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The radial Poisson equation with constant source one has $\psi =r^2/6+a+b/r$. The boundary conditions giveSubtracting gives $b=1$, then $a=-1/6$. HenceIts radial Laplacian is one and both endpoint values are one. The prescribed values ensure continuity with the adjacent region solutions, but do not impose derivative continuity.

The radial Laplace equation has solution $a+b/r$. Boundedness at the origin requires $b=0$, and the value at $r=1$ fixes $a=1$. Thus

The homogeneous sequences are again $(-2{)}^k$ and $(-3{)}^k$. A particular solution $D{2}^k$ has left-hand side $10D{2}^k$, so $D=1/10$. Consequently $y_k=A(-2{)}^k+B(-3{)}^k+2^k/10$. The two initial conditions require $A+B=9/10$ and $-2A-3B=4/5$, giving $A=7/2$ and $B=-13/5$. ThusThe linear recurrence relation can be rearranged as $y_{k+1}=2^k-5y_k-6y_{k-1}$. Starting with the two integer values, mathematical induction proves that every $y_k$ for $k\ge 0$ is an integer. For $n\ge 1$, the closed form givesThe first term on the right is divisible by ten; the second is too, because $(-2{)}^n$ supplies a factor two and $35$ supplies a factor five. Therefore $10$ divides $2^n-26(-3{)}^n$ for every positive integer $n$. The positivity condition matters: at $n=0$ the expression is $-25$.

For the homogeneous linear recurrence relation, a trial $u_k=r^k$ gives the characteristic equation of a linear recurrence $r^2+5r+6=(r+2)(r+3)=0$. Thus the two homogeneous sequences are $(-2{)}^k$ and $(-3{)}^k$. Seek a particular solution $u_k=ak+b$; substitution gives $12ak+12b-5a=12k+1$, so $a=1$ and $b=1/2$. HenceThe two independent constants supply arbitrary initial values for this second-order linear recurrence relation.

Multiply the first-order linear differential equation by the integrating factor $e^{t/\alpha }$. ThenIf $y\to 0$ in the remote past, then $e^{t/\alpha }y\to 0$ there. Integration from $-\infty$ gives the necessary formulaWhen this integral exists and tends to zero as $t\to -\infty$, differentiating verifies the equation and the past condition. Two such solutions differ by $C{e}^{-t/\alpha }$, which diverges at $-\infty$ unless $C=0$, proving uniqueness. This is the causal response of a first-order relaxation equation, a convolution with its decaying impulse response.

The phrase “general function” implicitly requires compatibility with the past condition: it does not guarantee existence for arbitrary $f$. A sufficient condition is bounded forcing with $f(t)\to 0$ as $t\to -\infty$; after setting $r=t-s$, the integral is bounded by the supremum of $|f|$ before $t$. For example, $f\equiv 1$ gives $y=1+C{e}^{-t/\alpha }$, so no solution has the stated past limit. Both forcings below vanish in the past and meet the intended causal condition, the impulse being interpreted distributionally.

For $z\ne 0$, division by $4z$ gives $p(z)=1/(2z)$ and $q(z)=1/(4z)$. Thus zero is singular, but $zp=1/2$ and $z^{2}q=z/4$ are analytic. Zero is a regular singular point.

Apply the Frobenius method with $y=z^r{\sum }_{n=0}^{\infty }{a}_n{z}^n$, $a_0\ne 0$. The lowest power $z^{r-1}$ has coefficient $[4r(r-1)+2r]a_0$, so the indicial equation is $2r(2r-1)=0$. Its characteristic exponents at a regular singular point are $0$ and $1/2$. At the next powers,With $a_0=1$, the two Frobenius solutions are thereforeBoth analytic power-series factors converge everywhere by the ratio test. Recognizing the trigonometric series givesOn a punctured neighbourhood choose a consistent square-root branch for the second solution; the first is entire in $z$. These solutions are independent, since their leading orders are $1$ and $z^{1/2}$. A direct square-root reduction of a regular-singular differential equation also verifies the closed forms: for $s=\sqrt{z}$ and $Y(s)=y(s^2)$, $4z{y}^{\prime \prime }=Y_{ss}-Y_s/s$ and $2y^{\prime }=Y_s/s$, so the equation reduces to $Y_{ss}+Y=0$.

