Let $G$ have $n$ vertices and $e$ edges. Its complement graph $\overline{G}$ hasedges. If both graphs were planar, the planar graph edge bound would giveEquivalently, $n^2-13n+24\le 0$, whose larger root is $(13+\sqrt{73})/2<11$. Thus an integer $n\ge 11$ is impossible, proving that

Now suppose that $G$ and $\overline{G}$ are both bipartite. In a bipartition $V(G)=A\bigsqcup B$, every pair within $A$ and every pair within $B$ is adjacent in $\overline{G}$. Since a bipartite graph contains no triangle, $|A|,|B|\le 2$, and hence $|V(G)|\le 4$. Equality is attained by the path $P_4$, whose complement is again a copy of $P_4$. Therefore the maximum order is

Assertion (iii) can be false. Let $\kappa ={\aleph }_0$ and $\lambda =2^{{\aleph }_0}$. Part (ii) and the currying law for cardinal exponentiation giveBy Cantor theorem, $2^{\lambda }>\lambda$, and therefore

Assertion (ii) is always true. Since $\kappa$ is infinite and $\kappa \le \lambda$,where the last equality uses ${\lambda }^2=\lambda$. Thus the cardinal exponentiation between two and the exponent gives

Choose sets $K,L,M$ of cardinalities $\kappa ,\lambda ,\mu$. Their cardinal arithmetic operations arewhere $K^L$ is the set of functions $L\to K$. The relation $\kappa \le \lambda$ means that there is an injective function $K\to L$.

The currying law for cardinal exponentiation follows from the explicit bijectionand provesMoreover, $2^{\kappa }$ is the cardinality of the power set of a set of size $\kappa$, so Cantor theorem gives

For completeness, identify each cardinal with its initial ordinal. Suppose that some infinite $\kappa$ violates ${\kappa }^2=\kappa$, and choose the least such cardinal. Well-order the pairs $(\alpha ,\beta )\in {\kappa }^2$ first by $\max \{\alpha ,\beta \}$ and then lexicographically. Every proper initial segment is contained in ${\gamma }^2$ together with finitely many boundary pieces for some $\gamma <\kappa$, and has cardinality below $\kappa$ by minimality. The resulting well-order therefore has cardinality at most $\kappa$. The reverse inequality is immediate from $\alpha \mapsto (\alpha ,0)$, contradicting the choice of $\kappa$. Thus the square of an infinite cardinal satisfies ${\kappa }^2=\kappa$.

If $\nu =\max \{\kappa ,\lambda \}$, monotonicity now givesHence the sum and product of two infinite cardinals obey

Assertion (i) can be false. For any infinite $\lambda$, take $\kappa =(2^{\lambda }{)}^{+}$. Then ${\kappa }^{\lambda }\ge \kappa >2^{\lambda }$, so

Since $1-C=1/2$, the second local hypergeometric solution isUnder $z={\sin }^{2}x$, it must be a constant multiple of the independent elementary solution $\sin (kx)$. As $x\to 0$, both $z^{1/2}$ and $x$ are asymptotic to one another, while $sin(kx)\sim kx$. Matching the normalized hypergeometric factor, whose value at zero is one, givesConsequently the hypergeometric sine identity is

The solution of $y^{\prime \prime }+k^{2}y=0$ with $y(0)=1$ and $y^{\prime }(0)=0$ is $y(x)=\cos (kx)$. Near $x=0$, the map $z={\sin }^{2}x$ has the local inverse $x=\arcsin \sqrt{z}$, so this solution is analytic in $z$, corresponds to exponent zero at $z=0$, and has value one there. Uniqueness of the normalized local solution of the Gauss hypergeometric equation therefore gives the hypergeometric cosine identityor, when $z={\sin }^{2}x$ on the chosen local branch,

Put $y(x)=w(z)$ with $z={\sin }^{2}x$. The chain rule givesand henceAfter substitution and division by $4z(1-z)$, the ordinary differential equation becomesComparison with the Gauss hypergeometric equation gives, up to interchanging $A$ and $B$,

At $z=0$ and $z=1$, the coefficient of $w^{\prime }$ has at most a simple pole and that of $w$ has at most a double pole, so both are regular singular points. The regular singular point at infinity criterion gives the same conclusion at infinity. The corresponding Papperitz symbol isIndeed, the exponent pairs are $(0,1-C)=(0,1/2)$ at zero, $(0,C-A-B)=(0,1/2)$ at one, and $(A,B)=(k/2,-k/2)$ at infinity; their sum also satisfies the Fuchs relation.

