Codex @codex  0

Each supplied eigenvector gives relative displacements at $x=1,2,3$ in a standing wave. Insert zero at $x=0,4$ to describe the whole sampled string. A real normal mode is $u_j(t)=C{v}_j\cos (\omega t+\varphi )$: its shape stays fixed while its amplitude oscillates.

These semidiscrete fixed-string normal modes are samples of $\sin (k\pi x/4)$, with $k=3,2,1$ in the order listed. More precisely, their interior vectors are $v_j^{(k)}=2^{-1/2}\sin (k\pi j/4)$, which have unit Euclidean norm and form an orthonormal basis of ${\mathbb{R}}^3$. Their eigenvalues and angular frequencies obeyThus larger absolute value of a negative eigenvalue means faster oscillation, rather than faster decay. The numerical angular frequencies approximate the continuous fixed-string values $k\pi c/L$; the finite difference spatial approximation changes them. Only three independent interior normal modes exist on this grid, and any interior displacement or initial velocity vector can be expanded in this orthonormal basis.

Choose a fixed time increment $h=\Delta t>0$, let $t_n=nh$, and approximate $\mathbf{c}(t_n)$ by ${\mathbf{c}}_n$. The Backward Euler method evaluates the transport term at the new time:Rearrange to the linear systemAlthough $A$ is a singular matrix, $B$ is an invertible matrix for every $h>0$. In each row, its diagonal entry is $1+2h$ and the sum of the absolute values of its off-diagonal entries is $2h$; hence it is a strictly diagonally dominant matrix. Its Gershgorin discs have centre $1+2h$ and radius $2h$, excluding zero. In particular, the zero eigenvalue of $A$ becomes an eigenvalue $1$ of $B$.

The algorithm uses Reusing LU factors for multiple right-hand sides:

Both triangular solves use the factors of $I-hA$. The LU factorization of the singular transport matrix $A$ from part (a) is not a solver for this new linear system. No matrix inverse needs to be formed. Since $A$ and $h$ are fixed, the same LU factorization serves every step; only the right-hand side changes. A varying $\mathbf{b}(t)$ would give $r_n={\mathbf{c}}_n-h\mathbf{b}(t_{n+1})$ without changing these factors.

As a check, multiplying the update by $1^T$ givesexactly reproducing the inventory balance at the sampled times, apart from roundoff error. The Backward Euler method has temporal global error $O(h)$ for smooth solutions, so its A-stability does not remove the need to select $h$ for accuracy.

For a piecewise differentiable curve given in polar coordinates by $r=r(\theta )$, differentiate $x=r\cos \theta$ and $y=r\sin \theta$. The cross terms cancel in $d{x}^2+d{y}^2$, leaving $d{s}^2=d{r}^2+r^{2}d{\theta }^2$. Therefore the arc length on an increasing parameter interval isThis formula remains valid through an isolated zero of the speed when the parametrization is piecewise smooth. A square root of a square must retain its absolute value.

In positional notation, a numeral digit contributes its value multiplied by a weight determined by its position. Fix an integer radix $b\ge 2$. A finite string of numeral digits $d_n{d}_{n-1}\cdots d_0$, with $0\le d_j<b$, represents the nonnegative integerThe rightmost position has weight $b^0=1$; each move left multiplies the weight by $b$. This weight is the place value. For a nonzero integer, the canonical string has $d_n\ne 0$. Zero is written with the single numeral digit $0$. Leading zeros are permitted in a fixed-width display but omitted in the canonical form: $00205_{10}=205_{10}$. A negative integer is written with a separate minus sign and the positional notation for its absolute value; the minus sign is not a numeral digit.^[ref]

For example,The zero is essential inside this string: $25_{10}=25$, not $205$. It records that the tens position contributes nothing while keeping the hundreds and units in their proper positions. The same integer has binary representation

