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intuition module | Taking derivatives | Differential Calculus | Khan AcademyStart with a curve, then ask how steep it is at each point. Sal Khan demonstrates this idea using f(x)=2x^3: seven orange controls encode tangent-slope estimates, and the revealed derivative graph is 6x^2. The slopes decrease toward the origin, reach 0, and then increase again.https://www.youtube-nocookie.com/embed/HtvikVD9aa0?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02011-10-20T13:16:01+00:00238https://math-video.com/zh/videos/derivative-intuition-khan-academy2026-10-12T03:04:50.510Zhttps://i.ytimg.com/vi/HtvikVD9aa0/maxresdefault.jpg导数的直观理解：Khan Academy 演示先看一条曲线，再问它在每一点有多陡。Sal Khan 用 f(x)=2x^3 演示这个想法：七个橙色控制点表示切线斜率的估计值，最后揭示的导函数图像是 6x^2。斜率先随横坐标趋向原点而减小，在原点达到 0，随后再次增大。读导函数图像还能估计原先采样点之外的斜率，例如 x=-0.5 与 x≈-2.25。本片建立图形直觉，没有证明幂法则。本站提供原创双语笔记，并明确标记补充的数学核验。视频演示的是历史互动模块，无法保证原练习目前仍可使用。https://www.youtube-nocookie.com/embed/HtvikVD9aa0?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02011-10-20T13:16:01+00:00238https://math-video.com/videos/common-probability-distributions-kai2026-10-12T03:06:59.158Zhttps://i.ytimg.com/vi/wMBBdS7n6hk/maxresdefault.jpgCommon probability distributions in seven minutes | Kai博士This Chinese-language video begins with uncertainty and compares discrete and continuous probability distributions. A binomial example uses an 82-game season, assumes a win probability of 0.7 per game, and asks for exactly 60 wins.https://www.youtube-nocookie.com/embed/wMBBdS7n6hk?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02022-08-14T19:31:08+00:00449https://math-video.com/zh/videos/common-probability-distributions-kai2026-10-12T03:06:59.158Zhttps://i.ytimg.com/vi/wMBBdS7n6hk/maxresdefault.jpg七分钟理解常见概率分布｜Kai博士这条中文视频从不确定事件出发，比较离散型与连续型概率分布。二项分布以一个82场赛季为例，假设每场胜率为0.7，讨论恰好赢60场的概率，并解释独立性、固定胜率与组合数的作用。随后以均匀、三角与右偏长尾图形说明密度形状怎样联系观测值，借收入数据区分平均数与中位数。最后介绍正态模型和成绩按秩配对：目标平均分70、标准差10，生成同人数的随机分数并分别排序后配对。本站详细笔记补充模型条件：连续分布的概率看区间面积，长尾未必有有限均值，正态近似须用实际数据验证；按秩变换不保证原始算术均值映射到目标均值，有限样本的均值、标准差和人数比例会波动，并列与舍入也需规则。https://www.youtube-nocookie.com/embed/wMBBdS7n6hk?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02022-08-14T19:31:08+00:00449https://math-video.com/videos/matrix-addition-subtraction-khan-academy2026-10-12T03:05:41.028Zhttps://i.ytimg.com/vi/WR9qCSXJlyY/maxresdefault.jpgMatrix addition and subtraction | Matrices | Precalculus | Khan AcademyMatrix arithmetic begins with matching positions. Check that both matrices have the same number of rows and columns, then add or subtract entries at the same row and column.https://www.youtube-nocookie.com/embed/WR9qCSXJlyY?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02013-02-22T01:21:52+00:00335https://math-video.com/zh/videos/matrix-addition-subtraction-khan-academy2026-10-12T03:05:41.028Zhttps://i.ytimg.com/vi/WR9qCSXJlyY/maxresdefault.jpg矩阵加法与减法：Khan Academy矩阵加减的关键是对应位置。先检查两个矩阵的行数、列数是否一致，再把同一行、同一列上的元素相加或相减。视频用完整例题追踪每个结果，解释加法为何满足交换律，并把减法改写为加上第二个矩阵的负数。最后通过维度不匹配的例子说明：不能仅因某些元素可以配对，就对两个不同大小的矩阵进行标准加减。https://www.youtube-nocookie.com/embed/WR9qCSXJlyY?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02013-02-22T01:21:52+00:00335https://math-video.com/videos/determinant-geometric-meaning-data-science2026-10-12T03:05:21.174Zhttps://i.ytimg.com/vi/PZvn6x-xnu0/maxresdefault.jpgDeterminants and geometric meaning | Engineers and Little PotatoA determinant records geometric scale and orientation for the linear transformation represented by a real square matrix. The video places vectors (3,0) and (3,4) in columns, computes a determinant of12, and connects it with a parallelogram of base3 and height4.https://www.youtube-nocookie.com/embed/PZvn6x-xnu0?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02021-02-07T05:32:54+00:00277https://math-video.com/zh/videos/determinant-geometric-meaning-data-science2026-10-12T03:05:21.174Zhttps://i.ytimg.com/vi/PZvn6x-xnu0/maxresdefault.jpg矩阵行列式的几何意义｜工程师和小土豆行列式不仅是算式，也记录实方阵对应线性变换的几何尺度与定向。视频把二维向量(3,0)、(3,4)作为矩阵的列，算得行列式12，并用底3、高4的平行四边形解释它。换一个向量顺序，行列式变号，但普通面积不变，因此面积应取行列式的绝对值。随后把矩阵改成[[3,4],[0,0]]：列向量共线，行列式为零，二维图形退化为线。本站补充明确表示方式：用列表示与转置成行表示可比较同样的向量，但不能把两种写法混作原矩阵的直接换行。对n阶实方阵，行列式为零意味着秩小于n，像空间可能降低多个维数；三维落在平面中的直观示例也可包含更低维子空间。这个条件讨论方阵的线性变换，不概括所有机器学习降维方法。https://www.youtube-nocookie.com/embed/PZvn6x-xnu0?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02021-02-07T05:32:54+00:00277https://math-video.com/videos/continuous-random-variable-expectation-intuition2026-10-10T04:44:33.633Zhttps://i0.hdslb.com/bfs/archive/cff0122b179190802872acb108f4b23b790112c0.jpgThe expectation of a continuous random variable | Teacher ChangjiangExpectation is an average weighted by probability. The video starts with a discrete weighted sum, then defines a continuous expectation for a random variable X with probability density f, explicitly requiring absolute convergence.https://player.bilibili.com/player.html?bvid=BV1LTbXz8ECN&p=1&autoplay=0&t=02025-07-25T08:56:39+00:00423https://math-video.com/zh/videos/continuous-random-variable-expectation-intuition2026-10-10T04:44:33.633Zhttps://i0.hdslb.com/bfs/archive/cff0122b179190802872acb108f4b23b790112c0.jpg连续型随机变量的期望怎么理解？直观意义是什么？数学期望是按概率加权的平均。视频先从离散取值的加权求和出发，再给出有概率密度f的随机变量X的连续期望公式，并明确要求绝对收敛。随后用定积分的分割、取样与矩形和解释积分：密度乘小区间宽度近似区间概率，再乘代表取值，得到连续加权和的一个近似贡献。原片教材明确写的是小区间概率；后续“取到代表点”的口头简写不能理解为点有非零概率。本站补充：有密度时单点概率为零，有限实数期望要求绝对可积，密度面积为1与期望积分是不同的量，负取值可作负贡献。全实轴手写求和提供直觉，严谨处理还需截断与尾部控制。连续分布并不必然有密度；允许正负无穷的扩展期望也须与有限值区分，正负部分都无限时未定义，不能用相消或主值代替期望。https://player.bilibili.com/player.html?bvid=BV1LTbXz8ECN&p=1&autoplay=0&t=02025-07-25T08:56:39+00:00423https://math-video.com/videos/nonsquare-matrices-3blue1brown2026-10-10T05:20:56.236Zhttps://math-video.com/zh/videos/nonsquare-matrices-3blue1brown2026-10-10T05:20:56.236Zhttps://math-video.com/videos/higher-order-derivatives-3blue1brown2026-10-10T05:21:04.410Zhttps://math-video.com/zh/videos/higher-order-derivatives-3blue1brown2026-10-10T05:21:04.410Zhttps://math-video.com/videos/one-sided-limits-asymptote-khan2026-10-12T03:04:16.315Zhttps://i.ytimg.com/vi_webp/5f1-Rg3MmKs/maxresdefault.webpOne-sided limits from graphs: asymptote | Limits and continuity | AP Calculus AB | Khan AcademyRead one-sided limits directly from a graph. Approaching x=6 from the left, the plotted branch grows without bound; from the right, it approaches the open circle at y=-3. The instructor uses nearby inputs and graphical tracing to distinguish these two behaviors.https://www.youtube-nocookie.com/embed/5f1-Rg3MmKs?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02016-07-11T18:55:19+00:00146https://math-video.com/zh/videos/one-sided-limits-asymptote-khan2026-10-12T03:04:16.315Zhttps://i.ytimg.com/vi_webp/5f1-Rg3MmKs/maxresdefault.webp从渐近线图像判断单侧极限｜Khan Academy从图像判断单侧极限：当 x 从左侧趋近 6 时，曲线向上无界增长；从右侧趋近时，函数值靠近空心圆所在的 y=-3。视频用相邻输入和沿曲线追踪的方法，对照这两种行为，并解释无穷大的含义：“正无穷极限”描述向上发散，并不是一个有限实数答案，因此它与“不存在有限极限”的说法并不矛盾。https://www.youtube-nocookie.com/embed/5f1-Rg3MmKs?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02016-07-11T18:55:19+00:00146https://math-video.com/videos/conditional-probability-occupancy-without-replacement2026-10-10T07:05:11.037Zhttps://i1.hdslb.com/bfs/archive/5e9b1ff6479b1c000098ab5e36c66fa02011976c.jpgConditional probability: occupied boxes and sampling without replacementConditional probability restricts attention to an event already known to occur, then measures the desired event within it. The video solves two complete examples. Four distinct balls enter four distinct boxes: given that at least one box is empty, the probability that exactly two are empty is 21/58.https://player.bilibili.com/player.html?bvid=BV1Jy4y1E7a4&p=1&autoplay=0&t=02021-03-17T16:48:42+00:00381https://math-video.com/zh/videos/conditional-probability-occupancy-without-replacement2026-10-10T07:05:11.037Zhttps://i1.hdslb.com/bfs/archive/5e9b1ff6479b1c000098ab5e36c66fa02011976c.jpg条件概率：小球入盒与不放回抽样条件概率先限定已经发生的条件，再在这个范围内计算目标事件的比例。视频用两个完整例题练习：四个不同小球投入四个不同盒子，在至少一个盒子为空的条件下，恰好两个为空的概率为21/58；20件产品中有4件次品，不放回抽取，在第一件为次品的条件下，第二件仍为次品的概率为3/19。第一个例题用事件包含关系简化交集，再按3加1、2加2两类分配计数，并用补事件求分母。本站明确建模条件：各球独立、等概率选择盒子，等可能的是每个球的具体去向，不是不同占用形状。原片中间的3/32是没有空盒的概率，条件分母为其补值29/32；不应把中间板书当成最终P(A)。第二个题重新定义事件，不能沿用小球例题的含义。https://player.bilibili.com/player.html?bvid=BV1Jy4y1E7a4&p=1&autoplay=0&t=02021-03-17T16:48:42+00:00381https://math-video.com/videos/sine-derivative-from-limit-definition2026-10-12T03:06:29.122Zhttps://i.ytimg.com/vi/lD0LPDRba0k/maxresdefault.jpgThe derivative of sine from the limit definitionThe video derives the derivative of sine at an arbitrary fixed real x from the limit definition. After substituting sine into the difference quotient, a sum-to-product identity expresses the numerator as a half-angle sine times a cosine.https://www.youtube-nocookie.com/embed/lD0LPDRba0k?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02022-11-23T22:00:20+00:00240https://math-video.com/zh/videos/sine-derivative-from-limit-definition2026-10-12T03:06:29.122Zhttps://i.ytimg.com/vi/lD0LPDRba0k/maxresdefault.jpg由导数定义推导正弦函数的导数视频从导数定义出发，推导正弦函数在任意实数x处的导数。把函数代入差商后，用正弦的和差化积公式，把分子写成半角正弦与余弦的乘积；再把外面的倍数并入分母，使正弦比值具有已知基本极限的形式。结合余弦的连续性，最后得到正弦函数的导数为余弦。本站补充适用范围：角度使用弧度，x保持固定，h从非零值趋于0；正弦基本极限与余弦连续性是本推导采用的先修结论，视频没有在这里证明它们。画面0/0只表示直接代入得到不定式，不能视为极限值或合法除法。https://www.youtube-nocookie.com/embed/lD0LPDRba0k?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02022-11-23T22:00:20+00:00240https://math-video.com/videos/matrix-multiplication-associativity-worked-example2026-10-12T03:05:42.739Zhttps://i.ytimg.com/vi/WTRWqIQkos0/maxresdefault.jpgWorked example: matrix multiplication associativityGiven three square matrices, compute the product by grouping the first pair, then repeat by grouping the last pair. The explanation expands each row-by-column dot product and reaches the same final matrix by both routes, illustrating associativity in this example.https://www.youtube-nocookie.com/embed/WTRWqIQkos0?