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The Laplace transform of a function on $t\ge 0$ isfor values of $s$ for which the integral converges. Integration by parts gives the Laplace transform of a derivative

Write ${\hat{N}}_i(s)=\mathcal{L}\{N_i\}(s)$. Transforming the first two equations and using the initial data givesThe third equation then givesand henceA partial fraction decomposition therefore yields the sequential radioactive decay formula

Now let ${\lambda }_1={\lambda }_2=\lambda$ and put $\Delta ={\lambda }_3-\lambda >0$. The transformed contribution originating from the initial $N_1$ population becomesEquivalently, solve the middle equation first to obtainand use an integrating factor in the final equation:Evaluating the elementary integral gives

An abstract smooth surface is a Hausdorff second-countable topological space with an atlas of charts to open subsets of ${\mathbb{R}}^2$ whose transition maps are smooth. It is orientable when it has such an atlas for which every transition map has positive Jacobian determinant. A map $f:S_1\to S_2$ is a smooth map when every coordinate representationis smooth wherever it is defined.

The map $a:C\to C$ is a smooth involution with no fixed point: the equationwould force $x=y=0$, which is impossible on $x^2+y^2=1$. For each $p\in C$, choose a sufficiently small coordinate neighbourhood $U_p$ such thatThen the quotient projection restricts to a homeomorphismTransporting a smooth chart ${\varphi }_p:U_p\to {\mathbb{R}}^2$ across this homeomorphism gives the quotient chartOn an overlap, a lift lies either in another chosen neighbourhood or in its image under $a$. The corresponding transition map is therefore a transition map on $C$, possibly composed with the diffeomorphism $a$, and is smooth. Since this is a free action of the finite group $\{1,a\}$, the quotient is Hausdorff and second countable. This constructs the smooth quotient by a free finite group action, and in these charts $\pi$ is locally the identity. Hence $\pi$ is a local diffeomorphism, in particular smooth.

To test orientability, parametrize the cylinder byThe involution acts in these coordinates aswhose derivative has determinant $-1$. Thus $a$ is an orientation-reversing diffeomorphism of the cylinder.

If the quotient surface $S$ were orientable, its orientation would pull back through the local diffeomorphism $\pi$ to an orientation of $C$. The identity $\pi \circ a=\pi$ would then force $a$ to preserve that pulled-back orientation, contradicting the negative determinant above. Therefore

The function $f$ has zero mean and its cosine coefficients vanish:Its sine coefficients areso

Away from the corners, $F^{\prime }=f$. Differentiating the Fourier series shows that $n{B}_n=a_n$ and $-n{A}_n=b_n$. Also,Consequentlyand

Evaluating the series for $f$ at $\theta =\pi /2$ gives the Leibniz formula for π,Evaluating the continuous Fourier series for $F$ at $\theta =0$ givesand hence

Forthe Euler-Lagrange equation isorThe general solution of this inhomogeneous linear differential equation isThe condition $y(0)=-1$ gives $C+D=0$, and the condition at $x=1$ gives $C=1$. Therefore

Computing the Laplacian givesandThus Laplace's equation holds exactly when

For $z=x+iy$,soOne suitable analytic function is therefore

For a geodesic triangle with interior angles $\alpha ,\beta ,\gamma$, the local Gauss-Bonnet theorem isbecause its geodesic sides have zero geodesic curvature.

Triangulate a closed oriented surface into $F$ geodesic triangles, with $E$ edges and $V$ vertices. Summing the local formula, the angles around each vertex total $2\pi$, soSince every triangular face has three edges and every edge belongs to two faces, $3F=2E$. Hencewhich is the global Gauss-Bonnet theorem.

For the sphere $S_r$, the unit normal is $N(p)=p/r$. Its shape operator is, up to the conventional sign, $(1/r)I$ on each tangent plane. Both principal curvatures therefore have magnitude $1/r$, and the Gaussian curvature is

An octant has one eighth of the sphere's area:Thus ${\int }_{T}KdA=\pi /2$. Its three great-circle sides meet at three right angles, soThe two sides of the local Gauss-Bonnet formula agree directly.

Let $G$ act by left multiplication on the set $G/H$ of $n$ left cosets. This gives a group action homomorphismIts kernelis a normal subgroup of $G$, and the first isomorphism theorem gives $G/K\cong \mathrm{im}\phi$. By Lagrange's theorem, $|G/K|$ divides $n!$. The image acts transitively on the $n$ cosets, so the orbit-stabilizer theorem shows that $n$ divides $|G/K|$; in particular,

Now suppose $G$ is nonabelian and simple. Since $H$ is proper, the coset action is nontrivial, so $K\ne G$. Simplicity gives $K=\{e\}$, and $\phi$ embeds $G$ into $S_n$. Composing with the sign homomorphism givesIts kernel is normal. A nontrivial map would embed the simple group $G$ into the abelian group of order two, which is impossible because $G$ is nonabelian. The sign is therefore always $+1$, so

An $m\times n$ matrix game with payoff matrix $A$ is a two-player zero-sum game: player I chooses a row and receives $A_{ij}$, while player II chooses a column and loses the same amount. For mixed strategies $p$ and $q$, the expected payoff to player I isPlayer I's optimal mixed strategy maximizes the payoff guaranteed against every $q$, while player II's minimizes the largest payoff obtainable by any $p$. Thusby the minimax theorem. Equivalently, optimal strategies satisfy

Here $A^T=-A$, so this is an antisymmetric zero-sum game. For every probability vector $r$,It follows that the row player's guaranteed payoff cannot exceed zero and the column player's worst loss cannot be below zero. Minimax therefore givesIf $p$ is optimal for player I, thenTransposing and using $A^T=-A$ giveswhich is precisely the optimality condition for player II. Thus every optimal strategy for player I is also optimal for player II.

The condition $Ap\le 0$ explicitly readsThe probability vectorsatisfiesso it is optimal for both players.

To prove uniqueness, let $q$ be any optimal strategy. Since $p$ is optimal for player II and $q$ is optimal for player I,and hence $q_3=0$. Writing $q=(a,b,0,d{)}^T$, the first, second, and fourth inequalities in $Aq\le 0$ giveConsequentlyso equality holds throughout: $a=2d$ and $b=4d$. The normalization $a+b+d=1$ yields $d=1/7$. Therefore the displayed $p$ is the unique optimal strategy.