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Open water absorbs more sunlight than bright sea ice, and melt ponds also lower the surface albedo. More absorbed heat promotes further melt and can be stored in the upper ocean, delaying autumn freezing. This ice-albedo feedback amplifies an initially forced loss during the sunlit season. Perovich and collaborators' 2008 measurements identify solar heating of the upper ocean as the main heat source for enhanced bottom melting in their observed Beaufort Sea case. Feedback is especially important locally after open water develops; it does not supply solar heat during polar winter.

Long-term thinning lowers the latent heat inventory that must be supplied before a patch of sea ice disappears. Thin ice is also more mobile and vulnerable to mechanical breakup. A smaller fraction of multi-year sea ice therefore increases susceptibility to a warm or windy summer. Thin ice grows relatively rapidly in winter because it conducts heat more readily, giving a stabilizing response that partly opposes continued thinning without cancelling sustained warming.

The Arctic Oscillation changes pressure patterns and winds, redistributing sea ice and altering export through Fram Strait. Removing thick old ice and replacing it with thinner new ice preconditions greater summer loss; winds also open leads and change ocean circulation. This is an important dynamic contributor and source of variability, rather than a universal one-to-one relation between the oscillation index and total summer ice extent.

Pacific inflow carries seasonally warm water through the Bering Strait into the western Arctic. Much of this heat lies higher in the water column than the warm Atlantic layer, making it more accessible to the ice in the Pacific sector. Its path, timing, temperature and volume transport affect basal melting and delayed freeze-up. Shimada and collaborators' 2006 observations associate warm Pacific summer water with regional ice loss; they do not imply that it explains all Arctic loss.

Erosion or displacement of the cold halocline weakens the barrier separating sea ice from warmer water below. It can therefore increase the upward ocean heat flux without requiring a large new warming of that deeper water. Wind-driven mixing, changes in shelf-water supply, and freshwater redistribution can alter the halocline. This mechanism couples the Atlantic heat reservoir to atmospheric and circulation changes.

Warmer Atlantic water increases the reservoir of heat beneath the Arctic surface layer. However, stable density stratification across the cold halocline often prevents that heat from directly reaching the ice. The effect on sea ice depends on upward mixing and ocean heat flux, with greater direct access in some Atlantic-influenced regions; a temperature increase at depth alone is insufficient to quantify basal melting.

Warmer air increases surface melt during the warm season, lengthens the melt season, and reduces the accumulated freezing degree-days available for winter growth. Enhanced downward heat transfer can therefore thin the cover even without extra ocean heat. It provides a broad thermodynamic influence on Arctic sea ice decline, while regional winds and cloud conditions determine part of the local response.

Apply the method to the Dahlquist test equation, put $z=h\lambda$, and retain every root of its amplification polynomial of a multistep method:An A-stable convergent method must satisfy the root condition for a multistep method for every $\mathrm{Re}z\le 0$.

First take $-1<a<0$. As negative real $z$ tends to minus infinity, one root tends to the nonzero root of $\sigma$,It follows by root continuity that some sufficiently negative finite $z$ is unstable. At $a=-1$, the recurrence has polynomial $w^2-2zw-1$; for any negative real $z$, its root $z-\sqrt{z^2+1}$ is below $-1$. Thus all negative convergent parameters are excluded.

For $0\le a<1$, use the boundary-locus test for multistep A-stability. A unit root $w=e^{i\theta }$ with $\sigma (w)\ne 0$ would require $z=\rho (w)/\sigma (w)$. Direct multiplication by the conjugate denominator givesThere is therefore no unit-root crossing in the open left half-plane. The leading coefficient of $P_z$ never vanishes there, since its only zero is the positive real value $2/(1+a)$. For a small negative real $z$, the simple root initially at one moves to $1+z+O(z^2)$, inside the disk, and the other root is close to $a$, also inside. Continuity of polynomial roots and connectedness of the open left half-plane then keep both roots strictly inside throughout that half-plane. Possible zeros of $\sigma$ cause no omitted crossing: for $a>0$ its other root lies strictly inside the unit disk, while at $a=0$ its unit-circle zero $w=-1$ has $\rho (-1)\ne 0$.

On the imaginary axis, if $a>0$ the displayed real part vanishes only at $w=1$, which corresponds to $z=0$ and is simple. Other imaginary-axis parameters retain strictly interior roots. If $a=0$, the two roots are zero and the stability function $(1+z/2)/(1-z/2)$, whose modulus is one on that axis; the unit root is simple. Thus the required boundary root condition for a multistep method holds as well.

