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.