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Infinite walls impose the Dirichlet boundary conditions $\psi (-a)=\psi (a)=0$. Inside the infinite square well, the time-independent Schrödinger equation is $-{\hslash }^2{\psi }^{\prime \prime }/(2m)=E\psi$. Its nonzero solutions satisfying both endpoint conditions have wave numbers $k_n=n\pi /(2a)$, $n\ge 1$. Sincethe normalized energy eigenstates are the sine modes with prefactor $1/\sqrt{a}$. They are orthonormal by sine orthogonality and complete by the Fourier sine series on an interval of length $2a$. Differentiating twice gives

Let $I$ denote the identity endomorphism. The two-root minimal polynomial gives $(\alpha -{\lambda }_{1}I)(\alpha -{\lambda }_{2}I)=0$. Every vector has the decompositionThe first summand lies in the eigenspace $E_1=\ker (\alpha -{\lambda }_{1}I)$, and the second in $E_2$. If a vector belongs to both, subtracting its two eigenvalue equations gives $({\lambda }_1-{\lambda }_2)v=0$, hence $v=0$. Thus $V=E_1\oplus E_2$.

For distinct real roots ${\lambda }_1,\dots ,{\lambda }_m$, define polynomial projectorsThe Lagrange interpolation polynomials in this expression sum to one: their sum has degree at most $m-1$ and equals one at the $m$ distinct roots. Hence ${\sum }_i{P}_i=I$. Also $(\alpha -{\lambda }_{i}I)P_i=0$, because its numerator is the minimal polynomial evaluated at $\alpha$. Thus every vector is a sum of eigenvectors, $v={\sum }_i{P}_{i}v$. On $E_j$, $P_i$ acts as ${\delta }_{ij}I$, so applying $P_i$ to any zero sum of vectors from the eigenspaces shows that every summand is zero. ThereforeChoosing a basis in each eigenspace proves diagonalizability over the reals. Conversely, if a real matrix is similar over the reals to a diagonal matrix, a polynomial annihilates it exactly when it vanishes at each distinct real diagonal entry. Its minimal polynomial is therefore the product of those distinct linear factors. We have proved real diagonalizability is equivalent to a minimal polynomial with distinct real roots.

Write $J_k(\lambda )$ for the Jordan block with diagonal $\lambda$ and ones on the superdiagonal. Its characteristic polynomial and minimal polynomial are both $(x-\lambda {)}^k$. Therefore one answer to the first request isFor the second request, takeThe algebraic multiplicities are five and two, and the largest block at each eigenvalue has size two, so both prescribed polynomials hold. They are not similar matrices: the eigenspace at three has dimensions three and four respectively, and similarity preserves eigenspace dimension.

There is no third similarity class. At eigenvalue one, the total size two and maximal block size two force the single block $J_2(1)$. At eigenvalue three, the blocks partition five, each has size at most two, and at least one has size two. The only partitions are $2+2+1$ and $2+1+1+1$. The uniqueness of Jordan normal form up to block order therefore leaves exactly the two classes already exhibited.

If $x^2\equiv -1(\mathrm{mod}p)$ with $p$ an odd prime, then $x^4\equiv 1$ but $x^2\not\equiv 1$. The multiplicative order consequently divides four and is neither one nor two, so it is four. The preceding divisibility result makes $4\mid p-1$ necessary.

For sufficiency, suppose $p\equiv 1(\mathrm{mod}4)$ and put $r=(p-1)/2$, an even integer. Pair opposite factors in the factorial:Wilson's theorem gives $(r!{)}^2\equiv -1(\mathrm{mod}p)$, explicitly producing a square root of minus one modulo a prime. Thus

Let $d$ be the multiplicative order of $a$ modulo $p$. Use Euclidean division to write $x=qd+r$, $0\le r<d$. Since $a^d\equiv 1$,If $a^x\equiv 1$, minimality of $d$ forces $r=0$, so $d$ divides $x$. Conversely, $d\mid x$ immediately gives $a^x\equiv 1$. For negative integers $x$, powers are interpreted using the multiplicative inverse and the same argument applies.

