The given square wave agrees almost everywhere with an odd function on $(-\pi ,\pi )$; its assigned value at the single point $0$ does not affect any Fourier coefficient. Thus $a_0=a_n=0$, whileThe Fourier series is thereforeThe Dirichlet convergence theorem for Fourier series applies because the periodic extension is piecewise smooth. It gives $f(x)$ at each continuity point and the midpoint $0$ at the jumps, including $x=0$; it does not reproduce the assigned value $f(0)=-1$ there.

At the continuity point $x=\pi /2$, $\sin ((2r+1)\pi /2)=(-1{)}^r$ and $f(\pi /2)=1$. ThusNo evaluation at a discontinuity or rearrangement of the conditionally convergent series is involved.

Divide $[a,b]$ into $N$ equal subintervals of width $h=(b-a)/N$, with $x_j=a+jh$. Linear interpolation between adjacent values of $f$ gives a trapezoid of signed area $h(f(x_{j-1})+f(x_j))/2$. Adding these areas produces the composite trapezoidal ruleFor continuous $f$, this tends to the integral as $h\to 0$; if $f$ has a bounded continuous second derivative, the error is at most $(b-a)h^2\max |f^{\prime \prime }|/12$.

The original PDF prints $n\to \infty$ while using $n$ as the bound summation index and $N$ as the upper limit. Literally this does not specify a varying upper limit. Interpreting the intended limit as $N\to \infty$, and relabelling the summation index $j$, gives the right Riemann sumFor $f(x)=x^4$, $a=0$, and $b=1$, the only difference from the composite trapezoidal rule is the endpoint weighting: $S_N=T_N+1/(2N)$. Hence

The substitution $y=xv$ gives $y^{\prime }=v+x{v}^{\prime }$. Cancelling $x^{3}v$ on both sides of the differential equation, for $x>0$, givesThe initial condition is $v(1)=1$. This selects a positive solution locally, so separation of variables givesUsing $v(1)=1$ gives $C=-1/2$, and continuity selects the positive square root:The real solution containing $x=1$ has maximal positive domain $0<x<\sqrt{e}$. Its denominator tends to zero and $y\to +\infty$ as $x\uparrow \sqrt{e}$, a finite-time blow-up of an ordinary differential equation in the independent variable $x$. Thus the source's wording “for $x>0$” needs this domain restriction; the prescribed initial value admits no real solution on the whole positive half-line. The division by $v^3$ loses the separate solution $v=0$, but that solution does not meet the initial condition.

On an interval avoiding $x=0$, divide by $x$ to obtain the first-order linear ordinary differential equationAn integrating factor is $\mu (x)=\exp \left(\int -2/xdx\right)=x^{-2}$. Multiplying through and using the product rule givesBy integration by parts, $\int x\cos xdx=x\sin x+\cos x+C$. ConsequentlyThe formula differentiates back to the original equation. Although division excluded $x=0$, this formula has a smooth extension there with $y(0)=0$, satisfying the original equation at zero as well.

Seek a particular solution $y_p=ax+b$. Substitution into the linear ordinary differential equation gives $6ax+2b=6x+2$, so $a=b=1$. Add the complementary solution from part (a):The two prescribed values yield $A+B=5$ and $A/2+B/4=3/2$. Thus $2A+B=6$, so $A=1$ and $B=4$. ThereforeDirect substitution gives $y(1)=7$ and $y(2)=9/2$, and the homogeneous solution contributes zero to the left-hand differential operator.

Write $Y(t)=y(e^t)$, so $x=e^t>0$. The chain rule givesHence $x^2{y}^{\prime \prime }=Y^{\prime \prime }-Y^{\prime }$ and the Cauchy-Euler equation becomesIts characteristic equation is $r^2+3r+2=(r+1)(r+2)=0$. The distinct roots give $Y(t)=A{e}^{-t}+B{e}^{-2t}$. Returning to the original independent variable givesThe same form holds on any negative interval avoiding $0$, using $t=\log |x|$; the stated substitution covers the positive interval. At $x=0$ the differential equation is singular, so crossing it is a separate extension problem.

