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y2jake215
certified Flat Birther theoristthe Last Good Boy onlineRegistered User regular

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!

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!

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

0

## Posts

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

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.

y2jake215onmaybe i'm streaming terrible dj right now if i am its here

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.

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...

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

SavantonSo 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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If anyone was wondering, it was for the exposure rate of radiation sources of different geometries.

This can be locked now!

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