May 2, 2015
 #16
avatar+105894 
+5

Can you work out why it has displayed incorrectly MG ?

 May 2, 2015
 #17
avatar+11885 
+5

whats the problem in mine MElody?

 May 2, 2015
 #18
avatar+4691 
+10

$$one\;more\;time\; // \frac{1}{2}$$after ''that'' i never did //

 May 2, 2015
 #19
avatar+105894 
+5

$$$ I didnt $$$know that$$$ Melody!

 

i don't know rosala,  I do not fully understand what the $ does.

the only time I have ever used a $ for is if I want text.

See this

 

\\I want this in text\\  

$I want this in text$

 

$$\\I want this in text\\
$I want this in text$$$

 

The \\ at the very beginning stopped the first line from being indented.

The \\ at the end of the first line caused a line break.

When i enclosed the text in $ signs it made it display as text.    

 May 2, 2015
 #20
avatar+4691 
+10

$$\frac{1}{2}//\frac{2}{5}$$

.
 May 2, 2015
 #21
avatar+105894 
+5

MG, one of your main problems is that you are using  forward slashes// instead of back slashes\\

 May 2, 2015
 #22
avatar+4691 
+15

But you said to skip lines use forward slashes.?

 May 2, 2015
 #23
avatar+105894 
+5

So I did, I am sorry they are meant to be back slashes - I shall go back and fix that post :(

 May 2, 2015
 #24
avatar+4691 
+15

It's ok everyone mistakes :)

 

It's the first time i spotted a mistake from melody, it's a miracle!

 

Joking.

 

$$\frac{1}{2}\\\frac{2}{5}$$

.
 May 2, 2015
 #25
avatar+105894 
+5

ok Mg   

Now put \\ at the front of your first line to get rid of that unwanted indent 

 May 2, 2015
 #26
avatar+4691 
+5

$$\\\frac{1}{2}\\\frac{2}{5}$$

.
 May 2, 2015
 #27
avatar+4691 
+5

Yes!

 

$$\\\\\frac{1}{2}\\\\\frac{2}{5}$$

.
 May 2, 2015
 #28
avatar+4691 
+10

$$\frac{1}{2}+\frac{1}{2}=1$$

 

$$4\frac{1}{2}+4\frac{1}{2}=9$$

.
 May 2, 2015
 #29
avatar+4691 
+5

What else can i learn

 May 2, 2015
 #30
avatar+105894 
+10

\\\sqrt9\\

\sqrt{144}\\

\sqrt[3]{64}\\

 

$$\\\sqrt9\\
\sqrt{144}\\
\sqrt[3]{64}\\$$

 

There is something else to digest :)

 May 2, 2015
 #31
avatar+4691 
+5

Why do you put { } on  144 but not 9

 

$$\\\sqrt16=4\\\\\[2]sqrt[-2]$$ 

 

hmmm.

 May 2, 2015
 #32
avatar+4691 
+5

OH!

 

 

$$\sqrt[2]{36}\\$$

 

On the first line of what you showed me.

 

i have two questions.

 

Could you give me some questions to try please so i can write it in Latex? If possible 

 

and

you did three backslash\\\ for the new line.

 

A new line also appeared on the second bit but why is there only one backslash\ when you still got a new line..?

 May 2, 2015
 #33
avatar+105894 
+5

Umm

2 backslashes for a new line

1 backslash belongs to the sqrt 

 

I write a lot of LaTex in many posts - you can try and copy it.

I just did a stack here

http://web2.0calc.com/questions/is-any-real-number-positive#r1

try reproducing that :)

 May 2, 2015
 #34
avatar+11885 
0

thanks to you mathsgod.....even ive learnt some of this!

 May 2, 2015
 #35
avatar+4691 
0

Thanks Melody, and I've learnt things from you too Rosala :)

 May 2, 2015
 #36
avatar+11885 
0

That's good but indirectly your post was a really good way to teach me ! Very impressive!

 May 2, 2015
 #37
avatar+4691 
0

Thanks.

 

I don't know if just copying is gonna help me.

 

I'm gonna look at the latex sticky notes

 May 2, 2015
 #38
avatar+11885 
0

Whenever I look there I never understand! I'm just don't get it and get confused!

 

 

Maybe it it can help you! 

