# Confusing expression in a paper

1. Mar 19, 2007

### natski

Hi all,

Whilst reading http://www.jhuapl.edu/techdigest/td1703/thomas.pdf [Broken] I found one of the equations a little difficult to interpret. On page 3, there is a function defined as f. On the denominator there it appears that they have differentiated n(z''(Z')) as a function of z'' and then evaluated it for the case where Z' goes to Zi'.

As far as I can tell, n is only a function of z'' and z'' is not a function of anything else, so the n(z''(Z')) doesn't make much sense to me. I interpretted it as meaning that I write down the expression n(z'') then replace all my z'' s with Z' using an appropriate equation of the form:
Z'=some function of z''.

However, I cannot even write down a clean expression for this! The form of Z', as seen just below eqn 4, cannot be easily rewritten to express z'' as a function of a Z'.

Can anyone help me interpretting this confusing expression?

Thanks!

Last edited by a moderator: May 2, 2017
2. Mar 19, 2007

### Crosson

Reading the text above equation (4), it appears that Z'(z'') originates from a "simple variable substitution" which they then detail, below equation (4).

3. Mar 19, 2007

### natski

Yes you are right but when I tried to rewrite Z'(z'') as z''(Z') I could not and it seemed to require solving in some kind of non-algebraic way.

4. Mar 19, 2007

### Crosson

The only true occurence of z'' is in the denominator, just supress n(z'') to n.

5. Mar 19, 2007

### natski

What do you mean suppress? n(z'') is quite a complex function of z''.

6. Mar 19, 2007

### Crosson

What I meant was to differentiate implicitly.

7. Mar 19, 2007

### natski

Umm ok, so I can write:

z''=[n(h) (R+h) Cos(Delta)]/(n(z'') Z') - R

Then are you thinking I should replace all my z''s in n(z'') with this equation? If I do this, I can't differentiate by z'' since now I have got Z' in there too.

I think I should differentiate first n(z'') by z'' first since that's not too hard. I then need to replace z'' with the equation above. However, when I then replace Z' with Zi' I still have z'' floating about which I can't get rid off!

8. Mar 29, 2007

### natski

Differentiating implicity or using some kind of chain rule or anything will not make life any easier as far as I can tell. The differentation part is the one part that is easily done, it's the replacing z''->Z'->Zi' that's the problem.