Interpret ${\mathbb{R}}_{+}$ as the strictly positive reals, as in the domain $x>0$ in question 1(e). A bound valid for every $n$ is immediately an eventual bound, by taking $m=0$ (or the least natural number under the other convention).

Conversely, suppose the eventual big O notation bound holds with constants $m,c$. Since $g(j)>0$, all the finitely many ratios $|f(j)|/|g(j)|$ with $j<m$ are finite. Chooseomitting the set of ratios when it is empty. For $n<m$ the definition of $C$ gives $|f(n)|\le C|g(n)|$, and for $n\ge m$ the original bound does so. Hence the two definitions are equivalent for the stated strictly positive codomain.

Strict positivity matters. If ${\mathbb{R}}_{+}$ were instead intended to include zero, an eventual bound need not extend to all indices: $g(0)=0$, $f(0)=1$, and $f(n)=g(n)=1$ for $n\ge 1$ give a counterexample. More generally, extension is possible exactly when each initial zero of $g$ is also a zero of $f$.

This is the pointwise order on a function space. Each of the three partial order axioms follows from the corresponding axiom in $B$.

For reflexivity, $f(x){⪯}_{B}f(x)$ for every $x$, so $f{⪯}_{\mathcal{F}}f$. For antisymmetry, if $f{⪯}_{\mathcal{F}}g$ and $g{⪯}_{\mathcal{F}}f$, then $f(x)=g(x)$ for every $x\in A$; equality of functions therefore gives $f=g$. For transitivity, if $f{⪯}_{\mathcal{F}}g{⪯}_{\mathcal{F}}h$, then $f(x){⪯}_{B}g(x){⪯}_{B}h(x)$ for every $x$, so $f{⪯}_{\mathcal{F}}h$.

Thus $(\mathcal{F},{⪯}_{\mathcal{F}})$ is a partially ordered set. The argument also covers an empty domain $A$, since then there is exactly one function $A\to B$.

A preorder is a binary relation that is reflexive and transitive: $a⪯a$, and $a⪯b⪯c$ implies $a⪯c$.

A partial order is a preorder that is also an antisymmetric relation: $a⪯b$ and $b⪯a$ together imply $a=b$.

A linear order, also called a total order, is a partial order in which every two elements are comparable: $a⪯b$ or $b⪯a$ for all $a,b\in A$.

Fix $m\ge 0$, using the convention for binomial coefficients that $\left(\genfrac{}{}{0ex}{}{k}{m}\right)=0$ when $m>k$. At $n=0$, both $\left(\genfrac{}{}{0ex}{}{0}{m}\right)$ and $\left(\genfrac{}{}{0ex}{}{1}{m+1}\right)$ equal $1$ if $m=0$ and $0$ otherwise.

For the successor step, apply the induction hypothesis, then Pascal's rule:Thus mathematical induction proves the hockey-stick identityThe zero convention also covers the case $n<m$, without a separate starting index.

For positive integer pairs, reflexivity is $ab=ba$. If $ad=bc$, reversing the equality gives $cb=da$, establishing symmetry. For transitivity, suppose $(a,b)R(c,d)$ and $(c,d)R(e,f)$. Then $ad=bc$ and $cf=de$, whenceSince $d>0$, cancellation gives $af=be$. Thus $(a,b)R(e,f)$, and $R$ is an equivalence relation.

Membership in the equivalence class of $(4,6)$ means $4b=6a$, or $2b=3a$. Since $2$ and $3$ are coprime, $a=2r$ and $b=3r$ for a positive integer $r$. Conversely every such pair satisfies the equality. ThereforeThese pairs are precisely the positive numerator-denominator representations of the rational number $2/3$.