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https://www.csdn.net/tags/MtTaEg0sMDUzOTQtYmxvZwO0O0OO0O0O.html | [
"• 贝叶斯优化 贝叶斯优化学习代码 什么是贝叶斯优化？ 一种使用高斯过程优化黑盒函数 f (x) 的技术（可能） 我想有效地搜索并找到 x_opt = argmax_x f (x)。 假设评估 f (x) 需要时间。程序大致如下。 t = 0，D_t = ...",
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"Python\n• BoTorch是一个基于PyTorch的贝叶斯优化库。 BoTorch目前处于测试阶段，并且正在积极开发中！ 为什么选择BoTorch？ BoTorch为BoTorch提供了模块化且易于扩展的界面，BoTorch是基于PyTorch构建的贝叶斯优化库。 ...\n• 贝叶斯优化是机器学习中一部分，由于网格搜索法效果不好，才选择贝叶斯优化，涉及到参数寻优的都可以用此方法，效率高并且有效性强。\n• BoTorch是一个基于PyTorch的贝叶斯优化库。 BoTorch目前处于测试阶段，并且正在积极开发中！ 为什么选择BoTorch？ BoTorch 提供一个模块化且易于扩展的界面，以构成贝叶斯优化原语，包括概率模型，采集函数和优化...",
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"Python\n• 公共贝叶斯优化库（COMBO） 贝叶斯优化已被证明是加速科学发现的有效工具。 但是，标准实施方式（例如scikit-learn）只能容纳少量的培训数据。 COMBO具有高效的协议，因此具有很高的可扩展性，该协议采用了Thompson...",
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"Python\n• Matlab实现的最大值熵搜索的多目标贝叶斯优化方法。 要运行代码，您应该从 GP-stuff软件包位于.zip文件中。 或者，您可以从以下位置获取最新版本的GP-stuff： 部分代码取自，它实现了有效熵贝叶斯优化的最大值熵...\n• 通过贝叶斯优化对硬件进行感知的超参数搜索 + 神经网络。 描述 神经网络（NN）的超参数优化已成为一个具有挑战性的过程。 如果我们发现最佳的（根据分类错误）NN配置也满足硬件约束，例如最大推理运行时间，最大GPU...",
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"Python\n• 贝叶斯优化 Hyperband 超参数优化 实施 要求 - numpy - scipy - statsmodels - dask - torch (example) 安装 pip3 install bohb-hpo 用法 from bohb import BOHB import bohb . configspace as cs def objective ...\n• GeoBO：用于地球科学中多目标贝叶斯优化和联合反演的Python包 GeoBO基于使用高斯过程（GP）先验的概率框架共同解决多线性正向模型。 该软件可从2D勘测数据（例如磁学和重力）以及任何先前存在的钻芯测量结果中生成多...",
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"Python\n• 贝叶斯优化最小二乘向量机，很好的优化方法，也比较少见",
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"lssvm\n• matlab代码粒子群算法BayesOptMat：MATLAB的贝叶斯优化 这是使用修正的高斯过程优化。 使用不同的采集功能执行贝叶斯全局优化。 除其他功能外，还可以使用BayesOptMat优化物理实验并调整机器学习算法的参数。 该代码...\n• LSTM Time Series Prediction with Bayesian optimization.zip",
null,
"LSTM\n•",
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"研究论文\n• 采用的是美国西储大学轴承数据中心的滚动轴承数据，贝叶斯优化后的准确率高达99%，也包含了和遗传算法以及网格搜索优化支持向量机的对比！希望可以帮助到大家！！！给两个积分意思一下就行了\n• 贝叶斯优化\n\n2020-12-23 16:33:43\n文起本篇文章记录通过 Python 调用第三方库，从而调用使用了贝叶斯优化原理的 Hyperopt 方法来进行超参数的优化选择。具体贝叶斯优化原理与相关介绍将在下一次文章中做较为详细的描述，可以参考这里。Hyperopt 是 ...",
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"万壑松风知客来，摇扇抚琴待留声\n\n1. 文起\n\n本篇文章记录通过 Python 调用第三方库，从而调用使用了贝叶斯优化原理的 Hyperopt 方法来进行超参数的优化选择。具体贝叶斯优化原理与相关介绍将在下一次文章中做较为详细的描述，可以参考这里。\n\nHyperopt 是 Python 的几个贝叶斯优化库中的一个。它使用 Tree Parzen Estimator(TPE)，其它 Python 库还包括了 Spearmint(高斯过程代理)和 SMAC(随机森林回归)。贝叶斯优化问题有四个部分：目标函数：使用的机器学习模型调用该组超参数在验证集上的损失。\n\n域空间：类似于网格搜索，传入超参数的搜索范围。\n\n参数搜索：构造替代函数并选择下一个超参数值进行评估的方法。\n\n存储结果：最小化函数在评估每组测试后的最优超参数存储结果。\n\n2. Hyperopt 简单样例说明：最简单的流程，实现以 XGBoost 作为调参模型，通过 hyperopt 完成上述贝叶斯优化的四个部分。\n\n一：定义目标函数1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n9\n\n10train_X, test_X, train_y, test_y = train_test_split(df_scaler, df_y, test_size=0.3, random_state=999)\n\ndata_train =xgb.DMatrix(train_X, train_y, silent=False)\n\n# 定义目标函数\n\ndef (params, n_folds=10):\n\ncv_results =xgb.cv(params, data_train, num_boost_round=1000, nfold=n_folds, stratified=False, shuffle=True, metrics='mae', early_stopping_rounds=10)\n\nmae = max(cv_results['test-mae-mean'])\n\nloss = mae\n\nreturn loss\n\nobjective() 是目标函数(黑盒函数)，作用是返回一个损失值，Hyperopt 也是根据这个损失值来逐步选择超参数的。后续的 fmin() 函数会最小化这个损失，所以你可以根据是最大化还是最小化问题来改变这个返回值。此处的目标函数十分基础并没有做过多的调整，可根据实际情况来做修改目标函数。\n\n二：设置域空间1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n9from hyperopt import hp\n\nspace = {\n\n'learning_rate': hp.loguniform('learning_rate',np.log(0.01),np.log(0.4)),\n\n'max_depth': hp.choice('max_depth',range(1,8,1)),\n\n'min_child_weight': hp.choice('min_child_weight',range(1,5,1)),\n\n'reg_alpha': hp.uniform('reg_alpha',0.0,1.0),\n\n'subsample': hp.uniform('subsample',0.5,1.0),\n\n'colsample_bytree': hp.uniform('colsample_bytree',0.6,1.0)\n\n}\n\n域空间，也就是给定超参数搜索的范围。这个可以通过 hyperopt 的 hp 方法实现，hp 方法有很多个参数选择的方式如：hp.loguniform(对数空间搜索)、hp.lognormal(对数正态分布)、hp.normal(正态分布)、hp.choice(列表选项搜索)、hp.uniform(连续均匀分布搜索)、hp.quniform(连续均匀分布整数搜索)。\n\n这里也只是简单的设置域空间，比如还可以通过 choice 方法构建列表从而实现对多个不同模型的筛选，这个在后续会有介绍。强调：如果使用参数名相同会报错 “hyperopt.exceptions.DuplicateLabel”，例如域空间中定义了 XGBoost、LightGBM 两种模型，由于它们的部分参数名相同所以如果使用同名参数将报错。解决方法：可以修改内层名称，这个在后续代码中会说明。\n\n三：参数搜索1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8# 参数搜索算法\n\nfrom hyperopt import tpe\n\ntpe_algorithm = tpe.suggest\n\n# 寻找目标函数的最小值\n\nfrom hyperopt import fmin\n\nMAX_EVALS = 500\n\nbest = fmin(fn=objective, space=space, algo=tpe.suggest, max_evals=MAX_EVALS)\n\nHyperopt 中可以使用几种搜索算法：tpe(tpe.suggest)、随机 rand(rand.suggest)、模拟退火 anneal(anneal.suggest)。同时定义用来寻找目标函数返回最小损失值的 fmin() 函数，其中 max_evals 参数表示迭代次数。\n\n四：存储结果1\n\n2# 优化结果参数\n\nprint(best)\n\nbest 就是 Hyperopt 返回最佳结果的参数。\n\n3. Hyperopt 可以变复杂\n\n上面使用几处简单的代码，实现了 Python 中调用 Hyperopt 来完成超参数选择的功能。任何方法都可以变得复杂，更何况 Hyperopt ，所以在上面代码的基础上稍加改变，实现一种稍微复杂点的 Hyperopt，从而完成较为强大的功能。说明：以下代码将实现，从 SVM、XGBoost、LightGBM、KNN、Linear 等多个不同模型中选择超参数，最终找到 Hyperopt 认为的最优参数。其中损失值为偏离度，并可以查看黑盒函数中的运行过程。\n\n一：定义目标函数1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n9\n\n10\n\n11\n\n12\n\n13\n\n14\n\n15\n\n16\n\n17\n\n18\n\n19\n\n20\n\n21\n\n22\n\n23\n\n24\n\n25\n\n26\n\n27\n\n28\n\n29\n\n30\n\n31\n\n32\n\n33\n\n34\n\n35\n\n36\n\n37\n\n38\n\n39\n\n40\n\n41\n\n42\n\n43\n\n44\n\n45\n\n46\n\n47\n\n48\n\n49\n\n50\n\n51\n\n52\n\n53\n\n54\n\n55\n\n56\n\n57\n\n58\n\n59\n\n60\n\n61\n\n62\n\n63\n\n64\n\n65\n\n66import pandas as pd\n\nimport numpy as np\n\nfrom sklearn.preprocessing import StandardScaler\n\nfrom sklearn.model_selection import train_test_split\n\nfrom sklearn.model_selection import cross_val_score\n\nfrom sklearn.metrics import mean_absolute_error\n\nfrom sklearn.svm import SVR\n\nimport xgboost as xgb\n\nimport lightgbm as lgb\n\nfrom sklearn.neighbors import KNeighborsRegressor\n\nfrom sklearn.linear_model import LinearRegression\n\nfrom hyperopt import STATUS_OK\n\nimport warnings\n\nwarnings.filterwarnings('ignore')\n\nfeatures = df.iloc[:,:-1]\n\ntarget = df.iloc[:,-1]\n\nscaler = StandardScaler()\n\nfeatures = scaler.fit_transform(features)\n\ntrain_X, test_X, train_y, test_y = train_test_split(features, target, test_size=0.25, random_state=999)\n\n# 计算偏离度\n\ndef dod(real_values, pred_values):\n\ndod_result = np.around((np.absolute(real_values - pred_values) / real_values).mean(), 4)\n\nreturn dod_result\n\n# 计算传入模型的损失\n\ndef (params):\n\nmodel_name = params['model_name']\n\ndel params['model_name']\n\nif model_name == 'svm':\n\nclf = SVR(**params)\n\nelif model_name == 'xgboost':\n\nclf = xgb.XGBRegressor(**params)\n\nelif model_name == 'lightgbm':\n\nclf = lgb.LGBMRegressor(**params)\n\nelif model_name == 'knn':\n\nclf = KNeighborsRegressor(**params)\n\nelif model_name == 'linear':\n\nclf = LinearRegression(**params)\n\nelse:\n\nreturn 0\n\nclf.fit(train_X, train_y)\n\npred = clf.predict(test_X)\n\nloss = dod(test_y, pred)\n\n# 偏离度是自己定义的方法，当然可以返回交叉验证的结果(loss是负值需要注意)\n\n# loss = cross_val_score(clf, train_X, train_y, cv=10, scoring='neg_mean_absolute_error').mean()\n\nreturn {'loss':loss, 'params':params, 'status':STATUS_OK}\n\n# 定义总的目标函数\n\ncount = 0\n\nbest_score = np.inf\n\nmodel_name = None\n\ndef fn(params):\n\nglobal model_name, best_score, count\n\ncount +=1\n\nscore = objective(params.copy())\n\nloss = score['loss']\n\nif loss < best_score:\n\nbest_score = loss\n\nmodel_name = params['model_name']\n\nif count % 50 == 0:\n\nprint('iters:{0}, score:{1}'.format(count, score))\n\nreturn loss\n\n二：设置域空间说明：两个树模型有相同的参数名，所以在设置参数时需要区别参数名。1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n9\n\n10\n\n11\n\n12\n\n13\n\n14\n\n15\n\n16\n\n17\n\n18\n\n19\n\n20\n\n21\n\n22\n\n23\n\n24\n\n25\n\n26\n\n27\n\n28\n\n29\n\n30\n\n31\n\n32\n\n33\n\n34\n\n35\n\n36\n\n37\n\n38\n\n39\n\n40\n\n41\n\n42from hyperopt import hp\n\n# 设置域空间\n\nspace = hp.choice('regressor_type',[\n\n{\n\n'model_name': 'svm',\n\n'C': hp.uniform('C',0, 10.0),\n\n'kernel': hp.choice('kernel', ['linear', 'rbf']),\n\n'gamma': hp.uniform('gamma', 0, 20.0)\n\n},\n\n{\n\n'model_name': 'xgboost',\n\n'n_estimators': hp.choice('xgb_n_estimators', range(50,501,2)),\n\n'learning_rate': hp.uniform('xgb_learning_rate', 0.01, 0.3),\n\n'max_depth': hp.choice('xgb_max_depth', range(2,8,1)),\n\n'min_child_weight': hp.choice('xgb_min_child_weight', range(1,5,1)),\n\n'reg_alpha': hp.uniform('xgb_reg_alpha', 0, 1.0),\n\n'subsample': hp.uniform('xgb_subsample', 0.5, 1.0),\n\n'colsample_bytree': hp.uniform('xgb_colsample_bytree', 0.6, 1.0)\n\n},\n\n{\n\n'model_name': 'lightgbm',\n\n'n_estimators': hp.choice('lgb_n_estimators', range(50,501,2)),\n\n'learning_rate': hp.uniform('lgb_learning_rate', 0.01, 0.3),\n\n'max_depth': hp.choice('lgb_max_depth', range(2,8,1)),\n\n'num_leaves': hp.choice('lgb_num_leaves', range(20, 50, 1)),\n\n'min_child_weight': hp.choice('lgb_min_child_weight', [0.001,0.005,0.01,0.05,0.1]),\n\n'min_child_samples': hp.choice('lgb_min_child_samples', range(5,51,5)),\n\n'subsample': hp.uniform('lgb_subsample', 0.5, 1.0),\n\n'colsample_bytree': hp.uniform('lgb_colsample_bytree', 0.6, 1.0),\n\n'reg_alpha': hp.uniform('lgb_reg_alpha', 0, 1.0)\n\n},\n\n{\n\n'model_name': 'knn',\n\n'n_neighbors': hp.choice('n_neighbors', range(2,11)),\n\n'algorithm': hp.choice('algorithm', ['auto','ball_tree','kd_tree','brute'])\n\n},\n\n{\n\n'model_name': 'linear',\n\n'normalize': hp.choice('normalize', [False, True])\n\n}\n\n])\n\n三：参数搜索1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n9\n\n10\n\n11\n\n12# 设定搜索算法\n\nfrom hyperopt import tpe\n\n# 如果有有必要，可以设置查看黑盒函数fn中的搜索情况(每次选择参数等)\n\nfrom hyperopt import Trials\n\ntrials = Trials()\n\n# 定义搜索目标函数最小值函数\n\nfrom hyperopt import fmin\n\nMAX_EVALS = 1500\n\nbest_params = fmin(fn=fn, space=space, algo=tpe.suggest, max_evals=MAX_EVALS,trials=trials)\n\nprint('model_name: {0}, best_score: {1}'.format(model_name, best_score))\n\n四：存储结果1\n\n2\n\n3\n\n4\n\n5# 最佳参数\n\nprint(best_params)\n\n# 查看黑盒函数中的变化\n\nprint(trials)# 可以通过循环依次打印结果\n\n4. 文末\n\nHyperopt 通过不同的设置可以变得更加完善和准确，从而在搜索参数上取得进一步成功。\n\n你需要注意一点，Hyperopt 并不是一个完全没有缺陷的方法，它也可能从一开始就陷入局部最优，所以相比于暴力的网格搜索它只是一种具有数学理论支撑的高级暴力搜索方法。\n\n展开全文",
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"• Dragonfly：用于可扩展贝叶斯优化的开源python库\n• 表格基线 不同的表格基线算法与超频带加贝叶斯优化（BOHB）相结合，用于超参数优化。",
null,
"Python\n• 深度学习模型优化 使用深度学习和贝叶斯优化进行超参数调整\n• 贝叶斯优化算法 贝叶斯优化算法(BOA)是由美国UIUC大学的Pelikan等在2000年前后提出 的在贝叶斯优化算法中根据均匀分布随机产生初始种群 然后采用进化算法 的各种选择方法比如二进制锦标赛选择比例选择截断选择等从...\n• 提出一种基于自适应代理模型的并行贝叶斯优化方法,用于求解计算成本高的复杂优化问题.该方法基于多点期望改进判据,通过批次采样实现并行优化.针对并行优化产生的大量历史数据会导致全局代理模型建模成本高的问题,...\n• 压缩包为贝叶斯优化在机器学习和深度学习中应用的小案例，里包含： (1)data:iris.csv和mnist.npz; (2)贝叶斯优化_ML.py; (3)贝叶斯优化_DL.py。\n• 快速信息理论贝叶斯优化 这是我们的论文提出的方法的MATLAB代码存储库。 我们开发了一种称为FITBO的新颖的信息理论贝叶斯优化方法，该方法将用于全局极小值的昂贵采样减少到一个附加超参数的更有效采样，从而显着...",
null,
"MATLAB\n• 这是使用高斯过程假设进行贝叶斯优化算法的matlab演示。 该算法在以下内容中有完整描述 在强盗环境中使用高斯过程进行优化估计（Zi Wang，Bolei Zhou，Stefanie Jegelka），在国际人工智能与统计会议（AISTATS）中，...\n• 针对传统物体形状分类算法中图像的空间结构特征表示不够准确, 以及分类器模型参数易陷入局部最优的问题, 提出结合重叠金字塔与贝叶斯优化神经网络的物体形状分类方法。首先, 将物体轮廓分割为不同长度的轮廓片段作为...\n• 通过Botorch为Echo状态网优化例程的贝叶斯优化（GPU优化）。 通过优化模型对例程进行聚类以得出时间序列。 参考： 2018年国际神经网络联合会议（IJCNN），第1-7页。 IEEE，2018年 使用范例 # Load data data = np . ...\n•",
null,
"贝叶斯优化\n• 通过贝叶斯优化进行实验设计： edbo是贝叶斯优化用于化学合成的实际实现。 参考：希尔兹（Shields），本杰明（Benjamin J.）； 史蒂文斯，杰森； 李俊Maven（Marvin） 达玛尼，法罕，珍妮，雅各布； 亚当斯（Ryan P...\n• 用于模块化贝叶斯优化的Python软件包。 该程序包提供了对可能的噪声损坏函数f进行优化的方法。 特别是，此程序包使我们可以优先考虑f的可能行为并选择点，以便收集有关函数及其最大值的信息。 安装 安装此软件包的...",
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https://paperzz.com/doc/8666417/semisimple-maximal-quotient-rings | [
"### SEMISIMPLE MAXIMAL QUOTIENT RINGS\n\n```SEMISIMPLE MAXIMAL QUOTIENT RINGS\nBY\nFRANCIS L. SANDOMIERSKK1)\nNotation and Introduction. R denotes an arbitrary associative ring. A right\nF-module A over R will be denoted AR. BR is a large submodule of AR (AR is an\nessential extension of BR), if BR is a submodule of AB having nonzero intersection\nwith every nonzero submodule of AR. A right ideal 7 of F is a large right ideal, if\nTBis a large submodule of RR.\nGiven AR, Z(AR) is the singular submodule of AR , which consists of all those\nelements of A whose annihilators in R are large right ideals.\nFollowing Johnson , g is a right quotient ring of R if Q is a ring with identity\ncontaining F as a subring (the identity of Q is the identity of F if F has one)\nand RR is a large submodule of QR.\nThe quotient rings considered by Goldie in , will be called classical quotient\nrings. Q is a classical right quotient of F if every regular element (nonzero divisor)\nof F is a unit in Q and every element of Q is of the form ab'1, a, b e R, b regular\nin R. In general, a ring F need not possess a classical right quotient ring.\nGoldie , has given necessary and sufficient conditions that a ring possess a\nclassical right quotient ring which is semisimple. Here semisimple means semisimple with minimum condition .\nThis paper is concerned with the question of characterizing those rings which\nhave a semisimple maximal right quotient ring , , , and in this case\ngeneralizing some simple well-known results about commutative integral domains,\ntheir quotient rings and modules over these domains. Johnson has shown that\nF has a regular maximal right quotient ring Q if and only if Z(FB) = 0, where Q\nis a regular ring if every finitely generated right (left) ideal of Q is generated\nby an idempotent. In this case QR is injective as a right F-module, hence the\ninjective hull of F .\nA ring F has a semisimple maximal right quotient ring Q if and only if Z(FB) = 0\nand dim RR is finite, where a right F-module M is of finite dimension if every direct\nsum of submodules of M has only finitely many nonzero summands. This is the\nmain result of §1. In addition another characterization is given for rings which\npossess a semisimple classical right quotient ring, namely, F has a semisimple\nclassical right quotient ring if and only if Z(Fs) = 0 and if T is a large right ideal\nof F, then there is an element ae I such that aR is a large right ideal of R.\nIf F has a semisimple classical right quotient ring Q, then it is known , that\nReceived by the editors February 10, 1966.\n(*) Supported in part by National Science Foundation\nGrant GP-3993.\n112\nSEMISIMPLE MAXIMAL QUOTIENT RINGS\n113\ng is the maximal right quotient ring of R. The converse is not valid, since there\nare rings with or without identity that have a semisimple maximal right quotient\nring g and g is not the classical right quotient ring of R. Let g be the ring of\n« x « matrices over a division ring A and R the set of upper triangular (strictly\nupper triangular) matrices of g. It is easily verified that g is a right quotient ring\nof 7?but g is not a classical right quotient ring of 7?. Since g is semisimple g is the\nmaximal right quotient ring of R.\nIt is also shown in §1 that if 7? has a semisimple maximal right quotient ring g,\nthen Z(A¡Z(A)) = 0 for every right 7?-module A. This generalizes the analogous\nresult if R is a commutative integral domain, since then Z(AB) is the torsion\nsubgroup of A.\nIn §2 rings with identity are considered.\nIn this case the following generalizations of results known when R is a commutative integral domain hold, thus extending some of the results of Gentile \nalso.\n1. R has a semisimple maximal right quotient ring g if and only if A -> A <g)Bg\nhas Z(AR) for its kernel for every unitary right 7?-module A.\nIf R has a semisimple maximal right quotient ring g, then\n2. Bg is flat as a left 7?-module.\n3. Every unitary left g-module is flat as a left 7?-module.\n4. If Z(AR)= 0, then 0 -> A -> A <g»B\ng is the injective hull of A, a unitary\n7?-module.\n5. Tori(^, Q¡R)^Z(AR) for every unitary right 7f-module A.\nAnother result with weaker hypothesis is valid.\nAny direct sum of injective right 7?-modules, each with zero singular submodule\nis injective if dim 7?Bis finite.\nThe following generalizes a result of Matlis . If 77Bis an epimorphic image\nof an injective right 7?-module ER and Z(77/Z(77))=0, then Z(77)) is a direct summand of 77 with complementary summand injective. The proof given here is\nsimpler in that it does not appeal to any quotient ring of R as was done in Matlis\n[12, Theorem 1.1] when 7? is a commutative integral domain. Also Proposition 2.1,\nProposition 2.2, and Proposition 2.4 of can be generalized to a noncommutative ring R which has a semisimple maximal right quotient ring utilizing identical\nproofs.\n1. Arbitrary rings.\nDefinition 1.1. If M is a right 7?-module, then the set of all large submodules\nof M is denoted by L(MR).\nIt is useful to recall the following results, which are essentially in .\nProposition 1.2.\n1. If A, Be L(MR), then A+Be L(MR) and An Be L(MR).\n2. If A e L(MR) and B a submodule of M containing A, then B e L(MR).\n114\nF. L. SANDOMIERSKI\n[July\n3. If A is a submodule of M, then there exists a submodule B of M maximal with\nrespect to the property that A n F=0 and consequently A + B eL(MR).\n4. Iffe HomB(M,TV)and A e L(NR),thenf' \\A)={xeM\\\nCorollary.\nIf Ax,...,\nAn eL(AR) and xx,...,\nf(x) e A}e L(MB).\nxne A, then I={r e R \\ xtr e At\nfor all i} e L(RR).\nProof, {re R\\ xtre Ai}= It for each i is the counter-image of Ai by the right R\nhomomorphism from F into A given by left multiplication by xi; hence 7 is the\nintersection of finitely many large right ideals of R, so le L(RR).\nGoldie calls a right F-module M of finite dimension, dim MR finite, if every\ndirect sum of nonzero submodules of M has only a finite number of direct summands which are nonzero.\nIf M is a right F-module and xx,..., xn e M, then [x1(..., x„] will denote the\nsubmodule of M generated by {x1;..., xn}. For xeM, xR={xr | r e R}. If R is a\nring with identity and M a unitary F-module, then clearly [x]=xR. A module MR\nwill be called regular if for O^x e M, xF^O. Clearly if F is a ring with identity\nevery unitary F-module is regular. Also if F is arbitrary then MR is regular if\nZ(MR) = 0.\nTheorem 1.3. Let MR be a right R-module, and consider the following conditions.\n(a) dim MR is finite.\n(b) If K e L(MR), there are xx,...,xneK\nsuch that [xx,..., xn] e L(MR).\n(c) If K is a submodule of M, then there are xx,..., xn e K such that [xx,..., xn]\neL(KR).\n(b*) If Ke L(MR), there are xx,..., x„ e K such that 2 *¡F e L(MR).\n(c*) If K is a submodule of M, then there are xx,...,xneK\nsuch that 2 x{R\neL(KR).\nIf MR is any R-module then (a), (b), and (c) are equivalent. If MR is a regular\nR-module then (a), (b*), and(c*) are equivalent, hence all the statements are equivalent.\nProof. Only the equivalence of (a), (b), and (c) will be shown as the equivalence\nof (a), (b*), and (c*) when MR is regular has an analogous proof.\n(a) implies (b). Let KeL(MR) and suppose (b) is denied. Let 0/xx e K, then\n[xx] <£L(KR), hence there is 0 ^x2 e K such that [xx] n [x2]=0. Suppose xx,..., xn\ne K such that 0 ^ [x(] and the sum 2 [*t] is direct. Since 2 [*«]= ixi, • • •, x„],\n2 [x¡] \\$ L(KR), hence there is 0^xn+1eF\nsuch that [x1;..., xn] n [xn +1]=0.\nThus, there is an infinite sum [x^ © • • • © [xn] © • • • £ Fs M contradicting (a), so\nfor some n [xx,..., xn]eL(KR), hence [xx,..., xn] eL(MR) since KeL(MR).\n(b) implies (c). Let F be a submodule of MR. By Zorn's lemma there is a submodule L of M which is maximal with respect to the property that K n F=0 and\nconsequently K+L e L(MR). By (b) there exist finitely many ax,..., an e K+L such\nthat [ax,..., an] e L(MR). Now aj = xi+>'(, kteK, yteL. The counter image of\n[X(,jilLi eL(MR) by the inclusion map K-^ K+L is [xx,..., xn], hence is large\nin K, so (c) follows.\n1967]\n115\nSEMISIMPLE MAXIMAL QUOTIENT RINGS\n(c) implies (a). Given K a direct sum of nonzero submodules of M, then there\nare finitely many xu ..., xneK such that [xt,..., xn]eL(KR). Now [xi,..., xj\nis contained in the direct sum of finitely many of the submodules of K and\n[xi,..., xn] thus has zero intersection with the others so the others are zero and\n(a) follows.\nSimilar proofs show the equivalence of (a), (b*), and (c*).\nProposition 1.4. For a unitary module AR, the following statements are equivalent.\n(a) A is semisimple (sum of simple submodules),\n(h) A is a direct sum of simple submodules.\n(c) Every submodule of AR is a direct summand of A.\n(d) L(AR) = {AR}.\nProof. The equivalence of (a), (b), and (c) is well known, e.g., . Clearly (c)\nimplies (d). Conversely, if B is a submodule of A, then there is a submodule C of\nA such that B n C=0 and B+C e L(AB),by (d) A=B © C so (c) follows.\nLemma 1.5. If g is a right quotient ring of R with Z(RR)= 0 and A, B R-submodules of gB such that A n B=0, then AQ n BQ = 0.\nProof. If xe AQ n BQ, then x=2 uí?í= 2 ¿>f/>i,\nax,...,aneA,\nbx,-..,bneB,\nPx, ■■-,Pn, (¡i, - • ■,In e g- I={r e B I qs g R,ptr e R for all i'}eL(RR) and x7=0 so\nx g Z( gB)=0, since Z(RR) = 0.\nCorollary.\nB n L(RR).\nIf g is a right quotient ring of R, Z(RR)=0 and BeL(QQ), then\nProof. If B nj=0 for a right ideal J of R, then Zfg n Jg=0\nhence .7=0 since BQ = B.\nby the lemma,\nTheorem 1.6. Let Z(RR) = 0, and Q the maximal right quotient ring of R, then the\nfollowing statements are equivalent.\n(a) IQ=Q far every IeL(RR).\n(b)\n(c)\n(d)\n(e)\nFor I e L(RR) there are alt...,anel\nsuch that 2 aiR e L(RR).\ndim RR is finite.\nIf I is a right ideal of R, then there are alf..., an e I such that 2 aiReL(IR).\nQ is a semisimple ring.\nProof. The equivalence of (b), (c), and (d) follows from Theorem 1.3 since\nZ(Z?B)= 0.\n(a) implies (b). If 7 g L(Rr), then 7g = g, hence there are au ..., an e /, qu..., qn\ne Q such that 2 a* = 1. ^={r g 7? | ^¡r g 7Î for all ¿}g Z,(7?b)and clearly 7s 2 aiR\nso (b) follows.\n(b) implies (e). If B e L(QQ), then B r\\ Re L(RR)by the corollary to Lemma 1.5.\nSo B n R has elements aly...,an such that Z= 2 ^R e L(RR). IQ is a finitely\ngenerated right ideal of g. Since Z(Z?B)= 0, g is a regular ring, hence Zg = eg,\ne=e2Gg. However, (l-e)Z=0\nso l-eGZ(gB)=0\nso Zg = g, but IQ=B so\nB=Q, that is Z.(g0)={g<j} so g is a semisimple ring.\n116\nF. L. SANDOMIERSKI\n[July\n(e) implies (a). If le L(RR), then IQ = eQ for some e = e2e Q since Q is semisimple. Since Q is the maximal right quotient ring of F and Q is semisimple, then\nQ is a regular ring so Z(QR) = 0 by . Therefore, since (1 -e)I=0, (l-e) eZ(QR)\n= 0, so l=e and IQ=Q.\nNow the case of a semisimple classical right quotient ring of F will be considered.\nTheorem 1.7. For a ring R, the following statements are equivalent.\n(a) F has a semisimple classical right quotient ring.\n(b) Z(RR) = 0 and for I e L(RR) there is a el such that aR e L(RR).\n(c) F is a semiprime ring, dim RR is finite and R satisfies the ascending chain\ncondition on right annihilators.\nProof. The equivalence of (a) and (c) was shown by Goldie .\n(a) implies (b)(2). Let Q be a semisimple classical right quotient ring of F, then\nQ is the maximal right quotient ring of F. By Theorem 1.6 Z(FB) = 0 and for\nIeL(RR), there are ax,..., ane I, qx,.. .,qne Q such that 2 añ\\ = L Since qx e Q,\nqx = cxdx~1,cx, dx e R, dx regular in F, hence <Ti= 2i>i aiqidx+axcx. Since q2dx e Q,\nq2dx= c2d21, c2, d2 e R, d2 regular in F, so dxd2=axcxd2+a2c2 + 2i>2 «W^Con-\ntinuing in this fashion it follows that there exist regular elements dx,...,\ndne R\nsuch that d=dx- • dne2a¡FsT.\nIf dRnJ=Q for a right ideal J of F, then\ndRQ r\\JQ=0, but since ReL(RR), RQ=Q so dRQ=Q since d is regular so\n/=0, hence dR e L(RR).\n(b) implies (a). Let Q be the maximal right quotient ring of F and q e Q, then\nI={r e R I qr e R}eL(RR). By (b) there is a e I such that aR e L(RR). By Theorem\n1.6 Q=aRQ = aQ so a has right inverse. Since Q is semisimple and a has a right\ninverse a has a left inverse so a is a regular element of F and q = ba'1. If a e R and\na is regular, then the right annihilator of a in F is zero, hence in Q also. Since Q\nis semisimple a is regular in Q and (a) follows.\nIt is not valid in general that for a right F-module AR, Z(A¡Z(AR))=0. Let F be\na local ring with Jacobson radical TV#0 such that N2=0. For instance T/(/»2),Tthe\nring of integers and /» a prime. Since TVis the unique maximal right ideal of F,\nNeL(RR), hence it follows that Z(RR)= N. Since (R/N)N=0, Z(F/TV)= F/TV^0.\nTheorem 1.8. If Z(RR)=0 and dimFB is finite, then Z(AIZ(A))=0 for every\nright R-module AR.\nProof. If x+Z(A) e Z(A¡Z(A)), then T={r e R \\ xr e Z(A)} e L(RR)by definition.\nBy Theorem 1.6, Q, the maximal right quotient ring of F is semisimple so IQ=Q,\nhence there are ax,...,anel,\nqx,.. .,qne Q such that 2a><7i= l- F°r eacn '.\nxa¡ e Z(A) so T¡,the annihilator of xat in F, is in L(RR). By the corollary to Proposi-\ntion 1.2/={r e F I qtr e T¡for each /'}e L(RR).For reJ, xr=x(2 afar)) = 2 xafar)\n= 0, so x e Z(^) and the theorem follows.\n(2) Goldie has shown this implication also.\n1967]\nSEMISIMPLE MAXIMAL QUOTIENT RINGS\n117\nThis theorem raises the question of whether or not the condition Z(A\\Z(A)) = 0\nfor every 7?-module is sufficient for R to possess a semisimple maximal ring of\nquotients.\n2. Rings with identity.\n7?-modules are unitary.\nIn this section R is a ring with unity 1, and all right\nLemma 2.1. Let Q be a right quotient ring of R and A a right R-module, then if\na ® 1=0 in A ®RQ, there are finitely many qt e Q, a, g A, a = ax, {rtj}= R such that\n2 rijCfi= 51; (Kronecker delta)\ni\nand 2j o/ij = 0 for all i.\nProof. Let F be a free right 7?-module with basis {x„ : a e A}, then the sequence\nO^K^F^A^O\nof right 7?-modules is exact, where (F -> A)(xa) = a, 7v= Ker(F-> A). Tensoring\nover R with g we have the exact sequence\nK® Q-*-F® Q-+A ® Q-+&\nIf a <g>1 =0 in A <g>g, then x0 (g>1 i s the image of an element from K <g>Q.\nxa ® l=2i^i ®<7¡- Since k^eKQF, ki = Jij x^Xa for each i, a finite sum. Now\nxa ® 1 = 2i Œi xa¡\\¡) ® 4i = 2í Xa, ® (2i Kad- Since representation in F <g>g is\nunique with respect to basis elements xa = xaj for some y say y'= 1, hence 2¡ Kñí —^i¡\nand from Ar¡= 2/ xa,\\j, 0 = 2; aj\\j for all i, and the lemma follows.\nProposition\n2.2. If Q is a right ring of quotients of R, A a right R-module, then\nthe kernel of the map A<giR->AtgiQis\ncontained in Z(AR).\nProof. If a <g)1=0 in A <g>g, then by Lemma 1.3 there exist finitely many\n{4,}Ç g, {a,)^A, ax = a, {\\¡,) = R such that\n2 XuQi\n= S«> 2 fl'A\"= ° for a11L\ni\ni\nLet 7={r e R \\ q¡r e R for each /}, then 7 is a large right ideal of R by the corollary\nto Proposition 1.2 and for Àg 7\n0 - 2(2 \"AW) = 2 MoteA)\n= 2 aí(2 A¡x?iA))\n= 2 a/snA)= aiA= aX>\nhence a is annihilated by 7, so a g Z(Ar).\nTheorem 2.3. If Q is the maximal right quotient ring of R, then (a), (b), (c), (d),\n(e) of Theorem 1.6 are all equivalent to (f) Ker(A <g>B\n7?-> A (g>BQ)=Z(AR) for\nevery right R-module A.\n118\nF. L. SANDOMIERSKI\nProof, (a) implies (f). By Proposition 2.2, Ker(^ ® R^A\n[July\n® Q)^Z(AR). If\na e Z(AR) then 7, the annihilator of a in 7?, is in L(RR) so 7g = g, hence there are\n«!, ...,«,€/,\nj!.?»eg\nsuch that 2atfi = l- In ¿1 (g)g, a <g>l=a <g>\n(2 ai?i)\n= 2 aai ® 9i=0 so (f) follows.\n(f) implies (a). If 7 g L(Rr), then from the exact sequence 7 ® g -> 7? (g) g\n-»■7?/7 ® g -> 0 we have that 7Î/7 <g>g is isomorphic to g/7g. Since IeL(RR)\nZ(R¡I) = R¡I, so Rjl <g>\n7?-> 7?/7<g>g is the zero map hence Ï (2)1 =0 in 7Î/7 <g>g.\nHowever, R/I ® Q=Q/IQ is a right g-module generated by Î <g>1 so g/7g = 0,\nhence 7 <g>g -»■g is onto. It is now clear that (a) follows since the image of\n7® g^gin\ngis7g.\nAn immediate consequence of the notion of singular submodule is\nProposition 2.4. [email protected], Et right R-modules then Z(E) = @¡Z(Et).\nIf R is a commutative integral domain, then any direct sum of torsion free\ninjective 7?-modules is injective, since it is torsion-free and divisible, hence injective,\n[1, Proposition VII. 1.3]. A generalization holds.\nTheorem 2.5. 7/dim RR is finite and ER is the direct sum of injectives which have\nzero singular submodule, then E is injective.\nProof. By Proposition 2.3 Z(E)=0. It is sufficient to show that every 7?-homomorphism from a large right ideal of 7? into E can be extended to R. Let /\ng Hom(7B, ER), ZBg L(RR). By Theorem 1.3, there exist finitely many a1;.. .,ane I\nsuch that J=^üiReL(RR).\nLet/' be the restriction of/to J. Since J is finitely\ngenerated, f'(J) is contained in a finite direct sum of injectives, hence /' has an\nextension/* g Hom(7?B, ER). The assertion is that/* is an extension off. Let xe I,\nthen K={reR\n\\ xr e J} e L(RR). Now for reK, (f(x)-f*(x))r=f(xr)-f*(xr)\n=f'(xr)-f'(xr)=0\nsof(x)-f*(x)eZ(E)=0,\nhencef*(x)=f(x).\nIt is known , that if Z(7?B)= 0, then g the maximal right ring of R is injective\nas a right Z?-module.\nTheorem 2.6. If Q is the maximal right quotient ring of R, Z(RR)= 0, dim RR\nfinite and AR a right R-module such that Z(AR)= 0, then the map 0 -*■A -> A ®B g\nii an injective hull of A as a right R-module.\nProof. The map is a monomorphism by Theorem 2.3.\nNow A ®B g is a right g module, hence semisimple since g is, so A ®B g is a\ndirect sum of direct summands of g. Since Z(gB)=0, Z(A ®B g)=0 regarding\nA <g)g as a right 7?-module and by Theorem 2.5 A ®B g is injective as a right\n7?-module.\nIf 0/x=2\na¡ ® q{g A ® g, then 7={r g R \\ q¡r e R) e L(RR). Now 0^x7 since\nZ(A ® g) = 0 and xI^lm(A -> A ® g) so 0 -> ^4 -> A <g>g is an essential monomorphism, i.e., Im(y4 -> ^4 ® g) is a large right 7? submodule of A ® g and the\ntheorem follows.\n1967]\nSEMISIMPLE MAXIMAL QUOTIENT RINGS\n119\nTheorem 2.7. If Q is a semisimple maximal right quotient ring of R, then Q is\nflat as a left R-module.\nProof. It is sufficient to show that I (g>Bß -> F <g)Bß is a monomorphism\nfor\nevery right ideal of F. As before I çg>B\nß is a right ß-module, then Z(I ® 0=0\nregarding T (8)g as a right F-module. If x = 2 a¡ <8>\n<7¡s Ker(T (g>ß -> F (8)0, then\nT={r e F | a/ e/i}eL(RR)and 2a*=0 in Q. Clearly xT=0, so x eZ(T ® 0=0,\nso x=0, hence T(g)ß->F<g)ßisa\nmonomorphism.\nCorollary.\n7/ ß is the maximal right quotient ring of R, Z(RR)=0, dim RRfinite,\nthen Torf(y4, QIR)^Z(AR) for every right R-module AR.\nProof. It follows from the exact sequence Tori(^, Q) -*■Torf (A, ß/F) -y A\n®RR^A®RQ\nsince by the theorem lor^(A, Q) = 0 and Ker(A ®RR^A®RQ)\n^Z(AR) by Theorem 2.3.\nCorollary.\nIf Q is a semisimple maximal right quotient of F, then every left\nQ-moduleis flat as a left R-module.\nProof. Every left ß-module is a direct sum of direct summands of Q, hence is\nflat as a left F-module since Tor\" commutes with direct sum s.\nMatlis [12, Theorem 1.1] has shown that if F is a commutative integral domain\nand TTan F-module, then the torsion submodule of T7 is a direct summand of T7,\nif TTis an epimorphic image of an injective F-module. This result is generalized\nand the proof does not appeal to the quotient ring of F.\nFirst, the notion of a closed submodule of a module will be considered and some\nconsequences. Johnson and Wong considered the notion of a closed submodule.\nDefinition 2.8. A submodule F of a module A is closed if F has no essential\nextension in A; i.e., C a submodule of A such that F is a large submodule of C\nimplies B=C.\nRemark. If F is an injective F-module and A a submodule of F, then A is\nclosed if and only if A is a direct summand of F. This follows from the fact that\nevery submodule of F is a large submodule of its injective hull in F, which is a\ndirect summand of F.\nLemma 2.9. If fe Hom(A7,A), B a submodule of A such that Z(A/B) = 0, then\nf~ 1(B) is a closed submodule of M.\nProof. Let F be a submodule of M containing /\" X(B) as a large submodule.\nIf deD, then I={r e R \\ dr ef'1(B)}eL(RR). Now f(d)I=f(dI)^B, hencef(d)\n+ B=[f(d)Y eZ(A/B)=0 so f(d) e B and def'\\E) and the lemma follows.\nTheorem 2.10. IfE ->f H -y 0 is an exact sequence of right R-module, E injective,\nZ(H/Z(H)) = 0, then H=Z(H) © F and F is injective.\nProof. Since Z(T7/Z(77))=0, by Lemma 2.9 f'\\Z(H))\nis closed in E, hence a\ndirect summand of E=f'\\Z(H))\n120\nF. L. SANDOMIERSKI\nCorollary.\nIf R is a ring such that Z(RR)=0, dim RRfinite, then every epimorphic image of an injective R-module has its singular submodule as a direct summand.\nProof. An immediate consequence of the theorem and Theorem 1.8.\nIt is interesting to note that some of the propositions of admit generalizations\nto noncommutative rings and their maximal right quotient rings, where torsion\nsubmodule is replaced with singular submodule and ß the maximal right quotient\nring of F.\nIf in addition to the hypothesis of [12, Proposition 2.1] we assume Z(FB) = 0,\ndim FB finite, then Proposition 2.1 is valid with the same proofs using the fact that\nZ(ßB)=0 and a direct sum of copies of ß is injective by Theorem 2.5. Similarly\na generalization of [12, Proposition 2.2] is valid in view of the corollary to Theorem\n2.11, as well as [12, Corollary 2.3] of [12, Proposition 2.2].\nBibliography\n1. H. Cartan and S. Eilenberg, Homological algebra, Princeton Univ. Press, Princeton, N. J.,\n1956.\n2. B. Eckmann and A. Schopf, Über injektive Modulen, Arch. Math. 4 (1953), 75-78.\n3. C. Faith, Injective modules and quotient rings, Lecture Notes, Rutgers, The State\nUniversity, New Brunswick, N. J., 1964.\n4. G. D. Findlay and J. Lambek, A generalized ring of quotients. I, II, Canad. Math. Bull. 2\n(1958), 77-85, 155-167.\n5. E. Gentile, On rings with one-sided field of quotients, Proc. Amer. Math. Soc. 11 (1960),\n380-384.\n6. A. W. Goldie, 77ie structure of prime rings under ascending chain conditions, Proc. Lond.\nMath. Soc. 8 (1958), 589-608.\n7. -,\nSemi-prime rings with maximum condition, Proc. London Math. Soc. 10 (1960),\n201-220.\n8. N. Jacobson,\nStructure of rings, Colloq. Publ., Vol. 37, Amer. Math. Soc, Providence,\nR. I., 1956.\n9. R. E. Johnson,\nThe extended centralizer of a ring over a module, Proc. Amer. Math. Soc.\n2 (1951), 891-895.\n10. R. E. Johnson and E. T. Wong, Self-injectiverings, Canad. Math. Bull. 2 (1959), 167-173.\n11. -,\nQuasi-injective modules and irreducible rings, J. London\nMath. Soc. 36 (1961),\n260-268.\n12. E. Matlis, Divisible modules, Proc. Amer. Math. Soc. 11 (1960), 385-391.\n13. J. von Neumann, On regular rings, Proc. Nat. Acad. Sei. U. S. A. 22 (1936), 707-713.\n14. Y. Utumi, On quotient rings, Osaka Math. J. 8 (1956), 1-18.\nUniversity of Wisconsin,"
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https://goprep.co/ex-5-q15-find-the-hcf-of-1620-1725-and-255-by-euclid-s-i-1nkxbm | [
"# Find the HCF of 1620 1725 and 255 by Euclid’s division algorithm.\n\nWe know Euclid's division lemma:\n\nLet a and b be any two positive integers. Then there exist two unique whole numbers q and r such that\n\na = b q + r,\n\nwhere 0 ≤ r < b\n\nHere, a is called the dividend,\n\nb is called the divisor,\n\nq is called the quotient and\n\nr is called the remainder.\n\nAccording to the problem given,\n\nApply the division lemma on 1725 and 1620,\n\n1725 = (1620 × 1) + 105\n\nSince the remainder is not equal to zero,\n\nApply lemma again on 1620 and 105. We get,\n\n1620 = (105 × 15) + 45\n\nSince the remainder is not equal to zero, apply lemma again on 105 and 45. We get,\n\n105 = (45 × 2) + 15\n\nSince the remainder is not equal to zero, apply lemma again on 45 and 15. We get,\n\n45 = (15 × 3) + 0\n\nThe remainder has now become zero.\n\nHCF (1620, 1725) = 15\n\nNow we have to find HCF of 255 and 15.\n\nSimilarly, apply lemma on 225 and 15.\n\nWe get,\n\n225 = (15 × 15) + 0\n\nSince, the remainder is equal to 0.\n\nHCF (225, 15) = 15\n\nTherefore, HCF (1620, 1725, 225) = 15.\n\nRate this question :\n\nHow useful is this solution?\nWe strive to provide quality solutions. Please rate us to serve you better.\nRelated Videos",
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http://clintonriverkings.org/epub/carbonated-hydroxyapatite-materials-synthesis-and-applications | [
"# Carbonated hydroxyapatite : materials, synthesis, and by Michael E. Fleet",
null,
"Posted by",
null,
"By Michael E. Fleet\n\ncontent material: 1. advent --\n2. Apatite-type constitution --\nthree. Crystal chemistry and geochemistry --\nfour. Synthesis of carbonate apatites --\nfive. X-ray constructions --\n6. Chemical spectroscopy --\n7. Carbonate apatite crystal chemistry --\neight. organic apatites.\n\nRead Online or Download Carbonated hydroxyapatite : materials, synthesis, and applications PDF\n\nBest mineralogy books\n\nHigh Resolution Morphodynamics and Sedimentary Evolution of Estuaries (Coastal Systems and Continental Margins)\n\nThis publication makes a speciality of using high-resolution geophysical recommendations, box observations and modeling to enquire the morphodynamics of estuaries on either glaciated and non-glaciated coasts and on diverse time scales. Papers during this e-book supply a brand new method of nearshore and estuary reports, with an emphasis on multidisciplinary concepts and knowledge integration.\n\nLayered Intrusions\n\nThis edited paintings comprises the newest advances on the topic of the learn of layered intrusions and cumulate rocks formation. the 1st a part of this publication provides reports and new perspectives of procedures generating the textural, mineralogical and geochemical features of layered igneous rocks. the second one half summarizes development within the examine of chosen layered intrusions and their ore deposits from diverse components of the area together with Canada, Southwest China, Greenland and South Africa.\n\nExtra info for Carbonated hydroxyapatite : materials, synthesis, and applications\n\nExample text\n\nThe hardness and chemical resistance of dental enamel is commonly attributed in part to the addition of minor and trace amounts of fluoride into the hydroxylapatite nanocrystals. More loosely bound anions, vacancies, and complex anions have a tendency to be locally ordered in c-axis columns, forming onedimensional domains. The domain ordering of chlorapatite (CLAP) has already been discussed (Fig. 4). 754 A) a formula of Ca15 (PO4 )9 IO (Z = 2). 75. 2 Summary of some apatite substitutions relative to ideal FAP, Ca10 (PO4 )6 F2 1 Channel (X ) anion X − = F− = 2F− X 2− + A+ + = Ca2+ + F− +2 = Ca2+ + 2F− = PO4 3− + F− A + + BO4 2− = Ca2+ + PO4 3− = 2F− BO4 2− + BO4 4− = 2PO4 3− CrO4 2− + SiO4 4− = 2PO4 3− +2 = Ca 2+ = Ca 2+ − + 2F A+ + = Ca2+ + F− 3+ = 2Ca A + BO4 2− = Ca A +A + + PO4 2A + BO4 = 2Ca 2+ 2A 3+ + = 3Ca2+ 2RE3+ + = 3Ca2+ + PO4 3− RE3+ + X 2− = Ca2+ + F− RE3+ + A + = 2Ca2+ RE3+ + SiO4 4− = Ca2+ + PO4 3− 2RE3+ + = PO4 3− + F− = 3Ca2+ Bi3+ + O2− = Ca2+ + F− VO4 3− = PO4 3− SO4 2− + SiO4 4− = 2PO4 3− A 3+ + BO4 4− = Ca2+ + PO4 3− 5− + A 3+ + BO4 4− = Ca2+ + PO4 3− AsO4 3− = PO4 3− 3− Bi3+ + O2− = Ca2+ + F− 3+ = PO4 3− + F− 4− 2A 3+ + BO4 5− = 2Ca2+ + PO4 3− 2+ 2+ BO4 4− + SiO4 + 2BO4 2− = Ca2+ + 2PO4 3− + = 2Ca2+ BO4 3− = PO4 3− Large ( A) cation A U4+ + + 2BO4 2− = Ca2+ + 2PO4 3− A 3+ + X 2− = Ca2+ + F− 2+ = 2Ca2+ = 2Ca2+ Phosphate group (BO4 ) BO4 4− + CO3 2− + A 4+ + Th4+ + Na+ + SO4 2− = Ca2+ + PO4 3− K+ + SeO4 2− = Ca2+ + PO4 3− SiO4 4− + (SO4 ,CO3 )2− = 2PO4 3− 2RE3+ + BO4 5− = 2Ca2+ + PO4 3− Type B carbonate + 2CO3 2− = Ca2+ + 2PO4 3− CO3 2− + SiO4 4− = 2PO4 3− Na+ + CO3 2− = Ca2+ + PO4 3− H3 O+ + CO3 2− = Ca2+ + PO4 3− 1.\n\n0 1. 1–5 recalculated from Table 1 of McClellan and Lehr:65 1. Florida; 2. Morocco; 3. North Carolina; 4. Tennessee; 5. Israel; 6. ideal FAP; 7. 5 35 September 23, 2014 16:41 PSP Book - 9in x 6in 03-Michael-Fleet-c03 36 Crystal Chemistry and Geochemistry bulk sample, introducing the possibility that they might be slightly overdetermined. 4z F2 . It is assumed that all carbonate substitutes for phosphate and that fluoride blocks the entry of carbonate into the apatite channel. 5. 1 are for natural apatites that only approximately correspond to ideal compositions.\n\n86 type AB CHAP was prepared by reaction of CaCO3 (calcite) with (NH4 )H2 PO4 solution under hydrothermal conditions. 2 M (NH4 )H2 PO4 solution in a stainless steel autoclave at 250◦ C and 1 kbar for 10 days. A minor amount of 2 M NH4 OH was added to the starting solution to bring the pH to 9. The reaction product of white color with a grain size of about 10 μm was washed with distilled water and then dried in air. 2 g CaHPO4 . Each solution was boiled at approximately 100◦ C for several hours.\n\nDownload PDF sample\n\nRated 4.67 of 5 – based on 45 votes"
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https://www.educationquizzes.com/us/middle-school-6th-7th-and-8th-grade/english-language-arts/words-used-in-maths-01/ | [
"Rapid Revision For School Students",
null,
"The circumference of a circle is the distance all round the edge of it.\n\n# Words Used in Math\n\nThis English Language quiz is called 'Words Used in Math' and it has been written by teachers to help you if you are studying the subject at middle school. Playing educational quizzes is a fabulous way to learn if you are in the 6th, 7th or 8th grade - aged 11 to 14.\n\nIt costs only \\$12.50 per month to play this quiz and over 3,500 others that help you with your school work. You can subscribe on the page at Join Us\n\nPerimeter and prime number are words predominantly used in Math. Math uses various English words that are rarely used elsewhere - in fact Math almost has a language all of its own!\n\nHave a go at this quiz to get you up to speed with mathematical terms and spelling.\n\n1.\nChoose the correct meaning for the following word(s).\nConvex\nCurving inwards\nCurving left\nCurving outwards\nCurving right\n2.\nChoose the correct meaning for the following word(s).\nConcave\nCurving inwards\nCurving left\nCurving outwards\nCurving right\n3.\nChoose the correct meaning for the following word(s).\nPerimeter\nAll round the edge of a shape\nHalfway round the edge of a shape\nPart of the way round the edge of a shape\nQuarter of the way round the edge of a shape\nThe greek prefix 'peri-' means 'round'\n4.\nChoose the correct meaning for the following word(s).\nQuotient\nResult of adding one number to another\nResult of dividing one number by another\nResult of multiplying one number by another\nResult of subtracting one number from another\nThe quotient of 4 divided by 2 is 2!\n5.\nChoose the correct meaning for the following word(s).\nPerpendicular\nAt a reflex angle to a line\nAt a right angle to a line\nAt an acute angle to a line\nAt an obtuse angle to a line\nMake sure you can spell 'perpendicular' correctly\n6.\nChoose the correct meaning for the following word(s).\nDiameter\nStraight line a part of the way across a circle\nStraight line a quarter of the way across a circle\nStraight line halfway across the middle of a circle\nStraight line right across the middle of a circle\nDon't confuse 'diameter' with 'radius' (half diameter)\n7.\nChoose the correct meaning for the following word(s).\nProduct\nResult of adding two or more numbers\nResult of dividing two or more numbers\nResult of multiplying two or more numbers\nResult of subtracting two or more numbers\nThe product of 2 x 2 is 4\n8.\nChoose the correct meaning for the following word(s).\nInteger\nOne letter\nOne whole number\nThe whole\nThe whole range of letters\n'Integer' means 'whole number' (not a fraction)\n9.\nChoose the correct meaning for the following word(s).\nCircumference\nDistance all round the edge of a circle\nDistance halfway round the edge of a circle\nDistance part of the way round the edge of a circle\nDistance quarter of the way round the edge of a circle\nThe latin prefix 'circum-' means 'round'\n10.\nChoose the correct meaning for the following word(s).\nPrime number\nNumber divisible only by itself and one\nNumber divisible only by itself and two\nNumber divisible only by itself and three\nNumber divisible only by itself and four\nSome examples of prime numbers are 3, 5, 7, 11, 13\nAuthor: Sue Daish"
]
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http://ksl-web.stanford.edu/KSL_Abstracts/KSL-92-29.html | [
"KSL-92-29\n\n## Order of Magnitude Reasoning Using Logarithms\n\nReference: Nayak, P. P. Order of Magnitude Reasoning Using Logarithms. 1992.\n\nAbstract: Converting complex equations into simpler, more tractable equations usually involves approximation. Approximation is usually done by identifying and removing insignificant terms, while retaining significant ones. The significance of a term can be determined by order of magnitude reasoning. In this paper we describe NAPIER, an implemented order of magnitude reasoning system. NAPIER defines the order of magnitude of a quantity on a logarithmic scale, and uses a set of rules to propagate orders of magnitudes through equations. A novel feature of NAPIER is the way it handles non-linear simultaneous equations, using linear programming in conjunction with backtracking. We show that order of magnitude reasoning in NAPIER is, in general, intractable and then discuss an approximate reasoning technique that allow it to run fast in practice. Some of NAPIER's inference rules are heuristic, and we estimate the error introduced by their use.\n\nFull paper available as ps.\n\nJump to... [KSL] [SMI] [Reports by Author] [Reports by KSL Number] [Reports by Year]\nSend mail to: [email protected] to send a message to the maintainer of the KSL Reports."
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https://users.rust-lang.org/t/a-question-about-variables-and-constants/94398 | [
"# A question about variables and constants\n\nHello,\nIn the Rust-Lang a variable is like a constant in other programming language and its value cannot be changed unless the word 'mut' is used. Why does code 1 allow changing the value of 'number1', but code 2 shows an error?\n\nCode 1:\n\n``````fn main() {\nlet number1;\nlet number2 = 22;\nnumber1 = number2;\nprint!(\"{}\", number1);\n}``````\n\nCode 2:\n\n``````fn main() {\nlet mut number1;\nlet mut number2 = 22;\nnumber1 = number2;\nprint!(\"{}\", number1);\n}``````\n\nThank you.\n\nFor code1, `let number1;` declares an uninitialized variable. Rust allows immutable variables to be declared uninitialized and then initialized once later. This is not strictly speaking a mutation. If you try to mutate number1 again after the first initialization it will be an error.\n\nFor code2, there is no error but you are probably seeing warnings about unnecessary `mut` modifiers. As explained above, because number1 was declared uninitialized, the assignment is actually initialization not mutation, so number1 is never actually mutated, which is why the compiler thinks the `mut` declaration is not needed.\n\n2 Likes\n\nSide note: `let x = y;` doesn't mean that `x` is constant. For example, if that line is part of a loop, then `x` will hold different values for each time the loop repeats.\n\nI acknowledge that in some languages it still is referred to as \"constant\". Just note that constants (`const`) in Rust are \"really\" constant at compile time.\n\n2 Likes\n\nHello,\nThank you so much for all replies.\nI have another question. As @jbe said, the number1 is not a constant really, but why its value can only be changed once.\nFor example, why code 1 shows me an error, but code 2 doesn't:\n\nCode 1:\n\n``````fn main() {\nlet number1;\nlet number2 = 22;\nnumber1 = number2;\nprint!(\"{}\", number1);\nnumber1 = 23;\nprint!(\"{}\", number1);\n}\n``````\n\nCode 2:\n\n``````fn main() {\nlet mut number1;\nlet number2 = 22;\nnumber1 = number2;\nprint!(\"{}\", number1);\nnumber1 = 23;\nprint!(\"{}\", number1);\n}\n``````\n\nThanks.\n\nDid you read the previous answers? Assigning the first value of an uninitialized variable is not \"changing\", it's initilaization. That's allowed for any variable.\n\nThe second assignment would be mutation, which is obviously not allowed for immutable bindings, this is completely logical.\n\nYou are basically asking \"why can't a non-mut variable be mutated while a mut one can?\", and the answer is \"because that's the whole point of `mut`.\"\n\n1 Like\n\nHello,"
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https://stat.ethz.ch/pipermail/r-help/2007-April/130765.html | [
"# [R] add arrows to barchart with groups\n\nDeepayan Sarkar deepayan.sarkar at gmail.com\nFri Apr 27 19:52:09 CEST 2007\n\n```On 4/27/07, GOUACHE David <D.GOUACHE at arvalisinstitutduvegetal.fr> wrote:\n> Hello Rhelpers,\n>\n> I am trying to represent the following data (hereafter named donnees) in a barchart with a grouping variable :\n>\n>\n> site traitement date s res\n> 1 NT 17/10/2005 normal 76.2\n> 1 T 17/10/2005 normal 103.2\n> 1 NT 23/11/2005 tardif 81.6\n> 1 T 23/11/2005 tardif 98\n> 2 NT 15/10/2005 normal 72.71\n> 2 T 15/10/2005 normal 94.47\n> 2 NT 15/11/2005 tardif 79.65\n> 2 T 15/11/2005 tardif 94.7\n>\n> barchart(res~s|site,groups=traitement,data=donnees)\n>\n> What I'd like to do is for each site represent with an arrow the difference in value of variable res between normal and tardif values of variable s.\n> I've found one way of doing it:\n>\n>\n> trellis.focus(\"panel\",1,1)\n> xx<-trellis.panelArgs()\\$x\n> yy<-trellis.panelArgs()\\$y\n> panel.arrows(as.numeric(xx)[c(1,3)]-0.1,yy[c(1,3)],as.numeric(xx)[c(1,3)]-0.1,yy[c(2,4)],lwd=2,code=3)\n> panel.text(as.numeric(xx)[c(1,3)]-0.35,c(87,87),paste(yy[c(2,4)]-yy[c(1,3)],\"\\nq/ha\"),font=2)\n> trellis.focus(\"panel\",2,1)\n> xx<-trellis.panelArgs()\\$x\n> yy<-trellis.panelArgs()\\$y\n> panel.arrows(as.numeric(xx)[c(1,3)]-0.1,yy[c(1,3)],as.numeric(xx)[c(1,3)]-0.1,yy[c(2,4)],lwd=2,code=3)\n> panel.text(as.numeric(xx)[c(1,3)]-0.35,c(87,87),paste(yy[c(2,4)]-yy[c(1,3)],\"\\nq/ha\"),font=2)\n> trellis.unfocus()\n>\n> But I would prefer doing this within a custom panel function so I can apply it more generally, and I haven't been able to figure out how...\n> Could anyone give me a hand?\n\nThe obvious analog (just copy/pasting your code) is:\n\nmy.panel <- function(x, y, ...)\n{\npanel.barchart(x, y, ...)\nxx <- x\nyy <- y\npanel.arrows(as.numeric(xx)[c(1,3)]-0.1, yy[c(1,3)],\nas.numeric(xx)[c(1,3)]-0.1, yy[c(2,4)],\nlwd = 2, code = 3)\npanel.text(as.numeric(xx)[c(1,3)] - 0.35, c(87,87),\npaste(yy[c(2,4)] - yy[c(1,3)], \"\\nq/ha\"),\nfont=2)\n}\n\nand this seems to work:\n\nbarchart(res~s|site,groups=traitement,data=donnees,\npanel = my.panel)\n\nbarchart(res~s|site,groups=traitement,data=donnees,\npanel = my.panel,\norigin = 0)\n\nI'm not sure what else you are looking for and what you mean by \"more\ngeneral\". For example, it's not clear what you want to happen If you\nhave more than 2 levels in 'groups', or if the second bar is not\nalways higher than the first.\n\n-Deepayan\n\n```"
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https://www.excelforum.com/excel-charting-and-pivots/1203574-adding-calculated-average-line-to-chart.html | [
"# Adding Calculated Average Line to Chart\n\n1. ##",
null,
"Adding Calculated Average Line to Chart\n\nI hope someone can help because I'm starting to hate charts. I'm basically trying to add an average line to my line-graph by using calculated averages instead of the trend-line tool.\n\nI have two tables. Table 1 has Column A (a number) and Column B (also a number). I've created a line graph of this table using Column A as the x-axis (horizontal) and Column B as the data points.\n\nI have another table that calculates the averages of Column B if Column A falls within a certain range.",
null,
"``Please Login or Register to view this content.``\n(simplified code since that's not the point of this post.)\n\nI want to add this line to the chart. It's basically the same as adding a trend line, except I need to have the calculated number since the ranges vary. I calculated the median of the ranges and used that as my x-axis and the averages as my data points.\n\nThe result:\nScreen Shot 2017-10-05 at 7.35.37 PM.png\n\nThe average line is the blue one. I think the issue is that I only have 20 or so \"average\" data points and thousands of the other data points. But I don't understand why the average line doesn't stretch across the graph since I've used the median of the ranges and those fall all along the x-axis (ie if I have an x-axis that goes from 1-10 and I plot points at 3, 5, and 7 I would think it should stretch out.)\n\nNote: I also plan to find the minimum of each average range and add that as well (but I will use a formula to find the exact point that instance happens rather than the median of the average range). So I hope the solution works in both instances.",
null,
"",
null,
"Register To Reply\n\n2. ## Re: Adding Calculated Average Line to Chart\n\nIt is difficult to see exactly what is going on based on a limited description of the data and a picture of the messed up chart.\n\nMy first thought, where column A is a number -- is there a reason you are using a line chart rather than an XY scatter chart? I find that it is often much easier to have different series with different x values on a scatter chart than other chart types. I wonder if your problem will resolve itself by simply changing chart types.",
null,
"",
null,
"Register To Reply\n\n3. ## Re: Adding Calculated Average Line to Chart\n\nYep, that fixed it! (with a bit of tweaking)\n\nThank you!! I was getting pretty frustrated. I'll remember to stay away from line-graphs in the future.",
null,
"",
null,
"Register To Reply\n\n4. ## Re: Adding Calculated Average Line to Chart\n\nI would not that it isn't about staying away from line-graphs. It's more about understanding the differences between line and scatter charts and when to use each one. If you are interested, this could be an interesting article to read: https://peltiertech.com/line-charts-vs-xy-charts/",
null,
"",
null,
"Register To Reply\n\n5. ## Re: Adding Calculated Average Line to Chart\n\nThank you, I'll check out the article.",
null,
"",
null,
"Register To Reply\n\nThere are currently 1 users browsing this thread. (0 members and 1 guests)",
null,
""
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https://eccc.weizmann.ac.il/report/2015/027/ | [
"",
null,
"",
null,
"Under the auspices of the Computational Complexity Foundation (CCF)",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"REPORTS > DETAIL:\n\n### Revision(s):\n\nRevision #1 to TR15-027 | 17th March 2015 09:06\n\n#### Cryptographic Hardness of Random Local Functions -- Survey",
null,
"Revision #1\nAuthors: Benny Applebaum\nAccepted on: 17th March 2015 09:06\nKeywords:\n\nAbstract:\n\nConstant parallel-time cryptography allows to perform complex cryptographic tasks at an ultimate level of parallelism, namely, by local functions that each of their output bits depend on a constant number of input bits. A natural way to obtain local cryptographic constructions is to use \\emph{random local functions} in which each output bit is computed by applying some fixed $d$-ary predicate $P$ to a randomly chosen $d$-size subset of the input bits.\n\nIn this work, we will study the cryptographic hardness of random local functions. In particular, we will survey known attacks and hardness results, discuss different flavors of hardness (one-wayness, pseudorandomness, collision resistance, public-key encryption), and mention applications to other problems in cryptography and computational complexity. We also present some open questions with the hope to develop a systematic study of the cryptographic hardness of local functions.\n\n### Paper:\n\nTR15-027 | 25th February 2015 09:39\n\n#### Cryptographic Hardness of Random Local Functions -- Survey\n\nTR15-027\nAuthors: Benny Applebaum\nPublication: 25th February 2015 09:46\nConstant parallel-time cryptography allows to perform complex cryptographic tasks at an ultimate level of parallelism, namely, by local functions that each of their output bits depend on a constant number of input bits. A natural way to obtain local cryptographic constructions is to use \\emph{random local functions} in which each output bit is computed by applying some fixed $d$-ary predicate $P$ to a randomly chosen $d$-size subset of the input bits."
