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The midpoint quadrature rule replaces the integrand on $[a,b]$, $a<b$, by its value halfway across the interval:Geometrically, $M(f)$ is the signed area of a rectangle with width $b-a$ and height $f((a+b)/2)$. Its single quadrature node is $(a+b)/2$ and its quadrature weight is $b-a$. Sampling in the interior is useful when endpoint evaluation is inconvenient, although an endpoint singularity can invalidate the quadrature error estimates below.

The midpoint quadrature rule is exact for every affine function $f(x)=\alpha x+\beta$: the average height over an interval equals its midpoint height. For example, on $[0,2]$, $f(x)=2x+1$ gives $I(f)=6$ and $M(f)=2f(1)=6$. Positive and negative area discrepancies cancel for a sloping straight line. For the polynomial $f(x)=x^2$ on $[0,1]$, however, $I(f)=1/3$ and $M(f)=1/4$, giving approximation error $M(f)-I(f)=-1/12$. Thus its degree of precision of a quadrature rule is exactly one.^[ref]

For the polynomial $f(x)=x^2$ on $[0,1]$, four panels give composite midpoint quadrature rule value $21/64$ and composite trapezoidal rule value $11/32$, bracketing the exact integral $1/3$. Individual midpoint rectangles can cross the curve; it is their total signed area that underestimates the integral.

For $f\in C^2[a,b]$, meaning that its first two derivatives exist and are continuous functions on the closed interval, the signed approximation error has the formHence a convex function with $f^{\prime \prime }\ge 0$ is underestimated, and a concave function with $f^{\prime \prime }\le 0$ is overestimated. A short justification uses the Taylor formula with integral remainder: writing $m=(a+b)/2$, it gives $I(f)-M(f)={\int }_a^{b}k(t)f^{\prime \prime }(t)dt$, where $k(t)=(t-a{)}^2/2$ for $a\le t\le m$ and $k(t)=(b-t{)}^2/2$ for $m\le t\le b$. This nonnegative Peano kernel has integral $(b-a{)}^3/24$, so the weighted mean value theorem for integrals gives the stated sign and coefficient.

For a function with continuous derivatives through order $m+1$ on the interval between $0$ and $x$, the Taylor formula with integral remainder isThis is Taylor's theorem about zero; translation gives the formula about any other point. To prove it, the case $m=0$ is the fundamental theorem of calculus. For $m\ge 1$, integration by parts givesSubstitution into the formula at order $m-1$ proves the formula at order $m$ by induction. The integral estimate, valid also for negative $x$, isFor the smooth function satisfying $f^{(3)}=f$, each fixed derivative is continuous on $[-R,R]$ and hence is bounded there by the extreme value theorem. Thus a bound $M_j$ exists for every $j$. Differentiating the differential equation repeatedly gives $f^{(j+3)}=f^{(j)}$. Only the three functions $f,f^{\prime },f^{\prime \prime }$ are needed: take $M$ to be the maximum of their absolute-value bounds on $[-R,R]$. Then $|f^{(j)}(x)|\le M$ simultaneously for every $j\ge 0$ and $|x|\le R$.

At zero the derivatives consequently repeat the values $1,0,0$. Taylor's theorem therefore givesThe last limit of a sequence follows because the ratio of consecutive majorants is $R/(m+2)$, eventually less than $1/2$. Since $R$ is arbitrary, Taylor expansion from a periodic derivative equation provesThis proves convergence to the function using a remainder bound; it does not assume beforehand that a smooth function equals its Taylor series.

The chain rule along the specified straight line givesHere equality of mixed partial derivatives follows from smoothness. Apply Taylor theorem in the single variable $t$:Writing $h_1=x_1-a=\alpha t$ and $h_2=x_2-b=\beta t$ produces the linear and quadratic termswith all displayed partial derivatives evaluated at $(a,b)$. For a remainder valid uniformly over directions, apply the Taylor formula with integral remainder to $s\mapsto f((a,b)+sh)$:Continuity of the Hessian matrix makes the difference between the integral and $\frac{1}{2}{h}^{T}H(a,b)h$ be $o(|h{|}^2)$. The printed ellipsis is therefore justified as a finite Taylor expansion with a remainder. Smoothness alone does not assert convergence of an infinite Taylor series.

At a stationary point, the linear term vanishes. The symmetric Hessian matrix satisfiesIf $H_{11}>0$ and $\det H>0$, this is a positive-definite quadratic form. Its minimum over the unit circle is a positive number $c$, so $h^{T}Hh\ge c|h{|}^2$. The Taylor remainder is smaller than $c|h{|}^2/4$ for sufficiently small $h$, proving $f((a,b)+h)>f(a,b)$ for nonzero nearby $h$. Thus the stationary point is a strict local minimum.

For the given polynomial, the gradient equations reduce toThe stationary points are $(0,0),(1,-1),(-1,1)$, andAt both points with $x_1=\pm 1$, we have $H_{11}=10>0$ and $\det H=16>0$. ConsequentlyAt the origin the determinant of the Hessian matrix is $-8$, giving a saddle point. An independent check is the identity $f=(x_1^2-1{)}^2+(x_1+x_2{)}^2-1$, which proves that the two displayed local minima are also the only global minima.