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02024-09-19T04:19:39+00:00198https://math-video.com/zh/videos/matrix-multiplication-associativity-worked-example2026-10-12T03:05:42.739Zhttps://i.ytimg.com/vi/WTRWqIQkos0/maxresdefault.jpg矩阵乘法结合律：两种括号位置的例题用三个给定方阵，分别计算先乘前两个矩阵、再乘最后一个，与先乘后两个矩阵、再从左边乘第一个的结果。讲解逐项展开横行与竖列的乘积和，得到相同的最终矩阵，展示这个例子中的结合性。本站补充：改变的是括号位置，因子始终按A、B、C的顺序排列；一般结合律适用于维数相容的矩阵，这个数值例题展示了性质，不能替代一般证明，也不意味着矩阵乘法满足交换律。https://www.youtube-nocookie.com/embed/WTRWqIQkos0?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02024-09-19T04:19:39+00:00198https://math-video.com/videos/row-operations-as-left-matrix-multiplication2026-10-12T03:06:36.027Zhttps://i.ytimg.com/vi/oNnujJFqLqU/maxresdefault.jpgRow operations as left matrix multiplicationExpress the specified row operations by left multiplication. X1 performs two row additions to obtain B; X2 then scales the second row of B to obtain C. Each new row is a linear combination of the old rows, whose coefficients form the operation matrix.https://www.youtube-nocookie.com/embed/oNnujJFqLqU?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02016-11-19T08:53:19+00:00205https://math-video.com/zh/videos/row-operations-as-left-matrix-multiplication2026-10-12T03:06:36.027Zhttps://i.ytimg.com/vi/oNnujJFqLqU/maxresdefault.jpg用矩阵左乘表示横行变换把指定的横行操作写成矩阵左乘：先用X1完成两次横行加法，得到B；再用X2缩放B的第二横行，得到C。每个新横行都是旧横行的线性组合，组合系数构成操作矩阵。两阶段合成时，后执行的操作写在左边，因此总矩阵为X2X1。本站符号核对：原画方程第三横行的y系数为-3，右上排版增广矩阵却显示+3；复述方程对应的矩阵时按方程纠正负号，并明确是本站修订。求总操作矩阵只依赖指定横行操作，不受这处局部排版差异影响。X1合成两次行加，是两个单步初等矩阵的乘积，不是一个单步初等矩阵。https://www.youtube-nocookie.com/embed/oNnujJFqLqU?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02016-11-19T08:53:19+00:00205https://math-video.com/videos/derivative-product-rule-khan2026-10-12T03:04:21.518Zhttps://i.ytimg.com/vi/79ngr0Bur38/maxresdefault.jpgProduct rule | Derivative rules | AP Calculus AB | Khan AcademyApply the product rule to differentiate a product of two functions: differentiate one factor, keep the other, then add the complementary term. Khan Academy demonstrates the complete example x² sin x, obtaining 2x sin x+x² cos x with consistent colors for the two factors.https://www.youtube-nocookie.com/embed/79ngr0Bur38?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02013-01-28T23:00:36+00:00159https://math-video.com/zh/videos/derivative-product-rule-khan2026-10-12T03:04:21.518Zhttps://i.ytimg.com/vi/79ngr0Bur38/maxresdefault.jpg导数的乘积法则｜Khan Academy用乘积法则对两个函数的乘积求导：一次对一个因子求导、保留另一个因子，再将对应的两项相加。Khan Academy 完整演算 x² sin x，得到 2x sin x+x² cos x，并用一致的颜色区分两个因子。原片侧重应用，不证明一般法则；配套说明补明可微性条件与三角函数求导所用的弧度制。https://www.youtube-nocookie.com/embed/79ngr0Bur38?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02013-01-28T23:00:36+00:00159https://math-video.com/videos/definite-integral-introduction-khan2026-10-12T03:04:05.424Zhttps://i.ytimg.com/vi_webp/0dDIPzqKgYk/maxresdefault.webpDefinite integrals intro | Accumulation and Riemann sums | AP Calculus AB | Khan AcademyConnect a nonnegative curve-area diagram to definite-integral notation. Khan Academy sketches f(x), marks a and b, shades the region above the horizontal axis and then writes ∫ from a to b of f(x) dx, explaining its bounds, integrand and integration variable.https://www.youtube-nocookie.com/embed/0dDIPzqKgYk?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02017-07-31T18:13:45+00:00148https://math-video.com/zh/videos/definite-integral-introduction-khan2026-10-12T03:04:05.424Zhttps://i.ytimg.com/vi_webp/0dDIPzqKgYk/maxresdefault.webp理解定积分与累积｜Khan Academy把非负曲线下的面积图形转为定积分记号。Khan Academy 画出 f(x)、标记 a 与 b、填充横轴上方的阴影区域，再写出从 a 到 b 的 f(x) 对 x 的积分，解释上下限、被积函数与积分变量。本完整短片讲清几何含义与记号，没有数值计算、黎曼和构造或微积分基本定理证明。配套说明明确普通面积解释的可积、非负条件，并区分一般的有符号累积。https://www.youtube-nocookie.com/embed/0dDIPzqKgYk?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02017-07-31T18:13:45+00:00148https://math-video.com/videos/definite-integral-from-animated-right-endpoint-sums2026-10-12T03:03:59.166Zhttps://i.ytimg.com/vi/-JSVueLK3fc/maxresdefault.jpgDefinite integrals from animated rectangle sumsThe animation begins with area under a curve, partitions [0,1] into equal pieces, takes right-endpoint function values as rectangle heights, and adds the rectangle areas. Increasing the partition count makes the rectangles thinner.https://www.youtube-nocookie.com/embed/-JSVueLK3fc?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02020-09-08T12:16:52+00:00237https://math-video.com/zh/videos/definite-integral-from-animated-right-endpoint-sums2026-10-12T03:03:59.166Zhttps://i.ytimg.com/vi/-JSVueLK3fc/maxresdefault.jpg用动画矩形和理解定积分动画从曲线下方的面积问题开始，把区间[0,1]等分，用右端点函数值作为矩形高度，再把各矩形的面积相加。分割数量增加时，矩形越来越细；随后板书把展开式写成求和，再取极限连接到定积分。本站明确数学范围：函数在该区间上黎曼可积，连续是一个充分条件；有限矩形和通常只近似积分，极限成立时与积分精确相等。函数非负时，积分等于图形面积；一般有符号函数的积分不等于总几何面积。动画的500指分割数量，不是面积值。原板书在极限前保留约等号，本站规范为有限和近似、极限与积分相等；视频提供直观解释，没有证明一般收敛，也不能据此断言每次增加分割数都会使误差单调减小。https://www.youtube-nocookie.com/embed/-JSVueLK3fc?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02020-09-08T12:16:52+00:00237https://math-video.com/videos/vector-magnitude-components-khan2026-10-12T03:06:39.506Zhttps://i.ytimg.com/vi/qMZnsj4ZO9g/maxresdefault.jpgFinding vector magnitude from componentsFind the Euclidean length of a=(5,-3) in a standard orthonormal Cartesian plane. Khan Academy first computes √(5^2+(-3)^2)=√(34) with color-coded components, then completes the geometric explanation: draw the displacement from the origin and form a right triangle with leg lengths 5 and 3.https://www.youtube-nocookie.com/embed/qMZnsj4ZO9g?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02016-01-07T20:45:16+00:00185https://math-video.com/zh/videos/vector-magnitude-components-khan2026-10-12T03:06:39.506Zhttps://i.ytimg.com/vi/qMZnsj4ZO9g/maxresdefault.jpg由分量计算向量模长｜Khan Academy在标准正交直角坐标平面中，求 a=(5,-3) 的欧氏长度。Khan Academy 先用颜色区分分量，算出 √(5^2+(-3)^2)=√(34)，再完整画出从原点出发的位移，构成边长为 5 与 3 的直角三角形，解释计算依据。本课应用已有的勾股定理；负分量表示方向，不表示负边长。配套说明补明坐标系条件，以及平移所讨论的是自由位移向量。https://www.youtube-nocookie.com/embed/qMZnsj4ZO9g?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02016-01-07T20:45:16+00:00185https://math-video.com/videos/vector-addition-arrows-coordinates-and-doubling2026-10-12T03:06:10.979Zhttps://i.ytimg.com/vi_webp/gfXju6Uc310/maxresdefault.webpVector addition: arrows, coordinates and doublingTwo presenters connect arrow diagrams, coordinates and vector addition on one whiteboard. They introduce magnitude and direction, write vectors starting at the origin as columns, translate arrows to form a head-to-tail sum, and verify the result with coordinates.https://www.youtube-nocookie.com/embed/gfXju6Uc310?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02021-02-17T02:11:14+00:00275https://math-video.com/zh/videos/vector-addition-arrows-coordinates-and-doubling2026-10-12T03:06:10.979Zhttps://i.ytimg.com/vi_webp/gfXju6Uc310/maxresdefault.webp向量加法：箭头、坐标与倍乘两位讲解者在同一块白板上，把向量的箭头表示、坐标与加法连起来。先从大小和方向理解几何向量，把从原点出发的两个向量写成列向量，再用平移与首尾相接找到和向量，并以坐标计算核对。广州经福州到上海的示意说明总位移与走过的路程不同。随后将水平向量乘以2，算出新的和向量，最后用并排的列向量引出矩阵。本站明确范围：这里讨论自由几何向量；一般起点下坐标应取终点减起点，坐标相加要求相同维度和同一基。原横幅把数乘方向说得过宽：正数乘非零向量保留方向，负数反向，乘数为0或原向量为零时没有方向。视频展示的是正数2倍的例子，未讨论负数或零；矩阵只作初步介绍，没有一般性质证明。https://www.youtube-nocookie.com/embed/gfXju6Uc310?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02021-02-17T02:11:14+00:00275https://math-video.com/videos/discrete-random-variable-expected-value-khan2026-10-12T03:06:41.206Zhttps://i.ytimg.com/vi_webp/qafPcWNUiM8/maxresdefault.webpMean (expected value) of a discrete random variable | AP Statistics | Khan AcademyKhan Academy computes the mean of a discrete workout-count distribution. Check that the probabilities are nonnegative and sum to 1, then multiply each value 0,1,2,3,4 by its probability 0.1,0.15,0.4,0.25,0.1 and add to obtain E(X)=2.1.https://www.youtube-nocookie.com/embed/qafPcWNUiM8?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02017-07-14T00:53:56+00:00271https://math-video.com/zh/videos/discrete-random-variable-expected-value-khan2026-10-12T03:06:41.206Zhttps://i.ytimg.com/vi_webp/qafPcWNUiM8/maxresdefault.webp离散随机变量的均值与期望｜Khan AcademyKhan Academy 完整计算每周锻炼次数这一离散分布的均值。先检查概率非负且总和为 1，再把取值 0,1,2,3,4 分别乘以概率 0.1,0.15,0.4,0.25,0.1，相加得到 E(X)=2.1。随后解释：每次结果都是整数，分布均值仍可以是非整数。配套说明区分分布期望、总次数的期望与有抽样条件的长期样本平均，避免将期望当作保证发生的实际结果。https://www.youtube-nocookie.com/embed/qafPcWNUiM8?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02017-07-14T00:53:56+00:00271https://math-video.com/videos/independent-dice-rolling-khan2026-10-12T03:04:07.110Zhttps://i.ytimg.com/vi/2MYA8Ba2PvM/maxresdefault.jpgDie rolling probability with independent events | Precalculus | Khan AcademyWork through a complete Khan Academy dice problem: on a fair six-sided die, the even faces 2, 4 and 6 give a single-roll probability of 3/6=1/2. Assuming three mutually independent rolls, multiply three one-half factors to obtain 1/8 for an even result on every roll.https://www.youtube-nocookie.com/embed/2MYA8Ba2PvM?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02010-06-27T20:39:32+00:00147https://math-video.com/zh/videos/independent-dice-rolling-khan2026-10-12T03:04:07.110Zhttps://i.ytimg.com/vi/2MYA8Ba2PvM/maxresdefault.jpg独立掷骰事件的概率计算｜Khan Academy跟随 Khan Academy 解完一道掷骰概率题：公平六面骰子的偶数点数为 2、4、6，单次概率是 3/6=1/2。在三次投掷相互独立的假设下，把三个二分之一相乘，得到每次都出现偶数的概率 1/8。配套说明区分公平性与独立性，并明确仅有两两独立不足以推出三事件的乘积公式。https://www.youtube-nocookie.com/embed/2MYA8Ba2PvM?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02010-06-27T20:39:32+00:00147https://math-video.com/videos/quick-proof-bayes-theorem-3blue1brown2026-10-12T03:05:31.974Zhttps://i.ytimg.com/vi/U_85TaXbeIo/maxresdefault.jpgA quick proof of Bayes’ theorem | 3Blue1BrownTwo ways to measure the same overlap lead directly to Bayes’ theorem. This complete visual proof starts with conditional proportions and rearranges the joint-probability identity, then tests the common shortcut of multiplying marginal probabilities.https://www.youtube-nocookie.com/embed/U_85TaXbeIo?