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At the origin the needed partial derivatives areThe Taylor series through total degree two is thereforewhich givesHere total degree is what matters: $x^2{y}^2$ is quartic, not quadratic. The exact omitted polynomial is $y^3+x^2{y}^2+x^2{y}^4$, which verifies the stated Big O notation remainder directly.

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.

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.

The number $K$ is fixed. By Big O notation, for each $i$ there are constants $C_i>0$ and $N_i$ such that $f_i(n)\le C_{i}g(n)$ when $n\ge N_i$. Take $N={\max }_i{N}_i$. For $n\ge N$, positivity givesThe constants in parentheses are independent of $n$, proving $\boxed{h=O(g),l=O(g^2)}$. If $K=1$, the second sum is empty and $l=0$, which satisfies the same upper bound.

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.

Suppose the definition of big O notation held with constants $C>0,N$. Choose an odd integer $n>\max (C,N)$. At this input, $f(n)=n$ and $g(n)=1$, so the required inequality would be $n\le C$, a contradiction. ThusThe infinitely many even inputs, where $f(n)/g(n)=1/n$, cannot repair the failure on odd inputs because the bound must hold for every sufficiently large input.

For $n\ge 1$, $\log 3<2$ implies $3^{\log n}=n^{\log 3}\le n^2$, and also $n\le n^2$. ConsequentlyThe definition of big O notation holds with $C=4,N=1$:

For functions $f,g:\mathbb{N}\to \mathbb{R}$, big O notation meansUsing the natural logarithm and exponential function, $3^{\log n}=n^{\log 3}$. Since $\log 3>1$, for every $n\ge 1$ we have $n\le n^{\log 3}$. The witnesses $C=1,N=1$ therefore prove

A direct mathematical induction proves $0\le F_n\le 2^n$. The base cases $F_0=0\le 1$ and $F_1=1\le 2$ hold. If the bound holds at $n-1$ and $n$, then the Fibonacci recurrence givesThus the definition of big O notation holds with $C=1$, proving

Count the final division producing zero as an iteration, and let the total be $n$. The terminal product $a_n{b}_n=0$ alone gives no useful bound. Instead apply (i) to each nonterminal product, which is a positive integer and hence at least $1$.

If $n\ge 2$, the penultimate product satisfiesAs $n$ is an integer, this impliesFor $n=1$, the same bound holds since $a>b\ge 1$ implies $ab\ge 2$. In particular the Euclidean algorithm uses $O(\log a+\log b)$ divisions, in big O notation. The convention that includes the final zero-producing division explains the one-step offset in the argument.

Use the Taylor series $e^u=1+u+u^2/2+\cdots$ with $u=x^3-x^5$. Since $u$ begins at cubic order and $u^2$ begins at sixth order, through cubic terms the Taylor series isThe linear and quadratic coefficients are zero. The stronger remainder $O(x^5)$, in big O notation, follows because the fourth-order term is also zero; the next term is $-x^5$.

Because $T$ is positive, the definition of big O notation becomesHere $C$ and $n_0$ are fixed independently of $n$.

The relevant derivatives at $x=3$ areThus the cubic Taylor polynomial isIn fact the exact expansion is $g(x)=32+4(x-3{)}^3+(x-3{)}^4$, so the omitted term has order $O((x-3{)}^4)$ in big O notation.

Take $f(n)=g(n)=n^2$, so $h(n)=n^2$ but $(f\circ g)(n)=n^4$. The ratio $(f\circ g)(n)/h(n)=n^2$ is unbounded, precluding any eventual constant upper bound in big O notation. False: in general $f\circ g\ne O(h)$. This is a counterexample involving function composition, not pointwise addition.

Positivity gives $\max \{f(n),g(n)\}\le f(n)+g(n)$. Thus $h(n)\le f(n)+g(n)$ for every positive $n$, meeting big O notation with constant $1$ and threshold $1$. True: $h=O(f+g)$.

Since $h(n)=\max \{f(n),g(n)\}$, both $f(n)\le h(n)$ and $g(n)\le h(n)$. Therefore $f(n)+g(n)\le 2h(n)$ for every positive $n$. The definition of big O notation is satisfied with constant $2$ and threshold $1$. True: $f+g=O(h)$.

For each fixed positive integer $k$, part (i) gives $(f(n)+g(n){)}^k\le (2h(n){)}^k=2^{k}h(n{)}^k$. This meets big O notation with constant $2^k$ and threshold $1$. True: $(f+g{)}^k=O(h^k)$ for fixed $k$. The constant may depend on $k$; no claim uniform in a varying $k$ is required.