I have to calculate the following expression:

$$f(A_1, A_2, B_1, B_2) = \frac{10^{-0.4A_1} + 10^{-0.4A_2}}{10^{-0.4B_1} + 10^{-0.4B_2}}$$

where $A_1, A_2, B_1, B_2$ are measurable quantities. The issue is that I can't measure those quantities as such. I can only measure them as $C_1, C_2$, where:

$$C_1 = A_1-B_1 \\ C_2 = A_2-B_2$$

Is there a way to express $f(A_1, A_2, B_1, B_2)$ as $f(C_1, C_2)$? If not (I'm pretty sure it can not be done), what is the closest expression I could get?

  • $\begingroup$ You can't replace four independent variables with two. Can you measure any other combination of $A_1,A_2,B_1,B_2$ such as $A_2-B_1$ or $B_2-B_1$? $\endgroup$ – John Wayland Bales Sep 30 '16 at 20:56
  • $\begingroup$ Unfortunately, no. I can only measure $C_1$ and $C_2$. Could it be approximated via Taylor expansion maybe? $\endgroup$ – Gabriel Sep 30 '16 at 23:22
  • $\begingroup$ So you can measure the values of $C_1,C_2$ and $f$ and you want the values of the $A$s and the $B$s? You need more information. $\endgroup$ – John Wayland Bales Sep 30 '16 at 23:38
  • $\begingroup$ No, I need to obtain $f$ but I can only measure $C_1$ and $C_2$, so I need to express $f$ as a function of those two variables. $\endgroup$ – Gabriel Sep 30 '16 at 23:43
  • $\begingroup$ You can express is as g(c1,c2) where the out come is going to be, not a value but a function that a takes two variables as its input. $\endgroup$ – fleablood Sep 30 '16 at 23:50

You can not express a function with four independent variables into a function with two.

What you can do though is convert it to a function which takes $C_1$ and $C_2$ as input and as output returns a function that takes two variables as input.

i.e $G: \mathbb R\times \mathbb R \rightarrow \mathbb R ^{\mathbb R\times \mathbb R}$

$G(C_1,C_2) = [h:\mathbb R\times \mathbb R \rightarrow \mathbb R;h(A_1, A_2) = \frac{10^{-0.4A_1} + 10^{-0.4A_2}}{10^{-0.4(A_1-C_1)} + 10^{-0.4(A_2-C_2)}}]$

So for example $G(6,5)= h$ such that $h(x,y) = \frac{10^{-0.4x} + 10^{-0.4x}}{10^{-0.4(x-6)} + 10^{-0.4(y-5)}}$

But that probably doesn't help you much, does it?

  • $\begingroup$ Nope, since I'd still need $A_i$ which I don't have. +1 for the answer anyway. $\endgroup$ – Gabriel Oct 1 '16 at 0:12
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    $\begingroup$ It can not be done. $\endgroup$ – fleablood Oct 1 '16 at 0:46

The fact that you represent a constant ( C) in the terms of what are variables given any particular 'C' measured at any time means that you need to keep the parameters as written. It looks like a differential guage measurement. Taking the LOG may be more difficult than converting to ln (change base of log).

  • $\begingroup$ This is a small part of the calculation I need to perform to obtain the combined magnitude of a binary star. $\endgroup$ – Gabriel Oct 1 '16 at 0:14

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