Apply the result to $x=p-1$, using Fermat's little theorem. Then $d\mid p-1$, equivalently

If $p$ is a prime number and $p\nmid a$, multiplication by $a$ permutes the nonzero residue classes modulo $p$: $ai\equiv aj$ implies $i\equiv j$ because $a$ has a multiplicative inverse. Multiplying all the permuted classes givesThe factorial is a nonzero element of the finite field ${\mathbb{F}}_p$, so cancellation is valid. Thereforewhich is Fermat's little theorem.

Write $r=(p-1)/2$, $O=1\cdot 3\cdots (p-2)$ and $E=2\cdot 4\cdots (p-1)$. The residues $p-2,p-4,\dots ,p-2r$ are exactly the odd factors of $O$, in reverse order. Henceby Wilson's theorem. It follows that the odd-residue squared product modulo a prime isThus it is $-1$ for $p\equiv 1(\mathrm{mod}4)$ and $1$ for $p\equiv 3(\mathrm{mod}4)$.

For a prime number $p$, the nonzero residue classes form the multiplicative group of the finite field ${\mathbb{F}}_p$. Every class has a unique multiplicative inverse. A class equals its own inverse precisely when $x^2=1$, so $(x-1)(x+1)=0$ in the field and $x=1$ or $x=-1$.

For an odd prime, pair each remaining class with its distinct inverse. Each pair contributes one to the product, leavingFor $p=2$, the same congruence is $1\equiv -1(\mathrm{mod}2)$. This proves Wilson's theorem.

No such set exists. If $R$ had precisely the sets $x$ satisfying $x\notin x$ as its elements, applying that membership criterion to $R$ itself would givea contradiction. This is Russell's paradox. It rules out this unrestricted universal collection; a set whose elements happen not to contain themselves is perfectly possible, for example the empty set.

Suppose $F:X\to \mathcal{P}(X)$ were a surjection onto the power set of the set $X$. Form the subsetSurjectivity would give $a\in X$ with $F(a)=D$. But then $a\in D$ holds exactly when $a\notin F(a)=D$, a contradiction. Hence no such surjection exists, and in particular there is no bijection between a set and its power set. This is the diagonal construction in Cantor's theorem; it also covers the empty set, whose power set has one element.

The direct product of groups uses componentwise multiplication:Associativity follows in each factor. The maps $g\mapsto (g,e_H)$ and $h\mapsto (e_G,h)$ are injective group homomorphisms onto subgroups, giving the required copies of $G$ and $H$.

Use the PDF's convention that the subscript in $D_n$ is the group order. Write the hexagon group as $D_{12}=⟨r,s\mid r^6=s^2=e,srs=r^{-1}⟩$ and the triangle group as $D_6=⟨a,b\mid a^3=b^2=e,bab=a^{-1}⟩$. Let $z$ generate $C_2$. In $D_{12}$, $r^2$ has order three and $s$ conjugates it to its inverse, so $⟨r^2,s⟩$ is a copy of $D_6$. The element $r^3$ has order two and is central, since $s{r}^{3}s=r^{-3}=r^3$.

DefineThe generator relations are preserved and $r^3$ commutes with the other images, so $\Phi$ is a group homomorphism from the direct product. The six residues $2i+3k$ are distinct modulo six: reduction modulo two first determines $k$, then reduction modulo three determines $i$. Together with $j$, they give all twelve distinct normal forms $r^m{s}^j$ in $D_{12}$. Thus $\Phi$ is bijective andThis is the dihedral splitting when the half-rotation order is odd with half-rotation order three.

