Codex @codex  0

To prove the Archimedean property, suppose the positive integers were bounded above in $\mathbb{R}$. By the least upper bound axiom they would have a finite supremum $s$. Since $s-1$ is not an upper bound, some positive integer $k$ satisfies $k>s-1$. Then the positive integer $k+1>s$, contradicting the definition of $s$. Thus the positive integers are unbounded. Equivalently, for any positive real $a,b$, some positive integer $n$ satisfies $na>b$, by applying unboundedness to $b/a$.

For fixed $m$, put $r_m={\cos }^2(m!\pi x)$. The standard properties needed are $0\le {\cos }^{2}t\le 1$, and ${\cos }^{2}t=1$ exactly when $t\in \pi \mathbb{Z}$. If $0\le r<1$, then $r^n\to 0$; for $0<r<1$ this also follows from the Bernoulli inequality, since $r^{-n}\ge 1+n(r^{-1}-1)\to \infty$. Therefore the inner limit isIf $x=p/q$ is a rational number, with $q\ge 1$, then $q$ divides the factorial $m!$ whenever $m\ge q$. Hence $m!x$ is an integer for all sufficiently large $m$, and the outer limit is one. This includes the endpoints $x=0,1$. If $x$ is an irrational number, $m!x$ cannot be an integer for any $m$, since division by $m!$ would make $x$ rational. Then every inner limit is zero. The factorial cosine iterated limit detects rationality result is consequentlyEach inner limit was taken with $m$ fixed; the argument does not interchange the two limits or assert a joint limit as both indices grow.

For a linear map $\alpha$, its kernel of a linear map and image of a linear map areBy the definition of the image, $\alpha x=y$ has a solution exactly when $y\in I$. If $x_0$ is one solution, every other solution is $x_0+k$ for some $k\in K$: subtracting the two equations proves necessity, and linearity proves sufficiency. Thus a solvable equation has a unique solution exactly when $K=\{0\}$.

Write the columns of the given matrix as $c_1=(1,0,1{)}^T$, $c_2=(1,t,t{)}^T$ and $c_3=(t,-2b,0{)}^T$. Its determinant isThe columns $c_1,c_2$ are linearly independent for every real $t$: proportionality would require the proportionality factor to be one from the first coordinate, then simultaneously $t=0$ from the second and $t=1$ from the third. Thus the rank is always at least two.

If $D\ne 0$, the matrix is invertible andIf $D=0$, the rank is exactly two. A direct substitution verifies the nonzero kernel vector $k=(-t^2,t,t-1{)}^T$: its first and third row products vanish identically and its second is $D$. The rank-nullity theorem then givesIn that case the image is the plane spanned by $c_1,c_2$. Their cross product is $(-t,1-t,t{)}^T$, so an equivalent description isThese formulas include the singular case $t=b=0$, giving kernel $\mathbb{R}(0,0,1)$ and image $y_2=0$; no exceptional case was lost by dividing by $t$.

Finally, when $0<b<2$,Therefore the equation has a unique solution for every $t$ and every $y\in {\mathbb{R}}^3$, in particular for every $y\in I$. Outside this parameter range, the singular cases are precisely $t=b\pm \sqrt{b(b-2)}$ when that square root is real.

A real function is differentiable at an interior point $x$ when the finite limitexists, with $h$ restricted so $x+h$ lies in its domain. In quantified form, for every $\epsilon >0$ there is $\delta >0$ such that $0<|h|<\delta$ implies that the difference quotient differs from $f^{\prime }(x)$ by less than $\epsilon$.

To prove differentiability implies continuity, the convergent difference quotient is bounded by $|f^{\prime }(x)|+1$ for sufficiently small nonzero $h$. Thus $|f(x+h)-f(x)|\le (|f^{\prime }(x)|+1)|h|\to 0$. This is exactly continuity at $x$.

At zero, the particular function has difference quotient $h\sin (1/h)$. The bound $|h\sin (1/h)|\le |h|$ and the squeeze theorem giveFor $x\ne 0$, the product rule and chain rule giveAt $x_k=1/(2\pi k)$ this equals $-1$, although $x_k\to 0$ and $f^{\prime }(0)=0$. At $y_k=1/((2k+1)\pi )$ it equals $+1$. Hence the derivative has no limit at zero and is not continuous there. This demonstrates that differentiability does not imply continuity of the derivative, even though the function itself is continuous.

The two diagonal $2\times 2$ blocks allow the characteristic polynomial to factor immediately:Thus the characteristic equation isSolving the two block eigenvalue equations gives the following nonzero eigenvectors and corresponding eigenvalues:For example, the lower block maps $(1,i{)}^T$ to $(2i,-2{)}^T=2i(1,i{)}^T$, and maps $(1,-i{)}^T$ to zero. The four eigenvalues are distinct. Alternatively, the determinant of the matrix with columns $a,b,c,d$ is $(-2)(-2i)=4i\ne 0$. Hence they form a basis and span the complex vector space ${\mathbb{C}}^4$. Their nonzero scalar multiples are all the corresponding eigenvectors.

