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Log expansion for infinite solenoid

  1. Feb 15, 2013 #1
    Hello, I found an approximation for this log function:

    [tex] log \Bigg(\frac{\Lambda}{\rho} + \sqrt{1 + \frac{\Lambda^2}{\rho^2}} \Bigg), [/tex]

    where [itex] \Lambda \rightarrow \infty [/itex]. The above is approximated to the following,

    [tex] -log \bigg(\frac{\rho}{\rho_o} \bigg) + log \bigg(\frac{2 \Lambda}{\rho_o} \bigg). [/tex]

    How is this done? I tried expanding the [itex] \sqrt{1 + x^2} [/itex] term, but I still don't get how they arrive to the above approximation.

    Any help would be greatly appreciated!

    Cheers!

    I have no idea why this was sent to linear algebra section . . . And I do not know how to move it to classical physics. . .
     
    Last edited: Feb 15, 2013
  2. jcsd
  3. Feb 15, 2013 #2

    jbunniii

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    What is ##\rho_0##? It appears in the second expression but not the first.
     
  4. Feb 15, 2013 #3
    Last edited: Feb 15, 2013
  5. Feb 15, 2013 #4
    Wow, never mind. Clearly I am being silly here, for [itex] \Lambda \rightarrow \infty [/itex].

    [tex] log\bigg( \frac{\Lambda}{\rho} + \sqrt{1 + \frac{\Lambda^2}{\rho^2}} \bigg) \rightarrow log \bigg( \frac{ 2 \Lambda}{\rho} \bigg) \rightarrow log(2 \Lambda) - log(\rho). [/tex]

    As for the [itex] \rho_o [/itex] I have no idea why that enters the equation.
     
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