Finally, $D_{12}$ contains an element of order six, namely $r$. The even cycle types on four symbols are the identity, a 3-cycle, and two disjoint transpositions; their orders are respectively one, three and two. Therefore $A_4$ has no element of order six. Since a group isomorphism preserves element orders,

Compose permutations from right to left. The indicated cycle lengths immediately give $X^3=P^2=Q^2=e$. Under conjugation, $X^{-1}(ab)X=(X^{-1}a{X}^{-1}b)$. Since $X^{-1}$ sends $1,2,3$ to $3,1,2$ and fixes $4$,Direct composition also gives $PQ=QP=(14)(23)$. Thus $V=\{e,P,Q,PQ\}$ is a Klein four-group. Conjugation by $X$ cycles its three nonidentity elements, so $X$ normalizes $V$.

The cosets $V,VX,V{X}^2$ have four elements each and are distinct. If two coincided, a nontrivial power of $X$ would belong to $V$, impossible because it has order three whereas every nonidentity element of $V$ has order two. Hence the subgroup generated by $X,P,Q$ contains at least twelve elements. These generators are even permutations, so the subgroup lies in the alternating group $A_4$, whose order is $4!/2=12$; multiplying by one fixed transposition pairs even and odd permutations to justify this count. ThereforeThe calculation also describes $A_4$ as the semidirect product $V⋊⟨X⟩$, with the order-three factor permuting the three involutions.

All three examples below have radius of convergence one, because the $n$th roots of their coefficient moduli tend to one.

For convergence everywhere on the boundary, take ${\sum }_{n=1}^{\infty }{z}^n/n^2$. On $|z|=1$ it converges absolutely: for $n\ge 2$, $1/n^2\le 1/[n(n-1)]=1/(n-1)-1/n$, whose sum telescopes to a finite value.

For divergence everywhere on the boundary, take ${\sum }_{n=0}^{\infty }{z}^n$. If $|z|=1$, every term has modulus one, so the term test for divergence applies.

For mixed boundary behavior, take ${\sum }_{n=1}^{\infty }{z}^n/n$. At $z=1$ it is the divergent harmonic series; each successive dyadic block contributes at least $1/2$. At every other point $|z|=1$, summation by parts proves convergence. Indeed put $B_k={\sum }_{j=m}^k{z}^j$, so the finite geometric series gives $|B_k|\le 2/|1-z|=:C_z$. ThenThus the partial sums form a Cauchy sequence and converge. This derives the relevant Dirichlet test rather than only naming it.

For $z\ne 0$,If $R|z|<1$, choose $q$ strictly between that limit and $1$. Eventually $|a_n{z}^n|\le q^n$, proving absolute convergence by the geometric series comparison. If $R|z|>1$, choose $q>1$ below the limit; eventually $|a_n{z}^n|\ge q^n$, so the terms do not tend to zero and the series diverges. The point $z=0$ is harmless. This proves the root test in the needed case and gives

Let $M$ bound all $|a_n{w}^n|$ and put $q=|z|/|w|<1$. ThenThe geometric series ${\sum }_{n=0}^{\infty }M{q}^n$ converges, so the comparison test for series proves absolute convergence of $\sum a_n{z}^n$. This is the bounded power-series terms force interior absolute convergence principle.

Define the radius of convergence byallowing $\rho =0$ and $\rho =\infty$. If $|z|<\rho$, the supremum supplies a point of convergence $w$ with $|w|>|z|$; convergence makes its terms bounded, so the argument above gives absolute convergence at $z$. If $|z|>\rho$, convergence is impossible by definition. At $|z|=\rho$, no general conclusion follows. Thus the convergence region has a circular interior centered at zero, with boundary behavior depending on the power series.

Put $G(t)={\int }_a^{t}f(x)dx$. At an interior point and for $h\ne 0$ small enough,By continuity at $t$, the modulus is bounded by the supremum of $|f(x)-f(t)|$ between $t$ and $t+h$, which tends to zero. Thus the fundamental theorem of calculus givesAt the interval endpoints the same argument supplies the corresponding one-sided derivatives.