For real functions $u,v$ that are continuously differentiable on a closed interval $[a,b]$, integration by parts statesIt follows from the product rule and the fundamental theorem of calculus. With the convention ${\int }_a^b=-{\int }_b^a$, the identity also applies to reversed limits. These regularity assumptions ensure that all the proper integrals exist.

For the smooth $f$ in the problem, define the integral remainder in Taylor theorem byThe fundamental theorem of calculus gives $f(t)=f(0)+R_1(t)$. Applying integration by parts to $R_n$, with $u=f^{(n)}(x)$ and antiderivative $v=-(t-x{)}^n/n!$ of $(t-x{)}^{n-1}/(n-1)!$, givesInduction starting from $n=1$ therefore provesThe same calculation holds for negative $t$: all integrals are oriented, and the segment joining $0$ and $t$ lies in $(-1,1)$. Thus the proof covers every stated $t$ and every positive integer $n$.

Now take $f(x)=\log (1-x)$. For $k\ge 1$, differentiation gives $f^{(k)}(x)=-(k-1)!/(1-x{)}^k$, and $f(0)=0$. With $n=N+1$ and $t=1/2$, the Taylor formula with integral remainder becomesFor $0\le x\le 1/2$, $0\le (1/2-x)/(1-x)\le 1/2$ and $(1-x{)}^{-1}\le 2$. Hence $0\le J_N\le 2^{-N}$ after integrating over an interval of length $1/2$. Thus the logarithmic series from an integral Taylor remainder satisfiesThe explicit vanishing remainder establishes convergence and identifies the sum, without using a pre-existing logarithmic power series.

For the first derivative, $f^{\prime }(x)=\frac{1}{2}(1+x{)}^{-1/2}$, which is the stated formula at $r=1$. If the formula holds at $r$, differentiating multiplies its coefficient by $\frac{1}{2}-r=-(2r-1)/2$. The factorial coefficients satisfyThe sign changes, proving by mathematical induction that

Apply the given Taylor formula with integral remainder at $x=1/2$. Its polynomial part is $1+2S_n$, where $S_n$ is the first $n$ terms of the requested series, sinceTo justify passage to the infinite series, for $0\le t\le 1/2$,Therefore the integral remainder obeysThe last inequality follows from $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)\le {\sum }_{j=0}^{2n}\left(\genfrac{}{}{0ex}{}{2n}{j}\right)=4^n$, a direct binomial theorem bound. Hence $\sqrt{3/2}=1+2\lim S_n$, and

The Taylor formula with integral remainder isFor $n=0$, this is the fundamental theorem of calculus. Integration by parts givesRearranging and applying mathematical induction proves the displayed expansion. The oriented integral also covers $x<a$.

For $x\ne 0$, the chain rule and product rule giveThusBecause Gaussian decay dominates every power, $p(1/x)e^{-1/x^2}/x\to 0$ as $x\to 0$. Hence $f^{\prime }(0)=0$, and the derivative formula extends continuously to zero. Repeating the argument replaces $p$ by another polynomial at each step: $p_{j+1}(t)=2t^3{p}_j(t)-t^2{p}_j^{\prime }(t)$. All derivative difference quotients at zero tend to zero, so the polynomial multiples of a Gaussian flat function are smooth functions withThe Taylor polynomial about zero is therefore identically zero for every degree, andA nonzero polynomial has only finitely many roots of a polynomial, so $p(1/x)\ne 0$ for every sufficiently small nonzero $x$. At each such fixed $x$, this remainder never tends to zero. The flat function is therefore smooth but not real analytic at zero.

Use the metric on $M$ induced by its given Euclidean embedding. For two smooth maps define the C1 metric on a smooth mapping space bywhere the differential has domain $T_{x}M$ with the induced inner product. Compactness makes both suprema finite. The norm properties give symmetry and the triangle inequality, and the first term makes distance zero imply equality of maps. Thus convergence in $d_1$ is uniform convergence of both maps and differentials.

Because $i$ is an immersion, $D{i}_x$ is injective at every point. The continuous function $|D{i}_{x}v|$ on the compact unit tangent bundle has a positive minimum $a$. ThereforeFor all sufficiently large $n$, ${\sup }_x\Vert D{f}_{n,x}-D{i}_x\Vert <a/2$, and thenConsequently every sufficiently late $f_n$ is an immersion. If $M$ has dimension zero it is a finite set, the immersion assertion is automatic, and eventual injectivity follows directly from separation of its finitely many image points; we henceforth treat positive dimension.

