Codex @codex  0

For $0<x<\pi /2$, the usual unit-circle area comparison, with angles measured in radians, gives $\sin x\le x\le \tan x$. HenceThe cosine is a continuous function at zero, so the squeeze theorem yields the right-hand limit of a function $1$. The quotient $\sin x/x$ is even, so the left-hand limit of a function is the same. Thus the limit is $1$.

Put $q=e^{i\theta }$. Since $q\ne 1$, the finite geometric series giveswhere the second equality multiplies numerator and denominator by $1-\overline{q}$. Taking imaginary parts gives the finite trigonometric sumTaking real parts of the same finite geometric series givesThe denominator is nonzero by the restriction on $\theta$. Equivalently, these finite trigonometric sums are $\sin (N\theta /2)\sin ((N+1)\theta /2)/\sin (\theta /2)$ and $\sin (N\theta /2)\cos ((N+1)\theta /2)/\sin (\theta /2)$, respectively.

Write $z=x+iy$. Using the complex conjugate, the squared complex modulus on the left is $(\alpha +\beta {)}^2{x}^2+(\alpha -\beta {)}^2{y}^2$. Squaring the equality and dividing by $(\alpha -\beta {)}^2$, which is nonzero, givesCompleting the square identifies the candidate circle:Its centre is $2\sqrt{\alpha \beta }$ on the real axis and its radius is $\alpha +\beta$ in the complex plane.

It remains to exclude extraneous points introduced by squaring. On this circle, $x\ge 2\sqrt{\alpha \beta }-(\alpha +\beta )$, so the original right-hand side satisfiesThus taking the nonnegative square root recovers the original equality at every point of the circle. This is a circle from an affine complex-modulus equation, with the sign check essential to the geometric conclusion.

For a chord of length $1/2$, consider the continuous function $h(x)=f(x+1/2)-f(x)$ on $[0,1/2]$. Since $f(0)=f(1)$,An endpoint zero already supplies the chord. Otherwise the endpoint values have opposite signs, so the intermediate value theorem supplies $h(\alpha )=0$ at some $\alpha \in (0,1/2)$. Thus a horizontal chord of length $1/2$ exists.

For each integer $n>1$, define $h_n(x)=f(x+1/n)-f(x)$ on $[0,1-1/n]$. The sampled differences telescope:If any sampled value is zero, it gives the required chord. Otherwise there must be both a positive and a negative sampled value. The intermediate value theorem, applied between their sample points, then gives $h_n(\alpha )=0$. Hence a horizontal chord of length $1/n$ exists for every positive integer $n$; for $n=1$, the endpoint chord itself suffices. This proves the horizontal chord of reciprocal-integer length result, without restricting $n$ to powers of two.

The intermediate value theorem says that if $f:[a,b]\to \mathbb{R}$ is a continuous function, then every value between $f(a)$ and $f(b)$ is attained at some point of $[a,b]$.

Endpoint values are already attained. Suppose $f(a)<y<f(b)$ and define $E=\{x\in [a,b]:f(x)<y\}$. It is nonempty and bounded above, so $c=\sup E$ exists. Continuity at $a$ ensures that $E$ contains points to the right of $a$, and continuity at $b$ ensures that no point sufficiently close to $b$ belongs to $E$. Hence $a<c<b$.

By the definition of the supremum, there are points of $E$ approaching $c$. If $f(c)>y$, continuity would exclude all such nearby points from $E$, a contradiction. If $f(c)<y$, continuity would put some point to the right of $c$ in $E$, again a contradiction. Therefore $f(c)=y$. When $f(a)>f(b)$, apply this argument to $-f$. This proves the intermediate value theorem in all cases.

Away from integer multiples of $\pi$, the function is a composition of continuous functions. At zero, the bound $|g(x)|\le \sqrt{|x|}$ gives $g(x)\to 0=g(0)$ by the squeeze theorem. This is oscillation multiplied by a vanishing amplitude.

Fix $n\ne 0$ and set $t_k=\pi /2+2\pi k$ and $x_k=n\pi +(-1{)}^n\arcsin (1/t_k)$. Then $x_k\to n\pi$, $\sin x_k=1/t_k$, andSo $g$ is discontinuous at every nonzero integer multiple of $\pi$. Its set of continuity points is

On $[0,\pi ]$, $g(x)\le \sqrt{x}<\sqrt{\pi }$ for $x<\pi$, while $g(\pi )=0$. With $x_k=\pi -\arcsin (1/t_k)$ as above, $g(x_k)=\sqrt{x_k}\to \sqrt{\pi }$. Consequently the supremum is $\sqrt{\pi }$ and is not attained. This illustrates that supremum can fail to be attained at an oscillatory endpoint.

On $[\pi ,3\pi /2]$, set $x_0=\pi +\arcsin (2/(3\pi ))$. It lies strictly inside the interval, and $1/\sin x_0=-3\pi /2$, so $g(x_0)=\sqrt{x_0}>\sqrt{\pi }$. Choose $0<\delta <x_0-\pi$. On $[\pi ,\pi +\delta ]$,On $[\pi +\delta ,3\pi /2]$, $g$ is a continuous function, so the extreme value theorem gives a maximum there, at least $g(x_0)$. It exceeds every value on the discarded interval. Therefore $g$ does attain its supremum on $[\pi ,3\pi /2]$. This is the interior-value criterion for an attained maximum.