]
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https://freefoodtips.com/how-to-measure-milk-volume-with-cups/ | [
"We will now calculate how many milliliters of milk are in one Metric, Imperial and US customary cup; how to measure necessary amount of milk by Metric, Imperial and US customary cups.\n\n## How many milliliters are in 1 Metric cup of milk?\n\nOne Metric cup holds 250 ml of milk.\n\n### How many fluid ounces of milk does 1 Metric cup hold?\n\nOne Metric cup holds 8.45 fl-oz of milk.\n\n### How many tablespoons of milk are in 1 Metric cup?\n\n1 Metric cup holds 16.5 tablespoons of milk.\n\n#### Milliliters\n\n• Three-fourths (¾) of a metric cup of milk is how much in milliliters? 0.75 of a Metric cup of milk = 188 ml of milk.\n• Two-thirds (2/3) of a metric cup of milk is how much in milliliters? 2/3 of a metric cup of milk = 167 ml of milk.\n• Half (½) of a metric cup of milk is how much in milliliters? 0.5 (½) of a metric cup of milk = 125 ml of milk.\n\n#### Ounces\n\n• Three-fourths (¾) of a metric cup of milk is how much in fluid ounces? 0.75 of a metric cup of milk = 6.34 fl-oz of milk.\n• Two-thirds (2/3) of a metric cup of milk is how much in fluid ounces? 2/3 of a metric cup of milk = 5.64 fl-oz of milk.\n• Half (½) of a metric cup of milk is how much in fluid ounces? 0.5 (½) of a metric cup of milk = 4.23 fl-oz of milk.\n\n### How to measure milk volume with Metric cups?\n\n• How many Metric cups are in 1 liter of milk? 1 liter (or 1000 milliliters) of milk = 4 Metric cups of milk.\n• How many Metric cups are in 900 milliliters of milk? 900 milliliters of milk = 3.5 Metric cups + 1.5 tablespoons of milk.\n• How many Metric cups are in 800 milliliters of milk? 800 milliliters of milk = 3 Metric cups + 3.5 tablespoons of milk.\n• How many Metric cups are in 700 milliliters of milk? 700 milliliters of milk = 2.5 Metric cups + 5 tablespoons of milk.\n• How many Metric cups are in 650 milliliters of milk? 650 milliliters of milk = 2.5 Metric cups + 1.5 tablespoons of milk.\n• How many Metric cups are in 600 milliliters of milk? 600 milliliters of milk = 2 Metric cups + 6.5 tablespoons of milk.\n• How many Metric cups are in 500 milliliters of milk? 500 milliliters of milk = 2 Metric cups of milk.\n• How many Metric cups are in 400 milliliters of milk? 400 milliliters of milk = 1.5 Metric cups + 3.5 tablespoons of milk.\n• How many Metric cups are in 300 milliliters of milk? 300 milliliters of milk = 1 Metric cup + 3.5 tablespoons of milk.\n• How many Metric cups are in 280 milliliters of milk? 280 milliliters of milk = 1 Metric cup + 2 tablespoons of milk.\n• How many Metric cups are in 250 milliliters of milk? 250 milliliters of milk = 1 Metric cup of milk.\n• How many Metric cups are in 200 milliliters of milk? 200 milliliters of milk = 0.5 Metric cup + 5 tablespoons of milk.\n• How many Metric cups are in 160 milliliters of milk? 160 milliliters of milk = 0.5 Metric cup + 2 tablespoons + 1 teaspoon of milk.\n• How many Metric cups are in 150 milliliters of milk? 150 milliliters of milk = 0.5 Metric cup + 1 tablespoon + 2 teaspoons of milk.\n• How many Metric cups are in 100 milliliters of milk? 100 milliliters of milk = 6.5 tablespoons of milk or 6 tablespoons + 2 teaspoons of milk.\n\n## How many milliliters of milk are in 1 Imperial cup?\n\nOne Imperial cup holds 284 ml of milk.\n\n### How many fluid ounces of milk does 1 Imperial cup hold?\n\nOne Imperial cup holds 9.60 fl-oz of milk.\n\n### How many tablespoons of milk are in 1 Imperial cup?\n\n1 Imperial cup holds almost 19 full tablespoons of milk.\n\n#### Milliliters\n\n• Three-fourths (¾) of an Imperial cup of milk is how much in milliliters? 0.75 of an Imperial cup of milk = 213 ml of milk.\n• Two-thirds (2/3) of an Imperial cup of milk is how much in milliliters? 2/3 of an Imperial cup of milk = 189 ml of milk.\n• Half (½) of an Imperial cup of milk is how much in milliliters? 0.5 (½) of an Imperial cup of milk = 142 ml of milk.\n\n#### Ounces\n\n• Three-fourths (¾) of an Imperial cup of milk is how much in fluid ounces? 0.75 of an Imperial cup of milk = 7.2 fl-oz of milk.\n• Two-thirds (2/3) of an Imperial cup of milk is how much in fluid ounces? 2/3 of an Imperial cup of milk = 6.4 fl-oz of milk.\n• Half (½) of an Imperial cup of milk is how much in fluid ounces? 0.5 (½) of an Imperial cup of milk = 4.8 fl-oz of milk.\n\n### How to measure milk volume with Imperial cups?\n\n• How many Imperial cups are in 1 liter of milk? 1 liter (or 1000 milliliters) of milk = 3.5 Imperial cups of milk.\n• How many Imperial cups are in 900 milliliters of milk? 900 milliliters of milk = 3 Imperial cups + 3 tablespoons of milk.\n• How many Imperial cups are in 800 milliliters of milk? 800 milliliters of milk = 2.5 Imperial cups + 6 tablespoons of milk.\n• How many Imperial cups are in 700 milliliters of milk? 700 milliliters of milk = 2.5 Imperial cups – 2 teaspoons of milk.\n• How many Imperial cups are in 650 milliliters of milk? 650 milliliters of milk = 2 Imperial cups + 5.5 tablespoons of milk.\n• How many Imperial cups are in 600 milliliters of milk? 600 milliliters of milk = 2 Imperial cups + 2 tablespoons of milk.\n• How many Imperial cups are in 500 milliliters of milk? 500 milliliters of milk = 1.5 Imperial cups + 8 tablespoons of milk.\n• How many Imperial cups are in 400 milliliters of milk? 400 milliliters of milk = 1 Imperial cup + 8 tablespoons of milk.\n• How many Imperial cups are in 300 milliliters of milk? 300 milliliters of milk = 1 Imperial cup + 1.5 tablespoons of milk.\n• How many Imperial cups are in 280 milliliters of milk? 280 milliliters of milk = 1 Imperial cup – 1 teaspoon of milk.\n• How many Imperial cups are in 250 milliliters of milk? 250 milliliters of milk = 0.5 Imperial cup + 7 tablespoons of milk.\n• How many Imperial cups are in 200 milliliters of milk? 200 milliliters of milk = 0.5 Imperial cup + 4 tablespoons of milk.\n• How many Imperial cups are in 160 milliliters of milk? 160 milliliters of milk = 0.5 Imperial cup + 1 tablespoon of milk.\n• How many Imperial cups are in 150 milliliters of milk? 150 milliliters of milk = 0.5 Imperial cup + 1 teaspoon of milk.\n• How many Imperial cups are in 100 milliliters of milk? 100 milliliters of milk = 6.5 tablespoons of milk or 6 tablespoons + 2 teaspoons of milk.\n\n## How many milliliters are in 1 US customary cup of milk?\n\nOne US cup holds 236.58 ml of milk.\n\n### How many fluid ounces of milk does 1 US customary cup hold?\n\nOne US cup holds 8 fl- oz of milk.\n\n### How many tablespoons of milk are in 1 US customary cup?\n\n1 US customary cup holds 15.5 tablespoons of milk.\n\n#### Milliliters\n\n• Three-fourths (¾) of a US cup of milk is how much in milliliters? 0.75 of a US cup of milk = 177 ml of milk.\n• Two-thirds (2/3) of a US customary cup of milk is how much in milliliters? 2/3 of a US cup of milk = 158 ml of milk.\n• Half (½) of a US cup of milk is how much in milliliters? 0.5 (½) of a US cup of milk = 118 ml of milk.\n\n#### Ounces\n\n• Three-fourths (¾) of a US cup of milk is how much in fluid ounces? 0.75 of a US cup of milk = 6 fl-oz of milk.\n• Two-thirds (2/3) of a US cup of milk is how much in fluid ounces? 2/3 of a US cup of milk = 5.34 fl-oz of milk.\n• Half (½) of a US cup of milk is how much in fluid ounces? 0.5 (½) of a US cup of milk = 4 fl-oz of milk.\n\n### How to measure milk volume with US customary cups?\n\n• How many US customary cups are in 1 liter of milk? 1 liter (or 1000 milliliters) of milk = 4 US cups + 3.5 tablespoons of milk.\n• How many US customary cups are in 900 milliliters of milk? 900 milliliters of milk = 3.5 US cups + 4.5 tablespoons of milk.\n• How many US customary cups are in 800 milliliters of milk? 800 milliliters of milk = 3 US cups + 6 tablespoons of milk.\n• How many US customary cups are in 700 milliliters of milk? 700 milliliters of milk = 3 US cups – 2 teaspoons of milk.\n• How many US customary cups are in 650 milliliters of milk? 650 milliliters of milk = 2.5 US cups + 4 tablespoons of milk.\n• How many US customary cups are in 600 milliliters of milk? 600 milliliters of milk = 2.5 US cups + 2 teaspoons of milk.\n• How many US customary cups are in 500 milliliters of milk? 500 milliliters of milk = 2 US cups + 2 tablespoons of milk.\n• How many US customary cups are in 400 milliliters of milk? 400 milliliters of milk = 1.5 US cups + 3 tablespoons of milk.\n• How many US customary cups are in 300 milliliters of milk? 300 milliliters of milk = 1 US cup + 4 tablespoons of milk.\n• How many US customary cups are in 280 milliliters of milk? 280 milliliters of milk = 1 US cup + 3 tablespoons of milk.\n• How many US customary cups are in 250 milliliters of milk? 250 milliliters of milk = 1 US cup + 1 tablespoon of milk.\n• How many US customary cups are in 200 milliliters of milk? 200 milliliters of milk = 0.5 US cup + 5.5 tablespoons of milk.\n• How many US customary cups are in 160 milliliters of milk? 160 milliliters of milk = 0.5 US cups + 2.5 tablespoons of milk.\n• How many US customary cups are in 150 milliliters of milk? 150 milliliters of milk = 0.5 US cups + 2 tablespoons of milk.\n• How many US customary cups are in 100 milliliters of milk? 100 milliliters of milk = 6.5 tablespoons of milk or 6 tablespoons + 2 teaspoons of milk.\n\nWe tried to give equivalent to most popular volumes of milk in recipes; do let us know if it was helpful and you used our calculations while cooking. Please, leave comments in the section below.\n\nPin It\nWe use cookies in order to give you the best possible experience on our website. By continuing to use this site, you agree to our use of cookies."
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https://answers.everydaycalculation.com/divide-fractions/15-9-divided-by-8-10 | [
"# Answers\n\nSolutions by everydaycalculation.com\n\n## Divide 15/9 with 8/10\n\n1st number: 1 6/9, 2nd number: 8/10\n\n15/9 ÷ 8/10 is 25/12.\n\n#### Steps for dividing fractions\n\n1. Find the reciprocal of the divisor\nReciprocal of 8/10: 10/8\n2. Now, multiply it with the dividend\nSo, 15/9 ÷ 8/10 = 15/9 × 10/8\n3. = 15 × 10/9 × 8 = 150/72\n4. After reducing the fraction, the answer is 25/12\n5. In mixed form: 21/12\n\nMathStep (Works offline)",
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"Download our mobile app and learn to work with fractions in your own time:\nAndroid and iPhone/ iPad\n\n#### Divide Fractions Calculator\n\n÷\n\n© everydaycalculation.com"
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https://www.hindawi.com/journals/jam/2014/419103/ | [
"#### Abstract\n\nBy the weighted ergodic function based on the measure theory, we study pseudo asymptotic behavior of mild solution for nonautonomous integrodifferential equations with nondense domain. The existence and uniqueness of -pseudo antiperiodic (-pseudo periodic, -pseudo almost periodic, and -pseudo automorphic) solution are investigated. Some interesting examples are presented to illustrate the main findings.\n\n#### 1. Introduction\n\nThe study of pseudo asymptotic behavior of solution is one of the most interesting and important topics in the qualitative theory of differential equations. Much work has been done to investigate the existence of pseudo antiperiodic, pseudo periodic, pseudo almost periodic, and pseudo almost automorphic solution for differential equations . Recently, Blot et al. [6, 7] used the results of the measure theory to establish -ergodic and introduce the new concepts of -pseudo almost periodic and -pseudo almost automorphic function, which are more general than pseudo almost periodic and pseudo almost automorphic function, respectively. They developed some results like completeness and composition theorems to investigate differential equations in Banach space.\n\nIntegrodifferential equations play a crucial role in qualitative theory of differential equations due to their application to physics, engineering, biology, and other subjects. This type of equations has received much attention in recent years and the general asymptotic behavior of solution is at present an active source of research.\n\nIn this paper, we study pseudo asymptotic behavior of solution to the following nonautonomous integrodifferential equations with nondense domain: where the linear operators have a domain not necessarily dense in Banach space and satisfy “Acquistapace-Terreni” conditions and , , and are continuous functions.\n\nSome recent contributions on almost periodic, almost automorphic, pseudo almost periodic, and pseudo almost automorphic solution to integrodifferential equations of the form (1) in the case are constant [4, 813]. However, for the nonautonomous case, that is, (1), the study of pseudo asymptotic behavior of solution is rare . In this paper, we will make use of the so-called “Acquistapace-Terreni” conditions associated with exponential dichotomy, fixed point theorem to derive some sufficient conditions to the existence and uniqueness of -pseudo antiperiodic (-pseudo periodic, -pseudo almost periodic, and -pseudo almost automorphic) mild solution to (1).\n\nThe paper is organized as follows. In Section 2, we recall some fundamental results about the notion of -class of functions including composition theorem. Section 3 is devoted to pseudo asymptotic behavior of mild solution to nonautonomous integrodifferential equations with nondense domain. In Section 4, we provide some examples to illustrate our main results.\n\n#### 2. Preliminaries and Basic Results\n\nLet and be two Banach spaces and , , , and stand for the set of natural numbers, integers, real numbers, and complex numbers, respectively. For being a linear operator on , , , , and stand for the domain, the resolvent set, the resolvent, and spectrum of . In order to facilitate the discussion below, we further introduce the following notations:(i) (resp., ): the Banach space of bounded continuous functions from to (resp., from to ) with the supremum norm;(ii) (resp. ): the set of continuous functions from to (resp., from to );(iii): the Banach space of bounded linear operators from to endowed with the operator topology; in particular, we write when ;(iv): the space of all classes of equivalence (with respect to the equality almost everywhere on ) of measurable functions such that .\n\n##### 2.1. Evolution Family and Exponential Dichotomy\n\nDefinition 1. A family of bounded linear operators on a Banach space is called a strong continuous evolution family if(i) and for all and ;(ii)the map is continuous for all and .\n\nDefinition 2. An evolution family on a Banach space is called hyperbolic (or has an exponential dichotomy) if there exist projections , , uniformly bounded and strongly continuous in and constants , such that(i) for and ,(ii)the restriction of is invertible for (and set ),(iii)for and . Here and next we set .\n\nRemark 3. Exponential dichotomy is a classical concept in the study of long-time behaviour of evolution equations. If for , then is exponentially stable. One can see for more details.\nIf is hyperbolic, then is called Green’s function corresponding to , and\n\n##### 2.2. μ-Ergodic and Functions by Measure Theory\n\ndenotes the Lebesgue -field of and stands for the set of all positive measures on satisfying and for all . We formulate the following hypothesis.For all , there exist and a bounded interval such that\n\nDefinition 4 (see ). Let . A function is said to be -ergodic if Denote by the set of such functions.\n\nLemma 5 (see ). Let and satisfy ; then is a translation invariant.\n\nDefinition 6. A function is said to be antiperiodic if there exists a with the property that for all . If there exists a less positive with this property, it is called the antiperiodic of . The collection of those functions is denoted by .\n\nDefinition 7. A function is said to be periodic if there exists a with the property that for all . If there exists a less positive with this property, it is called the periodic of . The collection of those -periodic functions is denoted by .\n\nDefinition 8 (see ). A function is said to be almost periodic if for each , there exists an , such that every interval of length contains a number with the property that for all . Denote by the set of such functions.\n\nDefinition 9 (see ). A function is said to be almost automorphic if for every sequence of real numbers , there exists a subsequence such that is well defined for each , and for each . Denote by the set of such functions.\n\nNext, we recall the -class of functions by the measure theory.\n\nDefinition 10. Let . A function is said to be -pseudo antiperiodic if it can be decomposed as , where and . Denote by the collection of such functions.\n\nDefinition 11. Let . A function is said to be a -pseudo periodic if it can be decomposed as , where and . Denote by the collection of such functions.\n\nDefinition 12 (see ). Let . A function is said to be -pseudo almost periodic if it can be decomposed as , where and . Denote by the collection of such functions.\n\nDefinition 13 (see ). Let . A function is said to be -pseudo almost automorphic if it can be decomposed as , where and . Denote by the collection of such functions.\n\nRemark 14. (i) If the measure is the Lebesgue measure, then , , , and are the following functions: pseudo antiperiodic ( ), pseudo periodic ( ), pseudo almost periodic ( ), and pseudo almost automorphic ( ), respectively. One can see [6, 7, 22] for more details.\n(ii) Let a.e. on for the Lebesgue measure. denotes the positive measure defined by where denotes the Lebesgue measure on ; then , , , and are the weighted class of functions: weighted pseudo antiperiodic ( ), weighted pseudo periodic ( ), weighted pseudo almost periodic ( ), and weighted pseudo almost automorphic ( ), respectively.\nLet It is not difficult to see that if and only if it can be decomposed as , where and .\n\nDefinition 15. Let ; is said to be equivalent to if there exist constants and a bounded interval (eventually ) such that\n\nSimilarly as the proof of [6, 7], we have the following results for the class of functions .\n\nLemma 16. Let ; then the following properties hold:(i) if ;(ii) if , ;(iii) is a Banach space with the supremum norm ;(iv).\n\nLemma 17. Let . If , then .\n\nTheorem 18. Let , and satisfy the following:(i) is uniformly continuous on each compact set in with respect to the second and third variables , ;(ii)for all bounded subsets of , is bounded on .Then if .\n\nCorollary 19. Let and , and there exists a constant such that then if .\n\n#### 3. Nonautonomous Integrodifferential Equations\n\nThis section is devoted to pseudo asymptotic behavior of mild solution to (1). In this section, we make the following assumptions.()There exist constants , , , and with such that for , .()The evolution family generated by has an exponential dichotomy with constants and ; dichotomy projections , ; and Green’s function .()Consider and for some positive constants .()There exists a constant such that ()There exists a constant such that () and satisfies .\n\nRemark 20. is usually called “Acquistapace-Terreni” conditions, which was first introduced in and widely used to study nonautonomous differential equations in [16, 17, 2325]. If holds, there exists a unique evolution family on , which governs the homogeneous version of (1) .\n\nBefore starting our main results, we recall the definition of the mild solution to (1).\n\nDefinition 21 (see ). A mild solution of (1) is a continuous function satisfying for all , .\n\nLemma 22. Assume that and , , and hold; then\n\nProof. First, we show that is well defined. In fact, if , so . By (2), it follows that is integrable over and is integrable over for .\nNote that . Next, we show that In fact, for , by using Fubini’s theorem, one has where Since and satisfies , by Lemma 5, it follows that , for each ; hence, and for all . Since and , then by using the Lebesgue dominated convergence theorem.\n\nLemma 23. Assume that and and hold; then\n\nProof. Similarly as the proof of Lemma 22, it is not difficult to see that is well defined and .\n(i) Note that , .\nLet , where and ; then where\nNote that ; then hence, .\nBy using Fubini’s theorem, one has Since , it follows that for each by Lemma 5; hence, by using the Lebesgue dominated convergence theorem. Hence, .\n(ii) Note that , .\nLet , where and ; then hence, . Since by (i), hence, .\n(iii) Note that , .\nBy , . Since by (i), hence, .\n(iv) Note that , .\nSimilarly as the proof of , . Since by (i); hence, .\n\n##### 3.1. Pseudo Almost Automorphic Perturbation\n\nIn this subsection, we investigated the existence and uniqueness of pseudo almost automorphic mild solution of (1).\n\nFirst, we introduce the concept of bi-almost automorphic function.\n\nDefinition 24 (see ). A function is called bi-almost automorphic if for every sequence of real numbers , there exists a subsequence such that is well defined for each , and for each . The collection of all such functions will be denoted by .\nNow, we make the following assumptions:() for each ;() and .\n\nLemma 25 (see ). Assume that , , , and hold; then\n\nTheorem 26. Suppose , , and hold; if , then (1) has a unique mild solution such that\n\nProof. First, we show that (1) admits a unique bounded solution given by (31), which is similar to the proof of [26, Theorem 3.3]. For , it is clear that by Lemma 23 and Corollary 19; then . By the definition of exponential dichotomy of , it is not difficult to see that (31) is well defined for each .\nFor all , , then that is, is a mild solution of (1). To prove the uniqueness, let be another mild solution of (1); then by the exponential dichotomy of , Similarly, So,\nNext, define the operator by By Lemma 22, Lemma 23, Lemma 25, and Corollary 19, maps into itself.\nFor any , By the Banach contraction mapping principle, has a unique fixed point in , which is the unique mild solution to (1).\n\nNext, consider the following nonautonomous Volterra integrodifferential equations: where the linear operators have a domain not necessarily dense in and satisfy “Acquistapace-Terreni” conditions and is a continuous function.\n\nFor the pseudo almost automorphy of (40), one has the following.\n\nTheorem 27. Suppose , , , , and hold, and satisfies() and , where is a constant.Then (40) has a unique mild solution if .\n\nProof. Let and define Similarly as the proof of Theorem 26, is well defined and (42) is a mild solution of (40).\nFor any , By the Banach contraction mapping principle, has a unique fixed point in , which is the unique mild solution to (40).\n\n##### 3.2. Pseudo Almost Periodic Perturbation\n\nIn this subsection, we investigated the existence and uniqueness of pseudo almost periodic mild solution of (1) and (40). We make the following assumptions:() for in ;() and .\n\nSimilarly as the proof of , we have the following results.\n\nLemma 28. Assume that , , , and hold; then\n\nBy Lemma 22, Lemma 23, and Lemma 28, similarly as the proof of Theorem 26, Theorem 27, the following results hold.\n\nTheorem 29. Suppose , , and hold; then (1) has a unique mild solution if .\n\nTheorem 30. Suppose , , , , and hold, and satisfies , then (40) has a unique mild solution if .\n\n##### 3.3. Pseudo Periodic (Antiperiodic) Perturbation\n\nIn this subsection, we investigated the existence and uniqueness of pseudo periodic (antiperiodic) mild solution of (1), (40). We make the following assumptions:()there exists"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.90095806,"math_prob":0.97225857,"size":12831,"snap":"2023-40-2023-50","text_gpt3_token_len":2972,"char_repetition_ratio":0.15560926,"word_repetition_ratio":0.21196893,"special_character_ratio":0.24214792,"punctuation_ratio":0.17142858,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9982449,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-25T00:47:07Z\",\"WARC-Record-ID\":\"<urn:uuid:25ca037c-9a61-4e73-ad35-82b6b762a428>\",\"Content-Length\":\"1057996\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:21e84ef5-68d2-4281-b513-c9d7d5a46753>\",\"WARC-Concurrent-To\":\"<urn:uuid:c5694962-f15e-4c61-a655-eeb8add6c643>\",\"WARC-IP-Address\":\"104.18.40.243\",\"WARC-Target-URI\":\"https://www.hindawi.com/journals/jam/2014/419103/\",\"WARC-Payload-Digest\":\"sha1:3U5BBEHGK34WWDCBLMYKLX2LXVCEIEAW\",\"WARC-Block-Digest\":\"sha1:V2USJRD3AVNPQZ323JGBSSNIPJUTVMMW\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506669.96_warc_CC-MAIN-20230924223409-20230925013409-00682.warc.gz\"}"} |
https://www.takshilalearning.com/ratio-and-proportion-methods-tricks/ | [
"",
null,
"# Ratio and Proportion Aptitude Questions and Answers Methods Shortcut Tricks",
null,
"## Ratio and Proportion Aptitude Questions and Answers\n\nBanking online Classes : Takshila Learning focusing mainly on providing online banking coaching to our students as well as online test series for bank po, mock test papers for IBPS PO, IBPS Clerk, SBI PO, SBI Clerk, RRB and many other competitive exams. We aim to provide Quality & result oriented education & guidance to the aspirants for better employment opportunities. You need not worry how to prepare for bank exams, we ensure that you get complete guidance for it.We regularly offer to our aspirant’s different blogs and articles on important topics, news, and information related to banking exams. In this article, the following topic ‘Ratio &Proportion’ from Quantitative Aptitude section is explained.\n\n## Ratio and Proportion Aptitude Questions and Answers\n\nRatio & Proportion is an important topic from Banking exams point of view as 1-2 questions are asked from this topic directly. Further indirect questions which make use of the concept of Ratio & Proportion are also frequently asked. Ratio & Proportion also plays an important role in D.I. based problems which makes this chapter very important for banking aspirants.\n\nQ:1 If A: B = 5: 7\n\nB: C = 3: 4, then what is A: B: C?\n\n(a) 4: 7: 9 (b) 15: 21: 28 (c) 3: 8: 15 (d) 4: 9: 15 (e) None of these\n\nSol: A: B = 5: 7……..× 3\n\nB: C = 3: 4……..×7 (We will make B equal in both the ratios)\n\nSo, A: B = 15: 21\n\nB: C = 21: 28\n\nTherefore, A: B: C = 15: 21: 28\n\nAns (b)\n\nQ:2 The ratio of two numbers is 5: 4. If their sum is 135, Find the smaller of the two numbers.\n\n(a) 30 (b) 40 (c) 55 (d) 65 (e) None of these\n\nSol: Let the two numbers be 5x and 4x.\n\nSo, 5x + 4x = 135\n\n9x = 135\n\nSo, x = 15\n\nTherefore, smaller number = 4 × 15 = 60\n\nAns (e)\n\nQ:3 An employer reduces the number of his employees in the ratio 9:8 & increases their wages in the ratio 14:15. If the original wage bill was Rs 18,900, Find the ratio in which the wage bill is decreased.\n\n(a) 21:20 (b) 36:41 (c) 5:8 (d) 32:37 (e) None of these\n\nSol: Let the initial no of employees be 9x& the employer gives Rs 14y as a wage to each.\n\nSo, 9x × 14y = 18900\n\nSo, xy = 150\n\nThe new bill is 8x × 15y = 120 xy\n\n= 120 × 150 = 18000\n\nSo, the required ratio 18900:18000\n\n= 21:20\n\nAns (a)\n\nFor online mock test papers of Quantitative Aptitude & more, click online bank coaching classes.\n\nQ:4 The cost of a piece of stone worth Rs 10,872 fell & broke into 3 pieces, the weights of which are proportional to 1:2:3. The value of each stone is directly proportional to the square of its weight. Find the loss in the value caused by the breakage.\n\n(a) 3624 (b) 4228 (c) 6644 (d) 7510(e) None of these\n\nSol: Let the initial weight of the stone (before breaking) be 6w\n\nSo, Weights of the broken stones are w, 2w & 3w resp.\n\nThe initial value of the stone is Rs. 10,872\n\n10,872 α (6w)2\n\nSo, 10,872= 36w2k\n\nW2k = 10,872 / 36 ……..1\n\nTotal value of the pieces of the stone = (12 + 22 + 32)w2k\n\n= 14w2k\n\nLoss in the value = 36w2k – 14 w2k\n\n= 22w2k\n\n= 22 × 10,872 / 36 ……From 1\n\n= 22 × 302\n\n= 6644\n\nAns (c)\n\nQ:5 If the income of Amit & Mohit is in the ratio 4:5 & their expenses are in the ratio of 6:7. Find the ratio of their savings, if Amit manages to save one-fourth of his salary.\n\n(a) 4:5 (b) 7:3 (c) 5:8 (d) 3:2 (e) None of these\n\nSol: Let the salary of Amit&Mohit be 4x & 5x resp.\n\n&their expenses be 6y & 7y resp.\n\nSavings of Amit = 4x- 6y\n\nSavings of Mohit = 5x- 7y\n\n1. T. Q\n\n4x- 6y = 4x / 4\n\n3x = 6y\n\nx = 2y\n\nRatio of their savings: 4(2y) – 6y / 5(2y) – 7y\n\n= 8y – 6y / 10y – 7y\n\n= 2y / 3y\n\n= 2: 3\n\nAns (e)\n\nFor more Ratio and Proportion Aptitude Questions and Answers, online bank exam coaching, SSC exam preparation, job alerts, study material,previous year papers and much more,register with Takshila Learning.\n\nFollow us on a Social media\n\nFollow us on Blogarama\n\n## NCERT Solutions for Class 6 Science Fun with Magnets Chapter 13",
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"https://i2.wp.com/www.takshilalearning.com/wp-content/uploads/2017/09/ratio-and-proportion-banking.jpg",
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"https://www.takshilalearning.com/wp-content/plugins/wp-contact-slider/img/delete-sign.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8913621,"math_prob":0.9709938,"size":3782,"snap":"2019-13-2019-22","text_gpt3_token_len":1289,"char_repetition_ratio":0.10693488,"word_repetition_ratio":0.028871391,"special_character_ratio":0.35589635,"punctuation_ratio":0.15172414,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9679046,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-20T00:52:29Z\",\"WARC-Record-ID\":\"<urn:uuid:9785ee7e-26ac-4bf2-a917-f1dfb117ae14>\",\"Content-Length\":\"161301\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:021ff367-814e-469b-9dfb-f779fbcfcccb>\",\"WARC-Concurrent-To\":\"<urn:uuid:91c072a2-766f-456c-acc4-81e1e032d8a8>\",\"WARC-IP-Address\":\"104.28.16.72\",\"WARC-Target-URI\":\"https://www.takshilalearning.com/ratio-and-proportion-methods-tricks/\",\"WARC-Payload-Digest\":\"sha1:E3R7X6EOZZXQVZYIT4WM3SE2YMVDGIQK\",\"WARC-Block-Digest\":\"sha1:F3BQ4PVXWIBWC5OERTNGKT4TPU7V2OBZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202188.9_warc_CC-MAIN-20190320004046-20190320030046-00517.warc.gz\"}"} |
https://www.clutchprep.com/chemistry/practice-problems/116716/an-aqueous-solution-contains-4-6-nacl-by-mass-calculate-the-molality-of-the-solu | [
"# Problem: An aqueous solution contains 4.6 % NaCl by mass. Calculate the molality of the solution.\n\n###### FREE Expert Solution\n\nWe’re being asked to calculate the molality of an aqueous solution that contains 4.6% NaCl by mass.\n\nWhen calculating for molality, we use the following equation:\n\nAqueous solution → water is the solvent\nSolute → NaCl\n\nWe will calculate the molality of the solution using the following steps:\n\nStep 1: Determine the composition of the solution.\nStep 2: Calculate the moles of the solute.\nStep 3: Calculate the mass of the solvent (in kg).\nStep 4: Calculate the molality of the solution.\n\nStep 1: Determine the composition of the solution.\n\n100% (236 ratings)",
null,
"###### Problem Details\nAn aqueous solution contains 4.6 % NaCl by mass. Calculate the molality of the solution.\n\nFrequently Asked Questions\n\nWhat scientific concept do you need to know in order to solve this problem?\n\nOur tutors have indicated that to solve this problem you will need to apply the Molality concept. You can view video lessons to learn Molality. Or if you need more Molality practice, you can also practice Molality practice problems."
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https://www.triangle-calculator.com/?q=32+32.8 | [
"# Triangle calculator\n\nPlease enter what you know about the triangle:\nYou have entered side a, b and c (hypotenuse-calculated).\n\n### Right scalene triangle.\n\nSides: a = 32 b = 32.8 c = 45.82440111732\n\nArea: T = 524.8\nPerimeter: p = 110.6244011173\nSemiperimeter: s = 55.31220055866\n\nAngle ∠ A = α = 44.29326806315° = 44°17'34″ = 0.77330531116 rad\nAngle ∠ B = β = 45.70773193685° = 45°42'26″ = 0.79877432152 rad\nAngle ∠ C = γ = 90° = 1.57107963268 rad\n\nHeight: ha = 32.8\nHeight: hb = 32\nHeight: hc = 22.90550223481\n\nMedian: ma = 36.49443831295\nMedian: mb = 35.95877529888\nMedian: mc = 22.91220055866\n\nInradius: r = 9.48879944134\nCircumradius: R = 22.91220055866\n\nVertex coordinates: A[45.82440111732; 0] B[0; 0] C[22.34663632664; 22.90550223481]\nCentroid: CG[22.72334581465; 7.63550074494]\nCoordinates of the circumscribed circle: U[22.91220055866; 0]\nCoordinates of the inscribed circle: I[22.51220055866; 9.48879944134]\n\nExterior(or external, outer) angles of the triangle:\n∠ A' = α' = 135.7077319369° = 135°42'26″ = 0.77330531116 rad\n∠ B' = β' = 134.2932680631° = 134°17'34″ = 0.79877432152 rad\n∠ C' = γ' = 90° = 1.57107963268 rad\n\n# How did we calculate this triangle?\n\n### 1. Input data entered: side a, b and c (hypotenuse-calculated).",
null,
"### 2. The triangle circumference is the sum of the lengths of its three sides",
null,
"### 3. Semiperimeter of the triangle",
null,
"### 4. The triangle area using Heron's formula",
null,
"### 5. Calculate the heights of the triangle from its area.",
null,
"### 6. Calculation of the inner angles of the triangle using a Law of Cosines",
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"https://www.triangle-calculator.com/tex/f20/f2049f11908a7.svg",
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"https://www.triangle-calculator.com/tex/b61/b61995916d667.svg",
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"https://www.triangle-calculator.com/tex/57a/57ab702fea06a.svg",
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"https://www.triangle-calculator.com/tex/c47/c4748b2287f95.svg",
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"https://www.triangle-calculator.com/tex/d22/d22bf341c4433.svg",
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"https://www.triangle-calculator.com/tex/51a/51a7847f746bc.svg",
null,
"https://www.triangle-calculator.com/tex/41c/41c3e001f92cd.svg",
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"https://www.triangle-calculator.com/tex/f3b/f3b30ea1d2d39.svg",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.642955,"math_prob":0.99783385,"size":1498,"snap":"2019-35-2019-39","text_gpt3_token_len":564,"char_repetition_ratio":0.14056225,"word_repetition_ratio":0.008658009,"special_character_ratio":0.52069426,"punctuation_ratio":0.22697368,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99990594,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-15T05:46:17Z\",\"WARC-Record-ID\":\"<urn:uuid:be492cc8-2de4-4ac4-b14f-a809214c0f07>\",\"Content-Length\":\"16846\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:513e975d-3568-4e0c-bd53-fdd1553b1429>\",\"WARC-Concurrent-To\":\"<urn:uuid:f3c54f3a-a895-4206-9ae9-5343c5ec2286>\",\"WARC-IP-Address\":\"104.28.12.22\",\"WARC-Target-URI\":\"https://www.triangle-calculator.com/?q=32+32.8\",\"WARC-Payload-Digest\":\"sha1:YYA3LNI5FN3BXVEMNTZFBIILHPQP3HSS\",\"WARC-Block-Digest\":\"sha1:J3JXU72APXNI6BTNJH3NLGBAIHVHQI46\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514570740.10_warc_CC-MAIN-20190915052433-20190915074433-00502.warc.gz\"}"} |
http://asq.org/statistics/2010/11/quality-tools/three-variable-scatter-plot-an-example.html | [
"",
null,
"ASQ - Statistics Division\n\n### Three Variable Scatter Plot: An Example – Basic Tool Taken from Spring 1995 Newsletter\n\nAbstract: This article shows how to construct a three variable scatter plot as an extension of a two variable scatter plot. Assign two variables to the x and y axes. Represent the third variable by coding its values and using symbols to represent the codes. The article gives an example portraying tennis ball for different producers due to two mechanisms.\n\nKeywords: Three Variable Scatter Plot\n\nAlready a member? Access this Content",
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null,
"https://www.facebook.com/tr",
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"http://asq.org/cms_prd_consump/groups/public/documents/web_asset/rating--white.gif",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8045418,"math_prob":0.6154969,"size":776,"snap":"2021-04-2021-17","text_gpt3_token_len":154,"char_repetition_ratio":0.11658031,"word_repetition_ratio":0.09375,"special_character_ratio":0.19072165,"punctuation_ratio":0.083333336,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.962637,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-21T04:57:40Z\",\"WARC-Record-ID\":\"<urn:uuid:3044d808-a39a-4f50-9397-c90081aa1b75>\",\"Content-Length\":\"45229\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:96c059cb-b6a5-4d8f-9015-26b11a841cc6>\",\"WARC-Concurrent-To\":\"<urn:uuid:ac094b7b-35ab-47d5-84b9-b70715503631>\",\"WARC-IP-Address\":\"206.128.156.25\",\"WARC-Target-URI\":\"http://asq.org/statistics/2010/11/quality-tools/three-variable-scatter-plot-an-example.html\",\"WARC-Payload-Digest\":\"sha1:OK5UVS2HNRPAX3YONBL3HHMPI3LXT7QD\",\"WARC-Block-Digest\":\"sha1:2K2NOQW2F5PI2YGHY73H3FEXRXVWLL57\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703522242.73_warc_CC-MAIN-20210121035242-20210121065242-00692.warc.gz\"}"} |
http://erlandsendata.no/?t=vbatips&c=other&p=2772 | [
"Retrieving formula results\n\n2009-02-18 Other 0 55\n\nIf you need to get the result from a formula in a cell, you would normally do something like this:\n\n`dblValue = Range(\"A1\").Value`\nIf you need to calculate parts of a formula or math expression then you can do something like this:\n```' assume that cell A1 contains this formula:\n' =(A2+A3)/(A4+A5)\nstrFormula = Range(\"A1\").Formula\nvarItems = Split(strFormula, \"/\")\ndblValueFirst = Application.Evaluate(varItems(0))\ndblValueSecond = Application.Evaluate(varItems(1))\n```\nThe evaluating expression can contain both cell references, functions and regular math expressions, all of the examples below will return a value:\n```dblValue = Application.Evaluate(\"(A4+A5)\")\ndblValue = Application.Evaluate(\"=AVERAGE(A1:A9)\")\ndblValue = Application.Evaluate(\"AVERAGE(A1:A9)\")\ndblValue = Application.Evaluate(\"=2+3*4\")\ndblValue = Application.Evaluate(\"2+3*4\")\n```"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.67320395,"math_prob":0.9956634,"size":897,"snap":"2019-26-2019-30","text_gpt3_token_len":222,"char_repetition_ratio":0.1881299,"word_repetition_ratio":0.0,"special_character_ratio":0.26421404,"punctuation_ratio":0.12903225,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999261,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-20T00:30:22Z\",\"WARC-Record-ID\":\"<urn:uuid:69814076-8e39-4981-a30d-ab2ad748dc16>\",\"Content-Length\":\"23881\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4b1f020e-85cd-4cee-8e54-cc99ae11abdc>\",\"WARC-Concurrent-To\":\"<urn:uuid:9bc74555-ceb1-4292-80f3-b88690ab28a4>\",\"WARC-IP-Address\":\"46.250.210.133\",\"WARC-Target-URI\":\"http://erlandsendata.no/?t=vbatips&c=other&p=2772\",\"WARC-Payload-Digest\":\"sha1:PYR76KYZSBZUKI3XPREV26ALS7J3ZX2Y\",\"WARC-Block-Digest\":\"sha1:Z3JCIX6MRAJQYIK2EHSA7BKX3GH6PKC6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999066.12_warc_CC-MAIN-20190619224436-20190620010436-00481.warc.gz\"}"} |
https://www.tec4tric.com/category/nn | [
"## How to Implement AND Function using McCulloch Pitts Neuron?\n\nImplement AND function using McCulloch Pitts Neuron. In the AND function the output will be High if both the inputs are high.\n\n## What is Activation Function in Neural Networks?\n\nActivation function in Neural Networks plays a major role. It decides a neuron will fire or not with the help of the Threshold value.\n\n## What is Bias in Artificial Neural Networks?\n\nThe Bias included in the network has its impact on calculating the net input. The bias is included by adding a value X0=1 to the input vector X. Hence W0=b.\n\n## What is Artificial Neural Networks(ANN)?\n\nAn Artificial Neural Network (ANN) is a computational model based on the structure and functions of a biological neural network. ANN is a core part of Deep Learning.\n\n## What is Perceptron Learning Algorithm in Neural Networks\n\nPerceptron Learning Algorithm is an algorithm or a way to train the weights of the Perceptron. This was proposed by Minsky and Papert in 1969."
]
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null
]
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https://customwritingreviews.com/math-225n-week-8-assignment-linear-regression-equations/ | [
"Do You Have an Online Class? Email us [email protected]\n\n# Math 225N Week 8 Assignment Linear Regression Equations\n\n## Math 225N Week 8 Assignment Linear Regression Equations\n\nWeek 8 Assignment Linear Regression Equations\n\nQuestion\n\nDaniel owns a business consulting service. For each consultation, he charges \\$95 plus \\$70 per hour of work. A linear equation that expresses the total amount of money Daniel earns per consultation is y=70x+95. What are the independent and dependent variables? What is the y-intercept and the slope?\n\nThe independent variable (x) is the amount, in dollars, Daniel earns for a consultation. The dependent variable (y) is the amount of time Daniel consults.\n\nDaniel charges a one-time fee of \\$95 (this is when x=0), so the y-intercept is 95. Daniel earns \\$70 for each hour he works, so the slope is 70.\n\nThe independent variable (x) is the amount of time Daniel consults. The dependent variable (y) is the amount, in dollars, Daniel earns for a consultation.\n\nDaniel charges a one-time fee of \\$95 (this is when x=0), so the y-intercept is 95. Daniel earns \\$70 for each hour he works, so the slope is 70.\n\nThe independent variable (x) is the amount, in dollars, Daniel earns for a consultation. The dependent variable (y) is the amount of time Daniel consults.\n\nDaniel charges a one-time fee of \\$70 (this is when x=0), so the y-intercept is 70. Daniel earns \\$95 for each hour he works, so the slope is 95.\n\nThe independent variable (x) is the amount of time Daniel consults. The dependent variable (y) is the amount, in dollars, Daniel earns for a consultation.\n\nDaniel charges a one-time fee of \\$70 (this is when x=0), so the y-intercept is 70. Daniel earns \\$95 for each hour he works, so the slope is 95.\n\nQuestion\n\nThe scatter plot below shows data relating competitive chess players’ ratings and their IQ. Which of the following patterns does the scatter plot show?\n\npositive linear pattern\n\npositive linear pattern with deviations\n\nnegative linear pattern\n\nnegative linear pattern with deviations\n\nno pattern\n\nQuestion\n\nThe scatter plot below shows data relating total income and the number of children a family has. Which of the following patterns does the scatter plot show?\n\nPositive linear pattern\n\nPositive linear pattern with deviations\n\nNegative linear pattern\n\nNegative linear pattern with deviations\n\nNo pattern\n\nQuestion\n\nJamie owns a house painting service. For each house, she charges \\$70 plus \\$40 per hour of work. A linear equation that expresses the total amount of money Jamie earns per house is y=70+40x. What are the independent and dependent variables? What is the y-intercept and the slope?\n\nThe independent variable (x) is the amount, in dollars, Jamie earns for a house. The dependent variable (y) is the amount of time Jamie paints a house.\n\nJamie charges a one-time fee of \\$40 (this is when x=0), so the y-intercept is 40. Jamie earns \\$70 for each hour she works, so the slope is 70.\n\nThe independent variable (x) is the amount, in dollars, Jamie earns for a house. The dependent variable (y) is the amount of time Jamie paints a house.\n\nJamie charges a one-time fee of \\$70 (this is when x=0), so the y-intercept is 70. Jamie earns \\$40 for each hour she works, so the slope is 40.\n\nThe independent variable (x) is the amount of time Jamie paints a house. The dependent variable (y) is the amount, in dollars, Jamie earns for a house.\n\nJamie charges a one-time fee of \\$40 (this is when x=0), so the y-intercept is 40. Jamie earns \\$70 for each hour she works, so the slope is 70.\n\nThe independent variable (x) is the amount of time Jamie paints a house. The dependent variable (y) is the amount, in dollars, Jamie earns for a house.\n\nJamie charges a one-time fee of \\$70 (this is when x=0), so the y-intercept is 70. Jamie earns \\$40 for each hour she works, so the slope is 40.\n\nQuestion\n\nGeorge is an avid plant lover and is concerned about the lack of daffodils that grow in his backyard. He finds the growth of the daffodils, G, is dependent on the percent of aluminium measured in the soil, x, and can be modelled by the function\n\nG(x)=16−4x.\n\nDraw the graph of the growth function by plotting its G-intercept and another point.\n\nQuestion\n\nWhat percent of aluminum in the soil must there be for the daffodils to grow only by 5 centimeters?\n\nQuestion\n\nThe number of questions marked incorrect on a statistics midterm, y, is dependent on the pages of notes a student wrote over the semester, x, and can be modeled by the function\n\n• y(x)=30−3.5x.\n\nDraw the graph of the function by plotting its y-intercept and another point.\n\nQuestion\n\nHow many pages of notes did a student take if they had 12 problems marked incorrect on the statistics midterm?\n\nThe number of incorrect problems on the statistics midterm is 12. So, we must find the number of pages of notes, x, so that y(x)=12.\n\n## Question\n\nA shoe designer explored the relationship between the percent of defects and the percent of new machines at various production facilities throughout the state. The designer collects information from 6 of their facilities, shown in the table below.\n\nUse the graph below to plot the points and develop a linear relationship between the percent of defects and the percent of new machines.\n\nThe percent of defects is the x-coordinate, while the percent of new machines is the y-coordinate. So, the table of values corresponds to the points\n\n(25,32), (20,40), (15,50), (10,65), (5,70), (0,85).\n\nQuestion\n\nUsing the linear relationship graphed above, estimate the percent of new machines if there is 12% defects in the shoes at various production facilities.\n\nQuestion\n\nAmelia plays basketball for her high school. She wants to improve to play at the college level. She notices that the number of points she scores in a game goes up in response to the number of hours she practices her jump shot each week. She records the following data:\n\nWhich scatter plot best represents Amelia’s data?\n\nQuestion\n\nThe scatter plot below shows data about the relationship between incomes and the number of years of education people have. Which of the following patterns does the scatter plot show?"
]
| [
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]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.91327804,"math_prob":0.9708935,"size":6580,"snap":"2022-27-2022-33","text_gpt3_token_len":1562,"char_repetition_ratio":0.14765815,"word_repetition_ratio":0.51632833,"special_character_ratio":0.24468085,"punctuation_ratio":0.103264995,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9984165,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-05T03:42:31Z\",\"WARC-Record-ID\":\"<urn:uuid:59d6ddf2-9bcc-4302-a929-e6ee6a59f90c>\",\"Content-Length\":\"892833\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ae875987-54b6-4269-9a87-72d00f031b36>\",\"WARC-Concurrent-To\":\"<urn:uuid:f4136cc0-c9fd-401e-bdde-9b0ee56f9851>\",\"WARC-IP-Address\":\"104.21.30.93\",\"WARC-Target-URI\":\"https://customwritingreviews.com/math-225n-week-8-assignment-linear-regression-equations/\",\"WARC-Payload-Digest\":\"sha1:Z3AZ5LJKIRQ4WCVSBSM5L636NRI2P5VW\",\"WARC-Block-Digest\":\"sha1:DGH5J7AO3XLOOAQSVJJISNAX2S4NXYFB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104512702.80_warc_CC-MAIN-20220705022909-20220705052909-00745.warc.gz\"}"} |
https://rbseguide.com/class-12-maths-chapter-16-ex-16-3-english-medium/ | [
"# RBSE Solutions for Class 12 Maths Chapter 16 Probability and Probability Distribution Ex 16.3\n\n## Rajasthan Board RBSE Class 12 Maths Chapter 16 Probability and Probability Distribution Ex 16.3\n\nQuestion 1.\nGiven there are two bags I and II. Bag I contains 3 red and 4 black bails while another bag II contains S red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from bag II.\nSolution:\nLet E1 be the event that the ball is drawn from the bag I and II and E2 be the event that it is drawn from the bag II and E be the event that the drawn ball is red.\nSince the drawn ball is found to be red, so we have to Find P($$\\frac { { E }_{ 2 } }{ A }$$) the probability of drawing red ball from bag II. since both the bags are equally likely to be selected to draw the ball, we have",
null,
"Probability of getting red ball from bag I",
null,
"Probability of getting red ball from bag II",
null,
"Probability of getting red ball from bag III",
null,
"Hence, using Baye’s Theorem,",
null,
"",
null,
"Question 2.\nA doctor has to visit a patient. From the past experience, it is known that the probabilities that he will come by train, bus, scooter or by other means of transport are respectively $$\\frac { 3 }{ 10 }$$,$$\\frac { 1 }{ 5 }$$,$$\\frac { 1 }{ 10 }$$ and $$\\frac { 2 }{ 5 }$$. The probabilities that he will be late are $$\\frac { 1 }{ 4 }$$,$$\\frac { 1 }{ 3 }$$ and $$\\frac { 1 }{ 12 }$$, if he comes by train, bus and scooter respectively, but if he comes by other means of transport, than he will not be late. When he arrives, he is late. What is the probability that he comes by train ?\nSolution:\nLet E be the event “Doctor comes late to visit a patient” and if Doctor comes by Train, Bus, Scooter and any-other means, then probabilities are T1, T2, T3 and T4 respectively.",
null,
"",
null,
"Question 3.\nBag I contains 3 red and 5 black balls and by II contains 4 red and 5 black balls. One ball transferred from bag I to bag II and then a ball is drawn from bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.\nSolution:\nBag I contain 3 red and 4 black balls, and Bag II contains 4 red and 5 black balls.\nLet, E1 = Event of drawing red ball from bag I\nand E2 = Event of drawing black ball from bag II\nand A : Event of drawing red ball after transfering one ball from I to II",
null,
"Question 4.\nA bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls, One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.\nSolution:\nLet E1 is the event of drawing from bag I\nand E2 of drawing from bag II\nand A represent the event of drawing red ball.\n∴ Probability of chosing one bag = $$\\frac { 1 }{ 2 }$$\nOr P(E1) = P(E2) = $$\\frac { 1 }{ 2 }$$\nThere are 4 red and 4 black balls in bag I.\n∴ Probability of drawing red ball from it",
null,
"In second bag, there are 2 red and 6 black balls.\n∴ Probability of drawing red ball from it",
null,
"Now probability of drawing red ball from first bag",
null,
"Question 5.\nThere are three coins. One is having head on both faces, another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin ?\nSolution:\nProbability of selecting one coin out of three = $$\\frac { 1 }{ 3 }$$\nE1 : event that coin having head on both faces.\nE2 : event that coin is biased that head come up 75% of the time.\nE3 : event that third is a unbiased,\nand A is the event of getting head",
null,
"Hence, required probability = $$\\frac { 4 }{ 9 }$$\n\nQuestion 6.\nA laboratory blood test is 99% effective in detecting a certain disease when it is if fact, present. However, the test also yields a false positive result for 0-5% of the healthy person tested. If 0.1 percent of the population actually has the disease. What is the probability that a person has the disease given that his test result is positive ?\nSolution:\nLet E1 : Event that patient is effected from desease.\nE2 = Event that patient is not effected from desease and A be the event that blood is tested.\n∴ Probability of the person effected from disease",
null,
"Probability of the person not affected from disease",
null,
"Probability of those who are patient and their blood are tested",
null,
"Probability of those who were tested but not patient",
null,
"∴ Probability that a person has the disease given that his test is positive",
null,
"Hence, the required probability = $$\\frac { 22 }{ 133 }$$\n\nQuestion 7.\nOf the sutdents in a college, it is known that 60% reside in hostel and 40% are day scholars. Previous year result report that 30% of all student who reside in hostel attain, A grade and 20% of day schloars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has on A grade, what is the probability that the student is a hostlier ?\nSolution:\nSolution : Let E1 : Event that students reside in hostel and E2 : Event that students do not reside in hostel\n∴ Probability of students reside in hostel",
null,
"Probability that students do not reside in hostel",
null,
"Probability of students securing grade A who reside in hostel",
null,
"Probability of students who do not reside in hostel but. secure grade A",
null,
"∴ Probability that student securing graded is a hostelier = P ($$\\frac { { E }_{ 1 } }{ A }$$)",
null,
"Hence, required probability : $$\\frac { 9 }{ 13 }$$\n\nQuestion 8.\nAn insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probabilities of an accident are 0-01, 0*03 and 0-15 respectively. One of the insured person meets with an accident. What is the probability that he is a scooter driver ?\nSolution:\nLet E1 : Event that scooter driver is insured\nE2 = Event that car driver is insured\nE3 = Event that truck driver is insured\nand A : Even that an accident occurs.\nA company insured 2000 scotter drivers, 4000 car drivers and 6000 truck drivers.\n∴ Total number of drivers insured = 2000 + 4000 + 6000 = 12000\n∴ Probability of scooter drivers insured",
null,
"Probability of car drivers insured",
null,
"Probability of truck drivers insured",
null,
"Probability of scooter drivers accidents",
null,
"Probability of car drivers accidents",
null,
"Probability of truck drivers accidents",
null,
"A person meets an accident who is insured.\n∴ Probability that the persons is a scooter driver",
null,
"Hence, required probability : $$\\frac { 1 }{ 52 }$$\n\nQuestion 9.\nIn answering a question on a multiple choice test, a student either knows the answer or guesses. Let $$\\frac { 3 }{ 4 }$$ be probability that he knows the answer and $$\\frac { 1 }{ 4 }$$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $$\\frac { 1 }{ 4 }$$. What is the probability that the student knows the answer given that he answered it correctly ?\nSolution:\nLet E1 : Event that student knows the answer\nE2 : Event that student guesses\nand A : Event that he answers correctly",
null,
"∴ Probability that he answers correctly",
null,
"Hence requried Probability = $$\\frac { 12 }{ 13 }$$\n\nQuestion 10.\nSuppose 5% men and 0.25% women has white hairs. A person with white hair is chosen at random. Find the probability that a male is selected. Assume that there are equal number of men and women.\nSolution:\nGiven that number of men and women are equal\nLet events E1 = To be a man\nE2 = To be a woman\nA = White hair",
null,
"",
null,
"Question 11.\nTwo groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducting a new product is 0.7 and the corresponding probability is 0-3 if the second group wins. Find the probability that the new product introduced was by the second group.\nSolution:\nLet events\nE1 : First group wins, introduce new product.\nE2 : Second group wins, introduce new product.\nGiven : Probability that first group wins = P(E1) = 0.6\nProbability that second group wins = P(E2) = 0.4\nProbability that first group wins introducting new product,",
null,
"Probability that second group wins introducting new product",
null,
"∴ Probability that new product introduced was by the second group",
null,
"",
null,
"Question 12.\nSuppose a girl throw a dice. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1,2,3 or 4 then tosses a coin once and notes whether a head or tall is obtained. If she obtained exactly one head, what is the probability that she threw 1,2,3 or 4 with the die ?\nSolution:\nNumber of possible results after throwing a die = (1,2, 3, 4, 5, 6) = 6\nLet E1 : Event she gets 5 or 6\nE2 : Event she gets 1, 2, 3, 4\nA : Event that she gets head if tosses a coin\n∴ Probability that 5 or 6 acheived",
null,
"Probability that 1, 2, 3,4 are acheived",
null,
"If she gets 5 or 6 she tosses a coin thrice and notes the number of heads.\n(HHH, HHT, HTH, THH, HIT, THT, TTH, TIT)\n∴ Number of ways to get one head = (HTT, THT, TTH) = 3\n∴ Probability to get one head = $$\\frac { 3 }{ 8 }$$",
null,
"If she gets 1, 2, 3, 4, then she tosses a coin once and notes whether a head or tail.\n∴ Probability to get one head = $$\\frac { 1 }{ 2 }$$",
null,
"If she obtained exactly one head, than probability that she threw 1,2, 3 or 4",
null,
"",
null,
"Question 13.\nA card from a pack of 52 card is lost From remaining cards of the pack, two cards are drawn found to be both diamonds. Find probability of the lost card being a diamond.\nSolution:\nLet Events E1 = Lost card is diamond\nE2 = Lost card is not diamond\nThe number of diamond card = 13 out of 52 cards.",
null,
"and 39 cards are not diamonds",
null,
"(i) When a card of diamond lost there remaining cards of diamond are 12 out of 52.",
null,
"Here A represents the cards lost.\n(ii) When diamonond card is not\nLost then probability of drawing 2 cards of diamond",
null,
"∴ Probability that lost card is a diamond",
null,
"",
null,
"Hence, required probability = $$\\frac { 11 }{ 50 }$$\n\nQuestion 14.\nA bag contains 3 red and 7 black. Two balls are drawn one by one at a time at random without replacement. If second drawn half is red, then what is the probability that first drawn ball is also red ?\nSolution:\nA = Event that drawn ball is Red first time.\nand B = Event that drawn ball is also red in second time.\nThen P(A ∩ B) = P( 1 red and 1 red)",
null,
""
]
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null,
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null,
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https://d2mvzyuse3lwjc.cloudfront.net/doc/de/OriginC/ref/vectorbase-Conjugate | [
"# 2.2.3.19.4 vectorbase::Conjugate\n\n## Description\n\nReplace vector with the conjugate of the vector. The method only available for complex or double data type vector.\n\n## Syntax\n\nint Conjugate( )\n\n## Return\n\nReturns 0 on success and -1 on failure which occurs if the vector is empty.\n\n## Examples\n\nEX1\n\nvoid vectorbase_Conjugate_ex1()\n{\n//Output Original vector\nvector<complex> vecC = {1+2i, 3-4i, 5+6i, 7-8i};\nprintf(\"Original vector:\\n\");\nfor( int ii = 0; ii < vecC.GetSize(); ii++)\nout_complex(\"\", vecC[ii]);\n\n//Replace vector with the conjugate of the vector\nvecC.Conjugate();\n\n//Output the result\nprintf(\"Conjugate vector:\\n\");\nfor( ii = 0; ii < vecC.GetSize(); ii++)\nout_complex(\"\", vecC[ii]);\n}\n\n## Remark\n\nReplace this vector with the conjugate of the vector. A runtime error occurs if the underlying base type of this vector is not double or Complex."
]
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https://developer.roblox.com/en-us/api-reference/class/BallSocketConstraint | [
"# BallSocketConstraint\n\nShow deprecated\n\nA BallSocketConstraint constrains its `Attachment|Attachments` so that they occupy the same position. By default it allows both attachments to freely rotate about all of their axes, but if `BallSocketConstraint/LimitsEnabled` is `true`, the attachments can only rotate in a limited cone.\n\nNote that if this constraint attaches one part (A) to another part (B) that is anchored or connected to an anchored part (Z), part A will not be locally simulated when interacting with a player.\n\n## Properties\n\n ``` bool``` ``` LimitsEnabled ``` Sets whether the BallSocketConstraint sets a limit on rotation based on `BallSocketConstraint/UpperAngle`.\n ``` float``` ``` Radius ``` The visualized radius of the BallSocketConstraint.\n ``` float``` ``` Restitution ``` How elastic `Attachment` connected by a BallSocketConstraint will be when they reach the end of the range specified by `BallSocketConstraint/UpperAngle` when `BallSocketConstraint/LimitsEnabled` is true. Constrained between 0 and 1.\n ``` bool``` ``` TwistLimitsEnabled ``` Sets whether the BallSocketConstraint sets a limit on twist rotation based on `BallSocketConstraint/TwistUpperAngle` and `BallSocketConstraint/TwistLowerAngle`.\n ``` float``` ``` TwistLowerAngle ``` Sets the lower twist rotation limit of the BallSocketConstraint, as long as `BallSocketConstraint/TwistLimitsEnabled` is `true`.\n ``` float``` ``` TwistUpperAngle ``` Sets the upper twist rotation limit of the BallSocketConstraint, as long as `BallSocketConstraint/TwistLimitsEnabled` is `true`.\n ``` float``` ``` UpperAngle ``` Sets the upper rotation limit of the BallSocketConstraint, as long as `BallSocketConstraint/LimitsEnabled` is `true`.\n\nInherited from Constraint:\n\n ``` bool``` ``` Active ``` ``` [ReadOnly] ``` ``` [NotReplicated] ``` Indicates if the constraint is currently active in the world\n ``` Attachment``` ``` Attachment0 ``` The `Attachment` that is connected to `Constraint/Attachment1`\n ``` Attachment``` ``` Attachment1 ``` The `Attachment` that is connected to `Constraint/Attachment0`\n ``` BrickColor``` ``` Color ``` The color of the constraint.\n ``` bool``` ``` Enabled ``` Toggles whether or not this Constraint is enabled.\n ``` bool``` ``` Visible ``` Toggles the visibility of this Constraint.\n\nInherited from Instance:\n\n ``` bool``` ``` Archivable ``` Determines if an `Instance` can be cloned using `/Instance/Clone` or saved to file.\n ``` string``` ``` ClassName ``` ``` [ReadOnly] ``` ``` [NotReplicated] ``` A read-only string representing the class this `Instance` belongs to\n ``` int``` ``` DataCost ``` ``` [ReadOnly] ``` ``` [NotReplicated] ``` ``` [Deprecated] ``` The cost of saving the instance using data persistence.\n ``` string``` ``` Name ``` A non-unique identifier of the `Instance`\n ``` Instance``` ``` Parent ``` Determines the hierarchical parent of the `Instance`\n ``` bool``` ``` RobloxLocked ``` If true, the `Instance` and its descendants cannot be indexed or edited by a `Script` or `LocalScript` and will throw an error if it is attempted\n ``` bool``` ``` archivable ``` ``` [Hidden] ``` ``` [NotReplicated] ``` ``` [Deprecated] ```\n ``` string``` ``` className ``` ``` [ReadOnly] ``` ``` [NotReplicated] ``` ``` [Deprecated] ```\n\n## Functions\n\nInherited from Instance:\n\n ``` void``` ``` ClearAllChildren ( ) ``` This function destroys all of an `Instance`'s children.\n ``` Instance``` ``` Clone ( ) ``` Create a deep copy of a Roblox instance and descendants where `Archivable = true`.\n ``` void``` ``` Destroy ( ) ``` Sets the `Instance/Parent` property to nil, locks the `Instance/Parent` property, disconnects all connections and calls Destroy on all children.\n ``` Instance``` ``` FindFirstAncestor ( string name ) ``` Returns the first ancestor of the `Instance` whose `Instance/Name` is equal to the given name.\n ``` Instance``` ``` FindFirstAncestorOfClass ( string className ) ``` Returns the first ancestor of the `Instance` whose `Instance/ClassName` is equal to the given className.\n ``` Instance``` ``` FindFirstAncestorWhichIsA ( string className ) ``` Returns the first ancestor of the `Instance` for whom `Instance/IsA` returns true for the given className.\n ``` Instance``` ``` FindFirstChild ( string name , bool recursive ) ``` Returns the first child of the `Instance` found with the given name.\n ``` Instance``` ``` FindFirstChildOfClass ( string className ) ``` Returns the first child of the `Instance` whose `Instance/ClassName|ClassName` is equal to the given className.\n ``` Instance``` ``` FindFirstChildWhichIsA ( string className , bool recursive ) ``` Returns the first child of the `Instance` for whom `Instance/IsA` returns true for the given className.\n ``` Variant``` ``` GetAttribute ( string attribute ) ```\n ``` RBXScriptSignal``` ``` GetAttributeChangedSignal ( string attribute ) ```\n ``` Dictionary``` ``` GetAttributes ( ) ```\n ``` Objects``` ``` GetChildren ( ) ``` Returns an array containing all of the `Instance`'s children.\n ``` string``` ``` GetDebugId ( int scopeLength ) ``` ``` [NotBrowsable] ``` Returns a coded string of the `Instance`s DebugId used internally by Roblox.\n ``` Array``` ``` GetDescendants ( ) ``` ``` [CustomLuaState] ``` Returns an array containing all of the descendants of the instance\n ``` string``` ``` GetFullName ( ) ``` Returns a string describing the `Instance`'s ancestry.\n ``` RBXScriptSignal``` ``` GetPropertyChangedSignal ( string property ) ``` Get an event that fires when a given property of an object changes.\n ``` bool``` ``` IsA ( string className ) ``` ``` [CustomLuaState] ``` Returns true if an `Instance`'s class matches or inherits from a given class\n ``` bool``` ``` IsAncestorOf ( Instance descendant ) ``` Returns true if an `Instance` is an ancestor of the given descendant.\n ``` bool``` ``` IsDescendantOf ( Instance ancestor ) ``` Returns true if an `Instance` is a descendant of the given ancestor.\n ``` void``` ``` Remove ( ) ``` ``` [Deprecated] ``` Sets the object’s Parent to nil, and does the same for all its descendants.\n ``` void``` ``` SetAttribute ( string attribute , Variant value ) ```\n ``` Instance``` ``` WaitForChild ( string childName , double timeOut ) ``` ``` [CustomLuaState] ``` ``` [CanYield] ``` Returns the child of the `Instance` with the given name. If the child does not exist, it will yield the current thread until it does.\n ``` Objects``` ``` children ( ) ``` ``` [Deprecated] ``` Returns an array of the object’s children.\n ``` Instance``` ``` clone ( ) ``` ``` [Deprecated] ```\n ``` void``` ``` destroy ( ) ``` ``` [Deprecated] ```\n ``` Instance``` ``` findFirstChild ( string name , bool recursive ) ``` ``` [Deprecated] ```\n ``` Objects``` ``` getChildren ( ) ``` ``` [Deprecated] ```\n ``` bool``` ``` isA ( string className ) ``` ``` [Deprecated] ``` ``` [CustomLuaState] ```\n ``` bool``` ``` isDescendantOf ( Instance ancestor ) ``` ``` [Deprecated] ```\n ``` void``` ``` remove ( ) ``` ``` [Deprecated] ```\n\n## Events\n\nInherited from Instance:\n\n ``` RBXScriptSignal``` ``` AncestryChanged ( Instance child , Instance parent ) ``` Fires when the `Instance/Parent` property of the object or one of its ancestors is changed.\n ``` RBXScriptSignal``` ``` AttributeChanged ( string attribute ) ```\n ``` RBXScriptSignal``` ``` Changed ( string property ) ``` Fired immediately after a property of an object changes.\n ``` RBXScriptSignal``` ``` ChildAdded ( Instance child ) ``` Fires when an object is parented to this `Instance`.\n ``` RBXScriptSignal``` ``` ChildRemoved ( Instance child ) ``` Fires when a child is removed from this `Instance`.\n ``` RBXScriptSignal``` ``` DescendantAdded ( Instance descendant ) ``` Fires when a descendant is added to the `Instance`\n ``` RBXScriptSignal``` ``` DescendantRemoving ( Instance descendant ) ``` Fires immediately before a descendant of the `Instance` is removed.\n ``` RBXScriptSignal``` ``` childAdded ( Instance child ) ``` ``` [Deprecated] ```"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5050212,"math_prob":0.49293423,"size":2977,"snap":"2019-51-2020-05","text_gpt3_token_len":767,"char_repetition_ratio":0.1782711,"word_repetition_ratio":0.26930693,"special_character_ratio":0.23177695,"punctuation_ratio":0.035460994,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9505421,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-22T18:12:12Z\",\"WARC-Record-ID\":\"<urn:uuid:b37261b4-7384-4927-8d12-db055aa82971>\",\"Content-Length\":\"581924\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a6e208d8-1307-4da4-8eb3-e0c11ff9c40e>\",\"WARC-Concurrent-To\":\"<urn:uuid:9608bbe6-9da9-489d-98aa-24e5685f42cd>\",\"WARC-IP-Address\":\"99.84.181.96\",\"WARC-Target-URI\":\"https://developer.roblox.com/en-us/api-reference/class/BallSocketConstraint\",\"WARC-Payload-Digest\":\"sha1:FSKFTNY7WNZY5VEMY7O45CEBMUE3OEMW\",\"WARC-Block-Digest\":\"sha1:ZMLDOJACHC5AOLI7GHB7BOUAM3N4QBEO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250607314.32_warc_CC-MAIN-20200122161553-20200122190553-00226.warc.gz\"}"} |
https://community.rapidminer.com/discussion/10846/howto-classification-via-multi-response-linear-regression | [
"# Howto: Classification via Multi response linear regression.\n\nMember Posts: 2",
null,
"Contributor I\nedited September 2019 in Help\nI have a classification problem. Instead of a model which tells me the most probable class of a sample i would like to have a model which tells me the chances the current sample has of being of each possible class.\nThis means that the labeled examples should have, instead of a column containing the given prediction, an additional column for each class containing a real value between 0 and 1 which approximate the likeliness the sample has of being of such class.\n\nIn the Weka community i'm told this is called \"Classification using multi response linear regression\" (or at least that MLR is the best or most common way to do that) and that the Weka component ClassificationViaRegression does exactly that but I didn't found such component in RapidMiner even if it does indeed contains most Weka components. I tried several components in RM for linear regression but all requires a single numerical label instead of my available nominal label, most probably because these do not support multi respose.\n\n1- If the above mentioned component or Weka is available in RM and where i can find it.\n2- If there are other component which does a similar thing (maybe even better).\n3- If there are other approaches to solve the problem in a similar way which you believe may work nicely.\n4- In case there aren't component which supports MLR natively, something (a guide, a tutorial, an example) about how to use single response linear regression components to approximate MLR in RapidMiner.\n\nThank you.\nTagged:\n\n• Member Posts: 2",
null,
"Contributor I\nI haven't understood exactly what \"Generate Prediction Ranking\" does but i believe that it's not what i'm looking for. The example you posted train a standard decision tree and then create a \"confidence\" for each class (i admit it's not clear to me how this is done), while i would like to have a model which actually produce \"confidence\" such that the error is minimized wrt the \"confidence\", not just the classification error.\n\nIn the example the decision tree somewhat minimize the error over the classification and then the confidence is somehow computed, while the criteria i'm refering to is actually a training heuristics which minimize the error over the \"confidence\", i.e. should be included in the labeled example \"by construction\".\n\n• RapidMiner Certified Analyst, RapidMiner Certified Expert, Member Posts: 2,531",
null,
"Unicorn\nHi,\nif you apply any classification model inside RapidMiner you will give so called confidence values which express the estimated probability that the example belongs to the referenced class. There's one confidence value for each class and they always sum up to 1.\nThe estimation of course depends on the classification model used and might differ from model to model.\n\nGreetings,\nSebastian"
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https://whatisconvert.com/352-cubic-feet-in-pints | [
"# What is 352 Cubic Feet in Pints?\n\n## Convert 352 Cubic Feet to Pints\n\nTo calculate 352 Cubic Feet to the corresponding value in Pints, multiply the quantity in Cubic Feet by 59.844155844181 (conversion factor). In this case we should multiply 352 Cubic Feet by 59.844155844181 to get the equivalent result in Pints:\n\n352 Cubic Feet x 59.844155844181 = 21065.142857152 Pints\n\n352 Cubic Feet is equivalent to 21065.142857152 Pints.\n\n## How to convert from Cubic Feet to Pints\n\nThe conversion factor from Cubic Feet to Pints is 59.844155844181. To find out how many Cubic Feet in Pints, multiply by the conversion factor or use the Volume converter above. Three hundred fifty-two Cubic Feet is equivalent to twenty-one thousand sixty-five point one four three Pints.\n\n## Definition of Cubic Foot\n\nThe cubic foot is a unit of volume, which is commonly used in the United States and the United Kingdom. It is defined as the volume of a cube with sides of one foot (0.3048 m) in length. Cubic feet = length x width x height. There is no universally agreed symbol but lots of abbreviations are used, such as ft³, foot³, feet/-3, etc. CCF is for 100 cubic feet.\n\n## Definition of Pint\n\nThe pint (symbol: pt) is a unit of volume or capacity in both the imperial and United States customary measurement systems. In the United States, the liquid pint is legally defined as one-eighth of a liquid gallon of precisely 231 cubic inches. One liquid pint is equal to 473.176473 milliliters (≈ 473 ml).\n\n## Using the Cubic Feet to Pints converter you can get answers to questions like the following:\n\n• How many Pints are in 352 Cubic Feet?\n• 352 Cubic Feet is equal to how many Pints?\n• How to convert 352 Cubic Feet to Pints?\n• How many is 352 Cubic Feet in Pints?\n• What is 352 Cubic Feet in Pints?\n• How much is 352 Cubic Feet in Pints?\n• How many pt are in 352 ft3?\n• 352 ft3 is equal to how many pt?\n• How to convert 352 ft3 to pt?\n• How many is 352 ft3 in pt?\n• What is 352 ft3 in pt?\n• How much is 352 ft3 in pt?"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8691478,"math_prob":0.9821515,"size":1958,"snap":"2020-45-2020-50","text_gpt3_token_len":528,"char_repetition_ratio":0.1888434,"word_repetition_ratio":0.08033241,"special_character_ratio":0.30847803,"punctuation_ratio":0.10810811,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99145275,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-30T04:48:57Z\",\"WARC-Record-ID\":\"<urn:uuid:9ee7ca97-f800-4ba7-9051-ac2bd2cabcb9>\",\"Content-Length\":\"30821\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9d075ebc-dba9-4b14-817e-23183b84823a>\",\"WARC-Concurrent-To\":\"<urn:uuid:cd3acc3d-a543-45ad-94ae-21c2b05380ce>\",\"WARC-IP-Address\":\"172.67.211.83\",\"WARC-Target-URI\":\"https://whatisconvert.com/352-cubic-feet-in-pints\",\"WARC-Payload-Digest\":\"sha1:YYMKBHVA45HMITUWGXHS7KYAKCATPZYA\",\"WARC-Block-Digest\":\"sha1:NH5IWI6DIPSZ7LDTQG47UB65V6D6ZWH3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141205147.57_warc_CC-MAIN-20201130035203-20201130065203-00004.warc.gz\"}"} |
https://karl.isenberg.us/2008/12/lua-deep-equals.php | [
"## Friday, December 19, 2008\n\n### Lua Deep Equals\n\nI was surprised when I couldn't easily google for a Lua deep compare function. It's main purpose is to compare tables and check if they have the same values, but of course it's recursive so it just uses ~= to check for difference in non-tables.\n\nAnyway, so I wrote my own and figured I'd post it here for posterity. Maybe you can come up with a shorter version?\n\n-- Deep Equals\nlocal function equals(t1, t2)\nif t1 == t2 then\nreturn true\nend\nif type(t1) ~= \"table\" or type(t2) ~= \"table\" then\nreturn false\nend\nlocal v2\nfor k,v1 in pairs(t1) do\nv2 = t2[k]\nif v1 ~= v2 and not equals(v1, t2[k]) then\nreturn false\nend\nend\nfor k in pairs(t2) do\nif t1[k] == nil then\nreturn false\nend\nend\nreturn true\nend"
]
| [
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]
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https://goprep.co/q5-if-the-difference-between-the-roots-of-the-equation-x-2-i-1nlqtd | [
"# If the difference between the roots of the equation",
null,
"is 2 write the values of a.\n\ngiven x2 + ax + 8 = 0 and α – β = 2\n\nAlso from given equation α β = 8\n\nAs α – β = 2\n\nThen",
null,
"α2 - 2 α – 8 =0\n\n(α-4) (α + 2)=0\n\nα = 4 and α = -2\n\nif α = 4 then substituting it in α – β = 2 we get ,\n\nβ = 2\n\nfrom the given equation,\n\nsum of roots = -a\n\nα + β = - a\n\n- a = 4+2\n\na = -6\n\nif α = - 2 then substituting it in α – β = 2 we get ,\n\nβ = - 4\n\nthen sum of roots α + β = - a\n\na = 6\n\ntherefore a = ± 6.\n\nRate this question :\n\nHow useful is this solution?\nWe strive to provide quality solutions. Please rate us to serve you better.\nRelated Videos",
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"",
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"",
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"",
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"",
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"",
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null,
"",
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"",
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null,
"",
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"",
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"Equations Of Motion from Calculus Method46 mins\nTry our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts\nDedicated counsellor for each student\n24X7 Doubt Resolution\nDaily Report Card\nDetailed Performance Evaluation",
null,
"view all courses",
null,
""
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| [
null,
"https://gradeup-question-images.grdp.co/liveData/PROJ33153/1560858713324536.png",
null,
"https://gradeup-question-images.grdp.co/liveData/PROJ33153/1560858714094975.png",
null,
"https://grdp.co/cdn-cgi/image/quality=90,fit=contain,width=350,f=auto/https://gs-post-images.grdp.co/2020/10/92d52ebd842616d8ad9af5461f7b5e56225f1ca34ea1a1e7b0ea1acd71df1b37poster-high-webp.png",
null,
"https://gs-post-images.grdp.co/2020/8/group-2x-img1598864374422-78.png-rs-high-webp.png",
null,
"https://grdp.co/cdn-cgi/image/quality=90,fit=contain,width=350,f=auto/https://gs-post-images.grdp.co/2020/10/a87eb8443257a3204e4f1e7aa894ca024f4f1d98b8a817c3b3a20c9375265f37poster-high-webp.png",
null,
"https://gs-post-images.grdp.co/2020/8/group-2x-img1598864374422-78.png-rs-high-webp.png",
null,
"https://grdp.co/cdn-cgi/image/quality=90,fit=contain,width=350,f=auto/https://gs-post-images.grdp.co/2020/8/b8024ff0c8d28f619e472ef4a2d4e133ca99557bf7fc6debd16d6d6d14b04713poster-high-webp.png",
null,
"https://gs-post-images.grdp.co/2020/8/group-2x-img1598864374422-78.png-rs-high-webp.png",
null,
"https://grdp.co/cdn-cgi/image/quality=90,fit=contain,width=350,f=auto/https://gs-post-images.grdp.co/2020/10/17968f6af90cbe679e39082ad57b7d5d14a1b0c0639cb737a0beca3a8e78d81eposter-high-webp.png",
null,
"https://gs-post-images.grdp.co/2020/8/group-2x-img1598864374422-78.png-rs-high-webp.png",
null,
"https://grdp.co/cdn-cgi/image/quality=90,fit=contain,width=350,f=auto/https://gs-post-images.grdp.co/2020/11/c4ab83cce8f5fc0c5b5a6dacba92f5527cfaae64a5bf9c05a72fd65f0b31710fposter-high-webp.png",
null,
"https://gs-post-images.grdp.co/2020/8/group-2x-img1598864374422-78.png-rs-high-webp.png",
null,
"https://grdp.co/cdn-cgi/image/quality=90,fit=contain,width=350,f=auto/https://gs-post-images.grdp.co/2020/10/cbcca27e2c6466c31bc325138f63a2d56c765004222747957b04e0c4ea8791f0poster-high-webp.png",
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null,
"https://grdp.co/cdn-cgi/image/quality=90,fit=contain,width=350,f=auto/https://gs-post-images.grdp.co/2020/8/061258e8ffc988707a061c4341e350fcbf9a2ddce74ccd1c1116adbce61d862aposter-high-webp.png",
null,
"https://gs-post-images.grdp.co/2020/8/group-2x-img1598864374422-78.png-rs-high-webp.png",
null,
"https://grdp.co/cdn-cgi/image/quality=90,fit=contain,width=350,f=auto/https://gs-post-images.grdp.co/2020/10/aeac18dd6041478ab09be4cf6646d589e3d066816cd3c16391b4ccb3c72f32ccposter-high-webp.png",
null,
"https://gs-post-images.grdp.co/2020/8/group-2x-img1598864374422-78.png-rs-high-webp.png",
null,
"https://grdp.co/cdn-cgi/image/quality=90,fit=contain,width=350,f=auto/https://gs-post-images.grdp.co/2020/10/8e91cdc558cdb5b70a370211c161d7dc3ea45570cdb2e897e4ff4393f8039229poster-high-webp.png",
null,
"https://gs-post-images.grdp.co/2020/8/group-2x-img1598864374422-78.png-rs-high-webp.png",
null,
"https://grdp.co/cdn-cgi/image/quality=90,fit=contain,width=350,f=auto/https://gs-post-images.grdp.co/2020/10/b18a714e618b47b96916ba3ce52775fc717ac207ee9c13a78c5c8e788f52bf02poster-high-webp.png",
null,
"https://gs-post-images.grdp.co/2020/8/group-2x-img1598864374422-78.png-rs-high-webp.png",
null,
"https://grdp.co/cdn-cgi/image/height=128,quality=80,f=auto/https://gs-post-images.grdp.co/2020/8/group-7-3x-img1597928525711-15.png-rs-high-webp.png",
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https://se.mathworks.com/help/comm/ref/pmdemod.html | [
"# pmdemod\n\nPhase demodulation\n\n## Syntax\n\n```z = pmdemod(y,Fc,Fs,phasedev) z = pmdemod(y,Fc,Fs,phasedev,ini_phase) ```\n\n## Description\n\n`z = pmdemod(y,Fc,Fs,phasedev)` demodulates the phase-modulated signal `y` at the carrier frequency `Fc` (hertz). `z` and the carrier signal have sampling rate `Fs` (hertz), where `Fs` must be at least `2*Fc`. The `phasedev` argument is the phase deviation of the modulated signal, in radians.\n\n`z = pmdemod(y,Fc,Fs,phasedev,ini_phase)` specifies the initial phase of the modulated signal, in radians.\n\n## Examples\n\ncollapse all\n\nSet the sample rate. To plot the signals, create a time vector.\n\n```fs = 50; t = (0:2*fs+1)'/fs;```\n\nCreate a sinusoidal input signal.\n\n`x = sin(2*pi*t) + sin(4*pi*t);`\n\nSet the carrier frequency and phase deviation.\n\n```fc = 10; phasedev = pi/2;```\n\nModulate the input signal.\n\n`tx = pmmod(x,fc,fs,phasedev);`\n\nPass the signal through an AWGN channel.\n\n`rx = awgn(tx,10,'measured');`\n\nDemodulate the noisy signal.\n\n`y = pmdemod(rx,fc,fs,phasedev);`\n\nPlot the original and recovered signals.\n\n```figure; plot(t,[x y]); legend('Original signal','Recovered signal'); xlabel('Time (s)') ylabel('Amplitude (V)')```",
null,
"## Version History\n\nIntroduced before R2006a"
]
| [
null,
"https://se.mathworks.com/help/examples/comm/win64/RecoverPhaseModulatedSignalAWGNChannelExample_01.png",
null
]
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https://boazcommunitycorp.org/1034-multiplication-factor.html | [
"# Multiplication factor\n\nIf, for example, there is an increase of 10% to a certain value, we can calculate the new value just by multiplying that value by 1,10, which is the multiplication factor. If the increase is 20%, we multiply by 1,20, and so on. See other examples in the table below:\n\n Increase or Profit Multiplication factor 10% 1,10 15% 1,15 20% 1,20 47% 1,47 67% 1,67\n\nExample: Increasing 10% in the amount of \\$ 10.00 we have: 10 * 1.10 = \\$ 11.\n\nIn case there is a decrease, we will have:\n\nMultiplication Factor = 1 - Discount Rate (in decimal form)\n\nSee examples in the table below:\n\n Discount Multiplication factor 10% 0,90 25% 0,75 34% 0,66 60% 0,40 90% 0,10\n\nExample: Discounting 10% of R \\$ 10,00 we have: 10 * 0,90 = \\$ 9.00\n\nNext content: Flat geometry"
]
| [
null
]
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https://www.secn.net/article/20258.html | [
"# JavaSE基础——（16）List集合\n\n三、栈和队列\n\n5.1泛型的基本概述\n\n5.2泛型的好处\n\n5.3泛型的使用\n\n5.4ArrayList存储字符串使用泛型遍历\n\n5.5泛型的由来\n\n6.1泛型类概述与使用\n\n6.2泛型方法概述与使用\n\n6.3泛型通配符\n\n7.1增强for循环的概述和使用\n\n7.2三种迭代是否能删除\n\n9.1可变参数概述与使用\n\n9.2Arrays工具类的asList()方法\n\nimport java.util.ArrayList;\nimport java.util.Iterator;\npublic class ListTest {\npublic static void main(String []args){\nArrayList list=new ArrayList();\nlist.add(\"a\");\nlist.add(\"a\");\nlist.add(\"b\");\nlist.add(\"b\");\nlist.add(\"b\");\nlist.add(\"c\");\nlist.add(\"c\");\nArrayList newList=getSingle(list);\nSystem.out.println(newList);\n}\npublic static ArrayList getSingle(ArrayList list){\nArrayList newList=new ArrayList<>();\nIterator it=list.iterator();\nwhile(it.hasNext()){\nObject obj=it.next();\nif(!newList.contains(obj)){\nnewList.add(obj);\n}\n}\nreturn newList;\n}\n}",
null,
"public void addFirst(E e)//将元素添加到链表头部\npublic void addLast(E e)//将元素添加到链表尾部\npublic E getFirst()//获取链表头部元素\npublic E getLast()//获取链表尾部元素\npublic E removeFirst()//删除链表头部元素，并返回删除的元素\npublic E removeLast()//删除链表尾部元素，并返回删除的元素\npublic E get(int index)//获取指定索引处的元素\n\n三、栈和队列\n\nimport java.util.LinkedList;\npublic class ListTest {\npublic static void main(String []args){\nStack stack=new Stack();\nstack.in(\"a\");\nstack.in(\"b\");\nstack.in(\"c\");\nwhile(!stack.isEmpty()){\nSystem.out.println(stack.out());\n}\n}\n}\nclass Stack{\nprivate LinkedList list=new LinkedList();\n//模拟进栈\npublic void in(Object obj){\nlist.add(obj);\n}\n//模拟出栈\npublic Object out(){\nreturn list.removeLast();\n}\n//判断栈元素是否为空\npublic boolean isEmpty(){\nreturn list.isEmpty();\n}\n}",
null,
"5.1泛型的基本概述\n\n5.2泛型的好处\n\n5.3泛型的使用\npublic class PatternTest {\npublic static void main(String []args){\nArrayList<Student> list=new ArrayList<Student>();\nlist.add(new Student(\"测试1\",20));\nlist.add(new Student(\"测试2\",21));\nlist.add(new Student(\"测试3\",22));\nlist.add(new Student(\"测试4\",23));\nIterator<Student> it=list.iterator();\nwhile(it.hasNext()){\nStudent stu=it.next();\nSystem.out.println(stu.getName()+\" \"+stu.getAge());\n}\n}\n}",
null,
"5.4ArrayList存储字符串使用泛型遍历\npublic class PatternTest {\npublic static void main(String []args){\nArrayList<String> list=new ArrayList<>();\nlist.add(\"a\");\nlist.add(\"b\");\nlist.add(\"c\");\nIterator<String> it=list.iterator();\nwhile(it.hasNext()){\nSystem.out.println(it.next());\n}\n}\n}",
null,
"5.5泛型的由来\n\n6.1泛型类概述与使用\n\nclass myPattern<T>{\nprivate T t;\npublic myPattern(T t) {\nthis.t=t;\n}\npublic T getT() {\nreturn t;\n}\npublic void setT(T t) {\nthis.t = t;\n}\n}\n\n6.2泛型方法概述与使用\n\npublic <泛型类型> 返回值类型 方法名(泛型类型 形参名)\n\npublic <T> void show(T t){\nSystem.out.println(t);\n}\n\n6.3泛型通配符\n\nList<?> list=new ArratList<String>();//当右边的泛型是不确定时，左边可以指定为？\n\n7.1增强for循环的概述和使用\n\n//遍历数组\nint arr[]={1,2,3};\nfor(int i:arr){\nSystem.out.println(i);\n}\n//遍历集合\nArrayList<String> list=new ArrayList<String>();\nlist.add(\"a\");\nlist.add(\"b\");\nlist.add(\"c\");\nfor(String i:list){\nSystem.out.println(i);\n}",
null,
"7.2三种迭代是否能删除\n\nArrayList<String> list=new ArrayList<String>();\nlist.add(\"a\");\nlist.add(\"b\");\nlist.add(\"b\");\nlist.add(\"c\");\n//for循环删除，索引要--\nfor(int i=0;i<list.size();i++){\nif(\"b\".equals(list.get(i))){\nlist.remove(i--);\n}\n}\n//迭代器删除，用迭代器自身的remove方法\nIterator<String> it=list.iterator();\nwhile(it.hasNext()){\nif(\"b\".equals(it.next())){\nit.remove();\n}\n}\n//增强for循环不可以删除，只能遍历\nfor(String str:list){\nif(\"b\".equals(str)){\nlist.remove(\"b\");\n}\n}\nSystem.out.println(list);\n\nimport static 包名.类名.方法名\n\nimport static java.util.Arrays.sort;\npublic class StaicImportTest {\npublic static void main(String []args){\nint arr[]={5,7,4,9,0,1,6,2,8,3};\nsort(arr);\nfor(int i=0;i<arr.length;i++){\nSystem.out.print(arr[i]+\" \");\n}\n}\n}",
null,
"9.1可变参数概述与使用\n\npublic static void main(String []args){\nint arr[]={5,7,4,9,0,1,6,2,8,3};\nprint(arr);\nprint(1,2,3);\n}\npublic static void print(int... arr){\nfor(int i=0;i<arr.length;i++){\nSystem.out.print(arr[i]+\" \");\n}\nSystem.out.println();\n}",
null,
"9.2Arrays工具类的asList()方法\nstatic List<T> asList(T... a)//返回一个受指定数组支持的固定大小的列表\n\nString str[]={\"a\",\"b\",\"c\"};\nList list= Arrays.asList(str);\nSystem.out.println(list.toString());",
null,
"ArrayList<ArrayList<Student>> gradeSeven=new ArrayList<>();\nArrayList<Student> classOne=new ArrayList<>();//一班\nclassOne.add(new Student(\"一班测试1\",20));\nclassOne.add(new Student(\"一班测试2\",21));\nclassOne.add(new Student(\"一班测试3\",22));\nArrayList<Student> classTwo=new ArrayList<>();//二班\nclassTwo.add(new Student(\"二班测试1\",20));\nclassTwo.add(new Student(\"二班测试2\",21));\n//将班级添加到学校中\ngradeSeven.add(classOne);\ngradeSeven.add(classTwo);\n//遍历年级集合，输出年级中的每个人\nfor(ArrayList<Student> i:gradeSeven){\nfor(Student j:i){\nSystem.out.println(j);\n}\n}",
null,
"## 指针连接两个字符串_极限Redis (3) 动态字符串\n\nRedis是使用C语言开发的，C语言中的字符串大都是通过字符数组实现的，但是使用字符数组有以下不足：字符数组的长度都是固定，容易发生空指针获取字...\n\n## 未定义与 char 类型的输入参数相对应的函数_MATLAB机器人仿真，调用link函数编译错误以及解决方法详解...\n\nMATLAB做机器人仿真，必须要使用的link函数，调用错误以及解决方法MATLAB比较著名的仿真就是simulink，这个仿真使用起来还是比较简单的\u000f..."
