The indicated integration by substitution givesPointwise, the argument of $\phi$ tends to $x$. Since the Gaussian integral is $\int e^{-{\eta }^2}d\eta =\sqrt{\pi }$, the expected limit is $\phi (x)$. To justify it for unbounded data, use the approximate identity property rather than assuming a uniform bound on $\phi$.

In the original variable, normalization givesFor $|\xi -x|<\delta$, continuity makes the difference smaller than any prescribed $\epsilon$ if $\delta$ is small enough, and the kernel mass is at most one. On the complement the Gaussian mass tends to zero; the data-weighted tail tends to zero by heat-kernel convergence for Gaussian-integrable initial data. This provesFor heat-kernel convergence from a conditionally convergent Gaussian integral, the same tail conclusion follows using the bounded primitive of $(\phi (\xi )-\phi (x))e^{-{\xi }^2/(4T)}$. The factor $q=K(x-\xi ,t)e^{{\xi }^2/(4T)}$ has its maximum at $\xi =xT/(T-t)$, approaching $x$. For sufficiently small $t$, it is monotone on each of the two tails outside $|\xi -x|<\delta$, and its values at the tail endpoints tend to zero exponentially. Integration by parts bounds each tail by a constant times its endpoint value, proving the same limit without an absolute-convergence assumption. The approach $t\downarrow 0$ is from positive time.

Write the kernel as $K(s,t)=(4\pi t{)}^{-1/2}{e}^{-s^2/(4t)}$, with $t>0$. Its partial derivatives areThus the heat kernel itself satisfies $K_t=K_{ss}$. The proposed convolution consequently givesThis verifies the heat equation. The parameter is positive time; the real integral formula is not defined at negative time.

The passage of derivatives through the integral needs justification. A sufficient condition is the Gaussian-weighted absolute integrability in heat-kernel convergence for Gaussian-integrable initial data; it holds, in particular, for data of polynomial growth. On compact positive-time intervals strictly below an integrable later time, the differentiated kernels have an integrable Gaussian bound, so differentiation under the integral sign applies.

Even if the PDF's assumed convergence is read as ordinary conditional convergence, one can justify the same calculation. Choose a later time $T>t$ and let $H(\xi )$ be a primitive of $\phi (\xi )e^{-{\xi }^2/(4T)}$. Convergence at spatial point zero and time $T$ makes $H$ bounded on both tails. Put $q(x,\xi ,t)=K(x-\xi ,t)e^{{\xi }^2/(4T)}$. This function, its spatial and time derivatives, and their $\xi$-derivatives decay as Gaussians when $t<T$. Integration by parts rewrites the convolution as $-\int H{q}_{\xi }d\xi$, with vanishing endpoint terms. Boundedness of $H$ now permits differentiation under this absolutely convergent integral; since $q_t=q_{xx}$, the same equation follows. Thus one need not silently equate conditional and absolute convergence.

For $0<x<1$,In particular its absolute value is bounded by $|b|x^{\min (b,0)}$, proving convergence when $b>-1$; the apparent endpoint singularity at one is removable, with value $b$. On any compact parameter interval with lower bound $b_0>-1$, the differentiated integrand is $x^b$, dominated by $x^{b_0}$. The dominated convergence theorem therefore justifies differentiation under the integral sign:Hence $J(b)=\ln (b+1)+C$. Since the integrand is identically zero at $b=0$, $J(0)=0$ fixes $C=0$. ThusThis proves the logarithmic integral representation of the natural logarithm, including its parameter range.

The Leibniz integral rule accounts for both the moving upper endpoint and the integrand's dependence on $a$:For the specified integrand this gives $I^{\prime }(a)=e^{-a^2}-{\int }_0^{a}x{e}^{-ax}dx$. When $a\ne 0$, integration by parts givessoFor the requested independent check, direct integration gives $I(a)=(1-e^{-a^2})/a$ at $a\ne 0$. Its derivative is $[2a^2{e}^{-a^2}-(1-e^{-a^2})]/a^2$, agreeing with the result above. At zero the apparent singularity is removable: $I(a)=a-a^3/2+O(a^5)$, so $\boxed{I(0)=0,I^{\prime }(0)=1}$.

Let $C$ denote a green pear on the first draw and $D$ a red apple on the second. The two categories are disjoint, so after $C$ there are still three red apples among seventeen remaining fruits. Thus $P(C\cap D)=(8/18)(3/17)$. Using $P(D)=1/6$ from (c), conditional probability givesEquivalently, conditioning on the second fruit being one particular category removes a red apple and leaves the other seventeen fruits symmetrically eligible for the first position.

Condition on whether the first selection was a red apple. Its probability is $3/18$. If it was, two red apples remain among seventeen fruits; otherwise three remain. The law of total probability givesThe same result follows from exchangeability: before either selection is observed, each ordered draw position has the original red-apple fraction $3/18$, despite dependence between the draws.

