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The error $e(f)=f^{\prime \prime }(0)-f(-1)+2f(0)-f(1)$ is linear and annihilates all degree-at-most-two polynomials. The Peano kernel theorem with derivative order three givesFor $t\ne 0$, the second derivative of the truncated quadratic at $x=0$ is $1_{\{t<0\}}$. Its values at $x=-1,0,1$ then giveThe value at the single point $t=0$ is irrelevant to the integral. This is the centered second-derivative Peano kernel. Its absolute integral isConsequentlyIts changing sign explains why cubic polynomials also have zero error: their third derivative is constant and ${\int }_{-1}^{1}K=0$. No fourth derivative is needed for the claimed bound.

Exactness on the polynomial basis $1,x,x^2$ requiresSolving givesThus the approximation is the unit-spacing centered second difference, $\mu (f)=f(-1)-2f(0)+f(1)$.

The set $Y$ is path-connected as the continuous image of $[1,\infty )$, hence connected. Its closure is $Y$ together with the unit circle. Indeed $g(k+\theta /(2\pi ))\to e^{i\theta }$ as $k\to \infty$, and any convergent sequence of spiral points either has bounded parameters, giving a point of $Y$, or has parameters tending to infinity along a subsequence, giving radius one. The closure of a connected set is connected: a separation of its closure would induce a separation of the dense original set, with neither open part missing that set. The closed disk is connected and meets this closure in the unit circle; their union $Y\cup \overline{D}$ is therefore connected.

To show it is not path-connected, suppose a path $\alpha :[0,1]\to Y\cup \overline{D}$ starts in $Y$ and ends in the disk. Let $u_{\ast }$ be its first contact with the disk. The disk is closed, so this first contact exists, and $u_{\ast }>0$. Before it, the path belongs to $Y$ and has a continuous uniquely determined parameterContinuity at contact forces $|\alpha (u_{\ast })|=1$, and hence $t(u)\to \infty$ as $u\uparrow u_{\ast }$. The intermediate value theorem supplies times approaching $u_{\ast }$ at which $t(u)$ is an integer, and other times at which it is an integer plus $1/2$. These times must approach contact because $t$ is bounded on every compact interval strictly before it. Along the first sequence $\alpha (u)\to 1$; along the second it tends to $-1$. This contradicts continuity at $u_{\ast }$. Thus no path joins the spiral to the disk, proving failure of path-connectedness for the spiral accumulating on a circle.

Define $F:[0,1]\to \mathbb{C}$ by $F(s)=s{e}^{2\pi i/s}$ for $s>0$ and $F(0)=0$. It is continuous away from zero, and $|F(s)-F(0)|=s\to 0$, so it is continuous at zero too. Its image is exactly $X\cup \{0\}$, using $s=1/t$. The interval is compact, so its continuous image is compact.

For any two image points $F(s_0),F(s_1)$, the explicit pathis continuous and joins them. Thus the reciprocal spiral with its endpoint is also path-connected. Infinite winding causes no discontinuity at the endpoint because its radius tends to zero.

Assume every continuous function $E\to \mathbb{R}$ is bounded. The Euclidean norm function $x\mapsto \Vert x\Vert$ is continuous, so its boundedness proves that $E$ is a bounded set.

If $E$ were not a closed set, there would be $a\in \overline{E}\setminus E$. Choose $x_j\in E$ with $\Vert x_j-a\Vert <1/j$. The function $f(x)=1/\Vert x-a\Vert$ is continuous everywhere on $E$, because its denominator is nonzero there. Yet $f(x_j)>j$, contradicting the hypothesis. Thus $E$ is closed. Together with the two previous implications this proves all three conditions are equivalent. The empty set satisfies all three conditions trivially. This is the Euclidean case of the bounded-continuous-function characterization of compact metric spaces.

Assume the subsequence property, and let $f:E\to \mathbb{R}$ be a continuous function. If $f$ were unbounded, for each positive integer $j$ we could choose $x_j\in E$ with $|f(x_j)|>j$. There is a subsequence $x_{j_k}\to x$ with $x\in E$. By continuity, $f(x_{j_k})\to f(x)$, so that real sequence is bounded. But $|f(x_{j_k})|>j_k\to \infty$, a contradiction. Hence every continuous real-valued function on $E$ is bounded. This is the bounded-function consequence of sequential compactness.