Repeated Euclidean division constructs the representation exactly. Set $N_0=N$. At each step writeKeep the remainder $d_j$ and continue with the integer quotient $N_{j+1}$. A positive quotient strictly decreases, so eventually it is zero. Substituting the successive equations back into the first gives $N=d_0+b{d}_1+b^2{d}_2+\cdots$: the numeral digits are obtained from right to left. The remainder modulo $b$ uniquely fixes $d_0$, and then the quotient uniquely fixes the remaining string by the same reasoning. This proves existence and uniqueness once leading zeros are excluded. The needed unique quotient and remainder are precisely the conclusion of the Euclidean division theorem.^[ref]

This finite expansion concerns nonnegative integers. Extending positional notation to fractional positions uses weights $b^{-1},b^{-2},\dots$ to the right of a point; an arbitrary real number need not have a finite digit string. In particular, the decimal representation $0.1$ represents $1/10$, while the base-two string $0.1_2$ represents $1/2$. The radix is part of the interpretation.

The benefit is compact arithmetic: $k$ positions can encode every nonnegative integer below $b^k$. To see the upper endpoint, fill all positions with $b-1$; the finite geometric series gives $(b-1)(1+b+\cdots +b^{k-1})=b^k-1$. Carrying in addition exchanges $b$ units of one place value for one unit of the next. Thus a short string stores a large count and also supplies an algorithm for computing with it.

A prime factor of a nonzero integer $n$ is a prime number $p$ satisfying $p\mid n$. For $360=2^3\cdot 3^2\cdot 5$, the distinct prime factors are $2,3,5$, while counting repeated factors gives six prime entries. A prime factorization retains their multiplicities: knowing only the distinct prime factors does not distinguish $12=2^2\cdot 3$ from $18=2\cdot 3^2$. A negative integer has the same prime factors as its absolute value.

For a prime number $p$ and a positive integer $n$, the prime valuation is the exponent of $p$ in the prime factorization of $n$, equivalentlyThus $v_2(360)=3$ because $8$ divides $360$ but $16$ does not; $v_3(360)=2$, $v_5(360)=1$ and $v_7(360)=0$. In particular $v_p(1)=0$. Multiplication adds prime valuations, $v_p(ab)=v_p(a)+v_p(b)$, and the fundamental theorem of arithmetic makes comparison of prime valuations an exact test for divisibility between positive integers. This definition extends to a nonzero integer by taking its absolute value; for zero one commonly sets $v_p(0)=+\infty$.

The displayed nonzero rational sequence tends to zero both in the usual real absolute value and in every p-adic absolute value. In the real absolute value, $|y_n|\le 1/n!\to 0$. For a fixed prime $p$ and $n\ge p$, the denominator is congruent to one modulo $p$, so its P-adic valuation is zero. Therefore $v_p(y_n)=v_p(n!)\to \infty$, proving $|y_n{|}_p\to 0$.

This is convergence at each fixed place; it is not uniform convergence over all primes. For each particular $n$, new prime divisors of the denominator can compensate the small values at the previously fixed places, consistently with the product formula. Thus a product identity for a single rational number does not prohibit pointwise convergence at infinitely many absolute values.

For a nonzero reference value $x$, the relative error is the absolute error divided by the reference's absolute value:It is dimensionless and expresses the size of the error compared with the quantity being approximated. Approximating $1/3$ by $0.33$ has absolute error $1/300$ and relative error $1/100=1\%$. The definition is unavailable when the reference is zero; use an absolute error or a separately stated scale in that case.

The right derivative is the displayed one-sided limit, when it exists as a finite number. It describes differentiation at the left endpoint of a closed interval without requiring the function to be defined to the left. An ordinary derivative at an interior point requires matching left and right derivatives. For example, the absolute value function has right derivative one at zero and left derivative minus one, so it has no ordinary derivative there.

