Some math met in EM hw solution

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In summary, the conversation discusses the steps involved in simplifying the equation 1 - z / (R^2 + Z^2) ^1/2 = 1- (1+ (R/Z)^2) ^-1/2, which involves dividing both the numerator and denominator by "z" and using the binomial theorem or a 2nd degree Taylor polynomial approximation. The last step may involve taking a limit to make the approximation more accurate. The conversation also includes a question about how to format text in a white block.
  • #1
tigigi
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I got this :

1 - z / (R^2 + Z^2) ^1/2 = 1- (1+ (R/Z)^2) ^-1/2
= 1 - 1 + (1/2) ( R/Z )^2

I'm confused why it got all these steps. it seems like taylor expansion ? or ?

Thank you.
 
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  • #2
First, it is confusing to me that you are using both Z and z. Are they the same thing?

I can't speak for WHY they are doing this but how they do the first line is evident: Divide both numerator and denominator of the second term by "z". In the numerator you get z/z= 1, in the denominator the "z" becomes "z2" inside the square root: [itex]\sqrt{R^2/z^2+ z^2/z^2}= \sqrt{1+ (R/z)^2}[/itex]. That is NOT exactly equal to the last line. The last line is an approximation. Yes, you could think of it as a 2nd[/b] degree Taylor polynomial approximation. You could also think of it as a special case of the binomial theorem- extended to fractional powers. Just as (1+ x)n= 1+ nx to first degree, [itex](1+ (R/z)^2)^(-1/2)[/itex] is [itex]1+ (-1/2)(R/z)[/itex]. Of course, now the "1" and "-1" will cancel. To second degree, 1- z/(R2+ z2)-1/2= (1/2)(R/z)2. I assume the next step will involve a limit as R goes to 0 or z goes to infinity or at least that (R/z) is small to make the approximation as accurate as possible.
 
  • #3
Thank you so much. I got it now.
btw, could I ask another question that how do you get those words in white block like that ? Thank you. I appreciate it.
 

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