Combining necessity and sufficiency givesIn particular, the third-order member $a=-1/5$ is convergent but not A-stable, in agreement with the Second Dahlquist barrier.

Write $S_n(t)={\sum }_{r=-n}^n{a}_r{e}^{irt}$. First we prove decay of coefficients of an almost-everywhere convergent trigonometric series rather than assuming the series can be integrated term by term. For $n\ge 1$, $T_n(t)=S_n(t)-S_{n-1}(t)=a_n{e}^{int}+a_{-n}{e}^{-int}$ tends to zero outside the finite set $E$. Suppose $M_n=|a_n|+|a_{-n}|$ does not tend to zero, and choose a subsequence with $M_n\ge \epsilon >0$. Then $|T_n/M_n|\le 1$ and $T_n/M_n\to 0$ almost everywhere. The dominated convergence theorem givesorthogonality of the two distinct Fourier modes instead makes the integral $(|a_n{|}^2+|a_{-n}{|}^2)/M_n^2\ge 1/2$, a contradiction. Thus $\boxed{a_n,a_{-n}\to 0}$.

Define on the real line the twice-integrated continuous functionThe series has both absolute convergence and uniform convergence because its coefficients are bounded. Its symmetric second quotient is therefore, for nonzero $h$,We next prove that this tends to zero whenever $t\notin E$, lifted periodically to the line. Let $b_0=a_0$ and $b_n=a_n{e}^{int}+a_{-n}{e}^{-int}$ for $n\ge 1$, so the partial sums of the $b_n$ are exactly $S_n(t)\to 0$. Put $k=|h|/2$ and $w_n=w(nk)$. Summation by parts givesThe boundary term vanishes because $S_n(t)$ is bounded and $w_n\to 0$ for fixed nonzero $k$. MoreoverIndeed $w^{\prime }(x)=O(x)$ near zero and $|w^{\prime }(x)|\le 2/x^2+2/x^3$ for $x\ge 1$. Each fixed weight difference tends to zero as $k\to 0$. Splitting the last series at an index beyond which $|S_n(t)|<\eta$ therefore bounds its upper limit by $C\eta$. Letting $\eta \to 0$ proves the claimed zero second quotient. This explicitly proves the needed Riemann summation of a convergent trigonometric series, despite the oscillations in its weights.

We also need the elementary fact that a continuous function with zero symmetric second derivative at every interior point of an interval is affine. For a real-valued function, subtract its endpoint chord on any compact subinterval $[a,b]$. If the difference were positive somewhere, adding $\epsilon (t-a)(t-b)$ for sufficiently small $\epsilon >0$ would retain a positive maximum at an interior point. At that maximum each symmetric second difference is nonpositive, while its quotient tends to $2\epsilon >0$, a contradiction. Thus the function lies below its chord. Apply the argument to its negative to get the reverse inequality. Applying this to real and imaginary parts proves the complex-valued case as well. This establishes that vanishing symmetric second derivative forces an affine function. Hence $F$ is an affine function on each open interval between the finitely many exceptional points in any period.

It remains to exclude corners at those points. Multiplying the formula for the second quotient by $h$, and using $h=2k$ and the evenness of $w$, givesfor every $t$, by the infinite-series version of part (i), since $a_n,a_{-n}\to 0$. continuity extends the adjacent affine pieces to each exceptional point, and the fact that symmetric second differences detect an affine corner makes their slopes and intercepts agree. Thus $F$ is globally an affine function on the real line.

The periodic part of its definition gives $F(t+2\pi )-F(t)=2\pi a_{0}t+2{\pi }^2{a}_0$. An affine function has constant increments, so $a_0=0$. The resulting $F$ is both an affine function and a periodic function, hence constant. Finally, absolute uniform convergence justifies integrating its defining series against $e^{-irt}$, and for $r\ne 0$ the resulting Fourier coefficient is $-a_r/r^2$. A constant has all these coefficients zero. Therefore $\boxed{a_r=0\text{ for every }r\in \mathbb{Z}}$, proving uniqueness of a trigonometric series outside a finite set with all exceptional points accounted for.

Put $w(x)=(\sin x/x{)}^2$, with $w(0)=1$, and $M={\sup }_r|A_r|<\infty$. Since $0\le w\le 1$, the finite-prefix bound is $\left|{\sum }_{r=1}^n{A}_{r}w(rk)\right|\le nM$. For $k\ne 0$, $w(rk)\le 1/(r^2{k}^2)$ gives absolute convergence of the tail and the correct estimateThe PDF omits the factor $k^{-2}$ from this bound. Its printed version is false: choose $A_4=1$ and all other $A_r=0$, take $n=3$, and let $k\to 0$. The left side tends to one while the printed right side is $2/3$. Thus the intended bound must be qualified as above. At $k=0$, even absolute convergence need not hold for a general sequence tending to zero.

