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The finite form of Taylor's theorem with the first $n$ terms iswhere $\xi$ lies strictly between $a$ and $x$ for $x\ne a$. A sufficient hypothesis is that $f$ has continuous derivatives through order $n-1$ on the interval joining the endpoints and an $n$th derivative in its interior. The displayed polynomial has degree at most $n-1$; its Taylor remainder uses the $n$th derivative, not the $(n+1)$st. At $x=a$ the remainder is zero. This is a finite Taylor polynomial identity and does not require an infinite Taylor series to converge.

For $f$ sufficiently differentiable on the interval between $a$ and $x$, Taylor's theorem with its first $n$ terms readswhere $\xi$ lies between $a$ and $x$. This is the Taylor theorem with Lagrange remainder. For example, $|f^{(n)}|\le M$ on that interval gives $|R_n(x)|\le M|x-a{|}^n/n!$. The finite formula does not assume that an infinite Taylor series converges to the function; that further conclusion requires the remainders to tend to zero.

Taylor's theorem gives a finite Taylor polynomial of degree $n$, consisting of the first $n+1$ terms, and a remainder:In the Taylor theorem with Lagrange remainder form, assuming the required derivatives exist on an interval containing $a$ and $x$,for some $\xi$ strictly between $a$ and $x$ when $x\ne a$; at $x=a$ the remainder is zero. In particular, if $|f^{(n+1)}|\le M$ on that segment, then $|R_{n+1}(x)|\le M|x-a{|}^{n+1}/(n+1)!$. This is a finite expansion; it does not assume equality with an infinite Taylor series.

Work on a fixed finite interval $[a,a+T]$. Write $Y_n=y(x_n)$, $t=x_n+h/2$, and use the paper's normalized local truncation errorAssume $y$ has bounded continuous derivatives through order three on this interval, with $\Vert y^{\prime \prime }\Vert \le M_2$ and $\Vert y^{\prime \prime \prime }\Vert \le M_3$. Assume also that the Jacobian matrix $f_y$ has operator norm at most $L$ on the segments joining $y(t)$ to the endpoint averages. For a scalar equation this last hypothesis is simply $|\partial f/\partial y|\le L$ there.

Two symmetric estimates separate the errors. By Taylor's theorem applied to $y^{\prime }$ about $t$, the linear term integrates to zero on $[-h/2,h/2]$, givingLikewise the two linear terms in the endpoint expansions cancel, and their second-order remainders giveThe derivative bound on $f$ gives a Lipschitz condition in its second variable. Since $y^{\prime }(t)=f(t,y(t))$, the triangle inequality now yieldsThe constant is independent of $n$ and $h$ as long as the steps remain in the fixed interval. This implicit midpoint consistency bound proves numerical consistency of order two. The corresponding unnormalized residual is $h{\tau }_n=O(h^3)$; the two powers conventionally called local truncation error differ because the paper divides by the step size. If local error instead means the error of an exactly solved numerical step started at $Y_n$, the small-step Lipschitz condition gives the bound $h\Vert {\tau }_n\Vert /(1-hL/2)=O(h^3)$ as well.

By the chain rule, $y^{\prime \prime }=f_x+f_{y}f$ along the exact solution. Applying Taylor's theorem to the exact increment and the two arguments of $f$ givesThe assumed differentiability makes the remainders uniform on $0\le x\le 1-h$. Subtracting gives the consistency errorHere $O$ denotes Big O notation. The general order of a numerical method is therefore one for $\beta \ne 1/2$ and at least two for $\beta =1/2$. For example, on $y^{\prime }=y$ with unit initial value, $y_N=(1+h+\beta h^2{)}^{1/h}$. Expanding its logarithm gives $y_N=e[1+(\beta -1/2)h+O(h^2)]$, so the first-order term is actually nonzero when $\beta \ne 1/2$.

To control accumulation, the mean value theorem and the bound on $f_y$ givePut $K_h=L(1+\beta hL)$, $q_h=1+h{K}_h$ and $e_n=y(x_n)-y_n$. The exact and numerical updates yield $|e_{n+1}|\le q_h|e_n|+h|T_n|$, with $e_0=0$. Iterating this recurrence, as in the Lipschitz one-step error bound, givesBecause $K_h\le L(1+L)$ for $h\le 1$, the prefactor is bounded independently of the mesh. When $\beta =1/2$, the consistency error is $O(h^2)$, hence the maximum global error is $O(h^2)$. This proves at least second-order numerical convergence of the explicit midpoint method.