]
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"https://img-blog.csdnimg.cn/20210115224353532.png",
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| {"ft_lang_label":"__label__zh","ft_lang_prob":0.544626,"math_prob":0.7317815,"size":5884,"snap":"2021-04-2021-17","text_gpt3_token_len":2800,"char_repetition_ratio":0.1680272,"word_repetition_ratio":0.07013575,"special_character_ratio":0.27974167,"punctuation_ratio":0.24910071,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.989355,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-16T16:50:02Z\",\"WARC-Record-ID\":\"<urn:uuid:6d21e210-7be0-4e7c-8ebd-8d8e696c2ac7>\",\"Content-Length\":\"35581\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:043dd608-179e-484b-9151-f73460572d78>\",\"WARC-Concurrent-To\":\"<urn:uuid:68b27a1f-2a93-45d1-9e8d-17b6ab135b60>\",\"WARC-IP-Address\":\"104.21.49.25\",\"WARC-Target-URI\":\"https://www.secn.net/article/20258.html\",\"WARC-Payload-Digest\":\"sha1:LVE53QRVTYQZEXCQETK3GEVC43AQ66W7\",\"WARC-Block-Digest\":\"sha1:6HUIM224EROBSDBH6GHRCOOESVG5AI2V\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038088245.37_warc_CC-MAIN-20210416161217-20210416191217-00247.warc.gz\"}"} |
https://www.itipapers.com/question/amplitude-factor-or-peak-factor-%E0%A4%95%E0%A5%8D%E0%A4%AF%E0%A4%BE-%E0%A4%B9%E0%A5%8B%E0%A4%A4%E0%A5%87-%E0%A4%B9%E0%A5%88/ | [
"# Amplitude Factor or Peak Factor क्या होते है\n\nDWQA QuestionsCategory: QuestionsAmplitude Factor or Peak Factor क्या होते है\n\nAmplitude Factor or Peak Factor क्या होते है Form Factor and peak factor formula Form factor and peak factor PDF What is peak Factor Form Factor and peak Factor Peak Factor Formula Form Factor formula What is form factor Value of peak factor"
]
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null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5005947,"math_prob":0.53068775,"size":755,"snap":"2023-40-2023-50","text_gpt3_token_len":263,"char_repetition_ratio":0.29294273,"word_repetition_ratio":0.11764706,"special_character_ratio":0.3986755,"punctuation_ratio":0.011235955,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9717656,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-06T01:02:51Z\",\"WARC-Record-ID\":\"<urn:uuid:13b3c7fd-d733-4bc1-be82-0f1940f975e6>\",\"Content-Length\":\"54919\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:71395d5f-94ef-4811-bc9a-5911102d8312>\",\"WARC-Concurrent-To\":\"<urn:uuid:d91bef85-a01b-4d1f-85cd-0a72a16a355c>\",\"WARC-IP-Address\":\"156.67.210.56\",\"WARC-Target-URI\":\"https://www.itipapers.com/question/amplitude-factor-or-peak-factor-%E0%A4%95%E0%A5%8D%E0%A4%AF%E0%A4%BE-%E0%A4%B9%E0%A5%8B%E0%A4%A4%E0%A5%87-%E0%A4%B9%E0%A5%88/\",\"WARC-Payload-Digest\":\"sha1:GTUR6HAL752RD6SFE2K2PGZOMQAQC4FQ\",\"WARC-Block-Digest\":\"sha1:BYUSHBXUBJADTZJ4SJRUVZAN6USTBTZL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100575.30_warc_CC-MAIN-20231206000253-20231206030253-00674.warc.gz\"}"} |
https://leverageedu.com/blog/waves-class-11/ | [
"# Waves Class 11 Notes\n\nFrom Gravitation and Thermal Properties to Kinetic Theory of Gases and Laws of Motion, the Physics syllabus for class 11th is structured not only to build a strong foundation of the subject but also to prepare students for various international and government exams after 12th Science. Though all are equally important, Waves is one of the fundamental topics of Physics and one needs to develop a stronghold over its concepts. So, to make your work easy, here is a blog that elucidates upon the important terms, definitions, as well as concepts included in the Waves class 11 chapter!\n\n## What is a Wave?\n\nA form of disturbance that travels through a material medium as a result of repeated periodic motion of the medium particles about their mean positions without any transportation of matter is called a wave.\n\n## Characteristics of Waves\n\nWaves have a set of characteristics that have been elucidated in the Waves class 11 chapter. Its summary has been given a rundown below:\n\n• Particles of a medium traversed by a wave execute comparatively smaller vibrations about their mean positions but aren’t permanently displaced in the direction of wave propagation.\n• The successive particles of a medium execute a motion quite similar to its predecessors, either along or perpendicular to the line of travel of the wave.\n• Only transfer of energy takes place during the motion of waves.\n\n## Different Types of Waves\n\nMajorly, there are 3 different types of Waves, i.e. Mechanical, Matter, and Electromagnetic waves. Though the concept has been clearly explained in the Waves class 11 chapter, here are the definitions of these types:\n\n#### Mechanical Waves\n\nThey are produced and propagated in material medium only and are governed by Newton’s laws of motion. Furthermore, this type of wave is bifurcated into 2 types, Transverse and Longitudinal. As explained in Wave class 11 chapter, here are the key details you need to know about types of Mechanical Waves.\n\n• Transverse wave motion- in these waves, medium particles vibrate at right angles to the direction of wave propagation. Waves on surface water, strings, and electromagnetic waves are its examples.\n• Longitudinal wave motion- the wave particles move to and fro about their mean positions along the direction of energy propagation. These are also called pressure waves. An example can be sound waves\n\n#### Electromagnetic Waves\n\nProduction and propagation of these waves require no material medium, and they can pass through the vacuum or any other material medium. The most prominent examples include:\n\n• Ultraviolet\n• Microwave\n• X-rays\n• Infrared\n• Gamma Rays\n\n#### Matter Waves\n\nThe third and the most important concept included in the Wave class 11 chapter is on the Matter waves. They are associated with moving material particles like protons, neutrons, electrons etc.\n\n## Waves Class 11 Chapter: Important Terms and Definitions\n\nIn the Waves class 11 chapter, there are several other terms that have been used to describe the essential properties of a wave. Understanding them is important to find solutions to numerical as well as theoretical problems. Here are some important terminologies which you must be aware of.\n\n##### Wavelength\n\nWavelength (ʎ) is the distance travelled by a disturbance during the period of one vibration of a particle of the medium. For a transverse wave, the wavelength can also be defined as the distance between two crests or two troughs. In contrast, for a longitudinal wave, the wavelength will be equal to the distance from the centre of one compression (or refraction) to another compression.\n\n##### Wave Velocity\n\nWave velocity is the rate of time of propagation of a wave in a medium. It is different from the particle velocity, and it depends on the nature of the wave. The formula for wave velocity is:\n\n##### Amplitude\n\nAmplitude of a wave is the maximum displacement of the constituent medium particles from their mean positions.\n\n##### Frequency\n\nFrequency is the number of vibrations made by a wave-particle in one second. In mathematical equations, it is represented by ‘v’. The S.I. unit for frequency is Hertz (Hz), and the formula is:\n\n##### Time Period\n\nThe time period refers to the time taken by a medium particle to complete one vibration. Its unit is seconds (s). The formula to calculate time period is:\n\n## Waves Class 11 Important Formulas\n\nNow that you are through with the basic concepts, let us take a quick look at some of the most important formulas included in Wave class 11 chapter.\n\n##### The velocity of the transverse wave in a stretched string\n\nHere ‘T’ is the tension in the string, and µ refers to the mass per unit length of the string. µ is also used to refer to the linear mass density of a string. he S.I. unit of µ is kg/m or kg m˄-1.\n\n##### The velocity of a longitudinal wave in an elastic medium\n\nHere ‘E’ is the modulus of elasticity of the medium, and ρ refers to the density of the medium. In the case of solids, E stands for Young’s modulus of elasticity (Y).\n\n## Waves Class 11 Notes for NEET & IIT JEE\n\nWaves Class 11 chapter is also needed to prepare for the physics section in the syllabus for NEET and IIT JEE. Here are the important topics you need to study for NEET and IIT JEE:\n\nWhat is a Wave?\n\nA wave can be simply described as disturbance that results in the propagation of energy from one place to another without the transport of matter.\n\nWhat are the two properties for the propagation of a wave in a medium?\n\n1. Elasticity\n2. Inertia\n\nWhat are the types of waves?\n\n1. Mechanical Waves\n2. Electromagnetic Waves\n3. Matter Waves\n\nWhat are the various terms related to waves?\n\n1. Amplitude (A): The maximum displacement of the constituent medium particles from their mean positions.\n2. Periodic Time (T): The time taken by a medium particle to complete one vibration. Its unit is seconds (s).\n3. Wave Frequency (n): The number of vibrations made by a wave-particle in one second.\n4. Wavelength (ʎ): The distance travelled by a disturbance during the period of one vibration of a particle of the medium.\n\nRight from class 11, be it the BiPC subjects or MPC subjects, the syllabus is so comprehensive and extensive that the foundation has to be a model of perfection. Apart from the Waves class 11, you also need to clear your concepts of other important chapters of the subject! Exploring education options abroad? Not sure where to study after 12th? Then take the assistance of the counsellors at Leverage Edu who will provide guidance not only in course selection but will also assist in completing the application formalities."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.92721236,"math_prob":0.9024727,"size":6395,"snap":"2022-27-2022-33","text_gpt3_token_len":1367,"char_repetition_ratio":0.12940072,"word_repetition_ratio":0.07855823,"special_character_ratio":0.20390931,"punctuation_ratio":0.095753536,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9634249,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-26T05:00:00Z\",\"WARC-Record-ID\":\"<urn:uuid:b36bc58e-f388-4cdd-b3d1-df7a1dc1b506>\",\"Content-Length\":\"405384\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:21fa4b69-d252-48bd-9b95-6cfceda096d5>\",\"WARC-Concurrent-To\":\"<urn:uuid:3ef7f2c7-a947-41f1-9e7a-e5a62aee4b83>\",\"WARC-IP-Address\":\"35.154.56.127\",\"WARC-Target-URI\":\"https://leverageedu.com/blog/waves-class-11/\",\"WARC-Payload-Digest\":\"sha1:3TJ57X2DRMW33N6MUEU734SFZ76EUPAT\",\"WARC-Block-Digest\":\"sha1:J55TWLA7HWBON7Q2MGSIVVL2CWZ25EO4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103037089.4_warc_CC-MAIN-20220626040948-20220626070948-00261.warc.gz\"}"} |
https://learn.microsoft.com/en-us/previous-versions//ff452070(v=technet.10)?redirectedfrom=MSDN | [
"# TOTALYTD Function (DAX)\n\nEvaluates the year-to-date value of the expression in the current context.\n\n``````TOTALYTD(<expression>,<dates>[,<filter>][,<year_end_date>])\n``````\n\n## Parameters\n\n Parameter Definition expression An expression that returns a scalar value. dates A column that contains dates. filter (optional) An expression that specifies a filter to apply to the current context. year_end_date (optional) A literal string with a date that defines the year-end date. The default is December 31.\n\n## Return Value\n\nA scalar value that represents the expression evaluated for the current year-to-date dates.\n\n## Remarks\n\nNote\n\nTo understand more about how context affects the results of formulas, see Context in DAX Formulas.\n\nThe dates argument can be any of the following:\n\n• A reference to a date/time column,\n\n• A table expression that returns a single column of date/time values,\n\n• A Boolean expression that defines a single-column table of date/time values.\n\nNote\n\nConstraints on Boolean expressions are described in the topic, CALCULATE Function (DAX).\n\nNote\n\nThe filter expression has restrictions described in the topic, CALCULATE Function (DAX).\n\nThe year_end_date parameter is a string literal of a date, in the same locale as the locale of the client where the workbook was created. The year portion of the date is ignored.\n\n### Example\n\nThe following sample formula creates a measure that calculates the 'year running total' or 'year running sum' for the Internet sales.\n\nTo see how this works, create a PivotTable and add the fields, CalendarYear, CalendarQuarter, and MonthNumberOfYear, to the Row Labels area of the PivotTable. Then add a measure, named Year-to-date Total, using the formula defined in the code section, to the Values area of the PivotTable.\n\n### Code\n\n``````=TOTALYTD(SUM(InternetSales_USD[SalesAmount_USD]),DateTime[DateKey])\n``````\n\n#### Other Resources\n\nALL Function (DAX)\n\nCALCULATE Function (DAX)\n\nDATESYTD Function (DAX)\n\nTOTALMTD Function (DAX)\n\nTOTALQTD Function (DAX)\n\nGet Sample Data for PowerPivot"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.73396856,"math_prob":0.8946882,"size":1822,"snap":"2023-40-2023-50","text_gpt3_token_len":401,"char_repetition_ratio":0.12871288,"word_repetition_ratio":0.022304833,"special_character_ratio":0.20581779,"punctuation_ratio":0.11006289,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9889015,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-03T11:29:22Z\",\"WARC-Record-ID\":\"<urn:uuid:f767bb0e-b3fd-43ad-af23-d3c86a339f28>\",\"Content-Length\":\"38093\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:35886fe7-ad6d-41d7-a0c6-a2a7124ac2cb>\",\"WARC-Concurrent-To\":\"<urn:uuid:f5a7a7e1-64fe-4b4b-8948-c02173730fe9>\",\"WARC-IP-Address\":\"23.50.126.168\",\"WARC-Target-URI\":\"https://learn.microsoft.com/en-us/previous-versions//ff452070(v=technet.10)?redirectedfrom=MSDN\",\"WARC-Payload-Digest\":\"sha1:VQ2FGB6UGMC2FXQBY2NM3L4XLCHDE5QS\",\"WARC-Block-Digest\":\"sha1:QMEFVVPSPFV3BDKGQASV3XDDNUR2QKHU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511075.63_warc_CC-MAIN-20231003092549-20231003122549-00645.warc.gz\"}"} |
http://www.leastcommonmultiple.net/lcm-of-179/ | [
"X\nX\n\n# Calculate the Least Common Multiple or LCM of 179\n\nThe instructions to find the LCM of 179 are the next:\n\n## 1. Decompose all numbers into prime factors\n\n 179 179 1\n\n## 2. Write all numbers as the product of its prime factors\n\n Prime factors of 179 = 179\n\n## 3. Choose the common and uncommon prime factors with the greatest exponent\n\nCommon prime factors: 179\n\nCommon prime factors with the greatest exponent: 1791\n\nUncommon prime factors: None\n\nUncommon prime factors with the greatest exponent: None\n\n## 4. Calculate the Least Common Multiple or LCM\n\nRemember, to find the LCM of several numbers you must multiply the common and uncommon prime factors with the greatest exponent of those numbers.\n\nLCM = 1791 = 179\n\nAlso calculates the:"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8786978,"math_prob":0.8755655,"size":726,"snap":"2021-31-2021-39","text_gpt3_token_len":171,"char_repetition_ratio":0.21191135,"word_repetition_ratio":0.2,"special_character_ratio":0.2493113,"punctuation_ratio":0.09022556,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.998843,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-21T16:29:59Z\",\"WARC-Record-ID\":\"<urn:uuid:33e3f036-1f31-47e1-b225-092aa436e8ee>\",\"Content-Length\":\"35052\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1e90ab88-0d52-4cac-bf2c-3d69aa343b2b>\",\"WARC-Concurrent-To\":\"<urn:uuid:e5ccad68-5bd7-49e7-8eb8-67b5dde28508>\",\"WARC-IP-Address\":\"107.170.60.201\",\"WARC-Target-URI\":\"http://www.leastcommonmultiple.net/lcm-of-179/\",\"WARC-Payload-Digest\":\"sha1:LICXGRFSK26TPEXTWOZJEB7JWQ6WSOOR\",\"WARC-Block-Digest\":\"sha1:KVG65DWF5NLELVJBGIIEPHOBCQUW2BPU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057225.57_warc_CC-MAIN-20210921161350-20210921191350-00615.warc.gz\"}"} |
http://ibpsexamguide.org/verbal-reasoning/daily-reasoning-quiz/reasoning-quiz-11.html | [
"# Reasoning Quiz – 11\n\nDirections (Q. 1-3) Study the following information to answer the given questions.\nL, M, N, O, P, Q and R are sitting around a circle facing the centre. O is sitting between L and R. Q is 2nd to the right of R and P is 2nd to the right of Q. N is not an immediate neighbour of R.\n\n1.Which of the following is not correct?\na) R is 2nd to the right of L\nb) M is 2nd to the left of N\nc) L sits exactly between O and P\nd) P and N are immediate neighbours\ne) P sits to the opposite of N\n\n2.How many persons are seated between L and Q if we count anticlockwise from L to Q?\na) One\nb) Two\nc) Three\nd) Four\ne) More than four\n\n3.Who is to the immediate left of P?\na) L\nb) N\nc) M\nd) O\ne) None of these\n\nDirections (Q. 4-5) Four of the following five are alike in a certain way based on their seating positions in the above arrangement and so form a group. Which is the one that does not belong to the group?\n\n4.\na) QPN\nb) MOR\nc) LRO\nd) RQM\ne) ROL\n\n5.\na) LP\nb) RM\nc) MQ\nd) NP\ne) OR\n\nDirections (Q. 6-10) Study the following information to answer the given questions.\nA, B, C, X, Y, Z are seated in a straight line facing North. C is 3rd to the right of Z and B sits 2nd to the right of C. X sits to the immediate right of A.\n\n6.Which of the following represents the pair of persons sitting exactly in the middle of the line?\na) XB\nb) ZB\nc) BX\nd) XC\ne) XY\n\n7.What is X’s position with respect to Z?\na) Immediate right of Z\nb) 2nd to the left\nc) 3rd to the right\nd) 2nd to the right\ne) None of these\n\n8.Four of the following five are alike in a certain way based on their seating positions in the above arrangement and so form a group. Which is the one that does not belong to the group?\na) ZA\nb) XC\nc) CY\nd) YB\ne) XA\n\n9.How many persons are seated between A and C?\na) One\nb) two\nc) Three\nd) Four\ne) None of these\n\n10.If A:X and Z:A, then Y:\na) Y\nb) B\nc) A\nd) X\ne) None of these\n\nDirections (Q. 11-15): In the following questions, assuming the given statements to be true find out which conclusion(s) is/are definitely true and give answer:\n1) If only conclusion I is true.\n2) If only conclusion II is true.\n3) If either conclusion I or II is true.\n4) If neither conclusion I nor conclusion II is true.\n5) If both conclusion I and II are true.\n\n11. Statements: K < L, K > M >=N > O\nConclusions:\nI. O < M\nII. O < K\n\n12. Statements: N >=J = M, S > B > R < O = M\nConclusions:\nI. N >=O\nII. S > R\n\n13. Statement: W <=Y > Z = X > P > J\nConclusions:\nI. Y > P\nII. Z < W\n\n14. Statements: N = P , P<=F, F >=L, L = K\nConclusions:\nI. F = K\nII. F > K\n\n15. Statements: A > B >=C, D > E >=F, D > C\nConclusions:\nI. E > C\nII. F > B\n\nDirections (Q. 16-21) Study the following information to answer the given questions.\nJ, P, Q, R, S, T, U and V are 4 married couples sitting in a circle facing the centre. The profession of the males within the group are Lecturer, Lawyer, Doctor and Scientist. Among the males, only R (the Lawyer) and V (the Scientist) are sitting together. Each man is seated besides his wife. U, the wife of the Lecturer is seated 2nd to the right of V. T is seated between U and V. P is the wife of the Doctor. Q is not the Doctor. S is a male.\n\n16.Which of the following is P’s position with respect to S?\na) 2nd to the right\nb) 2nd to the left\nc) Immediate right\nd) Immediate left\ne) 3rd to the left\n\n17.Which of the following is J’s position with respect to T?\na) 3rd to the left\nb) 4th to the right\nc) 3rd to the right\nd) Opposite T\ne) 2nd to the right\n\n18.Which of the following is not true regarding the couples?\na) P is the wife of S\nb) T is the wife of Q\nc) R is the husband of J\nd) J and S are seated adjacent to each other\ne) All are true\n\n19.The wives of which two husbands are immediate neighbors?\na) UT\nb) SR\nc) VQ\nd) RV\ne) None of these\n\n20.Four of the following are alike in a certain way based on their seating position in the above arrangement and so form a group. Which is the one that does not belong to the group?\na) RSJ\nb) TRV\nc) UTV\nd) SGP\ne) UPQ"
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https://www.javatpoint.com/program-to-print-the-elements-of-an-array-present-on-even-position | [
"## Q. Program to print the elements of an array present on even position.\n\n### Explanation\n\nIn this program, we need to print the element which is present in even position.\n\nEven positioned element can be found by traversing the array and incrementing the value of i by 2.",
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"In the above array, elements on even position are b and d.\n\n### Algorithm\n\n1. Declare and initialize an array.\n2. Calculate the length of the declared array.\n3. Loop through the array by initializing the value of variable \"i\" to 1 (because first even positioned element lies on i = 1) then incrementing its value by 2, i.e., i=i+2.\n4. Print the elements present in even positions.\n\n### Python\n\nOutput:\n\n```Elements of given array present on even position:\n2\n4\n```\n\n### C\n\nOutput:\n\n```Elements of given array present on even position:\n2\n4\n```\n\n### JAVA\n\nOutput:\n\n```Elements of given array present on even position:\n2\n4\n```\n\n### C#\n\nOutput:\n\n```Elements of given array present on even position:\n2\n4\n```\n\n### PHP\n\nOutput:\n\n```Elements of given array present on even position:\n2\n4\n```\n\nNext Topic#\n\n## Please Share",
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https://www.geeksforgeeks.org/how-to-calculate-percent-minus-px-in-sass/?ref=leftbar-rightbar | [
"# How to calculate percent minus px in SASS ?\n\nCalculating the difference between percent and px is not as simple as 50% – 30px. Obviously you’ll be getting incompatible unit error. This is because SASS cannot perform arithmetic operations on values that cannot be converted from one unit to another. SASS does not know exactly how wide a “percentage (%)” is in terms of pixels or any other unit. Only the browser is capable of knowing such things.\n\nSo here comes the need of calc() function.\n\nAbout calc(): The calc() function allows us to perform mathematical operations on property values. Instead of declaring, for example, static pixel values for an element’s width, we can use calc() to specify that the width is the result of the addition of two or more numeric values.\n\n• Example:\n\n `.class{ ` ` ``height``: calc(``30px` `+ ``50px``); ` `} `\n\nComplied file:\n\n```.class {\nheight: calc(30px + 50px);\n}```\n• But why do we need this here ?\nThe calc() function provides a better solution for the following reason. We can combine different units. Specifically, we can mix relative units such as percentages and viewport units, with absolute units such as pixels. For example, we can create an expression that will subtract a pixel value from a percentage value.\n\n• Example:\n\n `.class { ` ` ``width``: calc(``50%` `+ ``30px``); ` `} `\n\nComplied file:\n\n```.class {\nwidth: calc(50% + 30px);\n}```\n• Let’s come to our case now that is subtracting px from %. Using the calc() function in our SCSS code we can literally do the subtraction of two different units easily.\n\n• Example:\n\n `.class { ` ` ``height``: calc(``50%` `- ``20px``); ` `} `\n\nComplied file:\n\n```.class {\nheight: calc(50% - 20px);\n}```\n• Sometimes if your values are in variables, you may need to use interpolation to turn them into strings.\n\n• Example:\n\n `\\$x: 50%; ` `\\$y: 20px; ` ` ` `.class { ` ` ``width: calc(#{\\$x} - #{\\$y}); ` `} `\n\nComplied file:\n\n```.class {\nheight: calc(50% - 20px);\n}```\n\nNote: The calc() function can be used to perform addition, subtraction, multiplication, and division calculations with numeric property values. Specifically, it can be used with length, frequency, angle, time, number, or integer data types.\n\nexample:\n\n `.class { ` ` ``width``: calc(``50``vmax + ``3``rem); ` ` ``padding``: calc(``1``vw + ``1em``); ` ` ``transform: rotate( calc(``1``turn + ``28``deg)); ` ` ``background``: hsl(``100``, calc(``3` `* ``20%``), ``40%``); ` ` ``font-size``: calc(``50``vw / ``3``); ` `} `\n\nComplied file:\n\n```.class {\nwidth: calc(50vmax + 3rem);\ntransform: rotate(calc(1turn + 28deg));\nbackground: hsl(100, calc(3 * 20%), 40%);\nfont-size: calc(50vw / 3);\n}```\n\nMy Personal Notes arrow_drop_up",
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"Check out this Author's contributed articles.\n\nIf you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks.\n\nPlease Improve this article if you find anything incorrect by clicking on the \"Improve Article\" button below.\n\nArticle Tags :\n\n1\n\nPlease write to us at [email protected] to report any issue with the above content."
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https://studylib.net/doc/10281732/review-torque | [
"# Review Torque",
null,
"```Review Torque\nMC Section:\n1. A solid uniform sphere, of mass M and radius R is pivoted about an axis that is tangential to its\nsurface. It’s moment of inertia is:\na. MR2\nb. (2/5)MR2\nc. (3/5)MR2\nd. (5/2)MR2\ne. (7/5)MR2\n2. A disk with a rotational inertia of 5.0 kg m2 and a radius of 0.25 m rotates on a frictionless fixed\naxis perpendicular to the disk and through its center. A force of 8.0 N is applied in the plane of\nthe disk and tangent to the rim. If the disk starts at rest, then after it has turned through half a\nrevolution its angular velocity is:\n3. A 16 kg block is attached to a cord that is wrapped around the\nrim of a flywheel of diameter 0.4 m and hangs vertically as\nshown. The rotational inertia of the flywheel is 0.50 kg m2.\nThen the block is released and the cord unwinds, the\nacceleration of the block is:\na. 0.15g\nb. 0.56g\nc. 0.84g\nd. g\ne. 1.3g\nFor problems 4-6 refer to the picture below of a strut and hanging mass\n4. A 5.0 m weightless strut, hinged to the wall, is used to support an 800 N block as shown. What is\nthe tension force on the 3.0 m string?\na. 600 N\nb. 750 N\nc. 800 N\nd. 1000 N\n3m\ne. 1200 N\n5. If the strut has a weight of 400 N what is the tension on the string?\na. 600 N\nb. 750 N\nc. 800 N\nd. 1000 N\ne. 1200 N\n800N\n6. The horizontal force on the hinge for the strut is closest to:\na. 600 N\nb. 750 N\nc. 800 N\nd. 1000 N\ne. 1200 N\n7. A wheel rolls without slipping along a horizontal road as shown. The wheel has a center of mass\nvelocity of 12 m/s and a radius of 0.5 m. The instantaneous velocity of point P is:\nω\na. 6 m/s\nb. 12 m/s\nc. 24 m/s\nvcm\nd. zero\ne. not enough information\nP\n8. A man, holding a weight in each hand, stands at the center of a horizontal frictionless rotating\nturntable. The effect of the weights is to double the rotational inertia of the system consisting of\nthe man, the turntable, and the weights. As he is rotating, the man opens his hands and drops the\ntwo weights. They fall outside the turntable. As a result:\na. his angular velocity doubles\nb. his angular velocity remains the same\nc. his angular velocity is halved\nd. the direction of his angular momentum vector changes\ne. none of the above occur\nF\nF\nF\n9. A uniform disk, a thin hoop and a uniform sphere, all with the same mass and same radius are each\nfree to rotate about an axis through its center. Assume the hoop is connected to the central axis\nby thin spokes of negligible mass. Identical forces are simultaneously applied to the rims as\nshown. All objects start from rest. Rank the objects according to their angular momenta after a\ngiven time t, least to greatest.\na. All the same\nb. Disk, hoop, sphere\nc. Hoop, disk, sphere\nd. Hoop, sphere, disk\ne. Hoop, disk, sphere\n1. E 2. D 3. B 4. A 5. B 6. B 7. D 8. B 9. A\nFR Section:\n1. (1999 APC Mech Test) As shown above, a uniform disk is mounted to an axle and is free to\nrotate without friction. A thin uniform rod is rigidly attached to the disk so that it will rotate\nwith the disk. A block is attached to the end of the rod. Properties of the disk, rod, and block\nare as follows:\nDisk:\nRod:\nBlock:\nmass = 3m, radius = R, rotational inertia about center ID = 3/2mR2\nmass = m, length = 2R, rotational inertia about one end is IR = 4/3mR2\nmass = 2m\nThe system is held in equilibrium with the rod at an angle θ0 to the vertical, as shown above,\nby a horizontal string of negligible mass with one end attached to the disk and the other to a\nwall. Express your answers to the following in terms of m, R, θ0, and g\n(a) Determine the tension in the string.\nThe string is now cut, and the disk-rod-block system is free to rotate.\n(b) Determine the following for the instant immediately after the string is cut.\ni. The magnitude of the angular acceleration of the disk.\nii. The magnitude of the linear acceleration of the mass at the end of the rod.\nAs the disk rotates, the rod passes the horizontal position shown above.\n(c) Determine the linear speed of the mass at the end of the rod for the instant the rod is in\nthe horizontal position.\n2. A system is composed of two identical disks. Initially, the bottom disk is rotating with angular\nvelocity ω0 around a fixed frictionless axle, and the top disk is at rest. The top disk is now\ndropped onto the bottom one, and eventually both disks rotate with the same final angular\nvelocity ωf (around the same axis). There is no net external torque acting on the two-disk system.\na. Write an expression for ωf in terms of ω0.\nb. Has the rotational kinetic energy of the system changed? If so, how?\n3. A bowling ball is thrown with an initial speed of 9m/s. Initially it slides along the lane\nwithout spinning, but eventually catches and starts spinning. The coefficient of kinetic\nfriction between the ball and the lane is 0.24.\na. What is the balls linear acceleration and angular acceleration?\nvcm\nt=0\nω\nb. How long until the ball starts rolling without slipping?\nt=?\nc. How far has the ball slid?\nd. What is its linear speed and angular speed when the ball begins smooth rolling\n(i.e. rolling without slipping)\n```"
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https://www.slideserve.com/uriah-farmer/sections-4-1-4-2-4-3 | [
"",
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"Download Presentation",
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"Sections 4.1, 4.2, 4.3\n\n# Sections 4.1, 4.2, 4.3 - PowerPoint PPT Presentation\n\nSections 4.1, 4.2, 4.3. Important Definitions in the Text :. The definition of joint probability mass function (joint p.m.f.). Definition 4.1-1. The definitions of marginal probability mass function (marginal p.m.f.) and the independence of random variables. Definition 4.1-2.",
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"I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.\nDownload Presentation",
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"## Sections 4.1, 4.2, 4.3\n\nAn Image/Link below is provided (as is) to download presentation\n\nDownload Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.\n\n- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -\nPresentation Transcript\n1. Sections 4.1, 4.2, 4.3 Important Definitions in the Text: The definition of joint probability mass function (joint p.m.f.) Definition 4.1-1 The definitions of marginal probability mass function (marginal p.m.f.) and the independence of random variables Definition 4.1-2 If the joint p.m.f. of (X, Y) is f(x,y), and S is the corresponding outcome space, then the mathematical expectation, or expected value, of u(X,Y) is If the marginal p.m.f. of X is f1(x), and S1 is the corresponding outcome space, then E[v(X)] can be calculated from either An analogous statement can be made about E[v(Y)] .\n\n2. 1. Twelve bags each contain two pieces of candy, one red and one green. In two of the bags each piece of candy weighs 1 gram; in three of the bags the red candy weighs 2 grams and the green candy weighs 1 gram; in three of the bags the red candy weighs 1 gram and the green candy weighs 2 grams; in the remaining four bags each piece of candy weighs 2 grams. One bag is selected at random and the following random variables are defined: X = weight of the red candy , Y = weight of the green candy . 1/4 1/3 2 y 1 1/6 1/4 The space of (X, Y) is {(1,1) (1,2) (2,1) (2,2)}. 1 2 x 1 — 6 if (x, y) = (1, 1) 1 — 4 The joint p.m.f. of (X, Y) is f(x, y) = if (x, y) = (1, 2) , (2, 1) 1 — 3 if (x, y) = (2, 2)\n\n3. if x = 1 if x = 2 5 / 12 The marginal p.m.f. of X is f1(x) = 7 / 12 if y = 1 if y = 2 5 / 12 The marginal p.m.f. of Y is f2(y) = 7 / 12 A formula for the joint p.m.f. of (X,Y) is f(x, y) = x + y —— if (x, y) = (1, 1) , (1, 2) , (2, 1) , (2, 2) 12 A formula for the marginal p.m.f. of X is f1(x) = x + 1 x + 2 —— + —— = 12 12 2x + 3 ——— if x = 1, 2 12 f(x, 1) + f(x, 2) = A formula for the marginal p.m.f. of Y is f2(y) = 1 + y 2 + y —— + —— = 12 12 2y + 3 ——— if y = 1, 2 12 f(1, y) + f(2, y) =\n\n4. Sections 4.1, 4.2, 4.3 Important Definitions in the Text: The definition of joint probability mass function (joint p.m.f.) Definition 4.1-1 The definitions of marginal probability mass function (marginal p.m.f.) and the independence of random variables Definition 4.1-2 If the joint p.m.f. of (X, Y) is f(x,y), and S is the corresponding outcome space, then the mathematical expectation, or expected value, of u(X,Y) is E[u(X,Y)] = u(x,y)f(x,y) (x,y) S If the marginal p.m.f. of X is f1(x), and S1 is the corresponding outcome space, then E[v(X)] can be calculated from either or v(x)f1(x) v(x)f(x,y) x S (x,y) S 1 An analogous statement can be made about E[v(Y)] .\n\n5. 1. - continued E(X) = (1)(5/12) + (2)(7/12) = 19/12 E(X2) = (1)2(5/12) + (2)2(7/12)= 11/4 Var(X) = 11/4 – (19/12)2= 35/144 E(Y) = (1)(5/12) + (2)(7/12) = 19/12 E(Y2) = (1)2(5/12) + (2)2(7/12)= 11/4 Var(Y) = 11/4 – (19/12)2= 35/144 Since _________________________, then the random variables X and Y _______________ independent. f(x, y) f1(x)f2(y) are not Using the joint p.m.f., E(X + Y) = (1+1)(1/6) + (1+2)(1/4) + (2+1)(1/4) + (2+2)(1/3) = 19 / 6 Alternatively, E(X + Y) = 19/12 + 19/12 = 19 / 6 E(X) + E(Y) =\n\n6. Using the joint p.m.f., E(X–Y) = (1–1)(1/6) + (1–2)(1/4) + (2–1)(1/4) + (2–2)(1/3) = 0 Alternatively, E(X–Y) = 19/12 – 19/12 = 0 E(X + Y) can be interpreted as the mean of the total weight of candy in the bag. E(X–Y) can be interpreted as the mean of how much more the red candy in the bag weighs than the green candy. E(XY) = (1)(1)(1/6) + (1)(2)(1/4) + (2)(1)(1/4) + (2)(2)(1/3) = 5/2 Cov(X,Y) =\n\n7. 1. - continued = The least squares lines for predicting Y from X is The least squares lines for predicting X from Y is\n\n8. The conditional p.m.f. of Y | X = 1 is Y | X = 2 is For x = 1, 2, a formula for the conditional p.m.f. of Y | X = x is\n\n9. 1. - continued The conditional p.m.f. of X | Y = 1 is X | Y = 2 is For y = 1, 2, a formula for the conditional p.m.f. of X | Y = y is\n\n10. E(Y | X = 1) = E(Y2 | X = 1) = Var(Y | X = 1) = E(Y | X = 2) = E(Y2 | X = 2) = Var(Y | X = 2) = Is the conditional mean of Y given X = x a linear function of the given value, that is, can we write E(Y | X = x) = a + bx ?\n\n11. 1. - continued E(X | Y = 1) = E(X2 | Y = 1) = Var(X | Y = 1) = E(X | Y = 2) = E(X2 | Y = 2) = Var(X | Y = 2) = Is the conditional mean of X given Y = y a linear function of the given value, that is, can we write E(X | Y = y) = c + dy ?\n\n12. 2. An urn contains six chips, one \\$1 chip, two \\$2 chips, and three \\$3 chips. Two chips are selected at random and without replacement. The following random variables are defined: X = dollar value of the first chip selected , Y = dollar value of the second chip selected . The space of (X, Y) is {(1,2) (1,3) (2,1) (2,2) (2,3) (3,1) (3,2) (3,3)}. 1/10 1/5 1/5 3 y 2 1 1/15 1/15 1/5 1/15 1/10 1 — 5 if (x, y) = (2, 3) , (3, 2) , (3, 3) 1 2 3 x 1 — 10 if (x, y) = (1, 3) , (3, 1) The joint p.m.f. of (X, Y) is f(x, y) = 1 — 15 if (x, y) = (1, 2) , (2, 1) , (2, 2)\n\n13. 2. - continued if x = 1 if x = 2 if x = 3 1 / 6 The marginal p.m.f. of X is f1(x) = 1 / 3 1 / 2 if y = 1 if y = 2 if y = 3 1 / 6 The marginal p.m.f. of Y is f2(y) = 1 / 3 1 / 2 A formula for the joint p.m.f. of (X,Y) is f(x, y) = (There seems to be no easy formula.) A formula for the marginal p.m.f. of X is f1(x) = x / 6 if x = 1, 2, 3 A formula for the marginal p.m.f. of Y is f2(y) = y / 6 if y = 1, 2, 3\n\n14. E(X) = 7 / 3 E(X2) = 6 Var(X) = 6 – (7 / 3)2 = 5 / 9 E(Y) = 7 / 3 E(Y2) = 6 Var(Y) = 6 – (7 / 3)2 = 5 / 9 Since _________________________, then the random variables X and Y _______________ independent. f(x, y) f1(x)f2(y) are not P(X + Y < 4) = P[(X,Y) = (1,2)] + P[(X,Y) = (2,1)] = 1 / 15 + 1 / 15 = 2 / 15 Using the joint p.m.f., E(XY) = (1)(2)(2/30) + (1)(3)(3/30) + (2)(1)(2/30) + (2)(2)(2/30) + (2)(3)(6/30) + (3)(1)(3/30) + (3)(2)(6/30) + (3)(3)(6/30) = 16 / 3\n\n15. 2. - continued Cov(X,Y) = = The least squares lines for predicting Y from X is The least squares lines for predicting X from Y is\n\n16. The conditional p.m.f. of Y | X = 1 is Y | X = 2 is Y | X = 3 is For x = 1, 2, 3, a formula for the conditional p.m.f. of Y | X = x is\n\n17. 2. - continued The conditional p.m.f. of X | Y = 1 is X | Y = 2 is X | Y = 3 is For y = 1, 2, 3, a formula for the conditional p.m.f. of X | Y = y is\n\n18. E(Y | X = 1) = E(X | Y = 1) = E(Y2 | X = 1) = E(X2 | Y = 1) = Var(Y | X = 1) = Var(X | Y = 1) = E(Y | X = 2) = E(X | Y = 2) = E(Y2 | X = 2) = E(X2 | Y = 2) = Var(Y | X = 2) = Var(X | Y = 2) = E(Y | X = 3) = E(X | Y = 3) = E(Y2 | X = 3) = E(X2 | Y = 3) = Var(Y | X = 3) = Var(X | Y = 3) =\n\n19. 2. - continued Is the conditional mean of Y given X = x a linear function of the given value, that is, can we write E(Y | X = x) = a + bx ? Is the conditional mean of X given Y = y a linear function of the given value, that is, can we write E(X | Y = y) = c + dy ?\n\n20. 3. An urn contains six chips, one \\$1 chip, two \\$2 chips, and three \\$3 chips. Two chips are selected at random and with replacement. The following random variables are defined: X = dollar value of the first chip selected , Y = dollar value of the second chip selected . The space of (X, Y) is {(1,1) (1,2) (1,3) (2,1) (2,2) (2,3) (3,1) (3,2) (3,3)}. 3 y 2 1 1/12 1/6 1/4 1/18 1/9 1/6 1/36 1/18 1/12 1 2 3 x xy — 36 x = 1, 2, 3 if y = 1, 2, 3 The joint p.m.f. of (X, Y) is f(x, y) =\n\n21. 3. - continued if x = 1 if x = 2 if x = 3 1 / 6 The marginal p.m.f. of X is f1(x) = 1 / 3 1 / 2 if y = 1 if y = 2 if y = 3 1 / 6 The marginal p.m.f. of Y is f2(y) = 1 / 3 1 / 2 A formula for the joint p.m.f. of (X,Y) is f(x, y) = (The formula was found previously) A formula for the marginal p.m.f. of X is f1(x) = x / 6 if x = 1, 2, 3 A formula for the marginal p.m.f. of Y is f2(y) = y / 6 if y = 1, 2, 3\n\n22. E(X) = 7 / 3 E(X2) = 6 Var(X) = 6 – (7 / 3)2 = 5 / 9 E(Y) = 7 / 3 E(Y2) = 6 Var(Y) = 6 – (7 / 3)2 = 5 / 9 Since _________________________, then the random variables X and Y _______________ independent. f(x, y) =f1(x)f2(y) are P(X + Y < 4) = P[(X,Y) = (1,1)] + P[(X,Y) = (1,2)] + P[(X,Y) = (2,1)] = 1 / 36 + 1 / 18 + 1 / 18 = 5 / 36\n\n23. 3. - continued 3 3 (xy) (xy / 36) = x = 1 y = 1 3 3 (xy) (x / 6) (y / 6) = x = 1 y = 1 E(XY) = 3 3 (x) (x / 6) (y) (y / 6) = x = 1 y = 1 E(X) E(Y) = (7/3)(7/3) = 49 / 9 Cov(X,Y) = = The least squares lines for predicting Y from X is The least squares lines for predicting X from Y is\n\n24. For x = 1, 2, 3, the conditional p.m.f. of Y | X = x is E(Y | X = x) = Var(Y | X = x) = For y = 1, 2, 3, the conditional p.m.f. of X | Y = y is E(X | Y = y) = Var(X | Y = y) = Is the conditional mean of Y given X = x a linear function of the given value, that is, can we write E(Y | X = x) = a + bx ? Is the conditional mean of X given Y = y a linear function of the given value, that is, can we write E(X | Y = y) = c + dy ?\n\n25. For continuous type random variables (X, Y), the definitions of joint probability density function (joint p.d.f.), independence of X and Y, and mathematical expectation are each analogous to those for discrete type random variables, with summation signs replaced by integral signs. The covariance between random variables X and Y is The correlation between random variables X and Y is y Consider the equation of a line y = a + bx which comes “closest” to predicting the values of the random variable Y from the random variable X in the sense that E{[Y– (a + bX)]2} is minimized. x\n\n26. We let k(a,b) = E{[Y – (a + bX)]2} = To minimize k(a,b) , we set the partial derivatives with respect to a and b equal to zero. (Note: This is textbook exercise 4.2-5.) k — = a k — = b (Multiply the first equation by X, subtract the resulting equation from the second equation, and solve for b. Then substitute in place of b in the first equation to solve for a.)\n\n27. b = The least squares line for predicting Y from X is a = The least squares line for predicting Y from X can be written The least squares line for predicting X from Y can be written\n\n28. The conditional p.m.f./p.d.f. of Y given X = x is defined to be The conditional p.m.f./p.d.f. of X given Y = y is defined to be The conditional mean of Y given X = x is defined to be The conditional variance of Y given X = x is defined to be\n\n29. The conditional mean of X given Y = y and the conditional variance of X given Y = y are each defined similarly. For continuous type random variables (X, Y), the definitions of conditional mean and variance are each analogous to those for discrete type random variables, with summation signs replaced by integral signs. Suppose X and Y are two discrete type random variables, and E(Y | X = x) = a + bx. Then, for each possible value of x, Multiplying each side by f1(x), Summing each side over all x,\n\n30. Now, multiplying each side of by x f1(x), Summing each side over all x,\n\n31. The two equations and are essentially the same as those in the derivation of the least squares line for predicting Y from X. This derivation is analogous for continuous type random variables with summation signs replaced by integral signs. Consequently, if E(Y | X = x) = a + bx (i.e., if E(Y | X = x) is a linear function of x), then a and b must be respectively the intercept and slope in the least squares line for predicting Y from X. Similarly, if E(X | Y = y) = c + dy (i.e., if E(X | Y = y) is a linear function of y), then c and d must be respectively the intercept and slope in the least squares line for predicting X from Y.\n\n32. Suppose a set contains N = N1 + N2 + N3 items, where N1 items are of one type, N2 items are of a second type, and N3 items are of a third type; n items are selected from the N items at random and without replacement. If the random variable X1 is defined to be the number of selected n items that are of the first type, the random variable X2 is defined to be the number of selected n items that are of the second type, and the random variable X3 is defined to be the number of selected n items that are of the third type, then the joint distribution of (X1 , X2 , X3) is called a trivariate hypergeometric distribution. Since X3 = n – X1 – X2 , X3 is totally determined by X1 and X2 . The joint p.m.f. of (X1 , X2) is Each Xi has a distribution. If the number of types of items is any integer k > 1 with (X1 , X2 , … , Xk) defined in the natural way, then the joint p.d.f. is called a multivariate hypergeometric distribution.\n\n33. Suppose each in a sequence of independent trials must result in one of outcome 1, outcome 2, or outcome 3. The probability of outcome 1 on each trial is p1 , the probability of outcome 2 on each trial is p2 , and the probability of outcome 3 on each trial is p3 = 1 – p1 – p2 . If the random variable X1 is defined to be the number of the n trials resulting in outcome 1, the random variable X2 is defined to be the number of the n trials resulting in outcome 2, and the random variable X3 is defined to be the number of the n trials resulting in outcome 3, then the joint distribution of (X1 , X2 , X3) is called a trinomial distribution. Since X3 = n – X1 – X2 , X3 is totally determined by X1 and X2 . The joint p.m.f. of (X1 , X2) is Each Xi has a distribution. If the number of outcomes is any integer k > 1 with (X1 , X2 , … , Xk) defined in the natural way, then the joint p.d.f. is called a multinomial distribution.\n\n34. 4. (a) An urn contains 15 red chips, 10 blue chips, and 5 white chips. Eight chips are selected at random and without replacement. The following random variables are defined: X1 = number of red chips selected , X2 = number of blue chips selected , X3 = number of white chips selected . Find the joint p.m.f. of (X1 , X2 , X3) . (X1 , X2 , X3) have a trivariate hypergeometric distribution, and X3 = 8 –X1–X2 is totally determined by X1 and X2 . The joint p.m.f. of (X1 , X2) is 15 x1 10 x2 5 8 – x1 – x2 x1 = 0, 1, …, 8 if x2 = 0, 1, …, 8 f(x1, x2) = 30 8 3 x1 + x2 8\n\n35. (b) Find the marginal p.m.f. for each of X1 , X2 , and X3 . Each of X1 , X2 , and X3 has a hypergeometric distribution. 15 x1 15 8 – x1 f1(x1) = if x1 = 0, 1, …, 8 30 8 10 x2 20 8 – x2 f2(x2) = if x2 = 0, 1, …, 8 30 8\n\n36. 4. - continued (c) 5 x3 25 8 – x3 f3(x3) = if x3 = 0, 1, …, 5 30 8 Are X1 , X2 , and X3 independent? Why or why not? X1, X2, X3 cannot possibly be independent, because any one of these random variables is totally determined by the other two.\n\n37. (d) Find the probability that at least two of the selected chips are blue or at least two chips are white. P({X2 2} {X3 2}) = 1 – P({X2 1} {X3 1}) = 1 – [P(X2= 0 ,X3 = 0) + P(X2= 1 ,X3 = 0) + P(X2= 0 ,X3 = 1) + P(X2= 1 ,X3 = 1)] = 15 8 15 7 10 1 15 7 5 1 15 6 10 1 5 1 1 – + + + 30 8 30 8 30 8 30 8\n\n38. 4. - continued (e) (f) Find the conditional p.m.f. of X1 | x2 . X1 | x2 can be treated as “the number of red chips selected when For x2 = X1 | x2 has a distribution with p.m.f. E(X1 | x2) can be written as a linear function of x2 , since Therefore, the least squares E(X1 | x2) =\n\n39. line for predicting X1 from X2 must be (g) E(X2 | x1) can be written as a linear function of x2 , since E(X2 | x1) = Therefore, the least squares line for predicting X2 from X1 must be (h) Find the covariance and correlation between X1 and X2 by making use of the following facts (instead of using direct formulas): The slope in the least squares line for predicting X1 from X2 is The slope in the least squares line for predicting X2 from X1 is The product of the slope in the least squares line for predicting X1 from X2 and the slope in the least squares line for predicting X2 from X1 is equal to .\n\n40. Suppose a set contains N = N1 + N2 + N3 items, where N1 items are of one type, N2 items are of a second type, and N3 items are of a third type; n items are selected from the N items at random and without replacement. If the random variable X1 is defined to be the number of selected n items that are of the first type, the random variable X2 is defined to be the number of selected n items that are of the second type, and the random variable X3 is defined to be the number of selected n items that are of the third type, then the joint distribution of (X1 , X2 , X3) is called a trivariate hypergeometric distribution. Since X3 = n – X1 – X2 , X3 is totally determined by X1 and X2 . N1 x1 N2 x2 N – N1– N2 n – x1–x2 The joint p.m.f. of (X1 , X2) is if x1 and x2 are “appropriate” integers N n Each Xi has a distribution. If the number of types of items is any integer k > 1 with (X1 , X2 , … , Xk) defined in the natural way, then the joint p.d.f. is called a multivariate hypergeometric distribution. hypergeometric\n\n41. 5. (a) An urn contains 15 red chips, 10 blue chips, and 5 white chips. Eight chips are selected at random and with replacement. The following random variables are defined: X1 = number of red chips selected , X2 = number of blue chips selected , X3 = number of white chips selected . Find the joint p.m.f. of (X1 , X2 , X3) . (X1 , X2 , X3) have a trinomial distribution, and X3 = 8 –X1–X2 is totally determined by X1 and X2 . The joint p.m.f. of (X1 , X2) is x1 x2 8 –x1–x2 8! 1 — 2 1 — 3 1 — 6 f(x1, x2) = x1! x2! (8 –x1–x2)! x1 = 0, 1, …, 8 if x2 = 0, 1, …, 8 x1 + x2 8\n\n42. (b) Find the marginal p.m.f. for each of X1 , X2 , and X3 . Each of X1 , X2 , and X3 has a binomial distribution. 8 8! 1 — 2 f1(x1) = if x1 = 0, 1, …, 8 x1! (8 –x1)! x2 8 – x2 8! 1 — 3 2 — 3 f2(x2) = if x2 = 0, 1, …, 8 x2! (8 –x2)!\n\n43. 5. - continued (c) x3 8 – x3 8! 1 — 6 5 — 6 f3(x3) = if x3 = 0, 1, …, 8 x3! (8 –x3)! Are X1 , X2 , and X3 independent? Why or why not? X1, X2, X3 cannot possibly be independent, because any one of these random variables is totally determined by the other two.\n\n44. (d) Find the probability that at least two of the selected chips are blue or at least two chips are white. P({X2 2} {X3 2}) = 1 – P({X2 1} {X3 1}) = 1 – [P(X2= 0 ,X3 = 0) + P(X2= 1 ,X3 = 0) + P(X2= 0 ,X3 = 1) + P(X2= 1 ,X3 = 1)] = 8 7 8! 1 — 2 1 — 2 1 — 3 1 – + + 7! 1! 7 6 8! 8! 1 — 2 1 — 6 1 — 2 1 — 3 1 — 6 + 7! 1! 6! 1! 1!\n\n45. 5. - continued (e) (f) Find the conditional p.m.f. of X1 | x2 . X1 | x2 can be treated as “the number of red chips selected when For x2 = X1 | x2 has a distribution with p.m.f. E(X1 | x2) can be written as a linear function of x2 , since Therefore, the least squares E(X1 | x2) =\n\n46. line for predicting X1 from X2 must be (g) E(X2 | x1) can be written as a linear function of x2 , since E(X2 | x1) = Therefore, the least squares line for predicting X2 from X1 must be (h) Find the covariance and correlation between X1 and X2 by making use of the following facts (instead of using direct formulas): The slope in the least squares line for predicting X1 from X2 is The slope in the least squares line for predicting X2 from X1 is The product of the slope in the least squares line for predicting X1 from X2 and the slope in the least squares line for predicting X2 from X1 is equal to .\n\n47. Suppose each in a sequence of independent trials must result in one of outcome 1, outcome 2, or outcome 3. The probability of outcome 1 on each trial is p1 , the probability of outcome 2 on each trial is p2 , and the probability of outcome 3 on each trial is p3 = 1 – p1 – p2 . If the random variable X1 is defined to be the number of the n trials resulting in outcome 1, the random variable X2 is defined to be the number of the n trials resulting in outcome 2, and the random variable X3 is defined to be the number of the n trials resulting in outcome 3, then the joint distribution of (X1 , X2 , X3) is called a trinomial distribution. Since X3 = n – X1 – X2 , X3 is totally determined by X1 and X2 . The joint p.m.f. of (X1 , X2) is x1 x2 n–x1–x2 n! p1p2 (1 –p1–p2) x1! x2! (n–x1–x2)! if x1 and x2 are non-negative integers such that x1 + x2n Each Xi has a distribution. If the number of outcomes is any integer k > 1 with (X1 , X2 , … , Xk) defined in the natural way, then the joint p.d.f. is called a multinomial distribution. b( , ) n pi\n\n48. 6. One chip is selected from each of two urns, one containing three chips labeled distinctively with the integers 1 through 3 and the other containing two chips labeled distinctively with the integers 1 and 2. The following random variables are defined: X = largest integer among the labels on the selected chips , Y = smallest integer among the labels on the selected chips . The space of (X, Y) is {(1,1) (2,1) (3,1) (2,2) (3,2)}. 1/6 1/6 2 y 1 (Note: We immediately see that X and Y cannot be independent, since the joint space is not “rectangular”.) 1/6 1/3 1/6 1 2 3 x The joint p.m.f. of (X, Y) is f(x, y) = 1 — 6 if (x, y) = (1, 1) , (3, 1) , (2, 2) , (3, 2) 1 — 3 if (x, y) = (2, 1)\n\n49. if x = 1 if x = 2, 3 1 / 6 The marginal p.m.f. of X is f1(x) = 1 / x The marginal p.m.f. of Y is f2(y) = (3 – y) / 3 if y = 1, 2 E(X) = 13 / 6 E(X2) = 31 / 6 Var(X) = 31 / 6 – (13 / 6)2 = 17 / 36 E(Y) = 4 / 3 E(Y2) = 2 Var(Y) = 2 – (4 / 3)2 = 2 / 9\n\n50. 6. - continued Since _________________________, then the random variables X and Y _______________ independent f(x, y) f1(x)f2(y) are not (as we previously noted). Using the joint p.m.f., E(XY) = (1)(1)(1/6) + (3)(1)(1/6) + (2)(2)(1/6) + (3)(2)(1/6) + (2)(1)(1/3) = 3 Cov(X,Y) = = The least squares lines for predicting Y from X is"
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https://en.m.wikipedia.org/wiki/Overflow_flag | [
"# Overflow flag\n\nIn computer processors, the overflow flag (sometime called V flag) is usually a single bit in a system status register used to indicate when an arithmetic overflow has occurred in an operation, indicating that the signed two's-complement result would not fit in the number of bits used for the operation (the ALU width). Some architectures may be configured to automatically generate an exception on an operation resulting in overflow.\n\nAn illustrative example is what happens if we add 127 and 127 using 8-bit registers. 127+127 is 254, but using 8-bit arithmetic the result would be 1111 1110 binary, which is -2 in two's complement, and thus negative. A negative result out of positive operands (or vice versa) is an overflow. The overflow flag would then be set so the program can be aware of the problem and mitigate this or signal an error. The overflow flag is thus set when the most significant bit (here considered the sign bit) is changed by adding two numbers with the same sign (or subtracting two numbers with opposite signs). Overflow never occurs when the sign of two addition operands are different (or the sign of two subtraction operands are the same).\n\nInternally, the overflow flag is usually generated by an exclusive or of the internal carry into and out of the sign bit. As the sign bit is the same as the most significant bit of a number considered unsigned, the overflow flag is \"meaningless\" and normally ignored when unsigned numbers are added or subtracted.\n\nThe overflow flag is typically changed by all arithmetic operations, including compare instructions (equivalent to a subtract instruction without storing the result). In many processor architectures, the overflow flag is cleared by bitwise operations (and, or, xor, not), possibly including shifts and rotates, but it may also be left undefined by these. Instructions such as multiply and divide often leave the flag undefined, or affected by the last partial result.\n\nOn many processors (not only x86), addition and subtraction instructions affect both the carry/borrow and overflow flags, though only one of them will normally be of interest, depending on whether the operands represented signed or unsigned numbers."
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https://www.mathlearnit.com/what-is-1-287-as-a-percentage | [
"# What is 1/287 as a percentage?\n\n1 / 287 = 0.35%\n\n## Fraction to Percent Conversion Summary\n\nWe encourage you to check out our introduction to percentage page for a little recap of what percentage is. You can also learn about fractions in our fractions section of the website. Sometimes, you may want to express a fraction in the form of a percentage, or vice-versa. This page will cover the former case. Luckily for us, this problem only requires a bit of multiplication and division. We recommend that you use a calculator, but solving these problems by hand or in your head is possibly too! Here's how we discovered that 1 / 287 = 0.35% :\n\n• Step 1: Divide 1 by 287 to get the number as a decimal. 1 / 287 = 0.0\n• Step 2: Multiply 0.0 by 100. 0.0 times 100 = 0.35. That's all there is to it!\nNote that you can reverse steps 1 and 2 and still come to the same solution. If you multiply 1 by 100 and then divide the result by 287, you will still come to 0.35!\n\n## When are fractions useful?\n\nFractions are commonly used in everyday life. If you are splitting a bill or trying to score a test, you will often describe the problem using fractions. Sometimes, you may want to express the fraction as a percentage.\n\n## Convert 1 / 287 into a percentage or decimal\n\n### Fraction Conversion Table\n\nPercentage Fraction Decimal\n0.35% 1 / 287 0.0\nRemember: Converting a fraction to a percentage or decimal are all equivalent numbers. They are all used to represent a relationship between numbers. We can choose how we represent numbers by which style matches the appropriate situation.\n\n## Find the Denominator\n\nA percentage is a number out of 100, so we need to make our denominator 100!\n\nIf the original denominator is 287, we need to solve for how we can make the denominator 100.\n\nHint: percentages all have a denominator of 100!\n\nTo convert this fraction, we would divide 100 by 287, which gives us 0.35.\n\n## Find the Numerator\n\nNow, we multiply 0.35 by 1, our original numerator, which is equal to 0.35\n\n## Convert your Fraction to a Percent\n\n0.35 / 100 = 0.35%\n\nRemember, a percentage is any number out of 100. If we can balance 1 / 287 with a new denominator of 100, we can find the percentage of that fraction!\n\n## Help your students convert 1 / 287 through further understanding:\n\n• What is the numerator of 1 / 287?\n• What is the denominator of 1 / 287?\n• When would you use 1 / 287 as a fraction? Give examples\n• When would you use 1 / 287 as a decimal? Give examples\n• When would you use 1 / 287 as a percentage? Give examples\n• What are three other fractions that convert to 0.35%?\n• Ask your students to think of three real life examples of when to use fractions vs percentages.\n• Which fraction is larger: 1 / 287 or 11 / 133?"