 May 2, 2015
 #39
avatar+520 
+10

Notice that \sqrt 36 will be rendered as though you want (sqroot of 3) times 6

If you want sqroot (36) you need to use braces around all digits, e.g., \sqrt {36}

 

Let's try it.  \sqrt36 compared with \sqrt{36}

 

$$\sqrt 36\ \ \ \sqrt {36}$$

 

 🆗  🆗  🆗

 May 2, 2015
 #40
avatar+4691 
0

(didn't see this)

 

Someone gets me... This is the forth time i've gone there...Still don't get it.

 

 

It's too complex and complicated for my brain to take in.

(I wish i'd joined earlier so i coul've discussed every step)

 

That's why i made this post to take it in steps..

 

http://web2.0calc.com/questions/which-is-bigger-3-8-or-0-3

 

First use of Latex it isn't the best but it's all i had...

 

It's a whole lot complicated when you click the Latex and you seewhat was actually typed in.

 May 2, 2015
 #41
avatar+11885 
+5

Definitely! I wonder how other users r so good at it! I wish I too could be like that! Whenever I try to understand it's like a beehive! I always go to that place in a relaxed mind hoping to get some input but nothing gets in and I come back all confused! 

 May 2, 2015
 #42
avatar+4691 
+5

$$\sqrt36$$

 

$$\sqrt{36}$$

 

It's the same Badinage.

 May 2, 2015
 #43
avatar+4691 
+5

Just like me. In steps is a whole lot easier.

 May 2, 2015
 #44
avatar+3088 
0

latex formula? hmmmmm lets test it out

well anyway, the square root of 36 is 6

 May 2, 2015
 #45
avatar+4691 
0

Or in Latex 

 

$$\sqrt36=6$$

.
 May 2, 2015
 #46
avatar+11885 
0

The idea of staring this post is really wise you know! I was always too lazy to understand latex I thought it's never my cup of tea! So I just let it be! But the bit know I use it! Just coz of you even I've got some bits known!

 May 2, 2015
 #47
avatar+520 
+5

It's not the same. Can't you see that there is no line over top of the 6?

 

Try it with a larger number then, say sqroot of 1500000 and you'll see the difference!

 

 May 2, 2015
 #48
avatar+3088 
0

yes!!I GOT IT RIGHT!!!!!

I learned today that even I can learn from the Math God

 May 2, 2015
 #49
avatar+4691 
0

Yep.

 

This is the LaTeX sticky topic just in easier steps to talk through.

 

$$Yep$$

 

 

$$\\\Yep\\This\:is\:the\:LaTeX\:sticky\:topic\:just\:in\:easier\:to\:talk\:through$$

 

Forgot steps.

 May 2, 2015
 #50
avatar+11885 
0

Wow you've learnt a lot mathsgod! And that to too fast too! I wish to be like you! 

 May 2, 2015
 #51
avatar+3088 
0

yup, fast indeed

 May 2, 2015
 #52
avatar+4691 
0

I guess i do have a thing in fast learning.But if you were in my possition under Melody's teaching you would've learnt it in no time!

 

I do still have questions: Like how do you make an extra line longer (like pressing enter twice)

 May 2, 2015
 #53
avatar+3088 
0

nah

the most important question that can never have a answer is:

why do we exist?

or

which came first? chicken or egg?

 May 2, 2015
 #54
avatar+4691 
0

True ... Very true.

 May 2, 2015
 #55
avatar+11885 
0

I think we should first let our teacher melody wake and then we can learn a bit more!ill definitely try my best!

 May 2, 2015
 #56
avatar+3088 
0

some questions are just meant to have no answer

and another question:

does the Illuminati really exist?

 May 2, 2015
 #57
avatar+4691 
0

Yes we should let Melody Sensei awake.

 

And TR please no interruptions with "unnecessary things"

 May 2, 2015
 #58
avatar+4691 
0

Badinage,

 

Sorry I can't show the LaTeX on my device because it isn't working.

 

But the line does have a strong effect so you have to do the { } or it won't be long enough. Thanks for the tip.

 May 2, 2015
 #59
avatar+105894 
+5

TitaniumRome,

MathsGod1 specifically requested that no one add irrelevant post to this thread.  It is distracting and 

posts like this one    http://web2.0calc.com/questions/latex-form#r53   do not enhance learning.  

MG, if you do not want irrelevant play posts in your thread then do not reply to them because that will encouraging them.

There are plenty of play threads here.  If anyone wants to play that is where they need to be.

