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For $x,y\in [0,1]$, the binary entropy and the golden ratio $\phi$ satisfy

On the doubly infinite ladder graph, self-avoiding walks from a fixed graph vertex using only rightward or vertical steps are in bijection with words in $R,V$ containing no consecutive $V$ symbols. Their count is ${\sigma }_n=F_{n+2}$, since ${\sigma }_0=1$, ${\sigma }_1=2$, and ${\sigma }_n={\sigma }_{n-1}+{\sigma }_{n-2}$. Its exponential growth rate is the golden ratio.

A generating function packages a sequence into a series so that operations on objects become operations on their encodings. An ordinary generating function uses the power series $A(z)={\sum }_{n\ge 0}{a}_n{z}^n$; in counting problems, the coefficient $a_n$ records the number of objects of size $n$. The variable $z$ initially marks size: it need not be assigned a numerical value. Coefficient extraction writes this recovery as $[z^n]A(z)=a_n$. One compact expression can therefore answer infinitely many counting questions, while explaining how their answers fit together.

Start with two contributions: choose $i$ from $\{0,1,2\}$ and $j$ from $\{0,1\}$, with exactly one option at each allowed value. Their individual ordinary generating functions are the polynomials $P(x)=1+x+x^2$ and $Q(x)=1+x$. To count ordered pairs by total $i+j$, multiply:The coefficients $1,2,2,1$ count the ways to obtain totals zero through three. For total two, the possibilities are $(1,1)$ and $(2,0)$; the two corresponding monomials both contribute $x^2$. There are six ordered pairs altogether, agreeing with $P(1)Q(1)=3\cdot 2=6$.

If the first contribution has two distinguishable choices when $i=1$, replace $P(x)$ by $1+2x+x^2$. Its product with $Q(x)$ becomes $1+3x+3x^2+x^3$. The coefficients now carry multiplicities as well as possible totals; the exponent still records cost. This lets the same construction handle several actions with identical costs without losing their separate identities.

In general, the Cauchy product givesFor each split of the total size, the rule of product counts the choices of the two components; summing over those splits counts all possibilities. This requires each combined object to have a unique decomposition into the two components. The same idea works for costs, lengths, or other additive measures, not just the number of pieces. See MIT 18.310, generating functions, sections 1–2.

Now cover a board of length $n$ with tiles of lengths one and two, without gaps or overlaps. Let $t_n$ count these tilings. There is exactly one empty tiling, so $t_0=1$, and $t_1=1$. Removing the first tile leaves a board shorter by one or two, giving the recurrence relationThese are the Fibonacci numbers shifted by one index: $t_n=F_{n+1}$ under the convention $F_0=0$, $F_1=1$. Keeping the empty tiling is essential to both the counting rule and the resulting ordinary generating function.

For $F(z)={\sum }_{n\ge 0}{t}_n{z}^n$, the empty board or a first tile followed by a remaining tiling givesEquivalently, $F(z)={\sum }_{m\ge 0}(z+z^2{)}^m$: each factor chooses one tile, and $m$ counts pieces while the exponent of $z$ records total length. The same ordinary generating function counts ordered histories of two repeatable actions costing one and two units. Histories $(1,2)$ and $(2,1)$ differ despite having the same total cost. MIT's notes use dots and dashes with these costs, including the empty message. ^[ref]

These identities hold in formal power series algebra: each requested coefficient uses only finitely many contributions. The repeated-choice expansion is legitimate because $z+z^2$ has zero constant term; a zero-cost repeatable action would allow infinitely many histories of the same size. Numerical evaluation separately requires a convergent series. Here the radius of convergence is $(\sqrt{5}-1)/2$, the reciprocal of the golden ratio. At $z=1$, the rational function equals $-1$, but the counting series diverges: that value cannot represent its sum.

Other generating functions change the encoding: an exponential generating function divides by factorials, while a Dirichlet series uses $n^{-s}$. A counting generating function also differs from a probability generating function, whose coefficients must be nonnegative and sum to one.

