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A sufficient statistic is one for which the conditional distribution of the full sample, given its value, is independent of the parameter. For independent observations the likelihood is ${\prod }_{i=1}^6{p}_i^{n_i}$, depending on the data only through the counts. The factorization criterion for sufficiency therefore makes $(n_1,\dots ,n_6)$ sufficient. Directly, all ordered samples with given counts have the same probability, so their conditional distribution is uniform over the $n!/\prod n_i!$ possible arrangements, independently of the $p_i$.

The unrestricted maximum-likelihood estimator is ${\hat{p}}_i=n_i/n$. Thus the generalized likelihood-ratio test useswith zero-count terms interpreted by continuity. Reject for small $\Lambda$, equivalently large $D$.

Write $n_i=n/6+{\delta }_i$, so $\sum {\delta }_i=0$. A Taylor series expansion of $(n/6+{\delta }_i)\log (1+6{\delta }_i/n)$ gives ${\delta }_i+3{\delta }_i^2/n+O(|{\delta }_i{|}^3/n^2)$. Summing cancels the linear terms, and yieldsUnder the null hypothesis, the multinomial central limit theorem places the standardized count deviations in the five-dimensional subspace with coordinate sum zero, with identity covariance on that subspace. Their squared norm consequently tends to the chi-squared distribution with five degrees of freedom. This gives the Pearson chi-squared goodness-of-fit test.

For $T=8.12$, the asymptotic p-value is about $0.150$. The 5% critical value of ${\chi }_5^2$ is approximately $11.07$. Do not reject at the usual 5% level, or at 10%. A significance level is not specified in the paper, so an unconditional yes-or-no decision is not determined: using this approximation, rejection would require a chosen level at least about $15\%$.

For fixed $\mu \ne 0$, a large positive root must lie close to a pole of the tangent function: away from its poles, $\tan x$ cannot balance the unbounded right side. Label the adjacent poles by $a_n=(n+1/2)\pi$ and write $x_n=a_n+{\delta }_n$. The Taylor series of $-\cot \delta$ givesSeek the asymptotic expansion $\delta =A/a_n+B/a_n^3+\cdots$. The coefficients of $a_n$ and $a_n^{-1}$ giveThus the large roots near tangent poles satisfyThe first correction already supplies the requested dependence on $\mu$. For $\mu >0$ the root approaches the pole from below; for $\mu <0$ it approaches from above. This labels roots by their nearby poles, avoiding an irrelevant finite shift in the enumeration of positive roots.

No positivity restriction on $\mu$ is required. The expansion requires $\mu \ne 0$ fixed and $|\mu |a_n\gg 1$, so the displacement is small. It is not uniform as $\mu \to 0$. At $\mu =0$ the roots are exactly $n\pi$, a different leading sequence; these cannot be recovered by setting $\mu =0$ in the pole expansion. If $\mu$ varies with $n$, its size must be checked against the small-displacement and successive-term conditions rather than using the fixed-parameter remainder blindly.

The ordinary Fourier transform does not exist. There is a real simple pole at $x=-1$, withThe two separate improper integrals diverge logarithmically. Thus no ordinary large-$k$ expansion is defined under the stated integral convention. A Cauchy principal value or a contour prescription would be additional data, and must be stated rather than silently introduced.

For completeness, the symmetric Cauchy principal value has a precise answer. The real pole contributes the principal-value Fourier transform of a real poleThe other poles are $z_{\pm }=(1\pm i\sqrt{3})/2$ with residues $r_{\pm }=1/(3z_{\pm }^2)=(-1\mp i\sqrt{3})/6$. For $k>0$, close the contour integration in the lower half-plane, with clockwise orientation. The nonreal pole contributes $-2\pi i{r}_{-}{e}^{-ik{z}_{-}}$, while the real principal-value pole supplies the half-residue above. For $k<0$, the upper-half-plane contour gives the conjugate result. With $s=\mathrm{sgn}(k)$, the exact principal-value formula for $k\ne 0$ isThese are its two nonzero asymptotic contributions: a nondecaying oscillatory real-pole term and an exponentially small complex-pole term. There is no second nonzero inverse-power term. This formula is qualified by the principal-value choice; it is not the ordinary transform requested in the statement.

Other prescriptions change the leading term. For example, replacing the real pole by its upper or lower boundary value changes the distribution by $\mp i\pi \delta (x+1)/3$, and hence changes the transform by $\mp i\pi e^{ik}/3$. This demonstrates why a pole prescription is essential. The smooth Taylor series at the origin alone would miss the decisive real-pole contribution.

