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π > e {\displaystyle \pi >e\,}
π 4 = arctan ( 1 ) = 1 − 1 3 + 1 5 − 1 7 + ⋯ {\displaystyle {\frac {\pi }{4}}=\arctan(1)=1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+\cdots }
For an incompressible Newtonian fluid
∇ . U = 0 D ω D t − ( ω . ∇ ) U = ν ∇ 2 ω {\displaystyle {\begin{array}{rcl}{\boldsymbol {\nabla }}.\mathbf {U} &=&0\\{\frac {D{\boldsymbol {\omega }}}{Dt}}-({\boldsymbol {\omega }}.{\boldsymbol {\nabla }})\mathbf {U} &=&\nu {{\nabla }^{2}}{\boldsymbol {\omega }}\end{array}}}
