# How to adjust/increase the height of square root?

I defined this equation:

$$\delta {^s_i\lambda} = \begin{cases} \pm \dfrac{{^s\overline{\Delta T}}}{\sqrt[]{\delta {^s_1\bm{u}_l} \cdot \delta {^s_1\bm{u}_l} + \beta}} & \! \text{if} \; i = 1 \\[1em] - \dfrac{\delta{^s_1\bm{u} \cdot \delta{^s_1\bm{u}_r}}}{\delta {^s_1\bm{u}} \cdot \delta {^s_i\bm{u}_l} + \beta \delta {^s_1\lambda}} & \! \text{otherwise} \end{cases}$$


And I am using in preamble this:

\usepackage{amsmath,amsfonts,amssymb}
\usepackage{bm}


The result:

Although, I would like to avoid the horizontal trace of the square root to be so close to the terms inside of it. Does anyone know how to solve this?

• When I compile your code, it's not so close. You can try adding a \mathstrut in the square root. Oct 17, 2018 at 21:09
• Exactly @Bernard, I don't why this is happening, but '\mathstruct' works. Thanks for the suggestion!
– AXL
Oct 17, 2018 at 22:56

Getting square roots right requires a bit of manual intervention.

I essentially add a phantom \beta^K so TeX will see some more height, but not too much to make it choose the next size of the radical.

Also I add a phantom \Big| to the denominator, but smashed at the bottom so the apparent height will be increased, thus moving down the radical from the fraction line.

Important suggestion: {^s_i\lambda} and similar is not the best way to deal with prescripts; with \mathtools there is \prescript which will adapt the height of the prescripts to the object they apply to and also adds a thin space in order to make clear that the prescripts aren't scripts to the symbol at the left.

Finally, dcases* avoids the need for \dfrac and for \text.

\documentclass{article}
\usepackage{amsmath,mathtools,amssymb}
\usepackage{bm}

\begin{document}

The original equation is
$$\delta {^s_i\lambda} = \begin{cases} \pm \dfrac{{^s\overline{\Delta T}}}{\sqrt[]{\delta {^s_1\bm{u}_l} \cdot \delta {^s_1\bm{u}_l} + \beta}} & \! \text{if} \; i = 1 \\[1em] - \dfrac{\delta{^s_1\bm{u} \cdot \delta{^s_1\bm{u}_r}}}{\delta {^s_1\bm{u}} \cdot \delta {^s_i\bm{u}_l} + \beta \delta {^s_1\lambda}} & \! \text{otherwise} \end{cases}$$
and now the modified one
$$\delta\prescript{s}{i}{\lambda} = \begin{dcases*} \pm \frac{ \prescript{s}{}{\overline{\Delta T}} }{ \smash[b]{\vphantom{\Big|}} \sqrt{ \vphantom{\beta^K} \delta\prescript{s}{1}{\bm{u}_l} \cdot \delta\prescript{s}{1}{\bm{u}_l} + \beta } } & if i = 1 \\[1ex] -\frac{ \delta\prescript{s}{1}{\bm{u}} \cdot \delta\prescript{s}{1}{\bm{u}_r} }{ \delta\prescript{s}{1}{\bm{u}} \cdot \delta\prescript{s}{i}{\bm{u}_l} + \beta \delta\prescript{s}{1}{\lambda} } & otherwise \end{dcases*}$$

\end{document}


Quick hack to increase the height of the square root is to smuggle in a rule of zero width but with a height you can adjust to fit your taste:

\documentclass{article}

\usepackage{amsmath,amsfonts,amssymb}
\usepackage{bm}

\begin{document}

$$\delta {^s_i\lambda} = \begin{cases} \pm \dfrac{{^s\overline{\Delta T}}}{\sqrt{\rule{0pt}{2ex}\delta {^s_1\bm{u}_l} \cdot \delta {^s_1\bm{u}_l} + \beta}} & \! \text{if} \; i = 1 \\[2em] - \dfrac{\delta{^s_1\bm{u} \cdot \delta{^s_1\bm{u}_r}}}{\delta {^s_1\bm{u}} \cdot \delta {^s_i\bm{u}_l} + \beta \delta {^s_1\lambda}} & \! \text{otherwise} \end{cases}$$

\end{document}


If you want more or less distance between the terms inside the square root and the horizontal trace, adjust the value of \rule{0pt}{2ex}, this was just a quick guess.