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I don't understand an approximation (Taylor series related?)

  1. Jul 19, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    I've found in a physics book that when u <<L, we have [itex]\sqrt {u^2+L^2} -L = \frac{u^2}{2L}[/itex]. I don't understand why.


    2. Relevant equations
    Not sure.


    3. The attempt at a solution
    I've calculated the Taylor expansion of order 2 of [itex]\sqrt {u^2+L^2} -L[/itex] but I couldn't show the approximation. I get 0, which is "non sense".
    If [itex]f(u)=\sqrt {u^2+L^2} -L[/itex], then [itex]f'(u)=\frac{u}{\sqrt {u^2+L^2 }}[/itex] and [itex]f''(u)= \frac{1}{u^2+L^2} \left ( \sqrt {u^2+L^2}- \frac{u^2}{\sqrt {u^2+L^2} } \right )[/itex].
    So [itex]f(u)\approx f(0)+f'(0)u+ \frac{f''(0)u^2}{2}[/itex]. All of these terms are approximately worth 0 (the last one) or simply worth 0 (the 2 first).
    I'm totally stuck on this.

    Edit: Nevermind, I figured it out. The 3rd term (the one that isn't worth 0) is worth precisely [itex]\frac{u^2}{2L}[/itex]. I considered that u^2 was almost worth 0 at first, because in physics when we have a differential squared we consider it as 0... And here u is considered very small so I got confused, but I finally "understand" the solution now.
    Thanks for reading tough. Hope that can help someone else in future.
     
    Last edited: Jul 19, 2011
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  3. Jul 19, 2011 #2

    Pengwuino

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    Glad I could be of service.
     
  4. Jul 20, 2011 #3

    vela

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    Good work, Pengwuino!
     
  5. Jul 20, 2011 #4

    vela

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    One useful series expansion to know is the binomial series:[tex](1+x)^a = 1+ax+\frac{a(a-1)}{2!}x^2+\frac{a(a-1)(a-2)}{3!}x^3+\cdots[/tex]Using it can save you from having to take annoying derivatives. In this problem, you could have expanded the square root as follows:[tex]\sqrt{u^2+L^2} = L\sqrt{1+\frac{u^2}{L^2}}\approx L\left(1+\frac{1}{2}\frac{u^2}{L^2}\right)[/tex]
     
  6. Jul 20, 2011 #5

    Pengwuino

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    Just remember

    [tex]\sqrt{1+x^2} \approx 1 + {{x^2}\over{2}}[/tex]
    [tex]\sqrt{1+x} \approx 1 + {{x}\over{2}}[/tex]

    They're used so often in physics you should really be forced to tattoo them on your forearm.

    Of course, this is when x<1.

    Their inverses are also used a ton.


    [tex]{{1}\over{\sqrt{1+x}}} \approx 1 - {{x}\over{2}}[/tex]
    [tex]{{1}\over{\sqrt{1+x^2}}} \approx 1 - {{x^2}\over{2}}[/tex]
     
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