- Thread starter
- #1

#### Alexmahone

##### Active member

- Jan 26, 2012

- 268

Normal: $\sum\frac{n^5}{2^n}$

Using \displaystyle: $\displaystyle\sum\frac{n^5}{2^n}$

- Thread starter Alexmahone
- Start date

- Thread starter
- #1

- Jan 26, 2012

- 268

Normal: $\sum\frac{n^5}{2^n}$

Using \displaystyle: $\displaystyle\sum\frac{n^5}{2^n}$

We had this same issue on MHF as well.

Normal: $\sum\frac{n^5}{2^n}$

Using \displaystyle: $\displaystyle\sum\frac{n^5}{2^n}$

- Jan 26, 2012

- 890

The thing is that single dollar sign delimiters $\displaystyle \frac{x}{y}$ are for in-line maths-notation, making it larger means that the maths will foul the text in following lines.

Normal: $\sum\frac{n^5}{2^n}$

Using \displaystyle: $\displaystyle\sum\frac{n^5}{2^n}$

Or make the line spacing irregular, though you have too keep typing for a long time to get test wrapping to test this

More uses for \displaystyle . . .

\lim{x\to3}\frac{x^2-9}{x-3} . . . . . . . . . . . . . $\lim_{x\to3}\frac{x^2-9}{x-3}$

\displaystyle \lim_{x\to3}\frac{x^2-9}{x-3} . . . $\displaystyle \lim_{x\to3}\frac{x^2-9}{x-3} $

sum^{\infty}_{n=1}\frac{1}{n^2} . . . . . . . . . . . . . $\sum^{\infty}_{n=1} \frac{1}{n^2}$

\displaystyle \sum^{\infty}_{n=1}\frac{1}{n^2} . . . $\displaystyle \sum^{\infty}_{n=1}\frac{1}{n^2}$

\int^3_1 x^2\,dx . . . . . . . . . . . . $\int^3_1 x^2\,dx $

\displaystyle \int^3_1 x^2\,dx . . . $\displaystyle \int^3_1 x^2\,dx$

If \displaystyle makes a fraction too large:

. . \displaystyle \frac{x-3}{x+4} + \frac{1}{2}x^3 . . . . $\displaystyle \frac{x-3}{x+4} +\frac{1}{2}x^3$

it can be reduced with \tfrac:

. . \displaystyle \frac{x-3}{x+4} + \tfrac{1}{2}x^3 . . . $\displaystyle \frac{x-3}{x+4} + \tfrac{1}{2}x^3$

- Moderator
- #5

- Jan 26, 2012

- 995

If you want to displaystyle fractions, it's easier to use the \dfrac{}{} command.If \displaystyle makes a fraction too large:

. . \displaystyle \frac{x-3}{x+4} + \frac{1}{2}x^3 . . . . $\displaystyle \frac{x-3}{x+4} +\frac{1}{2}x^3$

it can be reduced with \tfrac:

. . \displaystyle \frac{x-3}{x+4} + \tfrac{1}{2}x^3 . . . $\displaystyle \frac{x-3}{x+4} + \tfrac{1}{2}x^3$

\frac{x+1}{x-1} yields $\frac{x+1}{x-1}$, but \dfrac{x+1}{x-1} yields $\dfrac{x+1}{x-1}$

So in the example you provided, we could say \dfrac{x-3}{x+4} +\frac{1}{2}x^3, which would give $\dfrac{x-3}{x+4}+\frac{1}{2}x^3$ (thus, \tfrac{}{} can be avoided in the inline case). I would say that the \tfrac{}{} option is most useful in scenarios when the rendered code is centered by $$ or \[ (the cases where the equations are automatically displaystyled).

I did know all that . . . and didn't get around to explain it,

. . but thanks for clarifying it.

I intended the \tfrac feature to be used when \displaystyle must be used.

$\displaystyle \text{Here's an example: }\:\int^5_3\frac{x^2-4}{x-2}\,dx$

To produce the large integral sign, I used \displaystyle.

. . It also produced the large fraction.

. . $\displaystyle \int^5_3\frac{(x-2)(x+2)}{x-2}\,dx \;=\;\int^5_3(x+2)\,dx \;=\;\frac{1}{2}x^2 + 2x\,\bigg]^5_3 $

To me, the $\dfrac{1}{2}$ seems awkwardly large (but maybe that's just me).

. . I would use \tfrac here.