Insert the values from (a) into Bayes' theorem:Thus the identity is satisfied. Equivalently, both $P(A\mid B)P(B)$ and $P(B\mid A)P(A)$ equal the joint probability $4/9$.

There are eighteen equally likely first choices. Counting the relevant subsets givesFor conditional probabilities, restrict the denominator to the conditioning subset:The five counts come from the same finite sample space, so no independence assumption is involved.

Factor $1+e^x=2e^{x/2}\cosh (x/2)$. This givesThe last function is even. Its Taylor series begins $\ln \cosh (x/2)=x^2/8+O(x^4)$, since $\cosh (x/2)=1+x^2/8+O(x^4)$ and $\ln (1+w)=w+O(w^2)$. ThereforeIn particular, the cubic coefficient is zero; the identity explains its cancellation before any lengthy differentiation.

The required derivatives are $f^{\prime }(x)=1/(2\sqrt{x})$, $f^{\prime \prime }(x)=-1/(4x^{3/2})$, $f^{(3)}(x)=3/(8x^{5/2})$, and $f^{(4)}(x)=-15/(16x^{7/2})$. The four-term Taylor polynomial about one, evaluated at two, isThe Taylor remainder is $R_3(2)=-5/(128{\xi }^{7/2})$ for some $1<\xi <2$. ThusIts negative sign also shows that this approximation is an overestimate. Bounding the fourth derivative throughout the interval establishes the requested error estimate without using a decimal value of $\sqrt{2}$.

A cubic version of Taylor's theorem requires the corresponding higher differentiability, rather than only one derivative. For example, if $f$ has four continuous derivatives on the interval between $a$ and $x$, thenwhere the Lagrange form of the Taylor remainder isfor some $\xi$ strictly between $a$ and $x$ when they differ. At $x=a$, the remainder is zero. These hypotheses apply to both functions used below.

With $a=2$, integrate the $dx$ coefficient with respect to $x$:Its other partial derivative is ${\Phi }_y=x^2/2-2xy+g^{\prime }(y)$. Matching the $dy$ coefficient gives $g^{\prime }=0$. Therefore the differential equation is $d\Phi =0$, and its implicit general solution isThese are level curves of the potential of a conservative vector field. Where $N\ne 0$, they give the usual graph solution $dy/dx=-M/N$. At $N=0$, the implicit description retains possible vertical tangents instead of dividing by zero.

Write the differential one-form as $Mdx+Ndy$. An exact differential form must satisfy equality of mixed partial derivatives, $M_y=N_x$. HereEquality at every $(x,y)$ holds precisely when $\boxed{a=2}$. Since these polynomials are smooth on the simply connected plane, the closed-form condition is also sufficient for exactness; the explicit potential of a conservative vector field constructed in (c) verifies that sufficiency directly.

The multivariable chain rule givesSubstituting the three parametrizing functions also writes the complete derivative asAt $t=\pi /2$, the point is $(0,3,\pi )$, its velocity is $(-3,0,2)$, and $\nabla F=(3,0,1)$. Hence

The condition at infinity gives $C_{\mathrm{out}}=0$. Matching the electric potential at $r=a$ givesConsequentlyThis function is finite and differentiable at the origin. At the joining radius, both $V$ and $V^{\prime }$ match, giving $V(a)=-k/a$ and $V^{\prime }(a)=k/a^2$. Thus the electric field is continuous there as well as being conservative on each side. This is the potential of a uniform spherical source, with source coefficient $3k/a^3$.

For a radial electric potential, $\mathbf{E}=-V^{\prime }(r)\mathbf{r}/r$. Comparing radial coefficients givesIntegrating these one-variable derivatives gives the general local expressionsThe two constants are initially independent because these integrations were performed in separate radial regions.

Inside the sphere, the electric field has the form $\mathbf{E}=\nabla [-k{r}^2/(2a^3)]$, including at the origin. Thus it is a conservative vector field. Outside, use the gradient of a radial function to obtainThis is also an explicit potential of a conservative vector field, single-valued throughout the exterior region. Equivalently, the curl vanishes in both regions: for a radial coefficient $h(r)$, $\nabla \times (h\mathbf{r})=\nabla h\times \mathbf{r}+h\nabla \times \mathbf{r}=0$ because $\nabla h$ is radial. The interior formula avoids any apparent $1/r$ issue at the origin. The vector-field potentials here have the opposite sign from the electric potential convention used in (b).

The volume of a parallelepiped is the absolute scalar triple product of its three edge vectors. Since $\overrightarrow{OP}\times \overrightarrow{OQ}=(0,0,3)$,The factor $1/6$ appropriate to a tetrahedron must not be applied to this parallelepiped.