For $x\le c$, the cumulative distribution function is zero. For $x>c$, integrate the normalized probability density function:It is continuous at $c$, increases to one, and is the cumulative distribution function of the stated Pareto distribution.

Call the normalization constant $K$ to distinguish it from the moment index used later. A probability density function must have total integral one, soTherefore $K=\theta c^{\theta }$, the value denoted $k$ in the density. This is the Pareto distribution with scale $c$ and shape $\theta$.

A small difference between successive secant method iterates alone does not certify a small absolute error. Instead use a residual root error bound. On $[2,3]$,and $f(2)<0<f(3)$, so there is a unique root $r\in (2,3)$. For an iterate $x\in [2,3]$, the mean value theorem givesTherefore stop when $x\in [2,3]$ and $|x{\log }_{2}x-4|\le (1+1/\ln 2)10^{-10}$. This guarantees the desired absolute error in exact arithmetic; the evaluated residual of a root equation should include an allowance for its much smaller roundoff error. A safeguarded implementation can retain the bracket $[2,3]$ and use bisection if a proposed secant method step leaves it.

Use $f(x)=x{\log }_{2}x-4$. Then $f(1)=-4$, $f(2)=-2$, and the secant method givesWriting $L={\log }_{2}3$, the next step isThus the requested approximation is $x_3=2.73$ to three significant figures. All intermediate logarithms should be retained at higher precision before this final rounding.

The secant method starts from two approximations $x_0,x_1$. Replace the function near its zero of a function by the secant line through $(x_{k-1},f(x_{k-1}))$ and $(x_k,f(x_k))$; the zero of that line is the next approximation:Stop immediately if $f(x_k)=0$; otherwise the denominator must be nonzero. Each further step needs one new function evaluation and no derivative. There is no automatic bracketing or global convergence guarantee.^[ref]

Write $P$ for the prime numbers in $A$ and $C=A\setminus P$ for its nonprime elements. Since $n\ge 4$, both antichains have at least two elements: $2,3\in P$ and $1,4\in C$. Every element of $P$ is below every element of $C$. We can classify a subset $S$ by its intersection with these two levels of the two-level partial order.

Consequently an upper bound exists exactly when $|S\cap C|\le 1$; a maximum or least upper bound exists exactly when $|S\cap C|=1$ or $S$ is a singleton subset of $P$. The exhaustive cases also justify why the last two existence criteria coincide here.

The prime numbers $2,3,5$ form the lower antichain; the nonprime elements $1,4$ form the upper antichain. In particular, $1$ is not a prime number. Each lower element is below each upper element, and no third element can lie strictly between them. Therefore the six cover relations are $2,3,5\prec 1,4$, represented by the following Hasse diagram. Upward movement represents increasing order; crossing lines do not add elements.

False. On every odd positive integer $n$, $f(n)/g(n)=n$. Given any proposed Big O notation constant $C$ and threshold $N$, choose an odd $n>\max (C,N)$. Then $f(n)>Cg(n)$, contradicting the required bound. The behavior on even integers cannot repair this failure on an unbounded subsequence.

True. The logarithm identity givesFor $n\ge 1$, $e^n\ge n$, so $0\le n{e}^{-n}\le 1$. By monotonicity of the natural logarithm, the expression lies between $0$ and $\ln 2$. It is therefore $O(1)$ in Big O notation.

True. Rationalizing the square root gives, for $n\ge 1$,The uniform bound is exactly the assertion $O(1)$ in Big O notation.

False. Set $a_n=4^n/n^4$. For $n\ge 4$,Iteration gives a growing geometric sequence as a lower bound for $a_n$. Hence $a_n\to \infty$ and no eventual constant can bound $4^n/n^4$, as Big O notation would require.

True. For every $n\ge 1$,These constants satisfy the definition of Big O notation.

Use mathematical induction with two consecutive base cases: $F_2=1=(3/2{)}^0$ and $F_3=2\ge (3/2{)}^1$. Suppose the bounds hold for exponents $n$ and $n+1$. The Fibonacci recurrence givesThus the consecutive pair of bounds advances one step. $F_{n+2}\ge (3/2{)}^n$ for every $n\ge 0$.

The base case $n=3$ isAssume the displayed formula holds at some $n\ge 3$. Adding the next term yieldsThe last step uses $(2n-1)(n+1)-2n=(2n+1)(n-1)$. This is the required formula at $n+1$, completing the proof by mathematical induction. The identity holds for all $n\ge 3$.

For each $y\in Y$, surjectivity makes the fiber of a function $f^{-1}(\{y\})$ nonempty. Choose one element $x_y$ of each such fiber of a function and define $g(y)=x_y$. Then $f(g(y))=y$ for every $y$, proving $f\circ g=I_Y$. For arbitrary sets, the simultaneous selection invokes the usual axiom of choice; this is the right-inverse characterization of the axiom of choice. For finite $Y$ only finitely many choices are needed. If $Y$ is empty, the unique empty function gives the same conclusion.

Suppose $g(y)=g(y^{\prime })$. Applying $f$ and using the right inverse identity givesThis is exactly injectivity, so every right inverse is injective.

Not injective; surjective. The equal values $f(0)=f(2)=1$ disprove injectivity. For any $y>0$, take $x=1+y^{-1/2}>1$. This lies in the stated domain of a function, andConsequently every member of the codomain has a preimage.

Not injective; its image of a function is exactly $\{1,2,3,\dots \}$. For example, both $x=e^{-1}$ and $x=e^{-1/2}$ give the ceiling function value $1$. Since $0<x<1$, $(\ln x{)}^2>0$, so the value $0$ is never attained. For every positive integer $j$,Each such interval is nonempty and contained in $(0,1)$, with the upper endpoint $1$ excluded when $j=1$. Thus the answer concerning surjectivity depends on the convention for the natural numbers: onto if $\mathbb{N}=\{1,2,\dots \}$, and not onto if $0\in \mathbb{N}$. The source does not settle that convention.

Both injective and surjective. The natural logarithm is strictly increasing on the positive real numbers, so it is injective. For every $y\in \mathbb{R}$, $x=e^y>0$ satisfies $\ln x=y$; hence the function is surjective, with the exponential function as its inverse function.

Not injective; surjective. Indeed, $f(0)=f(-1)=-9$ although $0\ne -1$. This polynomial is continuous, with $f(x)\to -\infty$ as $x\to -\infty$ and $f(x)\to \infty$ as $x\to \infty$. For any prescribed real $y$, choose $a<b$ with $f(a)<y<f(b)$; the intermediate value theorem supplies a preimage between $a$ and $b$.