The first swap necessarily removes one red ball from urn 1 and replaces it with a blue ball. Thus $\boxed{R_1=n-1}$ with probability one.

A partition of a sample space is a finite or countable family $(A_j)$ of pairwise disjoint events whose union is the whole sample space. If $X$ is integrable, the law of total expectation statesThis follows by summing $X{1}_{A_j}$ over the partition of a sample space; absolute integrability justifies the countable sum. For nonnegative $X$, the identity also holds with infinite expectations. Zero-probability parts contribute zero without requiring a conditional value on them.

For $\mathbb{P}(X=x)>0$, the conditional probability mass function isThe denominator is the sum of the numerator over all possible $y$, so these conditional probabilities sum to one.

The joint probability mass function isIt is nonnegative and its sum over $S\times S$ is one. Summing over one coordinate gives the other marginal probability mass function.

On the constraint $x^2-z^2=1$, substitute $x^2=1+z^2$ to obtainTherefore the two global minima areEvery pair $(y,z)$ is allowed on either smooth branch $x=\pm \sqrt{1+z^2}$, so these are also the only local extrema. There is no local maximum or global maximum: the displayed positive quadratic in $(y,z)$ has no other stationary point and is unbounded above. At the minima $F_z=-2z=1$, consistent with the hypothesis in part i.

Assume $f,F$ are continuously differentiable near the point and $F_z\ne 0$. The implicit function theorem supplies a local function $z(x,y)$ withAt an interior extremum of $g(x,y)=f(x,y,z(x,y))$, the multivariable chain rule givesMultiply these identities by $F_y$ and $F_x$, respectively, and subtract. The terms containing $f_z$ cancel, givingNo assumption $f_z\ne 0$ is needed for this elimination; dividing by $f_z$ would unnecessarily exclude valid cases. If $F_z=0$, this particular local graph argument is unavailable and a different chart or the full Lagrange multiplier condition is needed.

The tangent addition formula givesHence $\boxed{f(v)=\tan v}$. The restriction $xy\ne 1$ excludes the poles of tangent that can arise in the possible range $-\pi <v<\pi$. One should not replace this identity by $v=\arctan u$ globally: the principal inverse tangent can differ from $v$ by an integer multiple of $\pi$.

The partial derivatives areThus the Jacobian determinant isThis vanishing reflects the functional dependence found next.

No: the unbounded length of the domain matters. Let $\varphi (t)=\max \{1-|t|,0\}$ and define $f_n(x)=n^{-1}\varphi (x/n)$ on $\mathbb{R}$. These are continuous functions with compact support, and ${\sup }_{\mathbb{R}}|f_n|=1/n\to 0$, so the sequence has uniform convergence to zero. However, the change of variables $t=x/n$ givesThe integral of the limit is zero. The supports widen as the heights decrease; a uniform bound on the heights gives no bound on the total area over an infinite interval.

Take $f_n(x)=x^n$ on $[0,1]$. Each function is continuous, and the pointwise limit is $f(x)=0$ for $0\le x<1$, while $f(1)=1$. NeverthelessThe limiting function is Riemann integrable, since changing a value at one point does not change its Riemann integral. There is no uniform convergence: for every $n$, ${\sup }_{[0,1]}|f_n-f|=1$, with the supremum approached as $x\uparrow 1$ from below. Equivalently, the uniform limit theorem would require the limit to be continuous.

Let $f_n$ be the continuous triangular bumpand set $f_1=0$. At $x=0$ every term is zero, and each fixed $x>0$ eventually lies outside the support, so $f_n\to 0$ pointwise. The triangle has base $2/n$ and height $n$, henceThis concentrating triangular bump shows that pointwise convergence does not preserve integrals, even for continuous functions on a compact interval.