We prove the equivalence by a cycle of implications. First assume closedness and boundedness. Given a sequence $(x_j)$ in $E$, boundedness puts each coordinate in a fixed closed bounded real interval. Apply the one-dimensional Bolzano-Weierstrass theorem to its first coordinate to extract a convergent subsequence, then apply it to the second coordinate of that subsequence, and continue through the finitely many coordinates. The final subsequence converges coordinatewise to some $x\in {\mathbb{R}}^n$. It also converges in the Euclidean norm, since the sum of the squares of its finitely many coordinate differences tends to zero. Since $E$ is a closed set, this limit belongs to $E$. Thus closedness and boundedness imply the required subsequence property. No higher-dimensional compactness theorem has been assumed.

Inversion permutes the nonzero residues excluding $-1$. Hence the result of part (c) changes the requested sum intoThere are exactly $(p-1)/2$ nonzero squares: the map $x\mapsto x^2$ on nonzero residues identifies exactly the pairs $x,-x$, because $x^2=y^2$ implies $x=\pm y$ in the finite field. Thus there are equally many nonzero squares and nonsquares, and the complete Legendre-symbol sum is zero. Removing the $j=1$ term leaves

Since $n$ is nonzero modulo $p$, it has a unique multiplicative inverse $m$ in the nonzero residues. The inverse cannot be $-1$ because that would imply $n=-1$, excluded by the range. Thus $1\le m\le p-2$ as claimed. The congruence $mn=1$ givesThe Legendre symbol of the nonzero square $n^2$ is one. Using multiplicativity of the Legendre symbol therefore gives

Apply Euler's criterion to $-1$. It givesFor $p\nmid ab$, the same criterion gives $(ab/p)\equiv (ab{)}^{(p-1)/2}=a^{(p-1)/2}{b}^{(p-1)/2}\equiv (a/p)(b/p)(\mathrm{mod}p)$. Both sides are integers in $\{1,-1\}$, which are distinct modulo an odd prime, so congruence implies equality. This proves multiplicativity of the Legendre symbol:

For a nonzero residue $a$ modulo an odd prime $p$, the Legendre symbol $(a/p)$ equals $+1$ if $a$ is a nonzero square modulo $p$ and $-1$ otherwise. The extension to multiples of $p$ has value zero. Euler's criterion states

Use the standard basis $e_1,\dots ,e_n$. For the smallest possible rank, set $\alpha e_j=e_j$ for $j\le r$ and zero otherwise, and set $\beta e_j=-e_j$ for $j\le s$ and zero otherwise. Then the maps have ranks $r,s$, while their sum has nonzero diagonal entries only for $r<j\le s$. HenceFor full rank when $r+s\ge n$, instead take $\alpha e_j=e_j$ for $j\le r$ and zero otherwise, and $\beta e_j=e_j$ for $j\ge n-s+1$ and zero otherwise. Their supports cover every basis index because $n-s+1\le r+1$. The sum has diagonal entries one where just one map is nonzero and two where both are nonzero. None vanish over $\mathbb{R}$, so these linear maps have ranks $r,s$ and

Apply the preceding rank bounds for a sum of linear maps to $\alpha =(\alpha +\beta )+(-\beta )$. Since multiplying a linear map by $-1$ does not change its image, $\mathrm{rank}\alpha \le \mathrm{rank}(\alpha +\beta )+\mathrm{rank}\beta$. Thus $\mathrm{rank}(\alpha +\beta )\ge \mathrm{rank}\alpha -\mathrm{rank}\beta$. Interchanging the maps gives the other difference, proving

For every $v\in V$, $(\alpha +\beta )v=\alpha v+\beta v$, so $\mathrm{im}(\alpha +\beta )\subseteq \mathrm{im}\alpha +\mathrm{im}\beta$. Taking dimensions and applying the preceding dimension formula for a sum of subspaces givesThis proves the upper rank bounds for a sum of linear maps, without assuming the two images have zero intersection.

