brain

By Exercise 39, the inverse of the combinatorial matrix has entries b_{ij}=\frac{-y+\delta_{ij}(x+ny)}{x(x+ny)}.

Let $A = (a_{ik})$ be the Cauchy matrix defined by $a_{ik} = \frac{1}{x_i + y_k}, \qquad 1 \le i,k \le n,$ where all quantities $x_i + y_k$, $x_j - x_k$, and $y_i - y_k$ that appear below are nonzero.

Let $A = (a_{ij})_{1 \le i,j \le n}$ be the combinatorial matrix with entries $a_{ij} = x + y\,[i \ne j],$ equivalently $a_{ij} = y + x\,\delta_{ij}.$ Let $B = (b_{ij})_{1 \le i,j \le n}$ be defined b...

Let $V$ be the $n\times n$ Vandermonde matrix $V_{ij}=x_i^{\,j-1}$, with $x_1,\dots,x_n$ distinct.

Let D_n=\det\!

Let M=\sup_{R(i)} a_i,\qquad N=\sup_{S(j)} b_j.

Let $A = (a_{ij})$ be the $n \times n$ matrix defined by $a_{ij} = x_j^i$.

Let $A = (a_{ij})_{1 \le i,j \le n}$ be the combinatorial matrix defined by $a_{ij} = y + \delta_{ij} x,$ so that $a_{ii} = x+y$ and $a_{ij} = y$ when $i \ne j$.

For each integer $j$ with $1 \le j \le n$, define S_j = \sum_{i=1}^m a_{ij}.

Let $x_1, x_2, \ldots, x_n$ be distinct numbers and define $P(x)=\prod_{k=1}^n (x-x_k).$ Then for each $j$, $P'(x_j)=\prod_{\substack{1\le k\le n\\ k\ne j}}(x_j-x_k).$ For an integer $r\ge 0$, define...

Let T = \sum_{1 \le j < k \le n} (u_j - u_k)(v_j - v_k).

Let P(t)=\prod_{r=1}^n (t-r).

For $j \ge 2$, 1 - \frac{1}{j^2} = \frac{j^2 - 1}{j^2} = \frac{(j-1)(j+1)}{j^2} = \frac{j-1}{j} \cdot \frac{j+1}{j}.

The iterated sum \sum_{i=0}^n \sum_{j=0}^i \sum_{k=0}^j a_i a_j a_k ranges over all integer triples satisfying

Let $S=\sum_{j=1}^n a_j x_j,\qquad T=\sum_{j=1}^n b_j y_j.$ Then $ST=\left(\sum_{j=1}^n a_j x_j\right)\left(\sum_{k=1}^n b_k y_k\right)=\sum_{j=1}^n\sum_{k=1}^n a_j b_k x_j y_k. \eqno(1)$ Similarly, $...

Let P=\prod_{i=0}^n \prod_{j=0}^i a_i a_j.

Let $P_n = \prod_{j=1}^n (1 - a_j)$ and $S_n = \sum_{j=1}^n a_j$, where $0 < a_j < 1$ for each $j$.

The first equality \left(\sum_{i=1}^n a_i\right)\left(\sum_{j=1}^n \frac{1}{a_j}\right) = \sum_{1 \le i \le n} \sum_{1 \le j \le n} \frac{a_i}{a_j} follows directly from the distributive law (4), appl...

Let $J$ be the finite set of all integers $j$ for which $R(j)$ is true, and let $|J| = n$.

From (17), each restricted sum may be written as a full sum using Iverson’s convention: \sum_{R(j)} a_j = \sum_j a_j [R(j)], \qquad \sum_{S(j)} a_j = \sum_j a_j [S(j)].

When no integer satisfies $R(j)$, the expression $\sum_{R(j)} a_j$ must behave as an additive identity so that identities such as Eq.

The product notation is defined in Eq.

Let $k \ge 1$.

$\sum_{j=m}^n (a_j - a_{j-1}) = \sum_{j=m}^n a_j - \sum_{j=m}^n a_{j-1}.$ The first term is unchanged.

Let $R(i)$ be the relation “$i$ is a positive integer such that $n$ is a multiple of $i$,” equivalently $i \mid n$, and let $S(i,j)$ be the relation $1 \le j < i$.