Fix the following normalization of the local truncation error, also called the consistency error: insert a smooth exact solution into the linear multistep method and divide the residual by $h$,Thus the unscaled defect is $h{T}_n(h)$. Consistency of a numerical method means that $T_n(h)\to 0$ uniformly over mesh points in each fixed interval for every sufficiently smooth exact solution. Write the characteristic polynomials of a linear multistep method asBy Taylor's theorem, consistency of a numerical method is equivalent toor $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)$.

Second-order accuracy means $T_n(h)=O(h^2)$ for smooth solutions, equivalently an unscaled local truncation error $O(h^3)$. In addition to the two consistency conditions, its order conditions requireAn order of a numerical method of exactly two additionally requires that the next order condition fails. These statements concern the local defect; a second-order global error requires stability and suitably accurate starting values as well.

The coefficients of this linear multistep method are ${\alpha }_0=b$, ${\alpha }_1=-a$, ${\alpha }_2=1$ and ${\beta }_0={\beta }_1=0$, ${\beta }_2=c$. Substitution into the order of a numerical method conditions givesThe second equation gives $c=2-a$; substituting this into the third gives $-2+3a/2=0$. Consequently there is exactly one second-order choice,The resulting scheme is the BDF2 method,For a $C^4$ exact solution, Taylor's theorem in the original normalization gives the exact leading order defectIn particular the coefficient $-2/9$ is nonzero: the method has formal order exactly two, rather than three. Bounded third derivatives already suffice for the uniform estimate $T_n(h)=O(h^2)$; the stronger smoothness above only makes the displayed remainder immediate. Multiplying the recurrence by $3/2$ instead gives the common derivative normalization with leading defect $-h^2{y}^{\prime \prime \prime }(x_n)/3$.

The first of the characteristic polynomials of a linear multistep method isBoth roots are simple and lie in the closed unit disk, so the root condition for a multistep method proves zero-stability.

The Dahlquist equivalence theorem states that, for fixed-coefficient linear multistep methods on the usual class of initial value problems with a right-hand side continuous in $x$ and uniformly Lipschitz continuous in $y$, convergence of a numerical method for every such problem and every convergent set of starting values is equivalent to consistency of a numerical method and zero-stability. The statement is on fixed finite intervals where the exact solution exists, with implicit updates well defined. Its order estimate says that a zero-stable method with $T_n(h)=O(h^p)$ and starting errors $O(h^p)$ has global error $O(h^p)$ under the corresponding solution smoothness and Lipschitz continuity assumptions; merely convergent starting values do not ensure the order-$p$ rate.

For this application, sufficient assumptions are a unique solution $y\in C^3([x_0,x_0+T])$, a uniform Lipschitz continuity bound $L$ in $y$ on a neighborhood of its graph, and the nearby solution branch of each implicit update. For sufficiently small $h$, the mapis a contraction mapping with factor at most $2hL/3<1$ on a suitable small closed neighborhood. The Banach fixed-point theorem supplies local solvability of the nearby update. With starting errorsthe order estimate givesThus the BDF2 method is second-order convergent under these standard hypotheses. Exact $Y_0=y(x_0)$ together with one Euler method step for $Y_1$ already gives the required starting error when $y^{\prime \prime }$ is bounded, since only one starting step is taken.

The question's assumption of continuity of $f$ alone does not establish uniqueness, the needed solution smoothness, or this uniform convergence estimate. The added Lipschitz continuity condition is a sufficient hypothesis for the theorem used here; its failure does not imply that every particular initial value problem is nonunique or that its numerical approximations cannot converge.

On the vector space of nodal values with zero endpoints define the discrete Sobolev normIt is a norm because it is the Euclidean norm of the scaled list of nodal values and first finite differences, and vanishes only for the zero list. It is the discrete counterpart of the Sobolev space $H^1$ norm. Here $W=U-u$ has zero endpoints, so this definition applies directly.

Set $M_4={\max }_{0\le x\le 1}|u^{(4)}(x)|$, finite by the assumed continuity. Taylor's theorem with fourth-order remainders gives, for suitable ${\xi }_{i,+}\in (x_i,x_i+h)$ and ${\xi }_{i,-}\in (x_i-h,x_i)$,Adding the two expansions cancels the odd derivatives, soConsequently $\Vert \phi {\Vert }_h^2\le h(N-1)M_4^2{h}^4/144\le M_4^2{h}^4/144$. Doubling the bound in (c) gives $2\Vert D_x^{-}e]{\Vert }_h^2+\Vert e{\Vert }_h^2\le \Vert \phi {\Vert }_h^2$, and thereforeThus $M_4/12$ is a valid nonnegative constant. To satisfy the request for a strictly positive constant even when $u^{(4)}\equiv 0$, take $C=\max \{1,M_4\}/12$, independent of $h$. This proves second-order convergence of a numerical method in a discrete Sobolev norm, including control of the backward difference operator applied to the error.

