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http://forums.wolfram.com/mathgroup/archive/2006/Jun/msg00515.html	[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Help: ratio of integral of f(x)^2 to square of integral of f(x)\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg67352] Help: ratio of integral of f(x)^2 to square of integral of f(x)\n• From: ronnen.levinson at gmail.com\n• Date: Mon, 19 Jun 2006 00:01:20 -0400 (EDT)\n• Sender: owner-wri-mathgroup at wolfram.com\n\n```Hi.\n\nI'm trying to determine whether the following ratio\n\nr = (b-a) Integral[ f(x)^2 dx, {x, a, b} ] /\nIntegral[ f(x) dx, {x, a, b} ]\n\nis always greater than or equal to one for 0 < f(x) <= 1. All values\nall real.\n\nI've obtained r>=1 for all tested choices of f(x), but seek guidance to" ]	[ null, "http://forums.wolfram.com/mathgroup/images/head_mathgroup.gif", null, "http://forums.wolfram.com/mathgroup/images/head_archive.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/2.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/0.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/0.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/6.gif", null, "http://forums.wolfram.com/mathgroup/images/search_archive.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.78020877,"math_prob":0.9920198,"size":893,"snap":"2023-40-2023-50","text_gpt3_token_len":289,"char_repetition_ratio":0.20134982,"word_repetition_ratio":0.27564102,"special_character_ratio":0.33594626,"punctuation_ratio":0.1826484,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9984848,"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\":\"2023-09-30T05:33:59Z\",\"WARC-Record-ID\":\"<urn:uuid:04e29d83-5d9f-4953-9d27-db4e580a5820>\",\"Content-Length\":\"44447\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b0832958-b109-4177-9745-c0014b5c5a95>\",\"WARC-Concurrent-To\":\"<urn:uuid:ad7941ff-0f01-43de-a7ef-652ba4aea694>\",\"WARC-IP-Address\":\"140.177.205.73\",\"WARC-Target-URI\":\"http://forums.wolfram.com/mathgroup/archive/2006/Jun/msg00515.html\",\"WARC-Payload-Digest\":\"sha1:BTXKB445J25ILTVUHYKNIJW6N6DWTVWB\",\"WARC-Block-Digest\":\"sha1:Z56H44NC2E77USAXM4SIQNSSTDMSZJHR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510603.89_warc_CC-MAIN-20230930050118-20230930080118-00118.warc.gz\"}"}
https://wellsr.com/python/python-string-operations-and-string-formatting/	[ "## Introduction to Python String Operations and Formatting\n\nIn our previous tutorials, we used strings to supply input and output information to both the terminal and external files. This tutorial will take an in-depth look at Python strings, particularly Python string operations, and how we can format strings in output. If you follow the Python Tutorials Blog, you’re probably used to seeing the simple `print` function:\n\n``````s = 1/7\nprint(s)\n> 0.14285714285714285``````\n\nIn most settings, 17 decimals of precision would be misleading and/or unwise. Previously, we’ve used the `round` function to mathematically round these numbers to a set number of decimals:\n\n``````print(round(s, 4))\n> 0.1429``````\n\nBut what if we want to truncate this number instead? What if we want to represent this number in scientific notation? To do this without dedicated formatting, we would need to create a complex numeric function. Luckily for us, Python has a collection of string operations and formatting methods to solve this problem.\n\n## Python String Operations\n\nBefore we look at outputting strings, let’s look at a selection of Python strings methods and functions that can be used to change the string objects themselves. These methods will all take in a string and return a new string with the given Python operation performed. This is different from other languages that perform a function on the given string object as a side affect without a return value. Therefore, we will always need to “catch” the output of these functions in a new (or the existing) variable if we wish to save the results.\n\n### split Method\n\nThe `split` method is used to convert a Python string into a list of substrings, separated by a given delimiter. `split` has the format `[string].split([delimiter])`, where [delimiter] is the delimiting character (or substring of characters), and [string] is the string to be split. For example:\n\n``````s = \"The best of times.\"\nprint(s.split(\" \")) # Split with a space delimiter\n> ['The', 'best', 'of', 'times.']``````\n\nThe Python split operation is useful for splitting lines retrieved from a file with the `getline` function.