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In characteristic two, the binomial theorem gives $X^8-1=(X-1{)}^8$. The only root is already in ${\mathbb{F}}_2$, so the splitting field is the original field, despite the repeated root:

The binomial theorem isolates the only terms that can survive reduction modulo the prime number $p$:For $1\le r\le p-1$, neither $r!$ nor $(p-r)!$ is divisible by $p$, whereas $p!$ is. Consequently $p\mid \left(\genfrac{}{}{0ex}{}{p}{r}\right)$, and every intermediate term vanishes in modular arithmetic. ThusThis works for arbitrary integers $a,b$, including those divisible by $p$. It is the additive identity behind the Frobenius endomorphism in characteristic $p$.

Singularities. The complex exponential factors as $F(z)=e^{z/2}{e}^{1/(2z)}$. The second factor has an essential singularity at zero, and multiplication by the nonvanishing holomorphic function $e^{z/2}$ does not remove it. There are no other finite singularities. On the Riemann sphere, infinity is also an essential singularity, since $F(1/\zeta )=F(\zeta )$.

Symmetry and integral formula. The identity $F(1/z)=F(z)$ and uniqueness of the Laurent series on $0<|z|<\infty$ give $F_{-k}=F_k$ for every integer $k$. The Laurent coefficient formula, on the positively oriented unit circle, givesThe imaginary part vanishes by the substitution $\theta \mapsto 2\pi -\theta$. Hence

Polynomial contour integral. For an integer $n\ge 0$ and integer $p$, the binomial theorem gives $(1+z^2{)}^n={\sum }_{j=0}^n(\genfrac{}{}{0ex}{}{n}{j})z^{2j}$. The residue theorem selects $2j-n-p-1=-1$, that is $j=(n+p)/2$. Therefore the correct complete statement isThe printed nonnegative-integer condition alone misses $j\le n$. For example $n=0,p=2$ satisfies that condition, but the integral is zero and the displayed factorial $(-1)!$ is undefined.

Explicit coefficients. Both Taylor series inconverge absolutely and uniformly on every compact annulus away from zero, so their product can be regrouped. For $k\ge 0$, put $r=s+k$ and $s=m$. For arbitrary integer $k$, use the already proved symmetry:Here $I_j$ is a modified Bessel function. The printed series is directly meaningful for $k\ge 0$; for negative $k$ the absolute values, or an appropriately shifted lower summation limit, are essential.

Using $a=A$ and the end-face shear stress from part (c), the printed torque normalization givesThe binomial theorem expands the remaining integral into powers of $r$:Integrating each term gives an expression valid also at $\tau =0$:For example, $n=1$ gives $M=({\mu }_1+3{\mu }_2)A^4\tau /2+{\mu }_2{A}^6{\tau }^3/3$. The torque is an odd polynomial in the twist per unit reference length, as expected when reversing the twist reverses the end-face shear stress.

For a physical end-face torque $N$, integration over the azimuth normally givesThe original PDF explicitly defines $M$ without this $2\pi$, so the boxed answer retains its convention. Since $\mathbf{T}$ is the Cauchy stress tensor, its shear component here is spatial $T_{\theta z}$; the printed capital $Z$ in $T_{\theta Z}$ does not require a change to a nominal stress tensor.

Put $C=\cos (\chi /2)$ and $S=\sin (\chi /2)$. Commutativity of the two creation operators and the binomial theorem giveThusThe state is a fixed-number two-mode condensate with binomial mode occupations; it is not a product of two independent Poisson-distributed Glauber coherent states.

Let each step have length $a$, and let $K$ be the number of right steps among $N$ independent steps. Then $K$ has a binomial distribution and the physical displacement is $x=a[K-(N-K)]=a(2K-N)$. Therefore $K=(N+x/a)/2$, and the position probability mass isThis is zero unless $x/a$ is an integer, $|x/a|\le N$, and $N+x/a$ is even. The parity condition matters because the walk alternates between the two sublattices. If distance is measured in lattice spacings, replace $x/a$ by $x$.

Normalization follows from the binomial theorem. Its first two moments provide useful checks:These are exact discrete-time statements; a smooth continuum probability density requires coarse-graining over the lattice and its parity oscillation.

