text: # find maclaurin series
  
  • May 19th 2009, 02:24 PM
  cm3pyro
  find maclaurin series
  integral of 1/(2+x^3)dx bounds:0 to .5
  • May 19th 2009, 03:03 PM
  skeeter
  Quote:
  
  Originally Posted by cm3pyro
  integral of 1/(2+x^3)dx bounds:0 to .5
  
  $\frac{1}{2+x^3} = \frac{1}{2} \cdot \frac{1}{1 - \left(-\frac{x^3}{2}\right)} = \frac{1}{2}\left(1 - \frac{x^3}{2} + \frac{x^6}{4} - \frac{x^9}{8} + ... \right)$
  
  integrate ...
  
  $\frac{1}{2}\left[x - \frac{x^4}{8} + \frac{x^7}{28} - \frac{x^{10}}{80} + ... \right]_0^{0.5}$
  
  using the first three terms should get you within 0.00001 of the actual value
  • May 19th 2009, 03:10 PM
  cm3pyro
  Quote:
  
  Originally Posted by skeeter
  $\frac{1}{2+x^3} = \frac{1}{2} \cdot \frac{1}{1 - \left(-\frac{x^3}{2}\right)} = \frac{1}{2}\left(1 - \frac{x^3}{2} + \frac{x^6}{4} - \frac{x^9}{8} + ... \right)$
  
  integrate ...
  
  $\frac{1}{2}\left[x - \frac{x^4}{8} + \frac{x^7}{28} - \frac{x^{10}}{80} + ... \right]_0^{0.5}$
  
  using the first three terms should get you within 0.00001 of the actual value
  
  i need a maclaurin series
  • May 19th 2009, 04:18 PM
  skeeter
  Quote:
  
  Originally Posted by cm3pyro
  i need a maclaurin series
  
  what do you think I posted?
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