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https://bytes.com/topic/c/answers/63781-help-me-optimizing-speed | [
"469,329 Members | 1,470 Online\n\n# help me to optimizing speed\n\nHi there\nI added /O2 to compiler settings\nI added /Oe to compiler settings for accepting register type request , but\nit seems that is not allowed and if I remove register type for \"l\" , time of\ngenerating codes doesn't change\n\nthe original code makes some files , but I removed that section to make it\nmy system is Windows XP , 512 Mb ram , 1.6 Intel\nRegards\nArShAm\n\nCode is here :\n#include \"stdio.h\"\n#include \"stdlib.h\"\n#include \"string.h\"\n\nconst int NUM=7;//number of characters in each word\nconst int total=20000000;//total generation\nchar q[total][NUM+1];// the table\n\nvoid main(void)\n{\nint NUMBERS=26;//size of al array\nchar al[]=\"abcdefghijklmnopqrstuvwxyz\"; //array to generate a random code\nregister int l;//my windows XP doesn't respect this request\n\nfor(int i=1;i<total;i++)\n{\nfor(int j=0;j<NUM;j++)\n{\n}\n\nfor(l=0;l<i;l++)//comparing if it is unique or not\n{\nif(!strcmp(q[i],q[l]))\n{\nprintf(\" %d was equal to %d with %s:%s value\\n\",i,l,q[i],q[l]);\ni--;\nbreak;\n}\n}\nif(i%10000==0)printf(\"%d\\n\",i/10000);//each 10,000 times shows that the\nprogram is runing\n//printf(\"%s\\n\",q[i]);\n}\nprintf(\"\\r\\nDONE\");\ngetchar();\n}\nJul 19 '05 #1\n5",
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"3316",
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"ArShAm wrote:\nHi there\nOK.\nThe following information is off-topic here:\n<OT> I added /O2 to compiler settings\nI added /Oe to compiler settings for accepting register type request , but\nit seems that is not allowed and if I remove register type for \"l\" , time of\ngenerating codes doesn't change\n\nthe original code makes some files , but I removed that section to make it\nmy system is Windows XP , 512 Mb ram , 1.6 Intel <OT>\nCode is here :\n#include \"stdio.h\" #include <stdio.h>\n#include \"stdlib.h\" #include <stdlib.h>\n#include \"string.h\" #include <string.h>\n\nconst int NUM=7;//number of characters in each word\nconst int total=20000000;//total generation\nchar q[total][NUM+1];// the table\n\nvoid main(void) int main(void) // main() returns int...main() returns int...\n{\nint NUMBERS=26;//size of al array\nchar al[]=\"abcdefghijklmnopqrstuvwxyz\"; //array to generate a random code\nregister int l;//my windows XP doesn't respect this request\n\nfor(int i=1;i<total;i++)\n{\nfor(int j=0;j<NUM;j++)\n{\n}\n\nfor(l=0;l<i;l++)//comparing if it is unique or not\n{\nif(!strcmp(q[i],q[l]))\n{\nprintf(\" %d was equal to %d with %s:%s value\\n\",i,l,q[i],q[l]);\ni--;\nbreak;\n}\n}\nif(i%10000==0)printf(\"%d\\n\",i/10000);//each 10,000 times shows that the\nprogram is runing\n//printf(\"%s\\n\",q[i]);\n}\nprintf(\"\\r\\nDONE\");\ngetchar();\n}\n\nMicro-optimization -- or even compiler optimization -- is *not*\nan O(n log n) algorithm would be more appropriate.\n\nConsider using a std::set.\n\nHTH,\n--ag\n--\nArtie Gold -- Austin, Texas\nOh, for the good old days of regular old SPAM.\n\nJul 19 '05 #2\n\"ArShAm\" <ar************@hotmail.com> wrote...\nHi there\nI added /O2 to compiler settings\nI added /Oe to compiler settings for accepting register type request , but\nit seems that is not allowed and if I remove register type for \"l\" , time of generating codes doesn't change\n\nthe original code makes some files , but I removed that section to make it\nmy system is Windows XP , 512 Mb ram , 1.6 Intel\nRegards\nArShAm\n\nCode is here :\n[...]\n\nInstead of comparing each word with all others, sort them and then\nrun through the sorted array and see if any neighbours are equal.\n\nVictor\nJul 19 '05 #3\nIn article <bo*************@ID-89294.news.uni-berlin.de>,\nar************@hotmail.com says...\nHi there\nI added /O2 to compiler settings\nI added /Oe to compiler settings for accepting register type request , but\nit seems that is not allowed and if I remove register type for \"l\" , time of\ngenerating codes doesn't change\nThat almost certainly means that the compiler is putting your variable\nin a register automatically -- in fact, for most practical purposes, the\nregister keyword is obsolete.\nAs usual, the key to optimizing the code is not microscopic details like\nwhether a variable ends up in a register, but in improving the\nalgorithms and/or data structures involved.\n#include \"stdio.h\"\n#include \"stdlib.h\"\n#include \"string.h\"\nThese should really be enclosed in angle brackets instead of quotes\n(though changing that is extremely unlikely to change the speed at all\n-- and if it does, you're doing other things you really shouldn't (like\ncreating headers of your own with the same names as those supplied by\nthe system).\nconst int NUM=7;//number of characters in each word\nconst int total=20000000;//total generation\nchar q[total][NUM+1];// the table\n\nvoid main(void)\nmain always returns an int. Again, this won't really affect the speed,\nbut it's something you should do anyway.\nregister int l;//my windows XP doesn't respect this request\nXP, as such, has nothing to do with it one way or the other -- it's the\ncompiler, not the OS, that decides what goes into registers. In any\ncase, you probably have things backwards -- it isn't that it's ignoring\nyour request to put this in a register. Rather, it's putting it in a\nregister automatically, whether you ask for it or not.\nfor(int i=1;i<total;i++)\n{\nfor(int j=0;j<NUM;j++)\n{\n}\n\nfor(l=0;l<i;l++)//comparing if it is unique or not\n\nHere's where your real problem arises -- verifying uniqueness with a\nlinear search renders your overall algorithm O(N^2). Keeping the\nstrings sorted and doing a binary search will reduce this to O(N lg N)\ninstead -- a massive improvement when you're dealing with 20 million\nitems (see below for just how massive it really is).\n\nIn this case (creating 20 million _small_ strings) it's probably worth\nusing our own little string-like class instead of the full-blown\nstd::string, if only to save memory. Using std::map and my own pwd\nclass, I came up with this:\n\n#include <set>\n#include <string>\n#include <cstdlib>\n#include <iostream>\n#include <ctime>\n\nconst int total = 20000000;\n\nclass pwd {\nconst static int NUM = 7;\nconst static int NUMBERS = 26;\nchar data[NUM];\n\npublic:\n// generate a random string.\npwd() {\nstatic char letters[] = \"abcdefghijklmnopqrstuvwxyz\";\n\nfor (int i=0; i< NUM; i++)\ndata[i] = letters[std::rand() % NUMBERS];\n}\n\n// std::set requires ability to compare items.\nbool operator<(pwd const &other) const {\nreturn -1 == strncmp(data, other.data, NUM);\n}\n\n// support writing a pwd to a stream.\nfriend std::ostream &operator<<(std::ostream &os, pwd const &p) {\nreturn os.write(p.data, pwd::NUM);\n}\n};\n\nint main() {\n\nstd::srand(std::time(NULL));\n\nstd::clock_t start = clock();\nfor (unsigned long size=0; size<total; size++) {\nsize++;\nif ( size % 10000 == 0)\nstd::cout << '\\r' << size << std::flush;\n}\n\nstd::clock_t end = clock();\n\nstd::cout << \"\\nTime: \" << double(end-start)/CLOCKS_PER_SEC\n<< \"seconds\\n\";\n\n// show first and last passwords, so the optimizer won't eliminate\n// the loop above.\n\n#if 0\nstd::ostream_iterator<pwd>(std::cout, \"\\n\"));\n#endif\nreturn 0;\n}\n\nI modified your program slightly, so it would only produce 90 thousand\nstrings. I compiled that program with MS VC++ 7.1, using:\ncl /Oxb2 /G6ry pwds1.cpp\nand it ran in 56.4 seconds on my machine. Extrapolating from that, based\non an O(N^2) complexity, I estimate it would take around a month for\n\nCompiled the same way and run on the same machine, the code above\nproduces 20 million strings in about 58 seconds (i.e. 20 million strings\n_almost_ as fast as you were getting 90 thousand).\n\nA micro-optimization (like register) will rarely give an improvement\nmore than a few percent. If the compiler really screws up and you\nmanage to enregister something inside of a really tight loop, you might,\nconcievably get an improvement of, say, 50:1, but that's _extremely_\nrare (I don't think I've ever seen it). By contrast, this algorithmic\nimprovement gave an improvement of around fifty _thousand_ to 1, with\nonly minimal investment... :-)\n\n--\nLater,\nJerry.\n\nThe universe is a figment of its own imagination.\nJul 19 '05 #4\nThanks Dear Jerry,\nIt was gr8\nbefore I recieve your answer I tried to change my generation code , and I\nthink that will be much better\nbut there is a problem , and that is the program crashes with a huge number\nfor total\n\nhere is the code:\n\n#include \"stdio.h\"\n#include \"time.h\"\n#include \"stdlib.h\"\n#include \"string.h\"\n#include \"myqueue.h\"\n#include <sys/timeb.h>\n\nconst int NUM=10;\nchar element[NUM];\nint a[NUM]={0};\nchar al[]=\"ABCDEFGHIJKLMNOPQRSTUVWXYZ\";\nint size;\n\nvoid main(void)\n{\nconst int total=1000000;\nsize=sizeof(al)-2;\nvoid gen(void);\n\nmyQueue* table[total];\n\nfor(int r=0;r<NUM;r++)\nelement[r]='A';\nfor(int l=0;l<total;l++)\n{\ngen();\ntable[l]=new myQueue();\nstrcpy(table[l]->element,element);\n}\n\nsrand( (unsigned)time( NULL ) );\nint tot=total;\n\nFILE *file;\nfile=fopen(\"report.pin\",\"w\");\n\nint Sorted[total],shuffled[total];\nint i3, j3;\n\nfor ( i3 = 0; i3 < total; i3++ ) Sorted[i3] = i3;\n\nfor ( i3 = 0; i3 < total; i3++ )\n{\nj3 = rand() % (total-i3);\nshuffled[i3] = Sorted[j3];\n\nSorted[j3] = Sorted [ total-1-i3 ];\n}\n\nfor(register int i=0;i<total;i++)\n{\nfprintf(file,\"%06d:%s\\n\",i,table[shuffled[i]]->element);\n}\nfclose(file);\n\nprintf(\"Done\\n\");\nexit(0);\n}\n\nvoid gen(void)\n{\na++;\nfor(int k=0;k<NUM;k++)\n{\nif(a[k]>size)\n{\na[k]=0;\na[k+1]++;\n}\n}\n\nfor(int y=0;y<NUM;y++)\n{\nelement[y]=al[a[y]];\n}\n}\n\n//and the myQueue class :\nconst int NUM1=10;\nclass myQueue\n{\npublic:\nchar element[NUM1];\nmyQueue();\nvirtual ~myQueue();\n\nmyQueue(){\nstrcpy(element,\"\");}\n~myQueue(){}\n\n};\nRegards\nArShAm\nJul 19 '05 #5\nArShAm wrote:\nThanks Dear Jerry,\nIt was gr8\nbefore I recieve your answer I tried to change my generation code , and I\nthink that will be much better\nbut there is a problem , and that is the program crashes with a huge number\nfor total\n\nhere is the code:\n\n#include \"stdio.h\"\n#include \"time.h\"\n#include \"stdlib.h\"\n#include \"string.h\"\n#include \"myqueue.h\"\n#include <sys/timeb.h>\nI really don't understand.\nWhy are the standard header files using '\"'\nexcepth the last one? Be consistent and use\nangle brackets '<' and '>' instead of quotes (\").\n\nconst int NUM=10;\nchar element[NUM];\nint a[NUM]={0};\nchar al[]=\"ABCDEFGHIJKLMNOPQRSTUVWXYZ\";\nint size;\n\nvoid main(void)\nHmmm. How many people have told you that main()\nreturns int?\nMake the change.\n{\nconst int total=1000000;\nsize=sizeof(al)-2; Why the \"-2\"?\nI could understand a \"-1\" because of the terminating\nnul character, but why \"-2\". some comments would\nreally help.\n\nvoid gen(void); Better naming would help too.\nWhat does this function do?\n\nmyQueue* table[total]; The declaration for this class should be before\nits usage.\nWhy do you need 10,000,000 pointers to queues?\nSo far, this is a memory hog program. See below.\n\nfor(int r=0;r<NUM;r++)\nelement[r]='A'; Prefer library routines:\nmemset(element, 'A', sizeof(element));\nor\nstd::fill(element, element + num, 'A');\nthey may be optimized for your processor, where\na simple \"for\" loop isn't.\n\nfor(int l=0;l<total;l++)\n{\ngen();\ntable[l]=new myQueue();\nstrcpy(table[l]->element,element); In order for strcpy to work, there must be a\nterminating nul ('\\0') character within the\nstring.\nI didn't see any terminating null placed in the\n\"element\" array. The strcpy() function is probably\nthe cause of your crash. It will copy until it\n_finds_ a nul character or it accesses undefined or\nprotected memory.\nI think you want:\nmemcpy(table[l]->element, element, sizeof(element));\nor\nstd::copy(element, element + NUM, table[i]->element);\n}\nAt this point, you have allocated:\n10,000,000 pointers to myQueue objects\n10,000,000 myQueue objects.\nInsert these statements into your code:\ncout << \"current memory allocation: \"\n<< total * (sizeof (myQueue *) + sizeof(myQueue))\n<< \"\\n\";\nsrand( (unsigned)time( NULL ) ); Search the C FAQ and the C++ FAQ and the C language\nYou'll find some interesting information.\n\nint tot=total;\n\nFILE *file;\nfile=fopen(\"report.pin\",\"w\"); Are you programming in C or C++?\nostream file(\"report.pin\");\n\nint Sorted[total],shuffled[total]; Memory Hog!\nAt this point, you have asked the compiler to allocate\n20,000,000 integers in the \"automatic\" area (a.k.a stack).\n\ncout << \"Automatic variable allocation: \"\n<< 2 * total * sizeof(int)\n<< \"\\n\";\n\nint i3, j3;\n\nfor ( i3 = 0; i3 < total; i3++ ) Sorted[i3] = i3;\n\nfor ( i3 = 0; i3 < total; i3++ )\n{\nj3 = rand() % (total-i3);\nshuffled[i3] = Sorted[j3];\n\nSorted[j3] = Sorted [ total-1-i3 ];\n} I believe you would do better to use the shuffle algorithm\nof the STL.\n\nfor(register int i=0;i<total;i++)\n{\nfprintf(file,\"%06d:%s\\n\",i,table[shuffled[i]]->element);\n} Here is a big waste of time. Don't bother with register\nvariables. Formatted output requires a lot of time, so\nmaking the index variable as a register will not save any\nsignificant time.\nfclose(file);\n\nprintf(\"Done\\n\");\nexit(0);\n}\n\nvoid gen(void)\n{\na++;\nfor(int k=0;k<NUM;k++)\n{\nif(a[k]>size)\n{\na[k]=0;\na[k+1]++;\n}\n}\n\nfor(int y=0;y<NUM;y++)\n{\nelement[y]=al[a[y]];\n}\n} The above should not modify global variables. This poses\na hindrance when reading the code. Pass pointers to the\nvariables that will be modified. Also, use comments to\nexplain what this function is doing.\n\n//and the myQueue class :\nconst int NUM1=10;\nclass myQueue\n{\npublic:\nchar element[NUM1];\nmyQueue();\nvirtual ~myQueue();\n\nmyQueue(){\nstrcpy(element,\"\");}\n~myQueue(){}\n\n}; What is the purpose of this class?\nIt doesn't look like a queue container.\n\nRegards\nArShAm\n\nYou could gain a lot of speed by replacing the \"myQueue\"\nclass with a fixed size vector or array. The class\nprovides no useful functionality, so remove it.\n\nAlso, why do you need to store 10,000,000 random strings\nof 10 characters in length to a file. Your file will be\na minimum of 10,000,000 * 10 bytes long or 100MB.\n\nIf you're looking to write a program that generates\npossible passwords, a more successful approach is to\nuse common names, then common words. After those fail,\nthen generate the random list.\n\n--\nThomas Matthews\n\nC++ newsgroup welcome message:\nhttp://www.slack.net/~shiva/welcome.txt\nC++ Faq: http://www.parashift.com/c++-faq-lite\nC Faq: http://www.eskimo.com/~scs/c-faq/top.html\nalt.comp.lang.learn.c-c++ faq:\nhttp://www.raos.demon.uk/acllc-c++/faq.html\nOther sites:\nhttp://www.josuttis.com -- C++ STL Library book\n\nJul 19 '05 #6\n\n### This discussion thread is closed\n\nReplies have been disabled for this discussion."
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https://learn.careers360.com/ncert/question-i-purchased-a-hair-dryer-for-rs-5400-including-8-percent-vat-find-the-price-before-vat-was-added/ | [
"# 10 I purchased a hair-dryer for Rs 5,400 including 8% VAT. Find the price before VAT was added.\n\nGiven,\n\nVAT = 8%\n\nLet the original price be Rs. 100\n\nOriginal price + VAT = Rs. 100 + Rs.\n\nOriginal price + VAT = Rs. 100 + Rs. 8 = Rs. 108\n\nIf the price after VAT is Rs.5400, then the price before VAT is\n\nThe price before VAT was added is Rs. 5000.\n\nExams\nArticles\nQuestions"
]
| [
null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9232159,"math_prob":0.9836152,"size":609,"snap":"2020-24-2020-29","text_gpt3_token_len":196,"char_repetition_ratio":0.21652892,"word_repetition_ratio":0.796875,"special_character_ratio":0.3727422,"punctuation_ratio":0.1724138,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98899627,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-01T20:13:04Z\",\"WARC-Record-ID\":\"<urn:uuid:53137258-b14d-4ebe-b660-42df70890e85>\",\"Content-Length\":\"775880\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e2bc497a-b7f9-4bd5-9c33-4add72b29b96>\",\"WARC-Concurrent-To\":\"<urn:uuid:31977ef0-152e-4d08-bf78-cd9d094df777>\",\"WARC-IP-Address\":\"35.154.233.234\",\"WARC-Target-URI\":\"https://learn.careers360.com/ncert/question-i-purchased-a-hair-dryer-for-rs-5400-including-8-percent-vat-find-the-price-before-vat-was-added/\",\"WARC-Payload-Digest\":\"sha1:5JPY7DLAMG74XUCYL5YFL2OLALELMEZC\",\"WARC-Block-Digest\":\"sha1:XTMJS5MZORICMRGXKEO67RWNTRXGYKSN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347419593.76_warc_CC-MAIN-20200601180335-20200601210335-00254.warc.gz\"}"} |
https://fr.maplesoft.com/support/help/Maple/view.aspx?path=DifferentialGeometry/LieAlgebras/ChevalleyBasisDetails | [
"",
null,
"ChevalleyBasis Details - Maple Help\n Chevalley Basis Details The details for the construction of the Chevalley basis are as follows. Let be a real, split semi-simple Lie algebra. Start with a basis for $𝔤$, where ${H}_{i}$, is a basis for a Cartan subalgebra, and where , (the positive roots), gives a root space decomposition for $\\mathrm{𝔤}$. By definition of a split, semi-simple Lie algebra, the root vectors are all real. Let $B$ be the Killing form. Scale the vectorssuch that and set . Scale the vectors again (preserving ) so that the structure equations hold. Let be the simple roots, and set . This fixes the vectors in the Chevalley basis $\\mathrm{ℬ}$. Write . We need to make one final scaling of the vectors ${X}_{{\\mathrm{α}}_{}}$,for. We calculate the structure constants , for and and generate the system of quadratic equations . Here is the largest positive integer such that is not a root. Put for and solve for the remaining . Finally set and put and for . This completes the construction of the Chevalley basis $\\mathrm{ℬ}$' . We have for all , where the matrix ${a}_{\\mathrm{ij}}$ is the Cartan matrix for $\\mathrm{𝔤}$ and, also, where . Note that in the Chevalley basis all the structure constants are integers and that the transformation , is a Lie algebra automorphism. See N. Bourbaki, Lie Groups and Lie Algebras, Chapters 7-9, Section 4 for additional details."
]
| [
null,
"https://bat.bing.com/action/0",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.76490736,"math_prob":0.9985399,"size":1959,"snap":"2023-14-2023-23","text_gpt3_token_len":715,"char_repetition_ratio":0.12531969,"word_repetition_ratio":0.042056076,"special_character_ratio":0.3506891,"punctuation_ratio":0.25891182,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99679494,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-31T23:36:19Z\",\"WARC-Record-ID\":\"<urn:uuid:5105ab4f-6481-4f22-8b78-c3aa83117a21>\",\"Content-Length\":\"293097\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a97a8ebe-3af9-4ea4-b1a3-ee591e7dd662>\",\"WARC-Concurrent-To\":\"<urn:uuid:089d0556-219a-44d0-ba90-6edf07f25b67>\",\"WARC-IP-Address\":\"199.71.183.28\",\"WARC-Target-URI\":\"https://fr.maplesoft.com/support/help/Maple/view.aspx?path=DifferentialGeometry/LieAlgebras/ChevalleyBasisDetails\",\"WARC-Payload-Digest\":\"sha1:C6XL5MBSZKNL7B27S7JYQDSJ2ITYTR6X\",\"WARC-Block-Digest\":\"sha1:XJQJAXOZBPS4XECA3ODCPJYEVAEWUOGD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224647459.8_warc_CC-MAIN-20230531214247-20230601004247-00296.warc.gz\"}"} |
https://en.m.wikisource.org/wiki/Page:EB1911_-_Volume_03.djvu/290 | [
"Page:EB1911 - Volume 03.djvu/290\n\nThis page has been proofread, but needs to be validated.\n273\nBALLISTICS\nv. m. log k. Cr=gp=f(v)=vm/k.\n3600 1.55 2.3909520 $v^{1.55}\\times \\log ^{-1}{\\bar {3}}.6090480$",
null,
"2600 1.7 2.9038022 $v^{1.7}\\times \\log ^{-1}{\\bar {3}}.0961978$",
null,
"1800 2 3.8807404 $v^{2}\\times \\log ^{-1}{\\bar {4}}.1192596$",
null,
"1370 3 7.0190977 $v^{3}\\times \\log ^{-1}{\\bar {4}}.9809023$",
null,
"1230 5 13.1981288 $v^{5}\\times \\log ^{-1}{\\bar {14}}.8018712$",
null,
"970 3 7.2265570 $v^{3}\\times \\log ^{-1}{\\bar {8}}.7734430$",
null,
"790 2 4.3301086 $v^{2}\\times \\log ^{-1}{\\bar {5}}.6698914$",
null,
"The numbers have been changed from kilogramme-metre to poundfoot units by Colonel Ingalls, and employed by him in the calculation of an extended ballistic table, which can be compared with the result of the abridged table. The calculation can be carried out in each region of velocity from the formulae:—\n\n(25) $T(V)-T(v)=k\\int _{v}^{V}v^{-m}dv,\\ S(V)-S(v)=k\\int _{v}^{V}v^{m+1}dv$",
null,
"$I(V)-I(v)=gk\\int _{v}^{V}v^{-m-1}dv,$",
null,
"and the corresponding integration.\n\nThe following exercises will show the application of the ballistic table. A slide rule should be used for the arithmetical operations, as it works to the accuracy obtainable in practice.\n\nExample 1.- Determine the time t sec. and distance s ft. in which the velocity falls from 2150 to 1600 f/s\n\n(a) of a 6-in. shot weighing 100lb, taking n=0.96,\n(b) of a rifle bullet, 0.303-in. calibre, weighing half an ounce, taking n=0.8.\nV. v. T(V). T(v). t/C. S(V). S(v). s/C.\n2150 1600 28.6891 27.5457 1.1434 20700.53 18587.00 2113.53\nd. w. C. t/C. t. s/C. s.\n(a) 6 100 2.894 1.1434 3.307 2113.53 6114 (2038yds.)\n(b) 0.303 1/32 0.426 1.1434 0.486 2113.53 900 (300yds.)\n\nExample 2.- Determine the remaining velocity v and time of flight t over a range of 1000 yds. of the same two shot, fired with the same muzzle velocity V=2150f/s.\n\nS. s/C. S(V). S(v). v. T(V). T(v). t/C. t.\n(a) 3000 1037 20700.53 19663.53 1861 28.6891 28.1690 0.5201 1.505\n(b) 3000 7050 20700.53 13650.53 920* 28.6891 23.0803 5.6088 2.387\n* These numbers are taken from a part omitted here of the abridged ballistic table.\n\nIn the calculation of range tables for direct fire, defined officially as \"fire from guns with full charge at elevation not exceeding 15°,\" the vertical component of the resistance of the air may be ignored as insensible, and the actual velocity and its horizontal component, or component parallel to the line of sight, are undistinguishable.",
null,
"Fig. 1.\n\nThe equations of motion are now, the co-ordinates x and y being measured in feet,\n\n(26) ${\\frac {d^{2}x}{dt^{2}}}=-{\\mbox{r}}r=-{\\frac {gp}{C}},$",
null,
"(27) ${\\frac {d^{2}x}{dt^{2}}}=-g.$",
null,
"The first equation leads, as before, to\n\n(28) $t=C\\left\\{T(V)-T(v)\\right\\},$",
null,
"(29) :(28) $x=C\\left\\{S(V)-S(v)\\right\\},$",
null,
"The integration of (24) gives\n\n(30) ${\\frac {dy}{dt}}={\\mbox{constant}}-gt=g\\left({\\frac {1}{2}}T-t\\right),$",
null,
"if T denotes the whole time of flight from O to the point B (fig. 1), where the trajectory cuts the line of sight; so that ½T is the time to the vertex A, where the shot is flying parallel to OB.\n\nIntegrating (27) again,\n\n(31) $y=g\\left({\\frac {1}{2}}T-{\\frac {1}{2}}t^{2}\\right)=3/4gt(Tt)={\\frac {1}{2}}gt(T-t);$",
null,
"and denoting T-t by t', and taking g=32f/s2,\n\n(32) $y=16tt^{\\prime },\\,$",
null,
"which is Colonel Sladen's formula, employed in plotting ordinates of a trajectory.\n\nAt the vertex A, where y=H, we have t=t'=½T, so that\n\n(33) $H={\\frac {1}{8}}gT^{2},$",
null,
"which for practical purposes, taking g=32, is replaced by\n\n(34) $H=4T^{2},{\\mbox{ or }}(2T)^{2}.\\,$",
null,
"Thus, if the time of flight of a shell is 5 sec., the height of the vertex of the trajectory is about 100 ft.; and if the fuse is set to burst the shell one-tenth of a second short of its impact at B, the height of the burst is 7.84, say 8 ft.\n\nThe line of sight Ox, considered horizontal in range table results, may be inclined slightly to the horizon, as in shooting up or down a moderate slope, without appreciable modification of (28) and (29), and y or PM is still drawn vertically to meet OB in M.\n\nGiven the ballistic coefficient C, the initial velocity V, and a range of R yds. or X=3R ft., the final velocity v is first calculated from (29) by\n\n(35) $S(v)=S(V)-X/C,\\,$",
null,
"and then the time of flight T by\n\n(36) $T=C\\left\\{T(V)-T(v)\\right\\}.$",
null,
"Denoting the angle of departure and descent, measured in degrees and from the line of sight OB by φ and β, the total deviation in the range OB is (fig. 1)\n\n(37) $\\delta =\\phi +\\beta =C\\left\\{D(V)-D(v)\\right\\}.$",
null,
"To share the δ between φ and β, the vertex A is taken as the point of half-time (and therefore beyond half-range, because of the continual diminution of the velocity), and the velocity v0 at A is calculated from the formula\n\n(38) $T(v_{0})=T(V)-{\\frac {{\\frac {1}{2}}T}{C}}={\\frac {1}{2}}\\left\\{T(V)+T(v)\\right\\};$",
null,
"and now the degree table for D(v) gives\n\n(39) $\\phi =C\\left\\{D(V)-D(v_{0})\\right\\},$",
null,
"(40) $\\beta =C\\left\\{D(v_{0})-D(v)\\right\\},$",
null,
"This value of φ is the tangent elevation (T.E); the quadrant elevation (Q.E.) is φ-S, where S is the angular depression of the line of sight OB; and if O is h ft. vertical above B, the angle S at a range of R yds. is given by\n\n(41) $\\sin S=h/3R,\\,$",
null,
"or, for a small angle, expressed in minutes, taking the radian as 3438',\n\n(42) $S=1146h/R.\\,$",
null,
"So also the angle β must be increased by S to obtain the angle at which the shot strikes a horizontal plane — the water, for instance.\n\nA systematic exercise is given here of the compilation of a range table by calculation with the ballistic table; and it is to be compared with the published official range table which follows.\n\nA discrepancy between a calculated and tabulated result will serve to show the influence of a slight change in the coefficient of reduction n, and the muzzle velocity V.\n\nExample 3.- Determine by calculation with the abridged ballistic table the remaining velocity v, the time of flight t, angle of elevation φ, and descent β of this 6-in. gun at ranges 500, 1000, 1500, 2000 yds., taking the muzzle velocity V=2150 f/s, and a coefficient of reduction n=0.96. [For Table see p.274.]\n\nAn important problem is to determine the alteration of elevation for firing up and down a slope. It is found that the alteration of the tangent elevation is almost insensible, but the quadrant elevation requires the addition or subtraction of the angle of sight.\n\nExample.- Find the alteration of elevation required at a range of 3000 yds. in the exchange of fire between a ship and a fort 1200 ft. high, a 12-in. gun being employed on each side, firing a shot weighing 850 lb with velocity 2150 f/s. The complete ballistic table, and the method of high angle fire (see below) must be employed.",
null,
""
]
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https://en.m.wikipedia.org/wiki/Fixed_rate_mortgage | [
"# Fixed-rate mortgage\n\n(Redirected from Fixed rate mortgage)\n\nA fixed-rate mortgage (FRM) is a mortgage loan where the interest rate on the note remains the same through the term of the loan, as opposed to loans where the interest rate may adjust or \"float\". As a result, payment amounts and the duration of the loan are fixed and the person who is responsible for paying back the loan benefits from a consistent, single payment and the ability to plan a budget based on this fixed cost.\n\nOther forms of mortgage loans include interest only mortgage, graduated payment mortgage, variable rate mortgage (including adjustable-rate mortgages and tracker mortgages), negative amortization mortgage, and balloon payment mortgage. Unlike many other loan types, FRM interest payments and loan duration is fixed from beginning to end.\n\nFixed-rate mortgages are characterized by amount of loan, interest rate, compounding frequency, and duration. With these values, the monthly repayments can be calculated.",
null,
"Mortgage Loan. Total Payment (3 Fixed Interest Rates & 2 Loan Term) = Loan Principal + Expenses (Taxes & fees) + Total interest to be paid.\nThe final cost will be exactly the same:\n* when the interest rate is 2.5% and the term is 30 years than when the interest rate is 5% and the term is 15 years * when the interest rate is 5% and the term is 30 years than when the interest rate is 10% and the term is 15 years\n\n## Overview\n\nUnlike adjustable-rate mortgages (ARM), fixed-rate mortgages are not tied to an index. Instead, the interest rate is set (or \"fixed\") in advance to an advertised rate, usually in increments of 1/4 or 1/8 percent.\n\nThe fixed monthly payment for a fixed-rate mortgage is the amount paid by the borrower every month that ensures that the loan is paid off in full with interest at the end of its term.\n\n## Usage throughout the world\n\nThe United States Federal Housing Administration (FHA) helped develop and standardize the fixed rate mortgage as an alternative to the balloon payment mortgage by insuring them and by doing so helped the mortgage design garner usage. Because of the large payment at the end of the older, balloon-payment loan, refinancing risk resulted in widespread foreclosures. The fixed-rate mortgage was the first mortgage loan that was fully amortized (fully paid at the end of the loan) precluding successive loans, and had fixed interest rates and payments.\n\nFixed-rate mortgages are the most classic form of loan for home and product purchasing in the United States. The most common terms are 15-year and 30-year mortgages, but shorter terms are available, and 40-year and 50-year mortgages are now available (common in areas with high priced housing, where even a 30-year term leaves the mortgage amount out of reach of the average family).\n\nOutside the United States, fixed-rate mortgages are less popular, and in some countries, true fixed-rate mortgages are not available except for shorter-term loans. For example, in Canada the longest term for which a mortgage rate can be fixed is typically no more than ten years, while mortgage maturities are commonly 25 years. A fixed rate mortgage in Singapore has the interest rate fixed for only the first three to five years of the loan, and it then becomes variable.[citation needed] In Australia, \"honeymoon\" mortgages with introductory rates are common, but can last as short as a year, and may instead offer a fixed reduction in interest rate rather than a fixed rate itself. Furthermore, they are often combined with properties of flexible mortgages to create what is known as an Australian mortgage, which often allow borrowers to overpay to reduce interest charges and then draw on these overpayments in the future.\n\nThe mortgage industry of the United Kingdom has traditionally been dominated by building societies, whose raised funds must be at least 50% deposits, so lenders prefer variable-rate mortgages to fixed-rate mortgages to reduce asset–liability mismatch due to interest rate risk. Lenders, in turn, influence consumer decisions which already prefer lower initial monthly payments. Nationwide Commercial recently issued a 30-year fixed rate mortgage as bridging finance.\n\n## Comparisons\n\nFixed-rate mortgages are usually more expensive than adjustable rate mortgages. The inherent interest rate risk makes long-term fixed rate loans tend to have a higher interest rate than short-term loans. The relationship between interest rates for short and long-term loans is represented by the yield curve, which generally slopes upward (longer terms are more expensive). The opposite circumstance is known as an inverted yield curve and occurs less often.\n\nThe fact that a fixed-rate mortgage has a higher starting interest rate does not indicate that it is a worse type of borrowing than an adjustable-rate mortgage. If interest rates rise, the ARM will cost more, but the FRM will cost the same. In effect, the lender has agreed to take the interest rate risk on a fixed-rate loan.\n\nSome studies have shown that the majority of borrowers with adjustable rate mortgages save money in the long term but also that some borrowers pay more. The price of potentially saving money, in other words, is balanced by the risk of potentially higher costs. In each case, a choice would need to be made based upon the loan term, the current interest rate, and the likelihood that the rate will increase or decrease during the life of the loan.\n\n## Pricing\n\n• Note: Fixed-rate mortgage interest may be compounded differently in other countries, such as in Canada, where it is compounded every 6 months.\n\nThe fixed monthly payment for a fixed rate mortgage is the amount paid by the borrower every month that ensures that the loan is paid off in full with interest at the end of its term. This monthly payment $c$ depends upon the monthly interest rate $r$ (expressed as a fraction, not a percentage, i.e., divide the quoted yearly nominal percentage rate by 100 and by 12 to obtain the monthly interest rate), the number of monthly payments $N$ called the loan's term, and the amount borrowed $P_{0}$ known as the loan's principal; rearranging the formula for the present value of an ordinary annuity we get the formula for $c$ :\n\n$c={r \\over {1-(1+r)^{-N}}}P_{0}$\n\nFor example, for a home loan for \\$200,000 with a fixed yearly nominal interest rate of 6.5% for 30 years, the principal is $P_{0}=200000$ , the monthly interest rate is $r=6.5/100/12$ , the number of monthly payments is $N=30\\cdot 12=360$ , the fixed monthly payment $c=\\1264.14$ . This formula is provided using the financial function PMT in a spreadsheet such as Excel. In the example, the monthly payment is obtained by entering either of these formulas:\n\n{{{1}}}\n{{{1}}}\n${}=1264.14$\n\nThis monthly payment formula is easy to derive, and the derivation illustrates how fixed-rate mortgage loans work. The amount owed on the loan at the end of every month equals the amount owed from the previous month, plus the interest on this amount, minus the fixed amount paid every month.\n\nAmount owed at month 0:\n$P_{0}$\nAmount owed at month 1:\n$P_{1}=P_{0}+P_{0}*r-c$ ( principal + interest – payment)\n$P_{1}=P_{0}(1+r)-c$ (equation 1)\nAmount owed at month 2:\n$P_{2}=P_{1}(1+r)-c$\nUsing equation 1 for $P_{1}$\n$P_{2}=(P_{0}(1+r)-c)(1+r)-c$\n$P_{2}=P_{0}(1+r)^{2}-c(1+r)-c$ (equation 2)\nAmount owed at month 3:\n$P_{3}=P_{2}(1+r)-c$\nUsing equation 2 for $P_{2}$\n$P_{3}=(P_{0}(1+r)^{2}-c(1+r)-c)(1+r)-c$\n$P_{3}=P_{0}(1+r)^{3}-c(1+r)^{2}-c(1+r)-c$\nAmount owed at month N:\n$P_{N}=P_{N-1}(1+r)-c$\n$P_{N}=P_{0}(1+r)^{N}-c(1+r)^{N-1}-c(1+r)^{N-2}....-c$\n$P_{N}=P_{0}(1+r)^{N}-c((1+r)^{N-1}+(1+r)^{N-2}....+1)$\n$P_{N}=P_{0}(1+r)^{N}-c(S)$ (equation 3)\nWhere $S=(1+r)^{N-1}+(1+r)^{N-2}....+1$ (equation 4) (see geometric progression)\n$S(1+r)=(1+r)^{N}+(1+r)^{N-1}....+(1+r)$ (equation 5)\nWith the exception of two terms the $S$ and $S(1+r)$ series are the same so when you subtract all but two terms cancel:\nUsing equation 4 and 5\n$S(1+r)-S=(1+r)^{N}-1$\n$S((1+r)-1)=(1+r)^{N}-1$\n$S(r)=(1+r)^{N}-1$\n$S={{(1+r)^{N}-1} \\over r}$ (equation 6)\nPutting equation 6 back into 3:\n$P_{N}=P_{0}(1+r)^{N}-c{{(1+r)^{N}-1} \\over r}$\n$P_{N}$ will be zero because we have paid the loan off.\n$0=P_{0}(1+r)^{N}-c{{(1+r)^{N}-1} \\over r}$\nWe want to know $c$\n$c={{r(1+r)^{N}} \\over {(1+r)^{N}-1}}P_{0}$\nDivide top and bottom with $(1+r)^{N}$\n$c={r \\over {1-(1+r)^{-N}}}P_{0}$\n\nThis derivation illustrates three key components of fixed-rate loans: (1) the fixed monthly payment depends upon the amount borrowed, the interest rate, and the length of time over which the loan is repaid; (2) the amount owed every month equals the amount owed from the previous month plus interest on that amount, minus the fixed monthly payment; (3) the fixed monthly payment is chosen so that the loan is paid off in full with interest at the end of its term and no more money is owed.\n\nThe vendor may sell off the fix element as a \"fixed to floating\" derivative. Look up \"Black-Scholes\" to see how this is done."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9525069,"math_prob":0.9886458,"size":8433,"snap":"2021-31-2021-39","text_gpt3_token_len":1791,"char_repetition_ratio":0.19456638,"word_repetition_ratio":0.06504646,"special_character_ratio":0.21830903,"punctuation_ratio":0.09698138,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9976521,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-18T16:28:48Z\",\"WARC-Record-ID\":\"<urn:uuid:aab8fa43-9fbd-4044-99d3-56ea23aa70fe>\",\"Content-Length\":\"111098\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fd5f4c8b-00f9-4bb4-a2e8-16c0af387e58>\",\"WARC-Concurrent-To\":\"<urn:uuid:9120013b-d238-4407-abad-75572bfe14bf>\",\"WARC-IP-Address\":\"208.80.154.224\",\"WARC-Target-URI\":\"https://en.m.wikipedia.org/wiki/Fixed_rate_mortgage\",\"WARC-Payload-Digest\":\"sha1:QMMEE3DTDADUAGXN6YDIOCMKC2QSTUE2\",\"WARC-Block-Digest\":\"sha1:J6ZGBNWJPMGMAYNTSR7KGRMT7ME5VPD6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780056548.77_warc_CC-MAIN-20210918154248-20210918184248-00713.warc.gz\"}"} |
https://support.microsoft.com/en-us/office/percentile-exc-function-bbaa7204-e9e1-4010-85bf-c31dc5dce4ba?ui=en-us&rs=en-us&ad=us | [
"# PERCENTILE.EXC function\n\nThe PERCENTILE.EXC function returns the k-th percentile of values in a range, where k is in the range 0..1, exclusive.\n\n## Syntax\n\n=PERCENTILE.EXC(array,k)\n\nThe PERCENTILE.EXC function syntax has the following arguments:\n\nArgument\n\nDescription\n\narray\n\nRequired\n\nThe array or range of data that defines relative standing.\n\nk\n\nRequired\n\nThe array or range of data that defines relative standing.\n\nNotes:\n\n• If array is empty, PERCENTILE.EXC returns the #NUM! error value\n\n• If k is nonnumeric, PERCENTILE.EXC returns the #VALUE! error value.\n\n• If k is ≤ 0 or if k ≥ 1, PERCENTILE.EXC returns the #NUM! error value.\n\n• If k is not a multiple of 1/(n + 1), PERCENTILE.EXC interpolates to determine the value at the k-th percentile.\n\n• PERCENTILE.EXC will interpolate when the value for the specified percentile lies between two values in the array. If it cannot interpolate for the percentile, k specified, Excel will return #NUM! error.\n\n## Example",
null,
"## Need more help?\n\nYou can always ask an expert in the Excel Tech Community, get support in the Answers community, or suggest a new feature or improvement on Excel User Voice.",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.55249447,"math_prob":0.83653575,"size":1147,"snap":"2020-34-2020-40","text_gpt3_token_len":284,"char_repetition_ratio":0.15398075,"word_repetition_ratio":0.12765957,"special_character_ratio":0.22842197,"punctuation_ratio":0.16157205,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9566472,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,9,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-23T04:24:24Z\",\"WARC-Record-ID\":\"<urn:uuid:ba15e330-ea58-4796-909d-4bcfece8b7b7>\",\"Content-Length\":\"107403\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f48f66dd-ae53-4145-9939-9b650e992a91>\",\"WARC-Concurrent-To\":\"<urn:uuid:ba4366bd-3509-44cb-b0af-8dbf2485a80f>\",\"WARC-IP-Address\":\"184.25.98.190\",\"WARC-Target-URI\":\"https://support.microsoft.com/en-us/office/percentile-exc-function-bbaa7204-e9e1-4010-85bf-c31dc5dce4ba?ui=en-us&rs=en-us&ad=us\",\"WARC-Payload-Digest\":\"sha1:ACPYEEEUUFS4JIH2TJJMSCH3ZHVQ5Q7N\",\"WARC-Block-Digest\":\"sha1:GDDDWNYUPVWSNERE7WPLC6EAF5735WRZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400209665.4_warc_CC-MAIN-20200923015227-20200923045227-00338.warc.gz\"}"} |
https://it.mathworks.com/matlabcentral/cody/problems/14-find-the-numeric-mean-of-the-prime-numbers-in-a-matrix/solutions/2115866 | [
"Cody\n\n# Problem 14. Find the numeric mean of the prime numbers in a matrix.\n\nSolution 2115866\n\nSubmitted on 3 Feb 2020 by Hamza Maqbool\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\nx = 3; y_correct = 3; assert(isequal(meanOfPrimes(x),y_correct))\n\nout = 3\n\n2 Pass\nx = [1 2 3]; y_correct = 2.5; assert(isequal(meanOfPrimes(x),y_correct))\n\nout = 2.5000\n\n3 Pass\nx = [3 3; 3 3]; y_correct = 3; assert(isequal(meanOfPrimes(x),y_correct))\n\nout = 3\n\n4 Pass\nx = [7 3 8 8]'; y_correct = 5; assert(isequal(meanOfPrimes(x),y_correct))\n\nout = 5\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5121124,"math_prob":0.9978875,"size":644,"snap":"2020-45-2020-50","text_gpt3_token_len":219,"char_repetition_ratio":0.1578125,"word_repetition_ratio":0.0776699,"special_character_ratio":0.36645964,"punctuation_ratio":0.140625,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9973005,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-30T20:32:11Z\",\"WARC-Record-ID\":\"<urn:uuid:0da7d330-48fd-4f40-a3f7-7343bcaaea05>\",\"Content-Length\":\"80800\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5d522c2f-bf3e-430a-b809-06b56e6bc005>\",\"WARC-Concurrent-To\":\"<urn:uuid:01666fd7-92ee-41a5-878f-85687df730de>\",\"WARC-IP-Address\":\"23.223.252.57\",\"WARC-Target-URI\":\"https://it.mathworks.com/matlabcentral/cody/problems/14-find-the-numeric-mean-of-the-prime-numbers-in-a-matrix/solutions/2115866\",\"WARC-Payload-Digest\":\"sha1:ZOEX56E457LII677YZ2RLLJLMJ54X2XO\",\"WARC-Block-Digest\":\"sha1:F73HKQVVKW7YSKNLQRH2FBVOBXV3ENC4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107911229.96_warc_CC-MAIN-20201030182757-20201030212757-00252.warc.gz\"}"} |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=14&t=30786&p=111480 | [
"## Wavelength and frequency proportions [ENDORSED]\n\nBetty Wolkeba section 1L\nPosts: 32\nJoined: Fri Apr 06, 2018 11:05 am\n\n### Wavelength and frequency proportions\n\nCan someone clarify what Professor Lavelle was talking about in lecture about the wavelength and frequency relationships. For example, why is it when the frequency doubles, the wavelength halves? I'm still a little confused on this concept.\n\nKara Justeson 1B\nPosts: 31\nJoined: Fri Apr 06, 2018 11:03 am\n\n### Re: Wavelength and frequency proportions [ENDORSED]\n\nThe equation is c=hv (speed=wavelength x frequency). Since c (speed) is a constant if wavelength decreased, frequency would have to increase in order for the constant c to remain the same (3.00 x 10^8). Essentially, the wavelength and frequency have to equal the constant c when they're multiplied together, so their relationship works like that in order for the constant of speed to remain at it's value. Hope that makes sense!\n\narina_m 1A\nPosts: 30\nJoined: Fri Apr 06, 2018 11:04 am\n\n### Re: Wavelength and frequency proportions\n\nWhat Dr. Lavelle was drawing attention to was the relationship in which wavelength and frequency are inversely proportional to one another. This is why when one is equal to 2 the other is equal to 1/2. You can think of it as v=1/lamda (sorry I don't have the symbol) or frequency=1/wavelength.\n\nJulia Jones 1G\nPosts: 29\nJoined: Fri Sep 28, 2018 12:16 am\n\n### Re: Wavelength and frequency proportions\n\nwavelength and frequency are inversely proportional but energy and frequency are proportional to each other. the frequency can be thought of as 1/wavelength\n\nanishathomas\nPosts: 35\nJoined: Fri Sep 28, 2018 12:25 am\n\n### Re: Wavelength and frequency proportions\n\nWhen wavelength gets bigger, the frequency decreases. If you look at the equation v=f(lambda) and we know that there is constant speed(at least in the same medium) then we know that if one increases, the other has to decrease. Therefore, it is an indirect relationship.\n\nlukezhang2C\nPosts: 59\nJoined: Fri Sep 28, 2018 12:18 am\n\n### Re: Wavelength and frequency proportions\n\nThe two have an inversely proportional relationship, this property is just derived from the formula!"
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http://bahrainads.info/fifth-grade-division-worksheets/fifth-grade-division-worksheets-long-division-worksheets-grade-3-free-fifth-math-documents-grade-5-division-worksheets/ | [
"# Fifth Grade Division Worksheets Long Division Worksheets Grade 3 Free Fifth Math Documents Grade 5 Division Worksheets",
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"fifth grade division worksheets long division worksheets grade 3 free fifth math documents grade 5 division worksheets.\n\n3rd grade division word problems worksheets math 2 long 4th pdf,4th grade division worksheets common core 3rd introduction to long beautiful math printable of 6th,grade 5 multiplication and division worksheets pdf long worksheet with double digit divisors set 4 fifth 3rd word problems printable 6th decimal,fifth grade division worksheets printable math decimal for classy 5 3rd pdf 4th 2 digit,grade 2 math division worksheets pdf long 3 free fifth documents 4th multiplication 6th word problems,4th grade basic division worksheets printable for 5th long pdf 5 multiplication and,grade 2 division grouping worksheets decimal 6 4th coloring 6th,fifth grade division worksheets long with decimal 6th word problems 1 math 5th pdf,fifth grade division worksheets long worksheet with double 1 multiplication and 4th 6th,4th grade division facts worksheets fifth worksheet one page multiplication 5th word problems pdf 5 math fraction."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8053011,"math_prob":0.8757149,"size":1170,"snap":"2019-13-2019-22","text_gpt3_token_len":250,"char_repetition_ratio":0.27444252,"word_repetition_ratio":0.0,"special_character_ratio":0.18717949,"punctuation_ratio":0.057591625,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99333185,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-20T10:27:04Z\",\"WARC-Record-ID\":\"<urn:uuid:d1fdda0e-00f8-4c2a-b9f9-ec3d9258606b>\",\"Content-Length\":\"56667\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:88adda39-67d2-4f76-a95b-1c7ca4cb5ba4>\",\"WARC-Concurrent-To\":\"<urn:uuid:c964204f-54f9-4ed4-bafd-8d175345379d>\",\"WARC-IP-Address\":\"104.18.51.139\",\"WARC-Target-URI\":\"http://bahrainads.info/fifth-grade-division-worksheets/fifth-grade-division-worksheets-long-division-worksheets-grade-3-free-fifth-math-documents-grade-5-division-worksheets/\",\"WARC-Payload-Digest\":\"sha1:FTW62OJKY22PD65ZRFP4SLML2256R67F\",\"WARC-Block-Digest\":\"sha1:REYE5ZPPXFWLO2FMO5LQISCHIEM3Y2ZP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232255943.0_warc_CC-MAIN-20190520101929-20190520123929-00355.warc.gz\"}"} |
https://byjus.com/physics/unit-of-viscosity/ | [
"",
null,
"# Unit of Viscosity\n\n## What is Viscosity?\n\nViscosity is defined as the quantity that represents a fluid’s resistance to flow. To put it in simpler terms, viscosity is the friction between the fluid molecules. We usually relate to the viscosity of a liquid by measuring the “thickness”. For example, we can state that honey is “thicker” (meaning a higher viscosity) than normal water.\n\nViscosity is represented by the symbol η “eta”, and the use of the Greek letter mu (μ) is also common. Furthermore, viscosity is also known as dynamic viscosity or absolute viscosity. Apart from this, we also deal with terms like kinematic viscosity or momentum diffusivity. The main difference between kinematic and dynamic viscosity is that dynamic viscosity is the measurement of a fluid’s internal resistance to the flow. In contrast, kinematic viscosity is the ratio of dynamic viscosity to density. Now that we have learned some viscosity information let’s study the unit of viscosity below.\n\n Unit of Viscosity SI Unit Pascal seconds (Pa·s) or kg·m−1·s−1 CGS Unit Poise (P) Unit of Kinematic Viscosity Si Unit square metre per second or m2/s CGS Unit Stokes (St)\n\n### SI Unit of Viscosity\n\n• The SI unit of viscosity is the pascal second (Pa·s) or kg·m−1·s−1.\n• The SI unit of kinematic viscosity is square metre per second or m2/s.\n\n### CGS Unit of Viscosity\n\n• The cgs unit of viscosity is poise (P), which has been used in honour of French physiologist Jean Léonard Marie Poiseuille. It is also commonly used as centipoise (cP), particularly in ASTM standards.\n• The cgs unit for kinematic viscosity is stoke (St) and has been named after Irish mathematician and physicist Sir George Gabriel Stokes. Sometimes the unit centistokes (cSt) is also used.\n\n#### Other Common Viscosity Units\n\nSome non-standard units include the Reyn, which is a British unit of dynamic viscosity.\n\nEarlier, kinematic viscosity was measured using the Saybolt viscometer, expressing the units in Saybolt universal seconds (SUS). Other abbreviations such as SUV (Saybolt universal viscosity) or SSU (Saybolt seconds universal) are sometimes used.",
null,
"Related Articles:\n\n## Frequently Asked Questions\n\n### State Newton’s law of viscosity.\n\nNewton’s law of viscosity states that shear stress between the adjacent layers of the fluid is proportional to the velocity gradients between the two layers.\n\n### What will be the viscosity of fluid when the fluid is at rest?\n\nThe viscosity of fluid when it is at rest will be 1 Poise. Viscosity is not dependent on the state of the fluid. Therefore, if the fluid is at rest or in motion, 1 Poise will remain constant.\n\n### Mention the factors on which viscosity depends.\n\nFollowing are the factors on which viscosity depends:\n\n• Area of contact between the two adjacent layers\n• The velocity difference between the two adjacent layers\n• Distance between the two adjacent layers\n\n### Convert 1 Poise into N.s.m-2.\n\n$$\\begin{array}{l}1\\;Poise= 1\\;dyne\\;s\\;cm^{-2}=\\frac{1}{10}N\\;s\\;m^{-2}\\end{array}$$\n\n### Give examples of Newtonian fluids.\n\nWater, honey, and organic solvents are examples of Newtonian fluids.\n\nStay tuned with BYJU’S to learn more interesting Physics topics with the help of engaging video lessons.\n\nTest your knowledge on Unit Of Viscosity"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.91218215,"math_prob":0.83531636,"size":3106,"snap":"2022-40-2023-06","text_gpt3_token_len":719,"char_repetition_ratio":0.17375886,"word_repetition_ratio":0.021912351,"special_character_ratio":0.20476498,"punctuation_ratio":0.08787346,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.985747,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-05T15:53:11Z\",\"WARC-Record-ID\":\"<urn:uuid:7bed54d7-51e8-414a-9c19-15f7f43a58a7>\",\"Content-Length\":\"769912\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:00e2043c-7975-4fa0-912e-9fd932f29fcf>\",\"WARC-Concurrent-To\":\"<urn:uuid:12b8d56a-0bec-4f08-a7e0-52f86047f5ab>\",\"WARC-IP-Address\":\"162.159.129.41\",\"WARC-Target-URI\":\"https://byjus.com/physics/unit-of-viscosity/\",\"WARC-Payload-Digest\":\"sha1:LNWQE7QLUMOJ3QJPLOCFH3HPTTZLANDF\",\"WARC-Block-Digest\":\"sha1:YG72465KJKUKTB3CXS5ZYRGXMYHIL6KW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337631.84_warc_CC-MAIN-20221005140739-20221005170739-00440.warc.gz\"}"} |
https://www.qiansw.com/mongodb-basic-tutorial-find-query-condition-query-lt-gt-lte-gte-or-not.html | [
"## 查询条件\n\n• `\"\\$lt\"` `less than` `<`\n• `\"\\$gt\"` `greater than` `>`\n• `\"\\$lte\"` `less than or equal to ` `<=`\n• `\"\\$gte\"` `greater than or equal to` `>=`\n\n``> db.users.find({\"age\" : {\"\\$lte\" : 30, \"\\$gte\" : 18}})``\n\n``> SELECT * FROM `users` WHERE `age` <= 30 AND `age` >= 18;``\n\n``> db.users.find({\"username\" : {\"\\$ne\" : \"tom\"}})``\n\n``> SELECT * FROM `users` WHERE `username` != \"tom\";``\n\n## OR查询\n\nMongoDB 中有两种方式进行 OR 查询：`\"\\$in\"`可以用来查询一个键的多个值； `\"\\$or\"`更通用一些，可以在多个键中查询任意的给定值。\n\n``> db.users.find({\"username\" : {\"\\$in\" : [\"tom\", \"Lily\"]}})``\n\n``> SELECT * FROM `users` WHERE `username` IN (\"tom\", \"Lily\") ;``\n\n``> db.users.find({\"username\" : {\"\\$nin\" : [\"tom\", \"Lily\"]}})``\n\n``> SELECT * FROM `users` WHERE `username` NOT IN (\"tom\", \"Lily\") ;``\n\n`\"\\$or\"`接受一个包含所有可能条件的数组作为参数。\n\n``> db.users.find( {\"\\$or\" : [{\"username\" : \"tom\"}, {\"blog\" : \"https://www.qiansw.com\"}]})``\n\n``> SELECT * FROM `users` WHERE `username` = \"tom\" OR `blog` = \"https://www.qiansw.com\";``\n\n## \\$not取反查询\n\n`\"\\$not\"`是元条件句，即可以用在任何其他条件上。就拿取模运算符`\"\\$mod\"`来说。`\"\\$mod\"`会将查询的值初一第一个给定值，若余数等于第二个给定值则匹配成功：\n\n``> db.users.find({\"id_num\" : { \"\\$mod\" : [5, 1] }})``\n\n``> SELECT * FROM `users` WHERE MOD(id_num,5) = 1;``\n\n``> db.users.find({\"id_num\" : { \"\\$not\" : { \"\\$mod\" : [5, 1] }}})``\n\n``> SELECT * FROM `users` WHERE MOD(id_num,5) != 1;``\n\n`\"\\$not\"`与正则表达式联合使用时极为有用，用来查找哪些与特定模式不匹配的文档。"
]
| [
null
]
| {"ft_lang_label":"__label__zh","ft_lang_prob":0.78955215,"math_prob":0.7684421,"size":1771,"snap":"2022-27-2022-33","text_gpt3_token_len":976,"char_repetition_ratio":0.14544426,"word_repetition_ratio":0.18877551,"special_character_ratio":0.35460192,"punctuation_ratio":0.20274915,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.987193,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-29T04:32:52Z\",\"WARC-Record-ID\":\"<urn:uuid:4398bc95-953e-4cdb-93be-17617b8950df>\",\"Content-Length\":\"44096\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dc6b6a18-e43f-415e-9e68-3edaa7b280b5>\",\"WARC-Concurrent-To\":\"<urn:uuid:7cb3c9b2-29cd-4657-b5be-57f5abecdce8>\",\"WARC-IP-Address\":\"101.200.84.112\",\"WARC-Target-URI\":\"https://www.qiansw.com/mongodb-basic-tutorial-find-query-condition-query-lt-gt-lte-gte-or-not.html\",\"WARC-Payload-Digest\":\"sha1:42XGLXXFFQTCOSZ3BHGUFFAHZ5JOTBV2\",\"WARC-Block-Digest\":\"sha1:UJMDK3YUSM52EM7223NJ7CTIV6DUTRKG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103620968.33_warc_CC-MAIN-20220629024217-20220629054217-00265.warc.gz\"}"} |
https://www.brandonwalker.website/post/tutorial-ggplot-and-ggplotly/ | [
"# Tutorial: ggplot and ggplotly\n\n4 minutes\nJune 12, 2019\n\nIf you want to be considered good at R, it’s best you know ggplot. If you’re using base R graphics I get the impression other data scientists may look at your graphics as childish (though I think there is nothing wrong with using base R). I’ll give a quick walk through of ggplot and making use of ggplotly from the plotly package. We’re going to use the cars data set, which is comes with R so don’t worry about getting it. In this dataset we have speeds a car was going at and how many feet it moved once the brakes were applied before it stopped.\n\n### Using ggplot\n\nFirst load the package which is actually called ggplot2. Use the ggplot() function to make an empty graph, then put in your data as your first argument, and fill in the aesthetics of the graph by providing aes(x = , y = ). Finally to tell ggplot what type of graph you want use + geom_point(), geom_col(), geom_abline() and so on to get exactly what you want. To edit you axis labels just use + labs(x = , y =)\n\n``````library(ggplot2)\n\nplot <- ggplot(cars, aes(x = speed, y = dist)) + geom_point() +\nlabs(x = \"Speed\", y = \"Stopping Distance\")\n\nplot``````",
null,
"### Using ggplotly\n\nTo make any plot interactive add the ggplotly() function around any ggplot object! Working this into an Rmakrdown presentation takes your presentation one step further than powerpoint. It’s also really handy for data exploration. Hover your mouse over the chart to see what this does.\n\n``````library(plotly)\n\nggplotly(plot)``````"
]
| [
null,
"https://www.brandonwalker.website/post/2019-06-12-tutorial-ggplot-and-ggplotly_files/figure-html/unnamed-chunk-1-1.png",
null
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https://pastebin.com/PJEZc2fa | [
"",
null,
"Alphard\n\na guest\nApr 11th, 2015\n48\nNever\nNot a member of Pastebin yet? Sign Up, it unlocks many cool features!\n1. \"\"\"\n2. Created on Sun Apr 12 01:22:11 2015\n3.\n4. @author: Alphard\n5. \"\"\"\n6.\n7. import math\n8. import Image\n9. import ImageDraw\n10.\n11. def hexagonVertices(edge, center):\n12. x, y = center\n13. for angle in range(0, 360, 60):\n16. yield x, y\n17.\n18. def drawHexagon(draw, vertices, label):\n19. if not label:\n20. label = (150, 150, 150)\n21. draw.polygon(list(vertices), outline='black', fill=label)\n22.\n23. rows = 30\n24. cols = 6\n25. offsetX = 25;\n26. offsetY = 25;\n27. edge = 20;\n28.\n29. labels = [[0 for x in range(cols)] for x in range(rows)]\n30.\n31. for k in range(5, 13, 2):\n32. labels[k] = 'red'\n33. labels[k] = 'red'\n34.\n35. for k in range(4, 12, 2):\n36. labels[k] = 'red'\n37. labels[k] = 'red'\n38. labels[k] = 'red'\n39.\n40. for k in range(12, 28, 2):\n41. labels[k] = 'green'\n42.\n43. labels = 'green'\n44. labels = 'green'\n45. labels = 'green'\n46. labels = 'green'\n47. labels = 'green'\n48. labels = 'green'\n49.\n50.\n51. image = Image.new('RGB', (int(3*edge*cols+2*offsetX), int(edge*rows+2*offsetY)), 'white')\n52. draw = ImageDraw.Draw(image)\n53.\n54. for row in range(rows):\n55. for col in range(cols):\n56. oddOffset = 0 if row%2==1 else 1.5*edge\n57. x = 3*edge*col + oddOffset + offsetX\n58. y = 3**0.5*0.5*edge*row + offsetY\n59. drawHexagon(draw, hexagonVertices(edge, (x, y)), labels[row][col])\n60.\n61. image.show()\nRAW Paste Data"
]
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null,
"https://pastebin.com/themes/pastebin/img/guest.png",
null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5159912,"math_prob":0.9960239,"size":1343,"snap":"2022-05-2022-21","text_gpt3_token_len":496,"char_repetition_ratio":0.18894698,"word_repetition_ratio":0.072072074,"special_character_ratio":0.4534624,"punctuation_ratio":0.17985612,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99214727,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-17T05:39:53Z\",\"WARC-Record-ID\":\"<urn:uuid:6d912a5d-2849-4a24-9cbb-9bef0eccbaf6>\",\"Content-Length\":\"30240\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b11f6efe-da0c-4e2d-ad9b-4dc33659b2ed>\",\"WARC-Concurrent-To\":\"<urn:uuid:90735d6f-64e2-49e3-8f7b-0760e9cd412d>\",\"WARC-IP-Address\":\"104.23.99.190\",\"WARC-Target-URI\":\"https://pastebin.com/PJEZc2fa\",\"WARC-Payload-Digest\":\"sha1:6YRWT3MMGLHCMF46SYBBKT37JLI6I2LT\",\"WARC-Block-Digest\":\"sha1:WQLULKNQYIZK64K2XU426G5A2SZOJFTI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320300289.37_warc_CC-MAIN-20220117031001-20220117061001-00561.warc.gz\"}"} |
https://forum.image.sc/t/running-r-script-or-calling-r-commands-from-imagej-macro/23848 | [
"",
null,
"# Running R script or calling R commands from ImageJ macro\n\nI have an ImageJ macro that computes a chi-squared statistic as an indicator of clustering/self-avoidance. I’d like to convert it into a p-value. I’ve found an R command that generates the stat that I’d like to report (pchisq). When I call the pchisq command in an R script from the Macro Editor, it runs correctly.\n\nIs there a way to use the eval command or something similar in an ImageJ macro to call an R command or an R script? I tried the eval and exec commands, but, couldn’t get them to recognize R. When running the R script from the Macro Editor, the macro recorder will record a command, but, I can’t get it to run from a macro – it just generates an unrecognized command error… the command looks like… `run(\"script:D:\\code\\my1liner.R\");`.\n\nAny/all tips appreciated.\n\nthanks!\nJeff\n\nps - This code shows what I was hoping to do:\n\n``````vals = newArray(4);\nvals = 20; vals = 30; vals = 40; vals = 10;\n\nc = getChiSquared(vals);\np = getPfromChiSq(c,1);\nprint(\"Results {chisq,p} = {\"+c+\",\"+p+\"}\");\n\nfunction getPfromChiSq(chisq,df) {\np = eval(\"R\",\"pchisq(\"+chisq+\",df=\"+df+\",lower.tail=FALSE)\"); // THIS is what I wish I could do...\nreturn p;\n}\nfunction getChiSquared(myList) {\nm = myList.length;\nn = 0;\nfor (i = 0; i<m; i++) {\tn = n + myList[i];\t}\nchisq = 0;\nfor (i=0; i<m; i++) { chisq = chisq + pow(myList[i] - n/m,2)/(n/m); }\nreturn chisq;\n}\n``````\n1 Like\n\nHi Jeff,\n\nmight it be, that Rserve has the functionality you are looking for?\n\nHave a look here: http://wiki.cmci.info/documents/101105ij_r_jython\n\nCheers\nStefan\n\nThanks, Stefan. I was able to use the Renjin installed with FIJI. I couldn’t get the eval command to use the R engine. But, I was able to call the javascript engine and then call the R engine from javascript. It’s a little awkward, but, it does let me send a simple command to R and get an answer back.\n\nHere’s an example of what the code ended up looking like:\n\n``````ret = callRScriptviaJavaScript(\"3+5\");\nprint (\"According to R engine, 3+5 is \" + ret);\nchisq = 80; df = 63;\nret = callRScriptviaJavaScript(\"pchisq(\"+chisq+\",\"+df+\",lower.tail=FALSE)\");\nprint (\"P-Value for chi-square of \" + chisq + \" and df of \" + df + \" = \" + ret);\n\nfunction callRScriptviaJavaScript(script) {\n// Use javascript ability to call direct java commands to load R engine and send command to R\n// Note that anything because this is an ImageJ macro calling a javascript calling R...\n//\tpretty much anything with quotes in it is likely to fail.\njscode = \"importClass(javax.script.ScriptEngineManager); \";\njscode = jscode + \"sEM = new ScriptEngineManager();\";\njscode = jscode + \"engine = sEM.getEngineByName(\\\"Renjin\\\");\"\njscode = jscode + \"engine.eval(\\\"\"+script+\"\\\")\";\nret = eval(\"script\",jscode);\nreturn ret;\n}\n``````"
]
| [
null,
"https://aws1.discourse-cdn.com/business4/uploads/imagej/original/2X/a/a8bb0afa608549b70f9141516502fe7fb4693171.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.7696949,"math_prob":0.8463374,"size":1349,"snap":"2020-34-2020-40","text_gpt3_token_len":409,"char_repetition_ratio":0.10855018,"word_repetition_ratio":0.017391304,"special_character_ratio":0.31875464,"punctuation_ratio":0.17377049,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97380817,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-01T20:06:15Z\",\"WARC-Record-ID\":\"<urn:uuid:88ce436d-d129-4014-9034-cd39e7d3bcc8>\",\"Content-Length\":\"20014\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f6e4011f-a818-4502-80bd-63e2e3c76e5a>\",\"WARC-Concurrent-To\":\"<urn:uuid:29943a2f-2536-47a7-be83-a15f4a788516>\",\"WARC-IP-Address\":\"72.52.80.15\",\"WARC-Target-URI\":\"https://forum.image.sc/t/running-r-script-or-calling-r-commands-from-imagej-macro/23848\",\"WARC-Payload-Digest\":\"sha1:3TNFODWOM325C5MSTMH5INSIVIEKWBT4\",\"WARC-Block-Digest\":\"sha1:XSSSKZAQK2PJBR2GHHGVC2G6QNCVOWHQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600402131986.91_warc_CC-MAIN-20201001174918-20201001204918-00215.warc.gz\"}"} |
https://en.wikipedia.org/wiki/Cons | [
"# cons\n\nIn computer programming, `cons` (/ˈkɒnz/ or /ˈkɒns/) is a fundamental function in most dialects of the Lisp programming language. `cons` constructs memory objects which hold two values or pointers to two values. These objects are referred to as (cons) cells, conses, non-atomic s-expressions (\"NATSes\"), or (cons) pairs. In Lisp jargon, the expression \"to cons x onto y\" means to construct a new object with `(cons x y)`. The resulting pair has a left half, referred to as the `car` (the first element, or contents of the address part of register), and a right half, referred to as the `cdr` (the second element, or contents of the decrement part of register).\n\nIt is loosely related to the object-oriented notion of a constructor, which creates a new object given arguments, and more closely related to the constructor function of an algebraic data type system.\n\nThe word \"cons\" and expressions like \"to cons onto\" are also part of a more general functional programming jargon. Sometimes operators that have a similar purpose, especially in the context of list processing, are pronounced \"cons\". (A good example is the `::` operator in ML, Scala, F# and Elm or the `:` operator in Haskell, which adds an element to the beginning of a list.)\n\n## Use\n\nAlthough cons cells can be used to hold ordered pairs of data, they are more commonly used to construct more complex compound data structures, notably lists and binary trees.\n\n### Ordered pairs\n\nFor example, the Lisp expression `(cons 1 2)` constructs a cell holding 1 in its left half (the so-called `car` field) and 2 in its right half (the `cdr` field). In Lisp notation, the value `(cons 1 2)` looks like:\n\n```(1 . 2)\n```\n\nNote the dot between 1 and 2; this indicates that the S-expression is a \"dotted pair\" (a so-called \"cons pair\"), rather than a \"list.\"\n\n### Lists",
null,
"Cons cell diagram for the list (42 69 613), written with `cons`:\n```(cons 42 (cons 69 (cons 613 nil)))\n```\nand written with `list`:\n```(list 42 69 613)\n```\n\nIn Lisp, lists are implemented on top of cons pairs. More specifically, any list structure in Lisp is either:\n\n1. An empty list `()`, which is a special object usually called `nil`.\n2. A cons cell whose `car` is the first element of the list and whose `cdr` is a list containing the rest of the elements.\n\nThis forms the basis of a simple, singly linked list structure whose contents can be manipulated with `cons`, `car`, and `cdr`. Note that `nil` is the only list that is not also a cons pair. As an example, consider a list whose elements are 1, 2, and 3. Such a list can be created in three steps:\n\n1. Cons 3 onto `nil`, the empty list\n2. Cons 2 onto the result\n3. Cons 1 onto the result\n\nwhich is equivalent to the single expression:\n\n```(cons 1 (cons 2 (cons 3 nil)))\n```\n\nor its shorthand:\n\n```(list 1 2 3)\n```\n\nThe resulting value is the list:\n\n```(1 . (2 . (3 . nil)))\n```\n\ni.e.\n\n``` *--*--*--nil\n| | |\n1 2 3\n```\n\nwhich is generally abbreviated as:\n\n```(1 2 3)\n```\n\nThus, `cons` can be used to add one element to the front of an existing linked list. For example, if x is the list we defined above, then `(cons 5 x)` will produce the list:\n\n```(5 1 2 3)\n```\n\nAnother useful list procedure is `append`, which concatenates two existing lists (i.e. combines two lists into a single list).\n\n### Trees\n\nBinary trees that only store data in their leaves are also easily constructed with `cons`. For example, the code:\n\n```(cons (cons 1 2) (cons 3 4))\n```\n\nresults in the tree:\n\n```((1 . 2) . (3 . 4))\n```\n\ni.e.\n\n``` *\n/ \\\n* *\n/ \\ / \\\n1 2 3 4\n```\n\nTechnically, the list (1 2 3) in the previous example is also a binary tree, one which happens to be particularly unbalanced. To see this, simply rearrange the diagram:\n\n``` *--*--*--nil\n| | |\n1 2 3\n```\n\nto the following equivalent:\n\n``` *\n/ \\\n1 *\n/ \\\n2 *\n/ \\\n3 nil\n```\n\n## Use in conversation\n\nCons can refer to the general process of memory allocation, as opposed to using destructive operations of the kind that would be used in an imperative programming language. For example:\n\nI sped up the code a bit by putting in side effects instead of having it cons ridiculously.\n\n## Functional implementation\n\nSince Lisp has first-class functions, all data structures, including cons cells, can be implemented using functions. For example, in Scheme:\n\n```(define (cons x y)\n(lambda (m) (m x y)))\n(define (car z)\n(z (lambda (p q) p)))\n(define (cdr z)\n(z (lambda (p q) q)))\n```\n\nThis technique is known as Church encoding. It re-implements the cons, car, and cdr operations, using a function as the \"cons cell\". Church encoding is a usual way of defining data structures in pure lambda calculus, an abstract, theoretical model of computation that is closely related to Scheme.\n\nThis implementation, while academically interesting, is impractical because it renders cons cells indistinguishable from any other Scheme procedure, as well as introduces unnecessary computational inefficiencies.\n\nHowever, the same kind of encoding can be used for more complex algebraic data types with variants, where it may even turn out to be more efficient than other kinds of encoding. This encoding also has the advantage of being implementable in a statically typed language that doesn't have variants, such as Java, using interfaces instead of lambdas."
