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The PDF gives $V=H(t)$, the Heaviside step function; the TeX incorrectly changes this to $V=I(t)$. A unit voltage step has derivative the Dirac delta function, so the current is the causal Green function for $L{d}^2/d{t}^2+Rd/dt+1/C$.

The current must be continuous at zero: a jump would create a delta derivative through $L{I}^{\prime \prime }$, absent on the right-hand side. Integrating across zero therefore gives $L[I^{\prime }]+R[I]=1$, so $I(0^{+})=0$ and $I^{\prime }(0^{+})=1/L$. PutThese distinct negative characteristic roots give the overdamped RLC responseTogether with $I(t)=0$ for $t<0$, this solves the step response. The eventual current is zero: the capacitor charges and blocks steady current in the series RLC circuit.

With no electrical resistance or applied electric potential difference, divide the equation by the inductance $L$ to get a simple harmonic oscillator with angular frequency ${\omega }_0=(LC{)}^{-1/2}$. Its general real solution isEach nonzero solution has period $2\pi \sqrt{LC}$. Here ${\omega }_0$ is an angular frequency; the ordinary frequency in cycles per unit time is ${\omega }_0/(2\pi )$.

The conditional density givesusing either integration by parts or the expected value of an exponential distribution. HenceThis is a finite value for almost every realization of $X$, but not an integrable random variable: $\mathbb{E}[1/X]={\int }_0^{\infty }{e}^{-x}/xdx=\infty$. Since $Y\ge 0$, its conditional expectation remains well-defined in the nonnegative, extended-expectation sense; the tower property of conditional expectation then gives $\mathbb{E}Y=\infty$.

Integrate the joint probability density over $y$ to obtain the marginal density of $X$:Thus $X$ has an exponential distribution of rate one. Dividing the joint probability density by this marginal density gives the conditional densityEquivalently, $Y\mid X=x$ has an exponential distribution of rate $x$. The value at $x=0$ can be assigned arbitrarily: it is a probability-zero conditioning value and the density ratio there is not meaningful.

Put $U_0=1$, so the same definition $z_n=x_n/U_{n-1}$ includes $n=1$. Since $U_n=a_n{U}_{n-1}$ and every $a_n$ is nonzero,The factor $U_n^{-1}$ is a discrete integrating factor for this variable-coefficient affine recurrence. Sum the increments as a telescoping series and multiply back by $U_n$:The identity requires no assumption that the coefficients have the same sign.

Subtract successive terms and use a telescoping series:Thus the linear recurrence relation has solution

Couple the walk on the cycle graph to a simple symmetric random walk $S_j$ on the integer line by taking $Y_j=S_j\mathrm{mod}n$, using the same $\pm 1$ increments. The visited integer interval has distinct residues until its length reaches $n$; an interval of exactly $n$ consecutive integers contains every residue once. Consequently the cover time on the cycle is exactly $T_n$ in this coupling.

Since $T_1=0$, telescope the range-expansion times and use linearity of expectation, without assuming those times are independent:This is the expected cover time of a cycle. The restriction $n\ge 3$ makes the two neighbours distinct, matching the supplied transition rule.

The vertices visited by a simple symmetric random walk form an integer interval: nearest-neighbour steps cannot skip a vertex. At the stopping time $T_k$, write that interval as $[\ell ,r]$, with $r-\ell +1=k$. The walk is at one of its endpoints, because the $k$th new vertex must extend the earlier interval. This also covers $k=1$, when both endpoints coincide.

The Strong Markov property restarts the walk from this endpoint. Visiting a new vertex is exactly exiting $[\ell ,r]$, or reaching $\ell -1$ or $r+1$. Translate these absorbing boundaries to $0,k+1$. The starting point becomes either $1$ or $k$, so the expected duration of symmetric gambler's ruin is $k$ in both cases. Taking conditional expectations and then expected values gives the expected time to expand a random-walk range:

Write $K=k+1$ and $h_j={\mathbb{E}}_j{D}_j$. The expected duration of symmetric gambler's ruin satisfies $h_0=h_K=0$ and, by conditioning on the first step,These expected values are finite: from any interior state, $K$ consecutive right steps ensure absorption, an event of probability at least $2^{-K}$. Blocks of $K$ steps therefore bound the survival probability by a decreasing geometric sequence. The candidate $h_j=j(K-j)$ has the required boundary values and satisfies $h_{j+1}-2h_j+h_{j-1}=-2$. HenceFor completeness, the difference of two solutions of this linear recurrence relation has zero second difference and is affine in $j$; its two zero boundary values force it to vanish.

