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Math manipulation

  1. Nov 14, 2011 #1
    Why does
    86072215.png
    when l approaches infinity?
    This would mean that ef174ec5.png but I don't understand why that would be true.
    Thanks!!
     
  2. jcsd
  3. Nov 14, 2011 #2

    Dick

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    sqrt(1+r)=1+r/2-r^2/8+... That's the Taylor series expansion of sqrt(1+r). If r is small, then 1+r/2 is a good approximation. And as l->infinity then y^2/(l-x)^2 becomes very small. The sides aren't equal for any finite l. But the approximation becomes better and better as l->infinity.
     
  4. Nov 14, 2011 #3

    ehild

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    Try it: Calculate the square root of 1+x with smaller and smaller x-values. For example, √1.2=1.095, √1.02=1.0099, √1.002=1.000999....

    [itex]\sqrt{1+x} \approx 1+x/2[/itex]

    ehild
     
  5. Nov 15, 2011 #4
    Thank you Dick and ehild! :)
     
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