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Apply the preceding Taylor's theorem at $t=h$ and $t=-h$, with $h>0$. The even-order terms cancel in the central finite difference, givingBoth Lagrange remainders have magnitude at most $h^3/6$, so the triangle inequality gives the sharp central-secant derivative error

Substitution givesIn particular, $1$ is a stationary point. For $x$ sufficiently close to $1$, Taylor's theorem with fifth-order remainder giveswhere $\xi$ lies between $1$ and $x$. The fifth derivative is bounded near $1$, so the remainder is $O((x-1{)}^5)$ and is smaller in magnitude than half the positive quartic term for sufficiently small nonzero $x-1$. Therefore $x=1$ is a strict local minimum, despite the vanishing second derivative.

An exact algebraic check gives the stronger global conclusion:Equality holds only at $x=1$, so it is the unique global minimum on the actual domain.

Put $\mathbf{h}=\mathbf{y}-\mathbf{x}$. For $0<t\le 1$, convexity gives $f(\mathbf{x}+t\mathbf{h})\le (1-t)f(\mathbf{x})+tf(\mathbf{y})$, henceBy the first-order form of Taylor's theorem, the right side tends to ${\mathbf{h}}^{\mathsf{T}}\nabla f(\mathbf{x})$ as $t\downarrow 0$. ThereforeThis is the supporting inequality for a differentiable convex function. The stated partial derivatives are enough: coordinate derivatives give directional Taylor expansion for convex functions, so no unmentioned continuity assumption on the second partial derivatives is needed. The result in fact needs less smoothness than the question assumes.

Taylor's theorem at zero gives $f(x)=-1000+x+\frac{1}{2}{x}^2{f}^{\prime \prime }(\xi )$ for a point $\xi$ between zero and $x$. At $x=1000$, strict negativity of the second derivative gives $f(1000)<0$. At $x=2000$, its lower bound givesBy the intermediate value theorem there is a root strictly between these values. The given uniqueness identifies it as $1000<x^{\ast }<2000$.

Set $h=x_{n+1}-x_n=-f(x_n)/f^{\prime }(x_n)$. By Taylor's theorem,The first two terms cancel by the definition of the Newton root-finding iteration. Thuswith $\xi$ between the consecutive iterates. In particular, a nonzero step has negative new residual under the stated strict concavity.

First control the denominator. The mean value theorem applied to $f^{\prime }$ yields, for $0<x\le 2000$,Thus $f$ is strictly increasing on $[0,2000]$ and all proposed Newton steps within that interval are defined.

Assume $0\le x_n<x^{\ast }$. Then $f(x_n)<0$ and $f^{\prime }(x_n)>0$, so $x_{n+1}>x_n$. Apply Taylor's theorem instead at the root, expanding about $x_n$:After rearranging and substituting the Newton step, the signed error satisfiesConsequently $x_n<x_{n+1}<x^{\ast }$. Starting from $x_0=0$, mathematical induction proves this for every finite $n$, establishing that the signed errors are negative and strictly increasing. Since $f^{\prime \prime }({\xi }_n)>-1/4000$ and $2f^{\prime }(x_n)>1$,Writing $E_n=-{ϵ}_n$ gives $0<E_{n+1}\le E_n^2/4000$. Since $E_0=x^{\ast }<2000$, this also implies $E_n\le 4000(E_0/4000{)}^{2^n}\to 0$, proving quadratic convergence. This is the concave Newton iteration from below.

A dark, opaque circular planet of radius $R_p$ blocks specific intensity over its overlap with the projected star. With $k=R_p/R_{\star }$ and no transit dilution, the instantaneous exoplanet transit depth is exactlyWhen $k\ll 1$, the entire planetary disk is inside the stellar disk, and specific intensity varies little across that disk, small-planet transit weighting reduces toHere $z$ is the projected centre separation divided by $R_{\star }$. The numerator samples the local specific intensity; the denominator normalizes by the whole star's mean specific intensity. The small-size argument and the distinction between full overlap and ingress are explicit in the original analytic treatment. ^[ref]

For a short approximately straight crossing at constant transverse velocity $v_{\perp }$, write $x=v_{\perp }t/R_{\star }$ with $t=0$ at mid-transit, and let $b$ be the transit impact parameter. Then $z^2=x^2+b^2$ and ${\mu }_p=\sqrt{1-b^2-x^2}$. With stipulated $u=0.6$ and $k=0.05$, a central crossing has mid-transit exoplanet transit depth approximately $0.003125$, whereas $b=0.8$ gives $0.002375$. A uniform disk would give $k^2=0.0025$ for either fully overlapping crossing. The brighter central chord produces a deeper dip and the off-centre chord a shorter crossing at the same $v_{\perp }$.

At fixed $z<1$ away from the edge, a two-dimensional Taylor's theorem expansion over the symmetric planetary disk cancels the first derivatives; the first correction to the blocked radiant flux is of order $k^4$ if the second derivatives remain bounded. This error estimate is not uniform at the edge, where the radial derivative of the linear limb-darkening law diverges. Ingress, egress, a grazing exoplanet transit, and a crossing of a starspot require the overlap integral or a suitable more complete model. Inferring $k=\sqrt{\delta }$ without the local-to-mean specific intensity ratio can bias the estimated planet radius.

Taylor's theorem expresses a sufficiently differentiable function as a finite Taylor polynomial plus a Taylor remainder. One convenient version assumes $f$ is $n+1$ times continuously differentiable on the segment between $a$ and $x$. Thenfor some intermediate point $\xi$, when $x\ne a$. Thus a bound $|f^{(n+1)}|\le M$ on that segment gives $|R_{n,a}(x)|\le M|x-a{|}^{n+1}/(n+1)!$. The intermediate point usually need not be computed: bounding the derivative throughout the segment is enough. This is why local approximation can yield reliable numerical bounds. A finite Taylor polynomial does not require an infinite Taylor series to represent the function.

Historically, the Royal Society holds Brook Taylor's letter of 26 July 1712 to John Machin. Its catalogue connects his work on Kepler's second law with the expansion later published in the 1715 Methodus incrementorum directa et inversa. This is a documented route from an orbital calculation to a general approximation method, rather than a claim that every earlier use of such expansions began with Taylor. See Royal Society archive, letter MS/82/2.