MathML samples / test page

Various outstanding people

MathML

Some people have been known to claim that $\frac{x + y^{2}}{k + 1}$ , but I remain skeptical.

This formula has a lot to say.

(k_{0}, k_{1}, k_{2}, k_{3}, k_{4}, k_{5}, k_{6}) = (1, 0, 0, 0, 0, 0, 0)

μ (T) = max ({μ (T - r)} \cup {1 + μ (T_{j} - r_{j}) + \sum_{i \neq j} μ (T_{i}) : j \in {1, \dots, k}}),

This sample is from PMC3226351, and you can see what it looks like in classic PMC (as an image) here, or on the Elsevier site here.

\frac{m (T^{'})}{m (T)} = \frac{m_{0} (S^{'})}{m_{0} (S)} \cdot \frac{ρ (T_{1}) + ρ (C^{⌊ k / 2 ⌋ + 1} L) + ρ (C^{⌊ (k + 1) / 2 ⌋ + 1} L) + 1}{ρ (T_{1}) + ρ (C^{k_{0}} F) + ρ (C^{k_{1}} F) + ρ (C^{k_{2}} F)} = \frac{m_{0} (S^{'})}{m_{0} (S)} (1 + \frac{ρ (C^{⌊ k / 2 ⌋ + 1} L) + ρ (C^{⌊ (k + 1) / 2 ⌋ + 1} L) + 1 - ρ (C^{k_{0}} F) - ρ (C^{k_{1}} F) - ρ (C^{k_{2}} F)}{ρ (T_{1}) + ρ (C^{k_{0}} F) + ρ (C^{k_{1}} F) + ρ (C^{k_{2}} F)}) \geq \frac{m_{0} (S^{'})}{m_{0} (S)} (1 + \frac{ρ (C^{⌊ k / 2 ⌋ + 1} L) + ρ (C^{⌊ (k + 1) / 2 ⌋ + 1} L) + 1 - ρ (C^{k_{0}} F) - ρ (C^{k_{1}} F) - ρ (C^{k_{2}} F)}{\sqrt{3} - 1 + ρ (C^{k_{0}} F) + ρ (C^{k_{1}} F) + ρ (C^{k_{2}} F)})

See here for how this looks in classic PMC, and here for how it looks on the Elsevier site.

0 \leq \frac{m (T) - m (T^{'})}{M} = K (x + a, ρ_{1}, \dots, ρ_{k - 1}, y + b) - K (y + a, ρ_{1}, \dots, ρ_{k - 1}, x + b) = ((x + a) (y + b) - (y + a) (x + b)) K (ρ_{1}, \dots, ρ_{k - 1}) + ((x + a) - (y + a)) K (ρ_{1}, \dots, ρ_{k - 2}) + ((y + b) - (x + b)) K (ρ_{2}, \dots, ρ_{k - 1}) = (x - y) K (ρ_{1}, \dots, ρ_{k - 1}) (b - a + \frac{K (ρ_{1}, \dots, ρ_{k - 2})}{K (ρ_{1}, \dots, ρ_{k - 2}, ρ_{k - 1})} - \frac{K (ρ_{2}, \dots, ρ_{k - 1})}{K (ρ_{1}, ρ_{2}, \dots, ρ_{k - 1})}) = (x - y) K (ρ_{1}, \dots, ρ_{k - 1}) (b - a + \frac{1}{CF (ρ_{k - 1}, \dots, ρ_{1})} - \frac{1}{CF (ρ_{1}, \dots, ρ_{k - 1})}) = (x - y) K (ρ_{1}, \dots, ρ_{k - 1}) (CF (b, ρ_{k - 1}, \dots, ρ_{1}) - CF (a, ρ_{1}, \dots, ρ_{k - 1}))