When each variable appears in at most one normalized singleton clause, satisfy that favored literal with probability $p>1/2$ and choose independent variables. Every proper binary clause is then satisfied with probability at least $1-p^2$, and every singleton with probability $p$. Maximize the common lower bound by $p=1-p^2$, giving the reciprocal of the golden ratio. The method of conditional probabilities derandomizes the construction, obtaining approximation ratio $p$. Tautologies are harmless; the singleton restriction must be applied after removing repeated literals within clauses.

For $\phi =(1+\sqrt{5})/2$ and $\psi =(1-\sqrt{5})/2$, each power sequence satisfies the recurrence relation $x_{n+2}=x_{n+1}+x_n$, because both bases solve $t^2=t+1$. Their sum starts at $2,1$, so uniqueness of a recursively defined sequence proves that it is the Lucas number sequence. Alternatively, diagonalize the companion matrix $\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$ and use $(1,2{)}^T=(\phi ,1{)}^T+(\psi ,1{)}^T$. The formula explains the exponential growth governed by the golden ratio and the alternating decaying correction ${\psi }^n$.

For a closed right circular cylinder with radius $r>0$, height $h>0$, and total surface area $A=2\pi r(r+h)$, inscription in a sphere of radius $R$ gives $r^2+h^2/4=R^2$. The Lagrange multiplier equations imply $h^2+2rh-4r^2=0$, so the positive aspect ratio is $h/r=\sqrt{5}-1$. The dimensions are $r^2=R^2(5+\sqrt{5})/10$ and $h^2=2R^2(5-\sqrt{5})/5$.

There is also an eigenvalue proof of global maximality. Set $w=h/2$, so the constraint is $r^2+w^2=R^2$, and write $A/(2\pi )=r^2+2rw$. This quadratic form has symmetric matrix $\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$ with largest eigenvalue $(1+\sqrt{5})/2$, the golden ratio. Its maximizing eigenvector lies in the positive quadrant, yielding the displayed area and aspect ratio.

Represent the doubly infinite ladder graph by $\mathbb{Z}\times \{0,1\}$, with horizontal edges between successive columns and a vertical rung at each column. An allowed self-avoiding walk is coded by a word in $R,V$. Two successive $V$ steps revisit the preceding graph vertex and are forbidden. Conversely, every word without consecutive $V$ steps is a self-avoiding walk: horizontal steps strictly increase the column, and a column is visited vertically at most once. This establishes a bijection, not merely an upper bound.

For the directed ladder self-avoiding walk count, ${\sigma }_0=1$, ${\sigma }_1=2$, and for $n\ge 2$ a word either ends in $R$, or ends in $RV$. Removing that final block givesLet $\phi =(1+\sqrt{5})/2$ be the golden ratio, and $\psi =(1-\sqrt{5})/2$. The Binet formula for the Fibonacci number yieldsSince $|\psi |<\phi$, taking $n$th roots gives ${\lim }_{n\to \infty }{\sigma }_n^{1/n}=\phi$. This use of the golden-ratio symbol is independent of the connection rate in Question 1.

When $T\ll \Delta$, the argument of the inverse hyperbolic sine is large. Using $\mathrm{arsinh}z=\log (2z)+1/(4z^2)+O(z^{-4})$ givesThe mass remains the zero-temperature gap, with an exponentially small correction from thermally activated excitations.

When $T\gg \Delta$, expand around $z=1/2$. Ifis the golden ratio, then $\mathrm{arsinh}(1/2)=\log \phi$, andThe leading value $m/T=2\log \phi$ is universal. This is the quantum-critical regime: temperature is the only leading energy scale and the correlation length is of order $v/T$.

Letwhere $\phi$ is the golden ratio. If the union-closed family consists of one nonempty set, any element of that set has frequency one, so assume its cardinality exceeds one. Choose independent uniform members $A,B$ and let $X,Y\in \{0,1{\}}^n$ be their characteristic vectors of sets. Then $H(X)=H(Y)=\log |\mathcal{A}|>0$.

