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# first order approximation watch

1. for example in

where h>0 and is small

i can't quite think of a way to find a first order approximation for this other than by binomal theorem

i mean like it was worked out that the approx of cos was where cos is

the main thing actually is that im not sure what is approximately equal to, perhaps it's ?? it a complete guess but im sure that way would be quicker but is binomal theorem ok?
2. Use Taylor expansion up to terms in θ.
3. (Original post by B_9710)
Use Taylor expansion up to terms in θ.
i really wish i knew what this taylor expansion is lol, but is there no other way, because i've never done this taylor expansion or learnt it at all :/
4. 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 as (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 (to get the theta-dependent bit in the form , which is the easiest form to do a binomial expansion on).

You can then use 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...
5. (Original post by DFranklin)
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 as (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 (to get the theta-dependent bit in the form , which is the easiest form to do a binomial expansion on).

You can then use 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 is just x right?
6. (Original post by will'o'wisp2)
right.... i guess i'll go with the binomial expansion thanks
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 to 1st order.

last question is just x right
Generally speaking, it's |x|, although it depends on context. I don't see how the expression comes up, anyhow...
7. (Original post by DFranklin)
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 to 1st order.

Generally speaking, it's |x|, although it depends on context. I don't see how the expression comes up, anyhow...
ah ok
8. (Original post by DFranklin)
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 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

so would there be an x or mod x outside the brackets?
9. (Original post by will'o'wisp2)
for example taking out the factor making it suitable for binomial theorem perhaps

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...)
10. (Original post by DFranklin)
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
11. (Original post by will'o'wisp2)
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:

12. (Original post by DFranklin)
Not sure what you're asking, but here's an example:

ye that's what i mean like the jumbo ()

but like lyx won't do it .__. lemme show you(without showing you my homework lol)
13. I have no idea what lyx is; I just write straight TeX code.
14. (Original post by DFranklin)
I have no idea what lyx is; I just write straight TeX code.

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)
15. (Original post by will'o'wisp2)

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).
16. (Original post by DFranklin)
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).
Ah thank you so much man, life saver.

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