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Jyrki Lahtonen
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There is $\root{\displaystyle{3}}\of{\frac{a}{b}}$ that will make the $3$ larger:

$$\root{\displaystyle{3}}\of{\frac{a}{b}}.$$

Don't know if this substantially better. If your custom is to use the \sqrt[#1]{#2} construction instead of my preferred (straight from the $\TeX$book) \root{#1}\of{#2} that probably works too.

Undoubtedly other finetuning is also possible.


Experimenting with the \raise[n pt]{} construct:

\root{\raise2pt{\displaystyle{3}}}\of{\frac{a}{b}}

$$\root{\raise2pt{\displaystyle{3}}}\of{\frac{a}{b}},$$

\root{\raise4pt{3}}\of{\frac{a}{b}}

$$\root{\raise4pt{3}}\of{\frac{a}{b}}.$$

There is $\root{\displaystyle{3}}\of{\frac{a}{b}}$ that will make the $3$ larger:

$$\root{\displaystyle{3}}\of{\frac{a}{b}}.$$

Don't know if this substantially better. If your custom is to use the \sqrt[#1]{#2} construction instead of my preferred (straight from the $\TeX$book) \root{#1}\of{#2} that probably works too.

Undoubtedly other finetuning is also possible.

There is $\root{\displaystyle{3}}\of{\frac{a}{b}}$ that will make the $3$ larger:

$$\root{\displaystyle{3}}\of{\frac{a}{b}}.$$

Don't know if this substantially better. If your custom is to use the \sqrt[#1]{#2} construction instead of my preferred (straight from the $\TeX$book) \root{#1}\of{#2} that probably works too.

Undoubtedly other finetuning is also possible.


Experimenting with the \raise[n pt]{} construct:

\root{\raise2pt{\displaystyle{3}}}\of{\frac{a}{b}}

$$\root{\raise2pt{\displaystyle{3}}}\of{\frac{a}{b}},$$

\root{\raise4pt{3}}\of{\frac{a}{b}}

$$\root{\raise4pt{3}}\of{\frac{a}{b}}.$$

Source Link
Jyrki Lahtonen
  • 135.3k
  • 7
  • 59
  • 114

There is $\root{\displaystyle{3}}\of{\frac{a}{b}}$ that will make the $3$ larger:

$$\root{\displaystyle{3}}\of{\frac{a}{b}}.$$

Don't know if this substantially better. If your custom is to use the \sqrt[#1]{#2} construction instead of my preferred (straight from the $\TeX$book) \root{#1}\of{#2} that probably works too.

Undoubtedly other finetuning is also possible.