3

I've tried using the split environment as well as dfrac, however, I can't seem to be able to make this equation look nice at all.

\documentclass{article}
\usepackage{amsmath}
    \begin{document}
        \begin{equation*} 
            \begin{split}
            f_{xx}(t,x) = &\dfrac{1}{\underbrace{\left(2t^{5/2}  
         \left(e^{x^2/4t}+\sqrt{\dfrac{1}{t}}\right)^2\right)^2}_\text{c}}\\
         &\left(-\dfrac{x\left(e^{x^2/4t} \left(2t+x^2\right) \right)}{2t} 
            \cdot 2t^{5/2}\left(e^{x^2/4t}+\sqrt{\dfrac{1}{t}}\right)^2\right) - \\ 
            &\left(\left( e^{x^2/4t}\left(2t - x^2\right) +2\sqrt{t}\right) \cdot \left(2txe^{x^2/4t}\left(\sqrt{t}e^{x^2/4t}+1 \right)\right)\right)
            \end{split}
        \end{equation*}
    \end{document}

I took out the denominator and multiplied the entire numerator by 1 over the fraction, but the equation becomes quite unbalanced.

Edit: The denominator should be multiplied to all the terms that follow, I apologize if my code is unclear.

Thanks for your time.

  • You're using a split environment yet do not provide & alignment anchors. Is this intentional? – Mico Jul 11 '15 at 1:52
  • @Mico Thanks for pointing that out. Edited to include the alignment anchors. – Jun-Goo Kwak Jul 11 '15 at 1:56
1

I suggest you

  • indent the second line relative to the first and indent the third line even more to indicate the mathematical structure of the full formula; as per your follow-up comment, I've placed square brackets around the material in lines two and three;

  • replace the two \sqrt{\dfrac{1}{t}} subformulas with \sqrt{1/t};

  • not use \left( and \right to auto-size the parentheses; instead, use \big, \Big, and \bigg judiciously to size the "fences".

Note that since the material in a displayed equation is in so-called "display style", it doesn't make a difference whether one uses \frac or \dfrac for the two remaining fractional expressions.

enter image description here

\documentclass{article}
\usepackage{amsmath}
\begin{document}
before:
\begin{equation*} 
\begin{split}
    f_{xx}(t,x) = &\dfrac{1}{\underbrace{\left(2t^{5/2}  
 \left(e^{x^2/4t}+\sqrt{\dfrac{1}{t}}\right)^2\right)^2}_\text{c}}\\
 &\left(-\dfrac{x\left(e^{x^2/4t} \left(2t+x^2\right) \right)}{2t} 
    \cdot 2t^{5/2}\left(e^{x^2/4t}+\sqrt{\dfrac{1}{t}}\right)^2\right) - \\ 
    &\left(\left( e^{x^2/4t}\left(2t - x^2\right) +2\sqrt{t}\right) \cdot 
      \left(2txe^{x^2/4t}\left(\sqrt{t}e^{x^2/4t}+1 \right)\right)\right)
\end{split}
\end{equation*}

\bigskip
after:

\begin{equation*} 
\begin{split}
f_{xx}(t,x) 
&= \frac{1}{\underbrace{\Bigl(2t^{5/2}  
 \bigl(e^{x^2/4t}+\sqrt{1/t}\,\bigr)^2\Bigr)^2}_\text{c}}\\
 &\quad\times\biggl[-\frac{x\bigl(e^{x^2/4t} (2t+x^2) \bigr)}{2t} 
    \cdot 2t^{5/2}\bigl(e^{x^2/4t}+\sqrt{1/t}\,\bigr)^2 \\ 
    &\qquad\quad- \Bigl( e^{x^2/4t}(2t - x^2) +2\sqrt{t}\,\Bigr) 
    \Bigl(2txe^{x^2/4t}\bigl(\sqrt{t}\,e^{x^2/4t}+1 \bigr)\Bigr)\biggl]
    \end{split}
\end{equation*}

\end{document}

Addendum: Since you're assigning the symbol c to the denominator part, one might as well make use of it to simplify the equation still further (this is very similar to the idea Gonzalo Medina used in his answer):

enter image description here

\documentclass{article}
\usepackage{mathtools} % for "\shortintertext" macro
\begin{document}
\noindent
Making use of your ``$c$'' expression:

\begin{align*}
f_{xx}(t,x) 
&= c\times\biggl[-\frac{x\bigl(e^{x^2/4t} (2t+x^2) \bigr)}{2t} 
    \cdot 2t^{5/2}\bigl(e^{x^2/4t}+\sqrt{1/t}\,\bigr)^2 \\ 
    &\qquad\quad- \Bigl( e^{x^2/4t}(2t - x^2) +2\sqrt{t}\,\Bigr) 
    \Bigl(2txe^{x^2/4t}\bigl(\sqrt{t}\,e^{x^2/4t}+1 \bigr)\Bigr)\biggr]\\
\shortintertext{where}
c &= \Bigl(2t^{5/2}  
 \bigl(e^{x^2/4t}+\sqrt{1/t}\,\bigr)^2\Bigr)^{-2}
\end{align*}
\end{document}
  • @Jun-GooKwak - I've updated the code to incorporate your follow-up comment that the material on rows two and three belongs together. – Mico Jul 11 '15 at 2:29
3

I'd propose something like this:

enter image description here

The code:

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\begin{multline*} 
f_{xx}(t,x) = 
 \dfrac{1}{\alpha}\biggl(-\dfrac{x\left( \exp(x^2/4t) \left(2t+x^2\right) \right)}{2t} 
    \cdot 2t^{5/2} \bigl(\exp(x^2/4t)+\sqrt{1/t}\, \bigr)^2\bigr) \\ 
- \bigl( ( \exp(x^2/4t)  ( 2t - x^2 ) +2\sqrt{t} \bigr) \cdot   
  \bigl( 2tx\exp(x^2/4t) (\sqrt{t}\exp(x^2/4t)+1  ) \bigr) \biggr), 
\end{multline*}
where $\alpha=\bigl(\exp(x^2/4t)+\sqrt{1/t} )^2 \bigr)^2$.

\end{document}

Remarks

  • I took out the denominator and replaced it in the formula by \alpha

  • Instead of e^{<big exponent>} use the \exp notation.

  • Avoid, as much as possible, the \left, \right constructs. Use the \big, \Big, \bigg, \Bigg family of commands instead.

  • Being just one formula in, now, two lines, I'd rather use multline.

  • @GonzolaMedina Thank you for your time and solution. It is very elegant. – Jun-Goo Kwak Jul 11 '15 at 2:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.