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<math>\int_a^bf'(x)\,\mathrm{d}x=f(b)-f(a)</math>
 
<math>\int_a^bf'(x)\,\mathrm{d}x=f(b)-f(a)</math>
  
<math>f^{(n)}(z)=\frac{n!}{2\pi\mathrm{i}}\oint_\gamma\frac{f(t)}{(t-z)^{n+1}}\,\mathrm{d}t<\math>
+
<math>\Gamma(z)=\int_0^\infty t^{z-1}\mathrm{e}^{-t}\,\mathrm{d}t</math>
 +
 
 +
<math>f^{(n)}(z)=\dfrac{n!}{2\pi\mathrm{i}}\oint_\gamma\dfrac{f(t)}{(t-z)^{n+1}}\,\mathrm{d}t</math>
  
 
[[Category:MA181Fall2011Bell]]
 
[[Category:MA181Fall2011Bell]]

Latest revision as of 15:39, 24 August 2011

Homework 1 collaboration area

Here is an example of how to input math:

$ \int_0^\pi\sin x\,dx = \left[ -\cos x\right]_0^\pi = -(-1)-(-1)=2 $

How do you compute the derivative of

$ \frac{x^2+1}{3x^4-7}? $

just for testing:

$ \int_a^bf'(x)\,\mathrm{d}x=f(b)-f(a) $

$ \Gamma(z)=\int_0^\infty t^{z-1}\mathrm{e}^{-t}\,\mathrm{d}t $

$ f^{(n)}(z)=\dfrac{n!}{2\pi\mathrm{i}}\oint_\gamma\dfrac{f(t)}{(t-z)^{n+1}}\,\mathrm{d}t $

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Followed her dream after having raised her family.

Ruth Enoch, PhD Mathematics