Hi Janne,
Personnally I prefer to use the Plain TeX alternative \over (which works fine
in ConTeXt), that is
${a \over b}$
instead of
$\frac{a}{b}$
Compare the following two outputs in the example you want to typeset: I think
the second is more or less what you want
\starttext
Using \type{\frac} gives:
\startformula
f_{B_t | B_s = S, B_u = U}(x) = \frac{e^{-\frac{(u-s)x^2 - 2x(S(u-t) +
U(t-s)) + \frac{(S(u-t) +
U(t-s))^2}{(u-s)}}{2(t-s)(u-t)}}}{\sqrt{2\pi\frac{(t-s)(u-t)}{u-s}}}
\stopformula
\blank
Using \type{\over} gives:
\startformula
f_{B_t | B_s = S, B_u = U}(x) = {
e^{-{(u-s)x^2 - 2x(S(u-t) + U(t-s)) +
{(S(u-t) + U(t-s))^2 \over (u-s)} \over 2(t-s)(u-t)}} \over
\sqrt{2\pi {(t-s)(u-t) \over u-s}}}
\stopformula
\stoptext
Best regards: OK
On 28 janv. 2013, at 10:11, Janne Junnila <janne.junn...@gmail.com> wrote:
> Indeed it seems like the alignment is good with \dfrac, but this does
> not solve my problem, since I wish to also use fractions with
> script-size or scriptscript-size (\xfrac, \xxfrac). The specific
> formula I have is
>
> \startformula
> f_{B_t | B_s = S, B_u = U}(x) = \frac{e^{-\frac{(u-s)x^2 - 2x(S(u-t) +
> U(t-s)) + \frac{(S(u-t) +
> U(t-s))^2}{(u-s)}}{2(t-s)(u-t)}}}{\sqrt{2\pi\frac{(t-s)(u-t)}{u-s}}}
> \stopformula
>
> Thanks,
> Janne
>
> Roland wrote:
>> With \dfrac it looks good.
>> With \fraction the minus sign is on the top of the fraction.
>> Best regards, Roland
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