 

Badinage and Rosala,  thank you,  your posts would have helped MG a lot :)

 

Mathsgod1 sent me a private message with this comment:

For every thing (not numbers or letters) do you Have to start with \ 

e.g fraction

I shall try and explain.

\ indicates that a function is following.  A function means that LaTex is required to do something different from just printing the letter or symbols that it sees.

The function indicator \ is followed by the function name like frac or sqrt

The function name is followed by the function arguments.

Some functions have no arguments like \ (a second one) which indicates a new line.

Some functions only have one arguement and they can just be written as is.   eg      \sqrt6

Most functions have more then one argument and to indicate this the arguments must be put into parentheses { }   eg     \sqrt{144}

I expect that you have probably understood by now what Badinage was telling you.

\sqrt144    will only put the sqrt sign over the 1 because that is the only argument.

\sqrt{144} will puut the sqrt sign over all of 144 because the parentheses indicate that all of 144 is the argument.

 

From there the argument idea just grows.

\frac must have two sets of arguements.    eg   \frac{1}{4}    output is        $$\frac{1}{4}$$

so does binomials combinations     \binom{7}{3}    $$$$              output is       $$\binom{7}{3}$$

 

There, that should help a little more.  I am very impressed with how fast you are learning this MG.

I only started learnig LaTex about a year ago, there is a great deal I do not know.  I am constantly learning - mainly from Heureka and Nauseated.

Once you know a little., which you do now, it is easy to expand that knowledge.  

Getting started with new concepts is ALWAYS the hardest part :)

 May 3, 2015
 #60
avatar+105894 
+5

Have a look what i showed you here MG

http://web2.0calc.com/questions/does-1-1-0#r4

 

The Latex is not always displaying properly this morning - I do not know why  :/

 May 3, 2015
 #61
avatar+4691 
+5

$$\\$LaTeX form$\\The\sqrt{36}\;is\;6$\\In addition, $3\frac{1}{4}\;and\;4\frac{1}{4} = 7\frac{1}{2}...$$

 

 

^(The Latex was supposed to be appear under the text. I randomnly put together everything i've learnt).

 

Thank you for writing all of this Melody. I fully understand the reason of \ now. Much appreciated! 

 

:D

This is the what i wrote.

 

I have some questions,

\\$LaTeX form$\\The\sqrt{36}\;is\;6$\\In addition, $3\frac{1}{4}\;and\;4\frac{1}{4} = 7\frac{1}{2}...

 

The section i made bold, i don't understand the Fraction turned all squashed out, however once I'd imput the $ it was clear. I didn't think I'd have to add it, as I've already placed at the beggining.

($\\In addition)

 

\\$LaTeX form$\\The\sqrt{36}\;is\;6$\\In addition, $3\frac{1}{4}\;and\;4\frac{1}{4} = 7\frac{1}{2}...

 

Where the ''and'' is i had to add \; or it would've been squashed together. I didn't think i had to add it because of the $ at the beggining.

 May 3, 2015
 #62
avatar+105894 
+5

Hi MathsGod1

Okay I haven't digested all of your problems but see if this helps.

THIS IS YOUR CODE:

\\$LaTeX form$\\

The\sqrt{36}\;is\;6$\\

In addition, $3\frac{1}{4}\;and\;4\frac{1}{4} = 7\frac{1}{2}...

 

$$\\$LaTeX form$\\The\sqrt{36}\;is\;6$\\In addition, $3\frac{1}{4}\;and\;4\frac{1}{4} = 7\frac{1}{2}...$$

ok that looks perfect to me :/  you must have fixed your problems.

 

1) If you use the $s to show text they must be in pairs,  one at the start and one at the end.

2)  you cannot have any functions in tthe middle of the text or the code will do crazy things.

3) I am really surprised that this works because I thought that you would need  the $ on the other side of the new line function (just before the "in addition" ):/

 

This is was my code I would have used :

\\$LaTeX form$\\The\sqrt{36}\;is\;6\\$In addition, $3\frac{1}{4}\;and\;4\frac{1}{4} = 7\frac{1}{2}...

 

$$\\$LaTeX form$\\The\sqrt{36}\;is\;6\\$In addition, $3\frac{1}{4}\;and\;4\frac{1}{4} = 7\frac{1}{2}...$$

 

But yours works just as well so it doens't matter :))

 May 3, 2015
 #63
avatar+4691 
+5

\\$LaTeX form$\\The\sqrt{36}\;is\;6$\\In addition, $3\frac{1}{4}\;and\;4\frac{1}{4} = 7\frac{1}{2}...

