The conditional expectation given an event of positive probability isApply the law of total probability to each event $\{X=j\}$:This is the law of total expectation. Interchanging the sums is justified by nonnegative terms, for example by Tonelli theorem, so it remains valid when the expectation is infinite.

Because the $B_i$ partition the sample space, the events $A\cap B_i$ are disjoint and their union is $A$. Finite additivity and the definition of conditional probability give the law of total probabilityThe assumption $\mathbb{P}(B_i)>0$ ensures that each conditional probability is defined.

For a nonnegative integer-valued random variable, its expectation is the probability-weighted sumAll terms are nonnegative, so this is always defined as an extended nonnegative number, possibly $+\infty$.

Substitute the proposed Fourier sine series into the heat equation:Projecting onto each sine using orthogonality gives $T_n^{\prime }=-D{k}_n^2{T}_n$ for every mode with $b_n\ne 0$. The initial condition fixes $T_n(0)=1$, since $b_n$ already describes the initial amplitude. HenceFor a mode with $b_n=0$, the value of $T_n$ is immaterial and the same convention is convenient. The positive thermal diffusivity $D$ damps every mode, with higher frequencies decaying faster. For square-integrable initial data the damping justifies termwise differentiation for $t>0$ and recovery of the initial data in L2 norm convergence as $t\downarrow 0$.

The left panel illustrates the Fourier series in part (b). The right panel is a self-generated example with $D=L=1$ and ${\Theta }_0(x)=\sin (\pi x)+\frac{1}{2}\sin (3\pi x)$, showing the two normal modes decaying at rates ${\pi }^2$ and $9{\pi }^2$.

A Fourier sine series automatically meets the zero-temperature boundary conditions because $\sin 0=\sin (n\pi )=0$. Equivalently, extend ${\Theta }_0$ oddly from $(0,L)$ to $(-L,L)$ and then periodically with period $2L$. Its Fourier series has only sine terms, withThe coefficient formula follows from the orthogonality identityFor piecewise smooth initial data the series equals ${\Theta }_0$ at interior continuity points and the average of its one-sided values at interior jumps. More general square-integrable data have convergence in L2 norm convergence. A classical solution continuous at the initial corners additionally requires ${\Theta }_0(0)={\Theta }_0(L)=0$, in agreement with the prescribed boundary conditions.

Multiplying the recurrence relation by $n$ removes the average:For $n\ge 2$, subtract the same identity at $n-1$:Rearranging gives $T(n)=T(n-1)+2-1/n$. This also holds at $n=1$, because the original recurrence relation gives $T(1)=1$ and $T(0)=0$. Summing the differences therefore yieldsHere $H_n$ is a harmonic number. Since $H_n\le n$, we have $n\le T(n)\le 2n$ for $n\ge 1$, also establishing $\Theta (n)$ in big theta notation.

Unroll the recurrence relation down to its initial condition:ThusIndeed $T(n)/n^2\to 1/2$, so the sharper growth classification is $\Theta (n^2)$ in big theta notation.

For natural logarithms and sufficiently large $n$, take the logarithm of the positive expression:Set $t=\log \log n$. Then $\log n=e^t$. The exponential function grows faster than $t^2$: for example, its power series gives $e^t\ge t^3/6$ for $t>0$, so $t^2/e^t\le 6/t\to 0$. In particular, eventually $(\log \log n{)}^2\le \log n$. Exponentiating yieldsFor logarithms to a fixed base $b>1$, the natural logarithm of the left side is $(\log \log n-\log \log b{)}^2/\log b$, still smaller than $\log n$ eventually. The big O notation conclusion is therefore independent of that base choice.

Use natural logarithms; the same argument works for any fixed base greater than $1$. The function $\log \log x$ is increasing for $x>1$, since its derivative is $1/(x\log x)>0$. With natural logs the sole negative summand is at $i=2$. Put $C_0=|\log \log 2|$.

For all sufficiently large $n$ with $\log \log n\ge 1$, monotonicity and the triangle inequality giveThese are fixed constants and an eventual bound, provingFor another fixed logarithm base there are still only finitely many negative initial terms, which are absorbed in $C_0$.