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02019-12-22T14:45:25+00:00227https://math-video.com/zh/videos/quick-proof-bayes-theorem-3blue1brown2026-10-12T03:05:31.974Zhttps://i.ytimg.com/vi/U_85TaXbeIo/maxresdefault.jpg贝叶斯定理的简洁证明｜3Blue1Brown同一块交集面积，换个顺序计算，就能得到贝叶斯定理。视频从条件概率的比例解释出发，完整展示联合概率恒等式的变形，再检查“两个概率直接相乘”何时成立。硬币、骰子和兄弟的例子说明独立性需要依据模型判断。结尾用检测结果演示先验、似然和阳性结果总概率的关系。配套说明补充正概率条件，并区分示意模型与真实医疗数据。https://www.youtube-nocookie.com/embed/U_85TaXbeIo?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02019-12-22T14:45:25+00:00227https://math-video.com/videos/two-sided-limit-graph-khan2026-10-12T03:06:44.689Zhttps://i.ytimg.com/vi/rccqylTloMs/maxresdefault.jpgTwo-sided limit from graph | Limits | Differential Calculus | Khan AcademyCan a function have two different approached values at the same input? This graph exercise reads the left-hand limit at 3 as 4 and the right-hand limit as 1, then uses their disagreement to conclude that the two-sided limit does not exist.https://www.youtube-nocookie.com/embed/rccqylTloMs?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02015-06-18T22:07:01+00:00166https://math-video.com/zh/videos/two-sided-limit-graph-khan2026-10-12T03:06:44.689Zhttps://i.ytimg.com/vi/rccqylTloMs/maxresdefault.jpg从图像判断双侧极限｜Khan Academy同一个输入位置，左右趋近可以得到不同的值吗？这道图像练习先读出在 3 处的左极限为 4、右极限为 1，再由两者不同判断双侧极限不存在。移动的点与辅助线把曲线附近的行为和实心点的函数值 f(3)=1 区分开。配套说明提醒：有限采样用于理解图像，不能单独证明一般极限；空心、实心标记本身也不能决定极限。https://www.youtube-nocookie.com/embed/rccqylTloMs?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02015-06-18T22:07:01+00:00166https://math-video.com/videos/probability-worked-example-khan2026-10-09T23:18:46.255Zhttps://i.ytimg.com/vi/yUaI0JriZtY/maxresdefault.jpgFinding probability example | Probability and Statistics | Khan AcademyA single marble draw makes the counting rule concrete. The bag contains 3 yellow, 2 red, 2 green and 1 blue marble. The lesson names the yellow event, lists the individual possible draws, identifies the sample space, then counts 3 favorable outcomes among 8 total outcomes to obtain 3/8.https://www.youtube-nocookie.com/embed/yUaI0JriZtY?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02011-11-14T23:28:25+00:00175https://math-video.com/zh/videos/probability-worked-example-khan2026-10-09T23:18:46.255Zhttps://i.ytimg.com/vi/yUaI0JriZtY/maxresdefault.jpg概率计算例题｜Khan Academy用一次抽弹珠理解概率的计数方法。袋中有 3 颗黄色、2 颗红色、2 颗绿色和 1 颗蓝色弹珠。视频先命名黄色事件，再逐颗列出可能结果并引入样本空间，最后数出 3 个符合条件的结果和 8 个总结果，得到 3/8。配套说明明确模型前提：每颗实体弹珠被抽到的机会相同。板书中重复的颜色字母代表不同物体，并不是把相同字母重复作为形式集合的元素。https://www.youtube-nocookie.com/embed/yUaI0JriZtY?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02011-11-14T23:28:25+00:00175https://math-video.com/videos/determinant-area-meaning-khan2026-10-12T03:04:43.627Zhttps://i.ytimg.com/vi_webp/ER_96Mx-KEE/maxresdefault.webpInterpreting determinants in terms of area | Matrices | Precalculus | Khan AcademyInterpret A=[[3,1],[1,2]] through its two columns, (3,1) and (1,2). They span a parallelogram with area |det(A)|=5. The same matrix maps the standard unit vectors to these columns, so the unit square becomes a parallelogram of area 5.https://www.youtube-nocookie.com/embed/ER_96Mx-KEE?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02021-03-10T09:48:52+00:00259https://math-video.com/zh/videos/determinant-area-meaning-khan2026-10-12T03:04:43.627Zhttps://i.ytimg.com/vi_webp/ER_96Mx-KEE/maxresdefault.webp行列式的面积含义｜Khan Academy把 A=[[3,1],[1,2]] 的两列 (3,1)、(1,2) 看作平面向量，它们张成的平行四边形面积为 |det(A)|=5。这个矩阵把标准单位向量映到两列，因此单位正方形变为面积为 5 的平行四边形。完整后续讲解保持 A 不变，示意给定面积为 0.6 的弯曲小区域，其像面积为 0.6×5=3。小正方形近似提供统一面积缩放的直觉，不是正式证明。配套说明区分带符号的行列式与非负面积，并补明通常的坐标和面积条件。https://www.youtube-nocookie.com/embed/ER_96Mx-KEE?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02021-03-10T09:48:52+00:00259https://math-video.com/videos/limit-point-discontinuity-khan2026-10-12T03:06:20.364Zhttps://i.ytimg.com/vi_webp/hWJLd6bRthI/maxresdefault.webpGraphical limit at point discontinuityRead a graph where the curve has an open circle at (7,0) and a separate filled point at (7,3). The complete lesson approaches input 7 from both sides, finds nearby outputs tending to zero, then distinguishes the two-sided limit zero from the function value g(7)=3.https://www.youtube-nocookie.com/embed/hWJLd6bRthI?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02016-07-11T18:35:16+00:00158https://math-video.com/zh/videos/limit-point-discontinuity-khan2026-10-12T03:06:20.364Zhttps://i.ytimg.com/vi_webp/hWJLd6bRthI/maxresdefault.webp间断点处的图像极限｜Khan Academy图像在 (7,0) 处有空心点，在 (7,3) 处另有实心点。完整讲解从左右两侧趋近输入 7，观察附近输出趋近零，再区分双侧极限零与函数值 g(7)=3。例题依据局部图像与近似样本，没有给出解析公式，也不是 epsilon-delta 形式的严格证明。https://www.youtube-nocookie.com/embed/hWJLd6bRthI?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02016-07-11T18:35:16+00:00158https://math-video.com/videos/normal-distribution-mean-standard-deviation-and-probability2026-10-12T03:05:49.735Zhttps://i.ytimg.com/vi/YjvP0VYp2rU/maxresdefault.jpgNormal distributions: mean, standard deviation and interval probabilitySlides and interactive graphs introduce normal distributions: the mean sets the center, a positive standard deviation controls width, and interval probabilities are areas under the density.https://www.youtube-nocookie.com/embed/YjvP0VYp2rU?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02015-03-03T01:48:10+00:00276https://math-video.com/zh/videos/normal-distribution-mean-standard-deviation-and-probability2026-10-12T03:05:49.735Zhttps://i.ytimg.com/vi/YjvP0VYp2rU/maxresdefault.jpg正态分布：均值、标准差与区间概率视频用投影片和交互图形介绍正态分布：均值决定中心位置，正的标准差决定宽窄，区间概率对应密度曲线下的面积。随后调成标准正态，计算具体区间，并显示均值左右各一个、两个、三个标准差范围的概率，归纳成68–95–99.7的近似经验法则。本站明确：密度高度不是单点概率，正态变量落在任何指定单点的概率为零；自定义区间与标准差区域是界面中的不同积分选项。开头引用的百科简介省略了中央极限定理的条件与标准化，不能按字面推广。本站补充一个正确版本：独立同分布、有限且正的方差下，标准化后的和或均值趋于标准正态；二项情形还需固定成功概率严格介于零与一。参数随试验数变化的正态只是有限阶段近似，不是一个固定的极限分布。https://www.youtube-nocookie.com/embed/YjvP0VYp2rU?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02015-03-03T01:48:10+00:00276https://math-video.com/videos/determinant-notation-computation-khan2026-10-12T03:05:51.651Zhttps://i.ytimg.com/vi_webp/ZAkmBOCiNqU/maxresdefault.webpIntro to determinant notation and computation | Matrices | Precalculus | Khan AcademyIntroduce the symbolic matrix A=[[a,b],[c,d]] and three determinant notations: |A|, det(A), and bars around its entries. For this two-by-two matrix, the computation rule is ad-bc. The complete lesson applies it to B=[[1,-2],[3,5]], keeping the negative off-diagonal entry: det(B)=5-(-6)=11.https://www.youtube-nocookie.com/embed/ZAkmBOCiNqU?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02021-03-10T09:45:43+00:00206https://math-video.com/zh/videos/determinant-notation-computation-khan2026-10-12T03:05:51.651Zhttps://i.ytimg.com/vi_webp/ZAkmBOCiNqU/maxresdefault.webp行列式的记号与计算｜Khan Academy先用 A=[[a,b],[c,d]] 介绍行列式的三种写法：|A|、det(A)，以及元素两侧的竖线。这个二阶矩阵的计算规则是 ad-bc。完整讲解接着计算 B=[[1,-2],[3,5]]，保留副对角线元素的负号，得到 det(B)=5-(-6)=11。矩阵的行列式竖线与标量绝对值符号需要区分。本课给出并应用计算规则，不包含几何解释或从一般行列式公理出发的证明。https://www.youtube-nocookie.com/embed/ZAkmBOCiNqU?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02021-03-10T09:45:43+00:00206https://math-video.com/videos/defined-undefined-matrix-operations-khan2026-10-12T03:05:12.225Zhttps://i.ytimg.com/vi/O1-9f1g0OsI/maxresdefault.jpgDefined and undefined matrix operations | Matrices | Precalculus | Khan AcademyThree exercises distinguish the dimension tests for matrix operations. A 3-by-3 matrix cannot multiply a 2-by-2 matrix in that order, whereas two 2-by-1 column vectors can be added entry by entry. Finally, A is 2-by-2 and E is 1-by-2: AE is undefined, but EA is defined.https://www.youtube-nocookie.com/embed/O1-9f1g0OsI?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02013-08-06T22:21:21+00:00211https://math-video.com/zh/videos/defined-undefined-matrix-operations-khan2026-10-12T03:05:12.225Zhttps://i.ytimg.com/vi/O1-9f1g0OsI/maxresdefault.jpg矩阵运算何时有定义｜Khan Academy三道练习区分矩阵运算的维数条件。3 行 3 列矩阵不能按此顺序乘 2 行 2 列矩阵；两个 2 行 1 列的列向量则可以对应元素相加。最后，A 是 2 行 2 列，E 是 1 行 2 列：AE 无定义，EA 却有定义。加法检查整体形状，乘法按运算顺序检查内部维数。不同练习复用了字母 B，表示不同矩阵。https://www.youtube-nocookie.com/embed/O1-9f1g0OsI?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02013-08-06T22:21:21+00:00211https://math-video.com/videos/derivative-power-rule-khan2026-10-12T03:05:58.521Zhttps://i.ytimg.com/vi/bRZmfc1YFsQ/maxresdefault.jpgPower rule | Derivative rules | AP Calculus AB | Khan AcademyThe lesson recalls the derivative difference quotient, states the power rule, and applies it to x², x³, x^-100 and x^2.571. Bring the fixed exponent down as a coefficient and reduce it by one. The worked results are 2x, 3x², −100x^-101 and 2.571x^1.571. Proofs are reserved for later lessons.https://www.youtube-nocookie.com/embed/bRZmfc1YFsQ?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02013-01-09T23:57:06+00:00233https://math-video.com/zh/videos/derivative-power-rule-khan2026-10-12T03:05:58.521Zhttps://i.ytimg.com/vi/bRZmfc1YFsQ/maxresdefault.jpg导数的幂法则｜Khan Academy本课先回顾导数的差商极限，再给出幂法则，依次用于 x²、x³、x^-100 与 x^2.571：把固定指数写成系数，再将指数减一。结果分别为 2x、3x²、−100x^-101 与 2.571x^1.571。证明留待后续课程。编辑说明补充分清一般实数幂的正数定义域、整数幂的更大适用范围，以及指数为零的常数情形。https://www.youtube-nocookie.com/embed/bRZmfc1YFsQ?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02013-01-09T23:57:06+00:00233https://math-video.com/videos/binomial-graph-mode-variance-and-second-moment2026-10-12T03:05:08.609Zhttps://i.ytimg.com/vi/MX2LVliNfrc/maxresdefault.jpgA binomial graph: mode, variance and second momentRead a binomial graph, distinguish mode from mean, compare probability masses, then use a given standard deviation to recover the success probability and calculate a second moment.https://www.youtube-nocookie.com/embed/MX2LVliNfrc?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02015-02-16T00:00:00+00:00227https://math-video.com/zh/videos/binomial-graph-mode-variance-and-second-moment2026-10-12T03:05:08.609Zhttps://i.ytimg.com/vi/MX2LVliNfrc/maxresdefault.jpg二项分布图：众数、方差与二阶矩从二项分布柱状图读众数、比较概率，再用标准差反求成功概率，并计算二阶矩。