By independence, for $k\ge 0$,This is the geometric minimum law, givingThe support remains the positive integers; this parameter is a success probability, not a mean.

The expected value is the derivative of the probability generating function at one. The expression from part (ii) givesIts finiteness also follows directly from the exponentially decaying geometric probabilities.

Summing a geometric series gives the probability generating functionThe series converges absolutely for $|s|<1/(1-p)$. Differentiating the rational expression yields

The variable has positive-integer support, so $\mathbb{P}(X>0)=1$. Subtract consecutive values of the survival function:This is the positive-support convention for the geometric distribution.

Define the one-step local truncation error as the defect obtained by inserting the exact solution into an update:For a smooth solution, Taylor's theorem and $y^{\prime }=f(x,y)$ giveThus $m=2$ for the one-step defect convention. The leading derivative can also be written $y^{\prime \prime }=f_x+f_{y}f$ by the chain rule.

Some conventions instead call ${\tau }_{n+1}={\delta }_{n+1}/h$ the truncation error. That normalized local truncation error is $O(h)$, so $m=1$ under the normalized convention. Both describe the same Euler one-step defect; neither should be confused with an unqualified global error claim.

The Forward Euler method uses the current slope:Here $x_n=x_0+nh$ and the initial approximation normally equals the given initial value. Each step takes a tangent-line increment of width $h$.

No. On $\{0,1,2\}$ take $R=\{(0,1),(1,2),(2,0)\}$. This is an antisymmetric relation because it contains no opposite pair of distinct elements. Its reflexive transitive closure contains every ordered pair: following the directed three-cycle reaches any vertex from any other. In particular it contains both $(0,1)$ and $(1,0)$, violating antisymmetry.

This illustrates the reachability preorder: collapsing mutually reachable vertices into strongly connected components yields a partial order, but retaining all original vertices need not.

A preorder is reflexive and transitive: $xRx$ for every element, and $xRy,yRz$ imply $xRz$. A partial order additionally requires antisymmetry: $xRy$ and $yRx$ imply $x=y$. It need not compare every pair of distinct elements.

Beyond the same common threshold, the triangle inequality givesThus $f_1-f_2=O(g_1+g_2)$. The absolute value matters because the difference need not itself be positive, even though the four original functions are.

Choose constants $C_1,C_2>0$ and a common threshold beyond which $f_1\le C_1{g}_1$ and $f_2\le C_2{g}_2$. Positivity permits multiplication:The fixed constant $C_1{C}_2$ proves $f_1{f}_2=O(g_1{g}_2)$ by the definition of big O notation.

There are $6^3=216$ unrestricted placements, since one colour has $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ occupancy vectors. For any specified empty box, the two remaining boxes admit $3^3=27$ placements. For any specified pair of empty boxes, the one remaining box admits one placement. All three boxes cannot be empty.

The inclusion-exclusion principle therefore givesThe empty-box events are not disjoint events, so subtracting their counts without restoring their pairwise overlaps would be incorrect.

For one colour, the box occupancies form a weak composition of $k$ into $n$ nonnegative parts. For $n\ge 1$, stars and bars gives $\left(\genfrac{}{}{0ex}{}{k+n-1}{n-1}\right)$ choices. The three colours' occupancies can be selected independently, so the number isBoxes are labelled, while equal-colour balls contribute only occupancy counts; no extra factorial for permuting such balls is appropriate.

Modulo seven, $19\equiv -2$ and $(-2{)}^3=-8\equiv -1$. Therefore $19^9\equiv (-1{)}^3\equiv -1\equiv 6(\mathrm{mod}7)$. The remainder is $6$, using modular exponentiation.

Use modular arithmetic modulo eight: $801\equiv 1$, $32\cdot 19\equiv 0$, and $29\cdot 13\equiv 5\cdot 5\equiv 1$. The squared difference is consequently congruent to one, giving the remainder

The singleton set $\{\varnothing \}$ has cardinality one, so its power set has cardinality two. Taking the Cartesian product with the singleton $\{\varnothing \}$ leaves that cardinality equal to two. The outer power set therefore has $2^2=4$ elements.

A Cartesian product with the empty set is empty: there is no possible second coordinate. Thus the requested cardinality is $0$, independently of the first factor's power set.

The outermost braces form a singleton set, regardless of how its one element is nested. Its cardinality is therefore $1$. A set containing the empty set is different from the empty set itself.

If $x\in A\oplus B$, the definition of symmetric difference puts it in exactly one of $A,B$. It is therefore in their set union and outside their set intersection, giving $x\in (A\cup B)\setminus (A\cap B)$.

Conversely, an element of $(A\cup B)\setminus (A\cap B)$ lies in at least one of the two sets but not both. It consequently belongs to $A\setminus B$ or to $B\setminus A$, and hence to $A\oplus B$. Both set inclusions hold.

For the first set inclusion, take $x\in A\setminus (A\setminus B)$. Then $x\in A$ but $x\notin A\setminus B$. Since membership in $A$ is already known, failure to lie in that relative complement forces $x\in B$. Hence $x\in A\cap B$.

Conversely, if $x\in A\cap B$, then $x\in A$ and its membership in $B$ excludes it from $A\setminus B$. Therefore $x\in A\setminus (A\setminus B)$. The two inclusions prove the stated equality with the set intersection.

Under the negation of the desired disjunction, derive both negated propositions and contradict the premise. The natural deduction derivation isThe proof uses disjunction introduction, conjunction introduction and the classical RAA rule; it does not silently replace the required derivation by a truth-table check.