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

1. Jul 19, 2011

### fluidistic

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Not sure.

3. 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 $f'(u)=\frac{u}{\sqrt {u^2+L^2 }}$ and $f''(u)= \frac{1}{u^2+L^2} \left ( \sqrt {u^2+L^2}- \frac{u^2}{\sqrt {u^2+L^2} } \right )$.
So $f(u)\approx f(0)+f'(0)u+ \frac{f''(0)u^2}{2}$. 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.

Last edited: Jul 19, 2011
2. Jul 19, 2011

### Pengwuino

Glad I could be of service.

3. Jul 20, 2011

### vela

Staff Emeritus
Good work, Pengwuino!

4. Jul 20, 2011

### vela

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

5. Jul 20, 2011

### 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.

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