Let S_n = \sum_{k=1}^{n} k 2^k.

Let S=\sum_{j=0}^n jx^j.

Each term in $\sum_{j \in S} 1$ contributes the value $1$ once for every integer $j$ contained in $S$.

Let $S=\sum_{j=m}^n \sum_{k=r}^s jk.$ For fixed $j$, the factor $j$ does not depend on $k$, hence $\sum_{k=r}^s jk = j \sum_{k=r}^s k.$ Substituting into $S$ gives $S=\sum_{j=m}^n \left(j \sum_{k=r}^s...

From equation (15) with $a=0$ and $b=1$, $\sum_{0 \le j \le N} j = \frac{1}{2}N(N+1).$ This gives $\sum_{j=0}^n j = \frac{1}{2}n(n+1), \qquad \sum_{j=0}^{m-1} j = \frac{1}{2}(m-1)m.$ Since $m \le n$,...

Let S = 1 + \frac{1}{7} + \frac{1}{7^2} + \cdots + \left(\frac{1}{7}\right)^n.

No.

We consider the standard derivation of Eq.

If $x = 1$, each term $a x^j$ in the sum $\sum_{0 \le j \le n} a x^j$ becomes $a$.

Let $R,S \subseteq \mathbb{Z}$, and let $(a_j)$ be an arbitrary real-valued sequence.

Let A_n=\sum_{i=1}^n a_i,\qquad B_m=\sum_{j=1}^m b_j, and assume both series $\sum_i a_i$ and $\sum_j b_j$ converge, so

Let $R(i)$ and $S(j)$ both be the relation $i \ge 0$ and $j \ge 0$.

Define the transformation $p(j)=c-j$.

For $n = 3$, the left-hand side of Eq.

For the first sum, the condition $0 \le n \le 5$ restricts $n$ to the six integers $0,1,2,3,4,5$.

The notation $\sum_{1 \le j \le n} a_j$ means the sum of $a_j$ over all integers $j$ satisfying $1 \le j \le n$.

The expression $a_1 + a_2 + \cdots + a_0$ contains no indices $j$ satisfying $1 \le j \le 0$, so it is a sum over an empty set of indices and is defined to be $0$.

Let $t = \ln x$.

The goal is to compute $b^x$ for $0 \le x < 1$ using only shifting, addition, and subtraction, and to analyze accuracy.

For $b>1$, $b \log_b x = b \frac{\ln x}{\ln b} = (\ln x)\frac{b}{\ln b}.$ Since $x>1$, the factor $\ln x$ is positive and independent of $b$, so minimizing $b \log_b x$ is equivalent to minimizing $g(...

The key error in the previous solution is the assumption that a right shift implements exact division by a power of two.

We restart from a consistent floating-point model and propagate the error through the iterative state update.

The reviewer’s objections identify a real structural gap: the original proof implicitly assumed identical normalization behavior and uncontrolled error collapse.

Let $x > 0$.

For $x>0$, the binary, natural, and common logarithms are related by change of base: $\lg x = \frac{\ln x}{\ln 2}, \qquad \log_{10} x = \frac{\ln x}{\ln 10}.$ Hence $\ln x + \log_{10} x = \ln x \left(...

Let $\ln x$ be interpreted as the area described in Fig.

Assume $b>0$, $b\ne 1$, and $x>1$ so that all logarithms and iterated logarithms are defined.

Using the change of base formula (14) with base $2$, \log_8 x = \frac{\log_2 x}{\log_2 8}.

From equation (14), taking base $b=2$ and $c=10$, one obtains $\log_{10} 2 = \frac{\log_2 2}{\log_2 10}.$ Since $\log_2 2 = 1$ by equation (9), this simplifies to $\log_{10} 2 = \frac{1}{\log_2 10}.$...

$\lg 32 = \log_2 32 = 5$, because $2^5 = 32$.

A 14-digit decimal integer $n$ satisfies $10^{13} \le n \le 10^{14} - 1.$ A computer word with 47 bits for magnitude and one sign bit can represent integers in the range $- (2^{47} - 1) \le n \le 2^{4...

Let $x>0$ and let $n$ be a positive integer.