By the mean value theorem, $|f(x)-f(y)|\le A|x-y|$ for all real $x,y$. Thus $f$ is a contraction mapping on the complete metric space $\mathbb{R}$. Starting from arbitrary $x_1$, define $x_{n+1}=f(x_n)$. Induction gives $|x_{n+1}-x_n|\le A^{n-1}|x_2-x_1|$. For $m>n$, the triangle inequality and a geometric series yieldHence $(x_n)$ is a Cauchy sequence and completeness of the real numbers supplies a limit $x$. The Lipschitz continuity of $f$ implies $f(x)=\lim f(x_n)=\lim x_{n+1}=x$. Thus the limit is a fixed point, independently of the initial value. If $y$ were another, $|x-y|\le A|x-y|$ would force $x=y$. This proves the contraction mapping theorem in this setting, with error estimate

For every real $t$, $f(t)-t=1/(1+e^t)>0$, so there is no fixed point. Direct differentiation givesWriting $u=e^t>0$, the inequality $(1+u{)}^2\ge 4u$ showsIn particular $0<f^{\prime }(t)<1$, but $f^{\prime }(t)\to 1$ as $t\to \pm \infty$. There is no common contraction constant smaller than one. Strict pointwise bounds on the derivative therefore do not suffice for the contraction mapping theorem.

Suppose distinct fixed points $x<y$ existed. Then $h(t)=f(t)-t$ satisfies $h(x)=h(y)=0$. Rolle's theorem supplies $c\in (x,y)$ with $h^{\prime }(c)=0$, hence $f^{\prime }(c)=1$, contrary to the hypothesis. Thus there is at most one fixed point. The argument does not require $f^{\prime }$ to be continuous.

The derivative of $h(x)=f(x)-\sin x$ vanishes. By the constancy theorem on the interval $I$,This need not hold on a disconnected domain. On $I=(-\infty ,0)\cup (0,\infty )$, define $f(x)=\sin x$ for $x<0$ and $f(x)=\sin x+1$ for $x>0$. The function is differentiable throughout $I$ with $f^{\prime }(x)=\cos x$, but no single $C$ works on both components.

The constancy theorem states that a differentiable function on an interval whose derivative vanishes everywhere is constant. At any included endpoints, continuity there is understood.

Indeed, take arbitrary $x<y$ in the interval. The mean value theorem applies on $[x,y]$ and gives $f(y)-f(x)=f^{\prime }(c)(y-x)=0$. Every pair of values is equal, proving the assertion. The interval hypothesis ensures that the whole segment $[x,y]$ lies in the domain.

The mean value theorem states that if $f:[a,b]\to \mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$, where $a<b$, there is $c\in (a,b)$ such thatNo continuity assumption on the derivative is needed.

On $E=\mathbb{R}$, take $f_n(x)=x$ and $g_n(x)=1/n$. Then $f_n\to f(x)=x$ with uniform convergence, because there is no error, and $g_n\to 0$ with uniform convergence, with error $1/n$. Their products $f_n{g}_n=x/n$ converge pointwise to zero, but for every $n$ the choice $x=n$ gives $|f_n(x)g_n(x)|=1$. Thus the products do not converge uniformly. Here the first family consists of unbounded functions, exactly the missing hypothesis in the preceding estimate.

Write $f_n\to f$ and $g_n\to g$ with uniform convergence. The triangle inequality givesHence $f_n+g_n\to f+g$ with uniform convergence.

If every $f_n,g_n$ is a bounded function, part iii gives constants $M_f,M_g$ bounding the two families. Their pointwise limits satisfy $|f|\le M_f$ and $|g|\le M_g$. ConsequentlyThus both sums and products have uniform convergence under the stated hypotheses. The product estimate is the uniform convergence product rule for bounded functions.

Use uniform convergence to choose $N$ with $|f_n(x)-f_N(x)|\le 1$ for all $n\ge N$ and $x\in E$; this follows, for example, from the Uniform Cauchy criterion. Since each $f_j$ is a bounded function, put $M_j={\sup }_E|f_j|<\infty$ for $1\le j\le N$. Thenfor every $n$ and $x$. Thus $(f_n)$ is a uniformly bounded family of functions. The finite initial segment is essential: uniform convergence directly controls only a tail of a sequence.