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I have an expression I'm trying to simplify, following along in a certain book, and I can't see how to do it. Here's what I've done so far.

I have a smooth real-valued function $F(x,y,z)$, which by the implicit function theorem, defines a function $\newcommand{\what}{\hat{y}} \what(x,z)$ such that $F(x,\what,z)=0$ everywhere in some neighborhood, and such that:

$$\partial_x \what (x,z) \equiv -\frac{\partial_x F(x,\what,z)}{\partial_y F(x,\what,z)}.$$

So far so good. Now it turns out $F$ has a special form; it can be written as a determinant:

$$F(x,y,z) = \det\begin{bmatrix}f_1(x) & g_1(x) & 1\\ f_2(y) &g_2(y) & 1 \\ f_3(z) & g_3(z) & 1\end{bmatrix}.$$

Note that each row depends on only one variable. I can use this form to compute an expression for $\partial _x \what$ in terms of the $f_i,g_i$:

$$\partial_x \what = -\frac{f_1^\prime(g_2\circ \what -g_3) - g_1^\prime(f_2\circ\what - f_3)}{(f_2^\prime\circ \what)(g_1-g_3) - (g_2^\prime\circ\what)(f_1-f_3)}$$

But according to the book, we can rearrange this further. Using, I think, the fact that $F(x,\what,z)=0$ everywhere, we can write:

$$\partial_x \what = \frac{f_3-f_2}{f_1-f_3}\frac{f_1^\prime(g_2\circ \what -g_1) - g_1^\prime(f_2\circ\what - f_1)}{(f_2^\prime\circ \what)(g_2\circ\what -g_1) - (g_2^\prime\circ\what)(f_2-f_1)}$$

This looks like the original, except all subscripts dependent on 3 have been factored out into that term on the left. How did this happen? I've tried making the substitutions I could think of, but it gets messier rather than neater.

I'm following along in Ch 5 of The History and Development of Nomography by Evesham. I found a similar derivation step in James-Levy's paper Nomogram construction without quadratures.

user326210
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  • What happens if you tentatively set the two expressions equal to each other, then cross-multiply to get rid of denominators, and brazenly expand everything out and cancel every term that appears on both sides ... what's left, I wonder? – Greg Martin Jul 14 '21 at 00:31
  • @GregMartin Thanks. My question is about what operation transforms the first expression into the second. – user326210 Jul 14 '21 at 04:11
  • I'm not sure I'd say it's an "operation"; it might simply be that the two expressions are equal, as the algebra I suggested might verify, and that the author was more interested in using the fact that they were equal than showing the several algebraic steps that justified the equality. – Greg Martin Jul 14 '21 at 05:06
  • Oh, I did confirm that they're equal. When I say operation, I'm wondering what the several algebraic steps might be. – user326210 Jul 14 '21 at 08:21
  • Good job! I'm guessing that whatever your confirmation was, the steps involved in transforming the left side to the right side are pretty much the same steps as in your confirmation only in reverse. – Greg Martin Jul 14 '21 at 17:01

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