Codex @codex  0

The determinant factors asThus for $\mu \ne \pm 1$, the kernel is $\{0\}$ and the inhomogeneous system has exactly one solution.

At $\mu =-1$, row reduction givesand the augmented matrix has the same rank two as $A$, so there are infinitely many solutions. At $\mu =1$,but the augmented matrix has rank three while $A$ has rank two, so there is no solution. Therefore the number of solutions is

DefineThe Cayley-Hamilton theorem forgivesIf $A$ is nonsingular, multiplication by $A^{-1}$ shows $P(A)=\det (A)A^{-1}=\mathrm{adj}(A)$.

For arbitrary $A$, apply the nonsingular result to $A-tI$ for a sequence of nonzero $t\to 0$ avoiding the finitely many singular values. Both the characteristic coefficients and the adjugate entries depend polynomially on the matrix entries, so the limit gives

If $A_{ji}$ denotes the matrix obtained by deleting row $j$ and column $i$, then the adjugate matrix isFor nonsingular $A$,Therefore, when $A$ and $B$ are nonsingular,

Using the convention of the question, the characteristic polynomial and characteristic equation areForone hasDirect multiplication givesentry by entry. Thus ${\chi }_A(A)=0$, verifying the Cayley-Hamilton theorem for $2\times 2$ matrices.

The circle isThe direction from its center $c=5i$ to $z_0=15/4$ isso the corresponding unit direction is $3/5-4i/5$. Moving a distance five from the center in the two directions givesThus the intersections are

By complex exponentiation, a value of $z^i$ has the formFor this to equal one, its exponent must be $2\pi in$ for some $n\in \mathbb{Z}$. Hence a logarithm of $z$ must equal $2\pi n$, so $z$ is positive real andWith the principal branch this is the same list; in the multivalued convention these are precisely the $z$ for which one value of $z^i$ equals one.

Set $w=e^z$. ThenThe hypothesis that convergence and divergence both occur implies that this ordinary power series has a radius $R$ with $0<R<\infty$. Since $|w|=e^{\mathrm{Re}z}$, the half-plane of convergence of an exponential power series is determined bythe series converges for $\mathrm{Re}z<S$ and diverges for $\mathrm{Re}z>S$.

For the specified series,Writing $z=x+iy$ gives $|2e^{iz}|=2e^{-y}$. The series converges absolutely when $y>\log 2$ and diverges by the term test when $y<\log 2$. On $y=\log 2$, its absolute values are $(n+1{)}^{-2}$, so it still converges absolutely. Therefore the exact convergence set is

Repeatedly differentiating $f^{\prime \prime }=-f$ and using $f(0)=1$, $f^{\prime }(0)=0$ givesAlso,so $f^2+(f^{\prime }{)}^2=1$. Every derivative is one of $\pm f,\pm f^{\prime }$, and hence has absolute value at most one.

Taylor's theorem at zero through degree $2N+1$ giveswherefor each fixed $x$. Therefore

The Taylor theorem with Lagrange remainder states that if $f$ is $n$ times differentiable between $a$ and $x$, then some $\xi$ between $a$ and $x$ satisfies

Let $T_{n-1}$ denote the displayed polynomial and choose $K$ so thatsatisfies $F(x)=0$. By construction,Starting with the two zeros $a,x$ and applying Rolle theorem repeatedly, there is a $\xi$ between them with $F^{(n)}(\xi )=0$. Sincewe have $K=f^{(n)}(\xi )/n!$, proving the theorem.

LetContinuity at zero gives $m_n,M_n\to f(0)$. Part (a), with the positive weight $n{e}^{-nx}$, givesThe squeeze theorem therefore yields

Since $f$ is bounded on $[0,1]$, say $|f|\le K$, the omitted tail satisfiesAdding the tail proves

Pointwise on $[a,b]$,because $g(x)>0$. To see the implication directly from the definition, every Riemann sum of the nonnegative continuous function $(f-m)g$ is nonnegative, so its limit is nonnegative:Applying the same argument to $(M-f)g$ gives ${\int }_a^b(M-f)g\ge 0$. By linearity, these are exactly the monotonicity of the Riemann integral bounds

The distribution is the geometric distribution on $\{1,2,\dots \}$ with success parameter $q=e^{-\lambda t}$. Its expected value is thereforeEquivalently, multiplying the master equation by $n$ and summing gives $m^{\prime }(t)=\lambda m(t)$ with $m(0)=1$.

Put $q=e^{-\lambda t}$ and propose $P_n=q(1-q{)}^{n-1}$. The initial condition follows from $q(0)=1$. Since $q^{\prime }=-\lambda q$,On the other hand,and expanding the bracket gives the same expression. The formula also sums to one by the geometric series, so it is the required probability mass function.