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Absolute values are missing in the printed formula. For signed real blocks, the intended block mixed norm must beWithout them, at $q=1$ a block $(-1,1)$ has value zero although it is nonzero, and a negative scalar has negative value. For nonintegral $q$, a negative component may not even have a real power. The literal expression therefore is not a norm and does not define the claimed general convex problem. Even integer exponents do not fix all other $q$ in the stated range.

With the corrected norm, let $q^{\prime }$ be the Holder conjugate exponent: $q^{\prime }=q/(q-1)$ for $q>1$ and $q^{\prime }=\infty$ for $q=1$. Blockwise Holder inequality, with equality from a norming block, givesThe duality of block mixed norms identifies the closed compact convex set and the projection residual from the Moreau decompositionFor $q=2$, the dual blocks are Euclidean balls; for $q=1$, they are cubes given by componentwise bounds $|p_{i,j}^r|\le \lambda$. No injectivity or surjectivity of $A$ is required: compactness of the block product makes its linear image compact, and the strictly convex fidelity still gives a unique primal minimizer. For general $q^{\prime }$, the Euclidean projection onto a dual block ball is not generally obtained by radial scaling, so the special clipping formula from the isotropic case should not be reused without further analysis.

Under the usual definition, the null space property relative to $S$ is $\Vert h_S{\Vert }_1<\Vert h_{S^c}{\Vert }_1$ for every nonzero $h\in \ker A$. It characterizes recovery by basis pursuit of every vector supported in $S$. Indeed, if it holds, then for any such $x_0$ and any nonzero $h\in \ker A$,Conversely, failure for some $h$ makes $x_0=-h_S$ and $z=h_{S^c}$ feasible with the same measurements and $\Vert z{\Vert }_1\le \Vert x_0{\Vert }_1$. ThusFor the single fixed vector in the printing, necessity is false with that definition. For example,has $\ker A=\mathrm{span}\{(1,1,1)\}$. Along its feasible line, the objective is $|1+t|+|-1+t|+|t|$, uniquely minimized at $t=0$, although the null space property would require $2<1$.

The correct fixed-sign null space condition isSufficiency follows from the supporting-line inequality for the absolute value. Necessity follows by considering $x_0\pm th$ for small positive $t$, when all signs on $S$ remain unchanged: a nonpositive directional increase gives either a decrease or another minimizer. If “relative to $S$” was intended to include these signs, this is the precise condition needed.

It is useful to prove the contrapositive. Suppose $k$ is a peak of a sequence $(a_n)$ and $m$ is a peak of a sequence $(-a_n)$. The second condition says $a_j\ge a_m$ for every $j>m$. HenceOnly finitely many earlier terms remain, and their absolute values have a finite maximum. Together these give a bound for every term, so $(a_n)$ is a bounded sequence. Consequently,

All three examples below have positive integer coefficients, so they work under either convention for whether $0\in \mathbb{N}$.

For $R=1$, take $\boxed{a_n=1}$. This is the geometric series: it converges for $|z|<1$, and its terms fail to tend to zero for $|z|\ge 1$.

For $R=0$, take $\boxed{a_n=n!}$, using the factorial and $0!=1$. For every fixed $z\ne 0$, the absolute values of successive terms have ratio $(n+1)|z|$. This eventually exceeds two, so the terms grow without bound and the series cannot converge. At $z=0$ it is finite, giving radius of convergence zero.

For $R=1/\sqrt{2}$, takeusing the floor function. The coefficients satisfyThe upper bound and the comparison test for series give absolute convergence when $\sqrt{2}|z|<1$. The lower bound prevents the terms tending to zero when $\sqrt{2}|z|\ge 1$. Hence the radius of convergence is exactly $1/\sqrt{2}$.

