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Math Problem Simplification

y2jake215y2jake215 certified Flat Birther theoristthe Last Good Boy onlineRegistered User regular
edited November 2011 in Help / Advice Forum
I have an equation for a certain variable, and I'm trying to get it to simplify for long distances. It's proportional to

ln[(a^2 + b^2)/(a^2)]
where a is the distance, so far away, a>>b.

Wouldn't that give ln(1), which gives zero for the whole answer? I must be screwing up the simple assumption that
(a^2 + b^2)/(a^2) ~ (a^2 / a^2) = 1
so that ln( ) is 0 instead of 1.

I also have another equation where it's proportional to

[tan^-1(L1/d)+tan^-1(L2/d)]

again at long distance, where d>>L1 or L2.

I just can't seem to find any list of assumptions you can make through google, like 1/(1+x) being a Taylor series, or sin(x)~x for small angles or whatever.
Does anyone know a resource like that?

Thanks!

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maybe i'm streaming terrible dj right now if i am its here
y2jake215 on

Posts

  • KafkaAUKafkaAU Western AustraliaRegistered User regular
    I think the phrase you are looking for is an identity.

    e.g.

    Your second one is a trigonometric identity.

    Atan(x) + Atan(y) = Atan((x+y)/(1-xy))

    So that would be Atan((L1/d+L2/d)/(1-L1L2/d^2))

    = Atan((L1+L2)/(d-L1L2/d))

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  • y2jake215y2jake215 certified Flat Birther theorist the Last Good Boy onlineRegistered User regular
    edited November 2011
    I'm still not seeing how that cancels out into a point-source equation at large distances, though.

    Basically the equations are for various geometries, which I know all cancel down into a basic one for a point-source when taken at a large distance.

    So I know that
    Atan((L1/d+L2/d)/(1-L1L2/d^2))
    would have to simplify to 1/d.

    I'm just not sure which identities to use to get that.

    Same for the first, which needs to simplify to 1.

    y2jake215 on
    C8Ft8GE.jpg
    maybe i'm streaming terrible dj right now if i am its here
  • Dunadan019Dunadan019 Registered User regular
    from what I remember of derivations of physics equations from highschool.

    the idea was to take the variable you want to put at infinity, say x, and make every instance a dx and then isolate the dx and integrate it from 0 to x.

    unless this is some sort of relationship issue where it would be necessary to know that say, L1 + L2 always =1. there is usually some inherent geometric equation that you can use to simplify these things but we'd have to know more than just a proportional equation.

  • romanqwertyromanqwerty Registered User regular
    With regards to 1), are you sure you aren't making a mistake anywhere else or that it actually has to equal 1?

    http://www.wolframalpha.com/input/?i=series(ln((x^2+a^2)/x^2)) (look at the series about x=infinity)

    would imply that the taylor series about x=infinity doesn't approach 1 ever. The general way physicists deal with these sort of problems is to just make a taylor series and ignore the higher terms as they goto zero faster. Hence an argument could be said that for large distances it goes as 1/x^2, but that's obvious anyway.


    With regards to 2)

    http://www.wolframalpha.com/input/?i=series(arctan(a/x)+arctan(b/x))

    shows that it would go as (L1+L2)/x for large x.


    I just noticed too that you say 1 is supposed to resemble a point source? In that case, going as 1/x^2 is correct because that's how a point source diminishes with distance so i would check back at what answers you're expecting and why...


  • SavantSavant Simply Barbaric Registered User regular
    edited November 2011
    How much information do you want on the first one? It's been a little while since I've done these sorts of approximation tricks, but I think for the first one what you want to look at is the approximation through the Taylor expansion of e^x when x is very small, and try to see where that fits in what you have. You're dealing with the natural log, so that is a good place to look. The problem you seem to have is you are oversimplifying by approximating to zero when the approximation you should be looking for is simply a very small number that is close to zero at the scale provided. Since (a^2 + b^2)/a^2 = 1 + b^2 / a^2 = 1 + y, where y is really small.

    I can go through this in more detail if need be.

    Savant on
  • L|amaL|ama Registered User regular
    Yeah, factoring out through the big thing in functions of a sum is usually helpful, ie if you have f[x+y] where y>>x rewrite as f[(x)*(1+y/x)] and then try and take the x out as g[x]*f[1+y/x]. So for (a+b)^n we know that (1+x)^n ~=1+nx for small x, so (a+b)^n = a^n*(1+b/a)^n ~= a^n*(1+nb/a).

    So for your first one I'd do as Savant said and figure out the Taylor expansion of Ln[1+x] then set x=(a/b)^2 which is log[1+x] ~= x, to first order.

    For the second one I would look at the small angle approximation, sin(x)~=tan(x)~=x for small x, and so instead of inverse tan you'd have 1/x (I think, that seems kind of suspicious and I'm tired)

    I wouldn't expect to find tables of common approximations or anything, they're something you just learn to get a feel of as you do problems and think "hey, this is similar to that thing I did before and after a few pages of working and giving up on different approximations I found out that this one worked well"

  • y2jake215y2jake215 certified Flat Birther theorist the Last Good Boy onlineRegistered User regular
    Thanks all for the help - that Wolfram site was exactly what I needed, showing the series as x went to infinity.
    If anyone was wondering, it was for the exposure rate of radiation sources of different geometries.
    This can be locked now!

    C8Ft8GE.jpg
    maybe i'm streaming terrible dj right now if i am its here
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