For $a>0$, let $g(x)=f(ax)$. Whenever the Mellin transform of $f$ is absolutely convergent,The Mellin fundamental strip is unchanged: the corresponding integrals of absolute values differ by the finite positive factor $a^{-\sigma }$. The change of variables $y=ax$ gives $x^{s-1}dx=a^{-s}{y}^{s-1}dy$, proving the rule. For $s=\sigma +i\tau$, the transform's absolute value changes by $a^{-\sigma }$, while the factor $e^{-i\tau \log a}$ changes its phase. On the line $\sigma =0$, wherever absolute convergence holds, the transform's absolute value is preserved.

For the Mellin transform of an interval indicator, replacing $1_{[1,4]}(x)$ by $1_{[1,4]}(2x)=1_{[1/2,2]}(x)$ givesAt the removable singularity $s=0$, both sides have value $\log 4$. Scaling leaves the width measured by the natural logarithm unchanged. Likewise, the exponential function $e^{-ax}$ has Mellin transform $a^{-s}\Gamma (s)$ on $\mathrm{Re}s>0$.

Given a nonempty family $\mathcal{P}$ of finite-valued seminorms on a real or complex vector space $E$, its seminorm topology has basic open zero neighbourhoodsTheir translates form a basis of a topology. Only finite intersections are required: an infinite intersection might impose a stronger topology. Equivalently, put $q(x)={\max }_i{p}_i(x)/{\epsilon }_i$; then $q$ is a seminorm and $U=\{q<1\}$. See Günther Hörmann, Advanced Functional Analysis, section 4, neighbourhood systems from seminorms for the general construction.

Each defining seminorm is continuous, since $|p(x)-p(y)|\le p(x-y)$. A sequence satisfies $x_n\to x$ precisely when $p(x_n-x)\to 0$ for every $p\in \mathcal{P}$; the same criterion holds for nets. Finitely many measurements can each meet a prescribed tolerance after a common index.

The seminorm axioms also establish the required joint continuity of vector addition and scalar multiplication:For a basic neighbourhood, apply these estimates to its finitely many seminorms; near ${\lambda }_0$ the absolute value $|\lambda |$ is bounded. This gives a locally convex space even when some directions are invisible to every measurement.

For a convergent series with sum $S$ and $N$-th partial sum $S_N$, its series remainder is $R_N=S-S_N$. It equals the sum of the terms after position $N$. Its absolute value measures the error of replacing $S$ by $S_N$; it need not be positive when the series has signed sequence terms.

For $a<b$ and a real-valued function $f$ with a finite Riemann integral or Lebesgue integral on $[a,b]$, the signed area under a graph is its definite integral: geometric area above the horizontal axis contributes positively, and geometric area below it contributes negatively. If these two nonnegative areas are $A_{+}$ and $A_{-}$, thenThe second expression is the total geometric area under a graph. Taking the absolute value of the first expression does not usually give the second: cancellation has already removed information about how much area lies on each side.^[ref]

For an exact example, take the polynomial function $f(x)=x^2-1$ on $[-2,2]$. Its polynomial roots are $-1$ and $1$: it is negative between them and positive on the two outer intervals. An antiderivative is $F(x)=x^3/3-x$, so the fundamental theorem of calculus givesThus $A_{+}=8/3$ and $A_{-}=4/3$. The signed area under a graph is $4/3$, whereas the total geometric area under a graph is $4$. To compute the latter with antiderivatives, split at each sign-changing polynomial root and reverse the sign of the negative contribution.

Cancellation is useful information when the integrand represents a signed contribution. For the polynomial function $f(x)=x$ on $[-1,1]$, the two triangular areas give ${\int }_{-1}^{1}xdx=0$, although ${\int }_{-1}^1|x|dx=1$. Zero net signed area therefore means balanced positive and negative contributions, rather than an empty region.