原片先用主峰粗估参数，后结合新增方差条件确定精确解；配套说明明确区分众数与均值、标准差与方差、离散概率与正态密度，并补充二项模型假设和严格众数范围。https://www.youtube-nocookie.com/embed/MX2LVliNfrc?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02015-02-16T00:00:00+00:00227https://math-video.com/videos/definite-integral-graph-worked-khan2026-10-12T03:06:18.501Zhttps://i.ytimg.com/vi/hTTDI49LhjE/maxresdefault.jpgWorked example: Definite integral by thinking about the function's graph | Khan AcademyEvaluate the integral of √(9−x²) from −3 to 3 by recognizing its graph. Squaring identifies the circle relation, but the principal square root keeps only the nonnegative upper branch.https://www.youtube-nocookie.com/embed/hTTDI49LhjE?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02014-10-23T16:12:11+00:00210https://math-video.com/zh/videos/definite-integral-graph-worked-khan2026-10-12T03:06:18.501Zhttps://i.ytimg.com/vi/hTTDI49LhjE/maxresdefault.jpg从函数图像计算定积分例题｜Khan Academy通过识别 √(9−x²) 的图像，计算从 −3 到 3 的定积分。平方可以识别圆的关系，但主平方根只保留非负的上半分支。实数定义域为 [-3,3]，半径为 3；函数连续且非负，所以积分等于上半圆面积。π·3² 的一半给出精确值 9π/2。https://www.youtube-nocookie.com/embed/hTTDI49LhjE?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02014-10-23T16:12:11+00:00210https://math-video.com/videos/composing-three-by-three-matrices-khan2026-10-12T03:04:08.778Zhttps://i.ytimg.com/vi_webp/2O18RbWmrYA/maxresdefault.webpComposing 3x3 matrices | Matrices | Precalculus | Khan AcademyTwo real 3×3 matrices represent linear transformations in three-dimensional space. The composition A∘B applies B first and A second; in standard column-vector coordinates its matrix is the product AB.https://www.youtube-nocookie.com/embed/2O18RbWmrYA?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02021-03-21T11:00:40+00:00217https://math-video.com/zh/videos/composing-three-by-three-matrices-khan2026-10-12T03:04:08.778Zhttps://i.ytimg.com/vi_webp/2O18RbWmrYA/maxresdefault.webp三阶矩阵的复合运算｜Khan Academy两个实3×3矩阵表示三维空间中的线性变换。复合A∘B先作用B，再作用A；在标准列向量坐标下，其矩阵就是乘积AB。结果的第一列和第三列已给出，课程通过对B的向量[0,2,3]^T应用A，计算缺失的中间列。将该向量按标准基分解，系数为0,2,3；用这些系数对A的各列加权求和，得到[0,-16,9]^T。完整视频确实算出这一结果。圆圈在这里表示变换复合，不表示逐元素相乘；本课展开的是中间列的运算，而非乘积每个元素的完整计算。https://www.youtube-nocookie.com/embed/2O18RbWmrYA?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02021-03-21T11:00:40+00:00217https://math-video.com/videos/definite-integral-riemann-sum-limit-khan2026-10-12T03:04:39.492Zhttps://i.ytimg.com/vi/CXCtqBlEZ7g/maxresdefault.jpgDefinite integral as the limit of a Riemann sum | AP Calculus AB | Khan AcademyStarting from left-endpoint rectangles, this lesson builds the finite sum with width Δx=(b-a)/n and then adds the limit n→∞ to obtain the definite-integral notation. The same positive curve and interval remain visible as finer rectangles illustrate the limiting idea.https://www.youtube-nocookie.com/embed/CXCtqBlEZ7g?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02013-01-24T22:00:55+00:00266https://math-video.com/zh/videos/definite-integral-riemann-sum-limit-khan2026-10-12T03:04:39.492Zhttps://i.ytimg.com/vi/CXCtqBlEZ7g/maxresdefault.jpg定积分作为黎曼和的极限｜Khan Academy课程从左端点矩形出发，写出宽度Δx=(b-a)/n的有限和，再加入n→∞的极限，得到定积分记号。相同的正值曲线与区间始终可见，更细的矩形用于说明极限思想。后段将Δx与dx联系起来，并明确称其为不严谨的直观理解，同时讨论其他近似方法。补充适用条件：f在有限区间[a,b]上黎曼可积，a<b；连续是充分条件。采样点位于各自子区间，最大子区间宽度趋于零。显示的等分情形中，n→∞保证这一条件。非负f对应普通面积；变号f对应正负面积的净值。本课没有计算具体积分的数值。https://www.youtube-nocookie.com/embed/CXCtqBlEZ7g?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02013-01-24T22:00:55+00:00266https://math-video.com/videos/math-topic-a8l9a1ztmdy2026-10-12T03:05:56.703Zhttps://i.ytimg.com/vi/a8l9A1ztmDY/maxresdefault.jpgFind a Matrix Inverse Using the Classical AdjugateA complete worked3×3 inverse calculation: check the determinant, evaluate nine minors, apply cofactor signs, transpose to form the classical adjugate, and divide by the determinant.https://www.youtube-nocookie.com/embed/a8l9A1ztmDY?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02013-08-07T15:13:11+00:00156https://math-video.com/zh/videos/math-topic-a8l9a1ztmdy2026-10-12T03:05:56.703Zhttps://i.ytimg.com/vi/a8l9A1ztmDY/maxresdefault.jpg用古典伴随矩阵求逆矩阵：完整例题用一个3×3实方阵，完整演示行列式判定、九个二阶子式、余因子符号、转置构造伴随矩阵，以及由伴随矩阵求逆。配套说明区分子矩阵、子式行列式与带符号余因子，使用“横行、竖列”避免地区术语歧义，并独立核验两个方向的乘积均为单位矩阵；一般公式的证明不在本片范围内。https://www.youtube-nocookie.com/embed/a8l9A1ztmDY?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02013-08-07T15:13:11+00:00156https://math-video.com/videos/binomial-pdf-cdf-calculator-khan2026-10-12T03:04:56.011Zhttps://i.ytimg.com/vi_webp/IngLJs6T1Qw/maxresdefault.webpBinompdf and binomcdf functions | Random variables | AP Statistics | Khan AcademyUsing a free-throw count X, this complete lesson contrasts the TI-84 functions binompdf and binomcdf. The first question asks for exactly four successes and gives approximately 0.14; the second asks for fewer than five, which for an integer count means at most four, and gives approximately 0.94.https://www.youtube-nocookie.com/embed/IngLJs6T1Qw?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02017-10-09T17:36:14+00:00327https://math-video.com/zh/videos/binomial-pdf-cdf-calculator-khan2026-10-12T03:04:56.011Zhttps://i.ytimg.com/vi_webp/IngLJs6T1Qw/maxresdefault.webp二项分布概率与累积概率函数｜Khan Academy完整课程用罚球命中次数X比较TI-84的binompdf与binomcdf。首问求恰好命中四次的概率，结果约为0.14；第二问求命中少于五次的概率，对整数计数等价于至多四次，结果约为0.94。两个命令都使用n=7与p=0.35。模型假设：七次试验相互独立，每次只有成功或失败两种结果，成功概率同为0.35。仅有相同的单次命中概率不能保证各次试验独立。https://www.youtube-nocookie.com/embed/IngLJs6T1Qw?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02017-10-09T17:36:14+00:00327https://math-video.com/videos/math-topic-lzs35plrld42026-10-12T03:05:05.080Zhttps://i.ytimg.com/vi/LzS35plRLd4/maxresdefault.jpgWhy This Matrix Has No Inverse: A Zero-Column ProofA complete contradiction proves that the given3×3 real matrix has no inverse. For every candidate B, the third column of BA remains zero, whereas the identity has a nonzero third column.https://www.youtube-nocookie.com/embed/LzS35plRLd4?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02011-10-10T15:05:37+00:00168https://math-video.com/zh/videos/math-topic-lzs35plrld42026-10-12T03:05:05.080Zhttps://i.ytimg.com/vi/LzS35plRLd4/maxresdefault.jpg这个矩阵为什么没有逆矩阵：零列反证法通过一个第三竖列为零的3×3实方阵，完整证明逆矩阵不存在：假设存在任意候选B，计算BA，发现其第三竖列永远为零，而单位矩阵的对应竖列不是零。配套说明统一“横行、竖列”位置记号，不把地区术语差异当成原作者错误，并明确证明排除了所有B，而不是某一个候选。https://www.youtube-nocookie.com/embed/LzS35plRLd4?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02011-10-10T15:05:37+00:00168https://math-video.com/videos/definite-integral-area-formulas-khan2026-10-12T03:04:23.606Zhttps://i.ytimg.com/vi_webp/7R_DTzVHw5Y/maxresdefault.webpFinding definite integrals using area formulas | AP Calculus AB | Khan AcademyFour definite integrals are evaluated from the same graph using elementary geometry. The upper semicircle on [-6,-2] gives 2π, the lower trapezoid on [-2,1] gives −4, the upper triangle on [1,4] gives 6, and the lower semicircle on [4,6] gives −π/2.https://www.youtube-nocookie.com/embed/7R_DTzVHw5Y?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02018-05-04T16:09:26+00:00256https://math-video.com/zh/videos/definite-integral-area-formulas-khan2026-10-12T03:04:23.606Zhttps://i.ytimg.com/vi_webp/7R_DTzVHw5Y/maxresdefault.webp用面积公式计算定积分｜Khan Academy课程根据同一幅图，用初等几何求出四个定积分。[-6,-2]上的上半圆给出2π，[-2,1]上的下方梯形给出−4，[1,4]上的上方三角形给出6，[4,6]上的下半圆给出−π/2。普通几何面积始终非负，积分的正负取决于区域位于横轴的哪一侧。在有限有序区间a<b上，黎曼可积函数的定积分表示带符号面积：横轴上方的区域计为正，下方计为负。图中的分段曲线连续，这足以保证可积。两段弧线按本几何题给定的半圆处理，不能把任意相似曲线都认定为圆。https://www.youtube-nocookie.com/embed/7R_DTzVHw5Y?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02018-05-04T16:09:26+00:00256https://math-video.com/videos/math-topic-t-nqucmf-0k2026-10-12T03:06:50.428Zhttps://i.ytimg.com/vi/t_nqUCmf-0k/maxresdefault.jpgRecovering a Matrix and Swapping Its ColumnsRecover the two columns of an unknown3×2 matrix from products with two projection matrices, then right-multiply by a swap matrix to obtain the answer.https://www.youtube-nocookie.com/embed/t_nqUCmf-0k?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02016-11-19T08:53:19+00:0066https://math-video.com/zh/videos/math-topic-t-nqucmf-0k2026-10-12T03:06:50.428Zhttps://i.ytimg.com/vi/t_nqUCmf-0k/maxresdefault.jpg恢复未知矩阵，再交换两条竖列从两个投影矩阵的乘积结果读出未知3×2矩阵的两条竖列，再用右乘交换矩阵得到答案。原创说明用“横行、竖列”明确方向：原片台湾地区的“行运算”在这里对应竖列操作，属于术语差异；右乘组合竖列、左乘组合横行。https://www.youtube-nocookie.com/embed/t_nqUCmf-0k?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02016-11-19T08:53:19+00:0066https://math-video.com/videos/math-topic-w0pm8cafoni2026-10-12T03:05:37.321Zhttps://i.ytimg.com/vi/W0Pm8cAFOnI/maxresdefault.jpgMatrix Distributivity: Keep the Common Factor on the RightA complete2×2 worked example keeps the common factor C on the right, rewrites AC+BC as(A+B)C and finishes the addition and multiplication to obtain[[1,9],[3,-8]]. It applies distributivity rather than proving the general theorem.https://www.youtube-nocookie.com/embed/W0Pm8cAFOnI?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02024-09-19T04:19:25+00:00146https://math-video.com/zh/videos/math-topic-w0pm8cafoni2026-10-12T03:05:37.321Zhttps://i.ytimg.com/vi/W0Pm8cAFOnI/maxresdefault.jpg矩阵分配律例题：共同因子留在右边用具体2×2矩阵例题说明共同右因子C如何保留在右侧，将AC+BC合并为(A+B)C，再逐项完成加法和乘法。完整视频得到[[1,9],[3,-8]]。这是一道分配律应用例题，不是一般定理的证明；配套说明给出一般维数条件，并明确横行乘竖列的元素计算。https://www.youtube-nocookie.com/embed/W0Pm8cAFOnI?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02024-09-19T04:19:25+00:00146https://math-video.com/videos/random-variables-introduction-khan2026-10-12T03:04:14.399Zhttps://i.ytimg.com/vi/3v9w79NhsfI/maxresdefault.jpgRandom variables | Probability and Statistics | Khan AcademyA random variable assigns numerical values to outcomes of an experiment. The coin example uses heads1 and tails0, briefly replaces them with100 and703, then returns to1 and0.https://www.youtube-nocookie.com/embed/3v9w79NhsfI?