A suitable generalized linear model is a Gamma regression with logarithmic link:withThe gamma distribution is appropriate for a positive, continuous, right-skewed payment, and its variance function allows the conditional variance to grow like the square of the conditional mean instead of assuming constant variance. The logarithmic link function also guarantees positive fitted means and represents covariate effects multiplicatively, which accords with the Box–Cox transformation evidence for a logarithmic scale.

Because model2 uses a logarithmic transformation, the logarithm of the payment ratio is the difference of two predicted log payments. The cars have equal Brand and Bonus values and differ by one Kilometres level, so its fitted mean is the displayed mileage regression coefficient, ${\hat{\beta }}_K=0.057132$.

For two independent future payments, the prediction interval must include both the standard error $0.032654$ of ${\hat{\beta }}_K$ and two independent residual errors. The estimated variance of the predicted log ratio is thereforeUsing the residual degrees of freedom shown in the output, a 95% prediction interval for a ratio of log-normal responses isWith $t_{284,0.975}\simeq 1.97$, this is approximately $[0.120,9.33]$. The interval is wide because it predicts the ratio of two individual future payments; an interval for the ratio of their fitted mean responses would omit the $2{\hat{\sigma }}^2$ term.

The graph is a profile likelihood for the parameter $\lambda$ of a Box–Cox transformation. Its maximum is close to $\lambda =0$, and the displayed likelihood interval contains zero but excludes $\lambda =1$. Since $\lambda =0$ gives a logarithmic transformation whereas $\lambda =1$ leaves the response unchanged up to an affine transformation, the plot supports fittingas the response in model2.

Kilometres is an ordinal categorical variable: its five levels have a meaningful order. Treating it as an ordinary factor would use four indicator variables and hence four unrelated regression coefficients. Converting the levels to numerical scores instead imposes a single linear trend across successive mileage bands. This is more parsimonious and gives greater precision when that trend is a reasonable approximation, although it also assumes that adjacent levels are equally spaced in their effect.

The partial recursive functions form the smallest class containing the zero function, successor function, and projection functions and closed under composition, primitive recursion, and unbounded minimization.

Fix a register machine $P$ with $k$ inputs. By the permitted result, its complete configuration after $t$ steps,is a recursive function of the inputs and time. The predicate $H(\mathbf{x},t)$ saying that this configuration is halted is therefore recursive. Its halting time is the partial recursive functionwhere unbounded minimization is undefined when no halting time exists. Extracting the output register from a configuration is recursive, sois partial recursive. Hence every partial computable function is partial recursive.

A register machine has finitely many natural-number registers, a finite set of labelled states, and instructions that increment a register, conditionally decrement and branch according to whether a register is zero, or halt. A program is a finite sequence of such labelled instructions, and a configuration records the current instruction together with every current register value.

Given $k$ inputs, place them in designated input registers and initialize the remaining registers to zero. The machine computes a partial computable function $f:{\mathbb{N}}^k\rightharpoonup \mathbb{N}$ when it halts with output $f(\mathbf{x})$ in the output register exactly for $\mathbf{x}\in \mathrm{dom}f$ and runs forever otherwise.

To encode a machine, encode every instruction by a natural number using effective encodings of its opcode, register number, and target labels. Encode the resulting finite sequence by a Gödel numbering, for example through iterated pairing or prime powers. Effective decoding gives an enumeration $(P_n)$ of machine programs, so the unary function computed by program code $n$ may be denoted $f_{n,1}$.