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A math block that should render with KaTeX:

A math block that should be rendered with MathJax:

σ_{i,BTC} =min($9600,\text{ } $2500)

Another one with MathJax:

\text{Hourly Funding Payment} = \text{Position Notional Value} \times \text{Funding Rate}\

And a bigger one with KaTeX:

f(x)=x∗e2piiξxf(x) = x * e^{2 pi i \xi x}f(x)=x∗e2piiξx

A math inline with KaTex: f(x)=x∗e2piiξxf(x) = x * e^{2 pi i \xi x}f(x)=x∗e2piiξx and one with MathJax σ_{i,BTC} =min($9600,\text{ } $2500)

1(ϕ5−ϕ)e25π=1+e−2π1+e−4π1+e−6π1+e−8π1+⋯\displaystyle \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } }(ϕ5​​−ϕ)e52​π1​=1+1+1+1+1+⋯e−8π​e−6π​e−4π​e−2π​

Long inline math 1(ϕ5−ϕ)e25π=1+e−2π1+e−4π1+e−6π1+e−8π1+⋯1(ϕ5−ϕ)e25π=1+e−2π1+e−4π1+e−6π1+e−8π1+⋯1(ϕ5−ϕ)e25π=1+e−2π1+e−4π1+e−6π1+e−8π1+⋯1(ϕ5−ϕ)e25π=1+e−2π1+e−4π1+e−6π1+e−8π1+⋯1(ϕ5−ϕ)e25π=1+e−2π1+e−4π1+e−6π1+e−8π1+⋯1(ϕ5−ϕ)e25π=1+e−2π1+e−4π1+e−6π1+e−8π1+⋯\displaystyle \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } } \displaystyle \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } } \displaystyle \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } } \displaystyle \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } } \displaystyle \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } } \displaystyle \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } }(ϕ5​​−ϕ)e52​π1​=1+1+1+1+1+⋯e−8π​e−6π​e−4π​e−2π​(ϕ5​​−ϕ)e52​π1​=1+1+1+1+1+⋯e−8π​e−6π​e−4π​e−2π​(ϕ5​​−ϕ)e52​π1​=1+1+1+1+1+⋯e−8π​e−6π​e−4π​e−2π​(ϕ5​​−ϕ)e52​π1​=1+1+1+1+1+⋯e−8π​e−6π​e−4π​e−2π​(ϕ5​​−ϕ)e52​π1​=1+1+1+1+1+⋯e−8π​e−6π​e−4π​e−2π​(ϕ5​​−ϕ)e52​π1​=1+1+1+1+1+⋯e−8π​e−6π​e−4π​e−2π​