In the normalized second-order linear differential equation, an ordinary point $z_0$ is one at which both $p$ and $q$ are holomorphic functions in a neighbourhood of $z_0$. A singular point of a differential equation is a point that is not ordinary. A singular point $z_0$ is a regular singular point if $(z-z_0)p(z)$ and $(z-z_0{)}^{2}q(z)$ extend analytically to $z_0$. Equivalently, $p$ has at most a simple pole and $q$ at most a double pole there. If a singular point fails this test, it is an irregular singular point. The normalized form matters: zeros of a leading coefficient must first be assessed after division by that coefficient.

The constant solutions are fixed points of $G(u)=u^2(1+u)/2$. Solving $G(u)=u$ gives $u(u+2)(u-1)=0$, hence $u_{\ast }=0,-2,1$. For a discrete dynamical system, a small perturbation satisfies ${\eta }_{n+1}=G^{\prime }(u_{\ast }){\eta }_n+O({\eta }_n^2)$, whereA multiplier of magnitude less than one contracts perturbations; a multiplier greater than one expands them. Therefore $u=0$ is asymptotically stable, while $u=-2$ and $u=1$ are unstable. In particular, if $|u|\le 1/2$, then $|G(u)|\le (3/4)|u{|}^2\le (3/8)|u|$, proving convergence to zero without relying solely on the vanishing linear multiplier.

Put $F(y)=(y^2-4)\arctan y$. Its zeros, and hence the equilibrium points, are $-2,0,2$. For a perturbation $y=y_{\ast }+\eta$, linear stability analysis gives $\dot{\eta }=F^{\prime }(y_{\ast })\eta +O({\eta }^2)$, withAt zero, $F^{\prime }(0)=-4$, so sufficiently small perturbations decay and $y=0$ is asymptotically stable. At either outer equilibrium point, $F^{\prime }(\pm 2)=4\arctan 2>0$, so $y=-2$ and $y=2$ are unstable. The phase line confirms the result: $F$ has signs $-,+,-,+$ on the intervals cut out by $-2,0,2$, respectively. Flow points toward zero on both sides and away from both outer equilibria.

The statement is false. Take two independent fair bits $U,V$ on the four equally probable outcomes $00,01,10,11$, and put $A_1=\{U=0\}$, $A_2=\{V=0\}$, and $B=\{U=V\}$. Each of $A_1,A_2,B$ has probability $1/2$, andThus $B$ is independent of each event separately. But $A_1\cup A_2$ contains $00,01,10$, whereas its intersection with $B$ contains only $00$. ThereforeThis is a concrete failure of extending separate independence to the generated union. A counterexample at the permitted value $n=2$ disproves the universal claim; for any larger $n$ one can repeat $A_1$ as the additional events.

The statement is true. For $n\ge 2$, being in every pairwise intersection is equivalent to being in all the events:Indeed every index occurs in at least one pair. Apply the union bound to the complements of these $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ pairwise intersections to obtainThis is positive common intersection from pair-intersection probabilities. It remains valid for $n=2$, where it reduces to the asserted positivity of the single pairwise intersection.

The statement is true. Use the union bound on the complementary events:This proves the claim for arbitrary events, with no independence requirement.

The statement is true. Put $C_k={\bigcap }_{r=1}^k{A}_r$. The assumed $P(C_{n-1})>0$ ensures that every conditioning event $C_{k-1}$, $2\le k\le n$, has positive probability. By the definition of conditional probability,Multiplying makes the fractions telescope, givingThis proves the chain rule for probabilities here without any assumption of independence. The final numerator $P(C_n)$ may be zero; all denominators remain positive.