 

Ahh... I see the problem.

 

So after a new line you have start the $ and end it after the line is finished i thought the $ would continue after the 6$.

 

Thanks for clearing that up. :)

 May 3, 2015
 #64
avatar+4691 
0

What else can i digest?     :)

 May 3, 2015
 #65
avatar+105894 
0

okay MathsGod1 here is a really handy thing to know

 

 

y=8* (\frac{2}{4}-\frac{\frac{6^3}{2}}{15})

$$y=8* (\frac{2}{4}-\frac{\frac{6^3}{2}}{15})$$

 

Lets look at what i have written here.

I would like the times sign to be a proper times sign   that is   \times

Also, the brackets are too little - i want them to be bigger   I do this with  \left(   .......   \right)

like this

 

y=8\times \left(\frac{2}{4}-\frac{\frac{6^3}{2}}{15}\right)

$$y=8\times \left(\frac{2}{4}-\frac{\frac{6^3}{2}}{15}\right)$$

 

Now, why don't you try to solve this and show all of your working.  

You will need to know that \div is a division sign.       :)

 

This might challenge your latex as well as your maths :)   You can check your answer with the forum calc :)

 May 3, 2015
 #66
avatar+4691 
0

Should i work it out in LaTeX?

 May 3, 2015
 #67
avatar+105894 
0

Well MG you can present the whole answer in LaTex    

 May 3, 2015
 #68
avatar+4691 
0

$$\\y=8\times \left(\frac{2}{4}-\frac{\frac{6^3}{2}}{15}\right)\\(2 \div 4 =2)\\y=8\times \left(2-\frac{\frac{6^3}{2}}{15}\right)\\6^3=(6\times6\times\66)=216\\216 \div 2= 108\\y=8\times \left(2-7.2\right)\\2-7.2=5.2\\y=8\times-5.2=-41.6\\y=41.6$$

 

 

So many mistakes...

 

 

I meant -5.2.  2 - 7. 2 = -5.2. And i forgot to show 108/15=7.2

 

I think this is wrong ?..

 

But that's all my LaTeX!

 

CALCULATOR:

 

$${\mathtt{y}} = {\mathtt{8}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{2}}}{{\mathtt{4}}}}\right){\mathtt{\,-\,}}\left({\frac{{\frac{{{\mathtt{6}}}^{{\mathtt{3}}}}{{\mathtt{2}}}}}{{\mathtt{15}}}}\right) \Rightarrow {\mathtt{y}} = -{\mathtt{3.2}}$$

 

 

 

ooooohh.....Where did i go wrong...

 

 May 3, 2015
 #69
avatar+4691 
0

\\y=8\times \left(\frac{2}{4}-\frac{\frac{6^3}{2}}{15}\right)\\(2 \div 4 =2)\\y=8\times \left(2-\frac{\frac{6^3}{2}}{15}\right)\\6^3=(6\times6\times\66)=216\\216 \div 2= 108\\y=8\times \left(2-7.2\right)\\2-7.2=-5.2\\y=8\times-5.2=-41.6\\y=41.6

 

 

What i typed down.

 May 3, 2015
 #70
avatar+105894 
+10

MG,

 

$$2\div 4 \mathbf{\;\ne \; 2}$$

 

Your LaTex is great but you have to get the answer right too.     LOL   :)))

 May 3, 2015
 #71
avatar+4691 
0

Ahhh.... I was to focused on latex i thought it was 4/2... I kew tnere was something funky with that question!LOL

 May 3, 2015
 #72
avatar+105894 
0

So are you going to fix your post ?

 May 3, 2015
 #73
avatar
0

˙ˍ¯þ ʼß -¯þ´.

 May 3, 2015
 #74
avatar+4691 
+5

Ok then.

 

$$\\y=8\times \left(\frac{2}{4}-\frac{\frac{6^3}{2}}{15}\right)\\(2 \div 4 =0.5)\\y=8\times \left(0.5-\frac{\frac{6^3}{2}}{15}\right)\\6^3=(6\times6\times\66)=216\\216 \div 2= 108\\y=8\times \left(0.5-7.2\right)\\0.5-7.2=-6.7\\y=8\times-6.7=-41.6\\y=-53.6$$

 

 Keeping doing dumb mistakes... Meant 53.6 after 8*-6.7

What did i do wrong..?!  I don't know...

