At $x=0$, $f_n(x)=0$. For fixed $x>0$,Thus the pointwise limit is zero everywhere on $[0,1]$. However, with $t=nx$, the inequality $1+t^2\ge 2t$ gives $f_n(x)\le 1/2$, and equality occurs at $x=1/n\in [0,1]$. Henceso there is no uniform convergence. The moving evaluation point $1/n$ captures a peak missed by each fixed positive input.

Let $u<v$ lie in $f([a,b])$, and put $x=g(u)$ and $y=g(v)$. Equality $x=y$ would imply $u=v$. If $x>y$, strict monotonicity of $f$ would give $u=f(x)>f(y)=v$. Both are impossible, so $x<y$. Therefore the inverse function $g$ is strictly increasing.

If $x\ne y$, one of $x<y$ and $y<x$ holds. Since $f$ is strictly increasing, the corresponding values are strictly ordered, so $f(x)\ne f(y)$. Thus $f$ is an injection. Every element of the image of a function $f([a,b])$ equals $f(x)$ for some $x\in [a,b]$, which is precisely surjectivity onto that image of a function. Hence $f:[a,b]\to f([a,b])$ is a bijection.

Split the power series into even and odd indices:whenever the two series converge absolutely. If $|z|<1$, the even part is a convergent geometric series, while the odd part is bounded termwise in absolute value by ${\sum }_j|z{|}^{2j+1}$, also a convergent geometric series. Hence there is absolute convergence throughout $|z|<1$.

If $|z|>1$, the odd terms have absolute values $|z{|}^{2j+1}/(2j+1{)}^2$. Their successive ratio tends to $|z{|}^2>1$, so these terms grow eventually and do not tend to zero. The term test for divergence therefore forces divergence of the original power series. Consequently its radius of convergence is

For fixed $z\ne 0$, the ratio of successive absolute values isThe ratio test says that a series converges absolutely when this ratio has a limit of a sequence strictly below one. Thus this power series converges absolutely for every complex $z$; at zero the conclusion is immediate. Its radius of convergence is $\boxed{R=\infty }$.

Write $d=-c>0$. Suppose, for a contradiction, that only finitely many sign changes occur. Since every $a_n\ne 0$, the sign is then constant for $n\ge N$. Set $b_n=|a_n|>0$ on that tail. Multiplying the recurrence relation by this constant sign givesConsequently $b_{N+j}<2^{-j}{b}_N$ for $j\ge 1$, and $b_n\to 0$ by comparison with a geometric sequence. But positivity of $b_{n+1}$ requires $b_n^2>d$ at every step, which is incompatible with $b_n\to 0$. Thus the negative-parameter square-root iteration changes sign infinitely often.

Only (ii) is necessarily true. Every real $a_n$ satisfies $a_n^2\ge 0>c$, so (ii) holds automatically. For a single counterexample to (i) and (iii), choose $c=-3$ and $a_0=1$. The recurrence relation then gives the nonzero periodic sequenceHere $a_1<0$, disproving (i), and $a_2>a_1$, disproving (iii). This counterexample satisfies the requirement that no iterate is zero.

For $n\ge 1$, the preceding parts give a positive denominator and a nonpositive numerator inHence $\boxed{a_{n+1}\le a_n\text{ for every }n\ge 1}$. The restriction $n\ge 1$ matters: if $c>1$, then $a_1=(1+c)/2>a_0$.

For every $n\ge 0$, direct algebra givesThus $\boxed{a_n^2\ge c\text{ for every }n\ge 1}$. This identity is the key estimate in Newton iteration for a square root.

Rewrite the recurrence relation asSince $a_0=1>0$ and $c>0$, a positive $a_n$ gives a positive $a_{n+1}$. Mathematical induction therefore proves $\boxed{a_n>0\text{ for all }n\ge 0}$, and also shows that every division in the recurrence relation is defined.

False. Take $x_{1,n}=0$ for every $n$, and $x_{m,n}=1$ for every $m\ge 2$ and every $n$. Each row is constant, so its infimum and supremum coincide. Nevertheless,The two outer extrema can select different rows.