For real $x$ at which the denominator is nonzero, mapping to the unit circle gives $|ax+b{|}^2=|cx+d{|}^2$. There are infinitely many such real $x$, so the resulting quadratic polynomial identity has all coefficients zero:If $c=0$, the first equality gives $a=0$, contradicting the nonzero determinant $ad-bc$. Thus $a,c\ne 0$. Put $\lambda =a/c$, $k=b/a$ and $\ell =d/c$. The equalities becomeTwo complex numbers with equal real parts and equal moduli have imaginary parts equal in magnitude. Consequently $\ell =k$ or $\ell =\overline{k}$. The first option would give $ad-bc=ac(\ell -k)=0$, so it is excluded. The second givesMoreover $k$ must be nonreal, since otherwise the two options coincide and the map would be constant. This proves the Möbius transformations from the real axis to the unit circle representation directly from coefficients.

Conversely, for real $x$, the numbers $x+k$ and $x+\overline{k}$ are complex conjugates, and their moduli agree. Nonreal $k$ ensures the denominator is nonzero on the real line; multiplication by $\lambda$ preserves modulus. At the point at infinity on the Riemann sphere, the map takes the value $\lambda$. Thus the representation also respects the extended real axis.

The quotient is defined only when $z+\overline{u}\ne 0$. Its unit-modulus condition then gives $|z+u{|}^2=|z+\overline{u}{|}^2$. On expanding with complex conjugates,Hence at least one imaginary part is zero, and either $z$ or $u$ is real. Conversely, either reality condition makes the two squared moduli equal; provided the denominator is nonzero, the quotient has modulus one. No conclusion is drawn from an undefined quotient.

Write a straight line in Cartesian form $Ax+By+C=0$, where $A,B,C$ are real and $(A,B)\ne (0,0)$. With $z=x+iy$, set $c=(A+iB)/2$ and $r=C$. Using the complex conjugate,Thus the real linear equation of a line in the complex plane isConversely, writing $c=u+iv$ gives $2ux+2vy+r=0$, an ordinary straight line with normal vector $(u,v)$. The nonzero condition on $c$ matters: $c=0$ would give either the whole complex plane or the empty set.

Set $s=x+y+z$ and $Q=x^2+y^2+z^2$. The expression in the equation is $4(Q+4)-(s+2{)}^2$. Apply the Cauchy-Schwarz inequality in ${\mathbb{R}}^4$ toTheir dot product is $s+2$, and $|\mathbf{e}{|}^2=4$. Thus the expression is nonnegative, and it is zero precisely when equality holds in Cauchy-Schwarz. That requires $\mathbf{v}=\lambda \mathbf{e}$. The fourth coordinate forces $\lambda =2$, and the other three coordinates then giveSubstitution gives $3(4+4+4+4)-2(4+4+4)-4(2+2+2)=48-24-24=0$, so this unique possible choice really is a solution.

Apply the Cauchy-Schwarz inequality to $(x,y,z)$ and $(1,1,1)$ in the Euclidean inner product on ${\mathbb{R}}^3$. It givesAfter expansion and division by two,The equality criterion from the introductory proof says $x=y=z$. One can also see both the inequality and equality case from $x^2+y^2+z^2-xy-yz-zx=\frac{1}{2}[(x-y{)}^2+(y-z{)}^2+(z-x{)}^2]$.

Let $P\mathbf{x}=(\mathbf{n}\cdot \mathbf{x})\mathbf{n}$ be the axial orthogonal projection and $N\mathbf{x}=\mathbf{n}\times \mathbf{x}$. Then $NP=PN=0$, while the vector triple-product identity gives $N^2=P-I$. The map is $aP+a(I-P)+bN$. Its inverse is $P/a$ on the axis. On the perpendicular plane,Thus the inverse of an axis-angle linear map isor, in the requested vector form,The denominators are nonzero because $a,b>0$. Multiplication on each invariant component verifies the matrix inverse without choosing a coordinate frame.

In suffix notation, $x_i^{\prime }=a{x}_i+b{ϵ}_{ikj}{n}_k{x}_j$, so the matrix entries areHere $\delta$ is the Kronecker delta and $ϵ$ the Levi-Civita symbol. The diagonal determines $a=2$. Comparing the off-diagonal entries withgives $b\mathbf{n}=(1,2,2)$. As $\mathbf{n}$ is a unit vector and $b>0$, $b=3$ and $\mathbf{n}=(1,2,2)/3$. ThusIn this example the transverse scaling is $\sqrt{13}$ and the axial scaling is two; the rotation angle is $\arctan (3/2)$.