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02012-12-06T17:33:26+00:00331https://math-video.com/zh/videos/random-variables-introduction-khan2026-10-12T03:04:14.399Zhttps://i.ytimg.com/vi/3v9w79NhsfI/maxresdefault.jpg什么是随机变量｜Khan Academy随机变量为试验结果规定数值。抛硬币例子把正面记1、反面记0，暂时改用100和703，再恢复1和0。Y表示7个骰子的点数和；课程用P(Y\le30)简写事件概率，并提出点数和为偶数的概率，但没有计算数值。代数对比使用x+5=6，解为x=1，以及y=x+7。映射规则固定，试验结果具有不确定性；结果标签是取值，不是概率。这些有限结果的例子用来引出事件概率记号。随机变量也可以是常值，也能用等式定义并参与代数运算。https://www.youtube-nocookie.com/embed/3v9w79NhsfI?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02012-12-06T17:33:26+00:00331https://math-video.com/videos/math-topic-bltvvdji7wm2026-10-12T03:04:35.971Zhttps://i.ytimg.com/vi/BlTVVDJI7WM/maxresdefault.jpgMatrix Algebra: Counterexamples, Associativity and CancellationA3×3 multiple-choice example separates valid matrix laws from scalar rules that fail universally. One explicit example has nonzero factors with AB zero and BA nonzero, disproving universal commutativity and the scalar zero-product rule.https://www.youtube-nocookie.com/embed/BlTVVDJI7WM?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02016-11-19T08:53:18+00:00111https://math-video.com/zh/videos/math-topic-bltvvdji7wm2026-10-12T03:04:35.971Zhttps://i.ytimg.com/vi/BlTVVDJI7WM/maxresdefault.jpg矩阵代数多选题：反例、结合律与可逆消去通过3×3矩阵多选题区分哪些实数代数规则仍适用。一个具体反例同时否定“乘法恒可交换”和“积为零必有一个因子是零矩阵”：两个矩阵都非零，AB为零而BA非零。结合律成立；左侧共同因子可逆时可以左乘其逆消去；平方展开必须保留AB+BA，只有AB=BA时才能合并为2AB。完整原片正确答案是B、D，不把“不恒成立”说成“永不成立”。https://www.youtube-nocookie.com/embed/BlTVVDJI7WM?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02016-11-19T08:53:18+00:00111https://math-video.com/videos/math-topic-0hdzwxtwq182026-10-12T03:04:03.758Zhttps://i.ytimg.com/vi/0HdZWXTwq18/maxresdefault.jpgPopulation standard deviation: a worked exampleThis is an in-class exercise on univariate data analysis. The video first presents population data 6, 7, 8, 9, 10 and asks for the standard deviation.https://www.youtube-nocookie.com/embed/0HdZWXTwq18?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02024-03-25T04:23:05+00:0041https://math-video.com/zh/videos/math-topic-0hdzwxtwq182026-10-12T03:04:03.758Zhttps://i.ytimg.com/vi/0HdZWXTwq18/maxresdefault.jpg母体标准差：从平均数到方差的完整例题這是一段一維數據分析的隨堂練習。影片先給出母體資料 6、7、8、9、10，要求標準差；接著以三行白板式算式示範解法：先求平均數 8，再依母體變異數公式求得 σ₁²=2，最後取平方根得 σ₁=√2≈1.41。https://www.youtube-nocookie.com/embed/0HdZWXTwq18?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02024-03-25T04:23:05+00:0041https://math-video.com/videos/math-topic-w6fb2c5rcmg2026-10-12T03:05:39.226Zhttps://i.ytimg.com/vi/W6Fb2c5Rcmg/maxresdefault.jpgPermute Matrix Rows, Rearrange the Inverse ColumnsGiven an invertible A and its inverse K, reorder the rows of A as3,1,2 to form N. Its inverse reorders the columns of K as3,1,2, rather than its rows, giving option5.https://www.youtube-nocookie.com/embed/W6Fb2c5Rcmg?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02015-05-11T14:37:54+00:00124https://math-video.com/zh/videos/math-topic-w6fb2c5rcmg2026-10-12T03:05:39.226Zhttps://i.ytimg.com/vi/W6Fb2c5Rcmg/maxresdefault.jpg矩阵横行重排，逆矩阵竖列怎样排列已知可逆矩阵A及其逆K，将A的横行按3、1、2重排，得到新矩阵N。求逆时不能同样重排K的横行：应将K的竖列按3、1、2排列，得到第5项。原片用横行与竖列配对解释；本站补充N=PA、N⁻¹=KP⁻¹的置换记号核对全部乘积，避免只验证对角线。带撇号的小写字母是已知逆矩阵元素，不是求导符号。https://www.youtube-nocookie.com/embed/W6Fb2c5Rcmg?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02015-05-11T14:37:54+00:00124https://math-video.com/videos/vector-components-introduction-khan2026-10-12T03:06:16.438Zhttps://i.ytimg.com/vi_webp/hJkKADcQWj0/maxresdefault.webpVector components: magnitude and directionA vector records displacement rather than a fixed point. The lesson constructs horizontal and vertical components, then uses right-triangle ratios to find a=(3√3/2,3/2) from magnitude3 and direction30°, and converts b=(√2,√2) back to magnitude2 and direction45°.https://www.youtube-nocookie.com/embed/hJkKADcQWj0?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02021-04-28T11:35:22+00:00316https://math-video.com/zh/videos/vector-components-introduction-khan2026-10-12T03:06:16.438Zhttps://i.ytimg.com/vi_webp/hJkKADcQWj0/maxresdefault.webp向量分量：大小与方向的双向转换向量表示位移，不是固定点的位置。本课用直角三角形说明分量转换：先求得a=(3√3/2,3/2)，对应长度3、方向角30°；再由b=(√2,√2)反求长度2、方向角45°。计算使用互相垂直、单位尺度一致的坐标轴，分量表示有符号的坐标变化。https://www.youtube-nocookie.com/embed/hJkKADcQWj0?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02021-04-28T11:35:22+00:00316https://math-video.com/videos/math-topic-dr09wuz2k8c2026-10-12T03:04:41.616Zhttps://i.ytimg.com/vi/Dr09WUZ2K8c/maxresdefault.jpgPopulation Variance and Standard Deviation: Five WeightsThe five weights are67,79,55,61,88 kilograms, treated as the complete population. Find the mean70 kilograms, then calculate squared deviations from70 with sum720.https://www.youtube-nocookie.com/embed/Dr09WUZ2K8c?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02024-01-31T08:48:46+00:00118https://math-video.com/zh/videos/math-topic-dr09wuz2k8c2026-10-12T03:04:41.616Zhttps://i.ytimg.com/vi/Dr09WUZ2K8c/maxresdefault.jpg五人体重：总体方差与标准差怎么算五名队员的体重是67、79、55、61、88公斤，题目将他们作为完整总体。先求平均体重70公斤，再计算各体重与70的差的平方和720，除以人数5得总体方差144平方公斤；取非负平方根，得到总体标准差12公斤。本例使用总体定义，不能改成样本方差的分母。https://www.youtube-nocookie.com/embed/Dr09WUZ2K8c?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02024-01-31T08:48:46+00:00118https://math-video.com/videos/math-topic-ilbmib3hmeo2026-10-12T03:04:54.332Zhttps://i.ytimg.com/vi/ILbmiB3HmEo/maxresdefault.jpgNormal distribution: probability below a scoreUnder the stated normal model with mean70 and standard deviation10, score60 is one standard deviation below the mean. The diagram and68-95 rule give a rounded left-tail probability16%, selecting option0.16. This is an approximation: a precise normal-tail check gives approximately0.158655.https://www.youtube-nocookie.com/embed/ILbmiB3HmEo?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02016-12-08T03:34:24+00:0051https://math-video.com/zh/videos/math-topic-ilbmib3hmeo2026-10-12T03:04:54.332Zhttps://i.ytimg.com/vi/ILbmiB3HmEo/maxresdefault.jpg正态分布：低于60分的概率在题目给出的正态模型下，平均数为70、标准差为10，60比平均数低一个标准差。通过钟形图与68-95经验法则，左尾概率近似16%，选择最接近的0.16；补充精确正态概率核算约为0.158655。https://www.youtube-nocookie.com/embed/ILbmiB3HmEo?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02016-12-08T03:34:24+00:0051https://math-video.com/videos/math-topic-a1mg4ywncvc2026-10-12T03:05:55.032Zhttps://i.ytimg.com/vi/a1Mg4YWnCVc/maxresdefault.jpgSample Variance and Standard Deviation: Six Pulse CountsA class of 40 supplies a random sample of 6 students with pulse counts per minute 71, 83, 67, 74, 70, 79. The sample mean is 74 and the squared deviations sum to 180.https://www.youtube-nocookie.com/embed/a1Mg4YWnCVc?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02024-01-31T08:48:00+00:00141https://math-video.com/zh/videos/math-topic-a1mg4ywncvc2026-10-12T03:05:55.032Zhttps://i.ytimg.com/vi/a1Mg4YWnCVc/maxresdefault.jpg六人脉搏样本：方差与标准差怎么算从40人班级随机抽出6人，测得每分钟脉搏次数为71、83、67、74、70、79。样本均值是74，偏差平方和为180；本题按修正样本方差定义除以6−1=5，得样本方差36，取非负平方根得样本标准差6。40描述抽样背景，不能作为这组样本方差的分母；这些样本统计量也不等于已知全班总体参数。https://www.youtube-nocookie.com/embed/a1Mg4YWnCVc?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02024-01-31T08:48:00+00:00141https://math-video.com/videos/math-topic-vvowpultcag2026-10-12T03:06:55.640Zhttps://i.ytimg.com/vi/vVoWPUltcag/maxresdefault.jpgSpreadsheet mean and population standard deviationUse a spreadsheet to compute the mean and population standard deviation of12 monthly temperatures for Taipei, Kaohsiung and Hsinchu. The lesson enters AVERAGE and STDEVP with row ranges and copies their formulas down.https://www.youtube-nocookie.com/embed/vVoWPUltcag?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02015-06-10T15:26:27+00:00133https://math-video.com/zh/videos/math-topic-vvowpultcag2026-10-12T03:06:55.640Zhttps://i.ytimg.com/vi/vVoWPUltcag/maxresdefault.jpg试算表计算平均数与母体标准差用试算表计算台北、高雄、新竹各12个月温度的平均数与母体标准差，示范AVERAGE、STDEVP和向下复制行范围公式。把台北6月数据从28改为20后，显示平均数由23.48变为22.82，标准差由5.19变为5.08，折线图也同步更新。数值显示经过舍入；母体与样本的区别仅作概念提醒，不推导样本分母校正。https://www.youtube-nocookie.com/embed/vVoWPUltcag?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02015-06-10T15:26:27+00:00133https://math-video.com/videos/math-topic-myjijaiiix42026-10-12T03:06:32.553Zhttps://i.ytimg.com/vi/myJIJaIiiX4/maxresdefault.jpgNormal Curve or Sample Histogram? Probabilities and QuartilesThe question gives a histogram of 100 weights and a reference normal curve N with mean 55 and standard deviation 12.5. Keep them separate: the normal probability above 55 is 50%.https://www.youtube-nocookie.com/embed/myJIJaIiiX4?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02014-12-29T09:27:30+00:00158https://math-video.com/zh/videos/math-topic-myjijaiiix42026-10-12T03:06:32.553Zhttps://i.ytimg.com/vi/myJIJaIiiX4/maxresdefault.jpg正态曲线还是样本直方图：概率与四分位数题目给出100人的体重直方图，以及均值55、标准差12.5的参考正态曲线N。两者要分开判断：正态曲线超过55的概率为50%；超过80对应均值加2倍标准差，原片用95%经验近似估为2.5%，本站补充精确值约2.275%。样本直方图的区间是(35,45]、(45,55]等，前20%不超过45、累计53%不超过55，所以样本第一四分位数大于45，但中位数不大于55。样本超过80的比例只能确定在5%到11%之间，不对区间内部作均匀插值。完整答案为1、2、4、5。https://www.youtube-nocookie.com/embed/myJIJaIiiX4?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02014-12-29T09:27:30+00:00158https://math-video.com/videos/geometric-probability-bv1neaietekk2026-10-12T03:03:23.092Zhttps://i0.hdslb.com/bfs/archive/d987a8957005a01a025063d86a23c83ae9a114b8.jpgGeometric probability: length, area, and the meeting problemLearn geometric probability through two complete examples: a uniform point on [0,3] lies below 1 with probability 1/3, and two people arriving between 4 and 5 PM, with a 15-minute wait, meet with probability 7/16 under an independent uniform arrival model.https://player.bilibili.com/player.html?bvid=BV1neAieTEkK&p=1&autoplay=0&t=02025-02-19T01:49:56+00:00257https://math-video.com/zh/videos/geometric-probability-bv1neaietekk2026-10-12T03:03:23.092Zhttps://i0.hdslb.com/bfs/archive/d987a8957005a01a025063d86a23c83ae9a114b8.jpg几何概率：长度比、面积比与相见问题从两个完整例子理解几何概率：在 [0,3] 上均匀取点，坐标小于 1 的概率是 1/3；两人在下午 4–5 点到达、先到者等待 15 分钟，在独立均匀到达模型下，相见概率是 7/16。