The position vectors end at the midpoints of $BC$, $CA$ and $AB$, respectively. The drawing places $E$, $F$ and $G$ on those edges and shows the arrows from the origin. The volume of their parallelepiped is the absolute scalar triple product:Equivalently, this is the absolute determinant of the matrix whose columns are the three vectors. Since it is nonzero, the vectors are linearly independent. Three linearly independent vectors in the three-dimensional vector space ${\mathbb{R}}^3$ form a basis.

Choose Cartesian coordinates with unit vectors satisfying the right-hand rule, $\hat{\mathbf{x}}\times \hat{\mathbf{y}}=\hat{\mathbf{z}}$. Put $A$, $B$ and $C$ one unit along the positive $x$, $y$ and $z$ axes, respectively. The following original drawing also includes the vectors needed in the next part.

Use the Exact Newton error identity for a quadratic to expose the cancellation without a long quotient expansion. Put $e_k=x_k-x_{\ast }$. Since $f(x_{\ast })=0$, direct expansion givesFrom part (ii), $e_1=\frac{1}{2}{\epsilon }^2+O({\epsilon }^3)$ and $x_1-\epsilon \to 1$. It follows thatThus the stronger result isThe vanishing lower-order terms are a concrete instance of quadratic convergence: to leading order, this step squares the previous error.

The Newton root-finding iteration for this quadratic polynomial isAt $x_0=1$, it gives $x_1=1/(1-\epsilon )$. Its geometric series expansion is $1+\epsilon +{\epsilon }^2+{\epsilon }^3+O({\epsilon }^4)$. Subtracting the expansion of the polynomial root yields

The positive polynomial root is $x_{\ast }=\epsilon +\sqrt{1+{\epsilon }^2}$. Its Taylor expansion near zero isThe starting error is therefore $x_0-x_{\ast }=-\epsilon -\frac{1}{2}{\epsilon }^2+O({\epsilon }^4)$. ConsequentlyFor signed $\epsilon$, the magnitude scaling should be written with $|\epsilon |$; this is the precise interpretation of the proportionality in the question.

For each fixed $n$, the inner geometric series sums to $2^{-n}/(1-1/2)=2^{1-n}$. Summing this second geometric series givesAll terms are nonnegative, so there is no conditional-rearrangement issue; the iterated sum is finite and the double series converges. Equivalently, collecting equal values of $m$ gives ${\sum }_{m=0}^{\infty }(m+1)2^{-m}=4$.

For every integer $n$, $\cos ((2n-1)\pi )=-1$. The series is therefore the negative of the harmonic series:For a direct divergence argument, each block $2^{j-1}<n\le 2^j$ contributes at least $2^{j-1}/2^j=1/2$ to the positive sum. Its partial sums grow without bound, soDespite the odd multiples of $\pi$, the signs do not alternate.

Compute the two sides of the normal equations for linear least squares and divide by $n$:The first equation gives $\alpha =\overline{y}-\beta \overline{x}$. Substitute this into the second and solve:These simple linear regression coefficients are unique provided the observed distances are not all equal, sinceThe condition $n\ge 2$ alone does not guarantee this. If all $x_i=x_0$, the design matrix has matrix rank one and the valid alternative is the familyIn this rank-deficient ordinary least squares case, the fitted predictions are uniquely $\overline{y}$ but the two parameters cannot be separately determined.

Each prediction is a linear combination of the constant parameter and the coefficient of $x_i$. Thus the design matrix isIts $i$th row multiplied by the parameter vector $(\alpha ,\beta {)}^{\mathsf{T}}$ gives the required prediction $\alpha +\beta x_i$.

For either $A$ from part (i), chooseIndeed, using $A^2=-I$ and $A^{\mathsf{T}}=-A$,Its matrix trace is $-2/\sqrt{2}=-\sqrt{2}<0$. Explicitly, for the two $A$ matrices in the order displayed above, the corresponding orthogonal matrices areTheir negatives have positive matrix trace and would not meet the additional sign condition.

Since $A^2=-I$, the matrix inverse is $A^{-1}=-A$. The orthogonal matrix condition also gives $A^{-1}=A^{\mathsf{T}}$, so $A^{\mathsf{T}}=-A$. Thus $A$ is skew-symmetric, forcing $a=0$ and $c=-b$. Its columns must be unit vectors, giving $b^2=1$. Therefore the two possibilities areBoth satisfy the original conditions by direct multiplication. They are the two orthogonal skew-symmetric matrices in two dimensions, representing opposite quarter-turn rotations.

For integration by parts, take $u=\arctan x$ and $dv=x^{-2}dx$. Then $du=dx/(1+x^2)$ and $v=-1/x$, soUse partial fraction decomposition $1/[x(1+x^2)]=1/x-x/(1+x^2)$. ThereforeThe constant of integration may be chosen independently on the positive and negative intervals.

Expand hyperbolic sine in exponential functions:Integrating each term gives the indefinite integral

By the definition of conditional probability, or Bayes' theorem,Its denominator is positive throughout the specified range. Observing two blue balls raises the probability of replacement for $0<r<1$, because replacement makes a repeated colour more likely.

The event of two blue draws splits into disjoint replacement and no-replacement cases. Their probabilities from the sample space are $4r/25$ and $(1-r)/10$. Hence

Condition on whether replacement occurs. Both alternatives have blue second-draw probability $2/5$, so the law of total probability gives

For sampling with replacement, the box has its original contents at the second draw. Therefore

For sampling without replacement, sum the two outcomes with a blue second draw:The marginal blue probability equals the initial blue fraction even though the two draws are dependent.

Use the Maclaurin series of the logarithm and hyperbolic cosine:To retain terms through degree six, only the constant and quadratic terms of the reciprocal factor can multiply the leading cubic term. ThusThe numerator in the original PDF is $\ln (1+x^3)$; the cubic power is missing from the local TeX transcription.

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 $x=-2$.