After the first update, the sequence lies in $[\sqrt{a},x_1]$ and is nonincreasing. The monotone convergence theorem for sequences gives a limit $L\ge \sqrt{a}$. The update is continuous on this interval, so passing to the limit givesPositivity forces $\boxed{x_n\to \sqrt{a}\text{ for every }{x}_0>0}$. If an iterate reaches the root exactly, the constant tail gives the same conclusion. The Exact Newton error identity for a quadratic here is $x_n-\sqrt{a}=(x_{n-1}-\sqrt{a}{)}^2/(2x_{n-1})$, also showing quadratic convergence once the error is small.

Subtract the previous iterate in Newton's method:The denominator is positive and the numerator is negative. If $x_{n-1}=\sqrt{a}$, the difference is zero and all subsequent iterates equal the root.

The update in Newton's method is exactly $x_n=g(x_{n-1})$. Positivity is preserved, and the arithmetic-geometric mean inequality from part (a) applies to every update. ConsequentlyThis holds even when $0<x_0<\sqrt{a}$: the first update moves to or above the positive square root.

For $f(x)=x^2-a$, its derivative is $f^{\prime }(x)=2x$. Substitution into Newton's method givesEvery positive input gives another positive input, so a starting value $x_0>0$ keeps the denominator nonzero at every step.

Use the same integer rounding in a divide-and-conquer recurrence, now with root cost $n^2$. Since $1/4\le m/n\le 1/2$,Choose $C\ge \max \{T(1),8/3\}$ and use strong induction with inductive hypothesis $T(j)\le C{j}^2$. Both child sizes are positive and less than $n$, soThe same factor $5/8$ occurs for the exact fractional split. Therefore $\boxed{T(n)=O(n^2)}$ in Big O notation.

The printed fractional arguments require a size convention. To make the recurrence relation precise on positive integers, take a bounded base case $T(1)=t>0$ and use $m=\lceil n/4\rceil$, $n-m$ for $n\ge 2$. This integer rounding in a divide-and-conquer recurrence gives $1/4\le m/n\le 1/2$. The unrounded calculation is the special split fraction $1/4$.

Prove $T(n)\le An{\log }_{2}n+Bn$ by strong induction. Put $h_0=H_2(1/4)>0$, choose $A\ge 1/h_0$, and choose $B\ge t$. The base case follows. The entropy potential for a divide-and-conquer recurrence givesThe last step uses $H_2(m/n)\ge h_0$. This closes the inductive hypothesis and proves $\boxed{T(n)=O(n\log n)}$ in Big O notation.

For sufficiently large positive $n$, the logarithm identities giveAfter division by $n{\log }_{2}n$, the right-hand side tends to one. Thus $\boxed{\Theta (n\log n),\text{in particular }O(n\log n)}$ gives the asymptotic order in Big theta notation and Big O notation.

Here $a,b\ge 2$ are fixed integers. The term $a{b}^n$ has exponential growth, whereas $a{n}^b$ has fixed-degree polynomial growth and $n{b}^a$ is linear. Since a fixed-base exponential eventually dominates every fixed-degree polynomial,Hence the requested Big O notation bound, with its matching lower order, is $\boxed{\Theta (b^n),\text{in particular }O(b^n)}$.

For $n>0$,Consequently $\boxed{\sqrt{n^6+n^2}=\Theta (n^3),\text{in particular }O(n^3)}$ in Big theta notation and Big O notation.

Dividing by $n^2$ gives $10^5/n+2\to 2$. Thus the quadratic term controls the growth:The sharper Big theta notation also supplies the requested Big O notation bound.

Write $a_1-a_2=kn$ with $k\in \mathbb{Z}$. The difference of two squares factors asThe final factor is an integer, so $n$ divides the difference. By congruence modulo an integer, $\boxed{a_1^2\equiv a_2^2(\mathrm{mod}n)}$.

The notation means that $n$ divides $a_1-a_2$:This is congruence modulo an integer; equivalently, the two integers have the same remainder on division by the positive integer $n$.

For fixed $n\ge 0$, use mathematical induction on $k$. At $k=0$, both sides equal one. If the hockey-stick identity holds at $k$, then Pascal's identity givesThis is the required formula with $k$ replaced by $k+1$.

For $n\ge 2$ and $1\le k\le n-1$, the factorial formula for a binomial coefficient givesThis proves Pascal's identity. The boundary cases $k=0,n$ follow for $n\ge 1$ using the usual extension $\left(\genfrac{}{}{0ex}{}{m}{j}\right)=0$ for $j<0$ or $j>m$. The recurrence itself requires $n\ge 1$.

Set $y=f(x)$ and cube the equation. Solving for $x^3$ givesConsequently the inverse function isThe value $y=1$ is impossible in the original equation. Conversely, every real $y\ne 1$ gives exactly one real $x$ by the displayed cube root, and this $x$ cannot equal $-1$. Direct substitution verifies both inverse identities. Thus the corrected bijection is $f:\mathbb{R}\setminus \{-1\}\to \mathbb{R}\setminus \{1\}$.

The real cube root in the denominator vanishes at $x=-1$, so the printed rule needs the domain of a function $D_1=\mathbb{R}\setminus \{-1\}$. On its actual domain of a function, the second iterated function simplifies toThis suggestsFor the induction step, assume the formula at $n$ and restrict to arguments for which its value is not $-1$. Its cube is $x^3/(1+n{x}^3)$, so another application givesReal cube roots are multiplicative, including for negative arguments. The newly excluded point satisfies $x^3/(1+n{x}^3)=-1$, hence $x^3=-1/(n+1)$; all earlier exclusions remain. This proves both the formula and the domain of a function by mathematical induction. The inherited exclusions are essential for an iteration of a partial function: cancellation in the final expression cannot repair an undefined intermediate application.

Eliminate the existential premise with fresh witness $a$. Its first conjunct gives $Q(a)$ and its second gives $\forall y(Q(y)\to y=a)$. For a fresh arbitrary $b$, one direction of the desired logical biconditional follows by universal elimination. For the other, $b=a$ must be reversed before transporting $Q(a)$ to $Q(b)$ using the supplied oriented equality elimination rule.

Here $b$ is absent from every open assumption when line 12 uses universal introduction. In line 13 the existential quantifier binds the witness position occupied by $a$; the conclusion then has no free $a$, allowing line 14 to close its witness subproof. HenceThe premise says an object satisfying $Q$ exists and every object satisfying $Q$ equals it. The logical biconditional adds the converse explicitly: every object equal to that witness also satisfies $Q$.

Use the equality introduction rule to prove $t_1=t_1$. In the equality elimination rule, choose the first-order formula $\varphi (z)$ to be $z=t_1$. Substituting $t_2$ for $t_1$ in its first position proves symmetry:

Thus $\boxed{t_1=t_2⊢t_2=t_1}$. This uses only the two supplied equality rules; symmetry has not been assumed as an additional rule.

Extract the universally quantified logical implication and the existentially quantified logical conjunction from the premise. Use one fresh witness $a$ for both conjuncts of the latter, so that the resulting $R(a)$ and $\neg S(a)$ concern the same object:

The witness $a$ occurs in no undischarged premise or final conclusion. Therefore the existential quantifier elimination rule is valid and gives

The existential premise is $\exists x(Q(x)\to R(x))$, with the logical implication inside the existential quantifier. Eliminate it using a fresh witness $a$, specialize $\forall xQ(x)$ to that same witness, and introduce the required existential quantifier before closing the witness subproof:

The eigenvariable $a$ is absent from the premises and from $\exists xR(x)$. Consequently,