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# Math Problem Simplification

certified Flat Birther theoristthe Last Good Boy onlineRegistered User regular
edited November 2011
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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y2jake215 on

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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))

Origin: KafkaAU B-Net: Kafka#1778
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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

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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.

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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...

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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
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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"

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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!

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