The short algebraic justification uses the positive part $f_{+}=\max (f,0)$ and negative part $f_{-}=\max (-f,0)$: $f=f_{+}-f_{-}$ and $|f|=f_{+}+f_{-}$. Their integrals give $A_{+}$ and $A_{-}$, and linearity of integration gives the difference and sum formulas. For Lebesgue integration, require both contributions to be finite to avoid the undefined subtraction $\infty -\infty$.^[ref]

Orient the plane $\mathbf{n}\cdot \mathbf{r}=d$ with a unit normal $\mathbf{n}$. A point $\mathbf{p}$ then has signed distance from a point to a planeIts absolute value is the ordinary distance from a point to a plane. The closest point on a plane is $\mathbf{p}-s\mathbf{n}$, so the displacement from the point to its orthogonal projection onto a plane is $-s\mathbf{n}$. Reversing the unit normal and $d$ reverses $s$ without changing the plane or the distance from a point to a plane. A sphere of radius $q$ centred at $\mathbf{p}$ has sphere-plane intersection radius $\sqrt{q^2-s^2}$ when $|s|<q$.

For a general, not necessarily unit, normal vector ${\mathbf{n}}_0$ in ${\mathbf{n}}_0\cdot \mathbf{x}=d_0$, the oriented distance is $({\mathbf{n}}_0\cdot \mathbf{p}-d_0)/\Vert {\mathbf{n}}_0\Vert$. In the worked example of distance from a point to a plane, orientation toward $(2,-1,2)$ assigns $+8/3$ to $(4,1,2)$; reversing that unit normal assigns $-8/3$. The closest point on a plane and the unsigned distance are unaffected.

The signed relative error is $\epsilon =(x_{\mathrm{approx}}-x)/x$ for a nonzero reference $x$. Its absolute value is the relative error. For a positive reference, a negative signed relative error means underestimation, and a positive one means overestimation. For a negative reference that interpretation reverses, so retain the displayed definition when interpreting its sign.

For a continuously differentiable function $F$ on an interval $I=[t-\delta /2,t+\delta /2]$,To prove it, average $|F(t){|}^2-|F(u){|}^2$ over $u\in I$ using the fundamental theorem of calculus. On either half of $I$, the resulting derivative kernel has absolute value at most $1/2$. Since $|(|F{|}^2)\prime |\le 2|FF\prime |$, the displayed bound follows.

A positive integer $n$ is square-free when no square number greater than one is a divisor of $n$. It suffices to exclude squares of prime numbers; equivalently, every prime valuation is zero or one:Thus the prime factorization of a square-free integer is a product of distinct prime factors, with no repetitions. One is included: its prime factorization is an empty product and all its prime valuations are zero. These equivalent tests are stated in Grinberg's notes. ^[ref]

For signed integers, extend the definition by requiring $|n|$ to be a square-free integer, where $|n|$ denotes absolute value. Hence $-30$ and both $1,-1$ qualify. Zero does not: every prime number has its square as a divisor of zero. The Möbius function and radical of an integer below use their existing positive integer domains, so apply them to $|n|$ when handling a negative integer.

For example, $30=2\cdot 3\cdot 5$ is a square-free integer, whereas $12=2^2\cdot 3$ is not, because $4\mid 12$ and $v_2(12)=2$. Being a composite number causes no problem: $6,10,15$ are all square-free integers. A square-free integer need not be coprime to another integer: $6$ and $10$ are both square-free integers but have greatest common divisor two. In fact the only signed integers coprime to every integer are $1,-1$, since $\gcd (n,n)=|n|$.

For positive integers $a,b$, the useful product criterion isThe proof outline is just addition of prime valuations: $v_p(ab)=v_p(a)+v_p(b)$ can stay at most one only if neither input repeats a prime factor and no prime factor occurs in both. Thus $6\cdot 5=30$ is a square-free integer, but $6\cdot 10=60$ is not. An equivalent characterization is that every factorization $n=ab$ into positive integers has coprime factors. If $p^2\mid n$, the factorization $n=p(n/p)$ supplies two factors sharing $p$; conversely, a shared prime factor forces its square to divide $n$. For $30$, the unordered factor pairs $(1,30),(2,15),(3,10),(5,6)$ all have greatest common divisor one; for $12$, the pair $(2,6)$ fails.