视频依次介绍度量比公式、长度与面积及体积的选择、时间归一化、单位正方形中的事件带，并用补集三角形计算面积。这里明确补充可测、正且有限的空间度量，以及联合均匀分布的适用前提。https://player.bilibili.com/player.html?bvid=BV1neAieTEkK&p=1&autoplay=0&t=02025-02-19T01:49:56+00:00257https://math-video.com/videos/math-topic-g75thbxyzww2026-10-12T03:04:46.999Zhttps://i.ytimg.com/vi/G75thBxyZWw/maxresdefault.jpgNormal Curves: Compare Means, Widths and PeaksA score-distribution question separates the location and shape of normal curves. School A has mean60 and standard deviation10; School B has mean65 and standard deviation5. The thick line represents School A, and the thin line School B.https://www.youtube-nocookie.com/embed/G75thBxyZWw?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02015-03-03T01:48:10+00:00121https://math-video.com/zh/videos/math-topic-g75thbxyzww2026-10-12T03:04:46.999Zhttps://i.ytimg.com/vi/G75thBxyZWw/maxresdefault.jpg正态曲线：比较均值、宽窄与峰值用一道成績分布圖選擇題，區分正態曲線的位置與形狀。甲校均值60、標準差10，乙校均值65、標準差5；粗線代表甲、細線代表乙。因此甲的曲線中心偏左、較寬且峰值較低，乙的中心偏右、較窄且峰值較高，完整原片答案為A。題目說實際成績很接近正態分布，圖形判斷使用正態模型。面積均為1的直觀要放在同一歸一正態族和相同坐標尺度下理解，不能推成任意概率密度只要面積1便有固定寬高關係。https://www.youtube-nocookie.com/embed/G75thBxyZWw?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02015-03-03T01:48:10+00:00121https://math-video.com/videos/hypothesis-testing-p-values-khan2026-10-12T03:03:56.581Zhttps://i.ytimg.com/vi/-FtlH4svqx4/hqdefault.jpgHypothesis testing and p-values | Inferential statistics | Probability and Statistics | Khan AcademyThis 180-second segment introduces a one-sample hypothesis-testing example about whether a drug affects rat response time. The typed problem gives n=100 injected rats, an untreated mean of 1.2 seconds, an observed sample mean of 1.05 seconds, and a sample standard deviation of 0.5 seconds.https://www.youtube-nocookie.com/embed/-FtlH4svqx4?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02010-11-02T18:33:04+00:00686https://math-video.com/zh/videos/hypothesis-testing-p-values-khan2026-10-12T03:03:56.581Zhttps://i.ytimg.com/vi/-FtlH4svqx4/hqdefault.jpg假设检验与 p 值这个 180 秒的片段介绍了一个关于药物是否影响大鼠反应时间的单样本假设检验示例。打印的问题给出 n=100 只被注射的大鼠，未治疗平均值为 1.2 秒，观察到的样本均值为 1.05 秒，样本标准差为 0.5 秒。讲师首先将无影响主张形式化为 H0: \mu = 1.2 s，然后将竞争性影响主张形式化为双侧备择假设 H1: \mu \neq 1.2 s。片段的其余部分解释了基本决策逻辑：假设 H0 为真，询问在该假设下观察到的样本结果的可能性有多大，并将非常小的概率视为拒绝 H0 的理由。该片段在任何数值检验统计量或明确的 p 值计算进行之前停止。https://www.youtube-nocookie.com/embed/-FtlH4svqx4?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02010-11-02T18:33:04+00:00686https://math-video.com/videos/least-squares-approximation-khan2026-10-12T03:05:06.797Zhttps://i.ytimg.com/vi/MC7l96tW8V8/hqdefault.jpgLeast squares approximation | Linear Algebra | Khan AcademyThis 180-second whiteboard segment sets up the motivation for least squares by studying when Ax=b has no solution. It first fixes dimensions for an n×k matrix, then rewrites the equation as a linear combination of the columns of A.https://www.youtube-nocookie.com/embed/MC7l96tW8V8?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02009-11-10T22:15:13+00:00932https://math-video.com/zh/videos/least-squares-approximation-khan2026-10-12T03:05:06.797Zhttps://i.ytimg.com/vi/MC7l96tW8V8/hqdefault.jpg最小二乘近似这段 180 秒的白板片段通过研究 Ax=b 何时无解来建立最小二乘法的动机。它首先确定 n×k 矩阵的维度，然后将方程重写为 A 的各列的线性组合。由此，它将不相容性识别为陈述 b∉C(A)，并用列空间的平面和其外的向量在几何上加以说明。最后几秒通过引入候选 x^ 转向近似，但在提供的片段内未完成确切的标准。 这段 180 秒的白板片段为不相容线性系统 Ax=b 的最小二乘近似提供了动机。它从 b 不在列空间 C(A) 中的几何事实开始，因此不存在精确解。https://www.youtube-nocookie.com/embed/MC7l96tW8V8?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02009-11-10T22:15:13+00:00932https://math-video.com/videos/confidence-intervals-statquest2026-10-12T03:05:28.483Zhttps://i.ytimg.com/vi/TqOeMYtOc1w/hqdefault.jpgConfidence Intervals, Clearly Explained!!!This 180-second clip is an introductory StatQuest segment that previews confidence intervals and then spends most of its time refreshing bootstrapping.https://www.youtube-nocookie.com/embed/TqOeMYtOc1w?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02015-07-09T16:43:04+00:00401https://math-video.com/zh/videos/confidence-intervals-statquest2026-10-12T03:05:28.483Zhttps://i.ytimg.com/vi/TqOeMYtOc1w/hqdefault.jpg置信区间，清晰讲解这段 180 秒的片段是一个介绍性的 StatQuest 部分，预览了置信区间，然后大部分时间用于复习自助法。使用数轴上 12 个雌性小鼠重量的样本，视频区分了样本均值和总体均值，演示了一次允许重复的自助重采样，然后显示了许多重复的自助均值积累成分布。最后几秒将幻灯片重命名为“95% Confidence Intervals”，表明接下来将使用自助分布，但片段在区间本身构造之前结束。 这个 180 秒的片段通过数轴上的自助法图像教授置信区间。它首先将 95% 置信区间定义为覆盖 95% 自助法均值的区间，然后推广到更宽的 99% 区间。https://www.youtube-nocookie.com/embed/TqOeMYtOc1w?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02015-07-09T16:43:04+00:00401https://math-video.com/videos/introduction-to-functions-khan2026-10-12T03:05:33.700Zhttps://i.ytimg.com/vi/VhokQhjl5t0/hqdefault.jpgIntroduction to functionsThis 180-second whiteboard introduction presents functions first as black boxes that take an input x and produce an output f(x). It then works a concrete single-rule example, f(x)=x^2+1, computing f(2)=5 by substitution.https://www.youtube-nocookie.com/embed/VhokQhjl5t0?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02007-03-18T20:02:47+00:00573https://math-video.com/zh/videos/introduction-to-functions-khan2026-10-12T03:05:33.700Zhttps://i.ytimg.com/vi/VhokQhjl5t0/hqdefault.jpg函数入门这段 180 秒的白板入门先把函数描述为接收输入 x 并产生输出 f(x) 的黑箱。随后用一个具体的单规则示例 f(x)=x^2+1，通过代入算出 f(2)=5。最后一部分把概念推广到对偶数和奇数使用不同定义的分段函数，说明函数不必只由一个代数公式表示。分段设置已经完成，但该示例对 f(2) 的最终求值在片段结束前尚未完成。 这段 180 秒的介绍性代数片段首先使用分段规则 f(x)=x^2+1（当 x 为偶数）和 f(x)=x^2-1（当 x 为奇数）将函数解释为输入-输出机器，然后通过选择正确的分支评估 f(2)=5 和 f(3)=8。https://www.youtube-nocookie.com/embed/VhokQhjl5t0?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02007-03-18T20:02:47+00:00573https://math-video.com/videos/law-of-large-numbers-khan2026-10-12T03:05:35.458Zhttps://i.ytimg.com/vi/VpuN8vCQ--M/hqdefault.jpgLaw of large numbers | Probability and Statistics | Khan AcademyThis 180-second introductory whiteboard segment presents the law of large numbers first as a general statement and then as a fair-coin example.https://www.youtube-nocookie.com/embed/VpuN8vCQ--M?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02009-03-05T17:09:08+00:00539https://math-video.com/zh/videos/law-of-large-numbers-khan2026-10-12T03:05:35.458Zhttps://i.ytimg.com/vi/VpuN8vCQ--M/hqdefault.jpg大数定律这段 180 秒的白板入门先把大数定律表述为一般命题，再给出公平硬币的例子。在提醒这一定律直观但经常被误用后，讲解者定义随机变量 X、其期望值 E(x) 或总体均值 μ，以及样本均值 overlineXn=(x1+x2+cdots+xn)/n。黑板写出的核心非正式命题是 overlineXn→E(x)，等价地写成 overlineXn→μ，当 n→∞，并明确提醒这里是非正式地讨论收敛。随后黑板转向具体例子：X 表示公平硬币投掷 100 次后的正面次数，期望值算为 E(X)=100·0.5=50。片段在讲解者开始把这个例子联系到重复样本的平均过程时结束。https://www.youtube-nocookie.com/embed/VpuN8vCQ--M?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02009-03-05T17:09:08+00:00539https://math-video.com/videos/central-limit-theorem-statquest2026-10-12T03:05:46.229Zhttps://i.ytimg.com/vi/YAlJCEDH2uY/hqdefault.jpgThe Central Limit Theorem, Clearly Explained!!!This 180-second excerpt is an introductory visual explanation of the Central Limit Theorem. After naming prerequisites (the normal distribution and sampling from a statistical distribution), the presenter uses a uniform distribution on [0,1] as the population example.https://www.youtube-nocookie.com/embed/YAlJCEDH2uY?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02018-09-03T17:04:06+00:00455https://math-video.com/zh/videos/central-limit-theorem-statquest2026-10-12T03:05:46.229Zhttps://i.ytimg.com/vi/YAlJCEDH2uY/hqdefault.jpg中心极限定理，清晰讲解这段 180 秒的摘录是对中心极限定理的介绍性视觉解释。在命名先决条件（正态分布和从统计分布中抽样）后，演示者使用 [0,1] 上的均匀分布作为总体示例。他抽取了 20 个随机观测值，计算了它们的均值，并将该均值记录在直方图中。重复此过程直到收集了 100 个均值，直方图变得呈钟形。然后叠加一条蓝色正态曲线以强调主要观点：样本均值是正态分布的，即使原始数据来自均匀分布。片段在旁白完成最后一句话之前结束。 这段 180 秒的摘录是对中心极限定理的介绍性视觉解释。它首先回顾来自均匀总体的样本均值是正态分布的，然后通过大小为 20 的重复样本处理指数示例。https://www.youtube-nocookie.com/embed/YAlJCEDH2uY?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02018-09-03T17:04:06+00:00455https://math-video.com/videos/integration-by-parts-intro-khan2026-10-12T03:06:01.913Zhttps://i.ytimg.com/vi/dh__n9FVKA0/hqdefault.jpgIntegration by parts intro | AP Calculus BC | Khan AcademyThis 180-second whiteboard clip introduces integration by parts as a consequence of the product rule.https://www.youtube-nocookie.com/embed/dh__n9FVKA0?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02013-01-28T16:54:05+00:00232https://math-video.com/zh/videos/integration-by-parts-intro-khan2026-10-12T03:06:01.913Zhttps://i.ytimg.com/vi/dh__n9FVKA0/hqdefault.jpg分部积分入门这段 180 秒的白板片段介绍了分部积分作为乘积法则的结果。在写下标题“Product Rule -> Integration by Parts”后，讲师从 f(x)g(x) 开始，展开 d/dx[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)，然后对两边求原函数得到 f(x)g(x) = \int f'(x)g(x)\,dx + \int f(x)g'(x)\,dx。黄色框标记 \int f(x)g'(x)\,dx 为目标项，减去另一个积分得到重排的恒等式 \int f(x)g'(x)\,dx = f(x)g(x) - \int f'(x)g(x)\,dx。https://www.youtube-nocookie.com/embed/dh__n9FVKA0?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02013-01-28T16:54:05+00:00232https://math-video.com/videos/gaussian-elimination-row-echelon2026-10-12T03:06:03.690Zhttps://i.ytimg.com/vi/eDb6iugi6Uk/hqdefault.jpgGaussian Elimination & Row Echelon FormThis 180-second whiteboard segment introduces Gaussian elimination for a three-variable linear system. The presenter writes the system x+y-z=-2, 2x-y+z=5, and -x+2y+2z=1, converts it into an augmented matrix, marks target entries for row echelon form, and performs the row operation R1+R3\to R3.https://www.youtube-nocookie.com/embed/eDb6iugi6Uk?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02018-02-18T03:00:03+00:001120https://math-video.com/zh/videos/gaussian-elimination-row-echelon2026-10-12T03:06:03.690Zhttps://i.ytimg.com/vi/eDb6iugi6Uk/hqdefault.jpg高斯消元与行阶梯形这个 180 秒的白板片段介绍了针对三元线性方程组的高斯消元法。演示者写出方程组 x+y-z=-2, 2x-y+z=5, 和 -x+2y+2z=1，将其转换为增广矩阵，标记行阶梯形的目标元素，并执行行变换 R1+R3\to R3。得到的中间矩阵是 [1 1 -1 | -2; 2 -1 1 | 5; 0 3 1 | -1]。片段结束时，宣布但未完成消除第 2 行中 2 的下一步。 这段 180 秒的白板片段继续处理增广矩阵 [[1,1,-1|-2],[2,-1,1|5],[0,3,1|-1]] 的高斯消元示例。