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Put $a=\alpha \tilde{a}$ withThenFor $y={\tilde{a}}^2$ this becomesThe expanding big-bang branch hasDifferentiation givesThus, with $\omega =\sqrt{4\Lambda /3}$,where the initial derivative givesFor $\lambda >1$, $\tilde{a}$ expands forever and becomes asymptotically exponential. For $\lambda <1$, it reaches a maximum and recollapses to zero. At $\lambda =1$, $y$ increases monotonically to $3/(2\Lambda )$, the limiting static radius.

The positive-curvature term identifies $R$ as the curvature radius of the spatial three-sphere when the dimensionless scale factor is $a=1$; the physical curvature radius is $Ra(t)$.

Substituting $P=\rho /3$ into the continuity equation givessoThree powers arise from dilution of photon number in physical volume and one from the cosmological redshift of each photon's energy.

Since $C$ and $C^T$ have the same eigenvalues, an exponentially growing mode exists precisely when $L$ has an eigenvalue with real part greater than one. For the nonnegative contact matrix this iswhere $\rho (C)$ is the spectral radius.

With $S_i\simeq N_i$, the infective equations becomewhere reciprocity was used. For a column vector $\mathbf{I}$,Therefore

The term ${\lambda }_i{S}_i$ moves susceptible people in group $i$ into its infective class, while $\gamma I_i$ moves infectives into recovery. In$C_{ij}$ is the contact rate of a person in group $i$ with group $j$, and $I_j/N_j$ is the infectious fraction there. The reciprocity condition $N_i{C}_{ij}=N_j{C}_{ji}$ says that the total number of $i$--$j$ contacts agrees when counted from either group. Common $\beta$ and $\gamma$ mean age-independent transmission per infectious contact and recovery rate. There is no transmission between distinct age groups exactly when

The language $\{wabw:w\in \{a,b{\}}^{\ast }\}$ is not context free. Intersecting with a suitable regular language that forces both copies into long homogeneous blocks reduces to a marked copy language; the context-free pumping lemma then places the two pumped pieces within only finitely many local blocks and cannot change the two copies identically. Equivalently, the standard copy-language proof applies with the fixed marker $ab$.

The language $\{a^i{b}^j:i\ge j\ge 0\}$ is context free. For example the grammargenerates $a^{j+k}{b}^j$ with $j,k\ge 0$, and every required word has this form.

A parse tree has the start variable at its root; the ordered children of a variable spell the right side of the production used at that occurrence, and its leaves, read left to right, spell the derived word.

For a grammar in Chomsky normal form with $r$ variables, take pumping length $p=2^r$. A parse tree for a word of length at least $p$ has a root-to-leaf path containing one variable twice. The yield can therefore be writtenwhere $|vxy|\le p$, $|vy|>0$, and replacing the subtree between the repeated variables zero, one, or several times provesThis is the pumping lemma for context-free languages. Without Chomsky normal form but with no epsilon or unit productions, use the maximum production length to choose a larger exponential $p$; bounded branching and the same repeated-variable argument apply.

The language $\{a^i{b}^i{a}^i:i\ge 0\}$ is not context free. A pumped window of bounded length cannot alter all three long blocks equally, so pumping breaks one of the required equalities.

Write $n_b=n_e+n_H\simeq \eta n_{\gamma }$. Charge neutrality gives $n_p=n_e$, and thereforeSubstitution into the Saha ionization equation yieldsUsing the given photon density,

The baryon-to-photon ratio is extremely small. Even when $k_{B}T\ll I$, the high-energy tail of the enormous photon population contains enough ionizing photons per baryon to delay cosmological recombination. The exponential factor must become large enough to overcome both $\eta \sim 10^{-9}$ and the phase-space factor, which occurs near $k_{B}T\simeq 0.3\mathrm{eV}$ rather than near $I$.

Suppose there were no singular point in the Riemann sphere. Then every finite point would be ordinary, so $p$ and $q$ would be entire functions. Ordinary behavior at infinity would force $p(z)=2/z+O(z^{-2})$ and $q(z)=O(z^{-4})$. In particular $p(z)\to 0$, so Liouville theorem would give $p\equiv 0$, contradicting the required leading term $2/z$. Thus infinity cannot also be ordinary, and every such differential equation has at least one singular point in $\overline{\mathbb{C}}$.