Category: Algebra

Question Number 219940 by hardmath last updated on 03/May/25 $$\mathrm{Let}: \\ $$$$\mathrm{f}\::\:\left[\mathrm{n}−\mathrm{1}\:,\:\mathrm{n}\right]\:\rightarrow\:\left[\mathrm{n}\:,\:\mathrm{n}\:+\:\mathrm{1}\right] \\ $$$$\mathrm{be}\:\mathrm{a}\:\mathrm{continuous}\:\mathrm{function} \\ $$$$\mathrm{Such}\:\mathrm{that}: \\ $$$$\int_{\boldsymbol{\mathrm{n}}−\mathrm{1}} ^{\:\boldsymbol{\mathrm{n}}} \left(\mathrm{1}\:+\:\mathrm{xf}\:^{'} \left(\mathrm{x}\right)\right)\mathrm{dx}\:\leqslant\:\mathrm{nf}\left(\mathrm{n}\right)−\left(\mathrm{n}−\mathrm{1}\right)\mathrm{f}\left(\mathrm{n}−\mathrm{1}\right) \\ $$$$\mathrm{Then}\:\mathrm{prove}: \\ $$$$\int_{\boldsymbol{\mathrm{n}}−\mathrm{1}}…

Question Number 219935 by fantastic last updated on 03/May/25 Answered by MrGaster last updated on 04/May/25 Commented by MrGaster last updated on 04/May/25 $$\left(\mathrm{2}\right):\mathrm{Let}\:{b}_{{j}} =\sqrt{{a}_{{j}}…

Question Number 219869 by universe last updated on 02/May/25 Answered by MrGaster last updated on 03/May/25 Commented by MrGaster last updated on 03/May/25 Original text:\[ \begin{aligned} &\text{Solution:}\\ &\text{Let the equation of the circle be}C:x^2+y^2=n^2\left(2-\frac{2}{\sqrt{n}}+\frac{1}{n}\right).\text{As}n\to\infty,\text{the radius}R\sim n\sqrt{2}.\\ &\text{For a lattice line}l:ax+by+c=0,\text{its distance to the origin is}\frac{|c|}{\sqrt{a^2+b^2}}=R,\text{i.e.,}|c|=R\sqrt{a^2+b^2}.\\ &\text{Since}R\sim n\sqrt{2},\text{for sufficiently large}n,\text{we have}|c|\sim n\sqrt{2(a^2+b^2)}.\\ &\text{Consider that the line}l\text{must pass through at least two lattice points}(x_1,y_1),(x_2,y_2)\in S_n.\text{Their parameters satisfy:}\\ &\text{(1)The slope}m=\frac{y_2-y_1}{x_2-x_1}\text{is a rational number,and the intercept}c=y_1-mx_1\text{is an integer combination.}\\ &\text{(2)The intercept range is limited by}0\leq y_1,y_2\leq n,\text{so the upper bound of}|c|\text{is}O(n).\\ &\text{As}n\to\infty,\sqrt{a^2+b^2}\sim\sqrt{m^2+1}\text{(let}m=\frac{b}{a}\text{)},\text{then}|c|\sim n\sqrt{2(m^2+1)}.\\ &\text{However,the allowed range of}|c|\text{is}O(n),\text{so we need}\sqrt{2(m^2+1)}\leq 1,\text{i.e.,}m^2\leq\frac{1}{2}-1,\text{which has no real solutions.}\\ &\text{Therefore,the number of tangent lines}N{\text{tangent}}=o(N{\text{total}}).\\ &\text{The total number of lines}N{\text{total}}\sim\frac{3}{\pi^2}n^4\text{(based on the asymptotic estimate of lattice lines)},\text{and the number of tangent lines}N{\text{tangent}}\sim O(n^2)\text{(only a limited number of directions satisfy the intercept condition)}.\\ &\text{Thus,the probability}P_n=\frac{N{\text{tangent}}}{N{\text{total}}}\sim\frac{O(n^2)}{n^4}\to 0.\\ &\text{The final limit is:}\\ &\boxed{A} \end{aligned} \] Terms…

Question Number 219846 by hardmath last updated on 02/May/25 $$\mathrm{If}\:\:\:\mathrm{f}:\left[\mathrm{a},\mathrm{b}\right]\rightarrow\mathbb{R} \\ $$$$\:\:\:\:\:\:\mathrm{0}<\mathrm{a}\leqslant\mathrm{b} \\ $$$$\mathrm{f}\:-\:\mathrm{continuous} \\ $$$$\mathrm{Then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{b}^{\mathrm{4047}} \:−\:\mathrm{a}^{\mathrm{4047}} }{\mathrm{4047}}\:+\:\int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \:\mathrm{f}\:^{\mathrm{2}} \:\left(\mathrm{x}^{\mathrm{2024}} \right)\:\mathrm{dx}\:\geqslant\:\frac{\mathrm{1}}{\mathrm{1012}}\:\int_{\boldsymbol{\mathrm{a}}^{\mathrm{2024}} }…

Question Number 219696 by Spillover last updated on 01/May/25 Answered by SdC355 last updated on 01/May/25 $$\boldsymbol{\mathrm{E}}=\begin{pmatrix}{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{cus}\:\mathrm{A}=\begin{pmatrix}{\mathrm{3}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{5}}&{\mathrm{1}}&{\mathrm{2}}\\{\mathrm{4}}&{\mathrm{2}}&{\mathrm{7}}\end{pmatrix}\: \\ $$$$\mathrm{swap}\:\mathrm{row}\:\mathrm{1}\:\mathrm{and}\:\mathrm{row}\:\mathrm{2} \\ $$$$\begin{pmatrix}{\mathrm{5}}&{\mathrm{1}}&{\mathrm{2}}\\{\mathrm{3}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{4}}&{\mathrm{2}}&{\mathrm{7}}\end{pmatrix}\:\mathrm{and}\:\mathrm{Subtract}\:\frac{\mathrm{3}}{\mathrm{5}}×\:\mathrm{row}\:\mathrm{1}\:\mathrm{from}\:\mathrm{row}\:\mathrm{2} \\ $$$$\begin{pmatrix}{\mathrm{5}}&{\:\:\mathrm{1}}&{\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}}&{\frac{\mathrm{2}}{\mathrm{5}}}&{−\frac{\mathrm{1}}{\mathrm{5}}}\\{\mathrm{4}}&{\:\:\mathrm{2}}&{\:\:\:\:\:\mathrm{7}}\end{pmatrix}\:\:…