Suppose the continuous function $f$ did not have uniform continuity on $[a,b]$. Negating its definition gives ${\epsilon }_0>0$ such that for every $n\ge 1$ there are $x_n,y_n\in [a,b]$ withThe Bolzano-Weierstrass theorem, which states that every bounded sequence of real numbers has a convergent subsequence, gives $x_{n_k}\to p$. The inequalities $a\le x_{n_k}\le b$ imply $p\in [a,b]$. By the triangle inequality,Since continuous functions preserve convergent sequences, both $f(x_{n_k})$ and $f(y_{n_k})$ tend to $f(p)$. Their difference therefore tends to zero, contradicting its absolute value being at least ${\epsilon }_0$. This proves the Heine-Cantor theorem in the required setting: $f$ has uniform continuity on $[a,b]$.

On $(0,1)$, the product rule and chain rule give $f^{\prime }(x)=2x\sin (1/x)-\cos (1/x)$. On $(-1,0)$, the function is constant and has derivative zero. At zero,Its absolute value is at most $|h|$, so the squeeze theorem gives $f^{\prime }(0)=0$. ThereforeFor $x_n=1/(2\pi n)$, we have $x_n\to 0$ but $f^{\prime }(x_n)=-1$ for every $n$. This differs from $f^{\prime }(0)=0$, so $f^{\prime }$ is not continuous at zero. Indeed, $y_n=1/((2n+1)\pi )$ gives $f^{\prime }(y_n)=1$, so the right-hand limit of a function does not exist. Thus $f$ is differentiable everywhere on $(-1,1)$, while $f^{\prime }$ is discontinuous at zero.

A function $f:(a,b)\to \mathbb{R}$ is differentiable on $(a,b)$ when, at each $c\in (a,b)$, its difference quotient has a finite limit of a function:Here $h\ne 0$ and $c+h\in (a,b)$.

For sufficiently small nonzero $h$, the difference quotient has absolute value at most $|f^{\prime }(c)|+1$, because it tends to $f^{\prime }(c)$. ConsequentlyThis is continuity at $c$. Since $c$ was arbitrary, differentiability implies continuity on $(a,b)$.

The continuous function $|h|$ attains a finite maximum $M$ on $[a,b]$ by the extreme value theorem. Additivity and the triangle inequality for the Riemann integral imply, for any $x,y\in [a,b]$,For reversed endpoints, reverse the orientation of the integral before taking its absolute value. Given $\epsilon >0$, choose $\delta =\epsilon /(M+1)$. Then $|x-y|<\delta$ implies $|H(x)-H(y)|<\epsilon$. Thus $H$ is continuous on $[a,b]$, including one-sided continuity at the endpoints; it is even Lipschitz continuous. If $a=b$, the function is defined on one point and the conclusion is immediate.

Split the power series into even and odd indices:whenever the two series converge absolutely. If $|z|<1$, the even part is a convergent geometric series, while the odd part is bounded termwise in absolute value by ${\sum }_j|z{|}^{2j+1}$, also a convergent geometric series. Hence there is absolute convergence throughout $|z|<1$.

If $|z|>1$, the odd terms have absolute values $|z{|}^{2j+1}/(2j+1{)}^2$. Their successive ratio tends to $|z{|}^2>1$, so these terms grow eventually and do not tend to zero. The term test for divergence therefore forces divergence of the original power series. Consequently its radius of convergence is

For fixed $z\ne 0$, the ratio of successive absolute values isThe ratio test says that a series converges absolutely when this ratio has a limit of a sequence strictly below one. Thus this power series converges absolutely for every complex $z$; at zero the conclusion is immediate. Its radius of convergence is $\boxed{R=\infty }$.

The finite geometric series identity gives, for each $|x|<1$,For every fixed $n$, the absolute value of this error tends to infinity as $x\to 1^{-}$. Thusand there is no uniform convergence on $(-1,1)$.