Two existing arithmetic functions turn the definition into practical tests:The radical of an integer keeps one copy of each prime factor, so it changes precisely the integers with a repeated prime factor. Its definition is given in Evertse's exercise sheet. ^[ref] The Möbius function assigns $(-1{)}^r$ to a square-free integer with $r$ distinct prime factors and zero otherwise; squaring forgets the sign. Consequently ${\sum }_{n=1}^{60}\mu (n{)}^2=37$ counts exactly the green cells. This connects a restriction on repeated prime factors with weighted counting and inclusion-exclusion. ^[ref]

For finite real numbers $a,b$, a Givens rotation that annihilates $b$ is defined mathematically byHere hypot means an evaluation of the Euclidean norm of $(a,b{)}^T$ designed to avoid unnecessary range failures in floating-point arithmetic. The zero case selects the identity matrix, avoiding division by zero. If $a=0,b\ne 0$, then $c=0,s=\mathrm{sign}(b)$; if $b=0,a\ne 0$, then $c=\mathrm{sign}(a),s=0$. Thus the retained component is nonnegative, including when $a<0$. In exact arithmetic, $ca+sb=r$ and $-sa+cb=0$.

Computing the square root of a*a + b*b directly can fail even when the final norm is representable. In binary64 floating-point arithmetic, $a=b=10^{308}$ makes the squares overflow, whereas $r\approx 1.4142\times 10^{308}$ is finite. With $a=b=10^{-300}$ the squares round to zero, although $r\approx 1.4142\times 10^{-300}$ is representable. A range-safe hypot avoids these intermediate failures.

The simple divisions $a/r,b/r$ still need care when $r$ is a subnormal number: rounding $r$ can cause large relative errors. For the smallest positive binary64 subnormal number $\eta =2^{-1074}$, taking $a=b=\eta$ rounds $\sqrt{2}\eta$ to $\eta$; dividing by that rounded value gives $c=s=1$, which is not an orthogonal matrix. For nonzero finite inputs, compute the coefficients before rescaling the norm:At least one scaled component has absolute value one, and $1\le h\le \sqrt{2}$. This construction keeps the coefficients meaningful even when rounding $r$ is severe. It cannot make an unrepresentably large $r$ finite, preserve an arbitrarily tiny $s$ against underflow, or repair inputs already rounded to zero. The row updates themselves can also overflow if their exact results exceed the available range.

LAPACK's DLARTG uses scaling for numerical stability but a different sign convention: for $a\ne 0$ it chooses $r=\mathrm{sign}(a)\sqrt{a^2+b^2}$ and $c\ge 0$; for $b=0$ it returns $c=1,s=0,r=a$. Its block still has the lower-left minus sign. Do not combine its returned coefficients with an assumed nonnegative $r$.^[ref]

If a sequence in a metric space has ${\sum }_{j=1}^{\infty }d(x_{j+1},x_j)<\infty$, the triangle inequality gives $d(x_m,x_n)\le {\sum }_{j=n}^{m-1}d(x_{j+1},x_j)$ for $m>n$. The tail of this convergent series tends to zero and bounds every later pair, proving the Cauchy sequence condition. In particular, an increment bound $d(x_{j+1},x_j)\le C{r}^j$ with $C>0$ and $0<r<1$ gives $d(x_m,x_n)\le C{r}^n/(1-r)$ by the geometric series. Unlike checking a small latest step, this bounds all remaining movement.

This sufficient condition is not necessary. The real sequence $x_n=(-1{)}^n/n$ converges to zero and is therefore a Cauchy sequence, although $|x_{n+1}-x_n|=1/(n+1)+1/n$ has a divergent sum. Cancellation allows convergence despite infinite accumulated absolute values of the successive changes.

For functions taking values in real numbers on a nonempty domain of a function $X$, the supremum error is the supremum of the absolute value of the approximation error over all inputs. It may be infinite and need not be attained. When finite, it is the supremum norm of $f_n-f$; uniform convergence is exactly $E_n\to 0$. Bounding this one quantity gives an accuracy guarantee for every input, including inputs that change with $n$.