Write $n=2r$ and $k=2s$. Partition an $n$-element set into $r$ distinguished pairs, and let $\tau$ interchange the two elements of every pair. It induces an involution on the $k$-element subsets. Every nonfixed subset is paired with its distinct image, so nonfixed subsets contribute an even number to the binomial coefficient $\left(\genfrac{}{}{0ex}{}{2r}{2s}\right)$.

A fixed subset contains either both or neither of the elements of each distinguished pair. Since it contains $2s$ elements, it consists of exactly $s$ whole pairs. There are $\left(\genfrac{}{}{0ex}{}{r}{s}\right)$ such fixed subsets. Counting modulo $2$ therefore givesThe same involution explains part (i): a fixed subset cannot have odd size.

The stated identity $k\left(\genfrac{}{}{0ex}{}{n}{k}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)$ has an even right side because $n$ is even. Its left side is an odd integer $k$ times the integer binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$. An odd multiplier preserves integer parity, so that coefficient must be even. Equivalently, reducing the identity in modular arithmetic gives

Fix $n\ge 1$ and use mathematical induction on $m$ with two initial cases, as appropriate for the second-order Fibonacci recurrence. At $m=0$, the right side is $F_{n-1}{F}_0+F_n{F}_1=F_n$. At $m=1$, it is $F_{n-1}+F_n=F_{n+1}$.

If the identity holds at $m$ and $m+1$, thenHence the induction establishes

Use mathematical induction, starting at $n=1$. Since $F_0=0$ and $F_1=1$, the sum is $-1=(-1{)}^1{F}_0-1$.

Suppose the formula holds at $n\ge 1$. Adding the next Fibonacci number with its sign and using the Fibonacci recurrence givesThis is the required successor case. Thus

The algebraic identitysuggests $f(t)=t^3-3t$ for every $t>0$, and substitution verifies the functional equation.

In fact $x+1/x\ge 2$, since $(x-1{)}^2/x\ge 0$. Conversely, for each $t\ge 2$, the quadratic equation $x^2-tx+1=0$ has a positive root $(t+\sqrt{t^2-4})/2$. Thus the functional equation forces $f(t)=t^3-3t$ exactly on $[2,\infty )$; its values on $(0,2)$ are arbitrary. The displayed function is one convenient choice on the full required domain.

Replace $x$ by $1/x$, which stays in the domain of the functional equation. Writing $u=f(x)$ and $v=f(1/x)$ gives the linear systemTwice the second equation minus the first gives $3u=2/x-x$, so the only possible function isTo check existence as well as uniqueness, substitute this expression: $f(x)+2f(1/x)=(2/x-x+4x-2/x)/3=x$.

Let $h:B\to A$ be a two-sided inverse function of $f$. If $f(a)=f(a^{\prime })$, apply $h$ and use $h\circ f={\mathrm{id}}_A$ to obtain $a=a^{\prime }$, proving injectivity. For any $b\in B$, the other identity gives $b=f(h(b))$, proving surjectivity. Therefore $f$ is a bijection. Both inverse identities matter: a one-sided inverse alone would establish only the corresponding one of these two properties.

For each $c\in C$, surjectivity of the function composition gives $a\in A$ such that $g(f(a))=c$. Taking $b=f(a)\in B$ supplies a preimage of $c$ under $g$. Thus $g$ is a surjective function.

The function composition $g\circ f:A\to C$ is defined by $(g\circ f)(a)=g(f(a))$. Suppose $f(a)=f(a^{\prime })$. Applying $g$ gives $(g\circ f)(a)=(g\circ f)(a^{\prime })$. Since the composite is an injective function, $a=a^{\prime }$. Thus $f$ is an injective function.

Using set difference as intersection with a complement of a set, then De Morgan's laws and distribution, givesFor the third equality, expand the two set unions: the terms $A\cap C^c$ and $A\cap B$ are absorbed by $A$, and $A\cap A=A$. This proves the required identity entirely by set algebra.

Write $C^c=\mathcal{U}\setminus C$. The rules for set difference, set intersection, and complement of a set giveThe second step is De Morgan's laws; the third distributes set intersection over set union. The fourth uses $A\cap A^c=\varnothing$.