For an integrable discontinuous function, the single-point spike from part (i) has $G\equiv 0$, so $G^{\prime }(c)=0\ne f(c)=1$; this disproves the derivative identity even when the primitive is differentiable. To see that differentiability itself may fail, take the step function $f(x)=0$ for $x<c$ and $f(x)=1$ for $x\ge c$. It is Riemann integrable, but its primitive is $G(t)=\max (0,t-c)$ and has left derivative zero and right derivative one at $c$. Hence the full assertion is false for general Riemann-integrable functions.

Suppose the continuous nonnegative function has $f(t_0)>0$ somewhere. Continuity supplies a positive-length interval within $[a,b]$ on which $f\ge f(t_0)/2$; at an endpoint use a one-sided interval. By monotonicity of the Riemann integral, its contribution to the integral is strictly positive, a contradiction. Therefore $\boxed{f\equiv 0}$ under the stated continuous-function assumptions.

For the general integrable case, choose $c\in (a,b)$ and let $f(c)=1$ while $f(t)=0$ for $t\ne c$. This nonnegative indicator function is Riemann integrable with integral zero: isolate $c$ in partition intervals of arbitrarily small total length, making the upper sum arbitrarily small while the lower sum stays zero. Yet $f(c)=1$. This is the single-point changes preserve Riemann integrals phenomenon and supplies the counterexample.

The intermediate value theorem gives a root $r\in (a,b)$, unique because $f^{\prime }>0$ makes $f$ strictly increasing. We show that Newton root-finding iteration, started at $b$, always stays to the right of $r$. If $x>r$, the mean value theorem gives $f(x)=f^{\prime }(\xi )(x-r)$ for some $\xi \in (r,x)$. Since $f^{\prime \prime }>0$, $f^{\prime }$ is strictly increasing, so $0<f^{\prime }(\xi )<f^{\prime }(x)$. It follows thatInduction gives $r<x_{n+1}<x_n\le b$, so the bounded monotone sequence theorem supplies a limit $\alpha \ge r$. The continuous positive $f^{\prime }$ is bounded away from zero on $[a,b]$, making $F$ continuous. Passing to the limit in $x_{n+1}=F(x_n)$ gives $\alpha =F(\alpha )$ and hence $f(\alpha )=0$. Uniqueness implies $\boxed{\alpha =r}$.

For the rate, apply the Taylor theorem with Lagrange remainder to $f$ between $r$ and $x_n$:With $e_n=x_n-r>0$, rearrangement givesThus the monotone Newton convergence for a strictly convex function has quadratic convergence. More concretely, if $K={\max }_{[a,b]}{f}^{\prime \prime }/[2{\min }_{[a,b]}{f}^{\prime }]$, then $e_{n+1}\le K{e}_n^2$. Once $K{e}_N<1$, induction gives $e_{N+j}\le K^{-1}(K{e}_N{)}^{2^j}$, exhibiting the repeated squaring of small errors.

By continuity of $g^{\prime \prime }$ at $\alpha$, for any $ϵ>0$ choose $\delta (ϵ)>0$ such that $|g^{\prime \prime }(s)|\le |g^{\prime \prime }(\alpha )|+ϵ=:M$ whenever $|s-\alpha |<\delta (ϵ)$. Twice using the fundamental theorem of calculus gives the remainder asThe inner difference is ${\int }_{\alpha }^s{g}^{\prime \prime }(r)dr$ and has modulus at most $M|s-\alpha |$. Integrating this bound over the interval between $\alpha$ and $t$ proves, for either sign of $t-\alpha$,This is the local quadratic Taylor remainder estimate. The given $ϵ$ may enter the bound on the second derivative; no stronger differentiability assumption is needed.

The fixed-point property fails on an open interval. The continuous functionmaps $(a,b)$ into itself, because $a<t<f(t)<b$. Its only possible fixed point satisfies $t=(t+b)/2$, hence $t=b$, which is outside the domain. It is therefore a counterexample even with the self-map requirement retained.