https://www.youtube-nocookie.com/embed/eDb6iugi6Uk?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02018-02-18T03:00:03+00:001120https://math-video.com/videos/implicit-differentiation-khan2026-10-12T03:06:30.805Zhttps://i.ytimg.com/vi/mSVrqKZDRF4/hqdefault.jpgImplicit differentiation | Advanced derivatives | AP Calculus AB | Khan AcademyThis 180-second whiteboard excerpt introduces implicit differentiation through the unit circle equation x^2+y^2=1. The speaker first shows the circle graph and asks how to find the slope of the tangent line at any point.https://www.youtube-nocookie.com/embed/mSVrqKZDRF4?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02013-01-30T16:59:49+00:00481https://math-video.com/zh/videos/implicit-differentiation-khan2026-10-12T03:06:30.805Zhttps://i.ytimg.com/vi/mSVrqKZDRF4/hqdefault.jpg隐函数求导这段 180 秒的白板摘录通过单位圆方程 x^2+y^2=1 介绍了隐函数求导。演讲者首先展示圆图并询问如何找到任意一点切线的斜率。然后他解释说整个圆不是 x 的显式函数，指出了将其拆分为 y=√(1-x^2) 和 y=-√(1-x^2) 的诱人变通方法，转而将隐函数求导呈现为链式法则的直接应用。解题设置达到了 (d)/(dx)[x^2+y^2]=(d)/(dx)[1]，然后是 (d)/(dx)[x^2]+(d)/(dx)[y^2]=0，但片段在评估各个导数之前结束。 这段 180 秒的白板片段讲解了单位圆方程 x^2+y^2=1 的隐函数求导。https://www.youtube-nocookie.com/embed/mSVrqKZDRF4?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02013-01-30T16:59:49+00:00481https://math-video.com/videos/population-parameter-estimation-statquest2026-10-12T03:06:57.315Zhttps://i.ytimg.com/vi/vikkiwjQqfU/hqdefault.jpgPopulation and Estimated Parameters, Clearly Explained!!!This 180-second segment opens with a short ukulele joke, then introduces "Statistics Fundamentals: Population Parameters." After naming histograms, statistical distributions, and the normal distribution as prerequisites, it builds intuition with a counting example: mRNA transcripts of Gene X in…https://www.youtube-nocookie.com/embed/vikkiwjQqfU?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02019-07-01T17:30:00+00:00871https://math-video.com/zh/videos/population-parameter-estimation-statquest2026-10-12T03:06:57.315Zhttps://i.ytimg.com/vi/vikkiwjQqfU/hqdefault.jpg总体参数与参数估计这段 180 秒的片段以一个简短的尤克里里玩笑开场，随后介绍“Statistics Fundamentals: Population Parameters”。在列出直方图、统计分布和正态分布这些前置知识后，视频用一个计数示例建立直觉：统计肝细胞中 Gene X 的 mRNA 转录本数量，之后又推广到苹果或 T 恤。画面从点图读取五个示例值（3、13、19、24、29），再把范围扩展到设想的、含 240 billion 个肝细胞的完整总体。钟形直方图概括总体，显示大多数值落在 20 到 30 之间，小于 10 和大于 30 的较少。https://www.youtube-nocookie.com/embed/vikkiwjQqfU?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02019-07-01T17:30:00+00:00871https://math-video.com/videos/bilibili-bv1hp596ge4n2026-10-10T05:17:29.100Zhttps://math-video.com/zh/videos/bilibili-bv1hp596ge4n2026-10-10T05:17:29.100Zhttps://math-video.com/videos/bilibili-bv1vzzayieh22026-10-10T05:17:38.559Zhttps://math-video.com/zh/videos/bilibili-bv1vzzayieh22026-10-10T05:17:38.559Zhttps://math-video.com/videos/euclidean-algorithm-gcd-worked-example-michael-penn2026-10-12T03:05:53.347Zhttps://i.ytimg.com/vi/_EGMgO1F4RI/hqdefault.jpgNumber Theory: The Euclidean Algorithm Example 1Michael Penn applies the Euclidean algorithm to the positive integers 5295 and 4321. Eight divisions lead to a final nonzero remainder of 1, so their greatest common divisor is 1 and the pair is relatively prime.https://www.youtube-nocookie.com/embed/_EGMgO1F4RI?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02019-08-13T00:00:00+00:00163https://math-video.com/zh/videos/euclidean-algorithm-gcd-worked-example-michael-penn2026-10-12T03:05:53.347Zhttps://i.ytimg.com/vi/_EGMgO1F4RI/hqdefault.jpg欧几里得算法：最大公因数例题｜Michael PennMichael Penn 用欧几里得算法计算正整数 5295 与 4321 的最大公约数。八次带余除法后，最后一个非零余数为 1，因此两数的最大公约数是 1，两数互质。本课回顾停止规则并完整演算例题，不包含一般正确性证明。https://www.youtube-nocookie.com/embed/_EGMgO1F4RI?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02019-08-13T00:00:00+00:00163https://math-video.com/videos/graphs-incidence-matrices-and-kirchhoff-mit2026-10-12T03:05:26.462Zhttps://i.ytimg.com/vi/RwBCrVB98s8/maxresdefault.jpgGraphsGraphs here consist of nodes and edges, rather than plots of functions. A small example has 4 nodes and 5 edges. Its oriented incidence matrix records labeled connectivity; later, multiplying by node potentials gives edge potential differences.https://www.youtube-nocookie.com/embed/RwBCrVB98s8?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02016-05-06T00:00:00Z926https://math-video.com/zh/videos/graphs-incidence-matrices-and-kirchhoff-mit2026-10-12T03:05:26.462Zhttps://i.ytimg.com/vi/RwBCrVB98s8/maxresdefault.jpg图与关联矩阵：网络中的基尔霍夫定律这里的图由节点和边组成，不是函数曲线。示例包含4个节点与5条边。选定边方向后，关联矩阵记录带标签的连接关系；随后将它乘以节点电位向量，得到边上的电位差。 网页链接、电话连接和神经元连接说明图模型的用途。讲解随后回到示例：关联矩阵有5个边对应的横行和4个节点对应的竖列，并按所选边方向填写元素。这些是建模示例，不是完整的物理或生物理论。 一条边从节点i指向节点j时，对应横行在起点位置为−1、终点位置为+1，其余位置为零。讲解逐行填完示例图的五个横行。无向图也可以人为选定参考方向；关联矩阵不包含几何长度或材料参数。 完成的关联矩阵有5个横行与4个竖列。给每个节点指定电位后，就得到可与矩阵相乘的向量。https://www.youtube-nocookie.com/embed/RwBCrVB98s8?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02016-05-06T00:00:00Z926https://math-video.com/videos/euclidean-algorithm-example-socratica2026-10-12T03:06:09.298Zhttps://i.ytimg.com/vi/fwuj4yzoX1o/maxresdefault.jpgEuclidean Algorithm - An example ← Number TheoryThis educational video introduces the Euclidean algorithm as an efficient method for finding the greatest common divisor (GCD) of two integers without requiring prime factorization. It begins with a brief historical context, noting its origins in Euclid's Elements over 2,300 years ago.https://www.youtube-nocookie.com/embed/fwuj4yzoX1o?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02012-01-08T00:00:00Z123https://math-video.com/zh/videos/euclidean-algorithm-example-socratica2026-10-12T03:06:09.298Zhttps://i.ytimg.com/vi/fwuj4yzoX1o/maxresdefault.jpg欧几里得算法例题｜Socratica这段教学视频介绍欧几里得算法：它无需质因数分解，就能高效求两个整数的最大公约数。视频先回顾其历史，说明该算法在欧几里得《几何原本》中出现至今已有 2,300 多年。核心步骤是反复做长除法，把上一步的除数与余数用于下一步，直到余数为零；最后一个非零余数就是最大公约数。随后视频用动画完整演示求 1785 与 546 的最大公约数，并由每一步除法得到最终答案 21。https://www.youtube-nocookie.com/embed/fwuj4yzoX1o?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02012-01-08T00:00:00Z123https://math-video.com/videos/euclidean-algorithm-gcd-numa-zh2026-10-12T03:05:19.497Zhttps://i.ytimg.com/vi/PJt9c2R6O8U/maxresdefault.jpg輾轉相除法完整解析｜原理＋實作步驟＋範例教學｜最大公因數 GCD 快速求法This segment introduces the Euclidean algorithm for finding the greatest common divisor (GCD) of two positive integers.https://www.youtube-nocookie.com/embed/PJt9c2R6O8U?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02025-09-23T00:00:00Z164https://math-video.com/zh/videos/euclidean-algorithm-gcd-numa-zh2026-10-12T03:05:19.497Zhttps://i.ytimg.com/vi/PJt9c2R6O8U/maxresdefault.jpg辗转相除法完整解析：原理、步骤与例题本片段介紹了輾轉相除法（歐幾里得算法）用於求解兩個正整數的最大公因數。視頻首先通過計算 5911 和 4369 的具體例子展示了算法的操作流程：重複進行「大數除以小數取餘數」，直到餘數為零，最後的除數即為答案。隨後，視頻闡述了算法背後的數學原理——除法原理 (a=bq+r) 以及由此導出的最大公因數遞減性質 ((a,b)=(b,r))。雖然視頻中的證明過程略顯簡略（僅證明了單向整除關係），但清晰地解釋了為什麼該算法有效，並將理論應用回最初的數例進行驗證。https://www.youtube-nocookie.com/embed/PJt9c2R6O8U?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02025-09-23T00:00:00Z164https://math-video.com/videos/counting-with-bijections-zhuli2026-10-12T03:06:53.863Zhttps://i.ytimg.com/vi/uxjAHkn4VFI/maxresdefault.jpgThe Most Important Counting Concept You’ve (Probably) Never Heard OfThis introductory video segment defines the mathematical concept of a bijection as a one-to-one and onto pairing between two sets. Using an auditorium seating example, it demonstrates how a perfect matching allows us to determine the size of one set from another without direct counting.https://www.youtube-nocookie.com/embed/uxjAHkn4VFI?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02023-03-29T00:00:00Z611https://math-video.com/zh/videos/counting-with-bijections-zhuli2026-10-12T03:06:53.863Zhttps://i.ytimg.com/vi/uxjAHkn4VFI/maxresdefault.jpg计数中的双射：把困难问题变简单这段介绍性视频片段将双射的数学概念定义为两个集合之间的一对一且满射的配对。通过使用礼堂座位的例子，它展示了完美的匹配如何允许我们从另一个集合确定一个集合的大小，而无需直接计数。旁白随后确立标准计数从根本上是双射的应用，将对象映射到计数数。最后，该片段解释验证计数答案需要检查双射性质，直观对比多算（违反一对一）和少算（违反满射）作为枚举中的两个主要错误。 这段 180 秒的内容说明，正确计数必须同时排除重复计数与遗漏计数。圆圈到数字的映射表明，只修复遗漏仍可能保留重复；有效计数应形成双射，也就是与计数数的一一对应。https://www.youtube-nocookie.com/embed/uxjAHkn4VFI?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02023-03-29T00:00:00Z611https://math-video.com/videos/why-euclidean-algorithm-works-subtraction2026-10-12T03:06:25.675Zhttps://i.ytimg.com/vi/kVDlhQJkxZQ/maxresdefault.jpgWhy Does the Euclidean Algorithm Work? : Lessons in Applied MathematicsThis 94-second introductory clip explains the subtraction form of the Euclidean algorithm for finding the greatest common factor.https://www.youtube-nocookie.com/embed/kVDlhQJkxZQ?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02013-10-03T00:00:00Z94https://math-video.com/zh/videos/why-euclidean-algorithm-works-subtraction2026-10-12T03:06:25.675Zhttps://i.ytimg.com/vi/kVDlhQJkxZQ/maxresdefault.jpg欧几里得算法为什么成立：减法形式直观解释这段 94 秒的介绍性片段解释了用于寻找最大公因数的欧几里得算法的减法形式。在简短的标题和自我介绍之后，演示者在透明板上处理示例 12 和 8：他用较大的数减去较小的数得到 4，对 8 和 4 重复该过程，再次得到 4，并将 4 识别为最大公因数。然后他给出了正确性的非正式理由：最大公因数能整除，除法被视为重复减法，因此同一个约数也必须能整除由一步减法产生的差值。片段以品牌标识和字幕结束。数学内容清晰但基础，基于示例而非完全形式化。https://www.youtube-nocookie.com/embed/kVDlhQJkxZQ?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02013-10-03T00:00:00Z94https://math-video.com/videos/greatest-common-factor-math-antics2026-10-12T03:04:37.663Zhttps://i.ytimg.com/vi/CUEOL3_Wm3Y/maxresdefault.jpgGreatest Common Factor — Math AnticsA positive integer can have several factor pairs. For 12, the pairs 1\times12, 2\times6 and 3\times4 give the positive factors \1,2,3,4,6,12\. The lesson uses these lists to explain common factors and, later, the greatest common factor. Editorial scope: only positive-integer factors are considered.https://www.youtube-nocookie.com/embed/CUEOL3_Wm3Y?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02025-02-03T22:08:22+00:00265https://math-video.com/zh/videos/greatest-common-factor-math-antics2026-10-12T03:04:37.663Zhttps://i.ytimg.com/vi/CUEOL3_Wm3Y/maxresdefault.jpg最大公因数：因数列表与分数约分一个正整数可以有多对因数。12的因数对1\times12、2\times6、3\times4给出正因数集合\1,2,3,4,6,12\。