Condition on an apple occupying position two in a uniformly random ordering. The nine other positions then contain four apples and five bananas, each equally likely to occupy position three. The conditional probability is thereforeThis uses exchangeability of the unconditioned positions; there is no need to assume the first lunch contained a particular fruit.

Random sampling without replacement assigns each of the ten distinguishable pieces to each lunch with the same probability. Five pieces of each type therefore giveOne can check the second probability by conditioning on the first lunch:The identical marginal probabilities do not imply independence of the draws.

The definition of hyperbolic sine simplifies the composition exactly for $x>0$:Set $h=x-1$. The geometric series $1/(1+h)=1-h+h^2-h^3+h^4+O(h^5)$, valid for $|h|<1$, then gives

Divide numerator and denominator by $x$ before taking their Maclaurin series:For the reciprocal denominator, the geometric series gives $1-x^2/6+7x^4/360+O(x^6)$. Multiplying gives the Taylor expansionStrictly, the original quotient has a removable singularity at zero; this is the Taylor series of its continuous, analytic extension with value one there. The absence of odd powers also follows because the ratio of these two odd functions is even.

The superposition principle gives the general solution as $y_c+2y_{p,2}+y_{p,1}$:The first boundary condition gives $A=2/(3\sqrt{3})-1/3$. At $x=\pi$, the cosine vanishes and the sine equals $-1$. Comparing with the second boundary condition then givesThe unique solution satisfying both boundary conditions is thereforeUniqueness follows because a homogeneous solution zero at both endpoints would have $A=0$ from the first and $B=0$ from the second.

A linear polynomial trial $y=ux+v$ has $y^{\prime }=u$ and $y^{\prime \prime }=0$. Substituting gives $3ux+(\sqrt{3}u+3v)=x$, so $u=1/3$ and $v=-\sqrt{3}/9=-1/(3\sqrt{3})$. Thus a particular solution is

Use the method of undetermined coefficients with a particular solution $K{e}^{-\sqrt{3}x}$. The differential operator multiplies this exponential function by $3-3+3=3$. Matching the forcing coefficient gives $3K=1$, hence

Insert the exponential function trial $y=e^{rx}$ into the homogeneous linear differential equation. Its characteristic equation is $r^2+\sqrt{3}r+3=0$, with polynomial rootsTaking real combinations of the two complex exponential solutions produces the general solution

Multiplication by the integrating factor and the product rule turn the left side into a single derivative:Using integration by substitution $u=\cos x$ gives $y\sin x=-{\cos }^{3}x/3+C$. Therefore the general solution isThere is one arbitrary constant of integration, as expected for a first-order ordinary differential equation.

For a first-order linear ordinary differential equation $y^{\prime }+p(x)y=q(x)$, an integrating factor is $\exp (\int p(x)dx)$. Here $p(x)=\cot x$, soThe sine is positive on the specified interval, so this integrating factor never vanishes there. Any nonzero constant multiple is equally suitable.

In spherical coordinates, the cone condition forces $z\ge 0$ and becomes $0\le \vartheta \le \alpha$. The unit ball gives $0\le \rho \le 1$. HenceFor an independent calculation in cylindrical coordinates, the lower boundary is $z=r\cot \alpha$ and the upper boundary is $z=\sqrt{1-r^2}$. They meet at $r=\sin \alpha$, since $r^2{\cot }^2\alpha =1-r^2$. ThusThe equality uses ${\sin }^3\alpha \cot \alpha ={\sin }^2\alpha \cos \alpha$. The positive cosine on the stated range removes any square-root sign ambiguity.

At height $z>0$, the elliptic paraboloid has a circular cross-section $0\le r\le \sqrt{z}$, of area $\pi z$. The cylindrical coordinates volume element therefore reduces the volume integral toFor the specified $f$, the derivative is $e^{-z}(z^{-1/2}+\frac{1}{2}{z}^{-3/2})$. Thus $z{f}^{\prime }(z)$ is integrable at zero and decays exponentially at infinity. Apply integration by parts first on a truncated interval and then take the endpoint limits:Both boundary terms vanish because $zf(z)=-\sqrt{z}{e}^{-z}$. With $z=u^2$, integration by substitution gives the Gaussian integral:The integrand is singular at the apex but the resulting improper integral is convergent.

The cardioid has a nonnegative radial boundary, so each ray from the origin contributes $0\le r\le a(1+\cos \theta )$. Integrating the polar coordinates area element yieldsThe three angular integrals are $2\pi$, zero and $\pi$, respectively, using the double-angle formula for the last. Thus

Let $I$ be the positive Gaussian integral. Its improper integral converges because $e^{-x^2}\le e^{-|x|}$ for $|x|\ge 1$. Multiplying two copies and using the polar coordinates area element givesNonnegativity justifies the product double integral and the change of integration order. Since $I>0$,

The triangle inequality bounds the distance sum below by the separation of the two foci, which is two. Equality holds exactly for points on the closed segment joining them. Hence the locus isBoth endpoints are included. It is the segment on the positive imaginary axis shown in the right-hand panel above.

Write $z=x+iy$. Then $|z^{\ast }+2i|=\sqrt{x^2+(y-2{)}^2}$ and $|z|=\sqrt{x^2+y^2}$, so the condition fixes the sum of distances from $(0,2)$ and $(0,0)$. It is an ellipse with centre $(0,1)$, vertical semimajor axis $A=2$, and focal distance $c=1$. Its horizontal semiminor axis obeys $B^2=A^2-c^2=3$. Thus the locus isThe vertices are $(0,-1)$ and $(0,3)$; the horizontal extremities are $(\pm \sqrt{3},1)$. The complex conjugation places the second focus above the origin.

The literal expression in the original PDF is undefined. Indeed, hyperbolic tangent satisfies $\tanh w=\sinh w/\cosh w$, andAlready the $n=2$ summand contains $\tanh (i\pi /2)$, a pole. The shifted term has the same issue when $n\equiv 3(\mathrm{mod}4)$. Thus this is not a finite sum of defined complex numbers, and it has no literal real or imaginary part. Cancelling two undefined terms is not ordinary arithmetic.

There is nevertheless a natural interpretation by meromorphic telescoping cancellation. Insert the same small positive real shift $\epsilon$ into every hyperbolic tangent argument. Each term is then finite, and the telescoping sum isSince hyperbolic tangent has period $i\pi$, its first argument can be reduced to $\epsilon +i\pi /4$. Taking the limit givesThis is a common-shift continuation of the sum, plausibly the intended telescoping answer. It requires that additional convention; it is not the value of the expression as printed.

The exponential definitions of hyperbolic sine and hyperbolic cosine, together with Euler's formula, give $\sinh (iu)=i\sin u$ and $\cosh (iu)=\cos u$. The argument of the complex logarithm therefore reduces to $i\sin (\pi /2)+\cos (9\pi /2)=i$.

Using the principal logarithm, $\log i=i\pi /2$, soIf no branch of the complex logarithm is selected, its complete set of values is $i(\pi /2+2\pi k)$, $k\in \mathbb{Z}$. The real part remains zero, while the imaginary part is $\pi /2+2\pi k$. The original paper does not explicitly state a branch convention.

Powers of the imaginary unit repeat in blocks $i,-1,-i,1$, each with sum zero. The first $44$ terms comprise eleven complete blocks; the last is $i^{45}=i$. Applying Euler's formula to the resulting complex exponential gives $e^i=\cos 1+i\sin 1$. HenceThe angles here are in radians.

Multiply numerator and denominator by the complex conjugate $2+i$:Thus the real part and imaginary part are