No. On $(-1,1)$ take $c=0$ andAt zero, its difference quotient has absolute value $|x\sin (1/x)|\le |x|\to 0$, so $f^{\prime }(0)=0$. Away from zero, the product rule and chain rule apply. DefineThese are distinct positive inputs tending to zero. Since $\sin A_n=0$ and $\sin (A_n+B_n)=\sin B_n$,The even-indexed subsequence tends to $-2/\pi$, whereas the odd-indexed subsequence tends to $2/(3\pi )$. Thus the limit of a sequence does not exist. The two points lie on the same side of $c$, so the nonnegative-weight estimate in part (b) is unavailable. This is a concrete instance of differentiability does not imply continuity of the derivative.

The cardioid has reflection symmetry about the $x$ axis because $1+\cos (-\theta )=1+\cos \theta$. It passes through $(2,0)$ at $\theta =0$, through $(0,\pm 1)$ at $\theta =\pm \pi /2$, and has a cusp at the origin when $\theta =\pi$. These points and the following sketch show its orientation.

Using the polar coordinates formula for arc length,The absolute value changes sign at $\theta =\pi$, so

The log-likelihood function for the Laplace distribution isThus maximum likelihood estimation is equivalent to minimizing the sum of absolute values. Write $n=2m+1$ and sort the data as $Y_{(1)}\le \cdots \le Y_{(n)}$. Between observations, if $r$ values lie to the left of $\kappa$, the derivative of ${\sum }_i|Y_i-\kappa |$ is $r-(n-r)=2r-n$. It is negative before the middle observation and positive after it. The minimum occurs at the sample median:This minimizer remains unique for odd $n$ even when some observations coincide, since fewer than half the observations can lie strictly on either side of the middle value.

The euclidean algorithm computes the greatest common divisor of two integers by repeated division with remainder. Start with $r_0=|A|$, $r_1=|B|$, interchanging them if necessary, and repeatedly writeuntil a zero remainder occurs. The positive remainders strictly decrease, so the process terminates. At each step the common divisors of $r_{j-1},r_j$ are exactly the common divisors of $r_j,r_{j+1}$, because $r_{j+1}=r_{j-1}-q_j{r}_j$. HenceIf one input is zero, the answer is the absolute value of the other; the usual convention sets $\gcd (0,0)=0$. Back-substitution also expresses the greatest common divisor as an integer linear combination of the inputs, as in Bézout's identity.

A series ${\sum }_{n=1}^{\infty }{a}_n$ converges if its partial sums $s_N={\sum }_{n=1}^N{a}_n$ have a finite limit of a sequence $S$ as $N\to \infty$; its sum is then $S$. It has absolute convergence if the series ${\sum }_{n=1}^{\infty }|a_n|$ converges, where $|a_n|$ is the absolute value of the term.

A parametrization of the surface of revolution isIts tangent vectors are orthogonal, with norms $\sqrt{1+f^{\prime }(x{)}^2}$ and $|f(x)|$. ThusThe absolute value is necessary if the function graph passes below the axis. The formula gives the curved surface area; it adds no end discs. Zeros of $f$ can be treated by splitting the integral into regular portions and taking limits.

For fixed real $a\ne 0$, write the square root as $|a|\sqrt{1+x^2/a^2}$ and use the binomial series with exponent $-1/2$:The absolute values are necessary for negative $a$. This Taylor series is valid for $|x|<|a|$. If $a=0$, the quotient is $x/|x|$ for $x\ne 0$; it has no continuous extension at zero and therefore has no Taylor expansion about zero.

With $s=\sin {\theta }_0$ and $c=\cos {\theta }_0$, conversion from spherical coordinates gives $\mathbf{a}=r_0(s,0,c)$, $\mathbf{b}=r_0(-s/2,\sqrt{3}s/2,c)$ and $\mathbf{c}=r_0(-s/2,-\sqrt{3}s/2,c)$. The volume of the parallelepiped is the absolute value of their scalar triple product:The absolute value matters when the polar angle is in the lower hemisphere. At the poles or equator the three edge vectors are linearly dependent and the volume vanishes.

The geometric series has first term $1$ and common ratio $1/3$, whose absolute value is less than one. Therefore