 May 3, 2015
 #75
avatar+4691 
+5

Hmmm. I'm not sure why it's still wrong.

 

Melody could you give me a question similar to this question please? And could you write it in text. So I don't have the urge to copy and paste.  

 

:D

 May 3, 2015
 #76
avatar+4691 
+5

y=8\times \left(\frac{2}{4}-\frac{\frac{6^3}{2}}{15}\right)

 

I have some questions to do with this code:

 

  • Near the 2 (after 6^3) why is there two ]} ?
  • After the - (after {2}{4}) why is there two frac sign?

 

 

Sorry if it sounds rude, but I just want to know the reason why.

 

Thanks :)

 May 4, 2015
 #77
avatar+105894 
+5

Hi MG, 

It is not rude to ask for explanation.        

I am very pleased that you are learning from me.  That is what I am here for.  -   to teach people 

Plus intelligent, enthusiastic students like you are fun and rewarding to teach. 

 

 $$y=8\times \left(\frac{2}{4}-\frac{\frac{6^3}{2}}{15}\right)$$

 

\frac{\frac{6^3}{2}}{15}

$$\frac{\frac{6^3}{2}}{15}$$

I have 'nested' a fraction inside another fraction,  that is why there are two \frac   

The black parentheses belongs to the main fraction.  The red parentheses belong to the nested fraction.

 May 4, 2015
 #78
avatar+105894 
+10

Try reproducing this in Latex

 

 May 4, 2015
 #79
avatar+4691 
+5

$$x= \frac{-b ±\sqrt][b^2-4ac]}{2a}$$

 

 

x= \frac{-b ±\sqrt][b^2-4ac]}{2a}

It was hard but this is all i could've done.

 

I used this sign ± but some Á appeared...

 

And as for the two brackets after the sqrt.

 

This is what it looks like when take one out

 

 

$$x= \frac{-b ±\sqrt[b^2-4ac]}{2a}$$

.
 May 4, 2015
 #80
avatar+105894 
+5

Hi MathsGod1      

the \ indicates that a function follows,

then the arguments of the function must be in curly brackets  {parentheses}

The plus minus symbol is a new function for you.  It is  \pm

You should not be using any square brackets here at all.

 

Do you want to try and fix it and we will see how you go       You  are doing really well !

 May 4, 2015
 #81
avatar+4691 
+10

 

Ok

Ha!!!

 

I feel so proud! Thanks Melody, seriously thanks. I never though i'd take LaTeX this far (although this is nothing compared to what you do) I always thought I'd never understand these symbols everywhere.

 

I done it!    :D

 

PS:   ( Thanks for the compliments that gave me the Spirit to succeed  in this) Lol

 

 

 

 

 

 

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

 May 4, 2015
 #82
avatar+105894 
+10

hi MathsGod1 

 

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

 

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

 

so you have done it - Great job MG

 

If you want it to display bigger you can put the fraction in 'display' mode using  dfrac

like this

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

 

$$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$

.
 May 5, 2015
 #83
avatar+4691 
+15

Cool! :)

 

$$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$

.
 May 5, 2015
 #84
avatar+105894 
+5

That looks great MG  :)

 May 6, 2015
 #85
avatar+4691 
+10

Thank you :)

 

Now I have to go school to learn some more maths!!! :D

 

Bye Melody :)

 May 6, 2015
 #86
avatar+4691 
+15
Best Answer

Hello Melody,

 

I was wondering if there was anything else i could learn / practise on LaTeX.

 

Thanks.

 

 

 

EDIT: I've been trying to find an equation similar to that Quadratic formula...But I couldn't find anything.I want a challenge like that even if it doesn't make sense :D

 A challenge is always fun!

