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For a nonnegative integer-valued random variable, its probability generating function isThe convention $s^0=1$, including at $s=0$, gives $G(0)=p(0)$. The power series has absolute convergence for every real $|s|\le 1$, since ${\sum }_{j}p(j)|s{|}^j\le 1$. More generally its real domain as a finite expected value is $\{s\in \mathbb{R}:{\sum }_{j}p(j)|s{|}^j<\infty \}$, which may extend beyond $[-1,1]$. No extension beyond that guaranteed interval is automatic.

For separation of variables, set $u=X(x)Y(y)$, with $X^{\prime \prime }/X=-Y^{\prime \prime }/Y=-\lambda$. The side Neumann boundary conditions give $X^{\prime }(0)=X^{\prime }(\pi )=0$. As in the standard Sturm-Liouville problem, integration by parts shows $\lambda \ge 0$. The zero eigenvalue gives constant $X$, and $Y^{\prime \prime }=0$ together with $Y^{\prime }(\pi )=0$ makes $Y$ constant too.

For positive eigenvalues, the side conditions give $\lambda =n^2$, $X_n=\cos (nx)$, $n\ge 1$. The top condition chooses the hyperbolic cosine factor $\cosh (n(\pi -y))$. Normalize it to one at $y=0$, obtainingThe orthogonality of integer Fourier modes and the permitted interchange of summation and integration giveFor the remaining Fourier coefficients, integrate by parts twice. The sine boundary term vanishes, and with $f=x(\pi -x)$,The last integral vanishes because $f^{\prime \prime }=-2$ and $n$ is a positive integer. Thus $a_n=-2(1+(-1{)}^n)/n^2$: only even Fourier modes survive, as expected from reflection symmetry about $x=\pi /2$. Writing $n=2m$ givesThe factor involving hyperbolic cosine lies between zero and one on the closed rectangle. Comparison with $\sum m^{-2}$ proves uniform convergence, so the solution is continuous on the entire boundary and takes the prescribed bottom values. On every strip $y\ge \epsilon >0$, the Fourier coefficients have additional exponential decay, allowing differentiation and verifying Laplace's equation and all three Neumann boundary conditions.

To justify the endpoint calculation in b, differentiate at $0<x<\pi$, $y=0$ in the interior-edge sense:Subtracting $1$ from the hyperbolic tangent leaves an absolutely convergent series. For the remaining part, the power series $-\log (1-z)={\sum }_{m\ge 1}{z}^m/m$ gives, by taking real parts at $z=r{e}^{2ix}$ and then $r\uparrow 1$,The same formula follows as the limit of the differentiated interior solution, since its leading multiplier is $e^{-2my}$. This establishes the boundary normal derivative used in b. It also explains why the failure of the corner compatibility condition for mixed boundary data affects first derivatives at the bottom corners without spoiling continuity of $u$ itself.

Define the radius of convergence of the power series bywith $R=\infty$ when this set is unbounded. It is nonempty because the power series converges at zero.

If $|z|<R$, the definition of supremum supplies a point $w$ of convergence with $|z|<|w|$. The terms $a_n{w}^n$ of a convergent series tend to zero and are bounded, say $|a_n{w}^n|\le M$. ThereforeThe right-hand side has a convergent geometric series, since $|z|/|w|<1$. The comparison test proves absolute convergence at $z$.

If $|z|>R$ and the power series converged at $z$, then $|z|$ would belong to the set whose supremum is $R$, an immediate contradiction. ThusNo assertion about the boundary $|z|=R$ follows from this argument. If $R=0$ the first assertion is vacuous; if $R=\infty$ the second is vacuous.

Split the power series into even and odd indices:whenever the two series converge absolutely. If $|z|<1$, the even part is a convergent geometric series, while the odd part is bounded termwise in absolute value by ${\sum }_j|z{|}^{2j+1}$, also a convergent geometric series. Hence there is absolute convergence throughout $|z|<1$.

If $|z|>1$, the odd terms have absolute values $|z{|}^{2j+1}/(2j+1{)}^2$. Their successive ratio tends to $|z{|}^2>1$, so these terms grow eventually and do not tend to zero. The term test for divergence therefore forces divergence of the original power series. Consequently its radius of convergence is

For fixed $z\ne 0$, the ratio of successive absolute values isThe ratio test says that a series converges absolutely when this ratio has a limit of a sequence strictly below one. Thus this power series converges absolutely for every complex $z$; at zero the conclusion is immediate. Its radius of convergence is $\boxed{R=\infty }$.

By the definition of a probability generating function,Termwise differentiation of the power series is valid for $|s|<1$. At $s=1$, the equality extends by the finite mean assumption, and both sides equal one.

For a power series $\sum a_n{z}^n$, its radius of convergence $R\in [0,\infty ]$ separates absolute convergence for $|z|<R$ from divergence for $|z|>R$. One may define $R=\sup \{r\ge 0:\sum |a_n|r^n<\infty \}$, allowing $R=\infty$.

To justify convergence inside the radius, choose $|z|<r<R$ with $\sum |a_n|r^n<\infty$. In particular $|a_n|r^n\le M$ for some $M$, soThe comparison test with a geometric series proves absolute convergence. Conversely, convergence at a point of modulus $\rho$ makes $a_n{z}^n$ bounded there and gives absolute convergence at all smaller moduli by the same estimate. Consequently a convergent point of modulus greater than $R$ would contradict the definition of $R$.