Suppose for a contradiction that every element has frequency $p_i<c$. Put $q_i=1-p_i>1/\phi$, and let $Z=X\mathrm{OR}Y$, the characteristic vector of a set of $A\cup B$. The chain rule for information entropy and conditioning reduces entropy givebecause $Z_{<i}$ is a function of $(X_{<i},Y_{<i})$.

Fix the two prefixes and setThe two conditioned bits are independent random variables, and $Z_i=0$ exactly when both are zero. The supplied binary entropy product inequality therefore givesAveraging over the independent prefixes yieldswhenever either conditional entropy is positive. Summing and using $H(X)>0$ gives $H(Z)>H(X)$.

But set union keeps $A\cup B$ inside the union-closed family, so $Z$ is supported on $\mathcal{A}$. The maximum entropy distribution on a finite set gives $H(Z)\le \log |\mathcal{A}|=H(X)$, a contradiction. Some element must therefore occur in at least $c|\mathcal{A}|$ members, proving the entropy bound for a union-closed family.

Keep the constraint from part (b). For equal masses and spring constants, the generalized eigenvalue problem for small oscillations reduces toThe horizontal coordinate decouples, givingIn this normal mode, mass 1 oscillates horizontally, mass 2 has its horizontal coordinate fixed, and neither mass has a vertical displacement.

For the vertical block, write $\lambda =m{\omega }^2/k$. Its characteristic polynomial is $(2-\lambda )(1-\lambda )-1={\lambda }^2-3\lambda +1$. Thus ${\lambda }_{\pm }=(3\pm \sqrt{5})/2$, and the first vertical equation gives $y_2/y_1=2-{\lambda }_{\pm }$. If $\phi =(1+\sqrt{5})/2$ is the golden ratio, the two vertical modes areIn the lower-frequency mode, the masses move vertically in phase, with the lower mass having the larger amplitude. In the higher-frequency mode they move oppositely, with lower-to-upper amplitude ratio $1/\phi$. Their Euclidean inner product is zero because $1+\phi (-1/\phi )=0$; since $M=mI$, this is also mass-matrix orthogonality of normal modes.

The three frequencies satisfy ${\omega }_{-}<{\omega }_x<{\omega }_{+}$. Each mode has a scalar coordinate $A\cos \omega t+B\sin \omega t$, and their superposition gives the general constrained motion. No zero-frequency mode remains because the first spring anchors the system. Gravity fixes the equilibrium extensions but drops out of these frequencies for zero-natural-length springs.

Try a linear function $f(x)=cx$. The functional equation becomes $c^{2}x=(c+1)x$ for all $x$, so $c$ must be a root of a polynomial $c^2-c-1$. Its two roots of a polynomial giveFor either choice, $c^2=c+1$ proves the required equality directly. The positive coefficient is the golden ratio.

Let $\varphi =(1+\sqrt{5})/2$, the golden ratio. Since ${\varphi }^2=\varphi +1$, multiplication by ${\varphi }^n$ gives ${\varphi }^{n+2}={\varphi }^{n+1}+{\varphi }^n$. Adding this recurrence relation to the one for $A_n$ yieldsThus the extra exponential term changes the initial values, but not the linear recurrence relation.

The characteristic polynomial isIts two distinct real roots are the eigenvaluesTheir product is $-1$, so $\psi =-1/\phi$. Equivalently, ${\phi }^2=\phi +1$ implies $1/\phi =\phi -1$. The positive eigenvalue is the golden ratio.

Willem Abraham Wythoff, also spelled Willem Abraham Wijthoff, worked on combinatorial game theory and the geometry of polytopes. His1907 modification of Nim became Wythoff's game. His work also included a published proof of the triangle-free graph bound associated with Willem Mantel.^[ref] The game illustrates how a small change to legal moves can replace a binary-arithmetic strategy by one organized through Beatty sequences and the golden ratio. His geometric work concerns the symmetry and construction of higher-dimensional polytopes, rather than only recreational games.

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