With the gap-at-one convention, machine epsilon is the difference between $1$ and the next larger representable floating-point number. The double-precision floating-point format has 53 significant binary digits, including the implicit leading digit, soA related convention uses unit roundoff, the worst relative rounding scale for round-to-nearest: $u=2^{-53}\approx 1.11\times 10^{-16}$. Stating the convention distinguishes these two common values.^[ref]

Let $q=x^2+4y^2$. The objective has the exact sum of squares decompositionFor every feasible point, $q\le 8$, so this is at least $3-8/2=-1$. Equality requires simultaneously $q=8$ and $x+2y=0$. Substitution gives $y^2=1$ and $x=-2y$. Both resulting points are feasible. Therefore the global minimum is $-1$, attained exactly at $(2,-1)$ and $(-2,1)$. The lower bound and its equality conditions prove global attainment of the minimum, rather than merely identifying stationary points on the boundary.

Let $h(x,y)=2y^3+x{y}^2-y^2-4x$, so $g=e^h>0$. Its gradient is $\nabla g=g\nabla h$, andThus a stationary point must have $y=\pm 2$ and $x=1-3y$, giving $(-5,2)$ and $(7,-2)$. To classify them, useAt either stationary point, $\nabla h=0$ and $12y+2x-2=6y$. Hence $\det H_g=g^2(-4y^2)=-16g^2<0$. The Hessian has one positive and one negative eigenvalue, so both are saddle points. There are no other stationary points.

For $0<x<\pi$, apply Taylor's theorem to the sine about zero, retaining terms through degree four. Its fourth-order coefficient is zero, and its fifth derivative is the cosine function, givingThe cosine function is decreasing on $[0,\pi ]$, so $\cos x\le \cos \xi \le 1$. Multiplying by the nonnegative factor $x^5/120$ and adding $x-x^3/6$ proves both required bounds. At $x=0$ both are equalities.

One standard finite form of Taylor's theorem is the following. Suppose $f$ has continuous derivatives through order $m+1$ on an interval containing $a$ and $x$. Then, if $x\ne a$, there is some $\xi$ strictly between them such thatThe final term is the Lagrange remainder. At $x=a$ the identity holds directly with zero remainder. The finite expansion requires no assumption that an infinite Taylor series converges to $f$.

Pad every number with leading zeros to obtain exactly $n$ decimal digits. Its digits are a solution of $d_1+\cdots +d_n=9$ in nonnegative integers. Conversely every such solution is a valid digit string, since no individual digit can exceed $9$; its digit sum also excludes the number zero. The stars and bars count therefore givesFor $n\ge 1$, $n+j\le n(j+1)$, so $f(n)\le n^9{\prod }_{j=0}^8(j+1)/9!=n^9$. Thus $f(n)=O(n^9)$, with an explicit Big O notation constant $C=1$ and threshold $n=1$.

Let $E_i$ be the set of permutations fixing object $i$. If a specified collection of $k$ objects is fixed, the remaining $n-k$ objects can be arranged in $(n-k)!$ ways. There are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ such collections. Applying the inclusion-exclusion principle to exclude all fixed objects givesFor $n\ge 1$, the terms $k=0$ and $k=1$ cancel. Thus the number of derangements is $d_n=\frac{n!}{2!}-\frac{n!}{3!}+\cdots +(-1{)}^n\frac{n!}{n!}$, interpreting the remaining sum as empty when $n=1$. The full inclusion-exclusion formula also gives $d_0=1$.

There are $n$ choices for the first position, then $n-1$ choices for the second, and so on down to one choice. The multiplication principle therefore gives $n!$ permutations. This also agrees with the empty permutation convention $0!=1$.