For the binomial distribution, the binomial theorem givesDifferentiation of the probability generating function recovers the first expected value and second factorial moment:Since $X^2=X(X-1)+X$, the variance isThusThe formulas also hold for $n=0,1$ and the degenerate choices $p=0,1$, with derivatives read directly from the resulting polynomial.

Partition the power set into disjoint classes according to subset size. The class of size-$i$ subsets has cardinality $\left(\genfrac{}{}{0ex}{}{n}{i}\right)$, while the whole power set has cardinality $2^n$. Counting every subset once givesThis is also the binomial theorem evaluated at $1+1$, but the counting proof follows directly from the earlier parts.

We prove the stronger explicit bound $n!\le n^n/2^{n-1}$ by mathematical induction. The case $n=1$ is equality. The binomial theorem gives $(1+1/n{)}^n\ge 1+n/n=2$, so $(n+1{)}^n\ge 2n^n$. Consequently the induction hypothesis yieldsTherefore $n!=O(n^n/2^n)$, with constant $2$ and threshold $1$.

The second bound is asymptotically tighter. Using $n\le 2^n$,Thus its bounding function is negligible relative to the first in little o notation; this compares the bounds without needing Stirling's formula.

First prove $a^p\equiv a(\mathrm{mod}p)$ for every nonnegative integer $a$ by mathematical induction. The case $a=0$ is immediate. If it holds at $a$, the binomial theorem and the stated divisibility of the middle binomial coefficients giveThis establishes the induction. For an arbitrary integer $a$, take a nonnegative representative $r$ of its residue class; then $a^p\equiv r^p\equiv r\equiv a(\mathrm{mod}p)$. When $a\not\equiv 0(\mathrm{mod}p)$, part (b) supplies a multiplicative inverse modulo $p$. Multiplying the congruence by that inverse proves Fermat's little theorem:

Let $X_1,X_2$ be the two independent Poisson random variables. For $k\ge 0$, their discrete convolution iswhere the last equality is the binomial theorem. Thus $X_1+X_2\sim \mathrm{Poisson}(\lambda +\mu )$.

First $0\le M_n\le {\sum }_i{E}_i$, so linearity of expectation gives $\mathbb{E}{M}_n\le n<\infty$. The tail integral formula for expectation, followed by the finite binomial theorem expansion, givesThe exchange involves only a finite sum. As a useful check, substitution $t=e^{-x}$ rewrites the integral as ${\int }_0^1[1-(1-t{)}^n]/tdt={\sum }_{j=0}^{n-1}{\int }_0^1(1-t{)}^{j}dt={\sum }_{k=1}^{n}1/k$, the harmonic number $H_n$.

For a real or complex polynomial $p(t)={\sum }_{k\ge 0}{a}_k{t}^k$, this finite sum defines a norm: definiteness follows from uniqueness of polynomial coefficients, absolute homogeneity of a norm follows termwise, and the triangle inequality follows by summing $|a_k+b_k|\le |a_k|+|b_k|$. On $[0,1]$, its supremum norm satisfies $\Vert p{\Vert }_{\infty }\le \Vert p{\Vert }_{\mathrm{coef},1}$, since $|t^k|\le 1$. The reverse comparison has no constant independent of the degree of a polynomial: for $p_n(t)=(1-t{)}^n$, the two norms are $1$ and $2^n$ respectively, by the binomial theorem.

A polynomial identity is equality of two polynomials for every value of the variables, rather than an equation imposed only at selected roots. Over the real or complex numbers, this is equivalent to equality of corresponding coefficients: their difference, if nonzero, has only finitely many roots in one variable. The binomial theorem is an example. Coefficient comparison transfers a polynomial product identity into finite sums of binomial coefficients.

The binomial theorem for the commuting matrices $I,N$ proves the formula for nonnegative integers: every term containing $N^2$ vanishes. Since $(I+N)(I-N)=I$, the matrix inverse is $I-N$; applying the same argument to its powers gives the negative integers as well. The same identity shows that $I+N$ is invertible even when the nilpotent matrix $N$ is nonzero. In dimension two, the example $N=\left(\begin{array}{cc}0& p\\ 0& 0\end{array}\right)$ describes repeated shear mappings, whose displacement parameters add under composition.