MathsGod1 May 6, 2015
 #87
avatar+105894 
+10

I coded this today - you could try and repeat it :/      Have fun :)

 

 May 7, 2015
 #101
avatar+7747 
0

\(\left(z-\frac{2}{\sqrt{z}}\right)^9\)

\(\text{The general term is}\)

\((9Cr)(z)^{(9-r)}\left(\frac{-2}{\sqrt{z}}\right)^r\)

\(=(9Cr)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z})^r}\right)\)

\(=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(\sqrt{z})^r}\right)\)

\(=(9Cr)\left(\frac{z^{18/2}*(-2)^r}{z^{(2r/2)}*(z)^{(r/2)}}\right)\)

\(=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\)

\(=(9Cr)(z^{((18-3r)/2)}*(-2)^r)\)

\(\text{The constant term will be when}\)

\((18-3r)/2=0\)

\(18-3r=0\)

\(r=6\)

Just wanna try. This code is really crazy! Typing it twice takes me 30minutes. (I accidentally deleted when I have already finished)

MaxWong  Jul 16, 2016
 #89
avatar+4691 
+5

This seems hard! I'll try :)

 

hmmm... I done this code but it's not working it keeps buffering... This was my code 

(so far) it was looking good

 

\\\left(z-\frac{2}{\sqrt{z}}\right)\\$$\\The\:general\:term\:is\\$$\\=(9Cr)(z)(9-r)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z}^r}\right)$$\\$$=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(sqrt{z}^r)}\right)$$\\$$=(9Cr)\left(\frac{z^{18/2}*(-2)^r}{z^{(2r/2)}*(z)^{(r/2)}\right)$$\\$$=(9Cr)

 May 7, 2015
 #90
avatar+4691 
+10

I know there is different types of brackets...

 

{ }  =   Extra inside info ( \frac{2}{3} and sqrt{900})

( ) = Actual brackets for maths.

[ ] = But what are these for?

 May 8, 2015
 #91
avatar+4691 
+10

\\\left(z-\frac{2}{\sqrt{z}}\right)\\$$\\The\:general\:term\:is\\$$\\
=(9Cr)(z)(9-r)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z}^r}\right)$$\\$$
=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(sqrt{z}^r)}\right)$$\\$$
=(9Cr)\left(\frac{z^{18/2}*(-2)^r}{z^(2r/2)*(z)^(r/2)\right)$$\\$$
=(9Cr)\left(z^((18-2r-r))/2)*(-2)^r{1}\righr)$$\\$$
=(9Cr)(z^((12-3r)/2)*(-2)^r)$$\\$$
The\:constant\:term\:will\:be\:when$$\\$$

(18-3r)/2=0\\
18-3r=0\\
r=6

 

This was the code at the top of my head.

(It still buffers - crazy!)

 

The bit I'm confused in...

 

\\\left(z-\frac{2}{\sqrt{z}}\right)\\$$\\The\:general\:term\:is\\$$\\
=(9Cr)(z)(9-r)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z}^r}\right)$$\\$$
=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(sqrt{z}^r)}\right)$$\\$$
=(9Cr)\left(\frac{*(-2)^r}{z^(2r/2)*(z)^(r/2)\right)$$\\$$
=(9Cr)\left(z^((18-2r-r))/2)*(-2)^r{1}\righr)$$\\$$
=(9Cr)(z^((12-3r)/2)*(-2)^r)$$\\$$
The\:constant\:term\:will\:be\:when$$\\$$

(18-3r)/2=0\\
18-3r=0\\
r=6

 

The bold bit. Theres more of that...But why is does it come out normal.

 

$$z^{18/2}$$

 

When { } is taken out

 

$$z^18/2$$

 

 

I would just add { } but it affect my fractions...

 

 

EDIT: Ok, I'll do what you said.

 

It's been buffering for about 5 minutes i think it's because the code is wrong so its not responding but i don't know what's wrong.

 

http://i.imgur.com/lh7sCU8.png

 May 8, 2015
 #92
avatar+4691 
+5

Melody... Did you get this?

 May 9, 2015
 #93
avatar+105894 
0

Hi MG

I would like you to edit your post.

Put the ouput on the top using the Latex Box and put the code underneath so that I can see if it is working and if it is not I will help you sort out why :)

 May 9, 2015
 #94
avatar+105894 
0

Are you telling me that the code does not work but you do not know why?

 May 9, 2015
 #95
avatar+105894 
0

Hi MathsGod1    

 

Lets look at this one line at a time

LINE 1

\\\left(z-\frac{2}{\sqrt{z}}\right)\\$$\\The\:general\:term\:is\\$$\\

 

$$\\\left(z-\frac{2}{\sqrt{z}}\right)\\$$\\The\:general\:term\:is\\$$\\$$

Well the output is good but Iam wondering what those 2 double dollar signs  are for.  