True. For every $m,n$,Fix $m$. The number $\sup R_m$ is an upper bound for every column infimum, so ${\sup }_n\inf C_n\le \sup R_m$. This holds for every $m$, making ${\sup }_n\inf C_n$ a lower bound of the row suprema. Taking their infimum gives the weak minimax inequality

True. Set $l=\inf A$ and $v=\inf \{\inf R_m:m\ge 1\}$. Because $R_m\subseteq A$, every row infimum satisfies $\inf R_m\ge l$, so $v\ge l$. But $x_{m,n}\ge \inf R_m\ge v$ for all $m,n$, making $v$ a lower bound of $A$. Therefore $v\le l$, and

True. Write $s=\sup A$ and $u=\sup \{\sup R_m:m\ge 1\}$. All these suprema exist in $[0,1]$. Since $R_m\subseteq A$, we have $\sup R_m\le s$ for every $m$, hence $u\le s$. Conversely, every $x_{m,n}\le \sup R_m\le u$, so $u$ is an upper bound of $A$ and $s\le u$. Thus the iterated extrema of an array satisfy

Let $M=\max \{\sup A,\sup B\}$. Every element of the union $A\cup B$ is at most $M$, so $M$ is an upper bound. Conversely, every upper bound of $A\cup B$ is an upper bound of each of $A$ and $B$, hence is at least both suprema and therefore at least $M$. Consequently

Put $s=\sup (-A)$. For every $x\in A$, $-x\le s$, hence $-s\le x$, so $-s$ is a lower bound of $A$. If $l$ is any lower bound of $A$, then $-x\le -l$ for every $x\in A$, making $-l$ an upper bound of $-A$. The definition of supremum gives $s\le -l$, or $l\le -s$. Thus $-s$ is the greatest lower bound, and

For any integer $n\ge 1$, let $R=R_{2\pi /n}$ in the stabilizer subgroup from i. Since $R_{\theta }{R}_{\varphi }=R_{\theta +\varphi }$ and $R_{\theta }=I$ exactly when $\theta \in 2\pi \mathbb{Z}$, this rotation matrix has order exactly $n$. Thereforeis a subgroup of $G$ of order $n$, including the trivial case $n=1$.

For infinitely many non-Abelian examples, let $n\ge 3$, retain $R=R_{2\pi /n}$, and take the half-turnDirect multiplication gives $J^2=I$ and $JRJ=R^{-1}$. Every word in $R,J$ therefore reduces to $R^k$ or $R^{k}J$, $0\le k<n$. The $R^k$ are distinct, as are the $R^{k}J$, and no element from the first family equals one from the second: they send $N$ to $N$ and $-N$, respectively. Hence $⟨R,J⟩$ has exactly $2n$ elements and is a dihedral group.

It is non-Abelian because $R\ne R^{-1}$ for $n\ge 3$, so $JR\ne RJ$. Different values of $n$ give different group orders, and therefore different subgroups. Thus $G$ has infinitely many non-Abelian subgroups, already among its finite subgroups.

Let $v\in S$. Choose a unit vector $u$ perpendicular to $v$, and put $w=v\times u$. The cross product identities show that $(v,u,w)$ is an orthonormal basis with positive orientation: $u\times (v\times u)=v$, soThus the matrix $A_v$ with these columns belongs to the special orthogonal group $G$ and satisfies $A_{v}N=v$. For any $v,z\in S$, the element $A_z{A}_v^{-1}$ sends $v$ to $z$. Therefore the group action on the unit sphere is transitive.

For $X\in {\mathrm{SO}}_3(\mathbb{R})$, the defining identity $X^{T}X=I$ givesThus an orthogonal transformation preserves the Euclidean norm and sends the unit sphere to itself. Restricting the given group action to this invariant set preserves both action axioms.

If $XN=N$, the first column of $X$ is $N=e_1$. The other columns are unit vectors perpendicular to it, so the first row is also $(1,0,0)$. Hence $X=\mathrm{diag}(1,B)$ with $B^{T}B=I_2$ and $\det B=1$. Conversely, every such matrix lies in $G$ and fixes $N$. Since every element of the special orthogonal group in two dimensions has the planar rotation matrix form, the stabilizer subgroup isGeometrically these are exactly the rotations in three dimensions about the first coordinate axis.

Choose a normal subgroup $N$ of prime order $p\in \{5,7\}$ as in ii, and choose a subgroup $H$ of the other prime order $q$. Their intersection is trivial by Lagrange's theorem. Both are cyclic groups, and ii gives $|\mathrm{Aut}(N)|=p-1$.

The image of a group homomorphism $\pi :H\to \mathrm{Aut}(N)$ from iii has order dividing $q$, because its kernel of a group homomorphism partitions $H$ into equally sized cosets. Its order also divides $p-1$ by Lagrange's theorem. Butso $\pi$ is trivial in either case. By iii, $⟨N,H⟩\cong C_p\times C_q$ has order $pq=35$ and therefore equals $G$.

If $n$ and $h$ generate $N$ and $H$, the element $(n,h)$ has order $\mathrm{lcm}(p,q)=pq=35$, since its $k$th power is the identity exactly when both $p$ and $q$ divide $k$. Thus this direct product of groups is cyclic, and