]
| [
null,
"https://upload.wikimedia.org/wikipedia/commons/thumb/1/1b/Cons-cells.svg/350px-Cons-cells.svg.png",
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https://www.infoplease.com/videos?title=&page=4 | [
"",
null,
"Cite\n\n# Videos",
null,
"Dividing Three-Digit Numbers Using Models\nThis video uses a model to explain the step-by-step process of dividing a three-digit number by a one-digit number.",
null,
"Division as Repeated Subtraction\nIn this video, Emily learns how to divide by using repeated subtraction.",
null,
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null,
"Division With Even and Odd Numbers\nIn this video, Jackson learns about the relationship between division, and even and odd numbers.",
null,
"Do Seeds Need Water to Grow?\nWatch and learn how scientists use the scientific method to find answers. Do seeds need water to grow? To answer this question, students will…",
null,
"Elephant's Acute Sense of Hearing\nFind information and a video about the elephant, an exceptional listener and communicator that uses its ears, trunk, and feet to detect…",
null,
"Equivalent Fractions: Use Models\nIn this video, Carlos learns how to use fraction strips to compare the value of fractions.",
null,
"Equivalent Fractions: Use the Number Line\nIn this video, Carlos learns how to use a number line to compare the value of fractions.",
null,
"Estimate Decimal Quotients\nIn this video, Alex learns how to estimate a division problem when both numbers are decimals.",
null,
"Estimate Differences\nIn this video, Alex and Marta learn how to estimate differences by rounding numbers."
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https://www.classtools.net/QR/15-GEPJb | [
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null,
"# Print\n\n### Create a New QuizEdit this quizView a Sample QuizDetailed Case StudySearch the ArchiveFeedback",
null,
"## A. Prior to the lesson:\n\n1. Arrange students into groups. Each group needs at least ONE person who has a mobile device.\n\n2. If their phone camera doesn't automatically detect and decode QR codes, ask students to\n\n• Bring these devices into the lesson.\n\n4. Cut them out and place them around your class / school.\n\n## B. The lesson:\n\n1. Give each group a clipboard and a piece of paper so they can write down the decoded questions and their answers to them.\n\n2. Explain to the students that the codes are hidden around the school. Each team will get ONE point for each question they correctly decode and copy down onto their sheet, and a further TWO points if they can then provide the correct answer and write this down underneath the question.\n\n3. Away they go! The winner is the first team to return with the most correct answers in the time available. This could be within a lesson, or during a lunchbreak, or even over several days!\n\n## C. TIPS / OTHER IDEAS\n\n4. A detailed case study in how to set up a successful QR Scavenger Hunt using this tool can be found here.\n\n### Question\n\n1. A shape with 3 sides is called_____triangle\n2. A shape with 4 sides is called ____quadrilateral\n3. A number with 2 factors 1 and itself is called___prime number\n4. \\$20 - \\$7.50 = \\$12.50\n5. quarter of 40cm = 10cm\n6. True or false: 265 is a multiple of 3False\n7. What are the prime factors of 12?2x2x3\n8. What is 2 to the power of 5?32\n9. What is 1.23 x 10000?12300\n10. What is the value of 3 in 123456?3000\n11. 1/5 of ________is 630\n12. What is 55% as a fraction is simplest form?11/20\n13. What is 3/20 in percentage?15%\n14. Round \\$4.53 to the nearest 5 cents\\$4.55\n15. How many minutes is 25% of an hour?15\n16. The next prime number to 67 is__71\n17. 200 x 0.06 = 12\n18. 2+5-3x4 =-5\n19. 1/2 + 1/3 = 5/6\n20. 2.11+0.9+0.55=3.56\n21. 3 - 0.5 =2.5\n22. 5 divided by 0.01 = 500",
null,
"### Maths revision: QR Challenge",
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"Question 1 (of 22)",
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"### Maths revision: QR Challenge",
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"### Maths revision: QR Challenge",
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https://math.stackexchange.com/questions/712087/problems-with-a-proof-that-in-a-linear-order-a-minimal-element-is-the-smallest | [
"# Problems with a proof that -in a linear order- a minimal element is the smallest element\n\nI have a problem with a proof I found in Velleman's \"How to prove it\". This is sort of interesting, because it is the very first time I cannot see the structure of a proof presented in the book. The following are the theorem and Velleman's proof.\n\nTheorem:\nSuppose $R$ is a partial order on a set $A$, and $B \\subseteq A$. If $R$ is a total order and $b$ is a minimal element of $B$, then $b$ is the smallest element of $B$.\n\nVelleman's Proof:\nSuppose $R$ is a total order and $b$ is a minimal element of $B$. Let $x$ be an arbitrary element of $B$. If $x = b$, then since $R$ is reflexive, $bRx$. Now suppose $x \\neq b$. Since $R$ is a total order, we know that either $xRb$ or $bRx$. But $xRb$ can’t be true, since by combining $xRb$ with our assumption that $x \\neq b$ we could conclude that $b$ is not minimal, thereby contradicting our assumption that it is minimal. Thus, $bRx$ must be true. Since $x$ was arbitrary, we can conclude that $\\forall x \\in B(bRx)$, so $b$ is the smallest element of $B$.\n\nProblem\nMy problem is with the structure of the proof. In particular, I see a bit as a \"rabbit out of the hat\" the fact that we start by assuming that $x \\neq b$.\nIn the pages of the book before the the actual proof, the author presents the scratch work behind the proof. Now, I can see the logic behind it (the all problem about the $x$ being or not being equal to $b$), but I do not see how this smoothly enters in the picture of the proof. Just consider that using \"Proof Designer\" (the software that can be found on Velleman's site and that goes along the book), I could actually prove the theorem without starting with the assumption that $x \\neq b$. Indeed, the proof in words should go along the following lines:\n\nProof:\nAssume that $B \\subseteq A$, that $R$ is a total order, and that $b$ is a minimal element of $B$. Let $x \\in B$ be arbitrary. Thus, by the fact that $b$ is minimal, we have that if $(x,b) \\in R$, then $x=b$. Proceed by cases.\n1. $(x,b) \\notin R$:\nSince $x \\in B$ and $b \\in B$, we can conclude that $x \\in A$ and $b \\in A$. From the completeness of $R$ we have that either $(x,b) \\in R$ or $(b,x) \\in R$. Thus, by assumption that $(x,b) \\notin R$, we have that $(b,x) \\in R$.\n2. $x=b$:\nSince $x \\in B$ and $b \\in B$, we can conclude that $x \\in A$ and $b \\in A$. From the reflexivity of $R$ we have that $(b,b) \\in R$. But, since by assumption $x=b$, we can conclude that $(b,x) \\in R$. QED\n\nQUESTIONS:\nIs my proof sound?\nIs there somebody who can give me a feedback on what is actually the structure of Velleman's proof?\nI actually tried to replicated Velleman's structure of the proof with \"Proof Designer\", but I did not succeed.\n\nThanks a lot for any help.\n\nPS: Conditional on my proof being sound, any feedback on the writing is more than wellcome.\n\n• It's not so much \"rabbit out of the hat\" as you might think. What he does in the proof with $x$ is first noting that either $x = b$ or $x \\neq b$. Then he says that if $x = b$ we're done, so the only case that is left to check is if $x \\neq b$. He assumes that this is the case and goes on with the proof. – Arthur Mar 14 '14 at 12:53\n• Ok, I do see this. Maybe we can say that my way of looking at proofs is still a bit \"mechanical\"? [Even if I truly love \"Proof Designer\", and it has helped me a lot, I do think it makes users a bit too rigid] – Kolmin Mar 14 '14 at 13:02\n• Let's say that my problem is with the fact that he \"notes\" that either $x=b$ or $x \\neq b$. So the speak, the fact that he \"notes\" it lies outside the structure of the proof that comes from the way in which the problem is given. – Kolmin Mar 14 '14 at 13:04\n• Now, this is not a problem in itself. On the contrary. We could say that I would like to know when I can \"note\" things, so to say (...and maybe this last statement sounds like asking an advice on how to get intuition). – Kolmin Mar 14 '14 at 13:05\n• But what \"is behind\" your choice of 1 and 2 ? The fact that $(x,b) \\notin R$ or $(x,b) \\in R$ ... Then you exploit the fact that : \"if $(x, b) \\in R$, then $x=b$, and proceed with the two cases. Vellman's proof start from the same assumption and exploit also the fact that, \"if $(x, b) \\notin R$, then $x \\ne b$, because $R$ is total and if we have that $x = b$, then by reflexivity : $(b, b) \\in R$. – Mauro ALLEGRANZA Mar 14 '14 at 13:07\n\n## 1 Answer\n\nTo be honest I don't see a lot of difference between Velleman's proof and yours. Both are proofs by division into cases, and actually I would say yours is better written and better set out than his: you make the two cases clear, whereas he muddles them up together in one paragraph.\n\nYour choice of cases is $(x,b)\\in R$ or not; his is $x=b$ or not. To me, they are both sensible ideas to consider and neither is more \"rabbit-out-of-hatty\" than the other. And in fact logically they are identical: when $b$ is a minimal element in a partial order we have $(x,b)\\in R$ if and only if $x=b$.\n\nAs a final remark (though from one of your comments above I think you already realise this), don't forget that sometimes there is no alternative to pulling rabbits out of hats! I have seen various books on proof (though I don't know Velleman's) and many of them contain lots of good advice. But sometimes you just need a bright idea, and there really isn't any way to teach that.\n\n• I am starting to get closer to realizing what's going on. However, I have a problem with one of your statements, which -to me- seems to be the most important one, namely: \"when $b$ is a minimal element in a partial order we have $(x,b) \\in R$ if and only if $x=b$\". Is this really an if and only if? I didn't consider it such, because if it is the case, then my question is kinda trivial and it really seems it's a matter of taste. – Kolmin Mar 14 '14 at 13:20\n• @Kolmin: \"If\" is because $R$ is reflexive; the \"only if\" direction is the definition of \"minimal element\". – hmakholm left over Monica Mar 14 '14 at 13:23\n• Yes it is if and only if. \"If $(x,b)\\in R$ then $x=b$\" is true by definition of minimal, and the converse is true because $R$ is reflexive. In fact both of these remarks occur in your proof, and in Velleman's too! – David Mar 14 '14 at 13:26\n• Thus, I would say that the main thing that is not explicit in Velleman's proof is the fact that this is an iff and, from this, we can interchange $(x,b) \\notin R$ with $x \\neq b$, hence getting the cases he uses. – Kolmin Mar 14 '14 at 13:36\n• I would say that is a reasonable assessment. He does all this implicitly of course, but as you said, not explicitly. – David Mar 14 '14 at 13:38"
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https://www.thevedicmaths.com/post/vedic-maths-for-multiplication | [
"top of page\nSearch\n\n# Vedic Maths for Multiplication\n\nVedic Maths is a super-fast way of making all Mathematical calculations easy and gives accurate results. It is a domain of mathematical tricks written by an Indian monk, “Late Swami Sri Bharati Krishna Tirthaji'’\n\nHe has given us various tricks to solve any type of mathematical problem is hardly a few minutes.\n\nThese formulas are in the form of sutras, under each sutra we can see the mathematical operations hidden. Starting from addition, subtraction, multiplication, division, decimals, fractions, HCF/LCM, squares, cubes - its roots and along with Algebra, Trigonometry, Geometry, Calculus, etc,… When compared to our conventional method, the steps might be a little lengthier, but here the tricks can find solutions in a very simple way. That's the reason why most of the students are falling for the Vedic Maths Method.\n\nMostly under Arithmetic, Multiplication topics are more under Vedic Maths Sutras. You can check our Roadmap. I have explained all the Multiplication Topics under the Multiplication Mastery Course in detail.\n\nMultiplication is nothing but when two numbers are multiplied, any one of them is the multiplicand and the other is a multiplier.\n\nStudents generally learn multiplication by the conventional method, i.e., We just have multiplicand and multiplier and need to multiply given numbers and write the final answer, but under the Vedic Method, for every type of number or question, we have a method to calculate.\n\nExcited to know what they are???\n\nDon't worry I will explain it in detail.\n\nIn the Multiplication Mastery Course, we have interesting tricks which make multiplication very easy.\n\nMost of the students might have faced difficulty in solving bigger numbers right, here we have tricks to solve multiplication for nearly 5x5 digits too.\n\n## Now let’s see what are the topics under Multiplication:-\n\n1- Multiplication with 9, 99, 999,….\n\n2- Magic with 11\n\n3- Multiplication by 12,13,14,....19\n\n4- Multiplication of Numbers Near the Base\n\n5- Multiplication by 5,25,50,250,....\n\n6- Vertically and Crosswise Multiplication\n\n7- Times Tables\n\n8- Algebraic Multiplication etc,…\n\nLet’s understand the topics with examples:-\n\n## 1- Multiplication with 9, 99, 999,….\n\nTo multiply given numbers with 9999’s.., we have an interesting formula,\n\nStep-1:- By one less than one before\n\nStep-2:- All from nine and last from ten\n\nEg- 46 x 99:-\n\n46 is a 2 digit number and 99 also has two 9’s in it.\n\nFor LHS, subtract 1 from 46, i.e., 46 - 1 = 45\n\nFor RHS, write the complement of 46, i.e., (100-46) = 54",
null,
"Multiplication with 9, 99, 999\n\n## 2- Magic with 11:-\n\nThis method is also very interesting, same as multiplication by 999’s, here also we just have a one-line formula. I.e., just write the given numbers on both sides as it is and add both the numbers and write in between, check the below example.\n\nEg- 62 x 11:-\n\nWrite 6 and 2 on the corners, and the sum of 6 and 2, (6+2=8) in between. Therefore, the answer is 682.",
null,
"Magic with 11 | TheVedicMaths\n\n## 3- Multiplication by 12,13,14,....19:-\n\nYou need not memorize tables from 12...why because it can be found mentally using Vedic Tricks. Just learn tables from 2-9, the rest can be solved easily.\n\nEg-43 x 12:-\n\na- Dot sandwich method, put dot on both the sides of the given number = . 43 . (dot is considered as 0 )\n\nb- Starting from the RHS dot, taking 2 digits at a time, multiply the given number by 2 and add previous numbers.\n\n(2x3)+0 = 6\n\nc- Considering next pair, i.e., 43 (4 is second last digit\n\nand 3 is the last digit), we get: (2x4)+3 = 8+3 = 11, 1 carry over.\n\nd- Next pair, i.e., .4 (dot is the second last and 4 is the last digit), we get: (2x0)+4 = 4\n\n4+1 (carry) = 5\n\nSo, the answer is 43 x 12 = 516.\n\nSimilarly, we have tricks for 13...19, check out the Multiplication Mastery Course.",
null,
"Multiplication by 12,13,14,....19 | TheVedicMaths\n\n## 4- Multiplication of Numbers Near the Base:-\n\nThis comes under “Paravartya Yojayet Sutra”, here we have 3 cases under base method,\n\na- When the numbers are Above the Base\n\nb- When the numbers are Below the Base\n\nc- When the numbers are Mixed Base\n\nYou can check the 3rd chapter under Beginners guide, under Paravartya yojayet we have the above topics explained, and also under Multiplication Mastery, we have the above topics under chapter-6 in detail. The formula used here is “All from Nine and Last from Ten”.",
null,
"Multiplication of Numbers Near the Base | TheVedicMaths\n\n## 5- Multiplication by 5,25,50,250,...:-\n\nA unique advantage of Vedic maths is that many calculations can be done by mere observation only.\n\na:- Multiplying any number by 5\n\nSince, 5 = 10/2, so it means, first multiply the number by 10 and then half it.\n\nEg- 24 x 5\n\n24x5 = 240 / 2 = 120.\n\nb:- Multiplying any number by 50\n\nSince, 50=100/2, so when multiplied the number by 50, add two 0’s to its right and a half it.\n\nEg- 42 x 50\n\n42 x 50 = 4200 / 2 = 2100.",
null,
"Multiplication by 5,25,50,250 | TheVedicMaths\n\nc:- Multiplying any number by 25\n\nSince, 25=100/4, so when multiplying the number by 25, add two 0’s to its right and a half it twice.\n\nEg- 64 x 25\n\n64 x 25 = 6400 / 2x2 = 1600.\n\nSimilarly, we have formulas for 250, 500, etc.",
null,
"Multiplication by 5,25,50,250 | TheVedicMaths\n\n## 6- Vertically and Crosswise Multiplication:-\n\nThis is the general way of multiplying any two given numbers. The formula used is “Vertically and Crosswise”, this comes under Sutra “Urdhva-Tiryagbhyam”.\n\nHere we have formulas for multiplication up to 5 x 5 digits.\n\nEg- Find 47 x 28? ( | X | )\n\nStep-1:- Multiply vertically in the right-hand column\n\n7x8=56 (5 is carried over)\n\nStep-2:- Multiply crosswise and add the results for ten’s place. (4x8)+(7x2) = 46\n\n46 + 5(carry) = 51 (5 is carried over)\n\nStep-3:- Multiply vertically in the left hand column and add carry over. 4x2=8\n\n8+5 (carry) = 13.",
null,
"Vertically and Crosswise Multiplication | TheVedicMaths\n\n## 7- Times Tables upto 19:-\n\nFirstly, we need to know Tables from 2-9, then it's easy to calculate tables up to 99 too.\n\nEg-1:- 3 x 6 = 18 13 x 6 = 78 (6+1=7, take 8 from 18 as it is)\n\nEg-2:- 4 x 8 = 32 14 x 8 = 112 (8+3=11, take 2 from 32 as it is)\n\nEg-3:- 9 x 6 = 54 19 x 6 = 114 (6+5=11, take 4 as it is)\n\nTo get the answer, you need to add just the factor (of the corresponding table) to the ten’s place digit of the corresponding product. The one’s digit remains as it is. Please check the Multiplication Mastery Course for Tables upto 99.",
null,
"Times Tables upto 19 | TheVedicMaths\n\n## 8- Algebraic Multiplication:-\n\nThis chapter explains to us the Algebraic Expression, like binomials and trinomials in just one line, as compared to conventional maths where we usually do it in 4-5 steps.\n\nThe Formula used is “Vertically and Crosswise”, this is the same as Multiplication by vertically and crosswise method i.e., “Urdhva-Tiryagbhyam Sutra”.\n\nThe only difference is there we learned under Arithmetic Computation, here it is in Algebraic Expression.\n\nEg- (x+2) (3x+4)\n\nWe use the same Vertically and Crosswise formula, ( | X | )\n\nStep-1:- Write one binomial under the other as shown.\n\nMultiply vertically the digits in the right-hand column,\n\n2 * 4 = 8\n\nStep-2:- Multiply crosswise and add the products,\n\n(4 * x) + (2 * 3x) = 4x+6x = 10x.\n\nStep-3:- Multiply vertically in the left hand column,\n\nx * 3x = 3x²\n\nSo, the answer for (x+2) (3x+4) = 3x² + 10x + 8",
null,
"Algebraic Multiplication | TheVedicMaths\n\nBy now you would have had an idea of how Vedic Maths is used under Multiplication. So similarly we have few other courses which explain each topic in detail. Once you get into the courses you will have a clear picture and as it has lifetime access along with the Vedic Maths community for clearing doubts, you can excel in maths and throw away your maths phobia.\n\nGood Luck!"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.90120214,"math_prob":0.9881855,"size":7279,"snap":"2022-40-2023-06","text_gpt3_token_len":2083,"char_repetition_ratio":0.15175258,"word_repetition_ratio":0.031987812,"special_character_ratio":0.29907954,"punctuation_ratio":0.15239295,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.99911034,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-03T07:10:15Z\",\"WARC-Record-ID\":\"<urn:uuid:56246c7c-26ac-4938-8dcd-30fb2cf5b8b6>\",\"Content-Length\":\"1032087\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b1af7639-8087-477d-a1dc-363f539f7d5c>\",\"WARC-Concurrent-To\":\"<urn:uuid:7fa07bc3-6860-4c13-b6ea-724fbc8af017>\",\"WARC-IP-Address\":\"34.149.87.45\",\"WARC-Target-URI\":\"https://www.thevedicmaths.com/post/vedic-maths-for-multiplication\",\"WARC-Payload-Digest\":\"sha1:ZIJQ4TYSDVTT6LB2CZYOHVR3VPVVTRL6\",\"WARC-Block-Digest\":\"sha1:6565TAPG44TRE3F7FPFSZHMPUOBHDE3N\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500044.16_warc_CC-MAIN-20230203055519-20230203085519-00819.warc.gz\"}"} |
https://robotics.stackexchange.com/questions/1089/how-to-perform-this-reference-system-transformation | [
"How to perform this reference system transformation?\n\nI have two quaternions that indicate the initial orientation of a four wheel robot, each one in relative to one reference systems.\n\nThe robot's orientation given by a quaternion q is not the same in the two reference systems: For one reference system the quaternion q1 might refer to the robot looking at positive x while the same quaternion components q1 in the second reference system might refer to the robot looking at the negative x.\n\nI have two matrices which indicate the position of the robot in time in its correspondent reference system.\n\nI want to find the correspondent points of the first matrix in to the second reference system. Each matrix is built with a different sensor, so the results will be similar but not exactly the same.\n\nI think I should find the transformation from the first reference system to the second and then apply it for each point of the first matrix. How can I find this transformation? The translation part I think is clear, but the rotation not at all.\n\nEdit:\n\n@WildCrustacean\n\nThe solution proposed does not solve the problem, I think that the reason is because each system represents the robot in a different way.\n\nIn the initial one (A) the robot moving forward with no rotation would increase in the X axis. In the goal referential system (B) the robot moving forward with no rotation would increase in the Z axis. (See figure for more details)\n\nFirst system (A)\n______\n|\\ T \\\n| \\_____\\ z\n|B | | : y ^\n\\ | R | \\|\n\\|____| +--> x\n\nSecond system (B)\n______\n|\\ T \\\n| \\_____\\ x\n|B | | : ^\n\\ | R | |\n\\|____| +--> z\n\\\ny\n\nR: Right side\nB: Back side\nT: Top\n\nI think I have to apply an extra rotation to change the initial quaternion that belongs to the first system to be in accordance with the second system before applying the transformation of your post.\n\nOne rotation of 180 degrees around x followed by one of 90 around y. Would rotate from A to B\n\nThis is how I tried to solve it:\n\n# Quaternion to adjust reference system\nfirst_quat = make_quaternion(unitary_x, pi) # Generates the quaternion that rotates pi around X\nsecond_quat = make_quaternion(unitary_y, pi/2.0) # Generates the quaternion that rotates pi/2 around Y\ncomposed_fs_q = first_quat*second_quat\n\n# Quaternion to rotate from one reference system to the other\nquaternion_ini_A = quaternion_ini_A*composed_fs_q\nA2B_quaternion = quaternion_ini_B*(quaternion_ini_A.inverse())\n\nA2B_quaternion is the quaternion that I use for the rotation but still doesn't perform the right rotation. Any idea?\n\n• It looks like your two systems (A and B) have different \"handedness\"; that is, you can't get from one system to the other by simply using rotations. A is a left-handed system and B is right-handed. Mar 27 '13 at 17:33\n\nSo you need to find the rotation that gets you from the first coordinate system to the second, and then combine that with the translation to get the final transformation matrix.\n\nSince you have quaternions q1 and q2, you can write this rotation as:\n\n$q = q_2 * q_1^{-1}$\n\nYou can then convert from the quaternion form to a rotation matrix to make the final transformation matrix, there are equations for this.\n\nThe final transformation would be of the form:\n\n$\\begin{bmatrix}\\boldsymbol{R} & \\boldsymbol{t}\\\\\\boldsymbol{0} & 1\\end{bmatrix}$\n\nWhere $\\boldsymbol{R}$ is the 3x3 rotation matrix and $\\boldsymbol{t}$ is the 3x1 translation vector.\n\nIn your particular case, isn't it easiest to just transpose the axes:\n\nBx = Az\nBy = -Ay\nBz = Ax\n\nOr if you want to write that as a matrix transform:\n\nB = A * [ [ 0, 0, 1 ],\n[ 0, -1, 0 ],\n[ 1, 0, 0 ] ]"
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https://m.lanxicy.com/read/f220eb80d35212566a85e9a527573f2fd4593d8d.html | [
"# 等差数列求和第一课时 课件_图文\n\n1 + 2 + 3 +…+50+51+…+98+99+100\n? 101?50 ? 5050\n\n？ 如何求等差数列{an}的前n项和\nSn=a1+a2+…….+an-1+an\n\n(如果m ? n ? p ? q,那么am ? an ? ap ? aq.)\n\n1 2 3\n\ns100\n\n100 99 98\n\ns100\n100\n\n2S100 ? (1?100 ) ?100\n\ns100\n\n?\n\n(1?100 )?100 2\n\n?\n\n5050\n\n1\n\n： 推导方法二\nSn=a1+(a1+d)+ (a1+2d) +…….+ [a1+(n-1) d]\n\nSn =an+(an- d)+ (an-2d) +…… +[ an-(n-1) d ]\n\n2Sn=(a1+an)+(a1+an)+……+ (a1+an)+ (a1+an)\n\n=n (a1+an)\n\n? Sn\n\n?\n\nn(a1 ? an ) 2\n\n23+ 24+ ……+65=？\n\n43? (23 ? 65)\n\nS43 ?\n\n2\n\n=1892\n\na1\n\nSn\n\n?\n\nn(a1 ? 2\n\nan )\n\nn\n\nan\n\n1)前n个正整数的和：1+2+3+…+n=\n\nn(n ? 2\n\n1)\n\n.\n\n2)求正整数列中前n个偶数的和\n2+4+6+…+2n=\n\nn\n\n?\n\nn?(2 ? 2\n\n2n)\n\n?\n\nn(n\n\n?1).\n\nA)Sn\n\n?\n\nn(a1 ? an ) 2\n\nB)Sn\n\n?\n\nna1\n\n?\n\nn(n ?1) 2\n\nd\n\nC)Sn\n\n?\n\ndn2 2\n\n? (a1\n\n?\n\nd )n 2\n\nD)Sn ? a1 ? (n ?1)d\n\n?\n\nna1\n\n?\n\nn(n ?1) 2\n\nd\n\na1\n\nd\n\nn\n\nan\n\nsn\n\n5\n\n10\n\n10\n\n95\n\n500\n\n20\n\n-2\n\n8\n\n6\n\n104\n\n-38\n\n2\n\n15\n\n-10 -360\n\n(如果m ? n ? p ? q,那么am ? an ? ap ? aq.)\n\n1.公式的推导方法:倒序相加法求和。\n\n2.等差数列求和公式的特征。\n\nSn\n\n?\n\nn(a1 ? 2\n\nan )\n\nSn\n\n?\n\nna1\n\n?\n\nn(n ?1) 2\n\nd\n\n3.公式的灵活应用。\n\n2019SUCCESS\nPOWERPOINT\n2019/5/28\n\n2019SUCCESS\nTHANK YOU\n2019/5/28"
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| {"ft_lang_label":"__label__zh","ft_lang_prob":0.78588223,"math_prob":0.98851496,"size":1668,"snap":"2020-34-2020-40","text_gpt3_token_len":1404,"char_repetition_ratio":0.09375,"word_repetition_ratio":0.11935484,"special_character_ratio":0.56294966,"punctuation_ratio":0.19469027,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9924559,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-06T07:10:39Z\",\"WARC-Record-ID\":\"<urn:uuid:b4c3efe2-cbaf-4b3c-bbd2-f231e760bd0b>\",\"Content-Length\":\"8814\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6baa5fff-d888-4fab-bc68-0da87f8994c8>\",\"WARC-Concurrent-To\":\"<urn:uuid:6b455582-8a6a-4a0e-a689-b2490b723434>\",\"WARC-IP-Address\":\"137.220.245.67\",\"WARC-Target-URI\":\"https://m.lanxicy.com/read/f220eb80d35212566a85e9a527573f2fd4593d8d.html\",\"WARC-Payload-Digest\":\"sha1:YRH62N3AEVD7BEE4RWMY472OKCE3OOMK\",\"WARC-Block-Digest\":\"sha1:DENWC2GBI6ZWRTPX4EZXZG3FMKYDUJBK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439736883.40_warc_CC-MAIN-20200806061804-20200806091804-00401.warc.gz\"}"} |
https://stats.stackexchange.com/questions/136272/how-to-handle-data-normalization-in-knn-when-new-test-data-is-received | [
"# How to handle data normalization in kNN when new test data is received\n\nLets say we have 100 points of labeled data and we are using $k$-nearest neighbor method for prediction. So our data looks like this:\n\n$$X= \\left\\{(\\textbf{x}_1, y_1),(\\textbf{x}_2,y_2), \\,...,(\\textbf{x}_{100}, y_{100})\\right\\},$$\n\nwhere $\\textbf{x}_i\\in \\mathbb{R}^n,\\;y_i\\in\\mathbb{R}$.\n\nNow we use Leave-One-Out (or 10-fold etc.) cross validation for finding the best $k$-value. What we also do is that we normalize our data using z-score normalization. We do this at every iteration to prevent data snooping, i.e. at every iteration of Leave-One-Out we do the following:\n\n1. Drop the test point $(\\textbf{x}_t, y_t)$out from training data\n2. Now we find the mean $\\mu$ and standard deviation $\\sigma$ of the training set $X_0=X\\setminus (\\textbf{x}_t, y_t)$ and use them to normalize the training data set $X_0$\n\n3. Now we normalize the test point $(\\textbf{x}_t,y_t)$ with the mean $\\mu$ and standard deviation $\\sigma$ we got from $X_0$ and make a prediction for the test point.\n\n4. We repeat this process for whole data set and we get some error value $E$.\n\nLets say now that we found our optimal $k$-value to be $k=3$. So we have solved our optimal model to be $3$-nearest neighbor model and we prevented data snooping by not including the test point $(\\textbf{x}_t,y_t)$ in the calculations of $\\mu$ and $\\sigma$ we needed for the normalization.\n\nNow we receive new 50 unlabeled data points: $Z=\\{\\textbf{z}_1,...\\textbf{z}_{50}\\}$. We need to make predictions for these points. Now finally we get to the question:\n\nHow should we handle normalization of the test data now? Should we find the mean $\\mu$ and standard deviation $\\sigma$ of the set $X$ to normalize the new data points $Z$ OR should we use both of the sets $X$ and $Z$ to find out the $\\mu$ and $\\sigma$ used for normalization?\n\nSo my question is about how should the $\\mu$ and $\\sigma$ be calculated when we apply our model to the new set $Z$? Because we want to normalize the set $Z$...and for the normalization we need a mean $\\mu$ and standard deviation $\\sigma$...but the question is that what data should we use for calculating $\\mu$ and $\\sigma$? What would be the correct way to do this?\n\nP.S. note that my $\\mu$ and $\\sigma$ are vectors =) so each of the $n$ features has its own mean and std.\n\nHope my question is clear =) Thank you for any help! Please ask if my question is unclear\n\nUsing the same reasoning / model building process: After you have selected a k by validation, you build the final model using all the training data $X$ and calculate the mean and variance based only on $X$ since these values are also part of the model.\nAdditionally: A classification model should classify new unlabeled instances independent of each other. But if you calculate mean and variance also based on $Z$, then the prediction might change for the same unlabeled instance dependent on how the rest of $Z$ looks like. This is not correct."
]
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https://iblog.dearbornschools.org/fardous4/2019/03/20/march-18-march-22/ | [
"March 20\n\n# March 18-March 22\n\n### Monday March 18, 2019\n\nObjectives: I can recall information about finding different parts of a circle as well as Arc length and sector Area\n\nToday we worked on the followings:\n\n1- Bell work that helped us review Arc length and sector Area\n\n2- Went over our homework from the week beforeArc length and sector area homework answers\n\n3- Review all the concepts about circles that we did so far Putting all together answers\n\n_____________________________________________________________________________\n\n### Wednesday March 20, 2019\n\nObjectives:I can Find radius, center, and graph a circle by analyzing its equation\n\nToday we worked on the followings:\n\n1-Bell work that helped us review Arc length and sector Area\n\n2- Notes on equations of a circleequation of a circle notes filled\n\n3- Homework was assignedworksheet equation of a circle\n\n___________________________________________________________________________\n\n### Thursday March 21, 2019\n\nObjectives:I can Find radius, center, and graph a circle by analyzing its equation\n\nToday we worked on the followings:\n\n1-Bell work that helped us review central and inscribed angles\n\n2- Went over questions from homeworkequation of a circle homework answers\n\n3- Individual activity to practice graphing and writing an equation of a circle\n\n### 3- Quiz tomorrow Arc Length and Sector Area/SAT practice\n\n_____________________________________________________________________________\n\n### Friday March 22, 2019\n\nObjectives: I can practice for SAT\n\nToday we worked on the followings:\n\n1-Quiz\n\n2- SAT prep heart of Algebra SAT prep"
]
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https://mailman-1.sys.kth.se/pipermail/gromacs.org_gmx-users/2011-September/064805.html | [
"# [gmx-users] problem in binding free energy calculation and setting Lambda\n\nmeisam valizadeh kiamahalleh meisam488 at gmail.com\nTue Sep 27 10:49:54 CEST 2011\n\n```Dear GMX Users\n\nI would like to let you know that I have run MD simulation on a Carbon\nnanotube- drug system.\n\nAll steps (Energy minimization, Postion restrained, Equilibration with NVT\nand NPT ensembles and finally MD) have been done successfully. Currently if\nwould like to calculate binding free energy for the system (drug molecules\ninside Carbon nanotube). Here are my questions at this point;\n\n1.May I know if I must add the free energy stuff to all the mdp files for\neach steps of (Energy minimization, Position restrained, Equilibration with\nNVT and NPT ensembles and finally MD) and redo all the simulation steps? I\nam wondering because I have seen few people who only edited the final MD.mdp\nfile and add the free energy stuff.\n\n2. I know to get the binding energy curve; I need to run several\nsimulations, one for each lambda value. Hence, I need 21 point from\nLambda=0 to 1 to get the smooth curve. To do this, I have to start with\ninitial Lambda =0 as stated in the energy stuff below, in my mdp file. May\nI know which parameter(s) (init_lambda or foreign lambda or ...) need to be\nchanged in my mdp file to do the next simulation and until the last\nsimulation? I mean to go from Lambda=0.0 to the next lambda =0.05, 0.10,\n0.15, … and 1.00.\n\nThe free energy stuff in my mdp file is as below:\n\n; Free energy control stuff\n\nfree_energy = yes\n\ninit_lambda = 0.00\n\ndelta_lambda = 0\n\nforeign_lambda = 0.05\n\nsc-alpha = 0.5\n\nsc-power = 1.0\n\nsc-sigma = 0.3"
]
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https://www.jubiba.de/ball-mill/dj7jkbuh/how.html | [
"Online Service: Anytime: 7*24",
null,
"1. Home\n2. --\n3. News\nHot Products",
null,
"Ball Mill\n\nThe ball mill is one of the most widely used super fine grinding machine in the industry and it is the key grinding equipment after materials have been crushed.",
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"Cement Mill\n\nA cement mill is the equipment used to grind the hard, nodular clinker from the cement kiln into the fine grey powder that is cement.",
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"Sand Maker\n\nSand maker is suitable for the crushing of soft, hard and extremely hard material and reshape of those products.\n\nLatest News\n\n# how to calculate critical speed in ball mill process\n\nBall mills have been successfully run at speeds between 60 and 90 percent of critical speed but most mills operate at speeds between 65 and 79 percent of critical speed Rod mills speed should be limited to a maximum of 70 of critical speed and preferably should be in the 60 to 68 percent critical speed\n\nWhatever your requirements, you 'll find the perfect service-oriented solution to match your specific needs with our help.We are here for your questions anytime 24/7, welcome your consultation.\n\nChat Online",
null,
"Mill Critical Speed Calculation\n\nThe speed of the mill was kept at 63 of the critical speed The face angle was varied from 90 to 111 degrees for the three types of configuration 1 2 and 4 as shown in the figure Also the height of the lifter bar in configuration 3 was changed to observe the trajectory\n\nGet A Quote",
null,
"Mill Speed Critical Speed Paul O Abbe\n\nThe ideal mill speed is usually somewhere between 55 to 75 of critical speed Critical Mill Speed Critical Speed left is the speed at which the outer layer of media centrifuges against the wall Second Critical Speed middle is the speed at which the second layer of media centrifuges inside the first layer nth Critical speed right is the speed at which the nth layer of media centrifuges inside the n1 layer\n\nGet A Quote",
null,
"Ball Mill Calculations Critical Speed\n\nCalculate critical speed of ball ate critical speed of ball millrequest for quotation you can get the price calculations of ball mill Live Chat Ball Mill Load Calculations Gujaratgenomics Ball mill tion to calculate critical speed of ball mill design calculation of to calculate torque load\n\nGet A Quote",
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"how to calculate critical speed for a ball mill\n\nMill Speed Critical Speed Paul O Abbe Talk with the Experts at Paul O Abbe about your process requirements and Ball Mill Loading Wet Milling Size Reduction and Mill Speed Critical Speed needs\n\nGet A Quote",
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"Mill Critical Speed Formula Derivation Grinding\n\nThe formula to calculate critical speed is given below N c 42305 sqtDd N c critical speed of the mill D mill diameter specified in meters d diameter of the ball In practice Ball Mills are driven at a speed of 5090 of the critical speed the factor being influenced by economic consideration\n\nGet A Quote",
null,
"Mill Critical Speed Determination\n\nThe mill critical speed will be calculated based on the diameter above less twice this shell liner width\n\nGet A Quote",
null,
"critical speed of ball mill of raduis r\n\ncritical speed ball mill radius r The critical speed of a ball mill is calculated as 54 19 divided by the square root of the radius in feet ball mill calculation critical speed arhc Ball Mill Operating Speed Mechanical Operations The critical speed of ball mill is given by where R radius of ball mill r radius of ball For R 1000 mm and r 50 mm n c 30 7 rpm\n\nGet A Quote",
null,
"What it is the optimun speed for a ball mill\n\nOct 18 2006 · Congratulations on your ball mill Youll want to further reduce your motor speed by 31 to mill more efficiently Right now your RPM is just slightly under the speed where the media wont do any work The optimum speed for your jar to turn is around 90 RPM like Frozentech said\n\nGet A Quote",
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"Ball Mill DesignPower Calculation\n\nA motor with around 1400 Horse Power is calculated needed for the designed task Now we much select a Ball Mill that will draw this power The ball mill motor power requirement calculated above as 1400 HP is the power that must be applied at the mill drive in order to grind the tonnage of feed from one size distribution The following shows how the size or select the matching mill required to\n\nGet A Quote",
null,
"How To Calculate Critical Speed Of A Ball Mill\n\nThe critical speed of a ball mill is calculated as 5419 divided by the Chat Now The modelling of the mechanical alloying process in a planetary ball milling have been\n\nGet A Quote",
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null,
"critical speed of ball mill of raduis r\n\ncritical speed ball mill radius r The critical speed of a ball mill is calculated as 54 19 divided by the square root of the radius in feet ball mill calculation critical speed arhc Ball Mill Operating Speed Mechanical Operations The critical speed of ball mill is given by where R radius of ball mill r radius of ball For R 1000 mm and r 50 mm n c 30 7 rpm\n\nGet A Quote",
null,
"calculate critical speed grinder\n\ncalculate critical speed grinder critical speed of ball mill calculation grinder ball mill critical speed calculator Planetary Ball Mill Process in Aspect of Milling Energy Based on the Get Price And Support Online calculate critical speed grinder diavista to calculate critical speed of ball mill practical A ball mill is a type of grinder used to grind materials into\n\nGet A Quote",
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"Optimization of mill performance by using\n\nOptimization of mill performance by using online ball and pulp measurements Soon after a stop a mill is a dangerous place to enter for the personnel tasked with taking measurements or samples Crash stops are difficult to handle The mill and all feed streams should be stopped simultaneously but often they are stopped around about the same time\n\nGet A Quote",
null,
"CALCULATION OF THE POWER DRAW OF DRY\n\nfCs – Fraction of critical speed Ss – Ball size factor To determine the power that a dry grinding needs full grate discharge mill Equation 1 is multiplied by the factor of 108 A multicompartment ball mill consists of two or more grate discharge ball mills in series The same equation is used to\n\nGet A Quote",
null,
"Ball Screw Selection and Calculations University of Utah\n\nPermissible Speed • When the speed of a ball screw increases the ball screw will approach its natural frequency causing a resonance and the operation will become impossible π ρ λ π ρ λ E l d A EI l n b tr b c 2 2 2 2 2 15 2 60 nc Critical speed min1 lb Distance between supports m E Elastic modulus Pa I Second moment of\n\nGet A Quote",
null,
"Rod mills how they function Mine\n\nThe Critical Speed is used for the determination of ball mill ideal operating speed But for comparison rod mills would operate between 50 to 95 of the critical speed The faster the mill speed the greater the wear on the rods and liners\n\nGet A Quote",
null,
"AMIT 135 Lesson 6 Grinding Circuit – Mining Mill\n\nThe mill product can either be finished size ready for processing or an intermediate size ready for final grinding in a rod mill ball mill or pebble mill AGSAG mills can accomplish the same size reduction work as two or three stages of crushing and screening a rod mill and some or all of the work of a ball mill\n\nGet A Quote",
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https://siuba.org/develop/backend_pandas | [
"``````import pandas as pd\npd.set_option(\"display.max_rows\", 5)``````\n\nThe pandas backend has two key responsibilities:\n\n1. Implementing functions that enable complex and fast expressions over grouped data.\n2. Implementing verb behavior over pandas’ DataFrameGroupBy (e.g. a grouped mutate).\n\nWhile it might seem like a lot of work, this mostly involves using a few simple strategies to take advantage of logic already existing in the pandas library.\n\nThe strategy can be described as follows:\n\n• operations over grouped objects return grouped objects.\n• e.g. `add(SeriesGroupBy, SeriesGroupBy) -> SeriesGroupBy`.\n• operations that do aggregation return a subclass of SeriesGroupBy called GroupByAgg. This represents the case where each row is its own group. It also holds information about the original data, so it can be broadcast back to the original length.\n\nAs a final note, while the SQL backend uses a custom backend class (LazyTbl), the backend for table verbs in this case is just the pandas’ DataFrameGroupBy class itself.\n\n## Column op translation\n\n``````from siuba.ops import mean\nfrom siuba.data import mtcars\n\n# equivalent to mtcars.hp.mean()\nmean(mtcars.hp)``````\n``146.6875``\n``````import pandas as pd\nfrom siuba.ops.utils import operation\n\nmean2 = operation(\"mean\", \"Agg\", 1)\n\n# Series implementation just calls the Series' mean method\nmean2.register(\npd.Series,\nlambda self, *args, **kwargs: self.mean(*args, **kwargs)\n)``````\n``<function __main__.<lambda>(self, *args, **kwargs)>``\n``mean2(mtcars.hp)``\n``146.6875``\n``mean2.operation.name``\n``'mean'``\n\nThe purpose of the `.operation` data is to make it easy to generate new translations for functions. For example, if we want to translate pandas’ `ser.str.upper()` method, then it helps to know it uses the `str` accessor.\n\n### Using an existing translation\n\n``````from siuba.experimental.pd_groups.translate import method_el_op\n\ndf = pd.DataFrame({\n\"x\": [\"a\", \"b\", \"c\"],\n\"g\": [0, 0, 1]\n})\n\n# notice method_ag_op uses some details from .operation\nupper = method_el_op(\"upper\", is_property = False, accessor = \"str\")\nlower = method_el_op(\"lower\", is_property = False, accessor = \"str\")\n\ng_df = df.groupby(\"g\")\n\nres = upper(g_df.x)\n\n# note: .obj is how you \"ungroup\" a SeriesGroupBy\nres.obj``````\n``````0 A\n1 B\n2 C\nName: x, dtype: object``````\n``````# convert to uppercase and back to lowercase\n# equivalent to df.x.str.upper().str.lower()\nres2 = lower(upper(g_df.x))\nres2.obj``````\n``````0 a\n1 b\n2 c\nName: x, dtype: object``````\n``isinstance(res, pd.core.groupby.SeriesGroupBy)``\n``True``\n``lower(upper(g_df.x))``\n``<pandas.core.groupby.generic.SeriesGroupBy object at 0x7f66c8da1b20>``\n\nSee the internals of functions like `method_el_op` for details.\n\n## New verb implementations\n\nLike with other backends, verbs use single dispatch to register new backend implementations.\n\n``````from pandas import DataFrame\nfrom pandas.core.groupby import DataFrameGroupBy\nfrom siuba.dply.verbs import singledispatch2\n\n@singledispatch2(DataFrame)\ndef my_verb(__data):\nprint(\"Running default.\")\n\n# register grouped implementation ----\n@my_verb.register(DataFrameGroupBy)\ndef _my_verb_gdf(__data):\nprint(\"Running grouped!\")\n\n# test it out ----\nfrom siuba.data import mtcars\n\nmy_verb(mtcars.groupby(\"cyl\"))``````\n``Running grouped!``"
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https://austroads.com.au/publications/asset-management/ap-t201-12 | [
"# Asset management",
null,
"The Scoping and Development of Probabilistic Road Deterioration (RD) Models\n• Publication no: AP-T201-12\n• ISBN: 978-1-921991-29-5\n• Published: 16 May 2012\n\nThis report is a scoping document aimed at establishing how road deterioration (RD) models can be developed on a probabilistic basis. Deterministic RD models provide an average estimate to a future condition which is not necessarily likely to be achieved given the variability of pavement performance. A probabilistic approach can assign various probabilities to the future conditions of a pavement (the dependent variable). The report also considers the applicability of probabilistic RD model approaches to both a network and project level. A previous survey of practitioners, aimed at finding what type of RD models they wish to use in a pavement life-cycle costing context at a project level, found that most practitioners would consider using both probabilistic and deterministic RD models.\n\n• 1. INTRODUCTION\n• 1.1. Project Scope\n• 2. REVIEW OF PROBABILISTIC MODELLING APPROACHES\n• 2.1. Review of Probabilistic Approach\n• 2.1.1. Survivor Curves\n• 2.1.2. Markov Probabilistic Approach\n• 2.1.3. Semi-Markov Probabilistic Approach\n• 2.1.4. Continuous Probabilistic Approach\n• 2.2. Probabilistic Approaches Used in Australasia\n• 2.2.1. Markov and Semi-Markov Probabilistic Approaches\n• 2.2.2. Probability of Defect Initiation and Unacceptable Deterioration Rates\n• 2.2.3. Other Probability Approaches\n• 3. CONCLUSIONS\n• 3.1. General\n• 3.1.1. Survivor Curves\n• 3.1.2. Markov and Semi-Markov Approaches\n• 3.1.3. Continuous Probability Functions\n• 3.1.4. Other Probability Approaches\n• 4. FUTURE WORKS\n• 4.1. General\n• 4.1.1. Next Step\nRelated publications"
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https://math.stackexchange.com/questions/4337056/is-x3-ax-2a1-irreducible-over-bbb-z | [
"# Is $x^3-ax-(2a+1)$ irreducible over $\\Bbb Z$?\n\nIs the cubic polynomial $$x^3-ax-(2a+1)$$ irreducible over $$\\mathbb Z$$ for all positive integers $$a$$?\n\none way to prove irreducibility is to use eisenstein's criterion. we want to find prime $$p$$ st:\n$$p\\mid a$$\n$$p\\mid 2a+1$$\n$$p^2\\nmid 2a+1$$\nHowever, the first 2 conditions imply $$p\\mid 1$$, a contradiction. so this approach fails.\n\nanother approach is to use the rational root thm.\nlet $$x^3-ax-(2a+1)=(x+b)(x^2+cx+d)=x^3+(b+c)x^2+(bc+d)x+bd$$\nwhere b,c,d integers\n$$b+c=0$$\n$$bc+d=-a$$\n$$bd=-(2a+1)$$\n\nthere was a link to a similar problem, now deleted.\n\n• Your current question as well as the linked question do not have any context. Please try to improve your question by adding relevant context. Jan 12, 2022 at 2:56\n• No!! Context has to be part of the question body. I understand that you have acted here in good faith but note that while self answering the bar for standards is higher. For the current situation it is best to close the question unless improved. Jan 12, 2022 at 3:08\n\nIf a cubic is reducible, then it must have a linear factor $$x+b$$, so $$-b$$ is an integer root.\n\n$$-b^3+ab-(2a+1)=0$$ $$a(b-2)=b^3+1$$ $$a=\\frac{b^3+1}{b-2}$$ $$(b-2)\\mid 9$$ \\begin{align}b-2&=-9,-3,-1,1,3,9\\\\b&=-7,-1,1,3,5,11\\\\b&\\neq -1,1\\end{align} so $$(a,b,c)=(38,-7,11),(28,3,-19),(42,5,-17),(148,11,-27)$$.\n\n\\begin{align}(x-7)(x^2+7x+11)&=x^3-38x-77\\\\ (x+3)(x^2-3x-19)&=x^3-28x-57\\\\ (x+5)(x^2-5x-17)&=x^3-42x-85\\\\ (x+11)(x^2-11x-27)&=x^3-148x-297\\end{align}\n\nso there are 4 $$a$$'s such that the cubic is reducible. Simply stating (but not deriving) one of the 4 factorizations is enough to disprove the question.\n\nthanks to @dxiv for the suggestion.\n\n• Nice answer (+1), could use a bit more explanation at a few points. Dec 18, 2021 at 19:26\n• Can you elaborate on why $b-2$ must divide $9$? Dec 18, 2021 at 19:55\n• I see, because $a=\\frac{b^3+1}{b-2}=b^2+2b-4+\\frac{9}{b-2}$. Dec 18, 2021 at 20:01\n• You can shorten the proof somewhat by noting that any rational root must be an integer since the polynomial is monic. Let $n$ be such a root, then $a=\\frac{n^3-1}{n+2}$ and a similar argument follows.\n– dxiv\nDec 18, 2021 at 22:04\n• @cineel Regroup and collect: $\\,n^3-an-(2a+1)=0 \\iff n^3-1=(n+2)a\\,$.\n– dxiv\nDec 19, 2021 at 5:34"
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https://stacks.math.columbia.edu/tag/0FMC | [
"Lemma 50.8.3. Assume $X \\to S$ and $Y \\to S$ are smooth and quasi-compact and the morphisms $X \\to X \\times _ S X$ and $Y \\to Y \\times _ S Y$ are affine. Then the relative cup product\n\n$Ra_*\\Omega ^\\bullet _{X/S} \\otimes _{\\mathcal{O}_ S}^\\mathbf {L} Rb_*\\Omega ^\\bullet _{Y/S} \\longrightarrow Rf_*\\Omega ^\\bullet _{X \\times _ S Y/S}$\n\nis an isomorphism in $D(\\mathcal{O}_ S)$.\n\nProof. Immediate consequence of Lemma 50.8.2. $\\square$\n\nIn your comment you can use Markdown and LaTeX style mathematics (enclose it like $\\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar)."
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http://www.hsn.uk.net/forum/index.php?/topic/6122-who-belives-something-nothing/page__fromsearch__1 | [
"Warning: Illegal string offset 'html' in /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php on line 909\n\nWho belives SOMETHING = NOTHING ? - HSN forum",
null,
"# Who belives SOMETHING = NOTHING ?\n\n3 replies to this topic\n\n### #1cK_hoangan\n\n• Gender:Not Telling\n\nPosted 29 November 2007 - 02:04 PM\n\nSOMETHING = NOTHING, it's possible !!! Let see the magic of Maths now :\n\nConsider :\n\nS = 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 +.....\n=> 1/2 S = 1/4 + 1/6 + 1/8 + .....\n=>S - 1/2S = 1/2 + 1/3 + 1/5 + 1/7 + .... (canceled respectly 1/4; 1/6; 1/8; ... from S by 1/2 S)\n=> 1/2 S = 1/2 + 1/3 + 1/5 + 1/7 + ...\nNow we subtract the second 1/2 S for the first 1/2 S :\n\n1/2 S - 1/2 S = (1/2 + 1/3 + 1/5 + 1/7 + ... ) - ( 1/4 + 1/6 + 1/8 + ... )\n1/2 S - 1/2 S = 1/2 + 1/3 + 1/5 + 1/7 + ... - 1/4 - 1/6 - 1/8 - ...\n0 = 1/2 + ( 1/3 - 1/4) + ( 1/5 - 1/6) + ( 1/7 - 1/8 ) + ....\n0 = 1/2 + 1/12 + 1/30 + 1/56 + ....\n\nFrom now. obviously seeing that the right hand side would end up with a positive number which (of course) is greater than 0",
null,
". But on the left hand side, so funny, it's definetly 0.",
null,
"( what else can be if subtract 1/2 S by itself ?",
null,
")\n\nSo, can u now get the point ? Clearly, NOTHING = SOMETHING , MATHS is miracle !!!!",
null,
"### #2Marcus\n\nSite Swot\n\n•",
null,
"• Members\n•",
null,
"",
null,
"",
null,
"",
null,
"• 147 posts\n• Location:Aberdeen\n• Interests:Guitar - Passed my grade 6 :D :D<br />Write computer programs - won Heriot-Watt programming challenge 2006 and 2007 :D<br />\n• Gender:Male\n\nPosted 06 January 2008 - 01:07 AM\n\nThat puzzled me for quite a while, the problem is that your S changes its value, you have got an infinite series...but the number of terms you specify it to changes, therefore changing the value...\n\nkeeping it to 3 terms\n\nif S = 1/2 + 1/3 +1/4\nthen 1/2S = 1/4 + 1/6 + 1/8 = 13/24\n\nS-1/2S = 1/2 + 1/3 +1/4 - 1/4 - 1/6 - 1/8\n1/2S= 1/2 + 1/3 - 1/6 - 1/8 = 13/24\n\ntherefore if you subtracted the first 1/2S from the second 1/2S you would get 0 = 0\n\nQED\n=-=-=Marcus=-=-=\n\n### #3cK_hoangan\n\n• Gender:Not Telling\n\nPosted 17 January 2008 - 03:47 PM\n\nbut ... then, wat is the error ?",
null,
"If you agree that it is an infinite serie, how could you know the number of its terms is changed ?",
null,
"### #4Marcus\n\nSite Swot\n\n•",
null,
"• Members\n•",
null,
"",
null,
"",
null,
"",
null,
"• 147 posts\n• Location:Aberdeen\n• Interests:Guitar - Passed my grade 6 :D :D<br />Write computer programs - won Heriot-Watt programming challenge 2006 and 2007 :D<br />\n• Gender:Male\n\nPosted 20 January 2008 - 01:57 AM\n\nbut ... then, wat is the error ?",
null,
"If you agree that it is an infinite serie, how could you know the number of its terms is changed ?",
null,
"the problem is: you have solved an infinite series using a set number of terms, your S , and both",
null,
"s are all to different numbers of terms, it is the differences in these numbers of terms that the error arises...\n\nTo do this properly you would have to specify S in such a way that it can be infinite, ie S equals the recurrence relation:",
null,
"which you half:",
null,
"then you subtract them:",
null,
"now you should be able to see that if you subtract the two",
null,
"s now you would get",
null,
"=-=-=Marcus=-=-=\n\n#### 1 user(s) are reading this topic\n\n0 members, 1 guests, 0 anonymous users",
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https://homework.cpm.org/category/CCI_CT/textbook/int3/chapter/12/lesson/12.1.3/problem/12-49 | [
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"### Home > INT3 > Chapter 12 > Lesson 12.1.3 > Problem12-49\n\n12-49.\n\nSolve the system of equations below.\n\n$\\begin{array}{r} x+y=−3\\\\2x−y=−6\\\\3x−2y+5z=16 \\end{array}$\n\nAll of the strategies for solving systems of two variables still apply.\n\nUse the first two equations to solve for $x\\ \\text{and}\\ y$."
