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Assume the three standard dice are fair and independent. For one die $X$, a discrete uniform distribution on $1,\dots ,6$ givesThe variance is $\mathrm{Var}X=\mathbb{E}{X}^2-(\mathbb{E}X{)}^2=35/12$. For the sum $Q=X_1+X_2+X_3$, linearity of expectation and addition of variances for independent random variables give

Put $w=e^x-1=x+x^2/2+x^3/6+O(x^4)$. The local logarithmic series $\ln (1-w)=-w-w^2/2-w^3/3+O(w^4)$ givesFor the infinite Taylor series, it is necessary to distinguish the real domain of the logarithm, $x<\ln 2$, from its convergence interval about zero. Write $a=\ln 2$. The complex zeros of $2-e^z$ are $a+2\pi ik$ for integers $k$, so the nearest logarithmic singularity to zero is at $a$. The radius of convergence is therefore $a$, giving convergence for $|x|<a$ and divergence for $|x|>a$.

The two endpoints need separate checking. Factor off the nearest singularity:The quotient has a nonzero removable value $2a$ at $z=a$; its other zeros are at $a+2\pi ik$ with $k\ne 0$. Consequently it has a holomorphic logarithm on a zero-free disk larger than $|z|\le a$, which is a simply connected domain, and the Taylor series of $H$ converges absolutely at both real endpoints. At $x=-a$, the remaining series $-{\sum }_{n\ge 1}(-1{)}^n/n$ converges by the alternating series test. At $x=a$, it is $-{\sum }_{n\ge 1}1/n$ and diverges. Thus the full real convergence set isConvergence is absolute in the open interval and conditional at the left endpoint, where the sum equals $\ln (3/2)$. This endpoint distinction is an instance of isolating a logarithmic singularity to test a power-series endpoint.

For $x>0$, the definition of hyperbolic cosine simplifies the composite function exactly:Let $h=x-2$. The geometric series givesAdding $2+h$ and dividing by two gives the requested Taylor expansion:The underlying infinite geometric expansion has $|x-2|<2$.

The Maclaurin series of the hyperbolic cosine and cosine areBoth are even, so write their quotient as $1+a{x}^2+b{x}^4+O(x^6)$. Multiplication by the denominator gives coefficients $a-1/2$ and $b-a/2+1/24$. Matching the numerator yields $a=1$ and $b=1/2$. ThusThe denominator is nonzero near the expansion point, so division of the power series is valid there.

For $f$ sufficiently differentiable on the interval between $a$ and $x$, Taylor's theorem with its first $n$ terms readswhere $\xi$ lies between $a$ and $x$. This is the Taylor theorem with Lagrange remainder. For example, $|f^{(n)}|\le M$ on that interval gives $|R_n(x)|\le M|x-a{|}^n/n!$. The finite formula does not assume that an infinite Taylor series converges to the function; that further conclusion requires the remainders to tend to zero.

Put $D={\sigma }^2>0$ and $\xi =x-x_0$. Logarithmic differentiation of the positive Gaussian function givesDifferentiating once more in space givesMultiplying by $D$ provesThe assumption $\sigma \ne 0$ ensures both the normalization and exponential denominator are defined. This is the heat kernel, with diffusion coefficient $D$ and centre $x_0$.

The polar-coordinate Jacobian matrix and its Jacobian determinant areFor $r>0$, the corresponding area element is $dxdy=rdrd\theta$. The determinant vanishes at $r=0$, where the polar angle does not give a unique coordinate representation.

The original PDF specifies evaluation at $(1,1,1)$; the local TeX incorrectly gives $(1,1,0)$. Write the scalar field explicitly as $f=c/(x^2+y^2+z^2)$. Its gradient isA directional derivative measures rate per unit distance, so normalize the direction to the unit vector $\mathbf{n}=(1,1,0)/\sqrt{2}$. ThenUsing the unnormalized direction vector would instead differentiate with respect to a path parameter, not distance, and would give a different scale.

First retain the equilibrium solution $\boxed{y\equiv 0}$. Any nonzero real solution has constant sign on its interval of existence, since the polynomial right-hand side is locally Lipschitz and Picard-Lindelöf theorem prevents crossing the zero solution.

For a nonzero solution, the Bernoulli differential equation substitution $u=y^{-4}$ givesThis linear ordinary differential equation has solution $u+2=C{e}^{-8x}$. Therefore every nonzero real solution isEither sign may be chosen and remains fixed. The restriction ensures $y^{-4}>0$; nonpositive $C$ gives no nonzero real solution. These are maximal real intervals, and their restrictions are also solutions. As the right endpoint is approached, $|y|$ diverges in finite-time blowup. The substitution would have lost $y=0$ if it had not been checked separately.

Use cylindrical coordinates $x=r\cos \theta$, $y=r\sin \theta$. The Jacobian determinant givesThe inequalities imply $-d<z<0$ and $0\le r<a\sqrt{1+z/d}$. Thus the volume isEquivalently, integrating vertically first gives $V={\int }_0^{2\pi }{\int }_0^a{\int }_{d(r^2/a^2-1)}^{0}rdzdrd\theta$. The region is bounded by an elliptic paraboloid and its top disc; its volume is half that of the cylinder with base radius $a$ and height $d$. Including or excluding the boundary surfaces does not change the volume.