For $0<R<\infty$, the real domain of convergence is one of $(-R,R)$, $[-R,R)$, $(-R,R]$, or $[-R,R]$. At $R=1$ these are realized respectively byThe endpoint assertions follow from the harmonic series, alternating series test, and p-series. Replacing $z$ by $z/R$ realizes every finite positive radius. The degenerate domain $\{0\}$ occurs for $\sum n!z^n$, while the entire real line occurs for $\sum z^n/n!$. Their radii are zero and infinity by the ratio test.

For a geometric distribution on positive integers, the probability generating function is $ps/[1-(1-p)s]$. Conditional averaging over the red die yieldsAs a power series, its exact domain of absolute convergence isIndeed the slowest decaying component has failure probability $5/6$, so $\mathbb{P}(T=t)\ge (1/36)(5/6{)}^{t-1}$. At $|s|=6/5$, the terms of the power series fail to tend to zero; outside that disk they grow. All six component series converge inside it. In particular, the probability generating function is defined on the usual real interval $0\le s\le 1$, with $G_T(1)=1$. The rational expression beyond the convergence disk is only an algebraic continuation, not the defining expected value.

Adding and subtracting the exponential function power series cancels the appropriate odd or even terms:Both series converge absolutely for every complex number $x$, as follows from the ratio test.

Multiply the two power series and collect equal powers of $x$. The Cauchy product givesOnly terms whose two exponents add to the target exponent contribute.

The displayed power series is the matrix exponential $G=e^H$. From $H^3=-{\alpha }^{2}H$, its odd powers and nonconstant even powers sum separately:The scalar power series are $\sin \alpha /\alpha$ and $(1-\cos \alpha )/{\alpha }^2$, with continuous values $1$ and $1/2$ at zero. Substituting $H\mathbf{x}=\alpha \mathbf{n}\times \mathbf{x}$ and the preceding formula for $H^2\mathbf{x}$ yieldsThis is the Rodrigues rotation formula: the component parallel to $\mathbf{n}$ is fixed, while the perpendicular component undergoes a rotation through $\alpha$. For $\alpha =0$ the linear operator is the identity operator.

Since $g(0)=0$, only the first three positive powers of $g$ contribute through degree three. The composition coefficients of power series follow by using $g^2=g_1^2{x}^2+2g_1{g}_2{x}^3+O(x^4)$ and $g^3=g_1^3{x}^3+O(x^4)$ in $f_0+f_{1}g+f_2{g}^2+f_3{g}^3+\cdots$. HenceHere $f_m$ and $g_m$ are coefficients of the power series, rather than raw derivatives; their definitions already contain any factorial factors.

Put $d=a-b$. The rational factor has Taylor expansion $1+dx-bd{x}^2+b^{2}d{x}^3+O(x^4)$. Multiplication by the logarithmic power series makes the coefficients of $x^2,x^3$ equal to $d-1/2$ and $1/3-d/2-bd$. Setting both to zero gives $d=1/2$ and $b=1/6$, henceThe coefficient of $x^4$ is $-1/4+d/3+bd/2+b^{2}d=-1/36$. Thus the resulting expansion begins $x-x^4/36+O(x^5)$.

For $a_n=n^2/n!$, the ratio test givesThus the series converges absolutely. As a check, $n^2=n(n-1)+n$ and the exponential power series give the sum $e+e=2e$.

For $\lambda \ge 0$, the exponential power series gives the normalization of the Poisson distribution:For $\lambda =0$, the limiting law assigns probability one to $k=0$ and zero to every positive integer $k$. Negative $\lambda$ would not define the stated probability distribution.

Near zero the logarithm's argument is positive, so $\ln [(2+x{)}^3]=3\ln (2+x)=3\ln 2+3\ln (1+x/2)$. Use the Maclaurin series of the natural logarithm:The local calculation uses $x>-2$ for the real logarithm; the power series converges for $|x|<2$.

The generalized binomial series givesIn particular,The recurrence $u_{n+1}=u_n(1/3-n)/(n+1)$ follows directly from the coefficient product. The power series has radius of convergence one and represents the real cube root for $|x|<1$.

Both numerator and denominator vanish quadratically. Factor out $x^2$ before dividing the power series:The reciprocal of the denominator is $1+x^2/2-x^4/12+O(x^6)$. HenceThe expansion describes the continuous extension at the removable singularity $x=0$, where the original quotient is undefined but its limit is one. Odd powers vanish because the quotient is even.

Since $a_0=f(0)\ne 0$, write $1/f(x)=b_0+b_{1}x+b_2{x}^2+O(x^3)$ and multiply the two power series. Equating coefficients with the identity $f(x)/f(x)=1$ givesSolving successively yieldsThese are the constant, linear and quadratic terms; particular coefficients may vanish. For a convergent power series defining $f$ near zero, continuity and $a_0\ne 0$ ensure its reciprocal is also defined in a neighborhood of zero.

Set $h=x+1$, the displacement from the expansion point. The numerator has Taylor series $\sinh h=h+h^3/6+O(h^5)$, while the reciprocal denominator has geometric series $(1+h{)}^{-1}=1-h+h^2-h^3+h^4+O(h^5)$. Multiplication givesThe local power series converges for $|x+1|<1$, the distance to the denominator's zero at $x=-2$.