I doubt that they do anything

LINE 2

=(9Cr)(z)(9-r)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z}^r}\right)$$\\$$

 

$$=(9Cr)(z)(9-r)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z}^r}\right)$$\\$$

That looks good but you are missing a little bracket  :)

 

LINE 3

=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(sqrt{z}^r)}\right)$$\\$$

 

$$=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(sqrt{z}^r)}\right)$$\\$$$$

that is good - you are missing a backslash though.  :)

LINE 4

 

=(9Cr)\left(\frac{z^{18/2}*(-2)^r}{z^(2r/2)*(z)^(r/2)\right)$$\\$$

That one doesn't work at all,  I  can see straight off that you are missing a  }   

Once you have made one fatal error, nothing will work, and that is a fatal error.

------------

A quick look at the rest and I can see that right is spelt incorrectly. I do not know if there are any other problems.  but please try it without the double dollar signs - where did you get that idea from  


=(9Cr)\left(z^((18-2r-r))/2)*(-2)^r{1}\righr)$$\\$$
=(9Cr)(z^((12-3r)/2)*(-2)^r)$$\\$$
The\:constant\:term\:will\:be\:when$$\\$$

(18-3r)/2=0\\
18-3r=0\\
r=6

 

There you go - you try fixing this bit - I'll be back but I have to do some other work on the forum :)

 May 9, 2015
 #96
avatar+4691 
+5

Well yes, but this always happens to me when the code is incorrect,however when fixed the buffer goes...

 

But the thing is...I don't know whats wrong with the code..?

 

EDIT: Just refreshed page,you've already answered my question :)

 May 9, 2015
 #97
avatar+4691 
+5

The $$ at first i was trying to leave extra spaces between he words but I just stuck with \; .

I forgot to take them out also i was trying to leave bigger gaps between the equations I know you have to do \\ and with \\.

 

But how do you leave a slightly bigger gap.

 

LaTeX.

 

$$\\=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\
=(9Cr)(z^{((18-3r)/2)}*(-2)^r)\\
The\:constant\:term\:will\:be\:when\\
(18-3r)/2=0\\
18-3r=0\\
r=6$$

 

It's alot easier seeing the actual coding underneath

 May 9, 2015
 #98
avatar+4691 
+5

The code didn't display...

 

$$\\=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\
=(9Cr)(z^{((18-3r)/2)}*(-2)^r)
\\The\:constant\:term\:will\:be\:when\\
(18-3r)/2=0\\18-3r=0\\r=6$$

 

Here's the code Melody.

 

\\=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\
=(9Cr)(z^{((18-3r)/2)}*(-2)^r)
\\The\:constant\:term\:will\:be\:when\\
(18-3r)/2=0\\18-3r=0\\r=6

 May 9, 2015
 #99
avatar+4691 
+10

$$\\\left(z-\frac{2}{\sqrt{z}}\right)\\\\The\:general\:term\:is\\\\
=(9Cr)(z)(9-r)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z}^r}\right)\\
=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(sqrt{z}^r)}\right)\\
=(9Cr)\left(\frac{*(-2)^r}{z^(2r/2)*(z)^(r/2)}\right)\\
=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\
=(9Cr)(z^{((18-3r)/2)}*(-2)^r)
\\The\:constant\:term\:will\:be\:when\\
(18-3r)/2=0\\18-3r=0\\r=6$$

 

This is the entire coding correct...

 

Hopefully it shows...My laptop might not be able to take in all this LaTeX.

 

 

\\\left(z-\frac{2}{\sqrt{z}}\right)\\\\The\:general\:term\:is\\\\
=(9Cr)(z)(9-r)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z}^r}\right)\\
=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(sqrt{z}^r)}\right)\\
=(9Cr)\left(\frac{*(-2)^r}{z^(2r/2)*(z)^(r/2)}\right)\\
=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\
=(9Cr)(z^{((18-3r)/2)}*(-2)^r)
\\The\:constant\:term\:will\:be\:when\\
(18-3r)/2=0\\18-3r=0\\r=6

 May 9, 2015
 #100
avatar-59 
+5

ah, what's this?

oh it s the god of mathematics! i'm so scared! run before he blasts numbers in your brains! (sarcasm)

ah, we have some interesting crime activity to catch

 May 9, 2015
 #102
avatar+7747 
0

LOL

MaxWong  Jul 16, 2016
 #101
avatar+4691 
+10

Hahaha!

 

Lol, if I'm the god then I can't describe what Melody is, she taught me this!

 May 9, 2015

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