Choose a basis $e_1,\dots ,e_k$ for $U\cap U^{\prime }$. Extend it to a basis $e_1,\dots ,e_k,u_1,\dots ,u_r$ of $U$, and separately to a basis $e_1,\dots ,e_k,v_1,\dots ,v_s$ of $U^{\prime }$. The combined list spans $U+U^{\prime }$.

To prove linear independence, suppose a linear combination of the combined list is zero. Then the combination of the $v_j$ lies in $U\cap U^{\prime }$, so it can be expressed using the $e_i$. Independence of the chosen basis of $U^{\prime }$ forces every $v_j$ coefficient to vanish. Independence of the chosen basis of $U$ then forces the remaining coefficients to vanish. Thus the combined list is a basis, with $k+r+s$ elements. This proves the dimension formula for a sum of subspaces:

First use a homogeneous solution to meet the nonzero endpoint values:It satisfies $y_h^{\prime \prime }+k^2{y}_h=0$, $y_h(0)=a$ and $y_h(\pi )=b$. The Green-function representation then givesApplying the differential operator to the integral gives $f(x)$ by the defining Dirac delta property of $G$, and that integral vanishes at the endpoints. For continuous $f$ this is a classical solution; integrable $f$ also gives the corresponding weak solution. The same nonresonant homogeneous argument as in part (a) proves uniqueness.

Away from the source point, the Green function obeys the homogeneous equation and must satisfy the appropriate endpoint condition. Thus it has the form $A\sin (kx)$ to the left of $\xi$ and $B\sin (k(\pi -x))$ to the right. It must be continuous at $\xi$: otherwise its second distributional derivative would contain an unwanted derivative of a Dirac delta. Integrating across $\xi$ gives the jump condition $G_x({\xi }^{+},\xi )-G_x({\xi }^{-},\xi )=1$.

The left and right homogeneous solutions $u=\sin (kx)$, $v=\sin (k(\pi -x))$ have constant WronskianThis is nonzero for real noninteger $k$. Dividing the product of the left and right solutions by that Wronskian therefore gives the Dirichlet Helmholtz Green function on an intervalThe formula vanishes at both endpoints, is continuous at the source and has derivative jump one, verifying all distributional and boundary conditions. A difference of two such Green functions would be a homogeneous Dirichlet solution; it vanishes because $\sin (k\pi )\ne 0$, proving uniqueness. Although zero is excluded by the stated noninteger condition, the removable $k\to 0$ limit is $-\min (x,\xi )(\pi -\max (x,\xi ))/\pi$.

The same recurrence matrix hasTherefore ${\mathbf{v}}_{n+5}\equiv 3{\mathbf{v}}_n(\mathrm{mod}5)$, and in particular $F_{n+5}\equiv 3F_n(\mathrm{mod}5)$. Starting from $F_0=0$ and iterating,Consequently every nonnegative index divisible by five has a Fibonacci number divisible by five, including index zero. This directly proves the requested divisibility without assuming a general Fibonacci divisibility theorem.

Put ${\mathbf{v}}_n=(F_{n+1},F_n{)}^T$. The Fibonacci recurrence matrix modulo an integer is $A=\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$, so ${\mathbf{v}}_{n+r}=A^r{\mathbf{v}}_n$. Direct matrix multiplication givesTaking the second coordinate provesFor explicit arithmetic behind the second power, squaring $A^2=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$ gives $A^4=\left(\begin{array}{cc}5& 3\\ 3& 2\end{array}\right)$, whose square is the displayed $A^8$. Thus the modular period assertions follow from calculations and the recurrence, not from quoting a period theorem.

Write $F_n=x_n$. For $n\ge 1$, the recurrence and the previously proved greatest common divisor invariance giveRepeating reaches $\gcd (F_1,F_0)=\gcd (1,0)=1$. This also gives the $n=0$ case directly. Applying the same invariance once more,Thus both requested greatest common divisors equal one for every $n\ge 0$. This is the adjacent-index instance of Fibonacci gcd reduction, derived directly rather than assuming the stronger general formula.