Frequent assumption made (z R)

[ShizukaSm]

Frequently, when a length is bigger than another (say, an axial distance is bigger than the radius of something related) a certain assumption is made and the smaller length is "removed" from a formula. It's hard to describe, so let me show an example:

The magnetic field from a bobbin at an axial distance Z is:

B = \frac{\mu_0 i R^2}{2(R^2+z^2)^{3/2}}

But, according to my book, when z \gg R:

B = \frac{\mu_0 i R^2}{2z^3}

Even though this is just an example I've seen similar assumptions being made multiple times. How do I arrive at such conclusion? I hardly think that I must simply "delete" the smaller length (R^2) in the example, surely there must be a procedure to conclude that precisely?

 

[LCKurtz]

$$(R^2+z^2)^{\frac 3 2} = \left(z^2( \frac{R^2}{z^2}+1)\right)^{\frac 3 2}
\approx (z^2(0+1))^{\frac 3 2}=z^3$$

 

[kevinferreira]

Usually what is done, taking your example for clarity, is to write things in terms of the small quantity
<br /> \frac{R}{z}\ll 1.<br />
Then you do a Taylor expansion in terms of this small parameter (call it x). In this way you have a meaningful expansion that you can control. In your example:
<br /> B=\frac{\mu_0iR^2}{2z^3(1+x^2)^{3/2}}=\frac{\mu_0iR^2}{2z^3}+O(x).<br />

 

[HallsofIvy]

By the way, z\gg R, and similarly, z/R\ll 1, do NOT just mean "z is greater than R" and "z/R is less than 1". They mean "z is much greater than R" and "z/R is much less than 1". "much greater" and "much less" is generally interpreted as "one can be ignored, compared to the other". That is if z\gg R, z+ R is "indistinguishable" from z, just as if you have $100,000,000, you can "ignore" $1!

 

[haruspex]

<br /> B=\frac{\mu_0iR^2}{2z^3(1+x^2)^{3/2}}=\frac{\mu_0iR^2}{2z^3}+O(x).<br />

Do you mean
<br /> =\frac{\mu_0iR^2}{2z^3}(1+O(x^2))<br />?

 

[kevinferreira]

Do you mean
<br /> =\frac{\mu_0iR^2}{2z^3}(1+O(x^2))<br />?

It doesn't really make a difference.
Yes, indeed, the first derivative vanishes at 0, but it doesn't make a difference because we're interested only in the zero order, so what I wrote holds anyway (an O(x^2) is an O(x) ).
Regarding the parenthesis, it makes no difference whatsoever, the notation O(x^a) is sufficiently ambiguous to allow many different notations where you let constants get in or not. I'm using the standard Taylor expansion notation, f(x)=f(0)+xf&#039;(0)+....
So, what I wrote definitely holds, as what you wrote.