Question

Any \(\Latex\) experts online?

Answer 1

I'm working on a proof, and I have this section:
\[f(x) = \displaystyle\sum\limits_{n=0}^{\infty} \frac{(-1)^n(x^{2n})\displaystyle\prod\limits_{k=0}^{n-1}(2k-1)}{\sqrt{2\pi}(2n)!}\]
Anyone know how to make the sigma bigger, so it actually looks good?

Answer 2

Use the command \Large before the whole thing

Answer 3

And use the command \LaTeX to generate \(\LaTeX\)

Answer 4

My initial thought was to play with the size commands \Large and \small.
I don't think that will make it quite as elegant as you'd like, but I am not having any better ideas at the moment. :)

Answer 5

\[
f(x) = \displaystyle \Large \sum \limits_{n=0}^{\infty} \small \frac{(-1)^n(x^{2n})\displaystyle\prod\limits_{k=0}^{n-1}(2k-1)}{\sqrt{2\pi}(2n)!}
\]
Something like that...

Answer 6

\[\Large f(x) = \displaystyle\sum\limits_{n=0}^{\infty} \frac{(-1)^n(x^{2n})\displaystyle\prod\limits_{k=0}^{n-1}(2k-1)}{\sqrt{2\pi}(2n)!}\]

Answer 7

This is so elegant isn't it

Answer 8

but I only need the sigma bigger.
and yes it is elegant. It is the maclaurin series for the standard normal curve.

Answer 9

\[f(x) = \displaystyle \Huge \sum \limits_{n=0}^{\infty} \large \frac{(-1)^n(x^{2n})\displaystyle\prod\limits_{k=0}^{n-1}(2k-1)}{\sqrt{2\pi}(2n)!}\]

Answer 10

Winner

Answer 11

lol i just took access's code and changed a word

Answer 12

What about this

Answer 13

http://www.stat.berkeley.edu/~paciorek/computingTips/Type_sizes_changing_type_si.html

Answer 14

\[\Large f(x) = \displaystyle \Huge \sum \limits_{n=0}^{\infty} \left[\large \frac{(-1)^n(x^{2n})\displaystyle\prod\limits_{k=0}^{n-1}(2k-1)}{\sqrt{2\pi}(2n)!}\right]\]

Answer 15

Very nice!

I've had this reference sitting on a tab for a while:
http://estudijas.lu.lv/pluginfile.php/14809/mod_page/content/12/instrukcijas/matematika_moodle/LaTeX_Symbols.pdf
All the font sizes are at the bottom, and it has a bunch of other stuff if needed. :)

Answer 16

@myininaya thanks for your help but for some reason the Tex editor I'm using (writelatex.com) isn't recognizing you're design.

Answer 17

What is showing up instead? Is it all large font?

Answer 18

yeah.

Answer 19

Maybe enclose it in brackets
\Huge{ \sum \limits_{n=0}^{\infty} }
\Large{ \frac{(-1)^n(x^{2n})\prod\limits_{k=0}^{n-1}(2k-1)}{\sqrt{2\pi}(2n)!} }
\[ \displaystyle \Large f(x) = \Huge{ \sum \limits_{n=0}^{\infty}}
\Large{ \frac{(-1)^n(x^{2n})\prod\limits_{k=0}^{n-1}(2k-1)}{\sqrt{2\pi}(2n)!} } \]

Also removed the \displaystyle on product symbol because it is technically redundant. Other than that, if it doesn't want to cooperate I don't think there's a better way (I couldn't make a sum itself as a left delimiter...:c )

Answer 20

no change. i wish the delimiter would work.

Answer 21

\left\{ instead of \left{

Answer 22

\[\left\{\begin{matrix}\mbox{IT}\\\mbox{IS}\\\fbox{WORKING}\end{matrix}\right\}\]

Answer 23

will that let me use the sum as a delimiter?

Answer 24

What's a delimiter? Sorry English isn't my native language

Answer 25

it isn't really english...
it is the Latex function that makes the parentheses or brackets the size of the function. For example, instead of
\[(\frac{1}{2}\], it gives \[\left( \frac{1}{2} \right)\]

Answer 26

Testing:
\[\left\sum tesing \right\sum\]

Answer 27

No it won't

Answer 28

I don't know the particulars of your editor if you can use other packages, so I link this out of faith that maybe it is useful.
http://tex.stackexchange.com/questions/22773/making-a-big-summation-sign
I couldn't make any weird tricks work like \begin{array} with the sum before it, so I am out of ideas. Anyways, best of luck with this! I gotta go. :)

Answer 29

thanks for the help! I'll keep trying...