When the symbol in the \sqrt is too large, the \sqrt rendered in the article is too vertical and is not very beautiful. I want it to have a tilt angle, just like when there are few symbols inside. what should I do?

enter image description here

  • Are you aware of this question and the posts that are linked there? – user194703 Sep 19 '19 at 3:14
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    Just as going through the successive square-roots is boring and sometimes disheartening too, hyper-hyper-...-links are tiresome too (at least to myself) ! So dug out the answer deep-linked to @Schrödinger'scat that might be useful for this question. – Partha D. Sep 19 '19 at 4:17
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    Which font do you use? – Bernard Sep 19 '19 at 8:20
  • It would be helpful if you provided an MWE (from \documentclass... to \end{document}) showing what you did to produce your output. – Peter Wilson Sep 19 '19 at 17:06
  • See also tex.stackexchange.com/questions/127893/…. With the answer there, you could as an alternative, replace the \stretchrel with \scalerel[3ex] to scale the surd, with a 3ex width limit. – Steven B. Segletes Sep 19 '19 at 20:33

The fact that the steepness of the surds increases with the overall size of the square-root symbols is not a flaw in the design of the math font. Instead, it embodies a long-standing typographic tradition that has held up pretty well over the decades (and probably even centuries).

If you can't stand the "vertical look" of the taller surds, do contemplate some alternatives. Among them are (a) using \tfrac ("text style frac") rather than \frac; (b) using inline-fraction notation; (c) using reciprocal notation (e.g, 3^{-2} rather than \frac{1}{3^2}); and (d) parenthetic notation instead of \sqrt -- either with automatically sized parentheses or explicitly sized parentheses.

enter image description here

\usepackage{amsmath} % for 'align*' env. and '\tfrac' and '\dotsb' macros
\usepackage{newpxtext,newpxmath} % Palatino clone

&= \sqrt{\frac{1}{3^2}+\sqrt{\frac{1}{4^2}+\sqrt{\dotsb+\frac{1}{n^2}}}}\\
&= \sqrt{\tfrac{1}{3^2}+\sqrt{\tfrac{1}{4^2}+\sqrt{\dotsb+\tfrac{1}{n^2}}}}\\
&= \sqrt{1/3^2+\sqrt{1/4^2+\sqrt{\dotsb+1/n^2}}}\\
&= \sqrt{3^{-2}+\sqrt{4^{-2}+\sqrt{\dotsb+\mathstrut n^{-2}}}}\\
&= \left(\frac{1}{3^2}+\left(\frac{1}{4^2}+\left(\dotsb+\frac{1}{n^2}
      \right)^{\!\!-1/2\,} \right)^{\!\!-1/2\,} \right)^{\!\!-1/2}\\
&= \Biggl(\frac{1}{3^2}+\biggl(\frac{1}{4^2}+\Bigr(\dotsb+\frac{1}{n^2}
      \Bigr)^{\!\!-1/2\,} \biggr)^{\!\!-1/2\,} \Biggr)^{\!\!-1/2}

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