讲解以这些列表解释公因数，随后找出最大公因数。编辑补充：这里只讨论正整数的正因数。 比较12和42的正因数列表，可得到公因数集合\1,2,3,6\，最大公因数为6。列表法从共同因数中选择最大的数。共同兴趣的类比用于解释“共同”一词，本身不是整除性的证明。 最大公因数把因数列表与分数约分联系起来。对正整数12与42，先除以2得到6/21，两部分仍有公因数3；直接除以最大公因数6可得到2/7。编辑补充：正整数分子、分母具有大于1的公因数时，分数才可继续约分。https://www.youtube-nocookie.com/embed/CUEOL3_Wm3Y?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02025-02-03T22:08:22+00:00265https://math-video.com/videos/euclidean-algorithm-gcd-worked-examples-learn-math-tutorials2026-10-12T03:04:59.509Zhttps://i.ytimg.com/vi/JUzYl1TYMcU/maxresdefault.jpgEuclidean algorithm: finding the greatest common divisorLearn the Euclidean algorithm through two complete whiteboard examples: gcd(10,45)=5 and gcd(1701,3768)=3. Repeated integer division carries each divisor and remainder into the next line; the process stops at a zero remainder.https://www.youtube-nocookie.com/embed/JUzYl1TYMcU?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02012-12-17T05:16:40+00:00249https://math-video.com/zh/videos/euclidean-algorithm-gcd-worked-examples-learn-math-tutorials2026-10-12T03:04:59.509Zhttps://i.ytimg.com/vi/JUzYl1TYMcU/maxresdefault.jpg欧几里得算法：两个最大公因数例题从两道完整白板例题学会欧几里得算法：gcd(10,45)=5，gcd(1701,3768)=3。每次做整数除法，再用原除数和余数进行下一步，直到余数为零。原讲解混用了“分母”一词，这里应使用标准术语“最大公约数”。本站补充适用范围：输入为正整数，答案取最后一次整除的除数；在这两道例题中，它也是前一个非零余数。视频演示计算过程，下文的一般最大公约数不变性论证属于本站补充。https://www.youtube-nocookie.com/embed/JUzYl1TYMcU?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02012-12-17T05:16:40+00:00249https://math-video.com/videos/euclidean-algorithm-proof-and-cpp-implementation2026-10-10T09:09:30.105Zhttps://i1.hdslb.com/bfs/archive/a51d3a32fcf44d6c3c5ea6b071f0232e97d89635.jpgEuclidean Algorithm: Explanation, Proof and C++ ImplementationThis Mandarin lesson starts from the greatest common divisor and divisibility, then works with positive integers a≥b>0. For a=q1b+r1, the old gcd remains a common divisor of the new pair, and every common divisor of the new pair is bounded by it; this proves gcd(a,b)=gcd(b,r1).https://player.bilibili.com/player.html?bvid=BV1w5411L7aN&p=1&autoplay=0&t=02020-10-25T11:04:36+00:00488.341https://math-video.com/zh/videos/euclidean-algorithm-proof-and-cpp-implementation2026-10-10T09:09:30.105Zhttps://i1.hdslb.com/bfs/archive/a51d3a32fcf44d6c3c5ea6b071f0232e97d89635.jpg详细描述欧几里得算法（附证明）本课从最大公因数和整除关系出发，随后在正整数 a≥b>0 的范围内展开。对于 a=q1b+r1，原最大公因数仍是新数对的公因数，且新数对的任意公因数都不超过它，由此证明 gcd(a,b)=gcd(b,r1)。连续余数链进一步对应到迭代式 C++ 实现。完整例题用1160718174与316258250作输入，经过10次带余除法得到1078。笔记明确区分原片证明与本站补充的输入范围、立即整除边界及正整数下降的有限性说明。https://player.bilibili.com/player.html?bvid=BV1w5411L7aN&p=1&autoplay=0&t=02020-10-25T11:04:36+00:00488.341https://math-video.com/videos/greatest-common-divisor-recursion-subtraction-euclidean-algorithm2026-10-10T09:17:09.363Zhttps://i1.hdslb.com/bfs/archive/cda0d8ebc10dd28beccbf9bbe3e535b7cc7a3eef.jpgGreatest common divisor recurrences: subtraction and Euclidean algorithmsThe segment first uses a document to present the recurrence formula for the greatest common divisor of two numbers gcd(a,b)=gcd(b,a-k·b), explaining that the Subtraction Method and the Euclidean Algorithm are special cases for k=1 and k=lfloor a/brfloor respectively.https://player.bilibili.com/player.html?bvid=BV1eT4118729&p=1&autoplay=0&t=02023-05-08T22:34:50.000Z254https://math-video.com/zh/videos/greatest-common-divisor-recursion-subtraction-euclidean-algorithm2026-10-10T09:17:09.363Zhttps://i1.hdslb.com/bfs/archive/cda0d8ebc10dd28beccbf9bbe3e535b7cc7a3eef.jpg初级数论之：两个数的最大公约数的递推公式【更相减损术】【辗转相除法】片段先用文档给出两个数最大公约数的递推公式 gcd(a,b)=gcd(b,a-k·b)，并说明更相减损术和辗转相除法分别是 k=1 与 k=lfloor a/brfloor 的特例。随后切换到 C++ 代码，用 a=24、b=504 调用 gcd(a,b)，终端输出 24，验证 gcd(24,504)=24。讲解者还提醒演示环境因 C++ 版本较新可直接使用 gcd，考试环境可能需要自行实现。 片段先用 C++ 代码枚举 k=-1024 到 1024，检查 gcd(24,504) 是否总等于 gcd(504,24-k·504)；终端只输出星号，说明测试范围内没有反例。https://player.bilibili.com/player.html?bvid=BV1eT4118729&p=1&autoplay=0&t=02023-05-08T22:34:50.000Z254https://math-video.com/videos/euclidean-algorithm-gcd-softwareengenius2026-10-12T03:04:52.512Zhttps://i.ytimg.com/vi_webp/I-DClW_0bhU/maxresdefault.webpCalculate Greatest Common Divisors via the Euclidean AlgorithmThe lesson develops the Euclidean algorithm from positive common divisors, divisibility and congruence, then presents a recursive implementation and its call-count bound. Integer GCD inputs are not both zero.https://www.youtube-nocookie.com/embed/I-DClW_0bhU?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02020-06-26T16:15:01+00:00301https://math-video.com/zh/videos/euclidean-algorithm-gcd-softwareengenius2026-10-12T03:04:52.512Zhttps://i.ytimg.com/vi_webp/I-DClW_0bhU/maxresdefault.webp用欧几里得算法计算最大公因数本课从正公约数、整除与同余出发讲解欧几里得算法，随后展示递归实现与调用次数的界。整数最大公约数的输入不全为零。对有序正整数 a≥b>0，写成 a=bq+r 且 0≤r<b；恒等式 gcd(a,b)=gcd(b,r) 保持答案。除数达到零时，返回前一个正除数。编辑笔记补全原片未单独展示的反向公约数论证，并修正复杂度幻灯片的零商旁注笔误。该 O(\log2(N)) 界取 N=a+b，计算的是单位成本算术模型下的取余调用次数，而非所有位运算的实际耗时。https://www.youtube-nocookie.com/embed/I-DClW_0bhU?enablejsapi=1&origin=https%3A%2F%2Fmath-video.com&autoplay=0&hl=en&start=02020-06-26T16:15:01+00:00301https://math-video.com/videos/gradient-tipkfdc295m2026-10-12T03:06:48.196Zhttps://math-video.com/zh/videos/gradient-tipkfdc295m2026-10-12T03:06:48.196Zhttps://math-video.com/videos/arc-length-ohissmqv4-82026-10-12T03:05:13.944Zhttps://math-video.com/zh/videos/arc-length-ohissmqv4-82026-10-12T03:05:13.944Zhttps://math-video.com/videos/mean-value-theorem-6hri9k-2r8o2026-10-12T03:04:19.817Zhttps://math-video.com/zh/videos/mean-value-theorem-6hri9k-2r8o2026-10-12T03:04:19.817Zhttps://math-video.com/videos/partial-derivative-axqhweuetqu2026-10-12T03:04:31.618Zhttps://math-video.com/zh/videos/partial-derivative-axqhweuetqu2026-10-12T03:04:31.618Zhttps://math-video.com/videos/lagrange-multiplier-8mjcnxgmwfo2026-10-12T03:04:25.375Zhttps://math-video.com/zh/videos/lagrange-multiplier-8mjcnxgmwfo2026-10-12T03:04:25.375Zhttps://math-video.com/videos/line-integral-aff8fxxt5os2026-10-12T03:04:29.317Zhttps://math-video.com/zh/videos/line-integral-aff8fxxt5os2026-10-12T03:04:29.317Zhttps://math-video.com/videos/power-series-oxvbt83x8oc2026-10-12T03:05:15.773Zhttps://math-video.com/zh/videos/power-series-oxvbt83x8oc2026-10-12T03:05:15.773Zhttps://math-video.com/videos/taylor-series-9yaacea08ym2026-10-12T03:04:27.343Zhttps://math-video.com/zh/videos/taylor-series-9yaacea08ym2026-10-12T03:04:27.343Zhttps://math-video.com/videos/epsilon-n-definition-rfsqhvflxvq2026-10-12T03:05:24.697Zhttps://math-video.com/zh/videos/epsilon-n-definition-rfsqhvflxvq2026-10-12T03:05:24.697Zhttps://math-video.com/videos/epsilon-delta-definition-ejyeii0i5c2026-10-12T03:04:01.441Zhttps://math-video.com/zh/videos/epsilon-delta-definition-ejyeii0i5c2026-10-12T03:04:01.441Zhttps://math-video.com/videos/sequence-limit-wzw9ll80zbc2026-10-12T03:07:02.635Zhttps://math-video.com/zh/videos/sequence-limit-wzw9ll80zbc2026-10-12T03:07:02.635Zhttps://math-video.com/videos/comparison-test-lbxyq0tjxym2026-10-12T03:05:03.312Zhttps://math-video.com/zh/videos/comparison-test-lbxyq0tjxym2026-10-12T03:05:03.312Zhttps://math-video.com/videos/partial-sums-bh1urcgwz1k2026-10-12T03:06:00.212Zhttps://math-video.com/zh/videos/partial-sums-bh1urcgwz1k2026-10-12T03:06:00.212Zhttps://math-video.com/videos/error-estimate-gs9o255gdey2026-10-12T03:06:14.485Zhttps://math-video.com/zh/videos/error-estimate-gs9o255gdey2026-10-12T03:06:14.485Zhttps://math-video.com/videos/zeno-paradox-efqvnj-sgcc2026-10-12T03:04:45.296Zhttps://math-video.com/zh/videos/zeno-paradox-efqvnj-sgcc2026-10-12T03:04:45.296Zhttps://math-video.com/videos/limit-nonexistence-hxrnjxth8sa2026-10-12T03:06:22.107Zhttps://math-video.com/zh/videos/limit-nonexistence-hxrnjxth8sa2026-10-12T03:06:22.107Zhttps://math-video.com/videos/secant-line-680o1gl-lrg2026-10-12T03:04:18.078Zhttps://math-video.com/zh/videos/secant-line-680o1gl-lrg2026-10-12T03:04:18.078Zhttps://math-video.com/videos/local-linearization-u7dhn-hbhzq2026-10-12T03:06:52.097Zhttps://math-video.com/zh/videos/local-linearization-u7dhn-hbhzq2026-10-12T03:06:52.097Zhttps://math-video.com/videos/tangent-plane-wqtr4udz8ay2026-10-12T03:07:00.870Zhttps://math-video.com/zh/videos/tangent-plane-wqtr4udz8ay2026-10-12T03:07:00.870Zhttps://math-video.com/videos/normal-vector-2szkzhyaqj82026-10-12T03:04:12.616Zhttps://math-video.com/zh/videos/normal-vector-2szkzhyaqj82026-10-12T03:04:12.616Zhttps://math-video.com/videos/directional-derivative-n-zrclhenv02026-10-12T03:05:10.481Zhttps://math-video.com/zh/videos/directional-derivative-n-zrclhenv02026-10-12T03:05:10.481Zhttps://math-video.com/videos/total-differential-2bf6h-xu0ao2026-10-12T03:04:10.579Zhttps://math-video.com/zh/videos/total-differential-2bf6h-xu0ao2026-10-12T03:04:10.579Zhttps://math-video.com/videos/multivariable-function-trccbdwwcbc2026-10-12T03:05:30.235Zhttps://math-video.com/zh/videos/multivariable-function-trccbdwwcbc2026-10-12T03:05:30.235Zhttps://math-video.com/videos/multivariable-limit-ptwa4gkgzlm2026-10-12T03:05:22.939Zhttps://math-video.com/zh/videos/multivariable-limit-ptwa4gkgzlm2026-10-12T03:05:22.939Zhttps://math-video.com/videos/jacobian-p46qwyhqe6m2026-10-12T03:06:37.800Zhttps://math-video.com/zh/videos/jacobian-p46qwyhqe6m2026-10-12T03:06:37.800Zhttps://math-video.com/videos/iterated-integral-nveagsir4ic2026-10-12T03:06:34.307Zhttps://math-video.com/zh/videos/iterated-integral-nveagsir4ic2026-10-12T03:06:34.307Zhttps://math-video.com/videos/double-integral-gifqwttwqey2026-10-12T03:06:12.648Zhttps://math-video.com/zh/videos/double-integral-gifqwttwqey2026-10-12T03:06:12.648Zhttps://math-video.com/videos/moment-of-inertia-f7slkv01ax02026-10-12T03:06:05.730Zhttps://math-video.com/zh/videos/moment-of-inertia-f7slkv01ax02026-10-12T03:06:05.730Zhttps://math-video.com/videos/curved-trapezoid-bv1mj411j7e82026-10-11T12:19:09.762Zhttps://math-video.com/zh/videos/curved-trapezoid-bv1mj411j7e82026-10-11T12:19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