The extreme value theorem guarantees extrema on the closed interval. Compare its endpoints and the stationary points $0,\pm \sqrt{10/3}$:ThusThe two-point answer in each case also follows from the even function symmetry.

The first three derivatives factor asLet $a=\sqrt{10/3}$. Sign analysis of the factors givesAll displayed endpoints are excluded because a factor vanishes there; the extra zero of $f^{\prime \prime }$ at $0$ does not create a positive interval.

On the panel $[2j/n,2(j+1)/n]$, the maximum fourth derivative is at the left endpoint because it is decreasing. ThusThe geometric series and the sum of consecutive integers giveSubstitution provesFor $n>2$, at least one panel's maximum is smaller than the global maximum $1$, so this bound is strictly tighter than the global bound. For $n=2$, the two bounds coincide.

Take $n$ to be a positive even integer, as required by composite Simpson's rule. There are $n/2$ two-strip panels, each of width $2/n$. Applying the supplied panel bound with the global fourth-derivative minimum $0$ and maximum $1$, and summing, givesHere $E_n$ is the nonnegative absolute error. In this example Simpson's rule overestimates the integral, so $E_n=S_n-I$. Each panel has that same sign because its fourth derivative is nonnegative; cancellation is not being assumed.^[ref]

False. It suffices to find a subsequence along which the ratio is unbounded. Set $n=4m$. The last $3m$ factors of $(4m)!$ are at least $m$, while the first $m$ factors have product at least one. HenceThis contradicts the eventual uniform bound required by Big O notation. The proof is an elementary factorial comparison, independent of Stirling's approximation.

True. For every $n\ge 4$, $2\le \sqrt{n}$, and raising both nonnegative sides to the $n$th power givesThus the Big O notation condition holds with constant $C=1$ and threshold $n_0=4$.

False. For $n\ge 3$, every factor from $3$ through $n$ is at least $3$, soNo fixed multiple of $2^n$ eventually bounds the factorial, contradicting the definition of Big O notation. No use of Stirling's approximation is required.

The displayed relation is not a partial order on the printed $A$. Since the empty set belongs to $A$, vacuous truth makeshold for every finite $T$, including nonempty $T$. This violates the defining condition of an antisymmetric relation. It also violates transitivity: $\{2\}{⪯}_2\varnothing {⪯}_2\{1\}$ holds, but $\{2\}{⋠}_2\{1\}$. Consequently the usual unique-maximum and unique-least upper bound terminology for a partially ordered set does not apply as claimed.

The available inputs still determine what happens if one applies the formulas in (b) literally to this relation:

For $S_1$, no nonempty finite set can lie above all triples of arbitrarily large numbers, while $\varnothing$ does so vacuously. Thus $\varnothing$ is its sole literal upper bound, but is not a member of $S_1$. For each of the two bounded families, the literal upper bounds are $\varnothing$ and all nonempty finite sets whose elements are at least $100$. Among these, exactly $\varnothing$ and $\{100\}$ lie below every such upper bound. Both belong to $S_2$, but neither belongs to its three-element subfamily.

For comparison, a natural correction is to retain the given rule for nonempty sets and declare $\varnothing$ to be a bottom element: no nonempty set is then below $\varnothing$. Under this explicitly corrected partial order, $S_1$ has neither a maximum nor a least upper bound; $S_2$ has maximum and least upper bound $\{100\}$; and $S_1\cap S_2$ has no maximum but has least upper bound $\{100\}$. A nonempty upper bound must have every element at least as large as every element of the bounded family. These corrected answers are conditional on the stated repair, not answers obtained by silently excluding the empty set from the source.

Put $J=\{k\in \mathbb{N}:k\le 100\}$, avoiding any dependence on whether the convention includes $0$ among the natural numbers. Under set inclusion, the results are

The set union of all three-element subsets of $\mathbb{N}$ is $\mathbb{N}$. An upper bound for $S_1$ under set inclusion would have to contain this entire infinite set, whereas members of $A$ are finite. Thus $S_1$ has no upper bound and hence no maximum or least upper bound.

