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Guide to LaTeX (4th Edition) (Tools and Techniques

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12.2. St<strong>and</strong>ard features of A M S-L AT E X 273<br />

\begin{equation}\begin{split}<br />

H_c&=\frac{1}{2n}\sum_{l=0}ˆn (-1)ˆl (k-l)ˆ{p-2}<br />

\sum_{l_1+\dots+l_p=l} \prod_{i=1}ˆp \binom{n_i}{l_i}\\<br />

&\quad\times[(k-l) - (k_i-l_i)]ˆ{k_i-l_i}\times<br />

\Bigl[(k-l)ˆ2 - \sum_{j=1}ˆp (k_i-l_i)ˆ2\Bigr]<br />

\end{split}\end{equation}<br />

The alignment has been chosen <strong>to</strong> be just before the equals sign. The<br />

second line begins with the alignment marker & so that the equation<br />

continues below the = in the first line. A \quad has been inserted so<br />

that the × is not immediately below the equals sign. Alternatively, one<br />

could place the & after the = <strong>and</strong> dispense with \quad in the second line.<br />

However, the above is more suitable when there are continuation lines<br />

beginning with =. Note the centered equation number at the right.<br />

The gather environment<br />

The gather environment switches <strong>to</strong> math mode, centering each of its<br />

formula lines without any alignment. The formula lines are separated by<br />

\\ comm<strong>and</strong>s. Each line receives an equation number, unless the *-form<br />

has been used, or \notag has been issued in that line.<br />

1<br />

2 +<br />

4 9 n2 ∞ n2 2 3<br />

n<br />

n<br />

+ + · · · +<br />

+ · · · =<br />

(12.3)<br />

3 4<br />

n + 1<br />

n + 1<br />

n=1<br />

⎛<br />

<br />

n2 n n<br />

⎜<br />

converges since lim<br />

= lim ⎜ 1<br />

n→∞ n + 1 n→∞ ⎝<br />

1 + 1<br />

⎞n<br />

⎟<br />

⎠ =<br />

n<br />

1<br />

< 1<br />

e<br />

root condition<br />

2 + 3<br />

∞<br />

4 n + 1<br />

n + 1<br />

+ + · · · + + · · · =<br />

4 9 n2 n<br />

n=1<br />

2<br />

(12.4)<br />

∞<br />

<br />

x + 1<br />

diverges since<br />

dx = ln x −<br />

x2 1<br />

∞ = ∞ (integral condition)<br />

x<br />

c<br />

\begin{gather}<br />

\frac{1}{2} + \left(\frac{2}{3}\right)ˆ4 + \left(\frac{3}{4}<br />

\right)ˆ9 + \dots + \left(\frac{n}{n+1}\right)ˆ{nˆ2} + \dotsb<br />

= \sum_{n=1}ˆ\infty \left(\frac{n}{n+1}\right)ˆ{nˆ2} \\<br />

\text{converges since}\quad\lim_{n\<strong>to</strong>\infty}<br />

\sqrt[n]{\left(\frac{n}{n+1}\right)ˆ{nˆ2}} = \lim_{n\<strong>to</strong>\infty}<br />

\left(\frac{1}{1 + \dfrac{1}{n}}\right)ˆn = \frac{1}{\me} < 1<br />

\tag*{root condition}\\<br />

2 + \frac{3}{4} + \frac{4}{9} + \dots + \frac{n+1}{nˆ2} +<br />

\dotsb = \sum_{n=1}ˆ\infty \frac{n+1}{nˆ2}\\<br />

\text{diverges since}\quad\int_cˆ\infty \frac{x+1}{xˆ2}\,<br />

c

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