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",
null
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https://www.netlib.org/lapack/complex16/ | [
"# lapack/complex16\n\n```# ---------------------------------\n# Available SIMPLE DRIVER routines:\n# ---------------------------------\nfile zcgesv.f zcgesv.f plus dependencies\nprec complex16\nZCGESV computes the solution to a complex system of linear equations\nA * X = B,\nwhere A is an N-by-N matrix and X and B are N-by-NRHS matrices.\n\nZCGESV first attempts to factorize the matrix in COMPLEX and use this\nfactorization within an iterative refinement procedure to produce a\nsolution with COMPLEX*16 normwise backward error quality (see below).\nIf the approach fails the method switches to a COMPLEX*16\nfactorization and solve.\n\nThe iterative refinement is not going to be a winning strategy if\nthe ratio COMPLEX performance over COMPLEX*16 performance is too\nsmall. A reasonable strategy should take the number of right-hand\nsides and the size of the matrix into account. This might be done\nwith a call to ILAENV in the future. Up to now, we always try\niterative refinement.\n\nThe iterative refinement process is stopped if\nITER > ITERMAX\nor for all the RHS we have:\nRNRM < SQRT(N)*XNRM*ANRM*EPS*BWDMAX\nwhere\no ITER is the number of the current iteration in the iterative\nrefinement process\no RNRM is the infinity-norm of the residual\no XNRM is the infinity-norm of the solution\no ANRM is the infinity-operator-norm of the matrix A\no EPS is the machine epsilon returned by DLAMCH('Epsilon')\nThe value ITERMAX and BWDMAX are fixed to 30 and 1.0D+00\nrespectively.\n\nfile zcposv.f zcposv.f plus dependencies\nprec complex16\nZCPOSV computes the solution to a complex system of linear equations\nA * X = B,\nwhere A is an N-by-N Hermitian positive definite matrix and X and B\nare N-by-NRHS matrices.\n\nZCPOSV first attempts to factorize the matrix in COMPLEX and use this\nfactorization within an iterative refinement procedure to produce a\nsolution with COMPLEX*16 normwise backward error quality (see below).\nIf the approach fails the method switches to a COMPLEX*16\nfactorization and solve.\n\nThe iterative refinement is not going to be a winning strategy if\nthe ratio COMPLEX performance over COMPLEX*16 performance is too\nsmall. A reasonable strategy should take the number of right-hand\nsides and the size of the matrix into account. This might be done\nwith a call to ILAENV in the future. Up to now, we always try\niterative refinement.\n\nThe iterative refinement process is stopped if\nITER > ITERMAX\nor for all the RHS we have:\nRNRM < SQRT(N)*XNRM*ANRM*EPS*BWDMAX\nwhere\no ITER is the number of the current iteration in the iterative\nrefinement process\no RNRM is the infinity-norm of the residual\no XNRM is the infinity-norm of the solution\no ANRM is the infinity-operator-norm of the matrix A\no EPS is the machine epsilon returned by DLAMCH('Epsilon')\nThe value ITERMAX and BWDMAX are fixed to 30 and 1.0D+00\nrespectively.\n\nfile zgbsv.f zgbsv.f plus dependencies\nprec complex16\nZGBSV computes the solution to a complex system of linear equations\nA * X = B, where A is a band matrix of order N with KL subdiagonals\nand KU superdiagonals, and X and B are N-by-NRHS matrices.\n\nThe LU decomposition with partial pivoting and row interchanges is\nused to factor A as A = L * U, where L is a product of permutation\nand unit lower triangular matrices with KL subdiagonals, and U is\nupper triangular with KL+KU superdiagonals. The factored form of A\nis then used to solve the system of equations A * X = B.\n\nfile zgees.f zgees.f plus dependencies\nprec complex16\nZGEES computes for an N-by-N complex nonsymmetric matrix A, the\neigenvalues, the Schur form T, and, optionally, the matrix of Schur\nvectors Z. This gives the Schur factorization A = Z*T*(Z**H).\n\nOptionally, it also orders the eigenvalues on the diagonal of the\nSchur form so that selected eigenvalues are at the top left.\nThe leading columns of Z then form an orthonormal basis for the\ninvariant subspace corresponding to the selected eigenvalues.\n\nA complex matrix is in Schur form if it is upper triangular.\n\nfile zgeev.f zgeev.f plus dependencies\nprec complex16\nZGEEV computes for an N-by-N complex nonsymmetric matrix A, the\neigenvalues and, optionally, the left and/or right eigenvectors.\n\nThe right eigenvector v(j) of A satisfies\nA * v(j) = lambda(j) * v(j)\nwhere lambda(j) is its eigenvalue.\nThe left eigenvector u(j) of A satisfies\nu(j)**H * A = lambda(j) * u(j)**H\nwhere u(j)**H denotes the conjugate transpose of u(j).\n\nThe computed eigenvectors are normalized to have Euclidean norm\nequal to 1 and largest component real.\n\nfile zgegs.f zgegs.f plus dependencies\nprec complex16\nThis routine is deprecated and has been replaced by routine ZGGES.\n\nZGEGS computes the eigenvalues, Schur form, and, optionally, the\nleft and or/right Schur vectors of a complex matrix pair (A,B).\nGiven two square matrices A and B, the generalized Schur\nfactorization has the form\n\nA = Q*S*Z**H, B = Q*T*Z**H\n\nwhere Q and Z are unitary matrices and S and T are upper triangular.\nThe columns of Q are the left Schur vectors\nand the columns of Z are the right Schur vectors.\n\nIf only the eigenvalues of (A,B) are needed, the driver routine\nZGEGV should be used instead. See ZGEGV for a description of the\neigenvalues of the generalized nonsymmetric eigenvalue problem\n(GNEP).\n\nfile zgegv.f zgegv.f plus dependencies\nprec complex16\nThis routine is deprecated and has been replaced by routine ZGGEV.\n\nZGEGV computes the eigenvalues and, optionally, the left and/or right\neigenvectors of a complex matrix pair (A,B).\nGiven two square matrices A and B,\nthe generalized nonsymmetric eigenvalue problem (GNEP) is to find the\neigenvalues lambda and corresponding (non-zero) eigenvectors x such\nthat\nA*x = lambda*B*x.\n\nAn alternate form is to find the eigenvalues mu and corresponding\neigenvectors y such that\nmu*A*y = B*y.\n\nThese two forms are equivalent with mu = 1/lambda and x = y if\nneither lambda nor mu is zero. In order to deal with the case that\nlambda or mu is zero or small, two values alpha and beta are returned\nfor each eigenvalue, such that lambda = alpha/beta and\nmu = beta/alpha.\n\nThe vectors x and y in the above equations are right eigenvectors of\nthe matrix pair (A,B). Vectors u and v satisfying\nu**H*A = lambda*u**H*B or mu*v**H*A = v**H*B\nare left eigenvectors of (A,B).\n\nNote: this routine performs \"full balancing\" on A and B -- see\n\"Further Details\", below.\n\nfile zgels.f zgels.f plus dependencies\nprec complex16\nZGELS solves overdetermined or underdetermined complex linear systems\ninvolving an M-by-N matrix A, or its conjugate-transpose, using a QR\nor LQ factorization of A. It is assumed that A has full rank.\n\nThe following options are provided:\n\n1. If TRANS = 'N' and m >= n: find the least squares solution of\nan overdetermined system, i.e., solve the least squares problem\nminimize || B - A*X ||.\n\n2. If TRANS = 'N' and m < n: find the minimum norm solution of\nan underdetermined system A * X = B.\n\n3. If TRANS = 'C' and m >= n: find the minimum norm solution of\nan undetermined system A**H * X = B.\n\n4. If TRANS = 'C' and m < n: find the least squares solution of\nan overdetermined system, i.e., solve the least squares problem\nminimize || B - A**H * X ||.\n\nSeveral right hand side vectors b and solution vectors x can be\nhandled in a single call; they are stored as the columns of the\nM-by-NRHS right hand side matrix B and the N-by-NRHS solution\nmatrix X.\n\nfile zgelsd.f zgelsd.f plus dependencies\nprec complex16\nZGELSD computes the minimum-norm solution to a real linear least\nsquares problem:\nminimize 2-norm(| b - A*x |)\nusing the singular value decomposition (SVD) of A. A is an M-by-N\nmatrix which may be rank-deficient.\n\nSeveral right hand side vectors b and solution vectors x can be\nhandled in a single call; they are stored as the columns of the\nM-by-NRHS right hand side matrix B and the N-by-NRHS solution\nmatrix X.\n\nThe problem is solved in three steps:\n(1) Reduce the coefficient matrix A to bidiagonal form with\nHouseholder tranformations, reducing the original problem\ninto a \"bidiagonal least squares problem\" (BLS)\n(2) Solve the BLS using a divide and conquer approach.\n(3) Apply back all the Householder tranformations to solve\nthe original least squares problem.\n\nThe effective rank of A is determined by treating as zero those\nsingular values which are less than RCOND times the largest singular\nvalue.\n\nThe divide and conquer algorithm makes very mild assumptions about\nfloating point arithmetic. It will work on machines with a guard\ndigit in add/subtract, or on those binary machines without guard\ndigits which subtract like the Cray X-MP, Cray Y-MP, Cray C-90, or\nCray-2. It could conceivably fail on hexadecimal or decimal machines\nwithout guard digits, but we know of none.\n\nfile zgelss.f zgelss.f plus dependencies\nprec complex16\nZGELSS computes the minimum norm solution to a complex linear\nleast squares problem:\n\nMinimize 2-norm(| b - A*x |).\n\nusing the singular value decomposition (SVD) of A. A is an M-by-N\nmatrix which may be rank-deficient.\n\nSeveral right hand side vectors b and solution vectors x can be\nhandled in a single call; they are stored as the columns of the\nM-by-NRHS right hand side matrix B and the N-by-NRHS solution matrix\nX.\n\nThe effective rank of A is determined by treating as zero those\nsingular values which are less than RCOND times the largest singular\nvalue.\n\nfile zgelsy.f zgelsy.f plus dependencies\nprec complex16\nZGELSY computes the minimum-norm solution to a complex linear least\nsquares problem:\nminimize || A * X - B ||\nusing a complete orthogonal factorization of A. A is an M-by-N\nmatrix which may be rank-deficient.\n\nSeveral right hand side vectors b and solution vectors x can be\nhandled in a single call; they are stored as the columns of the\nM-by-NRHS right hand side matrix B and the N-by-NRHS solution\nmatrix X.\n\nThe routine first computes a QR factorization with column pivoting:\nA * P = Q * [ R11 R12 ]\n[ 0 R22 ]\nwith R11 defined as the largest leading submatrix whose estimated\ncondition number is less than 1/RCOND. The order of R11, RANK,\nis the effective rank of A.\n\nThen, R22 is considered to be negligible, and R12 is annihilated\nby unitary transformations from the right, arriving at the\ncomplete orthogonal factorization:\nA * P = Q * [ T11 0 ] * Z\n[ 0 0 ]\nThe minimum-norm solution is then\nX = P * Z' [ inv(T11)*Q1'*B ]\n[ 0 ]\nwhere Q1 consists of the first RANK columns of Q.\n\nThis routine is basically identical to the original xGELSX except\nthree differences:\no The permutation of matrix B (the right hand side) is faster and\nmore simple.\no The call to the subroutine xGEQPF has been substituted by the\nthe call to the subroutine xGEQP3. This subroutine is a Blas-3\nversion of the QR factorization with column pivoting.\no Matrix B (the right hand side) is updated with Blas-3.\n\nfile zgesdd.f zgesdd.f plus dependencies\nprec complex16\nZGESDD computes the singular value decomposition (SVD) of a complex\nM-by-N matrix A, optionally computing the left and/or right singular\nvectors, by using divide-and-conquer method. The SVD is written\n\nA = U * SIGMA * conjugate-transpose(V)\n\nwhere SIGMA is an M-by-N matrix which is zero except for its\nmin(m,n) diagonal elements, U is an M-by-M unitary matrix, and\nV is an N-by-N unitary matrix. The diagonal elements of SIGMA\nare the singular values of A; they are real and non-negative, and\nare returned in descending order. The first min(m,n) columns of\nU and V are the left and right singular vectors of A.\n\nNote that the routine returns VT = V**H, not V.\n\nThe divide and conquer algorithm makes very mild assumptions about\nfloating point arithmetic. It will work on machines with a guard\ndigit in add/subtract, or on those binary machines without guard\ndigits which subtract like the Cray X-MP, Cray Y-MP, Cray C-90, or\nCray-2. It could conceivably fail on hexadecimal or decimal machines\nwithout guard digits, but we know of none.\n\nfile zgesv.f zgesv.f plus dependencies\nprec complex16\nZGESV computes the solution to a complex system of linear equations\nA * X = B,\nwhere A is an N-by-N matrix and X and B are N-by-NRHS matrices.\n\nThe LU decomposition with partial pivoting and row interchanges is\nused to factor A as\nA = P * L * U,\nwhere P is a permutation matrix, L is unit lower triangular, and U is\nupper triangular. The factored form of A is then used to solve the\nsystem of equations A * X = B.\n\nfile zgesvd.f zgesvd.f plus dependencies\nprec complex16\nZGESVD computes the singular value decomposition (SVD) of a complex\nM-by-N matrix A, optionally computing the left and/or right singular\nvectors. The SVD is written\n\nA = U * SIGMA * conjugate-transpose(V)\n\nwhere SIGMA is an M-by-N matrix which is zero except for its\nmin(m,n) diagonal elements, U is an M-by-M unitary matrix, and\nV is an N-by-N unitary matrix. The diagonal elements of SIGMA\nare the singular values of A; they are real and non-negative, and\nare returned in descending order. The first min(m,n) columns of\nU and V are the left and right singular vectors of A.\n\nNote that the routine returns V**H, not V.\n\nfile zgges.f zgges.f plus dependencies\nprec complex16\nZGGES computes for a pair of N-by-N complex nonsymmetric matrices\n(A,B), the generalized eigenvalues, the generalized complex Schur\nform (S, T), and optionally left and/or right Schur vectors (VSL\nand VSR). This gives the generalized Schur factorization\n\n(A,B) = ( (VSL)*S*(VSR)**H, (VSL)*T*(VSR)**H )\n\nwhere (VSR)**H is the conjugate-transpose of VSR.\n\nOptionally, it also orders the eigenvalues so that a selected cluster\nof eigenvalues appears in the leading diagonal blocks of the upper\ntriangular matrix S and the upper triangular matrix T. The leading\ncolumns of VSL and VSR then form an unitary basis for the\ncorresponding left and right eigenspaces (deflating subspaces).\n\n(If only the generalized eigenvalues are needed, use the driver\n\nA generalized eigenvalue for a pair of matrices (A,B) is a scalar w\nor a ratio alpha/beta = w, such that A - w*B is singular. It is\nusually represented as the pair (alpha,beta), as there is a\nreasonable interpretation for beta=0, and even for both being zero.\n\nA pair of matrices (S,T) is in generalized complex Schur form if S\nand T are upper triangular and, in addition, the diagonal elements\nof T are non-negative real numbers.\n\nfile zggev.f zggev.f plus dependencies\nprec complex16\nZGGEV computes for a pair of N-by-N complex nonsymmetric matrices\n(A,B), the generalized eigenvalues, and optionally, the left and/or\nright generalized eigenvectors.\n\nA generalized eigenvalue for a pair of matrices (A,B) is a scalar\nlambda or a ratio alpha/beta = lambda, such that A - lambda*B is\nsingular. It is usually represented as the pair (alpha,beta), as\nthere is a reasonable interpretation for beta=0, and even for both\nbeing zero.\n\nThe right generalized eigenvector v(j) corresponding to the\ngeneralized eigenvalue lambda(j) of (A,B) satisfies\n\nA * v(j) = lambda(j) * B * v(j).\n\nThe left generalized eigenvector u(j) corresponding to the\ngeneralized eigenvalues lambda(j) of (A,B) satisfies\n\nu(j)**H * A = lambda(j) * u(j)**H * B\n\nwhere u(j)**H is the conjugate-transpose of u(j).\n\nfile zggglm.f zggglm.f plus dependencies\nprec complex16\nZGGGLM solves a general Gauss-Markov linear model (GLM) problem:\n\nminimize || y ||_2 subject to d = A*x + B*y\nx\n\nwhere A is an N-by-M matrix, B is an N-by-P matrix, and d is a\ngiven N-vector. It is assumed that M <= N <= M+P, and\n\nrank(A) = M and rank( A B ) = N.\n\nUnder these assumptions, the constrained equation is always\nconsistent, and there is a unique solution x and a minimal 2-norm\nsolution y, which is obtained using a generalized QR factorization\nof the matrices (A, B) given by\n\nA = Q*(R), B = Q*T*Z.\n(0)\n\nIn particular, if matrix B is square nonsingular, then the problem\nGLM is equivalent to the following weighted linear least squares\nproblem\n\nminimize || inv(B)*(d-A*x) ||_2\nx\n\nwhere inv(B) denotes the inverse of B.\n\nfile zgglse.f zgglse.f plus dependencies\nprec complex16\nZGGLSE solves the linear equality-constrained least squares (LSE)\nproblem:\n\nminimize || c - A*x ||_2 subject to B*x = d\n\nwhere A is an M-by-N matrix, B is a P-by-N matrix, c is a given\nM-vector, and d is a given P-vector. It is assumed that\nP <= N <= M+P, and\n\nrank(B) = P and rank( ( A ) ) = N.\n( ( B ) )\n\nThese conditions ensure that the LSE problem has a unique solution,\nwhich is obtained using a generalized RQ factorization of the\nmatrices (B, A) given by\n\nB = (0 R)*Q, A = Z*T*Q.\n\nfile zggsvd.f zggsvd.f plus dependencies\nprec complex16\nZGGSVD computes the generalized singular value decomposition (GSVD)\nof an M-by-N complex matrix A and P-by-N complex matrix B:\n\nU'*A*Q = D1*( 0 R ), V'*B*Q = D2*( 0 R )\n\nwhere U, V and Q are unitary matrices, and Z' means the conjugate\ntranspose of Z. Let K+L = the effective numerical rank of the\nmatrix (A',B')', then R is a (K+L)-by-(K+L) nonsingular upper\ntriangular matrix, D1 and D2 are M-by-(K+L) and P-by-(K+L) \"diagonal\"\nmatrices and of the following structures, respectively:\n\nIf M-K-L >= 0,\n\nK L\nD1 = K ( I 0 )\nL ( 0 C )\nM-K-L ( 0 0 )\n\nK L\nD2 = L ( 0 S )\nP-L ( 0 0 )\n\nN-K-L K L\n( 0 R ) = K ( 0 R11 R12 )\nL ( 0 0 R22 )\nwhere\n\nC = diag( ALPHA(K+1), ... , ALPHA(K+L) ),\nS = diag( BETA(K+1), ... , BETA(K+L) ),\nC**2 + S**2 = I.\n\nR is stored in A(1:K+L,N-K-L+1:N) on exit.\n\nIf M-K-L < 0,\n\nK M-K K+L-M\nD1 = K ( I 0 0 )\nM-K ( 0 C 0 )\n\nK M-K K+L-M\nD2 = M-K ( 0 S 0 )\nK+L-M ( 0 0 I )\nP-L ( 0 0 0 )\n\nN-K-L K M-K K+L-M\n( 0 R ) = K ( 0 R11 R12 R13 )\nM-K ( 0 0 R22 R23 )\nK+L-M ( 0 0 0 R33 )\n\nwhere\n\nC = diag( ALPHA(K+1), ... , ALPHA(M) ),\nS = diag( BETA(K+1), ... , BETA(M) ),\nC**2 + S**2 = I.\n\n(R11 R12 R13 ) is stored in A(1:M, N-K-L+1:N), and R33 is stored\n( 0 R22 R23 )\nin B(M-K+1:L,N+M-K-L+1:N) on exit.\n\nThe routine computes C, S, R, and optionally the unitary\ntransformation matrices U, V and Q.\n\nIn particular, if B is an N-by-N nonsingular matrix, then the GSVD of\nA and B implicitly gives the SVD of A*inv(B):\nA*inv(B) = U*(D1*inv(D2))*V'.\nIf ( A',B')' has orthnormal columns, then the GSVD of A and B is also\nequal to the CS decomposition of A and B. Furthermore, the GSVD can\nbe used to derive the solution of the eigenvalue problem:\nA'*A x = lambda* B'*B x.\nIn some literature, the GSVD of A and B is presented in the form\nU'*A*X = ( 0 D1 ), V'*B*X = ( 0 D2 )\nwhere U and V are orthogonal and X is nonsingular, and D1 and D2 are\n``diagonal''. The former GSVD form can be converted to the latter\nform by taking the nonsingular matrix X as\n\nX = Q*( I 0 )\n( 0 inv(R) )\n\nfile zhbev.f zhbev.f plus dependencies\nprec complex16\nZHBEV computes all the eigenvalues and, optionally, eigenvectors of\na complex Hermitian band matrix A.\n\nfile zhbevd.f zhbevd.f plus dependencies\nprec complex16\nZHBEVD computes all the eigenvalues and, optionally, eigenvectors of\na complex Hermitian band matrix A. If eigenvectors are desired, it\nuses a divide and conquer algorithm.\n\nThe divide and conquer algorithm makes very mild assumptions about\nfloating point arithmetic. It will work on machines with a guard\ndigit in add/subtract, or on those binary machines without guard\ndigits which subtract like the Cray X-MP, Cray Y-MP, Cray C-90, or\nCray-2. It could conceivably fail on hexadecimal or decimal machines\nwithout guard digits, but we know of none.\n\nfile zhbgv.f zhbgv.f plus dependencies\nprec complex16\nZHBGV computes all the eigenvalues, and optionally, the eigenvectors\nof a complex generalized Hermitian-definite banded eigenproblem, of\nthe form A*x=(lambda)*B*x. Here A and B are assumed to be Hermitian\nand banded, and B is also positive definite.\n\nfile zhbgvd.f zhbgvd.f plus dependencies\nprec complex16\nZHBGVD computes all the eigenvalues, and optionally, the eigenvectors\nof a complex generalized Hermitian-definite banded eigenproblem, of\nthe form A*x=(lambda)*B*x. Here A and B are assumed to be Hermitian\nand banded, and B is also positive definite. If eigenvectors are\ndesired, it uses a divide and conquer algorithm.\n\nThe divide and conquer algorithm makes very mild assumptions about\nfloating point arithmetic. It will work on machines with a guard\ndigit in add/subtract, or on those binary machines without guard\ndigits which subtract like the Cray X-MP, Cray Y-MP, Cray C-90, or\nCray-2. It could conceivably fail on hexadecimal or decimal machines\nwithout guard digits, but we know of none.\n\nfile zheev.f zheev.f plus dependencies\nprec complex16\nZHEEV computes all eigenvalues and, optionally, eigenvectors of a\ncomplex Hermitian matrix A.\n\nfile zheevd.f zheevd.f plus dependencies\nprec complex16\nZHEEVD computes all eigenvalues and, optionally, eigenvectors of a\ncomplex Hermitian matrix A. If eigenvectors are desired, it uses a\ndivide and conquer algorithm.\n\nThe divide and conquer algorithm makes very mild assumptions about\nfloating point arithmetic. It will work on machines with a guard\ndigit in add/subtract, or on those binary machines without guard\ndigits which subtract like the Cray X-MP, Cray Y-MP, Cray C-90, or\nCray-2. It could conceivably fail on hexadecimal or decimal machines\nwithout guard digits, but we know of none.\n\nfile zheevr.f zheevr.f plus dependencies\nprec complex16\nZHEEVR computes selected eigenvalues and, optionally, eigenvectors\nof a complex Hermitian matrix A. Eigenvalues and eigenvectors can\nbe selected by specifying either a range of values or a range of\nindices for the desired eigenvalues.\n\nZHEEVR first reduces the matrix A to tridiagonal form T with a call\nto ZHETRD. Then, whenever possible, ZHEEVR calls ZSTEMR to compute\neigenspectrum using Relatively Robust Representations. ZSTEMR\ncomputes eigenvalues by the dqds algorithm, while orthogonal\neigenvectors are computed from various \"good\" L D L^T representations\n(also known as Relatively Robust Representations). Gram-Schmidt\northogonalization is avoided as far as possible. More specifically,\nthe various steps of the algorithm are as follows.\n\nFor each unreduced block (submatrix) of T,\n(a) Compute T - sigma I = L D L^T, so that L and D\ndefine all the wanted eigenvalues to high relative accuracy.\nThis means that small relative changes in the entries of D and L\ncause only small relative changes in the eigenvalues and\neigenvectors. The standard (unfactored) representation of the\ntridiagonal matrix T does not have this property in general.\n(b) Compute the eigenvalues to suitable accuracy.\nIf the eigenvectors are desired, the algorithm attains full\naccuracy of the computed eigenvalues only right before\nthe corresponding vectors have to be computed, see steps c) and d).\n(c) For each cluster of close eigenvalues, select a new\nshift close to the cluster, find a new factorization, and refine\nthe shifted eigenvalues to suitable accuracy.\n(d) For each eigenvalue with a large enough relative separation compute\nthe corresponding eigenvector by forming a rank revealing twisted\nfactorization. Go back to (c) for any clusters that remain.\n\nThe desired accuracy of the output can be specified by the input\nparameter ABSTOL.\n\nFor more details, see DSTEMR's documentation and:\n- Inderjit S. Dhillon and Beresford N. Parlett: \"Multiple representations\nto compute orthogonal eigenvectors of symmetric tridiagonal matrices,\"\nLinear Algebra and its Applications, 387(1), pp. 1-28, August 2004.\n- Inderjit Dhillon and Beresford Parlett: \"Orthogonal Eigenvectors and\nRelative Gaps,\" SIAM Journal on Matrix Analysis and Applications, Vol. 25,\n2004. Also LAPACK Working Note 154.\n- Inderjit Dhillon: \"A new O(n^2) algorithm for the symmetric\ntridiagonal eigenvalue/eigenvector problem\",\nComputer Science Division Technical Report No. UCB/CSD-97-971,\nUC Berkeley, May 1997.\n\nNote 1 : ZHEEVR calls ZSTEMR when the full spectrum is requested\non machines which conform to the ieee-754 floating point standard.\nZHEEVR calls DSTEBZ and ZSTEIN on non-ieee machines and\nwhen partial spectrum requests are made.\n\nNormal execution of ZSTEMR may create NaNs and infinities and\nhence may abort due to a floating point exception in environments\nwhich do not handle NaNs and infinities in the ieee standard default\nmanner.\n\nfile zhegv.f zhegv.f plus dependencies\nprec complex16\nZHEGV computes all the eigenvalues, and optionally, the eigenvectors\nof a complex generalized Hermitian-definite eigenproblem, of the form\nA*x=(lambda)*B*x, A*Bx=(lambda)*x, or B*A*x=(lambda)*x.\nHere A and B are assumed to be Hermitian and B is also\npositive definite.\n\nfile zhegvd.f zhegvd.f plus dependencies\nprec complex16\nZHEGVD computes all the eigenvalues, and optionally, the eigenvectors\nof a complex generalized Hermitian-definite eigenproblem, of the form\nA*x=(lambda)*B*x, A*Bx=(lambda)*x, or B*A*x=(lambda)*x. Here A and\nB are assumed to be Hermitian and B is also positive definite.\nIf eigenvectors are desired, it uses a divide and conquer algorithm.\n\nThe divide and conquer algorithm makes very mild assumptions about\nfloating point arithmetic. It will work on machines with a guard\ndigit in add/subtract, or on those binary machines without guard\ndigits which subtract like the Cray X-MP, Cray Y-MP, Cray C-90, or\nCray-2. It could conceivably fail on hexadecimal or decimal machines\nwithout guard digits, but we know of none.\n\nfile zhesv.f zhesv.f plus dependencies\nprec complex16\nZHESV computes the solution to a complex system of linear equations\nA * X = B,\nwhere A is an N-by-N Hermitian matrix and X and B are N-by-NRHS\nmatrices.\n\nThe diagonal pivoting method is used to factor A as\nA = U * D * U**H, if UPLO = 'U', or\nA = L * D * L**H, if UPLO = 'L',\nwhere U (or L) is a product of permutation and unit upper (lower)\ntriangular matrices, and D is Hermitian and block diagonal with\n1-by-1 and 2-by-2 diagonal blocks. The factored form of A is then\nused to solve the system of equations A * X = B.\n\nfile zhpev.f zhpev.f plus dependencies\nprec complex16\nZHPEV computes all the eigenvalues and, optionally, eigenvectors of a\ncomplex Hermitian matrix in packed storage.\n\nfile zhpevd.f zhpevd.f plus dependencies\nprec complex16\nZHPEVD computes all the eigenvalues and, optionally, eigenvectors of\na complex Hermitian matrix A in packed storage. If eigenvectors are\ndesired, it uses a divide and conquer algorithm.\n\nThe divide and conquer algorithm makes very mild assumptions about\nfloating point arithmetic. It will work on machines with a guard\ndigit in add/subtract, or on those binary machines without guard\ndigits which subtract like the Cray X-MP, Cray Y-MP, Cray C-90, or\nCray-2. It could conceivably fail on hexadecimal or decimal machines\nwithout guard digits, but we know of none.\n\nfile zhpgv.f zhpgv.f plus dependencies\nprec complex16\nZHPGV computes all the eigenvalues and, optionally, the eigenvectors\nof a complex generalized Hermitian-definite eigenproblem, of the form\nA*x=(lambda)*B*x, A*Bx=(lambda)*x, or B*A*x=(lambda)*x.\nHere A and B are assumed to be Hermitian, stored in packed format,\nand B is also positive definite.\n\nfile zhpgvd.f zhpgvd.f plus dependencies\nprec complex16\nZHPGVD computes all the eigenvalues and, optionally, the eigenvectors\nof a complex generalized Hermitian-definite eigenproblem, of the form\nA*x=(lambda)*B*x, A*Bx=(lambda)*x, or B*A*x=(lambda)*x. Here A and\nB are assumed to be Hermitian, stored in packed format, and B is also\npositive definite.\nIf eigenvectors are desired, it uses a divide and conquer algorithm.\n\nThe divide and conquer algorithm makes very mild assumptions about\nfloating point arithmetic. It will work on machines with a guard\ndigit in add/subtract, or on those binary machines without guard\ndigits which subtract like the Cray X-MP, Cray Y-MP, Cray C-90, or\nCray-2. It could conceivably fail on hexadecimal or decimal machines\nwithout guard digits, but we know of none.\n\nfile zhpsv.f zhpsv.f plus dependencies\nprec complex16\nZHPSV computes the solution to a complex system of linear equations\nA * X = B,\nwhere A is an N-by-N Hermitian matrix stored in packed format and X\nand B are N-by-NRHS matrices.\n\nThe diagonal pivoting method is used to factor A as\nA = U * D * U**H, if UPLO = 'U', or\nA = L * D * L**H, if UPLO = 'L',\nwhere U (or L) is a product of permutation and unit upper (lower)\ntriangular matrices, D is Hermitian and block diagonal with 1-by-1\nand 2-by-2 diagonal blocks. The factored form of A is then used to\nsolve the system of equations A * X = B.\n\nfile zpbsv.f zpbsv.f plus dependencies\nprec complex16\nZPBSV computes the solution to a complex system of linear equations\nA * X = B,\nwhere A is an N-by-N Hermitian positive definite band matrix and X\nand B are N-by-NRHS matrices.\n\nThe Cholesky decomposition is used to factor A as\nA = U**H * U, if UPLO = 'U', or\nA = L * L**H, if UPLO = 'L',\nwhere U is an upper triangular band matrix, and L is a lower\ntriangular band matrix, with the same number of superdiagonals or\nsubdiagonals as A. The factored form of A is then used to solve the\nsystem of equations A * X = B.\n\nfile zposv.f zposv.f plus dependencies\nprec complex16\nZPOSV computes the solution to a complex system of linear equations\nA * X = B,\nwhere A is an N-by-N Hermitian positive definite matrix and X and B\nare N-by-NRHS matrices.\n\nThe Cholesky decomposition is used to factor A as\nA = U**H* U, if UPLO = 'U', or\nA = L * L**H, if UPLO = 'L',\nwhere U is an upper triangular matrix and L is a lower triangular\nmatrix. The factored form of A is then used to solve the system of\nequations A * X = B.\n\nfile zppsv.f zppsv.f plus dependencies\nprec complex16\nZPPSV computes the solution to a complex system of linear equations\nA * X = B,\nwhere A is an N-by-N Hermitian positive definite matrix stored in\npacked format and X and B are N-by-NRHS matrices.\n\nThe Cholesky decomposition is used to factor A as\nA = U**H* U, if UPLO = 'U', or\nA = L * L**H, if UPLO = 'L',\nwhere U is an upper triangular matrix and L is a lower triangular\nmatrix. The factored form of A is then used to solve the system of\nequations A * X = B.\n\nfile zspsv.f zspsv.f plus dependencies\nprec complex16\nZSPSV computes the solution to a complex system of linear equations\nA * X = B,\nwhere A is an N-by-N symmetric matrix stored in packed format and X\nand B are N-by-NRHS matrices.\n\nThe diagonal pivoting method is used to factor A as\nA = U * D * U**T, if UPLO = 'U', or\nA = L * D * L**T, if UPLO = 'L',\nwhere U (or L) is a product of permutation and unit upper (lower)\ntriangular matrices, D is symmetric and block diagonal with 1-by-1\nand 2-by-2 diagonal blocks. The factored form of A is then used to\nsolve the system of equations A * X = B.\n\nfile zstemr.f zstemr.f plus dependencies\nprec complex16\nZSTEMR computes selected eigenvalues and, optionally, eigenvectors\nof a real symmetric tridiagonal matrix T. Any such unreduced matrix has\na well defined set of pairwise different real eigenvalues, the corresponding\nreal eigenvectors are pairwise orthogonal.\n\nThe spectrum may be computed either completely or partially by specifying\neither an interval (VL,VU] or a range of indices IL:IU for the desired\neigenvalues.\n\nDepending on the number of desired eigenvalues, these are computed either\nby bisection or the dqds algorithm. Numerically orthogonal eigenvectors are\ncomputed by the use of various suitable L D L^T factorizations near clusters\nof close eigenvalues (referred to as RRRs, Relatively Robust\nRepresentations). An informal sketch of the algorithm follows.\n\nFor each unreduced block (submatrix) of T,\n(a) Compute T - sigma I = L D L^T, so that L and D\ndefine all the wanted eigenvalues to high relative accuracy.\nThis means that small relative changes in the entries of D and L\ncause only small relative changes in the eigenvalues and\neigenvectors. The standard (unfactored) representation of the\ntridiagonal matrix T does not have this property in general.\n(b) Compute the eigenvalues to suitable accuracy.\nIf the eigenvectors are desired, the algorithm attains full\naccuracy of the computed eigenvalues only right before\nthe corresponding vectors have to be computed, see steps c) and d).\n(c) For each cluster of close eigenvalues, select a new\nshift close to the cluster, find a new factorization, and refine\nthe shifted eigenvalues to suitable accuracy.\n(d) For each eigenvalue with a large enough relative separation compute\nthe corresponding eigenvector by forming a rank revealing twisted\nfactorization. Go back to (c) for any clusters that remain.\n\nFor more details, see:\n- Inderjit S. Dhillon and Beresford N. Parlett: \"Multiple representations\nto compute orthogonal eigenvectors of symmetric tridiagonal matrices,\"\nLinear Algebra and its Applications, 387(1), pp. 1-28, August 2004.\n- Inderjit Dhillon and Beresford Parlett: \"Orthogonal Eigenvectors and\nRelative Gaps,\" SIAM Journal on Matrix Analysis and Applications, Vol. 25,\n2004. Also LAPACK Working Note 154.\n- Inderjit Dhillon: \"A new O(n^2) algorithm for the symmetric\ntridiagonal eigenvalue/eigenvector problem\",\nComputer Science Division Technical Report No. UCB/CSD-97-971,\nUC Berkeley, May 1997.\n\nFurther Details\n1.ZSTEMR works only on machines which follow IEEE-754\nfloating-point standard in their handling of infinities and NaNs.\nThis permits the use of efficient inner loops avoiding a check for\nzero divisors.\n\n2. LAPACK routines can be used to reduce a complex Hermitean matrix to\nreal symmetric tridiagonal form.\n\n(Any complex Hermitean tridiagonal matrix has real values on its diagonal\nand potentially complex numbers on its off-diagonals. By applying a\nsimilarity transform with an appropriate diagonal matrix\ndiag(1,e^{i \\phy_1}, ... , e^{i \\phy_{n-1}}), the complex Hermitean\nmatrix can be transformed into a real symmetric matrix and complex\narithmetic can be entirely avoided.)\n\nWhile the eigenvectors of the real symmetric tridiagonal matrix are real,\nthe eigenvectors of original complex Hermitean matrix have complex entries\nin general.\nSince LAPACK drivers overwrite the matrix data with the eigenvectors,\nZSTEMR accepts complex workspace to facilitate interoperability\nwith ZUNMTR or ZUPMTR.\n\nfile zsysv.f zsysv.f plus dependencies\nprec complex16\nZSYSV computes the solution to a complex system of linear equations\nA * X = B,\nwhere A is an N-by-N symmetric matrix and X and B are N-by-NRHS\nmatrices.\n\nThe diagonal pivoting method is used to factor A as\nA = U * D * U**T, if UPLO = 'U', or\nA = L * D * L**T, if UPLO = 'L',\nwhere U (or L) is a product of permutation and unit upper (lower)\ntriangular matrices, and D is symmetric and block diagonal with\n1-by-1 and 2-by-2 diagonal blocks. The factored form of A is then\nused to solve the system of equations A * X = B.\n\n# ---------------------------------\n# Available EXPERT DRIVER routines:\n# ---------------------------------\nfile zgbsvx.f zgbsvx.f plus dependencies\nprec complex16\nZGBSVX uses the LU factorization to compute the solution to a complex\nsystem of linear equations A * X = B, A**T * X = B, or A**H * X = B,\nwhere A is a band matrix of order N with KL subdiagonals and KU\nsuperdiagonals, and X and B are N-by-NRHS matrices.\n\nError bounds on the solution and a condition estimate are also\nprovided.\n\nDescription\n===========\n\nThe following steps are performed by this subroutine:\n\n1. If FACT = 'E', real scaling factors are computed to equilibrate\nthe system:\nTRANS = 'N': diag(R)*A*diag(C) *inv(diag(C))*X = diag(R)*B\nTRANS = 'T': (diag(R)*A*diag(C))**T *inv(diag(R))*X = diag(C)*B\nTRANS = 'C': (diag(R)*A*diag(C))**H *inv(diag(R))*X = diag(C)*B\nWhether or not the system will be equilibrated depends on the\nscaling of the matrix A, but if equilibration is used, A is\noverwritten by diag(R)*A*diag(C) and B by diag(R)*B (if TRANS='N')\nor diag(C)*B (if TRANS = 'T' or 'C').\n\n2. If FACT = 'N' or 'E', the LU decomposition is used to factor the\nmatrix A (after equilibration if FACT = 'E') as\nA = L * U,\nwhere L is a product of permutation and unit lower triangular\nmatrices with KL subdiagonals, and U is upper triangular with\nKL+KU superdiagonals.\n\n3. If some U(i,i)=0, so that U is exactly singular, then the routine\nreturns with INFO = i. Otherwise, the factored form of A is used\nto estimate the condition number of the matrix A. If the\nreciprocal of the condition number is less than machine precision,\nINFO = N+1 is returned as a warning, but the routine still goes on\nto solve for X and compute error bounds as described below.\n\n4. The system of equations is solved for X using the factored form\nof A.\n\n5. Iterative refinement is applied to improve the computed solution\nmatrix and calculate error bounds and backward error estimates\nfor it.\n\n6. If equilibration was used, the matrix X is premultiplied by\ndiag(C) (if TRANS = 'N') or diag(R) (if TRANS = 'T' or 'C') so\nthat it solves the original system before equilibration.\n\nfile zgbsvxx.f zgbsvxx.f plus dependencies\nprec complex16\nZGBSVXX uses the LU factorization to compute the solution to a\ncomplex*16 system of linear equations A * X = B, where A is an\nN-by-N matrix and X and B are N-by-NRHS matrices.\n\nIf requested, both normwise and maximum componentwise error bounds\nare returned. ZGBSVXX will return a solution with a tiny\nguaranteed error (O(eps) where eps is the working machine\nprecision) unless the matrix is very ill-conditioned, in which\ncase a warning is returned. Relevant condition numbers also are\ncalculated and returned.\n\nZGBSVXX accepts user-provided factorizations and equilibration\nfactors; see the definitions of the FACT and EQUED options.\nSolving with refinement and using a factorization from a previous\nZGBSVXX call will also produce a solution with either O(eps)\nerrors or warnings, but we cannot make that claim for general\nuser-provided factorizations and equilibration factors if they\ndiffer from what ZGBSVXX would itself produce.\n\nDescription\n===========\n\nThe following steps are performed:\n\n1. If FACT = 'E', double precision scaling factors are computed to equilibrate\nthe system:\n\nTRANS = 'N': diag(R)*A*diag(C) *inv(diag(C))*X = diag(R)*B\nTRANS = 'T': (diag(R)*A*diag(C))**T *inv(diag(R))*X = diag(C)*B\nTRANS = 'C': (diag(R)*A*diag(C))**H *inv(diag(R))*X = diag(C)*B\n\nWhether or not the system will be equilibrated depends on the\nscaling of the matrix A, but if equilibration is used, A is\noverwritten by diag(R)*A*diag(C) and B by diag(R)*B (if TRANS='N')\nor diag(C)*B (if TRANS = 'T' or 'C').\n\n2. If FACT = 'N' or 'E', the LU decomposition is used to factor\nthe matrix A (after equilibration if FACT = 'E') as\n\nA = P * L * U,\n\nwhere P is a permutation matrix, L is a unit lower triangular\nmatrix, and U is upper triangular.\n\n3. If some U(i,i)=0, so that U is exactly singular, then the\nroutine returns with INFO = i. Otherwise, the factored form of A\nis used to estimate the condition number of the matrix A (see\nargument RCOND). If the reciprocal of the condition number is less\nthan machine precision, the routine still goes on to solve for X\nand compute error bounds as described below.\n\n4. The system of equations is solved for X using the factored form\nof A.\n\n5. By default (unless PARAMS(LA_LINRX_ITREF_I) is set to zero),\nthe routine will use iterative refinement to try to get a small\nerror and error bounds. Refinement calculates the residual to at\nleast twice the working precision.\n\n6. If equilibration was used, the matrix X is premultiplied by\ndiag(C) (if TRANS = 'N') or diag(R) (if TRANS = 'T' or 'C') so\nthat it solves the original system before equilibration.\n\nfile zgeesx.f zgeesx.f plus dependencies\nprec complex16\nZGEESX computes for an N-by-N complex nonsymmetric matrix A, the\neigenvalues, the Schur form T, and, optionally, the matrix of Schur\nvectors Z. This gives the Schur factorization A = Z*T*(Z**H).\n\nOptionally, it also orders the eigenvalues on the diagonal of the\nSchur form so that selected eigenvalues are at the top left;\ncomputes a reciprocal condition number for the average of the\nselected eigenvalues (RCONDE); and computes a reciprocal condition\nnumber for the right invariant subspace corresponding to the\nselected eigenvalues (RCONDV). The leading columns of Z form an\northonormal basis for this invariant subspace.\n\nFor further explanation of the reciprocal condition numbers RCONDE\nand RCONDV, see Section 4.10 of the LAPACK Users' Guide (where\nthese quantities are called s and sep respectively).\n\nA complex matrix is in Schur form if it is upper triangular.\n\nfile zgeevx.f zgeevx.f plus dependencies\nprec complex16\nZGEEVX computes for an N-by-N complex nonsymmetric matrix A, the\neigenvalues and, optionally, the left and/or right eigenvectors.\n\nOptionally also, it computes a balancing transformation to improve\nthe conditioning of the eigenvalues and eigenvectors (ILO, IHI,\nSCALE, and ABNRM), reciprocal condition numbers for the eigenvalues\n(RCONDE), and reciprocal condition numbers for the right\neigenvectors (RCONDV).\n\nThe right eigenvector v(j) of A satisfies\nA * v(j) = lambda(j) * v(j)\nwhere lambda(j) is its eigenvalue.\nThe left eigenvector u(j) of A satisfies\nu(j)**H * A = lambda(j) * u(j)**H\nwhere u(j)**H denotes the conjugate transpose of u(j).\n\nThe computed eigenvectors are normalized to have Euclidean norm\nequal to 1 and largest component real.\n\nBalancing a matrix means permuting the rows and columns to make it\nmore nearly upper triangular, and applying a diagonal similarity\ntransformation D * A * D**(-1), where D is a diagonal matrix, to\nmake its rows and columns closer in norm and the condition numbers\nof its eigenvalues and eigenvectors smaller. The computed\nreciprocal condition numbers correspond to the balanced matrix.\nPermuting rows and columns will not change the condition numbers\n(in exact arithmetic) but diagonal scaling will. For further\nexplanation of balancing, see section 4.10.2 of the LAPACK\nUsers' Guide.\n\nfile zgelsx.f zgelsx.f plus dependencies\nprec complex16\nThis routine is deprecated and has been replaced by routine ZGELSY.\n\nZGELSX computes the minimum-norm solution to a complex linear least\nsquares problem:\nminimize || A * X - B ||\nusing a complete orthogonal factorization of A. A is an M-by-N\nmatrix which may be rank-deficient.\n\nSeveral right hand side vectors b and solution vectors x can be\nhandled in a single call; they are stored as the columns of the\nM-by-NRHS right hand side matrix B and the N-by-NRHS solution\nmatrix X.\n\nThe routine first computes a QR factorization with column pivoting:\nA * P = Q * [ R11 R12 ]\n[ 0 R22 ]\nwith R11 defined as the largest leading submatrix whose estimated\ncondition number is less than 1/RCOND. The order of R11, RANK,\nis the effective rank of A.\n\nThen, R22 is considered to be negligible, and R12 is annihilated\nby unitary transformations from the right, arriving at the\ncomplete orthogonal factorization:\nA * P = Q * [ T11 0 ] * Z\n[ 0 0 ]\nThe minimum-norm solution is then\nX = P * Z' [ inv(T11)*Q1'*B ]\n[ 0 ]\nwhere Q1 consists of the first RANK columns of Q.\n\nfile zgesvx.f zgesvx.f plus dependencies\nprec complex16\nZGESVX uses the LU factorization to compute the solution to a complex\nsystem of linear equations\nA * X = B,\nwhere A is an N-by-N matrix and X and B are N-by-NRHS matrices.\n\nError bounds on the solution and a condition estimate are also\nprovided.\n\nDescription\n===========\n\nThe following steps are performed:\n\n1. If FACT = 'E', real scaling factors are computed to equilibrate\nthe system:\nTRANS = 'N': diag(R)*A*diag(C) *inv(diag(C))*X = diag(R)*B\nTRANS = 'T': (diag(R)*A*diag(C))**T *inv(diag(R))*X = diag(C)*B\nTRANS = 'C': (diag(R)*A*diag(C))**H *inv(diag(R))*X = diag(C)*B\nWhether or not the system will be equilibrated depends on the\nscaling of the matrix A, but if equilibration is used, A is\noverwritten by diag(R)*A*diag(C) and B by diag(R)*B (if TRANS='N')\nor diag(C)*B (if TRANS = 'T' or 'C').\n\n2. If FACT = 'N' or 'E', the LU decomposition is used to factor the\nmatrix A (after equilibration if FACT = 'E') as\nA = P * L * U,\nwhere P is a permutation matrix, L is a unit lower triangular\nmatrix, and U is upper triangular.\n\n3. If some U(i,i)=0, so that U is exactly singular, then the routine\nreturns with INFO = i. Otherwise, the factored form of A is used\nto estimate the condition number of the matrix A. If the\nreciprocal of the condition number is less than machine precision,\nINFO = N+1 is returned as a warning, but the routine still goes on\nto solve for X and compute error bounds as described below.\n\n4. The system of equations is solved for X using the factored form\nof A.\n\n5. Iterative refinement is applied to improve the computed solution\nmatrix and calculate error bounds and backward error estimates\nfor it.\n\n6. If equilibration was used, the matrix X is premultiplied by\ndiag(C) (if TRANS = 'N') or diag(R) (if TRANS = 'T' or 'C') so\nthat it solves the original system before equilibration.\n\nfile zgesvxx.f zgesvxx.f plus dependencies\nprec complex16\nZGESVXX uses the LU factorization to compute the solution to a\ncomplex*16 system of linear equations A * X = B, where A is an\nN-by-N matrix and X and B are N-by-NRHS matrices.\n\nIf requested, both normwise and maximum componentwise error bounds\nare returned. ZGESVXX will return a solution with a tiny\nguaranteed error (O(eps) where eps is the working machine\nprecision) unless the matrix is very ill-conditioned, in which\ncase a warning is returned. Relevant condition numbers also are\ncalculated and returned.\n\nZGESVXX accepts user-provided factorizations and equilibration\nfactors; see the definitions of the FACT and EQUED options.\nSolving with refinement and using a factorization from a previous\nZGESVXX call will also produce a solution with either O(eps)\nerrors or warnings, but we cannot make that claim for general\nuser-provided factorizations and equilibration factors if they\ndiffer from what ZGESVXX would itself produce.\n\nDescription\n===========\n\nThe following steps are performed:\n\n1. If FACT = 'E', double precision scaling factors are computed to equilibrate\nthe system:\n\nTRANS = 'N': diag(R)*A*diag(C) *inv(diag(C))*X = diag(R)*B\nTRANS = 'T': (diag(R)*A*diag(C))**T *inv(diag(R))*X = diag(C)*B\nTRANS = 'C': (diag(R)*A*diag(C))**H *inv(diag(R))*X = diag(C)*B\n\nWhether or not the system will be equilibrated depends on the\nscaling of the matrix A, but if equilibration is used, A is\noverwritten by diag(R)*A*diag(C) and B by diag(R)*B (if TRANS='N')\nor diag(C)*B (if TRANS = 'T' or 'C').\n\n2. If FACT = 'N' or 'E', the LU decomposition is used to factor\nthe matrix A (after equilibration if FACT = 'E') as\n\nA = P * L * U,\n\nwhere P is a permutation matrix, L is a unit lower triangular\nmatrix, and U is upper triangular.\n\n3. If some U(i,i)=0, so that U is exactly singular, then the\nroutine returns with INFO = i. Otherwise, the factored form of A\nis used to estimate the condition number of the matrix A (see\nargument RCOND). If the reciprocal of the condition number is less\nthan machine precision, the routine still goes on to solve for X\nand compute error bounds as described below.\n\n4. The system of equations is solved for X using the factored form\nof A.\n\n5. By default (unless PARAMS(LA_LINRX_ITREF_I) is set to zero),\nthe routine will use iterative refinement to try to get a small\nerror and error bounds. Refinement calculates the residual to at\nleast twice the working precision.\n\n6. If equilibration was used, the matrix X is premultiplied by\ndiag(C) (if TRANS = 'N') or diag(R) (if TRANS = 'T' or 'C') so\nthat it solves the original system before equilibration.\n\nfile zggesx.f zggesx.f plus dependencies\nprec complex16\nZGGESX computes for a pair of N-by-N complex nonsymmetric matrices\n(A,B), the generalized eigenvalues, the complex Schur form (S,T),\nand, optionally, the left and/or right matrices of Schur vectors (VSL\nand VSR). This gives the generalized Schur factorization\n\n(A,B) = ( (VSL) S (VSR)**H, (VSL) T (VSR)**H )\n\nwhere (VSR)**H is the conjugate-transpose of VSR.\n\nOptionally, it also orders the eigenvalues so that a selected cluster\nof eigenvalues appears in the leading diagonal blocks of the upper\ntriangular matrix S and the upper triangular matrix T; computes\na reciprocal condition number for the average of the selected\neigenvalues (RCONDE); and computes a reciprocal condition number for\nthe right and left deflating subspaces corresponding to the selected\neigenvalues (RCONDV). The leading columns of VSL and VSR then form\nan orthonormal basis for the corresponding left and right eigenspaces\n(deflating subspaces).\n\nA generalized eigenvalue for a pair of matrices (A,B) is a scalar w\nor a ratio alpha/beta = w, such that A - w*B is singular. It is\nusually represented as the pair (alpha,beta), as there is a\nreasonable interpretation for beta=0 or for both being zero.\n\nA pair of matrices (S,T) is in generalized complex Schur form if T is\nupper triangular with non-negative diagonal and S is upper\ntriangular.\n\nfile zggevx.f zggevx.f plus dependencies\nprec complex16\nZGGEVX computes for a pair of N-by-N complex nonsymmetric matrices\n(A,B) the generalized eigenvalues, and optionally, the left and/or\nright generalized eigenvectors.\n\nOptionally, it also computes a balancing transformation to improve\nthe conditioning of the eigenvalues and eigenvectors (ILO, IHI,\nLSCALE, RSCALE, ABNRM, and BBNRM), reciprocal condition numbers for\nthe eigenvalues (RCONDE), and reciprocal condition numbers for the\nright eigenvectors (RCONDV).\n\nA generalized eigenvalue for a pair of matrices (A,B) is a scalar\nlambda or a ratio alpha/beta = lambda, such that A - lambda*B is\nsingular. It is usually represented as the pair (alpha,beta), as\nthere is a reasonable interpretation for beta=0, and even for both\nbeing zero.\n\nThe right eigenvector v(j) corresponding to the eigenvalue lambda(j)\nof (A,B) satisfies\nA * v(j) = lambda(j) * B * v(j) .\nThe left eigenvector u(j) corresponding to the eigenvalue lambda(j)\nof (A,B) satisfies\nu(j)**H * A = lambda(j) * u(j)**H * B.\nwhere u(j)**H is the conjugate-transpose of u(j).\n\nfile zhbevx.f zhbevx.f plus dependencies\nprec complex16\nZHBEVX computes selected eigenvalues and, optionally, eigenvectors\nof a complex Hermitian band matrix A. Eigenvalues and eigenvectors\ncan be selected by specifying either a range of values or a range of\nindices for the desired eigenvalues.\n\nfile zhbgvx.f zhbgvx.f plus dependencies\nprec complex16\nZHBGVX computes all the eigenvalues, and optionally, the eigenvectors\nof a complex generalized Hermitian-definite banded eigenproblem, of\nthe form A*x=(lambda)*B*x. Here A and B are assumed to be Hermitian\nand banded, and B is also positive definite. Eigenvalues and\neigenvectors can be selected by specifying either all eigenvalues,\na range of values or a range of indices for the desired eigenvalues.\n\nfile zheevx.f zheevx.f plus dependencies\nprec complex16\nZHEEVX computes selected eigenvalues and, optionally, eigenvectors\nof a complex Hermitian matrix A. Eigenvalues and eigenvectors can\nbe selected by specifying either a range of values or a range of\nindices for the desired eigenvalues.\n\nfile zhegvx.f zhegvx.f plus dependencies\nprec complex16\nZHEGVX computes selected eigenvalues, and optionally, eigenvectors\nof a complex generalized Hermitian-definite eigenproblem, of the form\nA*x=(lambda)*B*x, A*Bx=(lambda)*x, or B*A*x=(lambda)*x. Here A and\nB are assumed to be Hermitian and B is also positive definite.\nEigenvalues and eigenvectors can be selected by specifying either a\nrange of values or a range of indices for the desired eigenvalues.\n\nfile zhesvx.f zhesvx.f plus dependencies\nprec complex16\nZHESVX uses the diagonal pivoting factorization to compute the\nsolution to a complex system of linear equations A * X = B,\nwhere A is an N-by-N Hermitian matrix and X and B are N-by-NRHS\nmatrices.\n\nError bounds on the solution and a condition estimate are also\nprovided.\n\nDescription\n===========\n\nThe following steps are performed:\n\n1. If FACT = 'N', the diagonal pivoting method is used to factor A.\nThe form of the factorization is\nA = U * D * U**H, if UPLO = 'U', or\nA = L * D * L**H, if UPLO = 'L',\nwhere U (or L) is a product of permutation and unit upper (lower)\ntriangular matrices, and D is Hermitian and block diagonal with\n1-by-1 and 2-by-2 diagonal blocks.\n\n2. If some D(i,i)=0, so that D is exactly singular, then the routine\nreturns with INFO = i. Otherwise, the factored form of A is used\nto estimate the condition number of the matrix A. If the\nreciprocal of the condition number is less than machine precision,\nINFO = N+1 is returned as a warning, but the routine still goes on\nto solve for X and compute error bounds as described below.\n\n3. The system of equations is solved for X using the factored form\nof A.\n\n4. Iterative refinement is applied to improve the computed solution\nmatrix and calculate error bounds and backward error estimates\nfor it.\n\nfile zhesvxx.f zhesvxx.f plus dependencies\nprec complex16\nZHESVXX uses the diagonal pivoting factorization to compute the\nsolution to a complex*16 system of linear equations A * X = B, where\nA is an N-by-N symmetric matrix and X and B are N-by-NRHS\nmatrices.\n\nIf requested, both normwise and maximum componentwise error bounds\nare returned. ZHESVXX will return a solution with a tiny\nguaranteed error (O(eps) where eps is the working machine\nprecision) unless the matrix is very ill-conditioned, in which\ncase a warning is returned. Relevant condition numbers also are\ncalculated and returned.\n\nZHESVXX accepts user-provided factorizations and equilibration\nfactors; see the definitions of the FACT and EQUED options.\nSolving with refinement and using a factorization from a previous\nZHESVXX call will also produce a solution with either O(eps)\nerrors or warnings, but we cannot make that claim for general\nuser-provided factorizations and equilibration factors if they\ndiffer from what ZHESVXX would itself produce.\n\nDescription\n===========\n\nThe following steps are performed:\n\n1. If FACT = 'E', double precision scaling factors are computed to equilibrate\nthe system:\n\ndiag(S)*A*diag(S) *inv(diag(S))*X = diag(S)*B\n\nWhether or not the system will be equilibrated depends on the\nscaling of the matrix A, but if equilibration is used, A is\noverwritten by diag(S)*A*diag(S) and B by diag(S)*B.\n\n2. If FACT = 'N' or 'E', the LU decomposition is used to factor\nthe matrix A (after equilibration if FACT = 'E') as\n\nA = U * D * U**T, if UPLO = 'U', or\nA = L * D * L**T, if UPLO = 'L',\n\nwhere U (or L) is a product of permutation and unit upper (lower)\ntriangular matrices, and D is symmetric and block diagonal with\n1-by-1 and 2-by-2 diagonal blocks.\n\n3. If some D(i,i)=0, so that D is exactly singular, then the\nroutine returns with INFO = i. Otherwise, the factored form of A\nis used to estimate the condition number of the matrix A (see\nargument RCOND). If the reciprocal of the condition number is\nless than machine precision, the routine still goes on to solve\nfor X and compute error bounds as described below.\n\n4. The system of equations is solved for X using the factored form\nof A.\n\n5. By default (unless PARAMS(LA_LINRX_ITREF_I) is set to zero),\nthe routine will use iterative refinement to try to get a small\nerror and error bounds. Refinement calculates the residual to at\nleast twice the working precision.\n\n6. If equilibration was used, the matrix X is premultiplied by\ndiag(R) so that it solves the original system before\nequilibration.\n\nfile zhpevx.f zhpevx.f plus dependencies\nprec complex16\nZHPEVX computes selected eigenvalues and, optionally, eigenvectors\nof a complex Hermitian matrix A in packed storage.\nEigenvalues/vectors can be selected by specifying either a range of\nvalues or a range of indices for the desired eigenvalues.\n\nfile zhpgvx.f zhpgvx.f plus dependencies\nprec complex16\nZHPGVX computes selected eigenvalues and, optionally, eigenvectors\nof a complex generalized Hermitian-definite eigenproblem, of the form\nA*x=(lambda)*B*x, A*Bx=(lambda)*x, or B*A*x=(lambda)*x. Here A and\nB are assumed to be Hermitian, stored in packed format, and B is also\npositive definite. Eigenvalues and eigenvectors can be selected by\nspecifying either a range of values or a range of indices for the\ndesired eigenvalues.\n\nfile zhpsvx.f zhpsvx.f plus dependencies\nprec complex16\nZHPSVX uses the diagonal pivoting factorization A = U*D*U**H or\nA = L*D*L**H to compute the solution to a complex system of linear\nequations A * X = B, where A is an N-by-N Hermitian matrix stored\nin packed format and X and B are N-by-NRHS matrices.\n\nError bounds on the solution and a condition estimate are also\nprovided.\n\nDescription\n===========\n\nThe following steps are performed:\n\n1. If FACT = 'N', the diagonal pivoting method is used to factor A as\nA = U * D * U**H, if UPLO = 'U', or\nA = L * D * L**H, if UPLO = 'L',\nwhere U (or L) is a product of permutation and unit upper (lower)\ntriangular matrices and D is Hermitian and block diagonal with\n1-by-1 and 2-by-2 diagonal blocks.\n\n2. If some D(i,i)=0, so that D is exactly singular, then the routine\nreturns with INFO = i. Otherwise, the factored form of A is used\nto estimate the condition number of the matrix A. If the\nreciprocal of the condition number is less than machine precision,\nINFO = N+1 is returned as a warning, but the routine still goes on\nto solve for X and compute error bounds as described below.\n\n3. The system of equations is solved for X using the factored form\nof A.\n\n4. Iterative refinement is applied to improve the computed solution\nmatrix and calculate error bounds and backward error estimates\nfor it.\n\nfile zpbsvx.f zpbsvx.f plus dependencies\nprec complex16\nZPBSVX uses the Cholesky factorization A = U**H*U or A = L*L**H to\ncompute the solution to a complex system of linear equations\nA * X = B,\nwhere A is an N-by-N Hermitian positive definite band matrix and X\nand B are N-by-NRHS matrices.\n\nError bounds on the solution and a condition estimate are also\nprovided.\n\nDescription\n===========\n\nThe following steps are performed:\n\n1. If FACT = 'E', real scaling factors are computed to equilibrate\nthe system:\ndiag(S) * A * diag(S) * inv(diag(S)) * X = diag(S) * B\nWhether or not the system will be equilibrated depends on the\nscaling of the matrix A, but if equilibration is used, A is\noverwritten by diag(S)*A*diag(S) and B by diag(S)*B.\n\n2. If FACT = 'N' or 'E', the Cholesky decomposition is used to\nfactor the matrix A (after equilibration if FACT = 'E') as\nA = U**H * U, if UPLO = 'U', or\nA = L * L**H, if UPLO = 'L',\nwhere U is an upper triangular band matrix, and L is a lower\ntriangular band matrix.\n\n3. If the leading i-by-i principal minor is not positive definite,\nthen the routine returns with INFO = i. Otherwise, the factored\nform of A is used to estimate the condition number of the matrix\nA. If the reciprocal of the condition number is less than machine\nprecision, INFO = N+1 is returned as a warning, but the routine\nstill goes on to solve for X and compute error bounds as\ndescribed below.\n\n4. The system of equations is solved for X using the factored form\nof A.\n\n5. Iterative refinement is applied to improve the computed solution\nmatrix and calculate error bounds and backward error estimates\nfor it.\n\n6. If equilibration was used, the matrix X is premultiplied by\ndiag(S) so that it solves the original system before\nequilibration.\n\nfile zposvx.f zposvx.f plus dependencies\nprec complex16\nZPOSVX uses the Cholesky factorization A = U**H*U or A = L*L**H to\ncompute the solution to a complex system of linear equations\nA * X = B,\nwhere A is an N-by-N Hermitian positive definite matrix and X and B\nare N-by-NRHS matrices.\n\nError bounds on the solution and a condition estimate are also\nprovided.\n\nDescription\n===========\n\nThe following steps are performed:\n\n1. If FACT = 'E', real scaling factors are computed to equilibrate\nthe system:\ndiag(S) * A * diag(S) * inv(diag(S)) * X = diag(S) * B\nWhether or not the system will be equilibrated depends on the\nscaling of the matrix A, but if equilibration is used, A is\noverwritten by diag(S)*A*diag(S) and B by diag(S)*B.\n\n2. If FACT = 'N' or 'E', the Cholesky decomposition is used to\nfactor the matrix A (after equilibration if FACT = 'E') as\nA = U**H* U, if UPLO = 'U', or\nA = L * L**H, if UPLO = 'L',\nwhere U is an upper triangular matrix and L is a lower triangular\nmatrix.\n\n3. If the leading i-by-i principal minor is not positive definite,\nthen the routine returns with INFO = i. Otherwise, the factored\nform of A is used to estimate the condition number of the matrix\nA. If the reciprocal of the condition number is less than machine\nprecision, INFO = N+1 is returned as a warning, but the routine\nstill goes on to solve for X and compute error bounds as\ndescribed below.\n\n4. The system of equations is solved for X using the factored form\nof A.\n\n5. Iterative refinement is applied to improve the computed solution\nmatrix and calculate error bounds and backward error estimates\nfor it.\n\n6. If equilibration was used, the matrix X is premultiplied by\ndiag(S) so that it solves the original system before\nequilibration.\n\nfile zposvxx.f zposvxx.f plus dependencies\nprec complex16\nZPOSVXX uses the Cholesky factorization A = U**T*U or A = L*L**T\nto compute the solution to a complex*16 system of linear equations\nA * X = B, where A is an N-by-N symmetric positive definite matrix\nand X and B are N-by-NRHS matrices.\n\nIf requested, both normwise and maximum componentwise error bounds\nare returned. ZPOSVXX will return a solution with a tiny\nguaranteed error (O(eps) where eps is the working machine\nprecision) unless the matrix is very ill-conditioned, in which\ncase a warning is returned. Relevant condition numbers also are\ncalculated and returned.\n\nZPOSVXX accepts user-provided factorizations and equilibration\nfactors; see the definitions of the FACT and EQUED options.\nSolving with refinement and using a factorization from a previous\nZPOSVXX call will also produce a solution with either O(eps)\nerrors or warnings, but we cannot make that claim for general\nuser-provided factorizations and equilibration factors if they\ndiffer from what ZPOSVXX would itself produce.\n\nDescription\n===========\n\nThe following steps are performed:\n\n1. If FACT = 'E', double precision scaling factors are computed to equilibrate\nthe system:\n\ndiag(S)*A*diag(S) *inv(diag(S))*X = diag(S)*B\n\nWhether or not the system will be equilibrated depends on the\nscaling of the matrix A, but if equilibration is used, A is\noverwritten by diag(S)*A*diag(S) and B by diag(S)*B.\n\n2. If FACT = 'N' or 'E', the Cholesky decomposition is used to\nfactor the matrix A (after equilibration if FACT = 'E') as\nA = U**T* U, if UPLO = 'U', or\nA = L * L**T, if UPLO = 'L',\nwhere U is an upper triangular matrix and L is a lower triangular\nmatrix.\n\n3. If the leading i-by-i principal minor is not positive definite,\nthen the routine returns with INFO = i. Otherwise, the factored\nform of A is used to estimate the condition number of the matrix\nA (see argument RCOND). If the reciprocal of the condition number\nis less than machine precision, the routine still goes on to solve\nfor X and compute error bounds as described below.\n\n4. The system of equations is solved for X using the factored form\nof A.\n\n5. By default (unless PARAMS(LA_LINRX_ITREF_I) is set to zero),\nthe routine will use iterative refinement to try to get a small\nerror and error bounds. Refinement calculates the residual to at\nleast twice the working precision.\n\n6. If equilibration was used, the matrix X is premultiplied by\ndiag(S) so that it solves the original system before\nequilibration.\n\nfile zppsvx.f zppsvx.f plus dependencies\nprec complex16\nZPPSVX uses the Cholesky factorization A = U**H*U or A = L*L**H to\ncompute the solution to a complex system of linear equations\nA * X = B,\nwhere A is an N-by-N Hermitian positive definite matrix stored in\npacked format and X and B are N-by-NRHS matrices.\n\nError bounds on the solution and a condition estimate are also\nprovided.\n\nDescription\n===========\n\nThe following steps are performed:\n\n1. If FACT = 'E', real scaling factors are computed to equilibrate\nthe system:\ndiag(S) * A * diag(S) * inv(diag(S)) * X = diag(S) * B\nWhether or not the system will be equilibrated depends on the\nscaling of the matrix A, but if equilibration is used, A is\noverwritten by diag(S)*A*diag(S) and B by diag(S)*B.\n\n2. If FACT = 'N' or 'E', the Cholesky decomposition is used to\nfactor the matrix A (after equilibration if FACT = 'E') as\nA = U'* U , if UPLO = 'U', or\nA = L * L', if UPLO = 'L',\nwhere U is an upper triangular matrix, L is a lower triangular\nmatrix, and ' indicates conjugate transpose.\n\n3. If the leading i-by-i principal minor is not positive definite,\nthen the routine returns with INFO = i. Otherwise, the factored\nform of A is used to estimate the condition number of the matrix\nA. If the reciprocal of the condition number is less than machine\nprecision, INFO = N+1 is returned as a warning, but the routine\nstill goes on to solve for X and compute error bounds as\ndescribed below.\n\n4. The system of equations is solved for X using the factored form\nof A.\n\n5. Iterative refinement is applied to improve the computed solution\nmatrix and calculate error bounds and backward error estimates\nfor it.\n\n6. If equilibration was used, the matrix X is premultiplied by\ndiag(S) so that it solves the original system before\nequilibration.\n\nfile zspsvx.f zspsvx.f plus dependencies\nprec complex16\nZSPSVX uses the diagonal pivoting factorization A = U*D*U**T or\nA = L*D*L**T to compute the solution to a complex system of linear\nequations A * X = B, where A is an N-by-N symmetric matrix stored\nin packed format and X and B are N-by-NRHS matrices.\n\nError bounds on the solution and a condition estimate are also\nprovided.\n\nDescription\n===========\n\nThe following steps are performed:\n\n1. If FACT = 'N', the diagonal pivoting method is used to factor A as\nA = U * D * U**T, if UPLO = 'U', or\nA = L * D * L**T, if UPLO = 'L',\nwhere U (or L) is a product of permutation and unit upper (lower)\ntriangular matrices and D is symmetric and block diagonal with\n1-by-1 and 2-by-2 diagonal blocks.\n\n2. If some D(i,i)=0, so that D is exactly singular, then the routine\nreturns with INFO = i. Otherwise, the factored form of A is used\nto estimate the condition number of the matrix A. If the\nreciprocal of the condition number is less than machine precision,\nINFO = N+1 is returned as a warning, but the routine still goes on\nto solve for X and compute error bounds as described below.\n\n3. The system of equations is solved for X using the factored form\nof A.\n\n4. Iterative refinement is applied to improve the computed solution\nmatrix and calculate error bounds and backward error estimates\nfor it.\n\nfile zsysvx.f zsysvx.f plus dependencies\nprec complex16\nZSYSVX uses the diagonal pivoting factorization to compute the\nsolution to a complex system of linear equations A * X = B,\nwhere A is an N-by-N symmetric matrix and X and B are N-by-NRHS\nmatrices.\n\nError bounds on the solution and a condition estimate are also\nprovided.\n\nDescription\n===========\n\nThe following steps are performed:\n\n1. If FACT = 'N', the diagonal pivoting method is used to factor A.\nThe form of the factorization is\nA = U * D * U**T, if UPLO = 'U', or\nA = L * D * L**T, if UPLO = 'L',\nwhere U (or L) is a product of permutation and unit upper (lower)\ntriangular matrices, and D is symmetric and block diagonal with\n1-by-1 and 2-by-2 diagonal blocks.\n\n2. If some D(i,i)=0, so that D is exactly singular, then the routine\nreturns with INFO = i. Otherwise, the factored form of A is used\nto estimate the condition number of the matrix A. If the\nreciprocal of the condition number is less than machine precision,\nINFO = N+1 is returned as a warning, but the routine still goes on\nto solve for X and compute error bounds as described below.\n\n3. The system of equations is solved for X using the factored form\nof A.\n\n4. Iterative refinement is applied to improve the computed solution\nmatrix and calculate error bounds and backward error estimates\nfor it.\n\nfile zsysvxx.f zsysvxx.f plus dependencies\nprec complex16\nZSYSVXX uses the diagonal pivoting factorization to compute the\nsolution to a complex*16 system of linear equations A * X = B, where\nA is an N-by-N symmetric matrix and X and B are N-by-NRHS\nmatrices.\n\nIf requested, both normwise and maximum componentwise error bounds\nare returned. ZSYSVXX will return a solution with a tiny\nguaranteed error (O(eps) where eps is the working machine\nprecision) unless the matrix is very ill-conditioned, in which\ncase a warning is returned. Relevant condition numbers also are\ncalculated and returned.\n\nZSYSVXX accepts user-provided factorizations and equilibration\nfactors; see the definitions of the FACT and EQUED options.\nSolving with refinement and using a factorization from a previous\nZSYSVXX call will also produce a solution with either O(eps)\nerrors or warnings, but we cannot make that claim for general\nuser-provided factorizations and equilibration factors if they\ndiffer from what ZSYSVXX would itself produce.\n\nDescription\n===========\n\nThe following steps are performed:\n\n1. If FACT = 'E', double precision scaling factors are computed to equilibrate\nthe system:\n\ndiag(S)*A*diag(S) *inv(diag(S))*X = diag(S)*B\n\nWhether or not the system will be equilibrated depends on the\nscaling of the matrix A, but if equilibration is used, A is\noverwritten by diag(S)*A*diag(S) and B by diag(S)*B.\n\n2. If FACT = 'N' or 'E', the LU decomposition is used to factor\nthe matrix A (after equilibration if FACT = 'E') as\n\nA = U * D * U**T, if UPLO = 'U', or\nA = L * D * L**T, if UPLO = 'L',\n\nwhere U (or L) is a product of permutation and unit upper (lower)\ntriangular matrices, and D is symmetric and block diagonal with\n1-by-1 and 2-by-2 diagonal blocks.\n\n3. If some D(i,i)=0, so that D is exactly singular, then the\nroutine returns with INFO = i. Otherwise, the factored form of A\nis used to estimate the condition number of the matrix A (see\nargument RCOND). If the reciprocal of the condition number is\nless than machine precision, the routine still goes on to solve\nfor X and compute error bounds as described below.\n\n4. The system of equations is solved for X using the factored form\nof A.\n\n5. By default (unless PARAMS(LA_LINRX_ITREF_I) is set to zero),\nthe routine will use iterative refinement to try to get a small\nerror and error bounds. Refinement calculates the residual to at\nleast twice the working precision.\n\n6. If equilibration was used, the matrix X is premultiplied by\ndiag(R) so that it solves the original system before\nequilibration.\n\n```"
