Understanding the Relationship Between A and B: A Variation Analysis

[ChrisVer]

Suppose I have such an equation:
A= f(B)
so A is a function of B.

Can I really use the fact that a variation of A is like taking the differential of it?
\delta A= dA

so that:

\delta A = \frac{d f(B)}{dB} \delta B
?

 

[mathman]

Superficially it looks OK. You need to look at the definition involved for variation.

 

[ChrisVer]

the reason I ask is because I have a confusion with one more way to see variation.
Suppose I vary B, so that A is going to be varied too:

A+ \delta A = f(B+\delta B)
And suppose f(x) = \frac{1}{\sqrt{x}}

Then:

A( 1+ \frac{\delta A}{A}) = \frac{1}{\sqrt{B+\delta B}}= \frac{1}{\sqrt{B}} \frac{1}{\sqrt{1+\frac{\delta B}{B}}}≈ A -A \frac{1}{2} \frac{\delta B}{B}

\frac{\delta A}{A}= - \frac{1}{2} \frac{\delta B}{B}This result would imply that:
\frac{d f(B)}{dB} = -\frac{1}{2}\frac{A}{B}

in my OP...

with a simple check, we have:

\frac{d (1/ \sqrt{B})}{dB} = -\frac{1}{2} \frac{1}{B^{3/2}}

This method works fine for such an easy function as f(x)=\frac{1}{\sqrt{x}}
Now suppose I have instead:
f(x)= \frac{1}{\sqrt{x}} + c

with c a constant. The same method won't work because I will end up with something like:

\frac{\delta A}{A} = - \frac{1}{A} \frac{1}{2} \frac{1}{\sqrt{B}} \frac{\delta B}{B}

unfortunately now \frac{1}{A} can't be canceled with the \frac{1}{\sqrt{x}} ... Only maybe, up to some error, if I consider c very small... The derivative however will kill c and give some other result (the same as for without c which I think is more physical- isn't it? a constant shouldn't destroy the variations)..

 

[ChrisVer]

I think I found my misconception... the variation is still independent of c... it's 1/A which gives me that dependence. So there's no way to escape from c rather than neglecting it :) that's fine...i guess

 

[Matterwave]

You can basically use Taylor expansion to show your equality.

If $A=f(B)$ and we add a small quantity $\delta B$ to $B$ then by Taylor's theorem we will have:

$$A'\equiv A+\delta A=f(B+\delta B)=f(B)+\frac{df}{dB}\delta B+\mathcal{O}(\delta B^2)$$

This gives as the limit $\delta B\rightarrow 0$:

$$\delta A=\frac{df}{dB}\delta B$$

Of course, this is usually much more interesting when we consider functionals rather than simple functions.