For $S_2$, every member is a subset of the finite set $J$, and $J$ itself belongs to $S_2$, making it both maximum and least upper bound. The union of the three-element subsets of $J$ is again $J$, so $J$ is the least upper bound of $S_1\cap S_2$. It is not a member of that family, and no three-element member contains all the others, so there is no maximum.

The initial values are $d_1=0$ and $d_2=1$: one element cannot move, while two elements must be swapped.

For $n\ge 3$, fix one distinguished element, called $n$, and choose its image $j\ne n$, in $n-1$ ways. If $n,j$ form a two-cycle, deleting them leaves an arbitrary derangement on the other $n-2$ elements, giving $d_{n-2}$ choices.

Otherwise the permutation cycle containing $n$ has length at least three. Remove $n$ from that cycle, joining its predecessor directly to $j$. No fixed point is created because the predecessor is not $j$. This gives a derangement on $n-1$ elements. Conversely, any such derangement uniquely recovers the original one by inserting $n$ immediately before $j$ in its cycle. This case contributes $d_{n-1}$ choices for each $j$.

The two cases are disjoint and exhaustive, so the derangement recurrence is

A derangement of a finite set $A$ is a permutation $\sigma :A\to A$ with no fixed point: $\sigma (a)\ne a$ for every $a\in A$. Thus every element is moved, although the permutation remains a bijection.

Induct on $n$, with the claim quantified over every $m\ge n$. At $n=0$,Suppose the identity holds for a fixed $n$ and every $m\ge n$. For $m\ge n+1$, the Fibonacci recurrence givesApplying the induction hypothesis with $m-1\ge n$ is legitimate. This proves $F_m{F}_{n+1}-F_n{F}_{m+1}=(-1{)}^n{F}_{m-n}$ throughout its domain, a case of the Fibonacci determinant identity.

For $n=0$, the difference is $F_1^2-F_0{F}_2=1=(-1{)}^0$. Define $D_n=F_{n+1}^2-F_n{F}_{n+2}$. The Fibonacci recurrence givesThe inductive step therefore changes $(-1{)}^n$ into $(-1{)}^{n+1}$. By mathematical induction, $D_n=(-1{)}^n$ for all $n\ge 0$, the requested form of Cassini's identity.

Let $\varphi =(1+\sqrt{5})/2$, the golden ratio. Since ${\varphi }^2=\varphi +1$, multiplication by ${\varphi }^n$ gives ${\varphi }^{n+2}={\varphi }^{n+1}+{\varphi }^n$. Adding this recurrence relation to the one for $A_n$ yieldsThus the extra exponential term changes the initial values, but not the linear recurrence relation.

Adding the Fibonacci recurrence and the identical recurrence relation for the Lucas numbers givesLinearity preserves this linear recurrence relation; the initial values come from $0+2$ and $1+1$.

Use $G_1=0$ and $G_{n+1}=n{G}_n+n-1$ for $n\ge 1$. Indeed,The initial value is $0!-1=0$, so the recurrence relation produces the specified factorial sequence.

The Fibonacci recurrence, with the supplied initial values, is $F_{n+2}=F_{n+1}+F_n$ for $n\ge 0$, with $F_0=0,F_1=1$. Two initial values are needed for this second-order linear recurrence relation.

It is surjective but not injective; there is no global inverse function. Noninjectivity follows fromFor surjectivity, restrict to $x<1$, whereThis restriction is a continuous function, tends to $-\infty$ as $x\to -\infty$ and to $+\infty$ as $x\to 1^{-}$. The intermediate value theorem therefore supplies a preimage for every real number. The additional branch $x>1$ destroys global injectivity but cannot destroy surjectivity.

It is neither injective nor surjective onto $\mathbb{R}$, so it has no inverse function on the stated domain and codomain. The distinct inputs $0$ and $\pi /3$ both lie in the domain and both give zero. All values lie in $[-1,1]$, so, for example, the real number $2$ has no preimage.

The function is injective and surjective, with inverse function $f^{-1}(y)=1-y$. Equality $1-x=1-x^{\prime }$ forces $x=x^{\prime }$. For any $y\in (0,1)$, the input $x=1-y$ also lies in $(0,1)$ and satisfies $f(x)=y$. Applying the function twice returns the original input.