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

[fluidistic]

Homework Statement

I've found in a physics book that when u <<L, we have \sqrt {u^2+L^2} -L = \frac{u^2}{2L}. I don't understand why.

Homework Equations

Not sure.

The Attempt at a Solution

I've calculated the Taylor expansion of order 2 of \sqrt {u^2+L^2} -L but I couldn't show the approximation. I get 0, which is "non sense".
If f(u)=\sqrt {u^2+L^2} -L, then f&#039;(u)=\frac{u}{\sqrt {u^2+L^2 }} and f&#039;&#039;(u)= \frac{1}{u^2+L^2} \left ( \sqrt {u^2+L^2}- \frac{u^2}{\sqrt {u^2+L^2} } \right ).
So f(u)\approx f(0)+f&#039;(0)u+ \frac{f&#039;&#039;(0)u^2}{2}. 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 \frac{u^2}{2L}. 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.

 

[Pengwuino]

Glad I could be of service.

 

[vela]

Good work, Pengwuino!

 

[vela]

One useful series expansion to know is the binomial series:(1+x)^a = 1+ax+\frac{a(a-1)}{2!}x^2+\frac{a(a-1)(a-2)}{3!}x^3+\cdotsUsing it can save you from having to take annoying derivatives. In this problem, you could have expanded the square root as follows:\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)

 

[Pengwuino]

Just remember

\sqrt{1+x^2} \approx 1 + {{x^2}\over{2}}
\sqrt{1+x} \approx 1 + {{x}\over{2}}

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.{{1}\over{\sqrt{1+x}}} \approx 1 - {{x}\over{2}}
{{1}\over{\sqrt{1+x^2}}} \approx 1 - {{x^2}\over{2}}