From $f^{\prime \prime }=f$, differentiating successively gives $f^{(2r)}=f$ and $f^{(2r+1)}=f^{\prime }$ whenever those derivatives have been constructed. Starting with twice differentiability, this relation supplies the next two derivatives at each stage. Thus $f$ is infinitely differentiable, and its origin series isFor a fixed $x$, both $f$ and $f^{\prime }$ are bounded on the compact interval between zero and $x$, say by $M_x$. Every higher derivative is one of these two functions. Taylor's theorem, with the remainder just proved, therefore bounds the order-$n$ remainder by $M_x|x{|}^n/n!$, which tends to zero: its consecutive positive-term ratio is $|x|/(n+1)$. Hence the series converges to $f(x)$ for every real $x$. It also gives $f(x)=A\cosh x+B\sinh x$ by the usual hyperbolic-function series.

At any other centre $a$, apply the proved origin result to $g(t)=f(a+t)$, which also satisfies $g^{\prime \prime }=g$. Its even derivatives at zero are $f(a)$, and its odd ones are $f^{\prime }(a)$. ConsequentlyThis proves convergence at every centre without assuming that smoothness alone implies equality to a Taylor series.

For an entire function, the Cauchy derivative formula on a positively oriented circle of radius $R$ about $z_0$ isThe resulting Cauchy estimate is $|f^{(n)}(z_0)|\le n!M_R/R^n$, where $M_R$ is the maximum of $|f|$ on that circle.

The Liouville theorem says that a bounded entire function is constant. Indeed, if $|f|\le M$ on the plane, the Cauchy estimate with $n=1$ gives $|f^{\prime }(z_0)|\le M/R$. Sending $R$ to infinity gives $f^{\prime }(z_0)=0$ at every point, hence $f$ is constant.

For the growth assumption, apply the Cauchy estimate on circles centred at zero. For every integer $n>k$,All sufficiently high Taylor series vanish. Since the Taylor series of an entire function converges everywhere, $f$ is a polynomial, of degree at most $\lfloor k\rfloor$ when $k\ge 0$. If a negative finite exponent is interpreted for nonzero $z$, the same estimate forces every coefficient to vanish, giving $f=0$.

The Taylor series definition says that $f\in C^{\infty }(\mathbb{R})$ and, for every $a$, there is $r>0$ such thatThe equivalent factorial derivative criterion for real analyticity says that, for every $a$, there are a neighborhood $I$ of $a$ and constants $C,A>0$ such thatThe uniformity over $I$ matters: bounds only at $a$ do not exclude a flat function.

Assume the factorial derivative criterion for real analyticity. The Taylor theorem with Lagrange remainder giveswhen the segment from $a$ to $a+h$ is contained in $I$. For sufficiently small $|h|<A^{-1}$ the Taylor remainder tends to zero, proving the Taylor series definition.

Conversely, write the convergent power series at $a$ as $\sum b_k{h}^k$. Choose $\rho$ strictly inside its radius of convergence; then $|b_k|\le M{\rho }^{-k}$ for some $M$. Termwise differentiation on $|h|\le \rho /2$ givesHere the sum is $(1-1/2{)}^{-n-1}$, obtained by differentiating the geometric series. This is the required locally uniform bound. The two definitions of a real analytic function are equivalent.

Consider the flat functionAway from zero every derivative has the form $P_n(1/x)e^{-1/x^2}$ for a polynomial $P_n$: differentiating preserves this form. For every $m\ge 0$,because an exponential function decays faster than any power. Inductively, extend each displayed derivative by zero at zero. It is continuous there, and its difference quotient at zero also tends to zero by the same estimate with one extra power of $|x{|}^{-1}$. Thus each extension is the derivative of the preceding extension. This proves $f\in C^{\infty }(\mathbb{R})$ and $f^{(n)}(0)=0$ for all $n$.

Its Taylor series at zero is identically zero, whereas $f(x)>0$ for every $x\ne 0$. It is smooth everywhere but not real analytic at zero.

The total differential order of ${\partial }_t-{\partial }_x^2$ is two, so its principal symbol is $p(\tau ,\xi )=-{\xi }^2$. The conormal to $t=0$ is $(1,0)$, on which $p$ vanishes. The initial line is a characteristic hypersurface for this total-order symbol; the first-order time derivative does not enter it.

Suppose a real analytic solution existed near $(0,0)$. Repeated use of the heat equation gives ${\partial }_t^{k}u={\partial }_x^{2k}u$. The initial power series is ${\sum }_{j\ge 0}(-1{)}^j{x}^{2j}$ near zero, henceThe time Taylor series at $x=0$ would therefore have coefficients $(-1{)}^k(2k)!/k!$. The ratio of successive absolute coefficients is $2(2k+1)\to \infty$, giving radius of convergence zero. This contradicts the assumed real analytic regularity. No such analytic local solution exists, although the initial function itself is real analytic.