Using the signs chosen above, explicit multiplication givesThus the nonzero cases obey $A{\mathbf{v}}_m=\sqrt{{\sigma }_m}{\mathbf{u}}_m$. This relates the right singular vectors and left singular vectors in a singular value decomposition. For example, if $U=({\mathbf{u}}_1{\mathbf{u}}_2)$ and $V=({\mathbf{v}}_1{\mathbf{v}}_2{\mathbf{v}}_3)$, then $A=U\Sigma V^{\mathsf{T}}$ with $\Sigma =\left(\begin{array}{ccc}1& 0& 0\\ 0& \sqrt{2}& 0\end{array}\right)$.

In $A^{\mathsf{T}}A$, every row sums to one, so $(1,1,1)$ is an eigenvector with eigenvalue $1$. The antisymmetric direction $(1,0,-1)$ gives eigenvalue $2$. A direction orthogonal to both is $(1,-2,1)$, and direct multiplication shows that it lies in the kernel of a linear map. Normalizing all three gives the complete systemThey form an orthonormal basis of ${\mathbb{R}}^3$. The extra zero eigenvalue reflects the one-dimensional kernel of a linear map for this rank-two rectangular matrix. The question's symbols ${\sigma }_m$ denote the eigenvalues of $A^{\mathsf{T}}A$; the nonzero singular values of $A$ are their square roots, namely $1$ and $\sqrt{2}$.

For a matrix with equal diagonal entries, the symmetric and antisymmetric directions $(1,1)$ and $(1,-1)$ are eigenvectors. Direct multiplication of $A{A}^{\mathsf{T}}$ gives eigenvalues $3/2-1/2=1$ and $3/2+1/2=2$. Dividing each direction by its Euclidean norm givesThe two normalized eigenvectors are orthogonal; either may independently be replaced by its negative.

A row's squared Euclidean norm is $[(1+\sqrt{3}{)}^2+1+(1-\sqrt{3}{)}^2]/6=3/2$, and the dot product of the two rows is $[2(1-3)+1]/6=-1/2$. Thus the row Gram matrix isFor the column Gram matrix, the first and third squared column norms are $4/3$, the middle one is $1/3$, and the respective cross-products are $1/3,-2/3,1/3$. HenceThe two products are square matrices of different sizes; their unequal dimensions are consistent with $A$ being rectangular.

Use integration by parts in $I_{n+1}$, writing the integrand as $x^{2n+1}(x{e}^{-x^2})$. Since an antiderivative of $x{e}^{-x^2}$ is $-e^{-x^2}/2$,The boundary term vanishes because the Gaussian exponential function decays faster than any fixed power. These improper integrals are absolutely convergent, justifying the limiting integration by parts.

Starting from the permitted Gaussian integral $I_0=\sqrt{\pi }$, the Gaussian moment recurrence gives $I_1=\sqrt{\pi }/2$ and thenMore generally, iteration gives $I_n=(2n-1)!!\sqrt{\pi }/2^n$, using the double factorial with $(-1)!!=1$.

Assume that wrong-note events are independent events and that each note has the same error probability $p$, with $0\le p\le 1$. Then their count $N$ has a binomial distribution. Choose the $n$ mistaken positions in $\left(\genfrac{}{}{0ex}{}{100}{n}\right)$ ways and multiply the probabilities of the errors and correct notes:The endpoint cases $p=0$ or $p=1$ are understood by continuity, giving deterministic counts. More than half means at least $51$ mistakes, soThe single-note error probability alone would not determine this count distribution if mistakes were correlated; independence is the additional standard modeling assumption.

A uniformly shuffled deck makes all unordered five-card hands equally likely, giving $\left(\genfrac{}{}{0ex}{}{52}{5}\right)$ possible hands. Choose the rank appearing four times in $13$ ways; all four suits of that rank are then fixed. The remaining card can be any of the other $48$ cards. No hand is counted twice, because five cards cannot contain two different complete four-card ranks. This use of combinatorics gives the probabilityHere $\left(\genfrac{}{}{0ex}{}{52}{5}\right)=2598960$ is a binomial coefficient; the probability also simplifies to $1/4165\approx 0.000240096$.

For $f(x)=x^{1/3}$, the required derivatives areAt $a=8$ these give $f(8)=2$, $f^{\prime }(8)=1/12$, and $f^{\prime \prime }(8)=-1/144$. Setting $x-a=1$, the three-term Taylor polynomial yieldsBy the Taylor theorem with Lagrange remainder, the signed error isIt is positive, and ${\xi }^{-8/3}<8^{-8/3}=1/256$. ConsequentlyThe last comparison is exact because $5\cdot 4000=20000<20736$. Thus the estimate is an underestimate with the required rigorous absolute error bound.

With the first $n$ terms counted from the constant term, Taylor's theorem giveswhere $\xi$ lies between $a$ and $x$. Continuous derivatives through order $n$ on the joining interval are a sufficient hypothesis for this Taylor theorem with Lagrange remainder. The Taylor remainder measures the error of the finite Taylor polynomial; convergence of the corresponding infinite Taylor series is a separate question.