A math block that should render with KaTeX:
f ( x ) = x ∗ e 2 p i i ξ x f(x) = x * e^{2 pi i \xi x} f ( x ) = x ∗ e 2 p ii ξ x A math block that should be rendered with MathJax:
A math inline with KaTex: f ( x ) = x ∗ e 2 p i i ξ x f(x) = x * e^{2 pi i \xi x} f ( x ) = x ∗ e 2 p ii ξ x and one with MathJax
Another one with MathJax:
And a bigger one with KaTeX:
1 ( ϕ 5 − ϕ ) e 2 5 π = 1 + e − 2 π 1 + e − 4 π 1 + e − 6 π 1 + e − 8 π 1 + ⋯ \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} } } } ( ϕ 5 − ϕ ) e 5 2 π 1 = 1 + 1 + 1 + 1 + 1 + ⋯ e − 8 π e − 6 π e − 4 π e − 2 π
Long inline math 1 ( ϕ 5 − ϕ ) e 2 5 π = 1 + e − 2 π 1 + e − 4 π 1 + e − 6 π 1 + e − 8 π 1 + ⋯ 1 ( ϕ 5 − ϕ ) e 2 5 π = 1 + e − 2 π 1 + e − 4 π 1 + e − 6 π 1 + e − 8 π 1 + ⋯ 1 ( ϕ 5 − ϕ ) e 2 5 π = 1 + e − 2 π 1 + e − 4 π 1 + e − 6 π 1 + e − 8 π 1 + ⋯ 1 ( ϕ 5 − ϕ ) e 2 5 π = 1 + e − 2 π 1 + e − 4 π 1 + e − 6 π 1 + e − 8 π 1 + ⋯ 1 ( ϕ 5 − ϕ ) e 2 5 π = 1 + e − 2 π 1 + e − 4 π 1 + e − 6 π 1 + e − 8 π 1 + ⋯ 1 ( ϕ 5 − ϕ ) e 2 5 π = 1 + e − 2 π 1 + e − 4 π 1 + e − 6 π 1 + e − 8 π 1 + ⋯ \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} } } } ( ϕ 5 − ϕ ) e 5 2 π 1 = 1 + 1 + 1 + 1 + 1 + ⋯ e − 8 π e − 6 π e − 4 π e − 2 π ( ϕ 5 − ϕ ) e 5 2 π 1 = 1 + 1 + 1 + 1 + 1 + ⋯ e − 8 π e − 6 π e − 4 π e − 2 π ( ϕ 5 − ϕ ) e 5 2 π 1 = 1 + 1 + 1 + 1 + 1 + ⋯ e − 8 π e − 6 π e − 4 π e − 2 π ( ϕ 5 − ϕ ) e 5 2 π 1 = 1 + 1 + 1 + 1 + 1 + ⋯ e − 8 π e − 6 π e − 4 π e − 2 π ( ϕ 5 − ϕ ) e 5 2 π 1 = 1 + 1 + 1 + 1 + 1 + ⋯ e − 8 π e − 6 π e − 4 π e − 2 π ( ϕ 5 − ϕ ) e 5 2 π 1 = 1 + 1 + 1 + 1 + 1 + ⋯ e − 8 π e − 6 π e − 4 π e − 2 π