I was trying to understand the solution of an exercise, when I read this passage. The book is able to expand:
$$ \frac{1}{\sqrt{1+(y'(x)+y''(x)dx)^2}} $$
in
$$ \frac{1}{\sqrt{1+y'(x)^2}}\left(1-\frac{y'(x)y''(x)dx}{1+y'(x)^2}\right) $$
Where $y$ is a function of $x$ and $dx$ is the differential of $x$. It only says: "Expanding the square root to first order in $dx$". I don't understand how.
Can someone explain this passage to me? Thanks a lot
Dave!
Here are all the references: The problem is from "David, Morin, Introductory classical mechanics", page I-19, that is problem 8 of chapter 1.
The text of the problem: (https://i.stack.imgur.com/n6GXR.png)
The solution of the problem: (https://i.stack.imgur.com/XJNrQ.png)(https://i.stack.imgur.com/KugIo.png) (https://i.stack.imgur.com/xx4AW.png)
However, I don't need someone explaining me the problem, but only the central passage of the expansion to the first order. I hope the pictures I added will be helpful.
(https://i.stack.imgur.com/KugIo.png)){kind=link}
You can manually compute the series for $f(u) = \frac{1}{\sqrt{a + u}}$ to first order: $$f(u) \sim f(0) + f'(0) u + O(u^2) = \frac{1}{\sqrt{a}} \left(1 - \frac{u}{2 a} + O(u^2) \right).$$
This is just the first-order truncation of the binomial series for $f$ (for appropriate choices of $a$ and $u$).