For injectivity, choose finitely many coordinate maps ${\varphi }_b$ on convex balls, with smaller coordinate neighborhoods covering $M$ and closures lying inside the larger balls. Write $I_b=i\circ {\varphi }_b$ and $F_{n,b}=f_n\circ {\varphi }_b$. On the finitely many relevant compact coordinate closures, the differentials of the $I_b$ have a common positive lower bound $s$, and convergence in $d_1$ givesfor all late $n$. The assumed uniform second-derivative bound gives a common bound $L$ for $D^2{F}_{n,b}$ on those larger coordinate neighborhoods. If the bound is expressed using an intrinsic Hessian, the coordinate chain rule also involves first derivatives and derivatives of the fixed charts; their bounds are uniform by C1 convergence and compactness. Thus $L$ can be larger than the printed constant $C$, but is independent of $n$.

For $u,v$ whose segment lies in one such coordinate ball, Taylor formula with integral remainder givesChoose $\delta >0$ with $L\delta \le s/2$. Then, whenever $0<|u-v|<\delta$,The Lebesgue number lemma applied to the finite smaller-chart cover, together with uniform continuity of its coordinate inverses on the relevant compact sets, gives $\rho >0$ such that every pair $x,y\in M$ with $0<|x-y|<\rho$ lies in one such chart with coordinate distance below $\delta$. The displayed estimate therefore excludes all sufficiently close pairs of equal images, uniformly in late $n$.

The remaining pairs form the compact setIf it is nonempty, injectivity of the smooth embedding $i$ gives a positive minimum $b$ of $|i(x)-i(y)|$ on it. Uniform convergence makes ${\sup }_x|f_n(x)-i(x)|<b/3$ for all sufficiently large $n$, and hence on $K_{\rho }$,If $K_{\rho }$ is empty there is nothing left to check. Combining the near-pair and distant-pair bounds makes $f_n$ injective for every sufficiently late $n$. It is already an immersion, and a continuous injection from compact $M$ into a Hausdorff space is a homeomorphism onto its image. By compact injective immersion is an embedding,This proves the requested stability of compact smooth embeddings using the stated second-derivative assumption. That assumption supplies a convenient Taylor constant; the stronger C1 openness result can replace it by the uniform continuity of the limiting differential on finitely many small coordinate balls.

The Peano kernel theorem says that if a continuous linear error functional $L$ annihilates all polynomials of degree less than $r$ and permits the required interchange with integration, then for $f\in C^r[a,b]$,It follows by applying $L$ to the integral-remainder Taylor formula with integral remainder. Here choose $r=2$ andThis functional annihilates constants and linear functions (indeed quadratics too), and its Peano kernel isThus ${\int }_0^3|K|d\theta =1/2$ and $\max |K|=1/3$. The integral representation givesFor the first inequality, continuous functions $f^{\prime \prime }$ with $\Vert f^{\prime \prime }{\Vert }_{\infty }\le 1$ can approximate $\mathrm{sgn}K$ away from its zeros, making $L(f)$ arbitrarily close to $\int |K|$. Such derivatives integrate twice to admissible $C^2$ functions, proving no smaller $\alpha$ works. For the second, use continuous $f^{\prime \prime }$ of $L^1$ norm one concentrated near $\theta =1$ or $2$, where $|K|=1/3$, with the sign of $K$ there. This makes $|L(f)|$ approach $1/3$, proving optimality of $\beta$.

For $f=x^3$, $L(f)=2$, $\max |f^{\prime \prime }|=18$ and $\Vert f^{\prime \prime }{\Vert }_1=27$. Both right-hand bounds equal nine, so both inequalities hold.

A kernel for density estimation is an integrable function $K$ with $\int K=1$. With $K_h(u)=h^{-1}K(u/h)$, the kernel density estimator isA kernel of order ell satisfies

Split the integrated $L^p$ risk into stochastic and bias terms:For fixed $x$, put $A_i(x)=K_h(x-X_i)$ and $W_i(x)=A_i(x)-\mathbb{E}{A}_i(x)$. The supplied Rosenthal inequality givesNowAlso $\mathbb{E}{W}_1(x{)}^2\le (K_h^2\ast f)(x)$. The Young convolution inequality with exponent $p/2$ givesAfter integration and multiplication by the outer factor $2^{p-1}$, the stochastic contribution is at most

It remains to control the bias. Here $\ell =\lceil \beta \rceil =m+1$. The Taylor formula with integral remainder, the vanishing kernel moments, and Minkowski integral inequality giveThe defining Nikol'skii smoothness bound is $L|hu{|}^{\beta -m}$, soRaising to the $p$th power and applying the outer factor $2^{p-1}$ proves

For a smooth real function satisfying the displayed equation with positive integer $r$, differentiating makes $f^{(j+r)}=f^{(j)}$. On any compact interval, all derivatives are therefore bounded by the maximum of the bounds of $f,f^{\prime },\dots ,f^{(r-1)}$. The Taylor formula with integral remainder consequently has error at most $M{R}^{n+1}/(n+1)!$, which tends to zero uniformly for $|x|\le R$. Hence the function equals its globally convergent Taylor series. For $r=3$ and initial derivatives $(1,0,0)$, it is ${\sum }_{k\ge 0}{x}^{3k}/(3k)!$.