October 20, 2022

KaTeX mathsf vs default

${\displaystyle \frac{\mathsf{1}}{(\sqrt{\mathsf{\varphi }\sqrt{\mathsf{5}}}-\mathsf{\varphi }){\mathsf{e}}^{\frac{\mathsf{2}}{\mathsf{5}}\mathsf{\pi }}}=\mathsf{1}+\frac{{\mathsf{e}}^{-\mathsf{2}\mathsf{\pi }}}{\mathsf{1}+\frac{{\mathsf{e}}^{-\mathsf{4}\mathsf{\pi }}}{\mathsf{1}+\frac{{\mathsf{e}}^{-\mathsf{6}\mathsf{\pi }}}{\mathsf{1}+\frac{{\mathsf{e}}^{-\mathsf{8}\mathsf{\pi }}}{\mathsf{1}+\cdots }}}}}$

${\displaystyle \frac{1}{(\sqrt{\varphi \sqrt{5}}-\varphi )e^{\frac{2}{5}\pi }}=1+\frac{e^{-2\pi }}{1+\frac{e^{-4\pi }}{1+\frac{e^{-6\pi }}{1+\frac{e^{-8\pi }}{1+\cdots }}}}}$

${\displaystyle {\left(\sum _{\mathsf{k}=\mathsf{1}}^{\mathsf{n}}{\mathsf{a}}_{\mathsf{k}}{\mathsf{b}}_{\mathsf{k}}\right)}^{\mathsf{2}}\le \left(\sum _{\mathsf{k}=\mathsf{1}}^{\mathsf{n}}{\mathsf{a}}_{\mathsf{k}}^{\mathsf{2}}\right)\left(\sum _{\mathsf{k}=\mathsf{1}}^{\mathsf{n}}{\mathsf{b}}_{\mathsf{k}}^{\mathsf{2}}\right)}$

${\displaystyle {\left(\sum _{k=1}^n{a}_k{b}_k\right)}^2\le \left(\sum _{k=1}^n{a}_k^2\right)\left(\sum _{k=1}^n{b}_k^2\right)}$

${\displaystyle \mathsf{1}+\frac{{\mathsf{q}}^{\mathsf{2}}}{(\mathsf{1}-\mathsf{q})}+\frac{{\mathsf{q}}^{\mathsf{6}}}{(\mathsf{1}-\mathsf{q})(\mathsf{1}-{\mathsf{q}}^{\mathsf{2}})}+\cdots =\prod _{\mathsf{j}=\mathsf{0}}^{\mathsf{\infty }}\frac{\mathsf{1}}{(\mathsf{1}-{\mathsf{q}}^{\mathsf{5}\mathsf{j}+\mathsf{2}})(\mathsf{1}-{\mathsf{q}}^{\mathsf{5}\mathsf{j}+\mathsf{3}})},\text{for }\mid \mathsf{q}\mid <\mathsf{1.}}$

${\displaystyle 1+\frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots =\prod _{j=0}^{\mathrm{\infty }}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})},\text{for }\mid q\mid <1.}$

Not much difference and certainly not true sans serif. arev has been mentioned. Fira Math is another possibility as is GFS Neohellenic. But how to get these into KaTeX/Blot? Maybe this can be some help.

I have heard that we now have inline math $$$$ like so. This is good news. () but it doesn’t work yet.

${\displaystyle f(x)=2x+y}$

TeX Math

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