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"Back to Main Page\n\nChoose Module Version:\n\n## Mathematics III / Applied Mathematics\n\n Module name (EN): Mathematics III / Applied Mathematics Degree programme: Mechatronics and Sensor Technology, Bachelor, ASPO 01.10.2011 Module code: MST.MA3 SAP-Submodule-No.: P231-0055 Hours per semester week / Teaching method: 4V+1U (5 hours per week, accumulated) ECTS credits: 6 Semester: 3 Duration: 2 semester Mandatory course: yes Language of instruction: German Assessment: Test + term paper [updated 10.05.2021] Applicability / Curricular relevance: MST.MA3 (P231-0055) Mechatronics and Sensor Technology, Bachelor, ASPO 01.10.2012, semester 3, mandatory course MST.MA3 (P231-0055) Mechatronics and Sensor Technology, Bachelor, ASPO 01.10.2011, semester 3, mandatory course Workload: 75 class hours (= 56.25 clock hours) over a 15-week period.The total student study time is 180 hours (equivalent to 6 ECTS credits).There are therefore 123.75 hours available for class preparation and follow-up work and exam preparation. Recommended prerequisites (modules): MST.NSW Numerical Software [updated 10.05.2019] Recommended as prerequisite for: MST.CVI Computer VisionMST.TMM Mechanical Engineering and Machine Dynamics [updated 10.05.2019] Module coordinator: N.N. Lecturer: N.N. [updated 01.10.2005] Learning outcomes: The objective of this module is to teach the applied mathematical methods required for engineering disciplines such as control engineering and physics. In addition, it is designed to teach applied methods of higher mathematics in a problem-oriented manner (in particular numerics and statistics) using practical examples. After successfully completing this module, students will be able to solve minor problems based on given measured data using MATLAB and SIMULINK, and present their solution in writing.Part I [updated 10.05.2021] Module content: 1 - Ordinary differential equationsĢ Separable DGĢ Linear differential equation with constant coefficients of the 1st orderĢ Linear differential equation with constant coefficients of the 2nd orderĢ Applications in technology 2 - The Fourier transformĢ Fourier series for periodic functionsĢ Fourier series for non-periodic functionsĢ Applications 3 - The Laplace transformĢ DefinitionĢ AlgorithmsĢ Inverse transformation methods (convolution, partial fraction decomposition)Ģ Applications 4 ¢ Functions with multiple variablesĢ Partial derivatives, tangent plane 3. Coordinate transformations, Ģ Multiple integrals and integral transform 5 ¢ Introduction to vector analysis Part II 1. Introduction to Matlab 1.1 Calculating with vectors and matrices, creating graphics2. Interpolations (Newton polynomials, Spline functions) 2.1 Mini-project3. Least squares fitting 3.1 Linear balance functions3.2 Non-linear balanced functions3.3 Mini-project4. Simple statistical measures for one, two, and more than two characteristics4.1 Mini-project5. Numerical differentiation and integration - Introduction 5.1 Mini-project6. Introduction to SIMULINK6.1 Solving initial value problems of ordinary differential equations with SIMULINK 6.2 Mini-project [updated 10.05.2021] Teaching methods/Media: The numerics part II will take place in the PC lab AMSEL (Angewandte Mathematik, Statistik, eLearning (Room 5306).All practical exercises for the lecture, as well as solving exercises, homework and case studies will be done with the e-learning system MathCoach (AMSEL lab: PC lab: \"Angewandte Mathematik, Statistik und eLearning\"). In addition, a performance-relevant midterm exam will be written as an online exam using the MathCoach elearning system. [updated 10.05.2021] Recommended or required reading: Part I0. B.Grabowski:\"Mathematik III für Ingenieure\", e-book mit MathCoach, 20111. L. Papula : \"Mathematik für Ingenieure\", Band 1-3 und Formelsammlungen, Vieweg, 20002. Engeln-Müllges, Schäfer, Trippler: \"Kompaktkurs Ingenieurmathematik\". Fachbuchverlag Leipzig im Carl Hanser Verlag: München/Wien, 1999.3. Brauch/Dreyer/Haacke, Mathematik für Ingenieure, Teubner, 2003 Part II0. B.Grabowski: \"E-Learning: Numerik mit MathCoach\", e-book, 20111. Preuss/Wenisch, Numerische Mathematik, Fachbuchverlag, 20012. Faires/Burden, Numerische methoden, Spektrum Akademischer Verlag, 20003. Gramlich/Werner, Numerische Mathematik mit MATLAB, dpunktverlag, 20004. Beucher, MATLAB und SIMULINK lernen, Addison-Wesley, 20005. Bartsch H.-J., Tachenbuch Mathematischer Formeln, Fachbuchverlag Leipzig, 2003 Materials1. www.htw-saarland.de/fb/gis/people/bgrabowski/vorles/mathe.htm (only available within the htw)2. www.htw-saarland.de/fb/gis/mathematik/ Script I and formula collection 1 for descriptive statistics [updated 10.05.2021]\n```[Thu Sep 29 16:55:03 CEST 2022, CKEY=ymmathe3, BKEY=yst, CID=MST.MA3, LANGUAGE=en, DATE=29.09.2022]\n```"
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https://optuna.readthedocs.io/en/latest/_modules/optuna/integration/cma.html | [
"# Source code for optuna.integration.cma\n\n```import math\nimport random\nfrom typing import Any\nfrom typing import Dict\nfrom typing import List\nfrom typing import Optional\nfrom typing import Sequence\nfrom typing import Set\n\nimport numpy\n\nimport optuna\nfrom optuna import distributions\nfrom optuna import logging\nfrom optuna._deprecated import deprecated\nfrom optuna._imports import try_import\nfrom optuna.distributions import BaseDistribution\nfrom optuna.distributions import CategoricalDistribution\nfrom optuna.distributions import DiscreteUniformDistribution\nfrom optuna.distributions import IntLogUniformDistribution\nfrom optuna.distributions import IntUniformDistribution\nfrom optuna.distributions import LogUniformDistribution\nfrom optuna.distributions import UniformDistribution\nfrom optuna.samplers import BaseSampler\nfrom optuna.study import Study\nfrom optuna.study import StudyDirection\nfrom optuna.trial import FrozenTrial\nfrom optuna.trial import TrialState\n\nwith try_import() as _imports:\nimport cma\n\n_logger = logging.get_logger(__name__)\n\n_EPS = 1e-10\n\n[docs]class PyCmaSampler(BaseSampler):\n\"\"\"A Sampler using cma library as the backend.\n\nExample:\n\nOptimize a simple quadratic function by using :class:`~optuna.integration.PyCmaSampler`.\n\n.. testcode::\n\nimport optuna\n\ndef objective(trial):\nx = trial.suggest_float(\"x\", -1, 1)\ny = trial.suggest_int(\"y\", -1, 1)\nreturn x ** 2 + y\n\nsampler = optuna.integration.PyCmaSampler()\nstudy = optuna.create_study(sampler=sampler)\nstudy.optimize(objective, n_trials=20)\n\nNote that parallel execution of trials may affect the optimization performance of CMA-ES,\nespecially if the number of trials running in parallel exceeds the population size.\n\n.. note::\n:class:`~optuna.integration.CmaEsSampler` is deprecated and renamed to\n:class:`~optuna.integration.CmaEsSampler`.\n\nArgs:\n\nx0:\nA dictionary of an initial parameter values for CMA-ES. By default, the mean of ``low``\nand ``high`` for each distribution is used.\nPlease refer to cma.CMAEvolutionStrategy_ for further details of ``x0``.\n\nsigma0:\nInitial standard deviation of CMA-ES. By default, ``sigma0`` is set to\n``min_range / 6``, where ``min_range`` denotes the minimum range of the distributions\nin the search space. If distribution is categorical, ``min_range`` is\n``len(choices) - 1``.\nPlease refer to cma.CMAEvolutionStrategy_ for further details of ``sigma0``.\n\ncma_stds:\nA dictionary of multipliers of sigma0 for each parameters. The default value is 1.0.\nPlease refer to cma.CMAEvolutionStrategy_ for further details of ``cma_stds``.\n\nseed:\nA random seed for CMA-ES.\n\ncma_opts:\nOptions passed to the constructor of cma.CMAEvolutionStrategy_ class.\n\nNote that ``BoundaryHandler``, ``bounds``, ``CMA_stds`` and ``seed`` arguments in\n``cma_opts`` will be ignored because it is added by\n:class:`~optuna.integration.PyCmaSampler` automatically.\n\nn_startup_trials:\nThe independent sampling is used instead of the CMA-ES algorithm until the given number\nof trials finish in the same study.\n\nindependent_sampler:\nA :class:`~optuna.samplers.BaseSampler` instance that is used for independent\nsampling. The parameters not contained in the relative search space are sampled\nby this sampler.\nThe search space for :class:`~optuna.integration.PyCmaSampler` is determined by\n:func:`~optuna.samplers.intersection_search_space()`.\n\nIf :obj:`None` is specified, :class:`~optuna.samplers.RandomSampler` is used\nas the default.\n\n.. seealso::\n:class:`optuna.samplers` module provides built-in independent samplers\nsuch as :class:`~optuna.samplers.RandomSampler` and\n:class:`~optuna.samplers.TPESampler`.\n\nwarn_independent_sampling:\nIf this is :obj:`True`, a warning message is emitted when\nthe value of a parameter is sampled by using an independent sampler.\n\nNote that the parameters of the first trial in a study are always sampled\nvia an independent sampler, so no warning messages are emitted in this case.\n\n.. _cma.CMAEvolutionStrategy: http://cma.gforge.inria.fr/apidocs-pycma/\\\ncma.evolution_strategy.CMAEvolutionStrategy.html\n\"\"\"\n\ndef __init__(\nself,\nx0: Optional[Dict[str, Any]] = None,\nsigma0: Optional[float] = None,\ncma_stds: Optional[Dict[str, float]] = None,\nseed: Optional[int] = None,\ncma_opts: Optional[Dict[str, Any]] = None,\nn_startup_trials: int = 1,\nindependent_sampler: Optional[BaseSampler] = None,\nwarn_independent_sampling: bool = True,\n) -> None:\n\n_imports.check()\n\nself._x0 = x0\nself._sigma0 = sigma0\nself._cma_stds = cma_stds\nif seed is None:\nseed = random.randint(1, 2 ** 32)\nself._cma_opts = cma_opts or {}\nself._cma_opts[\"seed\"] = seed\nself._cma_opts.setdefault(\"verbose\", -2)\nself._n_startup_trials = n_startup_trials\nself._independent_sampler = independent_sampler or optuna.samplers.RandomSampler(seed=seed)\nself._warn_independent_sampling = warn_independent_sampling\nself._search_space = optuna.samplers.IntersectionSearchSpace()\n\n[docs] def reseed_rng(self) -> None:\n\nself._cma_opts[\"seed\"] = random.randint(1, 2 ** 32)\nself._independent_sampler.reseed_rng()\n\n[docs] def infer_relative_search_space(\nself, study: Study, trial: FrozenTrial\n) -> Dict[str, BaseDistribution]:\n\nsearch_space = {}\nfor name, distribution in self._search_space.calculate(study).items():\nif distribution.single():\n# `cma` cannot handle distributions that contain just a single value, so we skip\n# them. Note that the parameter values for such distributions are sampled in\n# `Trial`.\ncontinue\n\nsearch_space[name] = distribution\n\nreturn search_space\n\n[docs] def sample_independent(\nself,\nstudy: Study,\ntrial: FrozenTrial,\nparam_name: str,\nparam_distribution: BaseDistribution,\n) -> float:\n\nself._raise_error_if_multi_objective(study)\n\nif self._warn_independent_sampling:\ncomplete_trials = [t for t in study.trials if t.state == TrialState.COMPLETE]\nif len(complete_trials) >= self._n_startup_trials:\nself._log_independent_sampling(trial, param_name)\n\nreturn self._independent_sampler.sample_independent(\nstudy, trial, param_name, param_distribution\n)\n\n[docs] def sample_relative(\nself, study: Study, trial: FrozenTrial, search_space: Dict[str, BaseDistribution]\n) -> Dict[str, float]:\n\nself._raise_error_if_multi_objective(study)\n\nif len(search_space) == 0:\nreturn {}\n\nif len(search_space) == 1:\n_logger.info(\n\"`PyCmaSampler` does not support optimization of 1-D search space. \"\n\"`{}` is used instead of `PyCmaSampler`.\".format(\nself._independent_sampler.__class__.__name__\n)\n)\nself._warn_independent_sampling = False\nreturn {}\n\ncomplete_trials = [t for t in study.trials if t.state == TrialState.COMPLETE]\nif len(complete_trials) < self._n_startup_trials:\nreturn {}\n\nif self._x0 is None:\nself._x0 = self._initialize_x0(search_space)\n\nif self._sigma0 is None:\nsigma0 = self._initialize_sigma0(search_space)\nelse:\nsigma0 = self._sigma0\n# Avoid ZeroDivisionError in cma.CMAEvolutionStrategy.\nsigma0 = max(sigma0, _EPS)\n\noptimizer = _Optimizer(search_space, self._x0, sigma0, self._cma_stds, self._cma_opts)\ntrials = study.trials\nlast_told_trial_number = optimizer.tell(trials, study.direction)\n\n@staticmethod\ndef _initialize_x0(search_space: Dict[str, BaseDistribution]) -> Dict[str, Any]:\n\nx0: Dict[str, Any] = {}\nfor name, distribution in search_space.items():\nif isinstance(distribution, UniformDistribution):\nx0[name] = numpy.mean([distribution.high, distribution.low])\nelif isinstance(distribution, DiscreteUniformDistribution):\nx0[name] = numpy.mean([distribution.high, distribution.low])\nelif isinstance(distribution, IntUniformDistribution):\nx0[name] = int(numpy.mean([distribution.high, distribution.low]))\nelif isinstance(distribution, (LogUniformDistribution, IntLogUniformDistribution)):\nlog_high = math.log(distribution.high)\nlog_low = math.log(distribution.low)\nx0[name] = math.exp(numpy.mean([log_high, log_low]))\nelif isinstance(distribution, CategoricalDistribution):\nindex = (len(distribution.choices) - 1) // 2\nx0[name] = distribution.choices[index]\nelse:\nraise NotImplementedError(\n\"The distribution {} is not implemented.\".format(distribution)\n)\nreturn x0\n\n@staticmethod\ndef _initialize_sigma0(search_space: Dict[str, BaseDistribution]) -> float:\n\nsigma0s = []\nfor name, distribution in search_space.items():\nif isinstance(distribution, UniformDistribution):\nsigma0s.append((distribution.high - distribution.low) / 6)\nelif isinstance(distribution, DiscreteUniformDistribution):\nsigma0s.append((distribution.high - distribution.low) / 6)\nelif isinstance(distribution, IntUniformDistribution):\nsigma0s.append((distribution.high - distribution.low) / 6)\nelif isinstance(distribution, (LogUniformDistribution, IntLogUniformDistribution)):\nlog_high = math.log(distribution.high)\nlog_low = math.log(distribution.low)\nsigma0s.append((log_high - log_low) / 6)\nelif isinstance(distribution, CategoricalDistribution):\nsigma0s.append((len(distribution.choices) - 1) / 6)\nelse:\nraise NotImplementedError(\n\"The distribution {} is not implemented.\".format(distribution)\n)\nreturn min(sigma0s)\n\ndef _log_independent_sampling(self, trial: FrozenTrial, param_name: str) -> None:\n\n_logger.warning(\n\"The parameter '{}' in trial#{} is sampled independently \"\n\"by using `{}` instead of `PyCmaSampler` \"\n\"(optimization performance may be degraded). \"\n\"`PyCmaSampler` does not support dynamic search space or `CategoricalDistribution`. \"\n\"You can suppress this warning by setting `warn_independent_sampling` \"\n\"to `False` in the constructor of `PyCmaSampler`, \"\n\"if this independent sampling is intended behavior.\".format(\nparam_name, trial.number, self._independent_sampler.__class__.__name__\n)\n)\n\n[docs] def after_trial(\nself,\nstudy: Study,\ntrial: FrozenTrial,\nstate: TrialState,\nvalues: Optional[Sequence[float]],\n) -> None:\n\nself._independent_sampler.after_trial(study, trial, state, values)\n\nclass _Optimizer(object):\ndef __init__(\nself,\nsearch_space: Dict[str, BaseDistribution],\nx0: Dict[str, Any],\nsigma0: float,\ncma_stds: Optional[Dict[str, float]],\ncma_opts: Dict[str, Any],\n) -> None:\n\nself._search_space = search_space\nself._param_names = list(sorted(self._search_space.keys()))\n\nlows = []\nhighs = []\nfor param_name in self._param_names:\ndist = self._search_space[param_name]\nif isinstance(dist, CategoricalDistribution):\n# Handle categorical values by ordinal representation.\n# TODO(Yanase): Support one-hot representation.\nlows.append(-0.5)\nhighs.append(len(dist.choices) - 0.5)\nelif isinstance(dist, (UniformDistribution, LogUniformDistribution)):\nlows.append(self._to_cma_params(search_space, param_name, dist.low))\nhighs.append(self._to_cma_params(search_space, param_name, dist.high) - _EPS)\nelif isinstance(dist, DiscreteUniformDistribution):\nr = dist.high - dist.low\nlows.append(0 - 0.5 * dist.q)\nhighs.append(r + 0.5 * dist.q)\nelif isinstance(dist, IntUniformDistribution):\nlows.append(dist.low - 0.5 * dist.step)\nhighs.append(dist.high + 0.5 * dist.step)\nelif isinstance(dist, IntLogUniformDistribution):\nlows.append(self._to_cma_params(search_space, param_name, dist.low - 0.5))\nhighs.append(self._to_cma_params(search_space, param_name, dist.high + 0.5))\nelse:\nraise NotImplementedError(\"The distribution {} is not implemented.\".format(dist))\n\n# Set initial params.\ninitial_cma_params = []\nfor param_name in self._param_names:\ninitial_cma_params.append(\nself._to_cma_params(self._search_space, param_name, x0[param_name])\n)\ncma_option = {\n\"BoundaryHandler\": cma.BoundTransform,\n\"bounds\": [lows, highs],\n}\n\nif cma_stds:\ncma_option[\"CMA_stds\"] = [cma_stds.get(name, 1.0) for name in self._param_names]\n\ncma_opts.update(cma_option)\n\nself._es = cma.CMAEvolutionStrategy(initial_cma_params, sigma0, cma_opts)\n\ndef tell(self, trials: List[FrozenTrial], study_direction: StudyDirection) -> int:\n\ncomplete_trials = self._collect_target_trials(trials, target_states={TrialState.COMPLETE})\n\npopsize = self._es.popsize\ngeneration = len(complete_trials) // popsize\nlast_told_trial_number = -1\nfor i in range(generation):\nxs = []\nys = []\nfor t in complete_trials[i * popsize : (i + 1) * popsize]:\nx = [\nself._to_cma_params(self._search_space, name, t.params[name])\nfor name in self._param_names\n]\nxs.append(x)\nys.append(t.value)\nlast_told_trial_number = t.number\nif study_direction == StudyDirection.MAXIMIZE:\nys = [-1 * y if y is not None else y for y in ys]\n\n# Calling `ask` is required to avoid RuntimeError which claims that `tell` should only\n# be called once per iteration.\nself._es.tell(xs, ys)\nreturn last_told_trial_number\n\ndef ask(self, trials: List[FrozenTrial], last_told_trial_number: int) -> Dict[str, Any]:\n\nindividual_index = len(self._collect_target_trials(trials, last_told_trial_number))\npopsize = self._es.popsize\n\n# individual_index may exceed the population size due to the parallel execution of multiple\n# trials. In such cases, `cma.cma.CMAEvolutionStrategy.ask` is called multiple times in an\n# iteration, and that may affect the optimization performance of CMA-ES.\n# In addition, please note that some trials may suggest the same parameters when multiple\n# samplers invoke this method simultaneously.\nwhile individual_index >= popsize:\nindividual_index -= popsize\n\nret_val = {}\nfor param_name, value in zip(self._param_names, cma_params):\nret_val[param_name] = self._to_optuna_params(self._search_space, param_name, value)\nreturn ret_val\n\ndef _is_compatible(self, trial: FrozenTrial) -> bool:\n\n# Thanks to `intersection_search_space()` function, in sequential optimization,\n# the parameters of complete trials are always compatible with the search space.\n#\n# However, in distributed optimization, incompatible trials may complete on a worker\n# just after an intersection search space is calculated on another worker.\n\nfor name, distribution in self._search_space.items():\nif name not in trial.params:\nreturn False\n\ndistributions.check_distribution_compatibility(distribution, trial.distributions[name])\nparam_value = trial.params[name]\nparam_internal_value = distribution.to_internal_repr(param_value)\nif not distribution._contains(param_internal_value):\nreturn False\n\nreturn True\n\ndef _collect_target_trials(\nself,\ntrials: List[FrozenTrial],\nlast_told: int = -1,\ntarget_states: Optional[Set[TrialState]] = None,\n) -> List[FrozenTrial]:\n\ntarget_trials = [t for t in trials if t.number > last_told]\ntarget_trials = [t for t in target_trials if self._is_compatible(t)]\nif target_states is not None:\ntarget_trials = [t for t in target_trials if t.state in target_states]\n\nreturn target_trials\n\n@staticmethod\ndef _to_cma_params(\nsearch_space: Dict[str, BaseDistribution], param_name: str, optuna_param_value: Any\n) -> float:\n\ndist = search_space[param_name]\nif isinstance(dist, (LogUniformDistribution, IntLogUniformDistribution)):\nreturn math.log(optuna_param_value)\nelif isinstance(dist, DiscreteUniformDistribution):\nreturn optuna_param_value - dist.low\nelif isinstance(dist, CategoricalDistribution):\nreturn dist.choices.index(optuna_param_value)\nreturn optuna_param_value\n\n@staticmethod\ndef _to_optuna_params(\nsearch_space: Dict[str, BaseDistribution], param_name: str, cma_param_value: float\n) -> Any:\n\ndist = search_space[param_name]\nif isinstance(dist, UniformDistribution):\n# Type of cma_param_value is np.floating, so cast it to Python's built-in float.\nreturn float(cma_param_value)\nif isinstance(dist, LogUniformDistribution):\nreturn math.exp(cma_param_value)\nif isinstance(dist, DiscreteUniformDistribution):\nv = numpy.round(cma_param_value / dist.q) * dist.q + dist.low\n# v may slightly exceed range due to round-off errors.\nreturn float(min(max(v, dist.low), dist.high))\nif isinstance(dist, IntUniformDistribution):\nr = numpy.round((cma_param_value - dist.low) / dist.step)\nv = r * dist.step + dist.low\nreturn int(v)\nif isinstance(dist, IntLogUniformDistribution):\nexp_value = math.exp(cma_param_value)\nv = numpy.round(exp_value)\nreturn int(min(max(v, dist.low), dist.high))\n\nif isinstance(dist, CategoricalDistribution):\nv = int(numpy.round(cma_param_value))\nreturn dist.choices[v]\nreturn cma_param_value\n\n[docs]@deprecated(\"2.0.0\", text=\"This class is renamed to :class:`~optuna.integration.PyCmaSampler`.\")\nclass CmaEsSampler(PyCmaSampler):\n\"\"\"Wrapper class of PyCmaSampler for backward compatibility.\"\"\"\n\ndef __init__(\nself,\nx0: Optional[Dict[str, Any]] = None,\nsigma0: Optional[float] = None,\ncma_stds: Optional[Dict[str, float]] = None,\nseed: Optional[int] = None,\ncma_opts: Optional[Dict[str, Any]] = None,\nn_startup_trials: int = 1,\nindependent_sampler: Optional[BaseSampler] = None,\nwarn_independent_sampling: bool = True,\n) -> None:\n\nsuper().__init__(\nx0=x0,\nsigma0=sigma0,\ncma_stds=cma_stds,\nseed=seed,\ncma_opts=cma_opts,\nn_startup_trials=n_startup_trials,\nindependent_sampler=independent_sampler,\nwarn_independent_sampling=warn_independent_sampling,\n)\n```"
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https://suburbanstylechallenge.com/dermatology/how-many-atoms-are-there-in-one-mole-of-gold.html | [
"How many atoms are there in one mole of gold?\n\nContents\n\nYou use the fact that 1 mole of any substance contains exactly 6.022⋅1023 atoms or molecules of that substance – this is known as Avogadro’s number. In your case, 1 mole of gold will have exactly 6.022⋅1023 atoms of gold.\n\nWhat is one mole of gold equal to?\n\nThis mass is given by the atomic weight of the chemical unit that makes up that substance in atomic mass units (amu). For example, gold has an atomic weight of 196.967 amu, so one mole of gold has a mass of 196.967 grams.\n\nHow many atoms are in 2 moles of gold?\n\nWe know we have 2 moles of Au, so multiply 6.02 x 2. Answer is 12.04 x 10^23 atoms of Au.\n\nWhat is the mass of 1 mole of gold atom?\n\nThe molar mass of gold = 196.96657 g/m. The molar mass of gold is defined as the mass of 1 mole of gold atoms – typically specified in grams. Therefore the molar mass of gold is stated in units of grams/mole or g/m. It is also frequently called the Molecular Weight of Gold.\n\nHow many atoms are in a gram of gold?\n\nGold (Au) has an atomic mass of nearly 197. It means that 1 mol Au weights 197 grams. 1 mol contains 6.022e+23 atoms (Avogadro’s number). Dividing Avogadro’s number by 197 yields 3.057e+21 Au atoms / gram, or 3,057 billions of billions of Au atoms per gram.\n\nHow many atoms are in a mole?\n\nThe mole is represented by Avogadro’s number, which is 6.022×1023 atoms or molecules per mol.\n\nHow many atoms are in a mole calculator?\n\nAvogadro’s number is a very important relationship to remember: 1 mole = 6.022×1023 6.022 × 10 23 atoms, molecules, protons, etc. To convert from moles to atoms, multiply the molar amount by Avogadro’s number.\n\nHow many atoms are in 2.5 moles of gold?\n\nThere are 15.055 X 1023 atoms of gold in a sample of 2.5 mol of Au, also known as gold.\n\nHow many atoms are in 0.9 mole of gold?\n\nThere are 6.022 X 1023 atoms in one mole of gold.\n\nHow many atoms are in one mole of bromine?\n\n1 mol Br contains 6.022×1023Br atoms;its mass is 79.90 g.\n\nWhat atomic number is gold?\n\nGold, Au, has an atomic number of 79. Gold is dense, soft, shiny, and most malleable of all the metals known. Gold is a transition metal that is attacked by chlorine, fluorine, and cyanide. Gold is the 79th element in the periodic table.\n\nHow many atoms are in a mole of carbon?\n\nThe value of the mole is equal to the number of atoms in exactly 12 grams of pure carbon-12. 12.00 g C-12 = 1 mol C-12 atoms = 6.022 × 1023 atoms • The number of particles in 1 mole is called Avogadro’s Number (6.0221421 x 1023).\n\nHow many moles are in copper?\n\nFrom your Periodic Table we learn that one mole of copper, 6.022×1023 individual copper atoms have a mass of 63.55⋅g .\n\nHow many atoms are in 1kg of gold?\n\n197 g of gold contains 6.023 x 1023 atoms of gold. ∴ 1000 g of gold will contain = 1000×6.023×1023197 1000 × 6.023 × 10 23 197 = 3.055 x 1024 atoms of Gold.\n\nHow many atoms are in a mole of silver?\n\nThis time we need to divide by Avogadro’s number. 3.01 x 1023 atoms of silver divided by 6.02 x 1023 atoms of silver per mole of silver gives 0.500 moles of silver.\n\nHow do I calculate moles?\n\nHow to find moles?\n\n1. Measure the weight of your substance.\n2. Use a periodic table to find its atomic or molecular mass.\n3. Divide the weight by the atomic or molecular mass.\n4. Check your results with Omni Calculator."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9011312,"math_prob":0.99028516,"size":3238,"snap":"2022-05-2022-21","text_gpt3_token_len":951,"char_repetition_ratio":0.19325912,"word_repetition_ratio":0.063897766,"special_character_ratio":0.3239654,"punctuation_ratio":0.13454075,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9970179,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-24T14:31:29Z\",\"WARC-Record-ID\":\"<urn:uuid:a17f60f5-8a23-48e8-b9a8-7c3afc01f7d0>\",\"Content-Length\":\"73296\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a22446bb-61fd-4d24-9eb6-c8ec48ef8c01>\",\"WARC-Concurrent-To\":\"<urn:uuid:7b39e7e5-d82a-45cf-8c4b-09ddf4cdab39>\",\"WARC-IP-Address\":\"144.126.128.2\",\"WARC-Target-URI\":\"https://suburbanstylechallenge.com/dermatology/how-many-atoms-are-there-in-one-mole-of-gold.html\",\"WARC-Payload-Digest\":\"sha1:OHWRNTYE6ANG5KOTRX6LAXN5ZO3NPEJC\",\"WARC-Block-Digest\":\"sha1:WQ3UV33QWOECXD2MITB4EJAQPPDCECN4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304570.90_warc_CC-MAIN-20220124124654-20220124154654-00674.warc.gz\"}"} |
https://realcashier.com/quick-ratio/ | [
"# Quick Ratio – Meaning, Formula, Ideal quick ratio\n\n#### Meaning of Quick Ratio\n\nQuick Ratio is also known as the “Acid-test Ratio” and it is considered as the best measure of a company’s short-term liquidity position.\n\nThe quick ratio also indicates the company’s ability to quickly convert its current assets (Except Inventories) into cash and pay off the current liability of the organization.\n\nQuick ratio will help to answer the question: “Whether the organization will be able to pay off its current liabilities even if there is no sales revenue generated?”\n\n#### Formula for Quick Ratio or Acid – Test Ratio\n\nQuick Ratio = Quick Assets / Quick Liabilities\n\nWhereas, Quick assets consist of all the current assets as used for computation of current ratio except Inventories.\n\nQuick Assets = Current Assets – Inventories\n\nSimilarly, Quick Liabilities consists of all the current liabilities used for Computation of Current Ratio.\n\nQuick liabilities = All the current liabilities of the organization.\n\nBased on the above definition we can define the formula of Quick Ratio or Acid-test ratio as follows:\n\nQuick Ratio = Current assets – Inventories / Current liabilities\n\n#### Standard / Ideal Quick ratio\n\nGenerally, a Quick ratio of 1 to 1 is considered satisfactory. Unless Quick asset consists of higher account receivables or Organization is inefficient in the collection of amount from debtors.\n\n#### Why Inventory is not considered for computation of the Quick ratio?\n\nThe main objective of the quick ratio is to determine the company’s ability to pay off its current liabilities instantly. Hence, for computation of quick ratio, only those assets are considered which can be instantly converted into cash. However, Inventory is considered a “Non-cash asset” since it cannot be converted into cash instantly, and even though the organization might be able to sell the inventories instantly but the value or sales consideration might not be at par with the book value of the inventories.\n\nHope you find this article useful ! In case of any doubts or suggestions please let me know in comment section below or reach us on Twitter !"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.93397146,"math_prob":0.8469177,"size":2069,"snap":"2021-43-2021-49","text_gpt3_token_len":399,"char_repetition_ratio":0.16416465,"word_repetition_ratio":0.011976048,"special_character_ratio":0.18704689,"punctuation_ratio":0.06460674,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95548904,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-16T21:43:05Z\",\"WARC-Record-ID\":\"<urn:uuid:e3ce9bc3-6761-4500-85c1-f0f2e6cc215c>\",\"Content-Length\":\"85460\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:720fb284-9f66-4aa3-9339-cb63eca97e23>\",\"WARC-Concurrent-To\":\"<urn:uuid:3270901b-ecc4-429f-96e9-83893d9fce49>\",\"WARC-IP-Address\":\"104.21.60.83\",\"WARC-Target-URI\":\"https://realcashier.com/quick-ratio/\",\"WARC-Payload-Digest\":\"sha1:DJWRPW3S6QCRQNW2GXBZEFNIENG5IR57\",\"WARC-Block-Digest\":\"sha1:6TDAMCH6JUN657IPY5SXXCIUZE6LXR2H\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585025.23_warc_CC-MAIN-20211016200444-20211016230444-00251.warc.gz\"}"} |
https://universalium.academic.ru/90986/charge_conjugation | [
"# charge conjugation\n\n\ncharge conjugation\nthe mathematical operation of replacing every elementary particle by its antiparticle. Symbol: C\n\n* * *\n\nin particle physics, an operation that replaces particles with antiparticles (antiparticle) (and vice versa) in equations describing subatomic particles (subatomic particle). The name charge conjugation arises because a given particle and its antiparticle generally carry opposite electric charge. The positive electron, or positron, for example, is the antiparticle of the ordinary negative electron.\n\nUntil the mid-1950s, it was believed that nature would be unaffected if particles were replaced by their antiparticles. The operation of charge conjugation, in other words, would not change the mathematical descriptions of fundamental physical processes. Several experiments on weak nuclear interactions, undertaken after 1956, however, indicated that nature is not indifferent to the exchange of particles and antiparticles. Thus, antiparticle universes, popular in science fiction, would have physical laws slightly different from the laws for this universe. When charge conjugation C is combined with two other operations, parity reversal P and time reversal T, strong theoretical grounds (CPT theorem) indicate that the basic laws of physics do remain invariant. See also CP violation.\n\n* * *\n\nUniversalium. 2010.\n\n### Look at other dictionaries:\n\n• charge conjugation — krūvinė sąsaja statusas T sritis fizika atitikmenys: angl. charge conjugation vok. Ladungskonjugation, f rus. зарядное сопряжение, n pranc. conjugaison de charge, f … Fizikos terminų žodynas\n\n• charge conjugation — noun : an operation in mathematical physics in which each particle in a system is replaced by its antiparticle … Useful english dictionary\n\n• Charge (physics) — This article is about charge in the most general physics sense. For charge as relates to electrical phenomena, see electric charge. For other uses of charge, see Charge (disambiguation). In physics, a charge may refer to one of many different… … Wikipedia\n\n• Charge-transfer complex — A charge transfer complex (CT complex) or electron donor acceptor complex is an association of two or more molecules, or of different parts of one very large molecule, in which a fraction of electronic charge is transferred between the molecular… … Wikipedia\n\n• Charge transfer complex — A charge transfer complex (or CT complex, electron donor acceptor complex) is a chemical association of two or more molecules, or of different parts of one very large molecule, in which the attraction between the molecules (or parts) is created… … Wikipedia\n\n• conjugaison de charge — krūvinė sąsaja statusas T sritis fizika atitikmenys: angl. charge conjugation vok. Ladungskonjugation, f rus. зарядное сопряжение, n pranc. conjugaison de charge, f … Fizikos terminų žodynas\n\n• chargeconjugation — charge conjugation n. Symbol C 1. A mathematical operator that changes the sign of the charge and of the magnetic moment of every particle in the system to which it is applied. 2. The theoretical conversion of matter to antimatter or of… … Universalium\n\n• CP violation — In particle physics, the violation of the combined conservation laws associated with charge conjugation, C (the operation of turning a particle into its antiparticle), and parity, P, by the weak force. In 1957 it was discovered that parity is… … Universalium\n\n• C-symmetry — In physics, C symmetry means the symmetry of physical laws under a charge conjugation transformation. Electromagnetism, gravity and the strong interaction all obey C symmetry, but weak interactions violate C symmetry. Contents 1 Charge reversal… … Wikipedia\n\n• CP violation — In particle physics, CP violation is a violation of the postulated CP symmetry: the combination of C symmetry (charge conjugation symmetry) and P symmetry (parity symmetry). CP symmetry states that the laws of physics should be the same if a… … Wikipedia"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8146441,"math_prob":0.8068454,"size":3896,"snap":"2019-35-2019-39","text_gpt3_token_len":849,"char_repetition_ratio":0.17959918,"word_repetition_ratio":0.13675214,"special_character_ratio":0.19096509,"punctuation_ratio":0.1171875,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9501268,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-21T13:32:36Z\",\"WARC-Record-ID\":\"<urn:uuid:a44eeb23-1fa0-4b1c-b0b7-5fe01788e526>\",\"Content-Length\":\"47727\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4dbd3f0d-92bc-43ec-a38b-f28a9e4c9c97>\",\"WARC-Concurrent-To\":\"<urn:uuid:b03bf1dd-9cfd-47e0-b335-724c8658d100>\",\"WARC-IP-Address\":\"95.217.42.33\",\"WARC-Target-URI\":\"https://universalium.academic.ru/90986/charge_conjugation\",\"WARC-Payload-Digest\":\"sha1:KICVCAYDBQOEB6N56Y2SLNSJIBKSC4FI\",\"WARC-Block-Digest\":\"sha1:OUTEZHWURZEPWLPEZJGJIW343QGQ7XU2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514574501.78_warc_CC-MAIN-20190921125334-20190921151334-00236.warc.gz\"}"} |
http://www.athoughtabroad.com/2010/10/14/nearest-neighbor-resampling-in-matlab | [
"14. Oktober 2010\n\n# Nearest-Neighbor-Resampling in Matlab\n\nThis article shows the derivation of an algorithm for resizing images in Matlab, using nearest-neighbor interpolation. Scaling an image is mathematically just multiplying image coordinates by a scaling factor. Surprisingly, when dealing with digital images, the scaling process becomes a little bit more complex.\n\n`jea_pygments_txp: File 'resample.m' does not exist.`\n\nHow to find this arithmetic expression? Step by step.\n\nConvert a 1-based index i to a 0-based index i’",
null,
"",
null,
"Convert a 0-based index i’ to a Cartesian coordinate\n\nThe first pixel has coordinate 0.5, the next pixel 1.5 and so on. This provides\nconsistent spacing between pixels and follows the standard image representation\non screens and other media.",
null,
"",
null,
"Scale",
null,
"",
null,
"Chaining the transformations together\n\nNow, given an 1-based index i_f from the input image, how to calculate the corresponding 1-based index i_g in the output image? By chaining the transformations together.",
null,
"By inverting the composite function, we obtain the sought-after function",
null,
"We can now plug in the function definitions",
null,
"Compare this expression with the algorithm."
]
| [
null,
"http://www.athoughtabroad.com/latex2image.php",
null,
"http://www.athoughtabroad.com/latex2image.php",
null,
"http://www.athoughtabroad.com/latex2image.php",
null,
"http://www.athoughtabroad.com/latex2image.php",
null,
"http://www.athoughtabroad.com/latex2image.php",
null,
"http://www.athoughtabroad.com/latex2image.php",
null,
"http://www.athoughtabroad.com/latex2image.php",
null,
"http://www.athoughtabroad.com/latex2image.php",
null,
"http://www.athoughtabroad.com/latex2image.php",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.79357463,"math_prob":0.92687035,"size":1137,"snap":"2019-43-2019-47","text_gpt3_token_len":245,"char_repetition_ratio":0.117387466,"word_repetition_ratio":0.0,"special_character_ratio":0.19700967,"punctuation_ratio":0.11219512,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9835431,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-18T11:11:40Z\",\"WARC-Record-ID\":\"<urn:uuid:cf241892-41ec-4b38-8122-c1247763300d>\",\"Content-Length\":\"21298\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:429610f1-01f1-453c-8e34-730669867c73>\",\"WARC-Concurrent-To\":\"<urn:uuid:d95d7964-0568-4fd2-9dc3-b6367da481fd>\",\"WARC-IP-Address\":\"104.24.110.239\",\"WARC-Target-URI\":\"http://www.athoughtabroad.com/2010/10/14/nearest-neighbor-resampling-in-matlab\",\"WARC-Payload-Digest\":\"sha1:LUSKRF6IYKRR25WSQAQQWBC4CBATRI3V\",\"WARC-Block-Digest\":\"sha1:BU5YXTLEOQ3KLH6RON3MUDXCCCKF7VIK\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496669755.17_warc_CC-MAIN-20191118104047-20191118132047-00315.warc.gz\"}"} |
https://www.stata.com/statalist/archive/2012-01/msg00113.html | [
"",
null,
"Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.\n\n# st: xtlsdvc; Sargan test and heteroscedasticity\n\n From Magnus Gulbrandsen To [email protected] Subject st: xtlsdvc; Sargan test and heteroscedasticity Date Wed, 4 Jan 2012 09:11:02 +0100\n\n```Hi,\n\nI am running a FE regression with LDV, and thus need to correct for\nthe possible Hurwicz bias. I use the -xtlsdvc- command, with option on\ninitial regression -initial(ab)- . After the regressions I have a few\nquandaries:\n\n1) The Sargan test comes out:\n\nSargan test of over-identifying restrictions:\nchi2(54) = 185.87 Prob > chi2 = 0.0000\n\nand if I am correct, this means in plain words that the instruments in\nthe Arrelano-Bond estimation (i.e. first step regression) are not\nvalid(correct?). If so, does this\nmean that the estimates in the LSDVC regressions also are wrong? Are\nthere any solutions to this problem?\n\n2) Econometricians seem to agree that as time observations (T)\nincreases the Hurwicz bias becomes negligible, but identify different\nnumber of T where this is the case (ranging from about 10 to 20). Thus, my\ndataset, where T=12, is a border case. The question is: If the z-values and the\ndifferent coefficients are only marginally different in the pure\nFE+LDV approach and the LSDVC approach with bootstrapped standard\nerrors, should I\ntake this as a sign that the Hurwicz bias is so small that I just as\nwell can carry on with the FE+LDV model?\n\n3) I would like to test for between-group and within-group\nheteroskedasticity. After -xtreg, fe- (with LDV), -xttest3- indicates\ngroup-wise heteroscedasticity, even\nwhen using robust/cluster standard errors (-vce(robust)- or\n-vce(cluster clusterid)-). Any suggestions to how this can be\ncorrected? And, how can i check for within-group heteroscdasticity?\nAre there any parallel tests for\nafter the -xtlsdvc- command?\n\nAll responense will be greatly appreciated.\n\nBest regards,\n\nMagnus Gulbrandsen\n\nMagnus Gulbrandsen\n\nMagnus Gulbrandsen\n\n*\n* For searches and help try:\n* http://www.stata.com/help.cgi?search\n* http://www.stata.com/support/statalist/faq\n* http://www.ats.ucla.edu/stat/stata/\n```"
]
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"https://s7.addthis.com/static/btn/lg-share-en.gif",
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https://homework.cpm.org/category/CON_FOUND/textbook/gc/chapter/1/lesson/1.1.4/problem/1-34 | [
"",
null,
"",
null,
"### Home > GC > Chapter 1 > Lesson 1.1.4 > Problem1-34\n\n1-34.\n\nFor each of the equations below, solve for y in terms of $x$.\n\n1. $2x-3y=12$\n\n$2x−3y= 12$\n$−2x = −2x$\nSubtract $x$ from both sides.\n$−3y= −2x+12$\n\n$−3y=−2x+12$ Divide both sides by $−3$.\n\n$y=\\frac{2}{3}x-4$\n\n1. $5x + 2y = 7$\n\nFollow the same steps to solve for $y$ in terms of $x$ in part (b)."
]
| [
null,
"https://homework.cpm.org/dist/7d633b3a30200de4995665c02bdda1b8.png",
null,
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",
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https://plato.stanford.edu/entries/physics-interrelate/ | [
"# Intertheory Relations in Physics\n\nFirst published Tue Jan 2, 2001; substantive revision Mon Jul 18, 2016\n\nMany issues in the philosophy of science concern the nature of theories and certain relations that may obtain between them. Typically, one is interested in the degree to which a successor to a given theory “goes beyond” (both descriptively and explanatorily) the theory it succeeds. Most often these issues are framed in the context of reductive relations between theories. When does a theory $$T'$$ reduce to a theory $$T$$? How is one to understand the nature of this reduction relation? Interestingly, there are two distinct, yet, related ways of understanding the reductive relationship between $$T$$ and $$T'$$. Thomas Nickles noted this in a paper entitled “Two Concepts of Intertheoretic Reduction.” On the one hand, there is the “philosopher’s” sense of reduction on which the supplanted theory is said to reduce to the newer more encompassing theory. On the other hand, the “physicist’s” sense of reduction puts things the other way. The newer, typically more refined theory is said to reduce to the older typically less encompassing theory in some sort of limit. These two senses of reduction will be discussed in turn.\n\n## 1. Philosopher’s Sense of Reduction\n\nMost contemporary discussions of reductive relations between a pair of theories owe considerable debt to the work by Ernest Nagel. In The Structure of Science, Nagel asserts that “[r]eduction … is the explanation of a theory or a set of experimental laws established in one area of inquiry, by a theory usually though not invariably formulated for some other domain.” (Nagel 1961, 338) The general schema here is as follows:\n\n• $$T$$ reduces $$T'$$ just in case the laws of $$T'$$ are derivable from those of $$T$$.\n\nShowing how these derivations are possible for “paradigm” examples of intertheoretic reduction turns out to be rather difficult.\n\nNagel distinguishes two types of reductions on the basis of whether or not the vocabulary of the reduced theory is a subset of the reducing theory. If it is—that is, if the reduced theory $$T'$$ contains no descriptive terms not contained in the reducing theory $$T$$, and the terms of $$T'$$ are understood to have approximately the same meanings that they have in $$T$$, then Nagel calls the reduction of $$T'$$ by $$T$$ “homogeneous.” In this case, while the reduction may very well be enlightening in various respects, and is part of the “normal development of a science,” most people believe that there is nothing terribly special or interesting from a philosophical point of view going on here. (Nagel 1961, 339.)\n\nLawrence Sklar (1967, 110–111) points out that, from a historical perspective, this attitude is somewhat naive. The number of actual cases in the history of science where a genuine homogeneous reduction takes place are few and far between. Nagel, himself, took as a paradigm example of homogeneous reduction, the reduction of the Galilean laws of falling bodies to Newtonian mechanics. But, as Sklar points out, what actually can be derived from the Newtonian theory are approximations to the laws of the reduced Galilean theory. The approximations, of course, are strictly speaking incompatible with the actual laws and so, despite the fact that no concepts appear in the Galilean theory that do not also appear in the Newtonian theory, there is no deductive derivation of the laws of the one from the laws of the other. Hence, strictly speaking, there is no reduction on the deductive Nagelian model.\n\nOne way out of this problem for the proponent of Nagel-type reductions is to make a distinction between explaining a theory (or explaining the laws of a given theory) and explaining it away. (Sklar 1967, 112–113) Thus, we may still speak of reduction if the derivation of the approximations to the reduced theory’s laws serves to account for why the reduced theory works as well as it does in its (perhaps more limited) domain of applicability. This is consonant with more sophisticated versions of Nagel-type reductions in which part of the very process of reduction involves revisions to the reduced theory. This process arises as a natural consequence of trying to deal with what Nagel calls “heterogeneous” reductions.\n\nThe task of characterizing reduction is more involved when the reduction is heterogeneous—that is, when the reduced theory contains terms or concepts that do not appear in the reducing theory. Nagel takes, as a paradigm example of heterogeneous reduction, the (apparent) reduction of thermodynamics, or at least some parts of thermodynamics, to statistical mechanics. For instance, thermodynamics contains the concept of temperature (among others) that is lacking in the reducing theory of statistical mechanics.\n\nNagel notes that “if the laws of the secondary science [the reduced theory] contain terms that do not occur in the theoretical assumptions of the primary discipline [the reducing theory] … , the logical derivation of the former from the latter is prima facie impossible.” (Nagel 1961, 352) As a consequence, Nagel introduces two “necessary formal conditions” required for the reduction to take place:\n\n1. Connectability. “Assumptions of some kind must be introduced which postulate suitable relations between whatever is signified by ‘A’ [the term to be reduced, that is, an element of the vocabulary of theory $$T'$$] and traits represented by theoretical terms already present in the primary [reducing] science.”\n2. Derivability. “With the help of these additional assumptions, all the laws of the secondary science, including those containing the term ‘A,’ must be logically derivable from the theoretical premises and their associated coordinating definitions in the primary discipline.” (Nagel 1961, 353–354)\n\nThe connectability condition brings with it a number of interpretive problems. Exactly what is, or should be, the status of the “suitable relations,” often called bridge “laws” or bridge hypotheses? Are they established by linguistic investigation alone? Are they factual discoveries? If the latter, what sort of necessity do they involve? Are they identity relations that are contingently necessary or will some sort of weaker relation, such as nomic coextensivity, suffice? Much of the philosophical literature on reduction addresses these questions about the status of the bridge laws.\n\nThe consideration of certain examples lends plausibility to the idea, prevalent in the literature, that the bridge laws should be considered to express some kind of identity relation. For instance, Sklar notes that the reduction of the “theory” of physical optics to the theory of electromagnetic radiation proceeds by identifying one class of entities — light waves — with (part of) another class — electromagnetic radiation. He says “… the place of correlatory laws [bridge laws] is taken by empirically established identifications of two classes of entities. Light waves are not correlated with electromagnetic waves, for they are electromagnetic waves.” (Sklar 1967, 120) In fact, if something like Nagelian reduction is going to work, it is generally accepted that the bridge laws should reflect the existence of some kind of synthetic identity.\n\nKenneth Schaffner calls the bridge laws “reduction functions.” He too notes that they must be taken to reflect synthetic identities since, at least initially, they require empirical support for their justification. “Genes were not discovered to be DNA via the analysis of meaning; important and difficult empirical research was required to make such an identification.” (Schaffner 1976, 614–615)\n\nNow one problem facing this sort of account was forcefully presented by Feyerabend in “Explanation, Reduction, and Empiricism.” (Feyerabend 1962) Consider the term “temperature” as it functions in classical thermodynamics. This term is defined in terms of Carnot cycles and is related to the strict, nonstatistical second law as it appears in that theory. The so-called reduction of classical thermodynamics to statistical mechanics, however, fails to identify or associate nonstatistical features in the reducing theory, statistical mechanics, with the nonstatistical concept of temperature as it appears in the reduced theory. How can one have a genuine reduction, if terms with their meanings fixed by the role they play in the reduced theory get identified with terms having entirely different meanings? Classical thermodynamics is not a statistical theory. The very possibility of finding a reduction function or bridge law that captures the concept of temperature and the strict, nonstatistical, role it plays in the thermodynamics seems impossible.\n\nThe plausibility of this argument, of course, depends on certain views about how meaning accrues to theoretical terms in a theory. However, just by looking at the historical development of thermodynamics one thing seems fairly clear. Most physicists, now, would accept the idea that our concept of temperature and our conception of other “exact” terms that appear in classical thermodynamics such as “entropy,” need to be modified in light of the alleged reduction to statistical mechanics. Textbooks, in fact, typically speak of the theory of “statistical thermodynamics.” The very process of “reduction” often leads to a corrected version of the reduced theory.\n\nIn fact, Schaffner and others have developed sophisticated Nagelian type schemas for reduction that explicitly try to capture these features of actual theory change. The idea is explicitly to include in the model, the “corrected reduced theory” such as statistical thermodynamics. Thus, Schaffner (1976, 618) holds that $$T$$ reduces $$T'$$ if and only if there is a corrected version of $$T'$$, call it $$T'^*$$ such that\n\n1. The primitive terms of $$T'^*$$ are associated via reduction functions (or bridge laws) with various terms of $$T$$.\n2. $$T'^*$$ is derivable from $$T$$ when it is supplemented with the reduction functions specified in 1.\n3. $$T'^*$$ corrects $$T'$$ in that it makes more accurate predictions than does $$T'$$.\n4. $$T'$$ is explained by $$T$$ in that $$T'$$ and $$T'^*$$ are strongly analogous to one another, and $$T$$ indicates why $$T'$$ works as well as it does in its domain of validity.\n\nMuch work clearly is being done here by the intuitive conception of “strong analogy” between the reduced theory $$T'$$ and the corrected reduced theory $$T'^*$$. In some cases, as suggested by Nickles (1973) and Wimsatt (1976), the conception of strong analogy may find further refinement by appeal to what was referred to as the “physicist’s” sense of reduction.\n\n## 2. Physicist’s Sense of Reduction\n\nPhilosophical theories of reduction would have it that, say, quantum mechanics reduces classical mechanics through the derivation of the laws of classical physics from those of quantum physics. Most physicists would, on the other hand, speak of quantum mechanics reducing to classical mechanics in some kind of correspondence limit (e.g., the limit as Planck’s constant $$(h/2\\pi)$$ goes to zero). Thus, the second type of intertheoretic reduction noted by Nickles fits the following schema:\n\n$\\tag*{$${\\bf Schema\\ R}$$} \\lim_{\\varepsilon \\rightarrow 0} T_f = T_c$\n\nHere $$T_f$$ is the typically newer, more fine theory, $$T_c$$ is the typically older, coarser theory, and $$\\varepsilon$$ is a fundamental parameter appearing in $$T_f$$ .\n\nOne must take the equality here with a small grain of salt. In those situations where Schema R can be said to hold, it is likely not the case that every equation or formula from $$T_f$$ will yield a corresponding equation of $$T_c$$ .\n\nEven given this caveat, the equality in Schema R can hold only if the limit is “regular.” In such circumstances, it can be argued that it is appropriate to call the limiting relation a “reduction.” If the limit in Schema R is singular, however, the schema fails and it is best to talk simply about intertheoretic relations.\n\nOne should understand the difference between regular and singular limiting relations as follows. If the solutions of the relevant formulae or equations of the theory $$T_f$$ are such that for small values of $$\\varepsilon$$ they smoothly approach the solutions of the corresponding formulas in $$T_c$$, then Schema R will hold. For these cases we can say that the “limiting behavior” as $$\\varepsilon \\rightarrow 0$$ equals the “behavior in the limit” where $$\\varepsilon = 0$$. On the other hand, if the behavior in the limit is of a fundamentally different character than the nearby solutions one obtains as $$\\varepsilon \\rightarrow 0$$, then the schema will fail.\n\nA nice example illustrating this distinction is the following: Consider the quadratic equation $$x^2 + x - 9\\varepsilon = 0$$. Think of $$\\varepsilon$$ as a small expansion or perturbation parameter. The equation has two roots for any value of $$\\varepsilon$$ as $$\\varepsilon \\rightarrow 0$$. In a well-defined sense, the solutions to this quadratic equation as $$\\varepsilon \\rightarrow 0$$ smoothly approach solutions to the “unperturbed” $$(\\varepsilon = 0)$$ equation $$x^2 + x = 0$$; namely, $$x = 0, -1$$. On the other hand, the equation $$x^2\\varepsilon + x - 9 = 0$$ has two roots for any value of $$\\varepsilon \\gt 0$$ but has for its “unperturbed” solution only one root; namely, $$x = 9$$. The equation suffers a reduction in order when $$\\varepsilon = 0$$. Thus, the character of the behavior in the limit $$\\varepsilon = 0$$ differs fundamentally from the character of its limiting behavior. Not all singular limits result from reductions in order of the equations. Nevertheless, these latter singular cases are much more prevalent than the former.\n\nA paradigm case where a limiting reduction of the form $$\\mathbf{R}$$ rather straightforwardly holds is that of classical Newtonian particle mechanics (NM) and the special theory of relativity (SR). In the limit where $$(v/c)^2\\rightarrow 0$$, SR reduces to NM. Nickles says “epitomizing [the intertheoretic reduction of SR to NM] is the reduction of the Einsteinian formula for momentum,\n\n$p = \\frac{m_0 v}{\\sqrt{1 - (v/c)^2}}$\n\nwhere $$m_0$$ is the rest mass, to the classical formula $$p = m_0 v$$ in the limit as $$v\\rightarrow 0$$.” (Nickles 1973, 182)\n\nThis is a regular limit—there are no singularities or “blowups” as the asymptotic limit is approached. As noted one way of thinking about this is that the exact solutions for small but nonzero values of $$|\\varepsilon$$| “smoothly [approach] the unperturbed or zeroth-order solution [$$\\varepsilon$$ set identically equal to zero] as $$\\varepsilon \\rightarrow 0$$.” In the case where the limit is singular “the exact solution for $$\\varepsilon = 0$$ is fundamentally different in character from the ‘neighboring’ solutions obtained in the limit $$\\varepsilon \\rightarrow 0$$.” (Bender and Orszag 1978, 324)\n\nIn the current context, one can express the regular nature of the limiting relation in the following way. The fundamental expression appearing in the Lorentz transformations of SR, can be expanded in a Taylor series as\n\n$\\frac{1}{\\sqrt{1\\rightarrow(v/c)^2}} = 1 - \\frac{1}{2}(v/c)^2 - \\frac{1}{8}(v/c)^4 - \\frac{1}{16}(v/c)^6 - \\cdots$\n\nand so the limit is analytic. This means that (at least some) quantities or expressions of SR can be written as Newtonian or classical quantities plus an expansion of corrections in powers of $$(v/c)^2$$. So one may think of this relationship between SR and NM as a regular perturbation problem.\n\nExamples like this have led some investigators to think of limiting relations as forming a kind of new rule of inference which would allow one to more closely connect the physicists’ sense of reduction with that of the philosophers’. Fritz Rohrlich, for example, has argued that NM reduces (in the philosophers’ sense) to SR because the mathematical framework of SR reduces (in the physicists’ sense) to the mathematical framework of NM. The idea is that the mathematical framework of NM is “rigorously derived” from that of SR in a “derivation which involves limiting procedures.” (Rohrlich 1988, 303) Roughly speaking, for Rohrlich a “coarser” theory is reducible to a “finer” theory in the philosophers’ sense of being rigorously deduced from the latter just in case the mathematical framework of the finer theory reduces in the physicists’ sense to the mathematical framework of the coarser theory. In such cases, we will have a systematic explication of the idea of “strong analogy” to which Schaffner appeals in his model of philosophical reduction. The corrected theory $$T'^*$$ in this context is the perturbed Newtonian theory as expressed in the Taylor expansion given above. The “strong analogy” between Newtonian theory $$T'$$ and the corrected $$T'^*$$ is expressed by the existence of the regular Taylor series expansion.\n\nAs noted the trouble with maintaining that this relationship between the philosophical and “physical” models of reduction holds generally is that far more often than not the limiting relations between the theories are singular and not regular. In such situations, Schema R fails to hold. Paradigm cases here include the relationships between classical mechanics and quantum mechanics, the ray theory of light and the wave theory, and thermodynamics and statistical mechanics of systems in critical states.\n\n## 3. Hierarchies of Theories\n\nDespite the fact that limiting relations between theories may be singular in this way, it is (at times) useful and appropriate to think of physical theories as forming a hierarchy related by length or energy scales. The idea is that different theories may apply at different length or energy scales. If one takes this idea seriously, then it may very well be the case that each theory in this hierarchy will be phenomenological relative to those theories at higher energies or shorter distances. Equivalently, such a hierarchy may form a tower of effective theories. An effective theory is one that describes the relevant phenomena in a circumscribed domain—a domain characterized by a range of energies, for example.\n\nThe idea of effective theories is not new. In the 19th century and earlier, scientists developed continuum equations such as the Navier-Cauchy equations describing the behavior of isotropic elastic solids and the Navier-Stokes equations for incompressible viscous fluids. These equations were, and still are, remarkably safe. This means that once one inputs the appropriate values for a few phenomenological parameters (such as Young’s modulus and the sheer stress in the Navier-Cauchy equations), one arrives at equation models that allow us to build bridges and buildings that do not collapse. It is remarkable that a theory/model that almost entirely fails to refer to the details of the atomic and molecular structure of a steel beam, say, can be so successful and safe. A question of deep philosophical interest concerns how this can be the case. The phenomenological parameters must encode at least some details about the atomic and molecular make up of the beam. (Hence, the “almost” in the statement above.)\n\nHowever, this raised an important question: Can one tell a story bridging the models at the atomic scale and those at the continuum scale of centimeters and greater? Reductionists typically believe that it is possible to connect, and presumably to derive, the continuum models starting from atomic scale details. There has been a battle for two centuries, at least, between those who are persuaded that such a bottom-up story can be told, and those such as Duhem, Mach, and others who have championed a top-down modeling strategies. In the 19th century this took the form of a heated dispute between so-called rari-constancy and multi-constancy theorists who, respectively, tried to determine the continuum equations from top-down (ignoring unknown micro details) considerations, and theorists trying to determine the continuum equations with small scale atomic assumptions guiding the constructions. In fact, surprisingly, the former prevailed. (Todhunter and Pearson 1960; Batterman 2012)\n\nThe debate between bottom-up, reductionist modelers and top-down, continuum modelers receives its modern presentation, at least in part, in the debates about the existence and nature of emergent phenomena. One area of recent interest where this occurs is in our understanding of effective quantum field theories.\n\nIn quantum field theory, for instance, there has been considerable success in in showing how a theory appropriate for some range of energy scales is related to a theory for another range via a process of renormalization (Bain 2012). Renormalization provides a kind of limiting relationship between theories at different scales despite the fact that the reductive Schema R typically fails because of divergences related to singular limits. The physics at one scale is relatively independent of that at some higher energy (shorter length). In effect, renormalization is a mathematical scheme for characterizing how the structure of interactions changes with changing scale: it turns out that the domain characterized by some lower energy (or larger length) scale is surprisingly and remarkably decoupled from that of higher energies (or smaller lengths). In other words, the decoupling entails that the higher energy regime does not much effect the behaviors and character of the lower energy regimes.\n\nNew work, more generally on the problem modeling systems at widely different scales (10+ orders of magnitude), in nano chemistry and in materials science, brings some hope that the all-or-nothing dichotomy between reduction and emergence can be somewhat blunted. As noted, a question of real philosophical interest concerns how to understand the relative autonomy of theories and models at large scales. (Why, again, are the continuum equations so safe for large scale modeling?) Contemporary work in applied mathematics on so-called homogenization theory is beginning to provide interesting connections across these widely separated scales. (Torquato 2002; Phillips 2001)\n\nThe mathematics of renormalization is best understood as an instance of this general strategy for homogenization or upscaling. (Batterman 2012) It is crucial for a contemporary understanding of relations between theories. It is fair to say, however, that being able to understand such intertheoretic relations via homogenization and renormalization techniques does not entail the existence of reductive relations between the theories either in the philosophers’ or the physicists’ sense of the term. However, such an understanding may very well lead to a more nuanced and precise characterization of the debates about reduction and emergence.\n\n## 4. Intertheory Relations\n\nIt seems reasonable to expect something like philosophical reductions to be possible in those situations where Schema R holds. On the other hand, neither philosophical nor “physical” reduction seems possible when the limiting correspondence relation between the theories is singular. Perhaps in such cases it is best to speak simply of intertheoretic relations rather than reductions. It is here that much of philosophical and physical interest is to be found. This claim and the following discussion should not be taken to be anything like the received view among philosophers of science. Instead, they reflect the views of the author.\n\nNevertheless, here is a passage from a recent paper by Michael Berry which expresses a similar point of view.\n\nEven within physical science, reduction between different levels of explanation is problematic—indeed, it is almost always so. Chemistry is supposed to have been reduced to quantum mechanics, yet people still argue over the basic question of how quantum mechanics can describe the shape of a molecule. The statistical mechanics of a fluid reduces to its thermodynamics in the limit of infinitely many particles, yet that limit breaks down near the critical point, where liquid and vapour merge, and where we never see a continuum no matter how distantly we observe the particles … . The geometrical (Newtonian) optics of rays should be the limit of wave optics as the wavelength becomes negligibly small, yet … the reduction (mathematically similar to that of classical to quantum mechanics) is obstructed by singularities … .\n\nMy contention … will be that many difficulties associated with reduction arise because they involve singular limits. These singularities have both negative and positive aspects: they obstruct the smooth reduction of more general theories to less general ones, but they also point to a great richness of borderland physics between theories. (Berry 2001, 43)\n\nWhen Schema R fails this is because the mathematics of the particular limit $$(\\varepsilon \\rightarrow 0)$$ is singular. One can ask what, physically, is responsible for this mathematical singularity. In investigating the answer to this question one will often find that the mathematical blow-up reflects a physical impossibility. For instance, if Schema R held when $$T_f$$ is the wave theory of light and $$T_c$$ is the ray theory (geometrical optics), then one would expect to recover rays in the shortwave limit $$\\lambda \\rightarrow 0$$ of the wave theory. On the ray theory, rays are the carriers of energy. But in certain situations families of rays can focus on surfaces or lines called “caustics.” These are not strange esoteric situations. In fact, rainbows are, to a first approximation, described by the focusing of sunlight on these surfaces following its refraction and reflection through raindrops. However, according to the ray theory, the intensity of the light on these focusing surfaces would be infinite. This is part of the physical reason for the mathematical singularities. See also the discussion of the rainbow by Pincock 2011, and Belot 2005.\n\nOne is led to study the asymptotic domain in which the parameter $$\\varepsilon$$ in Schema R approaches 0. In the example above, this is the short wavelength limit. Michael Berry (1980; 1990; 1994a; 1994b) has done much research on this and other asymptotic domains. He has found that in the asymptotic borderlands between such theories there emerge phenomena whose explanation requires in some sense appeal to a third intermediate theory. This is a claim (Batterman 2002) that when taken literally, has raised a number of hackles in the literature. However, understood in terms of the mathematics of characteristics and wavefronts, as was originally intended, the current author believes some of the debates are misdirected. The emergent structures (the rainbow itself is one of them) are not fully explainable either in terms of the finer wave theory or in terms of the ray theory alone. Instead, aspects of both theories (through asymptotic investigation of the wave equations) are required for a full understanding of these emergent phenomena.\n\nThis fact calls into question certain received views about the nature of intertheoretic relations. The wave theory, for example, is surely the fundamental theory. Nevertheless, these considerations seem to show that that theory is itself explanatorily deficient. There are phenomena within its scope whose explanations require examining the asymptotics of the appropriate equation. This involves paying attention to mathematical structures called characteristics and wavefronts. See Bóna and Slawinski 2011. These mathematical investigations of the deep asymptotic structure of hyperbolic equations are not at all like the straightforward derivations from initial data that are typical of in principle derivations often referred to in carrying out the dictates of Nagel style explanatory reductions. A similar situation arises in the asymptotic domain between quantum mechanics and classical mechanics where Planck’s constant can be considered asymptotically small. (See Belot 2005 for an alternative point of view.)\n\nThere is much here worthy of further philosophical study. Some very recent work by Butterfield (2011), Butterfield and Bouatta (2011), Norton (2012), Menon and Callender (2012) challenges the point of view suggested by the above discussion. These authors address issues about the nature of infinite idealizations, reduction, and emergence. A common theme is that it is possible to reconcile emergence and reduction. By and large these authors adopt a Nagelian sense of reduction as definitional extension. For a contrary point of view one can see Batterman (2002; 2012).\n\n## Bibliography\n\n• Bain, Jonathan, 2012, “Effective Field Theories,” in Robert Batterman (ed.), The Oxford Handbook of Philosophy of Physics, Oxford: Oxford University Press, pp. 224–254.\n• Batterman, R.W., 1991, “Chaos, quantization, and the correspondence principle”, Synthese, 89: 189–227.\n• –––, 1993, “Quantum chaos and semiclassical mechanics”, in PSA 1992, volume 2, pages 50–65. Philosophy of Science Association.\n• –––, 1995, “Theories between theories: Asymptotic limiting intertheoretic relations”, Synthese, 103: 171–201.\n• –––, 2002, The Devil in the Details: Asymptotic Reasoning in Explanation, Reduction, and Emergence. Oxford University Press, New York.\n• –––, 2012 “The Tyranny of Scales,” in Robert Batterman (ed.), The Oxford Handbook of Philosophy of Physics, Oxford: Oxford University Press, pp. 255–286.\n• Belot, Gordon, 2005, “Whose Devil? Which Details?,” Philosophy of Science, 72: 128–153.\n• Bender, C.M., and Orszag, S.A., 1978, Advanced Mathematical Methods for Scientists and Engineers. McGraw-Hill, New York.\n• Berry, M.V., 1990, “Beyond rainbows”, Current Science, 59/(21–22): 1175–1191.\n• –––, 1991, “Asymptotics, singularities and the reduction of theories”, in Dag Prawitz, Brian Skyrms, and Dag Westerståhl (eds.), Logic, Methodology and Philosophy of Science, IX: Proceedings of the Ninth International Congress of Logic, Methodology and Philosophy of Science, Uppsala, Sweden, August 7–14, 1991 (Studies in Logic and Foundations of Mathematics: Volume 134), Amsterdam: Elsevier Science B. V., 1994, 597–607.\n• –––, 1994, “Singularities in waves and rays”, in R. Balian, M. Kléman, and J. P. Poirier (eds), Physics of Defects (Les Houches, Session XXXV, 1980), pages 453–543, Amsterdam, 1994. North-Holland.\n• –––, 2001, “Chaos and the Semiclassical Limit of Quantum Mechanics (Is the Moon There When Somebody Looks?)”, in Quantum Mechanics: Scientific perspectives on divine action (eds: Robert John Russell, Philip Clayton, Kirk Wegter-McNelly and John Polkinghorne), Vatican Observatory CTNS publications, pp. 41–54.\n• Berry, M. V., 2002 “Singular Limits” Physics Today, May pp 10–11.\n• Berry, M.V. and Upstill, C., 1980, “Catastrophe optics: Morphologies of caustics and their diffraction patterns”, in E. Wolf (ed), Progress in Optics, volume XVIII, pages 257–346, Amsterdam, 1980. North-Holland.\n• Bokulich, Alisa, (2008) “Can Classical Structures Explain Quantum Phenomena?”, The British Journal for the Philosophy of Science, 59, 217–235.\n• Bokulich, Alisa, (2008a) Reexamining the Quantum-Classical Relation: Beyond Reductionism and Pluralism, Cambridge: Cambridge University Press.\n• Bóna, Andrej and Slawinski, Michael A., 2011, Wavefronts and Rays as Characteristics and Asymptotics, World Scientific: Singapore.\n• Butterfield, Jeremy, 2011 “Less is Different: Emergence and Reduction Reconciled”, Foundations of Physics, 41(6): 1065–1135.\n• Butterfield, Jeremy and Bouatta, Nazim, 2011 “Emergence and Reduction Combined in Phase Transitions.” arXiv:1104.1371v2.\n• Menon, Tarun and Callender, Craig, 2012 “Turn and Face the Strange . . . Ch-ch-changes”, in Robert Batterman (ed.), The Oxford Handbook of Philosophy of Physics, Oxford: Oxford University Press, pp. 189–223.\n• Cao, Tian Yu, 1993 “New philosophy of renormalization: From the renormalization group equations to effective field theories”, in Laurie M. Brown, editor, Renormalization: From Lorentz to Landau (and Beyond). Springer-Verlag, New York.\n• Castellani, Elena, 2002 “Reductionism, emergence, and effective field theories”, Studies in History and Philosophy of Modern Physics, 33: 251–267.\n• Emch, Gerard G. and Liu, Chuang, 2002, The Logic of Thermostatistical Physics, Springer-Verlag, Berlin.\n• Feyerabend, P.K., 1962, “Explanation, reduction, and empiricism”, in H. Feigl and G. Maxwell (eds), Minnesota Studies in the Philosophy of Science, volume 3, pages 28–97. D. Reidel Publishing Company.\n• Nagel, E., 1961, The Structure of Science. Routledge and Kegan Paul, London.\n• Nickles, T., 1973, “Two concepts of intertheoretic reduction”, The Journal of Philosophy, 70/7: 181–201.\n• Norton, John, 2012, “Approximation and Idealization: Why the Difference Matters,” Philosophy of Science, 79: 207–232.\n• Phillips, Rob, 2001, Crystals, Defects, and Microstructures: Modeling Across Scales, Cambridge: Cambridge University Press.\n• Pincock, Christopher, 2011, “Mathematical explanations of the rainbow,” Studies in History and Philosophy of Modern Physics, 42(1) 13–22.\n• Rohrlich, F., 1988, “Pluralistic ontology and theory reduction in the physical sciences”, The British Journal for the Philosophy of Science, 39: 295–312.\n• Schaffner, K. 1976, “Reductionism in biology: Prospects and problems”, in R.S. Cohen, et al. (eds), PSA 1974, pages 613–632. D. Reidel Publishing Company.\n• Sklar, L., 1967, “Types of inter-theoretic reduction”, The British Journal for the Philosophy of Science, 18: 109–124.\n• –––, 1993, Physics and Chance: Philosophical Issues in the Foundations of Statistical Mechanics. Cambridge University Press, Cambridge.\n• Todhunter, Isaac and Karl Pearson (eds.), 1960, A History of the Theory of Elasticity and of the Strength of Materials from Galilei to Lord Kelvin, volume 1: Galilei to Saint-Venant 1639–1850, Dover.\n• Torquato, Salvatore, 2002, d Random Heterogeneous Materials: Microstructure and Macroscopic Properties, New York: Springer.\n• Wimsatt, W. C., 1976, “Reductive Explanation: A Functional Account”, in A. C. Michalos, C. A. Hooker, G. Pearce, and R. S. Cohen (eds.), PSA-1974 (Boston Studies in the Philosophy of Science, volume 30) Dordrecht: Reidel, pp. 671–710.",
null,
"How to cite this entry.",
null,
"Preview the PDF version of this entry at the Friends of the SEP Society.",
null,
"Look up this entry topic at the Indiana Philosophy Ontology Project (InPhO).",
null,
"Enhanced bibliography for this entry at PhilPapers, with links to its database."