If exactly $k$ spins equal $+1$, then $M=(2k-n)/n$. Therefore its possible values are $E_n$, with $k=n(1+m)/2$ for $m\in E_n$. Choosing the $k$ positive spins gives the binomial coefficientEvery such configuration has the same Curie–Weiss model weight because ${\sum }_{i,j}{s}_i{s}_j=({\sum }_i{s}_i{)}^2=n^2{m}^2$. The probability mass function of this spin magnetization is consequentlyand zero elsewhere. Grouping the original partition function by the same count also giveswhich verifies normalization without counting any configuration twice.

Let $C_{12}=\mathbb{E}[S_1{S}_2]\ge 0$, using the supplied correlation property of the Curie–Weiss model. Since $1_{\{S_i=1\}}=(1+S_i)/2$ and each spin has zero expected value,Divide by $\mathbb{P}(S_1=1)=1/2$ to obtainThe expression assumes $n\ge 2$, as required for $S_2$ to exist. It makes the connection between the spin correlation and the requested conditional probability explicit.

Use the positive exponent in the PDF. The TeX introduces a minus sign in the numerator while retaining the positive-sign normalizing constant, which would not define the stated probability distribution. The Curie–Weiss model assigns equal weights to $s$ and $-s$, since $({\sum }_j{s}_j{)}^2$ is unchanged by simultaneous reversal. This spin inversion symmetry pairs every configuration with one having the opposite value of $S_i$, so

Use the rate convention for the exponential distribution. By convolution of independent random variables, the probability density function of $S=X+Y$ is, for $s\ge 0$,It is zero for $s<0$. The expression is nonnegative whichever rate is larger, and defines a two-stage hypoexponential distribution. As a check, its limit as $\mu \to \lambda$ is ${\lambda }^{2}s{e}^{-\lambda s}$, the shape-two gamma distribution.

For $M=\min \{X,Y\}$, independence gives the survival function $\mathbb{P}(M>s)=e^{-\lambda s}{e}^{-\mu s}$ for $s\ge 0$. Differentiating its complement givesThus the minimum has an exponential distribution whose rate is the sum of the rates.

Apply Markov's inequality to the nonnegative random variable $e^{\lambda X}$, since $X>t$ implies $e^{\lambda X}>e^{\lambda t}$:This exponential Markov bound remains true, though uninformative, when the expected value is infinite. For a standard normal distribution, completion of the square gives its moment-generating functionMinimize ${\lambda }^2/2-\lambda t$ over positive $\lambda$; the choice $\lambda =t$ gives the Chernoff bound

Set $h(x)=f(x)-yx$. ThenThe extreme value theorem gives a minimum of $h$ on $[a,b]$. If it occurs in $(a,b)$, the two-sided difference quotient gives $h^{\prime }(c)=0$. If it occurs at $a$, either $h^{\prime }(a)=0$ or a negative right derivative contradicts minimality; the endpoint $b$ is analogous. Hence some $c\in [a,b]$ satisfiesThis proves the Darboux theorem for derivatives in the stated case without assuming that $f^{\prime }$ is continuous.

Because a differentiable function is continuous, defineThe fundamental theorem of calculus gives $F^{\prime }(x)=f^{\prime }(x)+f(x)$. At the endpoints,whileApplying the result just proved to $F$ yields $d\in [a,b]$ with

A function is differentiable at $a$ when the finite limitexists. A derivative need not be continuous. For example,is differentiable everywhere and $f^{\prime }(0)=0$, but for $x\ne 0$,which has no limit at zero.

The mean value theorem states that if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then some $c\in (a,b)$ satisfies

The intermediate value theorem states that if $f:[a,b]\to \mathbb{R}$ is continuous and $y$ lies between $f(a)$ and $f(b)$, then there exists $c\in [a,b]$ such that $f(c)=y$.

First computeThe vector $\ell =(1,-2,4{)}^T$ lies in the kernel of $(AB{)}^T$, since ${\ell }^{T}AB=0$. If $ABx=d$ is solvable, multiplying by ${\ell }^T$ gives the necessary compatibility conditionTherefore