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    for example in

    \sqrt{h^4 -cos^2 \theta}

    where h>0 and \theta 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 1-\dfrac{bh^2}{2} where cos is cos(hx)

    the main thing actually is that im not sure what \sqrt{h^4 -cos^2 \theta} is approximately equal to, perhaps it's u+\delta u ?? it a complete guess but im sure that way would be quicker but is binomal theorem ok?
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    Use Taylor expansion up to terms in θ.
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    (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 :/
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    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...
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    (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 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?
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    (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 \sqrt{h^4-1} to 1st order.

    last question (x^2)^{\frac{1}{2}} is just x right
    Generally speaking, it's |x|, although it depends on context. I don't see how the expression comes up, anyhow...
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    (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 \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...
    ah ok
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    (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 \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?
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    (Original post by will'o'wisp2)
    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...)
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    (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
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    (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:

    \left( 1 + \sqrt{\dfrac{1-x^2}{1+x^2}} \right)
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    (Original post by DFranklin)
    Not sure what you're asking, but here's an example:

    \left( 1 + \sqrt{\dfrac{1-x^2}{1+x^2}} \right)
    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)
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    I have no idea what lyx is; I just write straight TeX code.
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    (Original post by DFranklin)
    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)
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    (Original post by will'o'wisp2)
    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).
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    (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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