]
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"https://plato.stanford.edu/symbols/sepman-icon.jpg",
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"https://plato.stanford.edu/symbols/sepman-icon.jpg",
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"https://plato.stanford.edu/symbols/inpho.png",
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"https://plato.stanford.edu/symbols/pp.gif",
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https://codingthesmartway.com/tensorflow-js-crash-course-machine-learning-for-the-web-handwriting-recognition/ | [
"Machine Learning\n\n# TensorFlow.js Crash Course – Machine Learning For The Web – Handwriting Recognition\n\nWelcome to the second episode of the CodingTheSmartWay.com TensorFlow.js Crash Course for absolute beginners.\n\nIn the first part TensorFlow.js Crash Course – Machine Learning For The Web – Getting Started we’ve covered the following topics:\n\n• What TensorFlow.js is\n• How TensorFlow.js can be used to add machine learning capabilities to your web application\n\nIn this part we’re going one step further and will explore another use case: the recognition of handwritten digits. Therefore it is assumed that you’re familiar with the the basic building blocks of TensorFlow.js which have been introduced in the first episode.\n\nIf you like CodingTheSmartWay, then consider supporting us via Patreon. With your help we’re able to release developer tutorial more often. Thanks a lot!",
null,
"## What We’re Going To Build\n\nLet’s take a look at the application which we’re going to build in this tutorial. The application will use the MNIST data set train a neural network. The model is being built and trained when the website is loaded. The progress can be seen in the Log Output area:",
null,
"Once the training procedure is completed the user is informed with the message “Training complete” and the button in the Predict area is activated.",
null,
"Pressing the button randomly selects one dataset fromn the MNIST data source to perform a prediction with the trained model. The output looks like the following:",
null,
"The image of the handwritten digit is presented, the original value and the predicted value is outputted. If the prediction is correct the text “Value recognized successfully” is visible as well. This shows us, that the trained neural network was able to recognize the digit from the image correctly.\n\nThe user is able to use the button multiple times. The output is extended as you can see in the following screenshot:",
null,
"## The MNIST Database Of Handwritten Digits\n\nMNIST is a data set which contains the images of handwritten digits from 0–9. We’ll use that database of images to train the model of our application. Furthermore we’ll make use of randomly selected images from the MINSIT data set to test if the neural network is able to perform predictions.\n\n## Preparing The Project\n\nAgain let’s start with setting up the project by creating a new folder:\n\n`\\$ mkdir tfjs02`\n\nChange into that newly created project folder:\n\n`\\$ cd tfjs02`\n\nInside the folder we’re now ready to create a package.json file, so that we’re able to manage dependencies by using the Node.js Package Manager:\n\n`\\$ npm init -y`\n\nBecause we’ll be installing the dependencies (e.g. the Tensorflow.js library) locally in our project folder we need to use a module bundler for our web application. To keep things as easy as possible we’re going to use the Parcel web application bundler because Parcel is working with zero configuration. Let’s install the Parcel bundler by executing the following command in the project directory:\n\n`\\$ npm install -g parcel-bundler`\n\nNext, let’s create two new empty files for our implementation:\n\n`\\$ touch index.html index.js`\n\nFinally let’s add the Bootstrap library as a dependency because we will be using some Bootstrap CSS classes for our user interface elements:\n\n`\\$ npm install bootstrap`\n\nNow let’s add further depependencies to the project to make sure that we’re able to use latest EcmaScript features like async / await:\n\n`npm install --save-dev babel-plugin-transform-runtime babel-runtime`\n\nCreate a .babelrc file and add:\n\n```{\n\"plugins\": [\n[\"transform-runtime\",\n{\n\"polyfill\": false,\n\"regenerator\": true\n}]\n]\n}```\n\nLast but not least we do not need to forget to install TensorFlow.js as well:\n\n`\\$ npm install @tensorflow/tfjs`\n\n## Building The Convolutional Neural Network Model\n\n### Creating A Sequential Model Instance\n\nBefore we’re starting to built the convolutional neural network model we’re defining a variable model which will hold the model and a function createModel which will contain the code which is needed to create and compile the machine learning model:\n\n``````var model;\n\nfunction createModel() {\n// Insert the following pieces of code here\n}``````\n\nLet’s first create the sequential model instance as already learned in episode 1 of this series and insert the following code in function createModel:\n\n``````createLogEntry('Create model ...');\nmodel = tf.sequential();\ncreateLogEntry('Model created');``````\n\nAdditionally we’re making use of a function named createLogEntry. This function is will be implemented later on and is used to output text messages to the Log Output area.\n\nFirst, let’s add a two-dimensional convolutional layer by using the following code:\n\n``````createLogEntry('Add layers ...');\ninputShape: [28, 28, 1],\nkernelSize: 5,\nfilters: 8,\nstrides: 1,\nactivation: 'relu',\nkernelInitializer: 'VarianceScaling'\n}));``````\n\nThe layer is created via tf.layers.conv2d. The configure the layer a configuration object is passed as a parameter to this method. The new layer is added to the model by passing it into the call of the method model.add.\n\nThe configuration object which is passed to conv2d is containing six configuration properties in total:\n\n• inputShape: This is the shape of the input data of the first layer. The MNIST data is containing images of shape 28×28-pixels. The color of the pixels can just be black and write, so we’re assigning the shape [28, 28, 1] here.\n• kernelSize: The kernelSize value if the size of the filter window of the convolutional layer which is applied to the input data. We’re using the value 5 here to define a square filter windows of 5×5 pixels.\n• filters: This is the number of filter windows (of size kernelSize) which are applied to the input data.\n• strides: This value specifies by how many pixels the filter window is sliding over the input image.\n• activation: The activation function which is applied to the data once the filter windows have been applied. Here we’re using the Reactified Linear Unit (ReLU) funtion, which is a very common activation function in machine learning.\n• kernelInitializer: We’re using VarianceScaling (which is a common initializer) to initialize the model weight’s.\n\nThe next layer we’re going to add to our neural network model is a two dimensional max pooling layer. We’re using that layer to down-sample the image so it is half the size of the input from the previous layer by defining the max pooling layer in the following way:\n\n``````model.add(tf.layers.maxPooling2d({\npoolSize: [2, 2],\nstrides: [2, 2]\n}));``````\n\nThe layer is configured by passing over a configuration object with two configuration properties:\n\n• poolSize: This is the size of the sliding windows (2×2 pixels) which is applied to the input.\n• strides: This value specifies by how many pixels the filter window is sliding over the input image.\n\nSince both values are set to 2,2, the pooling windows is completely non-overlapping. As a result this will cut the size of the input from the previous layer in half.\n\nA common pattern in convolutional neural network models used for image recognition is to repeat the first convolutional layer and the second max pooling layer. So let’s add again a two dimensional convolutional layer as the third layer in our model:\n\n``````model.add(tf.layers.conv2d({\nkernelSize: 5,\nfilters: 16,\nstrides: 1,\nactivation: 'relu',\nkernelInitializer: 'VarianceScaling'\n}));``````\n\nThis time we do not need to define the input shape because the shape is determined by the output shape of the previous layer automatically.\n\nThe fourth layer is again a max pooling layer to further down-sample the result:\n\n``````model.add(tf.layers.maxPooling2d({\npoolSize: [2, 2],\nstrides: [2, 2]\n}));``````\n\nHaving repeated the pattern of a convolutional layer and a max pooling layer a second time brings us now to the point to add a flatten layer as the fifth layer in our model:\n\n``model.add(tf.layers.flatten());``\n\nThis layer will flatten the output from the previous layer to a vector.\n\n### Adding A Dense Layer (Fully Connected Layer) For Performing The Final Classification\n\nThe final layer which is added to out model is a dense layer (fully connected layer). This layer will perform the final classification:\n\n``````model.add(tf.layers.dense({\nunits: 10,\nkernelInitializer: 'VarianceScaling',\nactivation: 'softmax'\n}));\ncreateLogEntry('Layers created');``````\n\nThe dense layer configuration consists of the following properties:\n\n• units: The size of the output. As we’d like to do a 10-class classification to predict digitis between zero and nine (0-9) we’re setting the value to 10.\n• kernelInitializer: Set to VarianceScaling\n• activation: The activation function which is used for classification. The softmax activation function creates a propability distribution over the 10 classes.\n\n### Compiling The Model\n\nAll needed layers have been added to the model. Before we’re going to train the model with MNIST data sets we need to make sure that the model is compiled:\n\n``````createLogEntry('Start compiling ...');\nmodel.compile({\noptimizer: tf.train.sgd(0.15),\nloss: 'categoricalCrossentropy'\n});\ncreateLogEntry('Compiled');``````\n\nThe object which is passed to the call of model.compile is containing two properties:\n\n• optimizer: The convolutional neural network model will make use of a SGD (Stochastic Gradient Descent) optimizer with a learning rate of 0.15.\n• loss: As loss function we choose categoricalCrossentropy which is often used for classification tasks.\n\nLet’s start training the model with MNIST data sets of handwritten digits. To access the MNIST data from a remote server we’re using the MnistData class from the project https://github.com/tensorflow/tfjs-examples/tree/master/mnist. To make that class available just download the file data.js from that repository and insert that file in our project directory. In index.js use the following import statement to make the MnistData class available:\n\n``import {MnistData} from './data';``\n\nThe data should be kept in a variable named data. A load function is added to our application to load the data by calling the MnistData method load:\n\n``````let data;\ndata = new MnistData();\n}``````\n\nWith the MNIST data records available we’re now ready to prepare for training. Let’s first define two constants:\n\n``````const BATCH_SIZE = 64;\nconst TRAIN_BATCHES = 150;``````\n\nThe training will not be performed in one operation. Instead we’ll perform the training in batches of data. The size of the batch and the number of batches to be trained is defined by those constants. The training logic is encapsulated in function train:\n\n``````async function train() {\ncreateLogEntry('Start training ...');\nfor (let i = 0; i < TRAIN_BATCHES; i++) {\nconst batch = tf.tidy(() => {\nconst batch = data.nextTrainBatch(BATCH_SIZE);\nbatch.xs = batch.xs.reshape([BATCH_SIZE, 28, 28, 1]);\nreturn batch;\n});\n\nawait model.fit(\nbatch.xs, batch.labels, {batchSize: BATCH_SIZE, epochs: 1}\n);\n\ntf.dispose(batch);\n\nawait tf.nextFrame();\n}\ncreateLogEntry('Training complete');\n}``````\n\n## Implementing the User Interface\n\nIn the next step let’s add the HTML / CSS code which is needed to implement the user interface of our application in index.html:\n\n``````<html>\n<body>\n<style>\n.prediction-canvas{\nwidth: 100px;\nmargin: 20px;\n}\n.prediction-div{\ndisplay: inline-block;\nmargin: 10px;\n}\n</style>\n<div class=\"card\">\n<strong>TensorFlow.js Demo - Handwriting Recognition</strong>\n</div>\n<div class=\"card-body\">\n<div class=\"card\">\n<div class=\"card-body\">\n<h5 class=\"card-title\">Log Output:</h5>\n<div id=\"log\"></div>\n</div>\n</div>\n<br>\n<div class=\"card\">\n<div class=\"card-body\">\n<h5 class=\"card-title\">Predict</h5>\n<button type=\"button\" class=\"btn btn-primary\" id=\"selectTestDataButton\" disabled>Please wait until model is ready ...</button>\n<div id=\"predictionResult\"></div>\n</div>\n</div>\n</div>\n</div>\n</div>\n\n<script src=\"./index.js\"></script>\n</body>\n</html>``````\n\nHere we’re making use of various Bootstrap CSS classes.\n\nFor the output which is written to the log output area a <div> element is added as a placeholder with ID log. The button which the user can use to perform a prediction based on a randomly selected MINST data set gets assigned the id selectTestDataButton.\n\nThe output area for the prediction result is the <div> element with ID predictionResult.\n\nThe createLogEntry function has already been used several times to output messages in the log area. Now let’s add the missing implementation of that function in index.js as well:\n\n``````function createLogEntry(entry) {\ndocument.getElementById('log').innerHTML += '<br>' + entry;\n}``````\n\nFinally let’s bring everything in order and implement function main to call createModel, load and train.\n\n``````async function main() {\ncreateModel();\nawait train();\ndocument.getElementById('selectTestDataButton').disabled = false;\ndocument.getElementById('selectTestDataButton').innerText = \"Randomly Select Test Data And Predict\";\n}\n\nmain();``````\n\nFurthermore we’re making sure that the button is enabled after the training procedure has been performed successfully.\n\n## Prediction\n\nLet’s move on to the final task and add the code which is needed to perform the predict based on our trained convolutional neural network. Therefore we’re adding the predict function in the following way:\n\n``````async function predict(batch) {\ntf.tidy(() => {\nconst input_value = Array.from(batch.labels.argMax(1).dataSync());\n\nconst div = document.createElement('div');\ndiv.className = 'prediction-div';\n\nconst output = model.predict(batch.xs.reshape([-1, 28, 28, 1]));\nconst prediction_value = Array.from(output.argMax(1).dataSync());\n\nconst image = batch.xs.slice([0,0], [1, batch.xs.shape]);\n\nconst canvas = document.createElement('canvas');\ncanvas.className = 'prediction-canvas';\ndraw(image.flatten(), canvas);\n\nconst label = document.createElement('div');\nlabel.innerHTML = 'Original Value: ' + input_value;\nlabel.innerHTML += '<br>Prediction Value: ' + prediction_value;\nconsole.log(prediction_value + '-' + input_value);\nif (prediction_value - input_value == 0) {\nlabel.innerHTML += '<br>Value recognized successfully!';\n} else {\nlabel.innerHTML += '<br>Recognition failed!';\n}\n\ndiv.appendChild(canvas);\ndiv.appendChild(label);\ndocument.getElementById('predictionResult').appendChild(div);\n});\n}``````\n\nPart of the output is the image of the handwritten digit. The draw the image we’re making use of the custom draw function. The implementation of that function needs to be added to index.js as well:\n\n``````function draw(image, canvas) {\nconst [width, height] = [28, 28];\ncanvas.width = width;\ncanvas.height = height;\nconst ctx = canvas.getContext('2d');\nconst imageData = new ImageData(width, height);\nconst data = image.dataSync();\nfor (let i = 0; i < height * width; ++i) {\nconst j = i * 4;\nimageData.data[j + 0] = data[i] * 255;\nimageData.data[j + 1] = data[i] * 255;\nimageData.data[j + 2] = data[i] * 255;\nimageData.data[j + 3] = 255;\n}\nctx.putImageData(imageData, 0, 0);\n}``````\n\nFinally we need to add the click event handler function for the selectTestDataButton:\n\n``````document.getElementById('selectTestDataButton').addEventListener('click', async (el, ev) => {\nconst batch = data.nextTestBatch(1);\nawait predict(batch);\n});``````\n\nInside this function we’re using method nextTestBatch from the MnistData class to retrieve a batch of test data of size 1 (which means that only one data set is included). Next we’re calling the asynchronous predict function by using the keyword await and passing of the test data set.\n\n### TensorFlow Online Courses",
null,
"Learn how to use Google’s Deep Learning Framework – TensorFlow – with the best online courses available. Become a Deep Learning Guru today!\n\n• Understand how Neural Networks Work\n• Use TensorFlow for Classification and Regression Tasks\n\n• Use TensorFlow for Image Classification with Convolutional Neural Networks\n\nCheck out the top TensorFlow developer online courses!\n\nUsing and writing about best practices and latest technologies in web design & development is my passion.",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.73043245,"math_prob":0.8149797,"size":16058,"snap":"2019-51-2020-05","text_gpt3_token_len":3528,"char_repetition_ratio":0.122399405,"word_repetition_ratio":0.045321025,"special_character_ratio":0.22960518,"punctuation_ratio":0.1443922,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96699977,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-24T13:33:30Z\",\"WARC-Record-ID\":\"<urn:uuid:6d7ce464-7ee1-4c53-a75c-7165923755eb>\",\"Content-Length\":\"69823\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8c2c448c-c261-4eae-a4b4-42d5cc9560ac>\",\"WARC-Concurrent-To\":\"<urn:uuid:a943c4c1-1d7f-4b3f-9756-260bb4d99b08>\",\"WARC-IP-Address\":\"85.13.133.44\",\"WARC-Target-URI\":\"https://codingthesmartway.com/tensorflow-js-crash-course-machine-learning-for-the-web-handwriting-recognition/\",\"WARC-Payload-Digest\":\"sha1:HBA6K4Y76ULSPNESPX3BLQOKSUYBUYTI\",\"WARC-Block-Digest\":\"sha1:M5HHAVH2BSKRD4R4DFYHX43ZTOPO3CWC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250620381.59_warc_CC-MAIN-20200124130719-20200124155719-00315.warc.gz\"}"} |
https://forum.amibroker.com/t/help-with-automatic-fibonacci-exploration-code-tweak/18636 | [
"# Help with Automatic Fibonacci Exploration Code Tweak\n\nI am making small teaks to a common Automatic Fibonacci script to enable it too much much more effective with exploration. It works well for the most part but has one flaw that needs to be fixed when the higher pivot and low pivot to do coincide with the start of the range been above or below.\n\nIn order to place the internal retracement lines the correct way, the code says if(StartBar < FinishBar), then do it one way else the other way.\n\nHowever, It is not entirely about whether the start and end periods are above or below, but where the HHV and LLV are within that range.\n\nCORRECT",
null,
"INCORRECT",
null,
"So the code needs to factor in within the \"range\" StartBar to FinishBar, whether the HHV before or after the LLV, and then set the internal retracement accordingly.\n\n``````_SECTION_BEGIN(\"Fib_Levels\");\n\nSetChartOptions(0,chartShowArrows|chartShowDates);\n\nPlot(C,\"\", IIf(C>O,colorGreen,colorRed) , styleCandle);\n// Get values for fib levels\n\n//StartBar=SelectedValue(BarIndex());\nFinishBar = EndValue( BarIndex() );\nStartBar = FinishBar-80;\ni = startbar;\nperiod = FinishBar - StartBar;\n\nLo = LLV(L,period);\nHi = HHV(H,period);\nLine0 = 0;\nLine1 = 0;\nLine2 = 0;\nLine3 = 0;\nLine4= 0;\nLine100 = 0;\n\nfor( i = startbar; i < finishbar; i++ )\n{\nif(StartBar < FinishBar)\n{\nLine0 = EndValue(Lo);\nLine100 = EndValue(Hi);\nLine1 = Line0 + abs(Line100-Line0)*0.236;\nLine2 = Line0 + abs(Line100-Line0)*0.382;\nLine3 = Line0 + abs(Line100-Line0)*0.5;\nLine4 = Line0 + abs(Line100-Line0)*0.618;\n\n}\nelse\n{\nLine0 = EndValue(Hi);\nLine100 = EndValue(Lo);\nLine1 =Line0 - abs(Line100-Line0)*0.236;\nLine2 = Line0 - abs(Line100-Line0)*0.382;\nLine3 = Line0 - abs(Line100-Line0)*0.5;\nLine4 = Line0 - abs(Line100-Line0)*0.618;\n\n}\n}\n\n// external fib lines begining fom selecetdbarindex()\nfib100= LineArray(startbar, Line0, finishbar, Line0, 0, 1);\nfib0 = LineArray(startbar, Line100, finishbar, Line100, 0, 1);\n\n// depth of middle lines\nn= round((finishbar-startbar)/2);\n\n// middle lines\nfib1= LineArray((finishbar-n), Line1, finishbar, Line1, 0, 1);\nfib2= LineArray((finishbar-n), Line2, finishbar, Line2, 0, 1);\nfib3= LineArray((finishbar-n), Line3, finishbar, Line3, 0, 1);\nfib4= LineArray((finishbar-n), Line4, finishbar, Line4, 0, 1);\n\nPlot(fib0,\"Base\", colorGrey50);\nPlot(fib100,\"Top\", colorGrey50);\nPlot(fib1,\"23.6%\", colorPink);\nPlot(fib2,\"38.2%\", colorLightBlue);\nPlot(fib3,\"50%\", colorRed);\nPlot(fib4,\"61.8%\", colorGold);\n\nPlotText(\" 0.00% \",FinishBar+2, Line0,colorWhite);\nPlotText(\" 23.6% \",FinishBar+2, Line1,colorGreen);\nPlotText(\" 38.6% \",FinishBar+2, Line2,colorBlue);\nPlotText(\" 50% \",FinishBar+2, Line3,colorGold);\nPlotText(\" 61.8% \",FinishBar+2, Line4,colorAqua);\nPlotText(\" 100% \",FinishBar+2, Line100,colorRed);\n``````\n1 Like"
]
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null,
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null
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https://primenumbers.info/33856.htm | [
"# Prime Numbers\n\n## Is number 33856 a prime number?\n\nNumber 33856 is not a prime number. It is a composite number.\n\nA prime number is a natural number greater than 1 that is not a product of two smaller natural numbers. We do not consider 33856 as a prime number, because it can be written as a product of two smaller natural numbers (check the factors of number 33856 below).\n\n### Other properties of number 33856\n\nNumber of factors: 21.\n\nList of factors/divisors: 1, 2, 4, 8, 16, 23, 32, 46, 64, 92, 184, 368, 529, 736, 1058, 1472, 2116, 4232, 8464, 16928, 33856.\n\nParity: 33856 is an even number. It means it is divisible by 2 with no remainder.\n\nPerfect square: yes, because 33856=1842 (a square number or perfect square is an integer that is the square of an integer).\n\nPerfect number: no, because the sum of its proper divisors is 36375 (perfect number is a positive integer that is equal to the sum of its proper divisors).\n\n Number:Prime number: 33849no 33850no 33851yes 33852no 33853no 33854no 33855no 33856no 33857yes 33858no 33859no 33860no 33861no 33862no 33863yes"
]
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http://ayylm.io/flipcode/www.flipcode.com/archives/Rectangle_Placement.html | [
"",
null,
"This section of the archives stores flipcode's complete Developer Toolbox collection, featuring a variety of mini-articles and source code contributions from our readers.",
null,
"Rectangle Placement Submitted by",
null,
"You have a bunch of rectangular pieces. You need to arrange them in a rectangular surface so that they don't overlap, keeping the total area of the rectangle as small as possible. This is fairly common when arranging characters in a bitmapped font, lightmaps for a 3D engine, and I guess other situations as well.\n\nThe idea of this algorithm is that, as we add rectangles, we can pre-select \"interesting\" places where we can try to add the next rectangles. For optimal results, the rectangles should be added in order. I initially tried using area as a sorting criteria, but it didn't work well with very tall or very flat rectangles. I then tried using the longest dimension as a selector, and it worked much better. So much for intuition... These \"interesting\" places are just to the right and just below the currently added rectangle. The first rectangle, obviously, goes at the top left, the next one would go either to the right or below this one, and so on. It is a weird way to do it, but it seems to work very nicely. The way we search here is fairly brute-force, the fact being that for most off-line purposes the performance seems more than adequate. I have generated a japanese font with around 8500 characters and all the time was spent generating the bitmaps. Also, for all we care, we could grow the parent rectangle in a different way than power of two. It just happens that power of 2 is very convenient for graphics hardware textures. I'd be interested in hearing of other approaches to this problem. Make sure to post them on http://www.flipcode.com\n\nCurrently browsing [RectPlacement.zip] (6,056 bytes) - [RectPlacement.h] - (3,011 bytes)\n\n ```// -------------------------------------------------------------------------------- // Name : RectPlacement.h // Description : A class that allocates subrectangles into power-of-2 rectangles // (C) Copyright 2000-2002 by Javier Arevalo // This code is free to use and modify for all purposes // -------------------------------------------------------------------------------- #ifndef _RECT_PLACEMENT_H_ #define _RECT_PLACEMENT_H_#include // -------------------------------------------------------------------------------- // -------------------------------------------------------------------------------- class CRectPlacement { public: // Helper classes struct TPos { int x, y; TPos() { } TPos(int _x, int _y): x(_x), y(_y) { } bool operator ==(const TPos &p) const { return x == p.x && y == p.y; } }; struct TRect: public TPos { int w, h; TRect() { } TRect(int _x, int _y, int _w, int _h): TPos(_x, _y), w(_w > 0? _w : 0), h(_h > 0? _h : 0) { } bool Contains (const TPos &p) const { return (p.x >= x && p.y >= y && p.x < (x+w) && p.y < (y+h)); } bool Contains (const TRect &r) const { return (r.x >= x && r.y >= y && (r.x+r.w) <= (x+w) && (r.y+r.h) <= (y+h)); } bool Intersects (const TRect &r) const { return w > 0 && h > 0 && r.w > 0 && r.h > 0 && ((r.x+r.w) > x && r.x < (x+w) && (r.y+r.h) > y && r.y < (y+h)); } // static bool Greater(const TRect &a, const TRect &b) { return a.w*a.h > b.w*b.h; } // Greater rect area. Not as good as the next heuristic static bool Greater(const TRect &a, const TRect &b) { return (a.w > b.w && a.w > b.h) || (a.h > b.w && a.h > b.h); } // Greater size in at least one dim. }; // --------------------- typedef std::vector CPosArray; typedef std::vector CRectArray; // --------------------- CRectPlacement() { Init(); } ~CRectPlacement() { End(); } void Init (int w = 1, int h = 1); void End (); bool IsOk () const { return m_size.w > 0; } int GetW () const { return m_size.w; } int GetH () const { return m_size.h; } long GetArea () const { return m_area; } long GetTotalArea () const { return m_size.w*m_size.h; } bool AddAtEmptySpotAutoGrow (TRect *pRect, int maxW, int maxH); private: TRect m_size; CRectArray m_vRects; CPosArray m_vPositions; long m_area; // --------------------- bool IsFree (const TRect &r) const; void AddPosition (const TPos &p); void AddRect (const TRect &r); bool AddAtEmptySpot (TRect &r); };#endif //_RECT_PLACEMENT_H_ ```\n\nCurrently browsing [RectPlacement.zip] (6,056 bytes) - [Test.cpp] - (4,835 bytes)\n\n ```// ---------------------------------------------------------------------------------------- // Name : Test.cpp // Description : Test the RectPlacement class // (C) Copyright 2000-2002 by Javier Arevalo // This code is free to use and modify for all purposes // ---------------------------------------------------------------------------------------- #include #include #include \"RectPlacement.h\"#define MAX_TEXTURE_W 256 #define MAX_TEXTURE_H 256#define NUM_SUBRECTS 500 #define MIN_SUBRECT 15 #define MAX_SUBRECT 50// -------------------------------------------------------------------------------- // -------------------------------------------------------------------------------- struct TSubRect: public CRectPlacement::TRect { int n; // Original index of this subrect, before sorting int nTex; // Texture in which this subrect will be placed. TSubRect() { } TSubRect(int _w, int _h, int _n): TRect(0, 0, _w, _h), n(_n), nTex(0) { } };// -------------------------------------------------------------------------------- // -------------------------------------------------------------------------------- typedef std::vector CSubRectArray; typedef std::vector CTextureArray;// -------------------------------------------------------------------------------- // Name : CreateLetters // Description : Here's where you calculate your rectangles, // -------------------------------------------------------------------------------- void CreateLetters (CSubRectArray &vecSubRects) { vecSubRects.clear(); // Let's just fill this with random stuff for (int i = 0; i < NUM_SUBRECTS; i++) vecSubRects.push_back(TSubRect((rand() % MAX_SUBRECT) + MIN_SUBRECT, (rand() % MAX_SUBRECT) + MIN_SUBRECT, i)); }// -------------------------------------------------------------------------------- // Name : CreateTextures // Description : Create array of textures containing all subrects // -------------------------------------------------------------------------------- void CreateTextures (CTextureArray &vecTextures, CSubRectArray &vecSubRects, int maxTexW, int maxTexH) { // Sort the subRects based on dimensions, larger dimension goes first. std::sort(vecSubRects.begin(), vecSubRects.end(), CRectPlacement::TRect::Greater); // Generate the first texture vecTextures.clear(); vecTextures.push_back(CRectPlacement()); // Add all subrects for (CSubRectArray::iterator it = vecSubRects.begin(); it != vecSubRects.end(); ++it) { // We make sure we leave one pixel between subrects, so texels don't bleed with bilinear. CRectPlacement::TRect r(0, 0, it->w+1, it->h+1); // If an existing texture has actual space bool bPlaced = false; for (int i = 0; !bPlaced && i < vecTextures.size(); i++) { bPlaced = vecTextures[i].AddAtEmptySpotAutoGrow(&r, maxTexW, maxTexH); if (bPlaced) it->nTex = i; } // Try starting a new texture and fit the rect in there if (!bPlaced) { vecTextures.push_back(CRectPlacement()); bPlaced = vecTextures[vecTextures.size()-1].AddAtEmptySpotAutoGrow(&r, maxTexW, maxTexH); if (bPlaced) it->nTex = vecTextures.size()-1; else printf(\"ERROR: Subrect is too big to fit in texture!\", it->w, it->h); } // If correctly placed in a texture, the coords are returned in r.x and r.y // Store them. if (bPlaced) { it->x = r.x; it->y = r.y; } } }// -------------------------------------------------------------------------------- // Name : main // Description : heh // -------------------------------------------------------------------------------- int main() { CTextureArray vecTextures; CSubRectArray vecSubRects; CreateLetters(vecSubRects); CreateTextures(vecTextures, vecSubRects, MAX_TEXTURE_W, MAX_TEXTURE_H); { for (CSubRectArray::const_iterator it = vecSubRects.begin(); it != vecSubRects.end(); ++it) { printf(\"Subrect %d (originally %d), size %dx%d, goes into texture %d at pos %d,%d\\n\", it - vecSubRects.begin(), it->n, it->w, it->h, it->nTex, it->x, it->y); } } printf(\"Created a total of %d textures\\n\", vecTextures.size()); { for (CTextureArray::const_iterator it = vecTextures.begin(); it != vecTextures.end(); ++it) { printf(\" Texture %d, size %dx%d, Coverage %d / %d (%d%%)\\n\", it - vecTextures.begin(), it->GetW(), it->GetH(), it->GetArea(), it->GetTotalArea(), it->GetArea()*100/it->GetTotalArea()); } } return 0; } ```\n\nCurrently browsing [RectPlacement.zip] (6,056 bytes) - [RectPlacement.cpp] - (7,435 bytes)\n\n ```// ---------------------------------------------------------------------------------------- // Name : RectPlacement.cpp // Description : A class that fits subrectangles into a power-of-2 rectangle // (C) Copyright 2000-2002 by Javier Arevalo // This code is free to use and modify for all purposes // ---------------------------------------------------------------------------------------- /* You have a bunch of rectangular pieces. You need to arrange them in a rectangular surface so that they don't overlap, keeping the total area of the rectangle as small as possible. This is fairly common when arranging characters in a bitmapped font, lightmaps for a 3D engine, and I guess other situations as well. The idea of this algorithm is that, as we add rectangles, we can pre-select \"interesting\" places where we can try to add the next rectangles. For optimal results, the rectangles should be added in order. I initially tried using area as a sorting criteria, but it didn't work well with very tall or very flat rectangles. I then tried using the longest dimension as a selector, and it worked much better. So much for intuition... These \"interesting\" places are just to the right and just below the currently added rectangle. The first rectangle, obviously, goes at the top left, the next one would go either to the right or below this one, and so on. It is a weird way to do it, but it seems to work very nicely. The way we search here is fairly brute-force, the fact being that for most off- line purposes the performance seems more than adequate. I have generated a japanese font with around 8500 characters and all the time was spent generating the bitmaps. Also, for all we care, we could grow the parent rectangle in a different way than power of two. It just happens that power of 2 is very convenient for graphics hardware textures. I'd be interested in hearing of other approaches to this problem. Make sure to post them on http://www.flipcode.com */#include \"RectPlacement.h\"// -------------------------------------------------------------------------------- // Name : // Description : // -------------------------------------------------------------------------------- void CRectPlacement::Init (int w, int h) { End(); m_size = TRect(0, 0, w, h); m_vPositions.push_back(TPos(0,0)); m_area = 0; }// -------------------------------------------------------------------------------- // Name : // Description : // -------------------------------------------------------------------------------- void CRectPlacement::End () { m_vPositions.clear(); m_vRects.clear(); m_size.w = 0; }// -------------------------------------------------------------------------------- // Name : IsFree // Description : Check if the given rectangle is partially or totally used // -------------------------------------------------------------------------------- bool CRectPlacement::IsFree (const TRect &r) const { if (!m_size.Contains(r)) return false; for (CRectArray::const_iterator it = m_vRects.begin(); it != m_vRects.end(); ++it) if (it->Intersects(r)) return false; return true; }// -------------------------------------------------------------------------------- // Name : AddPosition // Description : Add new anchor point // -------------------------------------------------------------------------------- void CRectPlacement::AddPosition (const TPos &p) { // Try to insert anchor as close as possible to the top left corner // So it will be tried first bool bFound = false; CPosArray::iterator it; for (it = m_vPositions.begin(); !bFound && it != m_vPositions.end(); ++it) { if (p.x+p.y < it->x+it->y) bFound = true; } if (bFound) m_vPositions.insert(it, p); else m_vPositions.push_back(p); }// -------------------------------------------------------------------------------- // Name : AddRect // Description : Add the given rect and updates anchor points // -------------------------------------------------------------------------------- void CRectPlacement::AddRect (const TRect &r) { m_vRects.push_back(r); m_area += r.w*r.h; // Add two new anchor points AddPosition(TPos(r.x, r.y+r.h)); AddPosition(TPos(r.x+r.w, r.y)); }// -------------------------------------------------------------------------------- // Name : AddAtEmptySpot // Description : Add the given rectangle // -------------------------------------------------------------------------------- bool CRectPlacement::AddAtEmptySpot (TRect &r) { // Find a valid spot among available anchors. bool bFound = false; CPosArray::iterator it; for (it = m_vPositions.begin(); !bFound && it != m_vPositions.end(); ++it) { TRect Rect(it->x, it->y, r.w, r.h); if (IsFree(Rect)) { r = Rect; bFound = true; break; // Don't let the loop increase the iterator. } } if (bFound) { // Remove the used anchor point m_vPositions.erase(it); // Sometimes, anchors end up displaced from the optimal position // due to irregular sizes of the subrects. // So, try to adjut it up & left as much as possible. for (int x = 1; x <= r.x; x++) if (!IsFree(TRect(r.x - x, r.y, r.w, r.h))) break; for (int y = 1; y <= r.y; y++) if (!IsFree(TRect(r.x, r.y - y, r.w, r.h))) break; if (y > x) r.y -= y-1; else r.x -= x-1; AddRect(r); } return bFound; } // -------------------------------------------------------------------------------- // Name : AddAtEmptySpotAutoGrow // Description : Add a rectangle of the given size, growing our area if needed // Area grows only until the max given. // Returns the placement of the rect in the rect's x,y coords // -------------------------------------------------------------------------------- bool CRectPlacement::AddAtEmptySpotAutoGrow (TRect *pRect, int maxW, int maxH) { if (pRect->w <= 0) return true; int orgW = m_size.w; int orgH = m_size.h; // Try to add it in the existing space while (!AddAtEmptySpot(*pRect)) { int pw = m_size.w; int ph = m_size.h; // Sanity check - if area is complete. if (pw >= maxW && ph >= maxH) { m_size.w = orgW; m_size.h = orgH; return false; } // Try growing the smallest dim if (pw < maxW && (pw < ph || ((pw == ph) && (pRect->w >= pRect->h)))) m_size.w = pw*2; else m_size.h = ph*2; if (AddAtEmptySpot(*pRect)) break; // Try growing the other dim instead if (pw != m_size.w) { m_size.w = pw; if (ph < maxW) m_size.h = ph*2; } else { m_size.h = ph; if (pw < maxW) m_size.w = pw*2; } if (pw != m_size.w || ph != m_size.h) if (AddAtEmptySpot(*pRect)) break; // Grow both if possible, and reloop. m_size.w = pw; m_size.h = ph; if (pw < maxW) m_size.w = pw*2; if (ph < maxH) m_size.h = ph*2; } return true; } ```\n\nThe zip file viewer built into the Developer Toolbox made use of the zlib library, as well as the zlibdll source additions."
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https://blogs.sas.com/content/graphicallyspeaking/2016/01/12/candlestick-chart-with-sas-9-2/ | [
"Let us start the new year by taking a trip back in history to SAS 9.2, first released in 2008, and the first SAS release that included the new ODS Graphics software including GTL and SG procedures. While we have recently released the third maintenance on SAS 9.4 (SAS 9.40M3), many of you are using various maintenance releases of SAS 9.3, and some are still using SAS 9.2.\n\nOne such SAS 9.2 user recently saw my post on creating a CandleStick Chart using SAS 9.3 which included a new plot type called the HighLow plot. This is a versatile plot that can not only handle the \"Candlestick\" chart commonly used in the financial domain, but is also useful to create many different graphs as you can see in other articles in this blog.. This user wanted to create a similar chart using SAS 9.2.\n\nI first sent them a program to create a High-Low-Close type graph using the GPLOT procedure, but user wanted something similar the the graph shown the linked article.",
null,
"While I could not think of a way to create such a graph using SGPLOT, it was possible, with some effort to create one using GTL. The graph on the right is created using the GTL BoxPlotParm statement. This statement was originally added to provide the user a way to plot a custom box plot, where the values for the various features of the box are computed by the user. So, the data set provided by user needs to contain the various statistics like \"High\", \"Low\", \"Q1\", \"Q3\" and so on for each value of the category.",
null,
"The data set would look like the table on the right. In this example, for each value of Date, we have 4 observations, one for each named statistic. Here we have the \"Min\", \"Max\", \"Q1\" and \"Q3\" values computed for each value of Date. The column names can be anything, but the \"Stat\" values must have the text strings shown.\n\nIn my case, I used a data step to compute these values. The Q1-Q3 range is represented by the \"Open\" and \"Close\" value of the stock, and the \"Low\" and \"High\" are the low and high values for the stock for that day.\n\nproc sort data=sashelp.stocks\n(where=(stock='IBM' and date > '01Jan2004'd))\nout=ibm;\nby date;\nrun;\n\ndata boxParm;\nlength Group \\$4;\nformat DateUp DateDn date7.;\nkeep Date DateUp DateDn Stat Value Group Close2;\nset ibm;\n\nStat='Min'; Value=low; output;\nStat='Max'; Value=high; output;\nStat='Q1'; Value=min(open, close); output;\nStat='Q3'; Value=max(open, close); Close2=close; output;\nrun;\n\nNow, we create a template using the BoxPlotParm statement for the graph. Note we have also superposed a Series plot to connect the \"Close\" value for each day.\n\n/*--Template for OHLC plot--*/\nproc template;\ndefine statgraph OHLC;\nbegingraph;\nentrytitle 'Stock Chart for IBM';\nlayout overlay / xaxisopts=(display=(ticks tickvalues line)\ndiscreteopts=(tickvaluefitpolicy=thin));\nboxplotparm x=date y=value stat=stat;\nseriesplot x=date y=close2 / lineattrs=(color=gray);\nendlayout;\nendgraph;\nend;\nrun;\n\n/*--OHLC plot--*/\nproc sgrender data=boxParm template=OHLC;\nrun;\n\nUser also wanted to see the boxes colored by whether the price was up or down. This would be easy with a GROUP option for the BoxPlotParm. Unfortunately, the SAS 9.2 release does not support a Group option. However, the saving grace was that this was not a real group, where there could be one or more group per category. Instead it is really a single colored box by the group classifier.",
null,
"With some creative coding we can achieve this result. Can you guess how I might have done this?\n\nWhat I have done is displayed all the boxes by date using the green color. Then, I have overdrawn only the boxes for the days the stock value was down. This causes only some of the green boxes to be hidden by the red ones. See the program for the full data step and code.\n\n/*--Template for OHLC plot by group--*/\nproc template;\ndefine statgraph OHLC_Grp;\nbegingraph;\nentrytitle 'Stock Chart for IBM';\nlayout overlay / xaxisopts=(display=(ticks tickvalues line)\ndiscreteopts=(tickvaluefitpolicy=thin));\nboxplotparm x=date y=value stat=stat / fillattrs=graphdata2\nname='All' legendlabel='Up';\nboxplotparm x=datedn y=value stat=stat / fillattrs=graphdata1\nname='Dn' legendlabel='Down';;\nseriesplot x=date y=close2 / lineattrs=(color=gray);\ndiscretelegend 'All' 'Dn';\nendlayout;\nendgraph;\nend;\nrun;\n\nSAS 9.2 Code for CandleStick Chart: Stock_Plot_92\n\nShare",
null,
""
]
| [
null,
"https://blogs.sas.com/content/graphicallyspeaking/files/2016/01/Stock-Plot_GTL_92_a-300x180.png",
null,
"https://blogs.sas.com/content/graphicallyspeaking/files/2016/01/Data-242x300.png",
null,
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null,
"https://blogs.sas.com/content/graphicallyspeaking/files/userphoto/250.jpg",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.85149795,"math_prob":0.7513503,"size":4283,"snap":"2019-13-2019-22","text_gpt3_token_len":1093,"char_repetition_ratio":0.10329516,"word_repetition_ratio":0.03773585,"special_character_ratio":0.24071912,"punctuation_ratio":0.13824885,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96442515,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,2,null,2,null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-20T05:54:40Z\",\"WARC-Record-ID\":\"<urn:uuid:93622835-5bce-4252-b176-26f2cb91bbbf>\",\"Content-Length\":\"35064\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:680cd0da-6162-4b60-a816-fdfc6e697e7b>\",\"WARC-Concurrent-To\":\"<urn:uuid:3ce150c1-2e82-4b5f-87a4-64855cf16200>\",\"WARC-IP-Address\":\"149.173.160.44\",\"WARC-Target-URI\":\"https://blogs.sas.com/content/graphicallyspeaking/2016/01/12/candlestick-chart-with-sas-9-2/\",\"WARC-Payload-Digest\":\"sha1:AGJBARPL4MAIYK2GBREXHZFEALLZVG6U\",\"WARC-Block-Digest\":\"sha1:CBW6TL4WIBHMJZQ26VGUYWSGPGSHYILZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202299.16_warc_CC-MAIN-20190320044358-20190320070358-00004.warc.gz\"}"} |
https://www.convert-measurement-units.com/convert+Speed+of+light+to+Inch+per+second.php | [
" Convert c to ips (Speed of light to Inch per second)\n\n## Speed of light into Inch per second\n\nnumbers in scientific notation\n\nhttps://www.convert-measurement-units.com/convert+Speed+of+light+to+Inch+per+second.php\n\n## How many Inch per second make 1 Speed of light?\n\n1 Speed of light [c] = 11 802 852 677.165 Inch per second [ips] - Measurement calculator that can be used to convert Speed of light to Inch per second, among others.\n\n# Convert Speed of light to Inch per second (c to ips):\n\n1. Choose the right category from the selection list, in this case 'Velocity'.\n2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.\n3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Speed of light [c]'.\n4. Finally choose the unit you want the value to be converted to, in this case 'Inch per second [ips]'.\n5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.\n\nWith this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '301 Speed of light'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Speed of light' or 'c'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Velocity'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '41 c to ips' or '98 c into ips' or '35 Speed of light -> Inch per second' or '51 c = ips' or '13 Speed of light to ips' or '72 c to Inch per second' or '34 Speed of light into Inch per second'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.\n\nFurthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(74 * 76) c'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '301 Speed of light + 903 Inch per second' or '83mm x 53cm x 61dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.\n\nThe mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).\n\nIf a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 1.975 308 624 ×1020. For this form of presentation, the number will be segmented into an exponent, here 20, and the actual number, here 1.975 308 624. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 1.975 308 624 E+20. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 197 530 862 400 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.86353636,"math_prob":0.9736537,"size":3938,"snap":"2023-14-2023-23","text_gpt3_token_len":912,"char_repetition_ratio":0.13650228,"word_repetition_ratio":0.021929825,"special_character_ratio":0.25926867,"punctuation_ratio":0.1298377,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99159926,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-25T17:20:01Z\",\"WARC-Record-ID\":\"<urn:uuid:067acdd0-0b86-41fa-a5dd-cba44bbe1c1f>\",\"Content-Length\":\"50912\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b8de7ace-e07c-441f-8014-48fdbe8c51e2>\",\"WARC-Concurrent-To\":\"<urn:uuid:2b4d8b60-0194-4c2b-990f-8af9d883d80a>\",\"WARC-IP-Address\":\"135.181.75.227\",\"WARC-Target-URI\":\"https://www.convert-measurement-units.com/convert+Speed+of+light+to+Inch+per+second.php\",\"WARC-Payload-Digest\":\"sha1:J5L6UWE3C7WGVMC3AFGPUHB7PJL3Z2PP\",\"WARC-Block-Digest\":\"sha1:LWIBCWJR26KZFYI7DNXZEMXPS2LOAHWC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296945368.6_warc_CC-MAIN-20230325161021-20230325191021-00779.warc.gz\"}"} |
https://www.circuitbasics.com/what-are-square-wave-generators/ | [
"## Oscillators\n\nIt is an electronic circuit that changes state from positive to negative in a repeating cycle without any stimulus other than DC power. This produces an AC waveform at the output.\n\n## Square wave generators\n\nSquare wave generators are generally used in electronics and in signal processing. It is just like a Schmitt trigger circuit in which the reference voltage for the comparator depends on the output voltage. It is also said to be an astable multivibrator.\n\nA square-wave generator obviously produces a square wave. However, this may also be adjustable in mark-to-space ratio and is often used for timing, pulsing and clocking circuits. One of the easiest ways to generate a square-wave is by using a relaxation oscillator.\n\n## Relaxation oscillators\n\nRelaxation oscillators have two alternating states: a long relaxation period in which the system comes to rest and then a short change-over period in which the stable point flips over to a second stable state for a period then flips back again. The period is set by the time constant which is usually an RC or LC pair.\n\nSome sort of active switching device is needed, such as a transistor pair or a uni-junction transistor or an op-amp comparator, or a custom chip such as a 555 timer. The active device switches between charging and discharging modes, producing a repeating waveform.\n\nFor any oscillator to qualify as a relaxation oscillator, it must:\n\n• Produce a non-sinusoidal periodic waveform like triangular, square, or rectangular wave.\n• The circuit of a relaxation oscillator must be nonlinear. This means the design of the circuit must use a semiconductor device like a transistor, MOSFET, or op-amp.\n• The circuit design must use an energy-storing component like an inductor or capacitor that continuously charges and discharges to produce a cyclic waveform.\n\nSeesaw A shows the seesaw in a state of equilibrium and at “relaxation,” but as the bucket slowly fills up, a critical tipping point is reached. The state rapidly changes as the bucket end drops and the bucket tips out. As the bucket drains, the left-hand side is now suddenly much heavier and falls to the ground again, and then the bucket lifts and begins to fill again. (Let’s assume it corrects itself again). In an electronic circuit, this is what is happening: a capacitor gets slowly charged through a resistor until a non-linear part of the circuit is reached, causing a sudden discharge, and the cycle begins again.\n\nIn the waveform above and the multi-vibrator circuit below, the blue curve shows the voltage across one of the capacitors C1. It charges until the bias trigger point is reached, then suddenly turning the other transistor on, then discharges again. The black curve is the voltage at the collector, which is the output. In the multi-vibrator below, either collector can be used as the output. However, in this circuit, we are just flashing two LEDs alternately.\n\nShown below are the multivibrator circuit and a breadboard lash-up. The two LEDs flash alternately at about 1.5Hz. The transistors are any NPN GP transistors. The mark-space ratio can be varied by changing the C and/or R on one half.\n\nR1 and R4 are 560Ω, R2 and R3 are 47k, and C1 and C2 are 10uF.\n\nBelow is the collector voltage waveform.\n\nShown here is the square wave output of the above multivibrator circuit. You can see that the square wave is fairly good, but there is a slight charging delay.\n\nThe period of each half is 0.69CR. So if R2 is 47k and C1 is 10uF, that would be 0.32S per half or 0.64 together. Then f = 1/0.64 = 1.5Hz.\n\nA nice relaxation oscillator can be made from any inverting gates. Although two gates will work (NOR, NAND, OR, Schmitt), three gives a better startup. The frequency is set by R1 and C1:\n\nSo, here we have\n\nwhich gives 45Hz.\n\nThe frequency is adjustable over a 10:1 range and the output is set by R4. R3 is for feedback and is not involved in the timing. The waveform is nice and square.\n\nWith R1=100k, C=.004 f=1kHz, C=.04 f=100Hz, C=0.4 f=10Hz.\n\nThat’s all for square wave oscillators! A good oscillator can be made from the famous 555 timer and we will look at that in the next article on sawtooth and triangular wave generators. Leave a comment below if you have questions about anything!"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.92683643,"math_prob":0.9702923,"size":5230,"snap":"2023-40-2023-50","text_gpt3_token_len":1237,"char_repetition_ratio":0.11921164,"word_repetition_ratio":0.0021978023,"special_character_ratio":0.22829828,"punctuation_ratio":0.11484594,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98000866,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-30T05:21:03Z\",\"WARC-Record-ID\":\"<urn:uuid:3fc32b31-a9db-49fe-9140-00212afc6254>\",\"Content-Length\":\"170435\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:74300171-de79-467a-9e4a-013adab8519d>\",\"WARC-Concurrent-To\":\"<urn:uuid:d357f458-ac13-4e6c-9e64-bf2a1ef23a53>\",\"WARC-IP-Address\":\"108.179.209.177\",\"WARC-Target-URI\":\"https://www.circuitbasics.com/what-are-square-wave-generators/\",\"WARC-Payload-Digest\":\"sha1:EG3MJF67UMKET64RMGIGGZP2BHXBJYMU\",\"WARC-Block-Digest\":\"sha1:GDY3V7THAMCYLOC2CSDMLF3BCSL7WEEJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100164.87_warc_CC-MAIN-20231130031610-20231130061610-00450.warc.gz\"}"} |
https://answers.everydaycalculation.com/simplify-fraction/34-29 | [
"Solutions by everydaycalculation.com\n\n## Reduce 34/29 to lowest terms\n\n34/29 is already in the simplest form. It can be written as 1.172414 in decimal form (rounded to 6 decimal places).\n\n#### Steps to simplifying fractions\n\n1. Find the GCD (or HCF) of numerator and denominator\nGCD of 34 and 29 is 1\n2. Divide both the numerator and denominator by the GCD\n34 ÷ 1/29 ÷ 1\n3. Reduced fraction: 34/29\nTherefore, 34/29 simplified to lowest terms is 34/29.\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:"
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"https://answers.everydaycalculation.com/mathstep-app-icon.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.80799824,"math_prob":0.84138966,"size":417,"snap":"2021-21-2021-25","text_gpt3_token_len":125,"char_repetition_ratio":0.118644066,"word_repetition_ratio":0.0,"special_character_ratio":0.36690646,"punctuation_ratio":0.0952381,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95465267,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-09T02:33:17Z\",\"WARC-Record-ID\":\"<urn:uuid:c8a88640-8a95-47b7-9b90-ea8b5b0d7766>\",\"Content-Length\":\"6443\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:204ca38f-36c8-4b8d-9605-fda51296f14e>\",\"WARC-Concurrent-To\":\"<urn:uuid:35a9129a-a53e-461d-a80d-6783e639f623>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/simplify-fraction/34-29\",\"WARC-Payload-Digest\":\"sha1:7LG4ZLPKGGE447WWH6AHSZYGKGXSHYX4\",\"WARC-Block-Digest\":\"sha1:NK7IB57J3EYJNSG7H3ZKLASJ2H2WMIAU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988953.13_warc_CC-MAIN-20210509002206-20210509032206-00493.warc.gz\"}"} |
https://remysharp.com/2017/11/24/binary-strings | [
"Working with (quite) a few projects that required binary representations of data, I wanted to note down the right way to get a binary string. In particular when handling negative numbers…\n\nA cheap and easy way of getting a binary string for a number is to use `toString` using base 2:\n\n``````const n = 39; // a random int\nconsole.log(n.toString(2));\n// => 100111\n``````\n\nSince my value started as a byte, I'd rather see my binary as 8 bits (although the value is still good in it's 6 digit form):\n\n``````const n = 39; // a random int\n// => 00100111\n``````\n\nThe `String.prototype.padStart` is an ES6 method that's pretty handy. There's a polyfill that you might need (there's also a version on MDN).\n\n## Hex too\n\nThis same method works for hex too:\n\n``````const n = 39;\n// => 27\n``````\n\nI'd usually be inclined to prefix my hex values with `0x` just so it's completely clear (and binary can/should be `0b`).\n\n## Binary for strings\n\nStrings really aren't a big deal, since I can use the `charCodeAt` method to get the character representation.\n\n``````const name = 'Remy';\nconsole.log(\nname.split('').map(n =>\nn\n.charCodeAt(0)\n.toString(2)\n)\n);\n// => Array (4)[ \"01010010\", \"01100101\", \"01101101\", \"01111001\" ]\n``````\n\n## 🚨 Negatives and pandora's box\n\nFor most of my projects this earlier simple method worked fine, but it doesn't fly for negative numbers:\n\n``````const n = -39; // a random int\n// => 0-100111\n``````\n\nThe string result is completely corrupted. Honestly, I'm not sure what's going on in the internals of ECMAScript (I'm sure I'll land myself in the spec one day to learn), but I need to revise my code if the number is going to be correct.\n\nMy first run at solving this was to use a bitwise left shift and cycle through the entire byte length and test the value against a binary `and` operator. In code (since I'm not sure that sentence is entirely clear!):\n\n``````const toBinary = n => {\nreturn Array.from({ length: 8 }, (_, i) => {\nreturn ((n << i) & 128) === 128 ? 1 : 0;\n}).join('');\n};\n\nconsole.log(toBinary(-39));\n// => 11011001\n``````\n\nA note about 128: in the 8 bit value, 128 is the Most Significant Bit (MSb), as I'm shifting left, and forward through our bits, my code keeps checking the MSb. In the following screenshot I've shown how the number is tested against each index and the final binary string value is calculated (the MSb highlighted in grey in the screenshot):",
null,
"To test this value is correct, I'll put this binary value into an `Int8Array` and test the decimal value is -39:\n\n``````console.log(new Int8Array([0b11011001]));\n// => Int8Array (1)[ -39 ]\n``````\n\nBut here's where pandora's box opens up. This is binary for -39 in an 8 bit integer. This is not -39 in a 16 bit integer (that binary value as a 16 bit in results in 217). So it's important that I consider the type that I'm working with (equally my earlier `toString` example didn't consider anything other than 8 bit values).\n\nSo it's important to consider exactly the value type when this function is called. The first limitation in this code is the number of bits returned is hard coded to `8` values. There's also a killer bug in my code! If I try to get -1 for a 32 bit signed integer, my function returns 0 (or more precisely: 32 zeros!).\n\nThe problem comes from when -1 is shifted left by the MSb for a 32 bit value: 2,147,483,648 (which, as numbers go, is full of fun facts).\n\n## Final version: shifting to the right\n\nThe answer is to shift to the LSb, but this puts the actual bit order in reverse. So, when negative, the result is shifted right using the LSb (`1` in all cases), reversed and returned.\n\nUsing the right shift avoids the overflow issue that would trigger with the 32 bit signed int.\n\n``````const toBinary = (n, size = 8) => {\nif (n < 0) {\nreturn Array.from({ length: size }, (_, i) => {\nreturn ((n >> i) & 1) === 1 ? 1 : 0;\n})\n.reverse()\n.join('');\n}\n};\n\n// --- tests ----\n\n// 8 bit signed int\nconsole.assert(-39 === new Int8Array([parseInt(toBinary(-39), 2)]));\n// 32 bit signed int\nconsole.assert(-1 === new Int8Array([parseInt(toBinary(-1, 32), 2)]));\n// this is only reached if the tests are good\nconsole.log('tests passed');\n``````\n\nDemo\n\nIn JavaScript, it's a highly dynamic language. Some people don't like that about JavaScript. I kinda like it. It throws some fun challenges every now and then."
]
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null,
"https://remysharp.com/images/negative-39.gif",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.85339344,"math_prob":0.9175542,"size":4557,"snap":"2023-40-2023-50","text_gpt3_token_len":1230,"char_repetition_ratio":0.10652317,"word_repetition_ratio":0.027638191,"special_character_ratio":0.3166557,"punctuation_ratio":0.15,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9846418,"pos_list":[0,1,2],"im_url_duplicate_count":[null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-04T06:33:20Z\",\"WARC-Record-ID\":\"<urn:uuid:e58c1685-eeff-4c87-9e37-0c78583af45a>\",\"Content-Length\":\"37657\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7f21ec81-839d-4ffa-9a72-365ec12e3218>\",\"WARC-Concurrent-To\":\"<urn:uuid:83f029e4-bdba-4495-a6c9-45b7b78192f5>\",\"WARC-IP-Address\":\"44.217.161.11\",\"WARC-Target-URI\":\"https://remysharp.com/2017/11/24/binary-strings\",\"WARC-Payload-Digest\":\"sha1:RO6ZYYEK5DYLMUX6XBCOP75BWK4HIWUI\",\"WARC-Block-Digest\":\"sha1:XVMTYN4N6RV6LFV72MOOHZOKMQ6ZKLR5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511361.38_warc_CC-MAIN-20231004052258-20231004082258-00394.warc.gz\"}"} |
https://help.orionorigin.com/2022/08/04/partial-relationship-semi-partial-meaning-analogy-2/ | [
"Posted on\n\n# Partial Relationship & Semi-Partial: Meaning & Analogy\n\nPartial Relationship & Semi-Partial: Meaning & Analogy\n\nPartial relationship steps the effectiveness of a love between a couple variables, while you are dealing with on effectation of a minumum of one additional factors. Such as, you might find out if there can be a correlation ranging from number of dinner used and you will blood pressure, while dealing with to have lbs or level of do so. It’s possible to manage getting several variables (named control details or covariates). However, more than one otherwise several often is not recommended since additional control variables, this new reduced legitimate the test.\n\nLimited correlation keeps you to definitely continuous separate adjustable (this new x-value) and something proceeded created variable (the new y-value); Here is the just like during the normal correlation analysis. Regarding the blood circulation pressure analogy significantly more than, the fresh independent variable is actually “level of dinner taken” in addition to dependent changeable was “blood pressure”. The newest control variables – lbs and level of do it – should be continuous.\n\nNotation\n\nA period in the subscript separates the correlated variables and the controlled for variables. For example, correlating caloric intake (X1) against blood pressure (X2), while controlling for weight (Xstep three), is written as: rseveral.step 3 Alternatively, a bar is used instead of a period and subscript: r(1,2|3).\n\n## Running the exam\n\nThe correlation coefficient, r, is also used to show the results from partial correlation. Like the regular correlation coefficient, rpartial returns a value from -1 to 1.\n\nPartial correlation can often be carried out by powering several regression research. Some software packages were partial correlation. Such as for example, into the SPSS favor Get to know > Correlations > Limited.\n\n## Simple tips to Understand the effect\n\nIf the partial correlation, r12.3, is smaller than the simple (two-variable) correlation r12, but greater than 0, then variable 3 partly explains the correlation between X and Y.\n\n## Semi-Partial Relationship\n\nSemi-limited relationship is nearly the same as limited. Actually, of several experts utilize the a couple of terms so you can suggest exactly the same thing. not https://datingranking.net/latvian-dating/, anyone else create improve after the subtle huge difference:\n\nHaving semi-partial relationship, the third changeable keeps lingering to have often X or Y however, perhaps not both; which have partial, the 3rd changeable holds constant for X and you can Y.\n\nFor example, the latest partial limited relationship statistic can tell us the specific part out-of difference, one a specific independent varying shows you. They demonstrates to you just how one particular separate variable influences this new built adjustable, whenever you are additional factors was regulated to possess to end them getting back in the way in which.\n\nTo acquire it, estimate the fresh relationship between your created variable while the residual off the brand new anticipate of one independent changeable by the someone else.\n\n## Example\n\nSuppose we use a set of data (from a 2002 paper from Abdi et al.) which lists three variables over six children. Each child was tested for memory span (Y) and speech rate (X2), and their age was also noted. A correlation statistic was desired which predicts Y (memory span) from X1 and X2 (age and speech rate).\n\nNormally, in a situation where X1 and X2 were independent random variables, we’d find out how important each variable was by computing a squared coefficient of correlation between X1 and X2 and the dependent variable Y. We would know that these squared coefficients of correlation were equal to the square multiple coefficient of correlation. But in a case like ours, X1 and X2 are anything but independent. Speech rate is highly dependent on age, and so using the squared coefficient will count the contributions of each variable several times over.\n\n## Sources\n\n1. Abdi, Herve. Area (Partial Limited) and you will Limited Regression Coefficients. Recovered away from\n\nNeed help with a research otherwise attempt matter? Which have Chegg Study, you should buy action-by-step solutions to the questions you have from a specialist around. Very first half-hour that have a beneficial Chegg tutor is free!"
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https://developerpublish.com/python-program-to-check-if-a-number-is-even-or-odd/ | [
"# Python Program to Check If a Number is Even or Odd\n\nIn this python tutorial, you will learn how to Check If a Number is Even or Odd using the if and else statements along with the modulus operator of the python programming language.\n\n## How to Check If a Number is Even or Odd in Python?\n\nLet’s take a look at the first source code , here the values are assigned in the code, the == equality operator and the modulus operator along with the if and else statements carry out the function.\n\n```num = 12\nif (num % 2) == 0:\nprint(\"{0} is an Even number\".format(num))\nelse:\nprint(\"{0} is an Odd number\".format(num))```\n\nOUTPUT:\n\n`12 is an Even number`\n1. At the start, we declare the variable `num` and assign the integer value `12`.\n2. We declare the `if` statement with the condition `(num % 2) == 0` in which the `%` operator divides the integer value by `2` and if the value equals `0`, the condition is true, if it is not equal to `0`, then the condition is false.\n3. Now, we display the output value using the `print` and display the statement/string `(\"{0} is an Even number\".format(num))` followed by the `else` statement and the statement/string `(\"{0} is an Odd number\".foramt(num))`\n4. In the above statement/string, the variables `{0}` will hold the value `num` where the `format` function helps in variable substitution and data formatting.\n5. As given in the second point, with respect to whether the condition is true or false the respective print statement will be displayed with the help of the `if` and `else` statement.\n\nLet’s take a look at the second source code , here the values are given as input by the user in the code, the == equality operator and the modulus operator along with the if and else statements carry out the function.\n\n```num = int(input(\"Enter a number: \"))\nif (num % 2) == 0:\nprint(\"\\n{0} is an Even number\".format(num))\nelse:\nprint(\"\\n{0} is an Odd number\".format(num))```\n\nINPUT:\n\n`13`\n\nOUTPUT:\n\n```Enter a number:\n13 is an Odd number```\n1. Here we give the user the option to enter the values and the input values are scanned using the `input` function and are stored the variables `num` with the statements/strings `Enter a number:`.\n2. In the STDIN section of the code editor the input values are entered.\n3. Now, we display the output value using the `print` and display the statement/string `(\"{0} is an Even number\".format(num))` followed by the `else` statement and the statement/string `(\"{0} is an Odd number\".foramt(num))`\n4. In the above statement/string, the variables `{0}` will hold the value `num` where the `format` function helps in variable substitution and data formatting.\n5. As given in the second point, with respect to whether the condition is true or false the respective print statement will be displayed with the help of the `if` and `else` statement.\n\nNOTE:\n\n• The == equality operator is a comparison operator which returns True is the two items are equal and returns False if not equal.\n• The modulus operator % is used to compute the reminder.\n• The if and else statements evaluates whether an expression is true or false. If a condition is true, the “if” statement is executed otherwise, the “else” statement is executed.\n• The colon : at the end of the if and else statement tells Python that the next line of code should only be run if the condition is true.\n• The statement for the input function are enclosed in single quotes and parenthesis.\n• The \\n in the code indicates a new line or the end of a statement line or a string.\n• The variables used by the format function for substitution is enclosed in curly braces.\n• The print statement is followed by a period, to initiate the format function.\n• The print statement/string to be displayed in enclosed in double quotes."
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https://ch.mathworks.com/help/finance/opprofit.html | [
"# opprofit\n\n## Syntax\n\n``Profit = opprofit(AssetPrice,Strike,Cost,PosFlag,OptType)``\n\n## Description\n\nexample\n\n````Profit = opprofit(AssetPrice,Strike,Cost,PosFlag,OptType)` returns the profit of an option.```\n\n## Examples\n\ncollapse all\n\nThis example shows how to return the profit of an option. For example, consider buying (going long on) a call option with a strike price of \\$90 on an underlying asset with a current price of \\$100 for a cost of \\$4.\n\n`Profit = opprofit(100, 90, 4, 0, 0)`\n```Profit = 6 ```\n\n## Input Arguments\n\ncollapse all\n\nAsset price, specified as a scalar or a `NINST`-by-`1` vector.\n\nData Types: `double`\n\nStrike or exercise price, specified as a scalar or a `NINST`-by-`1` vector.\n\nData Types: `double`\n\nCost of the option, specified as a scalar or a `NINST`-by-`1` vector.\n\nData Types: `double`\n\nOption position, specified as a scalar or a `NINST`-by-`1` vector using the values `0` (long) or `1` (short).\n\nData Types: `logical`\n\nOption type, specified as a scalar or a `NINST`-by-`1` vector using the values `0` (call option) or `1` (put option).\n\nData Types: `logical`\n\n## Output Arguments\n\ncollapse all\n\nOption profit, returned as a scalar or a `NINST`-by-`1` vector.\n\n## Version History\n\nIntroduced before R2006a"
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https://flutterq.com/solved-typeerror-required-argument-mat-pos-2-not-found/ | [
"Hello Guys, How are you all? Hope You all Are Fine. Today I get the following error TypeError: Required argument ‘mat’ (pos 2) not found in python. So Here I am Explain to you all the possible solutions here.\n\n## How TypeError: Required argument ‘mat’ (pos 2) not found Error Occurs?\n\nToday I get the following error TypeError: Required argument ‘mat’ (pos 2) not found in python.\n\n## How To Solve TypeError: Required argument ‘mat’ (pos 2) not found Error ?\n\n1. How To Solve TypeError: Required argument 'mat' (pos 2) not found Error ?\n\nTo Solve TypeError: Required argument 'mat' (pos 2) not found Error `cv2` didn't the find the “mat” (matrix), because you passed the image as the first argument, but the first argument is supposed to be the window name.\n\nTo Solve TypeError: Required argument 'mat' (pos 2) not found Error `cv2` didn't the find the “mat” (matrix), because you passed the image as the first argument, but the first argument is supposed to be the window name.\n\n## Solution 1\n\nThe first argument to `cv2.imshow` is the window name, so it’s considering the second input `mat` (the image) as missing. If you don’t want to name the window, you can just give an empty string as the first input parameter.\n\n`cv2.imshow('', frame) `\n\n## Solution 2\n\n`cv2` didn’t the find the “mat” (matrix), because you passed the image as the first argument, but the first argument is supposed to be the window name.\n\n```cv2.imshow(winname, mat)\n```\n\nIn most cases, you don’t care about the window name, so use this:\n\n```cv2.imshow('', frame)\ncv2.waitKey(0)\n```\n\nThis is the result when you change the window name:\n\n```import numpy as np\nimport cv2\n\nimage = np.full((300, 300, 3), 255).astype(np.uint8)\n\ncv2.putText(image, 'some picture', (20, 60),\ncv2.FONT_HERSHEY_SIMPLEX, 1, [0, 0, 0])\n\ncv2.imshow('custom window name', image)\ncv2.waitKey(0)\n```\n\nOne of the reasons you might want to change window names is to plot many pictures at once, in different windows.\n\n## Summery\n\nIt’s all About this issue. Hope all solution helped you a lot. Comment below Your thoughts and your queries. Also, Comment below which solution worked for you? Thank You."
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https://www.studytonight.com/python/python-mysql-table-join | [
"Dark Mode On/Off\n\n# Python MySQL - Table Joins\n\nIn this tutorial, we will learn how to join two or more MySQL tables in Python.\n\nIn order to combine two or more tables in MySQL, `JOIN` statement is used. One thing to note here is that there must be a common column in both tables based on which this operation will be performed.\n\nWe have already created the students table in the studytonight database(from the Python MySQL create table tutorial). Let us create another table named results after that we will join both the tables.\n\nBelow we have code to create a table named results. Both results and students table have one field common that is rollno.\n\nYuo can directly run the query to create the new table in MySQL command line.\n\n## Python MySQL - Joining two tables\n\nBelow we have a code snippet where we will join two tables named results and students based on the column rollno:\n\n``````import mysql.connector as mysql\n\ndb = mysql.connect(\nhost = \"localhost\",\ndatabase = \"studytonight\"\n)\n\ncursor = db.cursor()\n\nsql = \"SELECT students.rollno, students.name,students.Branch,students.Address, results.status from students INNER JOIN results ON students.rollno=results.rollno;\"\n\ncursor.execute(sql)\n\nmyresult = cursor.fetchall()\n\nfor x in myresult:\nprint(x)\n``````\n\nThe output of the above code will be:\n\n(1, 'Ramesh', 'CSE', '149 Indirapuram', 'Pass') (2, 'Peter', 'ME', 'Noida', 'Fail') (3, 'Navin', 'CE', 'New Delhi', 'Pass')\n\nThe above output indicates that both tables are joined.\n\nHere is the snapshot of the actual output:\n\n## Python MySQL - Left Join\n\nIt is important to note that the `INNER JOIN`(which we covered in the example above) only shows the rows in the resultset when there is a match.\n\nIf you want to fetch all the records fromthea left-hand side table even if there is no match then `Left Join` will be used.\n\nLet us see an example given below for the Left Join:\n\n``````import mysql.connector as mysql\n\ndb = mysql.connect(\nhost = \"localhost\",\ndatabase = \"studytonight\"\n)\n\ncursor = db.cursor()\n\nsql = \"SELECT students.rollno, students.name,students.Branch,students.Address, results.status from students LEFT JOIN results ON students.rollno=results.rollno;\"\n\ncursor.execute(sql)\n\nmyresult = cursor.fetchall()\n\nfor x in myresult:\nprint(x)\n``````\n\nThe output of the above code is as follows. Let us see:\n\n(1, 'Ramesh', 'CSE', '149 Indirapuram', 'Pass') (2, 'Peter', 'ME', 'Noida', 'Fail') (3, 'Navin', 'CE', 'New Delhi', 'Pass') (5, 'suraj', 'ECE', 'Panipat', None) (6, 'Mukesh', 'CSE', 'Samalkha', None)\n\nIn the above output, the status of rollno 5 and 6 is None because their result is not mentioned in the results table. But as we have applied `LEFT JOIN` so the query selects all the rows from the left table even if there is no match.\n\nHere is the snapshot of the actual output:\n\n## Python MySQL - Right Join\n\nIf you want to fetch all the records from the right-hand side table even if there is no match then `Right Join` will be used.\n\nLet us see an example given below for the `Right Join`. The code given below will fetch all the rows from the right-hand side table. It will not return those rows with rollno 5 and 6:\n\n``````import mysql.connector as mysql\n\ndb = mysql.connect(\nhost = \"localhost\",\ndatabase = \"studytonight\"\n)\n\ncursor = db.cursor()\n\nsql = \"SELECT students.rollno, students.name,students.Branch,students.Address, results.status from students RIGHT JOIN results ON students.rollno=results.rollno;\"\n\ncursor.execute(sql)\n\nmyresult = cursor.fetchall()\n\nfor x in myresult:\nprint(x)\n``````\n\nThe output of the above code is as follows:\n\n(1, 'Ramesh', 'CSE', '149 Indirapuram', 'Pass') (2, 'Peter', 'ME', 'Noida', 'Fail') (3, 'Navin', 'CE', 'New Delhi', 'Pass')\n\nHere is the snapshot of the actual output:\n\nAnd with this we have covered all the basics of Python MySQL. If you have to develop an application in which you want to have a database with multipl tables, in which you want to store data and retrieve data from the tables, then we hope these tutorials help you."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.74733096,"math_prob":0.79940736,"size":4021,"snap":"2022-40-2023-06","text_gpt3_token_len":996,"char_repetition_ratio":0.12969878,"word_repetition_ratio":0.35837245,"special_character_ratio":0.2511813,"punctuation_ratio":0.17326732,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96091396,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-04T15:55:44Z\",\"WARC-Record-ID\":\"<urn:uuid:3c8497f9-db93-42f0-9f48-1435879dada7>\",\"Content-Length\":\"248859\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:af3dd5d6-b1c5-4351-aa46-9af015de0453>\",\"WARC-Concurrent-To\":\"<urn:uuid:5336d910-b7b7-4308-8666-2eb77d82e623>\",\"WARC-IP-Address\":\"65.2.112.116\",\"WARC-Target-URI\":\"https://www.studytonight.com/python/python-mysql-table-join\",\"WARC-Payload-Digest\":\"sha1:VI5P2FVU3L4KOL7HAK236Q3Y2DNNHGIL\",\"WARC-Block-Digest\":\"sha1:KZGBKBHDSAG4L2WL7ZBOSP2TEZCKMCLK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337516.13_warc_CC-MAIN-20221004152839-20221004182839-00147.warc.gz\"}"} |
https://docs.w3cub.com/nim/algorithm/ | [
"/Nim\n\nModule algorithm\n\nThis module implements some common generic algorithms.\n\nTypes\n\nSortOrder = enum\nDescending, Ascending\nsort order\n\nProcs\n\nproc `*`(x: int; order: SortOrder): int {...}{.inline, raises: [], tags: [].}\nflips x if order == Descending; if order == Ascending then x is returned. x is supposed to be the result of a comparator, ie < 0 for less than, == 0 for equal, > 0 for greater than.\nproc fill[T](a: var openArray[T]; first, last: Natural; value: T)\nfills the array a[first..last] with value.\nproc fill[T](a: var openArray[T]; value: T)\nfills the array a with value.\nproc reverse[T](a: var openArray[T]; first, last: Natural)\nreverses the array a[first..last].\nproc reverse[T](a: var openArray[T])\nreverses the array a.\nproc reversed[T](a: openArray[T]; first: Natural; last: int): seq[T]\nreturns the reverse of the array a[first..last].\nproc reversed[T](a: openArray[T]): seq[T]\nreturns the reverse of the array a.\nproc binarySearch[T, K](a: openArray[T]; key: K;\ncmp: proc (x: T; y: K): int {...}{.closure.}): int\n\nbinary search for key in a. Returns -1 if not found.\n\ncmp is the comparator function to use, the expected return values are the same as that of system.cmp.\n\nproc binarySearch[T](a: openArray[T]; key: T): int\nbinary search for key in a. Returns -1 if not found.\nproc smartBinarySearch[T](a: openArray[T]; key: T): int {...}{.deprecated.}\nDeprecated since version 0.18.1; Use binarySearch instead.\nproc lowerBound[T, K](a: openArray[T]; key: K; cmp: proc (x: T; k: K): int {...}{.closure.}): int\n\nReturns a position to the first element in the a that is greater than key, or last if no such element is found. In other words if you have a sorted sequence and you call insert(thing, elm, lowerBound(thing, elm)) the sequence will still be sorted.\n\nThe first version uses cmp to compare the elements. The expected return values are the same as that of system.cmp. The second version uses the default comparison function cmp.\n\nexample:\n\nvar arr = @[1,2,3,5,6,7,8,9]\narr.insert(4, arr.lowerBound(4))\n# after running the above arr is `[1,2,3,4,5,6,7,8,9]`\nproc lowerBound[T](a: openArray[T]; key: T): int\nproc upperBound[T, K](a: openArray[T]; key: K; cmp: proc (x: T; k: K): int {...}{.closure.}): int\n\nReturns a position to the first element in the a that is not less (i.e. greater or equal to) than key, or last if no such element is found. In other words if you have a sorted sequence and you call insert(thing, elm, upperBound(thing, elm)) the sequence will still be sorted.\n\nThe first version uses cmp to compare the elements. The expected return values are the same as that of system.cmp. The second version uses the default comparison function cmp.\n\nexample:\n\nvar arr = @[1,2,3,4,6,7,8,9]\narr.insert(5, arr.upperBound(4))\n# after running the above arr is `[1,2,3,4,5,6,7,8,9]`\nproc upperBound[T](a: openArray[T]; key: T): int\nproc sort[T](a: var openArray[T]; cmp: proc (x, y: T): int {...}{.closure.};\norder = SortOrder.Ascending)\nDefault Nim sort (an implementation of merge sort). The sorting is guaranteed to be stable and the worst case is guaranteed to be O(n log n). The current implementation uses an iterative mergesort to achieve this. It uses a temporary sequence of length a.len div 2. Currently Nim does not support a sensible default argument for cmp, so you have to provide one of your own. However, the system.cmp procs can be used:\nsort(myIntArray, system.cmp[int])\n\n# do not use cmp[string] here as we want to use the specialized\nsort(myStrArray, system.cmp)\n\nYou can inline adhoc comparison procs with the do notation. Example:\n\npeople.sort do (x, y: Person) -> int:\nresult = cmp(x.surname, y.surname)\nif result == 0:\nresult = cmp(x.name, y.name)\nproc sorted[T](a: openArray[T]; cmp: proc (x, y: T): int {...}{.closure.};\norder = SortOrder.Ascending): seq[T]\nreturns a sorted by cmp in the specified order.\nproc isSorted[T](a: openArray[T]; cmp: proc (x, y: T): int {...}{.closure.};\norder = SortOrder.Ascending): bool\nChecks to see whether a is already sorted in order using cmp for the comparison. Parameters identical to sort\nproc product[T](x: openArray[seq[T]]): seq[seq[T]]\nproduces the Cartesian product of the array. Warning: complexity may explode.\nproc nextPermutation[T](x: var openArray[T]): bool {...}{.discardable.}\nCalculates the next lexicographic permutation, directly modifying x. The result is whether a permutation happened, otherwise we have reached the last-ordered permutation.\nvar v = @[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]\nv.nextPermutation()\necho v # @[0, 1, 2, 3, 4, 5, 6, 7, 9, 8]\nproc prevPermutation[T](x: var openArray[T]): bool {...}{.discardable.}\nCalculates the previous lexicographic permutation, directly modifying x. The result is whether a permutation happened, otherwise we have reached the first-ordered permutation.\nvar v = @[0, 1, 2, 3, 4, 5, 6, 7, 9, 8]\nv.prevPermutation()\necho v # @[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]\nproc rotateLeft[T](arg: var openArray[T]; slice: HSlice[int, int]; dist: int): int\nPerforms a left rotation on a range of elements. If you want to rotate right, use a negative dist. Specifically, rotateLeft rotates the elements at slice by dist positions. The element at index slice.a + dist will be at index slice.a. The element at index slice.b will be at slice.a + dist -1. The element at index slice.a will be at slice.b + 1 - dist. The element at index slice.a + dist - 1 will be at slice.b.\nproc rotateLeft[T](arg: var openArray[T]; dist: int): int\ndefault arguments for slice, so that this procedure operates on the entire arg, and not just on a part of it.\nproc rotatedLeft[T](arg: openArray[T]; slice: HSlice[int, int]; dist: int): seq[T]\nsame as rotateLeft, just with the difference that it does not modify the argument. It creates a new seq instead\nproc rotatedLeft[T](arg: openArray[T]; dist: int): seq[T]\nsame as rotateLeft, just with the difference that it does not modify the argument. It creates a new seq instead\n\nTemplates\n\ntemplate sortedByIt(seq1, op: untyped): untyped\n\nConvenience template around the sorted proc to reduce typing.\n\nThe template injects the it variable which you can use directly in an expression. Example:\n\ntype Person = tuple[name: string, age: int]\nvar\np1: Person = (name: \"p1\", age: 60)\np2: Person = (name: \"p2\", age: 20)\np3: Person = (name: \"p3\", age: 30)\np4: Person = (name: \"p4\", age: 30)\npeople = @[p1,p2,p4,p3]\n\necho people.sortedByIt(it.name)\n\nBecause the underlying cmp() is defined for tuples you can do a nested sort like in the following example:\n\necho people.sortedByIt((it.age, it.name))\n\n© 2006–2018 Andreas Rumpf"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.62958497,"math_prob":0.9276066,"size":5506,"snap":"2019-43-2019-47","text_gpt3_token_len":1675,"char_repetition_ratio":0.13195202,"word_repetition_ratio":0.3641686,"special_character_ratio":0.31456593,"punctuation_ratio":0.28426397,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9782144,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-17T13:15:56Z\",\"WARC-Record-ID\":\"<urn:uuid:b444fc46-2ff0-4ce1-ba23-e2fb825a0845>\",\"Content-Length\":\"13731\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7af94d26-52c3-4afa-a83e-de67e98c586a>\",\"WARC-Concurrent-To\":\"<urn:uuid:d9f035bf-e0d7-40d4-b411-83c937791281>\",\"WARC-IP-Address\":\"185.199.108.153\",\"WARC-Target-URI\":\"https://docs.w3cub.com/nim/algorithm/\",\"WARC-Payload-Digest\":\"sha1:2PD63Z3ICZ6FQPXOH7MGINZEQXENNJUL\",\"WARC-Block-Digest\":\"sha1:QMMEDPDQRK7EHTTSKHTJGMMKX5BSINHE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986675316.51_warc_CC-MAIN-20191017122657-20191017150157-00040.warc.gz\"}"} |
http://www.useful-ai.xyz/2021/04/basis-function-regression.html | [
"Basis Function Regression - Tech It Yourself\n\n## Saturday, 24 April 2021\n\nCheck Linear Regression post here.\n\n1. Basis Function\n\nIn mathematics, a basis function is an element of a particular basis for a function space. Every continuous function in the function space can be represented as a linear combination of basis functions, just as every vector in a vector space can be represented as a linear combination of basis vectors.\n\nOne trick you can use to adapt linear regression to nonlinear relationships between variables is to transform the data according to basis functions.\n\nThis is still a linear model, the linearity refers to the fact that the coefficients an never multiply or divide each other.\nWhat we have effectively done is taken our one-dimensional x values and projected them into a higher dimension, so that a linear fit can fit more complicated relationships between x and y\nPolynomial basis functions\nThe PolynomialFeatures transformer has converted one-dimensional array into a three dimensional (polynominal) array by taking the exponent of each value. The new higher dimensional data representation can then be plugged into a linear regression.\nimport numpy as np\nimport matplotlib.pyplot as plt\nfrom sklearn.linear_model import LinearRegression\nfrom sklearn.preprocessing import PolynomialFeatures\n\nx = np.array([1, 2, 3, 4, 5])\ny = np.array([4, 2, 1, 3, 7])\nplt.scatter(x, y);\nplt.show()\nX = x[:, np.newaxis]\npoly = PolynomialFeatures(degree=3, include_bias=False)\nX2 = poly.fit_transform(X)\nmodel = LinearRegression().fit(X2, y)\nyfit = model.predict(X2)\nplt.scatter(x, y)\nplt.plot(x, yfit);\nplt.show()\n\nGaussian basis functions\nOther basis functions can be used. Instead of using a sum of polynomial try to use a sum of Gaussians.\nimport numpy as np\nimport matplotlib.pyplot as plt\nfrom sklearn.linear_model import LinearRegression\n\nrng = np.random.RandomState(1)\nx = 10 * rng.rand(50)\ny = np.sin(x) + 0.1 * rng.randn(50)\n\ncenter_of_gauss = np.linspace(x.min(), x.max(), 20).astype(np.float)\nstd = (center_of_gauss - center_of_gauss)*0.7\n\ndef get_gauss(x):\nX = x[:, np.newaxis]\ngauss = np.exp(-0.5 * (((X-center_of_gauss)/1.41*std) ** 2))\nreturn gauss\n\nmodel = LinearRegression().fit(get_gauss(x), y)\nxfit = np.linspace(0, 10, 1000)\nyfit = model.predict(get_gauss(xfit))\nplt.scatter(x, y)\nplt.plot(xfit, yfit)\nplt.xlim(0, 10);\nplt.show()"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7094072,"math_prob":0.99662024,"size":2289,"snap":"2021-43-2021-49","text_gpt3_token_len":560,"char_repetition_ratio":0.12166302,"word_repetition_ratio":0.08722741,"special_character_ratio":0.25644386,"punctuation_ratio":0.18082789,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999546,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-18T23:35:56Z\",\"WARC-Record-ID\":\"<urn:uuid:bba88ba5-930d-4bdb-b6cf-b25a4b10a1a9>\",\"Content-Length\":\"293621\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0d876b77-f1e6-47d1-9205-80606ff282a6>\",\"WARC-Concurrent-To\":\"<urn:uuid:80923a4c-6462-4458-a7bc-d48c20306514>\",\"WARC-IP-Address\":\"172.217.9.211\",\"WARC-Target-URI\":\"http://www.useful-ai.xyz/2021/04/basis-function-regression.html\",\"WARC-Payload-Digest\":\"sha1:DR35LLG3QBVPUORJUKUS4GZ6ZMI2JZNX\",\"WARC-Block-Digest\":\"sha1:UW5ZHA36WEGXEZ5PMITP5W6XROSPNL6R\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585215.14_warc_CC-MAIN-20211018221501-20211019011501-00696.warc.gz\"}"} |
https://www.tutorialspoint.com/c-program-to-construct-a-dfa-which-accepts-l-an-n-1 | [
"# C Program to construct a DFA which accepts L = {aN | N ≥ 1}\n\nLet us take a string S of size N, we have to design a Deterministic Finite Automata (DFA) for accepting the language L = {aN | N ≥ 1}.\n\nThe string accepting the language L is {a, aa, aaa, aaaaaaa…, }.\n\nNow the user has to enter a string, if that string is present in the given language, then print “entered string is Accepted”. Otherwise, print “entered string is not Accepted”.\n\nDFA transition diagram for the given language is",
null,
"## Example\n\nFollowing is the C program to construct DFA which accepts the language L = {aN | N ≥ 1}\n\n#include<stdio.h>\nint main() {\nchar S;\nint l,i;\nint count = 0;\nprintf(\"To implement DFA of language {aN | N ≥ 1} enter input string:\");\nscanf(\"%s\",S);\nl=strlen(S);\nfor (i=0;i<l;i++) {\nif(S[i]!='a') {\nprintf(\"entered string is NOT ACCEPTED\");\ngetch();\nexit(0);\n}\nif (S[i] == 'a')\ncount++;\nelse\nprintf(\"invalid input\");\n}\nif (count == l && count != 0) {\nprintf(\"entered string is accepted\");\n}\nreturn 0;\n}\n\n## Output\n\nYou will get the following output −\n\nRun 1:\nTo implement DFA of language {aN | N >= 1}\nenter input string:aaaa\nentered string is accepted\nRun 2:\nTo implement DFA of language {aN | N >= 1}\nentered string is NOT ACCEPTED"
]
| [
null,
"https://www.tutorialspoint.com/assets/questions/media/53182/DFA.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.69179714,"math_prob":0.83031446,"size":2714,"snap":"2023-14-2023-23","text_gpt3_token_len":853,"char_repetition_ratio":0.15313654,"word_repetition_ratio":0.1021611,"special_character_ratio":0.33824614,"punctuation_ratio":0.09213052,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9682237,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-20T16:14:16Z\",\"WARC-Record-ID\":\"<urn:uuid:36a66293-7b1a-457a-8842-b846e06843ec>\",\"Content-Length\":\"39796\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:09ac98b1-85d3-4f2c-8850-937f6cf8f4fd>\",\"WARC-Concurrent-To\":\"<urn:uuid:2832fc1c-15fc-440e-9ba4-efc1947f7797>\",\"WARC-IP-Address\":\"192.229.210.176\",\"WARC-Target-URI\":\"https://www.tutorialspoint.com/c-program-to-construct-a-dfa-which-accepts-l-an-n-1\",\"WARC-Payload-Digest\":\"sha1:WH6BHKAB4ZRIHISENGFY4OESKK2JH46S\",\"WARC-Block-Digest\":\"sha1:MESJXT775ROGMIQCDMWYKMENV6CPGZT5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296943484.34_warc_CC-MAIN-20230320144934-20230320174934-00602.warc.gz\"}"} |
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