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Expansion of arctanx

  1. Oct 4, 2007 #1
    Problem

    expansion
    [tex]arctanx=\frac{\pi}{2}-\frac{1}{x}+o(\frac{1}{x})[/tex]

    Attempt:
    arctanx=y
    tany=x
    for [tex]y=\frac{\pi}{2}-z[/tex] [tex]tan(\frac{\pi}{2}-z)=\frac{1}{tanz}=x[/tex]
    [tex]\frac{cosz}{sinz}=\frac{1+o(z^2)}{z+o(z^3)}=\frac{1}{z}(1+o(z^2))=x[/tex]*
    [tex]arctanx=y=\frac{\pi}{2}-z=\frac{\pi}{2}-\frac{1}{x}+o(\frac{1}{x})[/tex]
    But how we can convert [tex]\frac{1}{z}(1+o(z^2))=x[/tex]
    to [tex]z=\frac{1}{x}+o(\frac{1}{x})[/tex]
     
  2. jcsd
  3. Oct 5, 2007 #2
    i think i've found the answer.
    [tex]z=\frac{1}{x}+o(\frac{z^2}{x})[/tex]
    [tex]arctanx=y=\frac{\pi}{2}-z[/tex]
    [tex]z=\frac{\pi}{2}-arctanx=o(1)(x\rightarrow\infty)[/tex]
    so
    [tex]z=\frac{1}{x}+o(\frac{1}{x})[/tex]
     
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