User:Milton Beychok/Sandbox: Difference between revisions

Revision as of 18:23, 16 September 2011

As currently rendered in the EoE's "Velocity" article:

<math>mathbf{v} = lim_{Delta t to 0}{{mathbf{x}(t+Delta t)-mathbf{x}(t)} over Delta t}={mathrm{d}mathbf{x} over mathrm{d}t}.</math>

This is the correct version of the LaTeX coding:

<math>\mathbf{v} = \lim_{\Delta t \to 0}{{\mathbf{x}(t+\Delta t)-\mathbf{x}(t)} \over \Delta t}={\mathrm{d}\mathbf{x} \over \mathrm{d}t}.</math>

Note that the coding in EOE's "Velocity" article does not include any of the necessary reverse slashes \.

And this is how it should be rendered in any articles:

$\mathbf {v} =\lim _{\Delta t\to 0}{{\mathbf {x} (t+\Delta t)-\mathbf {x} (t)} \over \Delta t}={\mathrm {d} \mathbf {x} \over \mathrm {d} t}.$

@@ Line 1: / Line 1: @@
+'''As currently rendered in the EoE's "Velocity" article:'''
+<nowiki><math>mathbf{v} = lim_{Delta t to 0}{{mathbf{x}(t+Delta t)-mathbf{x}(t)} over Delta t}={mathrm{d}mathbf{x} over mathrm{d}t}.</math></nowiki>
+'''This is the correct version of the LaTeX coding:'''
+<nowiki><math>\mathbf{v} = \lim_{\Delta t \to 0}{{\mathbf{x}(t+\Delta t)-\mathbf{x}(t)} \over \Delta t}={\mathrm{d}\mathbf{x} \over \mathrm{d}t}.</math></nowiki>
+Note that the coding in EOE's "Velocity" article does not include any of the '''necessary''' reverse slashes '''\'''.
+And this is how it should be rendered in any articles:
+<math>\mathbf{v} = \lim_{\Delta t \to 0}{{\mathbf{x}(t+\Delta t)-\mathbf{x}(t)} \over \Delta t}={\mathrm{d}\mathbf{x} \over \mathrm{d}t}.</math>