Use the first 10, 20, 30 and 40 entries of the already computed list in a two-by-two array of axes. Each subplot needs an index sequence starting at one and having the same length as its data:

The panels use different vertical scales, which is why the exponential-looking growth remains visible at each range. The available values through 50 are retained in list50, even though the requested four panels stop at 40.

A list comprehension evaluates the function for the fifty required indices:

Python's upper range endpoint is excluded, so 51 includes index 50. The first entries are $1,1,2,3,5,8,13,21,34,55$, and the final entry is $F_{50}=12586269025$.

The expression is the Binet formula for the Fibonacci numbers. Compute the two roots and round the result to the nearest integer:

The returned object is an integer. This floating-point arithmetic implementation correctly rounds the requested range through $n=50$; it is not an arbitrary-precision method for indefinitely large $n$. An integer recurrence is a useful independent numerical check.

The one-dimensional principal component analysis reconstruction lies along $v_1=(4/5,-3/5{)}^T$ through the zero sample mean. Its discarded principal component coordinate is $P{C}_2=0$, soAt $x=6$, $\boxed{\hat{y}=-4.5}$. This intersects the retained principal component line with the vertical line at the supplied $x$. It is a reconstruction from that line, rather than an ordinary linear regression of $y$ on $x$.

The explained variance of a principal component is its eigenvalue divided by the sum of all eigenvalues of the covariance matrix. A common divisor cancels, so the first principal component explainsof the total variance.

Place observations in rows of the data matrix $X$. Since the coordinate sample means are zero, the Gram matrix of its columns isThe sample covariance matrix is $S=X^{T}X/(3-1)$; using divisor three instead changes eigenvalues by a common factor and leaves principal component loadings unchanged. For $X^{T}X$, the characteristic polynomial isFor the larger eigenvalue $150$, the eigenvector equation gives $-36\alpha -48\beta =0$, so $3\alpha +4\beta =0$. Normalize to unit Euclidean norm and choose $v_1=(4/5,-3/5{)}^T$. An orthogonal unit eigenvector for $50$ is $v_2=(3/5,4/5{)}^T$. ThereforeThe sample variances of these principal components are $75$ and $25$, respectively. Their principal component scores for observations $A,B,C$ areEither eigenvector may be multiplied by $-1$ without changing its principal component direction or its variance. No coordinate standardization is applied here: that would instead be principal component analysis on a correlation matrix.

Yes, the data are centred. Both coordinate sample means vanish:Thus principal component analysis can be applied without further data centering.

The conserved total molecule amount fixes the final common concentration:Solve $c_c-c_n=\Delta c(t)$ together with $V_c{c}_c+V_n{c}_n=(V_c+V_n)c_{\infty }$ to obtainSubstituting the decaying concentration difference from part (d) supplies both explicit time dependences. The concentrations approach the volume-weighted initial average, which need not be the arithmetic average if the compartment volumes differ.

Take the membrane particle flux density to be positive from cytoplasm into the cell nucleus. With membrane permeability $\alpha$,For fixed compartment volumes, conservation of mass and the absence of exterior flux giveSubtract the two concentration derivatives:This linear ordinary differential equation has solutionIts relaxation time is $\boxed{T=V_c{V}_n/[\alpha A_n(V_c+V_n)]}$. Here $\alpha$ has dimensions length/time, so $\alpha A_n/V$ has dimensions inverse time. A positive initial concentration difference decays monotonically toward zero; a negative one approaches zero from below.

On a moving step, the largest speed resolved by the discrete model is the distance per interval,This is a step-averaged speed; the model does not specify motion within the interval. Use the cell length $a=2r=20\mu \mathrm{m}=2.0\times 10^{-3}\mathrm{cm}$ in the stated Reynolds number formula. In consistent centimetre–gram–second units,Thus viscous effects dominate inertia overwhelmingly. The surrounding flow is in the Stokes flow regime: sustained movement needs applied forces, and inertial coasting is negligible. Using radius rather than the specified length as the characteristic size would halve the Reynolds number, without changing that physical conclusion.

For unbiased one-dimensional diffusion, the mean squared displacement is $⟨x^2(t)⟩=2Dt$. With $t=n\tau$, compare this to the mutant expression:Therefore the mutant diffusion coefficient is

Let ${\xi }_j$ be the displacement in step $j$, so the random walk position is $x(n)={\sum }_{j=1}^n{\xi }_j$. The single-step expected value and second moment areBy linearity of expected value, $\boxed{⟨x(n)⟩=0}$. The steps are independent random variables, so for $i\ne j$, $\mathbb{E}[{\xi }_i{\xi }_j]=\mathbb{E}[{\xi }_i]\mathbb{E}[{\xi }_j]=0$. Expanding the square gives the mean squared displacementThese are ensemble expected values; the empirical average over a finite number $N$ of independent cells fluctuates about them. The expected mean squared displacement itself does not depend on $N$.

The diagram has all possible undirected edges except $AB$: it is a complete graph on five vertices with one edge removed. Therefore $A$ and $B$ each have three neighbours, while $C,D,E$ each have four.

The neighbours of $A$ and $B$ form a triangle, so their clustering coefficients equal one. Each of $C,D,E$ has five of the six possible neighbour–neighbour edges, giving $5/6$. The nodal average shortest path length is ${\ell }_i=\frac{1}{4}{\sum }_{j\ne i}d(i,j)$. All graph distances equal one except $d(A,B)=d(B,A)=2$, henceAveraging the nodal quantities givesThe ordered-pair distance sum is $22$, and the graph diameter is two.

If the intended unnormalized “small-worldness” is the ratio of clustering to path length, the computed quantities giveThat ratio has no universal threshold for identifying a small-world network. The standard reference-normalized small-worldness, $\sigma =(\bar{C}/C_{\mathrm{ref}})/(L/L_{\mathrm{ref}})$, cannot receive a numerical value when no reference is specified and the question forbids using a network null model.^[ref]

The graph has high clustering and short paths in the descriptive sense. With nine of ten possible edges, however, those properties follow directly from its near completeness; they do not by themselves establish small-world network structure relative to a reference. This distinguishes the metrics that can be computed directly from the stronger classification that requires a stated convention.

The supplied adjacency matrix describes a four-vertex cycle graph, with every vertex degree equal to two. Degree-preserving edge rewiring is random, so it does not determine one unique labelled output adjacency matrix. For example, replace edges $AB$ and $CD$ by $AC$ and $BD$, retaining $AD$ and $BC$. This valid switch gives, in order $A,B,C,D$,Each row still sums to two. Under the simple graph convention, all allowed outputs have the same unlabelled four-cycle topology: there are three possible labelled four-cycles. A rejected switch or later return can also leave the original adjacency matrix.

It is not a scale-free network. Its degree distribution is concentrated entirely at degree two, with no power-law tail; degree-preserving edge rewiring cannot change that.