The characteristic equation of the constant-coefficient linear ordinary differential equation iswhose roots of a polynomial are $r=-1\pm i/2$. Taking real and imaginary parts of the corresponding exponential functions gives the real general solution

The endpoint boundary conditions areThe finite difference equation can be written $y_{n+1}-2y_n+y_{n-1}=2h^2$. Set $d_n=y_n-y_{n-1}$. Then $d_{n+1}-d_n=2h^2$, so $d_n=d_1+2h^2(n-1)$. Summing from $1$ to $n$ givesUsing $y_N=5$ and $h=2/N$ yields $N{d}_1=4-h^{2}N(N-1)=4/N$, hence $d_1=4/N^2=h^2$. ThereforeThis also covers $N=1$, where there are no interior equations. Uniqueness follows because the difference of two solutions has zero second finite difference, hence is linear in $n$, and vanishes at both endpoints. This is an exact finite difference solution for constant second derivative: the method is exact at the nodes here because the centered second finite difference differentiates this quadratic polynomial exactly.

For $y=1+x^2$, the derivatives are $y^{\prime }=2x$ and $y^{\prime \prime }=2$. Its endpoint values are $y(0)=1$ and $y(2)=5$, so it satisfies both the ordinary differential equation and the boundary conditions. Moreover any solution of $y^{\prime \prime }=2$ has the form $x^2+Ax+B$; the boundary conditions force $B=1$ and $A=0$. Thus the unique solution of the boundary value problem is

Apply the multivariable chain rule to a differentiable function $f(s,t)$ on the chosen local branch of the parabolic coordinates:Substituting part (ii) gives

The determinant of the Jacobian matrix is $s^2+t^2$. Wherever $(s,t)\ne (0,0)$, the inverse function theorem permits a local inverse of the parabolic coordinates transformation, andReading off its entries givesThese are local formulas on a chosen branch: $(s,t)$ and $(-s,-t)$ have the same image, and at the origin the Jacobian matrix is singular, so this inverse-derivative formula does not apply there.

Taking the partial derivatives of the parabolic coordinates transformation, holding the other coordinate fixed each time, givesThus its Jacobian matrix is $J=\left(\begin{array}{cc}t& s\\ -s& t\end{array}\right)$.

Let $S_N=N^{-3}{\sum }_{n=1}^N{n}^2$. The expression in part (ii) givesBecause $3x^2$ is continuous on $[0,1]$, the composite trapezoidal rule converges to its definite integral, which is $1$ by part (i). In fact its standard composite trapezoidal error bound is $|T_N-1|\le 1/(2N^2)$, since $f^{\prime \prime }=6$. Therefore

For the composite trapezoidal rule, the mesh width is $h=1/N$ and the nodes are $x_n=n/N$. The endpoint weights are halves and the interior weights are ones, soThis formula is an approximation to the definite integral, not an exact equality with it for finite $N$.

By the fundamental theorem of calculus,

Put $k=\lfloor y^{1/n}\rfloor$. Because $y>0$, $k$ is a nonnegative integer, andThe function $t\mapsto t^n$ is strictly increasing for $t\ge 0$, so $k^n\le y<(k+1{)}^n$. Both powers are integers. Taking the floor function therefore givesTaking nonnegative $n$th roots now gives $k\le \lfloor y{\rfloor }^{1/n}<k+1$. ConsequentlyThe argument also covers $0<y<1$, when both outer floors are zero.

Set $k=\lfloor x\rfloor$. The definition of the floor function gives $k\le x<k+1$. Adding the integer $m$ yieldsSince $k+m$ is an integer, it is exactly the floor of $x+m$. Thus

Using $a_i=b_i+n{u}_i$ with integers $u_i$,The quantity in parentheses is an integer, so the divisibility criterion for modular arithmetic gives

By modular arithmetic, $a_1-b_1=nu$ and $a_2-b_2=nv$ for some integers $u,v$. Adding givesSince $u+v$ is an integer, this proves

We prove the stronger bound $T(n)\le 2n^2$ for every $n\ge 1$ by strong induction. The initial case is $T(1)=1\le 2$. For $n\ge 2$, put $m=\lfloor n/2\rfloor$. Then $1\le m<n$, so the induction hypothesis and the defining recurrence relation giveThusThe recurrence relation must be read as applying for $n\ge 2$; applying it at $n=1$ would invoke $T(0)$ outside the printed domain. No value of $T(0)$ is needed.

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

For each unordered pair $\{a,b\}$ with $a\ne b$, antisymmetry allows three choices: neither direction, only $(a,b)$, or only $(b,a)$. This contributes $3^{\left(\genfrac{}{}{0ex}{}{n}{2}\right)}$. The diagonal is unrestricted by antisymmetry, but being not irreflexive requires a nonempty subset of the $n$ diagonal pairs. There are $2^n-1$ such choices. HenceThis allows some diagonal pairs to be absent; “not irreflexive” does not require a reflexive relation.

A symmetric relation has $n$ independent choices on the diagonal. For each of the $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ unordered pairs $\{a,b\}$ of distinct elements, either both ordered pairs are included or neither is. Thus there are $2^{n+\left(\genfrac{}{}{0ex}{}{n}{2}\right)}$ symmetric relations, so the requested number is

There are $n^2$ possible ordered pairs and hence $2^{n^2}$ binary relations. A reflexive relation must contain the $n$ diagonal pairs, while its remaining $n^2-n$ pairs are unrestricted. Subtracting the reflexive relations givesbinary relations that are not reflexive.

For the base case $n=2$, $1+1/\sqrt{2}>\sqrt{2}$, since multiplying by $\sqrt{2}$ reduces this to $\sqrt{2}+1>2$. Suppose the claimed inequality holds for some $n\ge 2$. The identitythen givesThus mathematical induction proves

Use mathematical induction. For $n=1$, the left side is $2^0=1=2^1-1$. If the identity holds at $n$, thenThe base case and induction step establish the geometric series identity