We use the Bolzano-Weierstrass theorem: every bounded real sequence has a convergent subsequence. If $f$ were unbounded on $[a,b]$, there would be $x_n\in [a,b]$ with $|f(x_n)|\ge n$. The Bolzano-Weierstrass theorem gives a subsequence $x_{n_k}\to c$, and the closed interval contains $c$. Since $f$ is a continuous function, $f(x_{n_k})\to f(c)$, contradicting $|f(x_{n_k})|\ge n_k\to \infty$. Thus $f$ is bounded.

Let $s={\sup }_{x\in [a,b]}f(x)$. By the definition of the supremum, choose $y_n$ with $s-1/n<f(y_n)\le s$. Again the Bolzano-Weierstrass theorem gives $y_{n_k}\to d\in [a,b]$. Continuity yields $f(d)=s$. Applying the same argument to $-f$ gives a point where $f$ attains its infimum. Therefore both the supremum and the infimum are attained. This proves the extreme value theorem for a closed bounded interval.

The Markov property and the Chapman-Kolmogorov equation giveFor each intermediate state, one first uses the next-step transition probability and then the following one; the sampled history adds no information once $X_{2n}$ is known. Thus $W$ is a homogeneous Markov chain with transition matrix $P^2$ and initial law $\lambda$. In particular,If zero is not a state of $S$, this probability is zero.

The sign of the Dirac delta distribution in the equation is positive after the outer minus sign is expanded. Integrating through zero, the nonsingular potential contributes a term tending to zero, whereas the delta term contributes $3k\psi (0)$. The jump condition for an impulse is thereforeContinuity at zero gives $C_{+}=C_{-}=C$. The two one-sided derivatives are $-3kC/2$ and $+3kC/2$, whose difference is exactly $-3kC$. Hence the matching condition introduces no further restriction on $C$, and the globally bounded solution isThis continuous solution has precisely the derivative jump required for its second distributional derivative to cancel the delta term. Its amplitude remains arbitrary because no normalization has been imposed.

Away from zero the Dirac delta distribution is absent. On either half-line, use $u=1+k|z|$ and the power-law ansatz $\psi =u^{\alpha }$. The Euler-Cauchy equation becomesThe growing term is excluded by boundedness at either end. Thus the independently bounded half-line solutions are

For real nonintegral $\alpha$, the Parseval identity and the coefficient formula giveSolving for the sum yieldsAt $\alpha =5/2$, the sine square is one and the cotangent is zero, giving $2{\pi }^2/25$.

Evaluate the absolutely convergent complex Fourier series at $x=\pi$. The factors $(-1{)}^n{e}^{in\pi }$ equal one, and the $n=0$ term is $1/{\alpha }^2$. Pairing positive and negative integers givesSince $\sin (\pi \alpha )\ne 0$ for nonintegral complex $\alpha$, rearranging provesThe last term means $\pi \cot (\pi \alpha )/(2\alpha )$, so half-integer arguments are included by the finite cotangent value zero.

Integrate the two exponentials in $\cos (\alpha x)$ against $e^{-inx}$. For $\alpha \notin \mathbb{Z}$ there are no zero denominators, andThus the complex Fourier series isThis is the periodic extension, not a globally nonperiodic cosine formula. For each fixed complex $\alpha$, the coefficients are $O(n^{-2})$, giving absolute and uniform convergence. The extension is continuous at the endpoints because the two endpoint values agree; piecewise smooth pointwise convergence of a piecewise smooth Fourier series identifies the sum with it, including those endpoints.

The Parseval identity for a complex Fourier series isIt holds for every square-integrable function, with complex conjugation expressed by the absolute squares. It is the Pythagorean theorem extended to the complete orthonormal system of complex exponentials.

The complex Fourier series of a $2L$-periodic function isThe coefficient formula follows from orthogonality of the exponential functions. For an arbitrary square-integrable function, this representation means convergence in the $L^2$ norm. With appropriate additional regularity it also gives pointwise convergence, to the mean of the one-sided limits at a jump. No pointwise representation is asserted for an unrestricted function merely from periodicity.

In $\mathbb{R}$, takeThe sets are nonempty, closed and strictly descending, but their intersection is empty. Their diameters are infinite, so the shrinking-diameter assumption is essential.

In $\mathbb{R}$, takeThese are nonempty descending sets of diameter $1/n$, with empty intersection. They are not closed sets, so there is no contradiction with the Cantor intersection theorem for the complete metric space $\mathbb{R}$.

Take the incomplete metric space $X=(0,1)$ with its usual distance, andEach is nonempty and closed relative to $X$: it is $X\cap (-\infty ,1/(n+1)]$. The sequence is strictly descending and its diameters are $1/(n+1)\to 0$. Its intersection is empty, since every positive point exceeds $1/(n+1)$ for sufficiently large $n$. The excluded limiting point is zero.