Set $z=\pi h/2$, so $0<z\le \pi /4$. The formulas in (a) and (b) giveFor $0\le t\le \pi /4$, $0\le \sin t\le t$ and $1-\cos t=2{\sin }^2(t/2)\le t^2/2$. ThereforeThese trigonometric inequalities implySubstitution yields the explicit eigenvalue error boundThis constant is asymptotically sharp: Taylor's theorem gives ${\sin }^{2}z=z^2-z^4/3+O(z^6)$, so $({\lambda }_{1,1}-{\Lambda }_{1,1})/h^2\to {\pi }^4/6$. Here $O$ denotes Big O notation.

Write $g(v)=-\cos (\pi v/2)$. For the stipulated $0<\delta <1$, the displayed tent is continuous at its junction, has zero endpoint values, and its slopes are $\delta$ and $\delta -4$. At both slopes $g^{\prime }$ has the same value $c=(\pi /2)\sin (\pi \delta /2)>0$, and $g^{\prime \prime }=({\pi }^2/4)\cos (\pi v/2)$ is strictly positive on common sufficiently small neighborhoods of them. Taylor's theorem gives, for small $\Vert {\varphi }^{\prime }{\Vert }_{\infty }$,Thus $u^{\delta }$ is a strict weak local minimizer of a variational integral. It is not a strong local minimizer of a variational integral: the global supporting-tangent requirement at either slope has nonzero slope $c$, whereas $g$ is bounded. Letting the trial slope tend to positive infinity violates that requirement. The Weierstrass necessary condition therefore fails.

The opening parameter range $\delta \in (0,1)$ is essential. The subsequent phrase “for each $\delta >0$” must be read within it; taken literally beyond that range it is false. At $\delta =2$ the tent slopes are $2,-2$, where $g=1$ and $g^{\prime \prime }<0$, so this is not a weak minimum.

Apply the superposition principle to the result of part (a). If $F(a,h)=-qh/[2\pi {ϵ}_0(a^2+h^2{)}^{3/2}]$, then the exact normal electric field from the two real point charges and their image charges isFor $d\ll h$, Taylor's theorem gives $E_z=-d{\partial }_{h}F+O(d^2)$. Sincethe leading electric dipole result isThe electric dipole moment is $\mathbf{p}=-qd\hat{\mathbf{z}}$. The circle $a=\sqrt{2}h$ is the zero of the leading normal electric field and hence of the induced surface charge density. Inside it, induced surface charge density is negative; outside it, positive. The tangential electric field is zero everywhere on the grounded electrical conductor, so the leading full electric field vanishes on that circle. The corresponding dipole-induced charge on a grounded plane integrates to zero over the whole plane: the negative central region is balanced by the positive exterior.

For finite $d$, $a=\sqrt{2}h$ is only the leading-order zero. With $u=h^{2/3}$ and $v=(h+d{)}^{2/3}$, the exact zero satisfies $a^2=uv(u+v)$, giving $a=\sqrt{2}(h+d/2)+O(d^2/h)$. For example, inserting any $d>0$ in the exact expression at $a=\sqrt{2}h$ does not give zero. This is a controlled electric dipole approximation, not an exact finite-separation identity.

Write $A_k=N_{2k}-N_k$ and $B_k=N_k$. The stationary increments and independent increments of the Poisson process make $A_k$ and $B_k$ independent random variables, each with Poisson distribution of parameter $k$. ThusMultiplication of the two moment-generating functions givesThe odd terms cancel in Taylor's theorem, so $\log M_k(t)=t^2+O(k^{-1})$. The limit $e^{t^2}$ is the moment-generating function of a normal distribution with mean zero and variance two. The same continuity theorem for moment-generating functions therefore yieldsIn particular, the two equal-length increments must be separated before using independence: $N_{2k}$ and $N_k$ themselves are dependent.

For fixed real $t$, Taylor's theorem for the exponential function givesSubstituting into the previous moment-generating function,The continuity theorem for moment-generating functions states that, if the moment-generating functions of $Z_n$ and $Z$ are finite on a common open interval containing zero and $M_{Z_n}(t)\to M_Z(t)$ throughout that interval, then $Z_n\stackrel{\mathrm{d}}{\to }Z$. Here finiteness holds for all real $t$, and the limit is the moment-generating function of a standard normal variable. ThereforeThis proves the required instance of the Poisson central limit theorem directly from moment-generating functions; no appeal to a general central limit theorem is needed.