For a prime $p$ and $k\ge 1$, write $a=b+p^{k}h$. The binomial theorem shows $a^p-b^p$ is divisible by $p^{k+1}$: the linear term contains the extra factor $p$ and every higher term contains at least $p^{2k}$. Iteration gives the displayed modular congruence for every $r\ge 0$, including $p=2$. Applied to Fermat's little theorem, it yields $a^{p^n}\equiv a^{p^{n-1}}(\mathrm{mod}{p}^n)$.

For sufficiency, let $N$ be composite and squarefree, with $p-1\mid N-1$ for every prime factor $p$. For any integer $a$ coprime to $N$, Fermat's little theorem gives $a^{N-1}\equiv 1(\mathrm{mod}p)$ for every such $p$. Their product is $N$, so the Chinese remainder theorem gives the required congruence modulo $N$.

For necessity, first rule out a repeated factor. If $N=p^{k}m$ with $k\ge 2$ and $\gcd (p,m)=1$, the Chinese remainder theorem supplies $a\equiv 1+p(\mathrm{mod}{p}^k)$ and $a\equiv 1(\mathrm{mod}m)$. This $a$ is coprime to $N$. But the binomial theorem givescontradicting the Carmichael number condition. Hence $N$ is squarefree.

Now choose a primitive root $g$ modulo a prime factor $p$. Lift it to a unit modulo $N$ by prescribing $a\equiv g(\mathrm{mod}p)$ and $a\equiv 1(\mathrm{mod}N/p)$. The Chinese remainder theorem applies because $N$ is squarefree. The Carmichael congruence implies $g^{N-1}\equiv 1(\mathrm{mod}p)$. Since $g$ has multiplicative order $p-1$, that order divides $N-1$. This proves the other condition.^[ref]

In a commutative ring, the radical of an ideal is monotone and idempotent: inclusion $I\subseteq J$ gives $\sqrt{I}\subseteq \sqrt{J}$, and $a^n\in \sqrt{I}$ means $a^{nm}\in I$ for some $m$, giving $\sqrt{\sqrt{I}}=\sqrt{I}$. Since $\sqrt{I+J}$ contains both $\sqrt{I}$ and $\sqrt{J}$, it contains their sum of ideals. Taking radicals of ideals gives $\sqrt{\sqrt{I}+\sqrt{J}}\subseteq \sqrt{I+J}$. The reverse inclusion follows from $I+J\subseteq \sqrt{I}+\sqrt{J}$.

In particular, $\sqrt{I}+\sqrt{J}=R$ implies $I+J=R$. One can also see this directly: write $1=a+b$ with $a^m\in I$ and $b^n\in J$, then expand $(a+b{)}^{m+n-1}$ by the binomial theorem. Each term belongs to $I$ or $J$, so $1\in I+J$.

Let $U=aT+bI$ on a nonzero finite-dimensional vector space over a field, with $a\ne 0$. An invariant block for $T$ is also invariant for $U$. If $v,Tv,\dots ,T^{d-1}v$ is its cyclic vector basis, then $v,Uv,\dots ,U^{d-1}v$ is another basis: the binomial theorem gives a triangular change-of-basis matrix with diagonal $1,a,\dots ,a^{d-1}$. The monic polynomial $g(t)=a^{d}f((t-b)/a)$ annihilates this block, and its degree is $d$, so it is the block's minimal polynomial. In the new cyclic basis the block is $C(g)$.

The substitution preserves divisibility of the monic invariant factors of a linear operator. Apply it to every block to obtain the rational canonical form of $U$. If $a=0$, then $U=bI$ is a scalar matrix instead: all invariant factors are $t-b$, one for each dimension. This exceptional case cannot be obtained by dividing by $a$.

Write $m_k=\mathbb{E}{S}^k$, $m_0=1$, and ${\mu }_j=\mathbb{E}{X}^j$. Apply the Poisson size-bias identity with $h(s)=s^{k-1}$, and expand $(S+X{)}^{k-1}$ using the binomial theorem and independence. A finite $k$th severity moment suffices. In particular, the aggregate expected value, variance, and third central moment are $\lambda {\mu }_1$, $\lambda {\mu }_2$, and $\lambda {\mu }_3$, respectively; the latter two use raw severity moments.