For each fixed $n$, the displayed finite-prefix limit follows immediately: $|k{\sum }_{r=1}^n{A}_{r}w(rk)|\le |k|nM\to 0$. The stronger statement needed for part (ii) isHere is a proof, so the fixed-prefix formula is not silently substituted for an infinite-series result. Set $N=\lfloor 1/|k|\rfloor$ for small nonzero $k$. The multiplied tail is at most $2{\sup }_{r>N}|A_r|/(N|k|)$, which tends to zero because $N|k|\to 1$. For the prefix, choose $J$ with $|A_r|<\eta$ for $r>J$. ThenIts upper limit is at most $\eta$, and $\eta$ is arbitrary. This proves sinc-squared summation of a sequence tending to zero, including approach from either sign of $k$.

For the unheaded piecewise-affine continuation, continuity gives $Ac+B=A^{\prime }c+B^{\prime }$. Evaluate the asserted symmetric quotient at $t=c$. For sufficiently small positive $h$,Its zero limit forces $\boxed{A=A^{\prime }}$, and the continuity relation then forces $\boxed{B=B^{\prime }}$. Only the condition at the corner is needed; at points away from it such a quotient is automatically zero even for unequal slopes.

Put $d=k-1$ and let $e_m$ be the elementary symmetric polynomial of degree $m$ in the $d$ interior spline knots $t_{i+1},\dots ,t_{i+d}$, with $e_0=1$. Expanding the knot polynomial givesCompare the coefficient of $x^{d-m}$ in the Marsden identity. The left coefficient is $(-1{)}^m(\genfrac{}{}{0ex}{}{d}{m})t^m$; the right coefficient is $(-1{)}^m{\sum }_i{e}_m(t_{i+1},\dots ,t_{i+d})N_i(t)$. Cancelling the common sign therefore gives the monomial B-spline coefficients:Equivalently, for $m\ge 1$ the numerator is ${\sum }_{1\le r_1<\cdots <r_m\le k-1}{t}_{i+r_1}\cdots t_{i+r_m}$. In particular $a_i^{(0)}=1$, so the order-zero case reproduces constants, and for $k\ge 2$ the linear coefficients are $a_i^{(1)}=(t_{i+1}+\cdots +t_{i+k-1})/(k-1)$, the Greville abscissae. At $m=k-1$, the coefficient is the product of all interior knots. Each formula holds on the basic knot interval used in the Marsden identity.

Fix $t$ and let $F_t(u)=(u-t{)}_{+}^{k-1}$. Initially assume distinct spline knots, so the ordinary polynomial interpolation definitions are unambiguous. The two interpolants ${\ell }_i(\cdot ,t)$ and ${\ell }_{i+1}(\cdot ,t)$ agree with $F_t$ at their $k-1$ shared nodes $t_{i+1},\dots ,t_{i+k-1}$. Their difference has degree at most $k-1$ and all these roots, henceFor $k=1$ the product is empty and the same statement concerns constants. The leading coefficient of an interpolating polynomial is its highest-order divided difference, so the divided difference recurrence givesThis proves Lee's formula:Summing over $i$ telescopes to ${\ell }_{n+1}(x,t)-{\ell }_1(x,t)$. If $t_k<t<t_{n+1}$, the first interpolation nodes $t_1,\dots ,t_k$ are all smaller than $t$, so their truncated-power data vanish and ${\ell }_1=0$. The last nodes $t_{n+1},\dots ,t_{n+k}$ are larger than $t$, so their data are values of the polynomial $(u-t{)}^{k-1}$ and uniqueness gives ${\ell }_{n+1}(x,t)=(x-t{)}^{k-1}$. Thus the telescoped identity isthe Marsden identity, valid for every real $x$.

If repeated spline knots are allowed, ordinary value interpolation at repeated nodes does not specify ${\ell }_i$ uniquely. The standard extension uses confluent divided differences and Hermite interpolation, or limits of distinct knots: the same polynomial identity then persists wherever these quantities are defined. At a maximally repeated knot where the truncated power lacks the required derivative, the appropriate one-sided spline convention must be fixed; an unspecified classical derivative is not part of the proof. For distinct knots, the printed identity requires no such qualification.