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.

One standard finite form of Taylor's theorem is the following. Suppose $f$ has continuous derivatives through order $m+1$ on an interval containing $a$ and $x$. Then, if $x\ne a$, there is some $\xi$ strictly between them such thatThe final term is the Lagrange remainder. At $x=a$ the identity holds directly with zero remainder. The finite expansion requires no assumption that an infinite Taylor series converges to $f$.

For $0<x<\pi$, apply Taylor's theorem to the sine about zero, retaining terms through degree four. Its fourth-order coefficient is zero, and its fifth derivative is the cosine function, givingThe cosine function is decreasing on $[0,\pi ]$, so $\cos x\le \cos \xi \le 1$. Multiplying by the nonnegative factor $x^5/120$ and adding $x-x^3/6$ proves both required bounds. At $x=0$ both are equalities.

The second-order remainder form of Taylor's theorem gives, for a point ${\xi }_x$ between $0$ and $x$,For $x\ne 0$, the given curvature bounds implyIn particular $f(1)<-1/2<0$ and $f(2)\ge 5>0$. Differentiability implies continuity, so the intermediate value theorem gives a root in $(1,2)$.

The mean value theorem applied to $f^{\prime }$ gives $f^{\prime }(x)=3+x{f}^{\prime \prime }({\eta }_x)\ge 3+2x>0$ for $x\in (1,2)$. Thus $f$ is strictly increasing there, and it cannot have two roots. There is exactly one root in $(1,2)$.

Apply the mean value theorem, the first-order case of Taylor's theorem, to $f^{\prime }$ between $0$ and $x>1$. It yieldsAlso $f(3/2)\ge -5+9/2+9/4=7/4>0$, so the initial point is to the right of the root $x^{\ast }$. For an increasing strictly convex function, its tangent at a point to the right of a root meets the axis between that root and the point. Hence the exact Newton root-finding iteration remains in $[x^{\ast },3/2]$ and decreases towards $x^{\ast }$.

On every relevant interval, $|f^{\prime \prime }|<3$ and $f^{\prime }>5$, so the stated quadratic convergence bound gives, with $e_n=|x_n-x^{\ast }|$,Iterating the bound givesThus four Newton corrections suffice to reduce the exact-iteration error bound below the floating-point spacing scale. The quadratic convergence doubles the approximate number of correct digits at each late step. Beyond this point further corrections cannot reliably improve a generic floating-point answer. This is a bound on useful accuracy under the usual stable-evaluation model, not a claim that the data specify the exact iteration at which every implementation must stagnate.

For negative $x$, the mean value theorem gives $f^{\prime }(x)=3+x{f}^{\prime \prime }({\xi }_x)$. In particular,Since $f^{\prime }$ is continuous and strictly increasing, it vanishes at exactly one point $x_m\in (-2,-1)$. The positive second derivative makes $f$ a strictly convex function on the entire real line. Its derivative changes from negative to positive at $x_m$, so $x_m$ is the unique global minimum. One may also use $f(x)\ge x^2+3x-5\to \infty$ to verify coercivity.

In the endpoint-bracketing form of golden-section search, begin with $[-2,-1]$, a width-one interval known to contain this unique minimum, and place the two standard interior probes. The attainable position accuracy is different from that for a simple root. Around $x_m$, Taylor's theorem givesHere the minimum value is nonzero and of order one: the lower quadratic bound gives $f(x_m)\ge -29/4$, while $f(x_m)\le f(-1)<-13/2$. Ordinary relative round-off error in these function values therefore hides changes of order $ϵ$, once $h^2$ has that size. The function-value resolution near a nondegenerate minimum is consequently of orderEach golden-section search contraction multiplies the bracket width by $\rho =(\sqrt{5}-1)/2$. Using $2^{-20}$ as the nominal target width givesUnknown evaluation-error constants can change the actual stopping point; $29$ is the standard scale-based count. Requiring $2^{-40}$ position accuracy from these function-value comparisons would disregard the quadratic flatness of the minimum. The same precision effect is documented for one-dimensional numerical minimization.^[ref]

By Taylor's theorem with Lagrange remainder, for an increment $t$,where ${\xi }_t$ lies between $x$ and $x+t$. The given third derivative bound implies $|R_x(t)|\le |t{|}^3/6$.