From equation (14), with $b = e$ and $c = 10$, one obtains $\log_{10} x = \frac{\log_e x}{\log_e 10}.$ By definition (15), $\log_e x = \ln x$ and $\log_e 10 = \ln 10$, hence $\log_{10} x = \frac{\ln x...

Let $b > 0$, $b \ne 1$, $c > 0$, and let $y$ be a real number.

For $x,y>0$, write $\frac{x}{y} = x \cdot \frac{1}{y}.$ Using Eq.

Let $y = 10^x$.

Assume that $\log_{10} 2$ is rational.

Let $m$ be a positive integer and let $u>0$.

Since $10^x$ is strictly increasing for real $x$ when $10>1$, the inequalities in (8) imply $10^{0.30102999} < 2 < 10^{0.30103000}.$ Applying the monotonicity of $10^x$ and the definition of logarithm...

Let $b>0$, and let $x=\frac{p}{q}$ and $y=\frac{r}{s}$, where $p,r\in\mathbb{Z}$ and $q,s$ are positive integers.

The issue is that decimal expansions are not unique.

A real number $x$ can be defined by a binary expansion $x = n + 0.b_1 b_2 b_3 \ldots,$ where $n$ is an integer and each $b_i \in {0,1}$, with the restriction that the sequence does not end in infinite...

We compute (0.

Let the exponentiation of an integer exponent be defined as in Eq.

The flaw in the previous solution is not the asymptotic analysis itself, but the unjustified introduction of the factor $(n/2)^{1-k/2}$.

Let $r$ be a positive rational number.

The expression $1 + 0.239999999\ldots$ equals the real number whose fractional part is $0.239999999\ldots$, hence it represents $1.239999999\ldots$.

Let $b^n$ be defined for integer $n$ by the rules in (4).

Write C_n=\int_0^r e^{-nx}f(x)\,dx+\int_r^\infty e^{-nx}f(x)\,dx = A_n+B_n.

The function $Q(n)$ is defined in this section so that its normalization removes the factor $n^n$ from sums of the form \sum_{k=0}^n \binom{n}{k} k^k (n-k)^{n-k}.

The previous argument fails because it never uses a valid definition of $Q(k)$ and therefore cannot justify any of the inversion steps.

For fixed integer $n>0$, consider the series S(n) = 1 + \frac{n}{n+1} + \frac{n}{n+1}\frac{n+1}{n+2} + \cdots = \sum_{k \ge 0} a_k,

We restart from the integrand and determine all contributions that survive after integration down to order $O(x^{-2})$.

Start from the identity \left(1+\frac{u}{x}\right)^x = \exp\!

Consider I=\int_0^\infty \left(1+\frac{z}{n}\right)^n e^{-z}\,dz.

The errors in the previous solution stem from treating the Euler–Maclaurin remainder and the size of $f^{(3)}$ in a purely global way, and from introducing an unsupported “cancellation” step.

Start from Eq.

From the definition, v = u - \ln(1+u).

Start from the exact factorization \frac{(n+\alpha)^n}{n^{n+\beta}} = n^{-\beta}\left(1+\frac{\alpha}{n}\right)^n.

Let \gamma(a,x)=\int_0^x e^{-t}t^{a-1}\,dt,\qquad a>0,\ x\ge 0.

Let S_n(x)=\sum_{k=0}^{n}\frac{x^k}{k!

Assume $g(n)=\Omega(f(n))$.

The previous solution fails because it treats a non-power-series manipulation as if it were coefficient algebra.

Let m = an^2 + bn.

Let \binom{2n}{n}=\frac{(2n)!

We are given a string s of length n, and we should think of every occurrence of a character as a position on a line. Moving between two positions has a cost that depends on whether the characters are the same.

Let $u(z)=O(z^m)$ as $z\to 0$.

From the definition of the exponential function, e^x = \sum_{k\ge 0}\frac{x^k}{k!

Assume n!

We restart the argument using a representation of $\Gamma$ that avoids any delicate extraction from the digamma kernel.

Let P_n = 1^1 2^2 3^3 \cdots n^n.

Let $f(x)=x^m$.

Equation (10) is obtained in Euler’s summation formula by repeated application of finite-difference identities that depend on a specific sequence of coefficients introduced in Eq.

Equation (7) in Section 1.

Let $m=2k>0$ and assume $f^{(m)}(x)$ has constant sign on $1\le x\le n$.