More specifically, was Newton aware that given an inverse pair of functions $f$ and $h$ such that

$$f(h(x)) = x = h(f(x))$$ about the origin that, for


the derivatives satisfy

$$f^{'}(x) = 1/h^{'}(y)$$


$$dy/dx = 1/(dx/dy)$$

near the origin?

Heuristically, this follows symbolically from

$$dy = f^{'}(x)dx = f^{'}(x)h^{'}(y)dy. $$

And it follows geometrically for a function whose graph lies in the first quadrant by reflection through the bisector of the first quadrant, the line $y=x$. Clearly, the slope for any tangent line is inverted by the reflection just as displacements along the $x-$axis and the $y-$axis are interchanged. d. In fact, it follows directly from the tangent line perspective since $$ y = m \; x + b$$ and $$y = \frac{1}{m}(x-b)$$ describe an inverse pair.

Surely, with Newton's mastery of geometric calculus, he was aware of these relationships. Is there evidence of this in Newton's work?

Related MO-Q by Ziegler.

Cross-posted from this MO-Q and an identical question on MSE.

Edit 6/12/17:

An example of a calculation incorporating the IFT that would have been obvious to Newton and plausible for him to have performed if only as a simple check of his general formulas:

It was known well before Newton that

$$\frac{d\tan(x)}{dx} = 1+ \tan^2(x),$$

or, with $y = \tan(x)$,

$$\frac{dy}{dx} = 1+ y^2.$$

In terms of fluxions and fluents, this could be put in the form of Newton's implicit function



$$\frac{\dot{x}}{\dot{y}}= \frac{1}{1+y^2}=\frac{dx}{dy}, $$

and application of the binomial theorem and integration would give the series

$$ \arctan(y) = x = y - \frac{y^3}3+\frac{y^5}5-\frac{y^7}7+\dots. \tag3 $$

Newton could then have derived a series expression for $\tan(x)$ using his series reversion formula (see Ferraro) for finding the series for the compositional inverse of a function from its power series. In fact, the same procedure is applied to finding a series for $\sin(x)$ in Ferraro on pages 76-78 following an alleged reconstruction by Horsley of Newton's derivation of the series.

  • $\begingroup$ The comments to the identical MSE and MO questions provide related references. $\endgroup$ – Tom Copeland Jun 12 '17 at 23:52

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