\begin{align}
x &= y
\end{align}
\marginnote[-\baselineskip]{This is because I said so.}
or
\marginnote[\baselineskip]{This is because I said so.}
\begin{align}
x &= y
\end{align}
The command marginnote has the full specification
\marginnote[<offset>]{Required argument}
where [<offset>] indicates that the margin note should be
shifted down (or up if negative) the given amount.
If your original attempt worked, then the number generated
by the sidenote might make the equation look wrong. That's
my primary reason for suggesting the use of marginnote. If
you instead try the above approach with sidenote, you will
need to do something like
\sidenote[][\baselineskip]{This is because I said so.}
\begin{align}
x &= y
\end{align}%
This is because sidenote has the full specification
\sidenote[<number>][<offset>]{Required argument}
where [<number>] supplies a different mark or number for
the sidenote. If you want to supply the offset, but still
use the default number, then you MUST supply an empty
first argument. The problem now, however, is that the
sidenote number is likely to be hanging out in space. So,
yet another reason to use marginnote.
Hope this helps!
Happy Holidays,
-Dan
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--
Dan T. Abell :: dabell at txcorp dot com :: 303.444.2452
Tech-X Corp., 5621 Arapahoe Ave, Ste A, Boulder CO 80303
http://www.txcorp.com :: 303.748.6894/c 303.448.7756/fx
On Thu, Dec 23, 2010 at 8:23 PM, hal <hald...@gmail.com> wrote:
> Hrmm.... That works in this case, but I guess I wasn't sufficiently
> specific. I really want to do something like:
>
> \begin{align}
> y &= (x-2z)^2 \marginnote{By definition} \\
> &= x^2 + 4z^2 - 4xz \marginnote{By quadratic rule} \\
> &\le x^2 + 4z^2 \marginnote{x,z assumed positive}
> \end{align}
>
> Or something like that. I suppose I could do it with a bunch of
> clever uses of offsets, but that's kind of a hassle.
>
> (Of course, alternative suggestions of how to typeset this are also
> welcome!)
Since you're already using the align environment, you could place your
explanatory text after &&:
\begin{align}
y &= (x-2z)^2 && \text{By definition} \\
&= x^2 + 4z^2 - 4xz && \text{By quadratic rule} \\
&\le x^2 + 4z^2 && \text{$x$, $z$ assumed positive}
\end{align}
(Note that the \text command is provided by the amsmath package. You
can replace it with \textrm if you prefer.)
--Kevin Godby