first order approximation

Use Taylor expansion up to terms in θ.

Problem here is you say "for example", but the appropriate thing to do depends a lot on the exact form of the expression (and which parts you expect to be big or small).

In this case, I would rewrite $h^4 - \cos^2 \theta$ as $h^4 -1 + \sin^2 \theta$ (to get a "theta-dependent" bit that's small for small theta), and then take out a factor of sqrt h^4-1 to get $\sqrt{h^4 - 1}\sqrt{1 + \frac{\sin^2 \theta}{h^4 - 1}}$ (to get the theta-dependent bit in the form $(1 + f(\theta))^\alpha$, which is the easiest form to do a binomial expansion on).

You can then use $\sin \theta \approx \theta$ and the binomial theorem to get a first order approximation.

In the real world, it's quite common for these first order approximations to have pretty messy derivations (and nasty final forms), as you can see...

right.... i guess i'll go with the binomial expansion thanks

last question $(x^2)^{\frac{1}{2}}$ is just x right

Should also be noted that this is in fact a 2nd order approximation.

I guess if you really knew a 1st order approximation was "good enough", then it's enough to observe that cos^2 theta = 1 to 1st order, and (unless h^4 = 1) sqrt(h^4 - cos^2 theta) has finite derivative. So you could just say it's $\sqrt{h^4-1}$ to 1st order.

Generally speaking, it's |x|, although it depends on context. I don't see how the expression comes up, anyhow...

Should also be noted that this is in fact a 2nd order approximation.

I guess if you really knew a 1st order approximation was "good enough", then it's enough to observe that cos^2 theta = 1 to 1st order, and (unless h^4 = 1) sqrt(h^4 - cos^2 theta) has finite derivative. So you could just say it's $\sqrt{h^4-1}$ to 1st order.

Generally speaking, it's |x|, although it depends on context. I don't see how the expression comes up, anyhow...

for example taking out the factor making it suitable for binomial theorem perhaps$(x^2+tan\theta )^{\frac{1}{2}}$

so would there be an x or mod x outside the brackets?

Normally, |x|. (Although if I'm being honest, it's probably more actually more common to take out x and ignore the fact that it's wrong if x < 0...)

thanks 1 more question

in lyx i'm trying to make the jumbo ( brackets work by putting in \left( but it doesn't seem to want to do it, i just keeps the \left part in red with a blue ( which annoys me because i can't make the jumbo brackets which auto fit to however tall the tallest thing inside the brackets is -__- i have to use the plebian handy toolbox of all those options

Not sure what you're asking, but here's an example:

$\left( 1 + \sqrt{\dfrac{1-x^2}{1+x^2}} \right)$

I have no idea what lyx is; I just write straight TeX code.

ok nvm about that

another 1 last time question

for a binomial expansion is each step there equal to each other or is it approximately equal?(i mean i am only taking the first x term so i miss out all the other x terms cus i don't need them for first order)

If you expand out (1+x)^a, you get something of the form 1 + Ax + Bx^2 + Cx^3 + ... where A, B, C etc. don't depend on x. So for small enough x, you can ignore all the terms apart from 1+Ax.

(i.e. yes, you don't need the other terms for 1st order).