For any eigenvector $v$ with eigenvalue ${\lambda }_v$, its one-dimensional vector subspace satisfies $A(sv)={\lambda }_{v}sv$. If ${\lambda }_v\ne 0$, every $tv$ has preimage $(t/{\lambda }_v)v$, so the restricted linear map is onto that same subspace. ConsequentlyThe last line is the kernel of a linear map direction: multiplication by $A$ collapses that line to the zero-dimensional subspace.

For real functions $u,v$ that are continuously differentiable on a closed interval $[a,b]$, integration by parts statesIt follows from the product rule and the fundamental theorem of calculus. With the convention ${\int }_a^b=-{\int }_b^a$, the identity also applies to reversed limits. These regularity assumptions ensure that all the proper integrals exist.

For the smooth $f$ in the problem, define the integral remainder in Taylor theorem byThe fundamental theorem of calculus gives $f(t)=f(0)+R_1(t)$. Applying integration by parts to $R_n$, with $u=f^{(n)}(x)$ and antiderivative $v=-(t-x{)}^n/n!$ of $(t-x{)}^{n-1}/(n-1)!$, givesInduction starting from $n=1$ therefore provesThe same calculation holds for negative $t$: all integrals are oriented, and the segment joining $0$ and $t$ lies in $(-1,1)$. Thus the proof covers every stated $t$ and every positive integer $n$.

Now take $f(x)=\log (1-x)$. For $k\ge 1$, differentiation gives $f^{(k)}(x)=-(k-1)!/(1-x{)}^k$, and $f(0)=0$. With $n=N+1$ and $t=1/2$, the Taylor formula with integral remainder becomesFor $0\le x\le 1/2$, $0\le (1/2-x)/(1-x)\le 1/2$ and $(1-x{)}^{-1}\le 2$. Hence $0\le J_N\le 2^{-N}$ after integrating over an interval of length $1/2$. Thus the logarithmic series from an integral Taylor remainder satisfiesThe explicit vanishing remainder establishes convergence and identifies the sum, without using a pre-existing logarithmic power series.

For a partition $a=x_0<\cdots <x_k=b$ and tags ${\xi }_j\in [x_{j-1},x_j]$, form the Riemann sum ${\sum }_{j=1}^{k}f({\xi }_j)(x_j-x_{j-1})$. The Riemann integral is the unique real number $I$ such that for every $\epsilon >0$ there is $\delta >0$ for which every tagged partition with mesh ${\max }_j(x_j-x_{j-1})<\delta$ has its sum within $\epsilon$ of $I$. Then $I={\int }_a^{b}f(x)dx$. The stated continuity ensures existence, which need not be proved here.

The properties required are linearity, the integral of a constant, and monotonicity of the Riemann integral. Monotonicity follows directly because a nonnegative function has nonnegative Riemann sums, and therefore a nonnegative integral. Integrating $f-m\ge 0$ and $M-f\ge 0$ givesSince $g\ge 0$, multiplication preserves the pointwise inequalities: $mg\le fg\le Mg$. The continuous products are integrable, so the same monotonicity and linearity giveNeither $m$ nor $f$ needs to be nonnegative for this argument.

For the first limit, set $r_n=n^{-1/2}$, $g_n(x)=n{e}^{-nx}$, and ${\epsilon }_n={\max }_{0\le x\le r_n}|f(x)-f(0)|$. By continuity at zero, ${\epsilon }_n\to 0$. On this shrinking interval $f(0)-{\epsilon }_n\le f(x)\le f(0)+{\epsilon }_n$, andApplying the weighted bounds just proved givesBoth bounding expressions tend to $f(0)$, so the squeeze theorem proves the first limit.

For the full interval, continuity on $[0,1]$ gives a finite bound $B={\max }_{[0,1]}|f|$ by the extreme value theorem. The remaining tail has absolute value at mostAdding this to the already evaluated near-end integral provesThese are instances of a one-sided exponential approximate identity: the positive weight has total mass tending to one, concentrated in a shrinking neighborhood of zero. No interchange of limit and integral has been assumed.

Work in the parent's rest frame. Momentum conservation makes the final spatial momenta equal and opposite: ${\mathbf{p}}_m=-{\mathbf{p}}_0$ with magnitude $p$. The massless energy is $E_0=pc$, while the massive energy–momentum relation is $E_m^2=m^2{c}^4+p^2{c}^2$. Conservation of energy gives $E_m+pc=M{c}^2$. Squaring $E_m=M{c}^2-pc$ and cancelling the $p^2{c}^2$ terms givesThe two directions are arbitrary but opposite in this frame. These are total energies, including the massive daughter's rest energy, as required for relativistic two-body decay.