\n\n### join Method\n\nThe `join` method performs the reverse of the `split` statement; it takes a list of Python strings and concatenates them together with a given delimiter. `join` has the format `[delimiter].join([list])` where [delimiter] is the delimiting character (or substring of characters), and [list] is the list of strings to be joined. Notice that the format has the opposite order of inputs as the `split` method.\n\n``````t = ['The', 'best', 'of', 'times.']\nprint(\" \".join(t)) # Joining with a space delimiter\n> The best of times.``````\n\nWhere the Python split method separates strings, the Python `join` method is the string operation to bring them back together.\n\n### count Method\n\nThe `count` method counts the number of non-overlapping instances of a substring within a given string. The `count` string operation is case sensitive and has the format `[string].count([substring])`, where [substring] is the substring to be counted, and [string] is the string from which substrings will be counted. This example will help you make more sense of that.\n\n``````s = \"It was the best of times.\"\nprint(s.count(\"t\")) # Count the number of t's in a string\n> 4``````\n\n### strip Method\n\nThe `strip` method removes all instances of a given substring from the beginning and end of a given string. `strip` has the format `[string].strip([substring])` where [substring] is the substring to be removed, and [string] is the string from which the substrings will be removed. For example, this command is useful for removing an unknown number of asterisks from a given string of currency:\n\n``````s = \"**10.00\"\nprint(s.strip(\"\"))\n> 10.00\ns = \"\\$***10.00\" # Note that strip does not remove internal substrings\nprint(s.strip(\"\"))\n> \\$**10.00 # The dollar sign prevented removal of the internal asterisks``````\n\n`strip` will strip the substring from both the left and right sides of the string, but will not touch strings in the middle. It’s a string cleanup operation. `lstrip` is used to only strip characters from the left side of the string, and `rstring` will do the same for the right side of your Python strings.\n\n``````s = \"10.00\"\nprint(s.lstrip(\"\"))\n> 10.00**\nprint(s.rstrip(\"\"))\n> **10.00``````\n\n### replace Method\n\nThe `replace` method will replace a given substring with another substring inside a given string. `replace` has the format `[string].replace([old],[new],[count])` where [string] is the string from which the substrings will be replaced, [old] is the substring to be replaced, [new] is the new substring to be inserted, and [count] is an optional specifier for the number of substrings you want to replace. The default for [count] is to replace all substrings.\n\n``````s = \"\\$*10.00\"\nprint(s.replace(\"\", \"\")) # Substitute asterisks with nothing\n> \\$10.00\ns = \"The best of times.\"\nprint(s.replace(\"best\", \"worst\"))\n> The worst of times.``````\n\nWhere the `strip` operation is used to clean up characters from the left and right side of your string, the `replace` method can be used to remove (or replace) characters throughout your string.\n\nCan't get enough Python?\n\nPython is powerful! Show me more free Python tips\n\n### upper and lower Methods\n\nThe `upper` and `lower` methods are used to change the capitalization of letters within a given string. These methods have the format `[string].upper()` and `[string].lower()`, where [string], of course, is the string to be modified.\n\n``````s = \"It was the BEST of times!\"\nprint(s.upper())\n> IT WAS THE BEST OF TIMES!\nprint(s.lower())\n> it was the best of times!``````\n\nThese methods can be handy when requesting strings from a user that may have unknown capitalization.\n\n``````s = input(\"Continue? Input y/n: \")\n> Continue? Input y/n:\n>> N\nif (s.lower() == \"n\"): # Now the user can enter n or N.\nexit()``````\n\n### find Method\n\nThe `find` method will search a string from left to right, and return the first (lowest) index of a given substring. It’s important to remember that index counting starts at 0 on the far left of your string. The `find` Python string operation both determines if the string contains the substring, and returns its location. `find` has the format `[string].find([substring])`, where [string] is the string to be searched and [substring] is the substring to be found. The method will return -1 if the substring cannot be found. This is similar to the VBA InStr function for my VBA readers out there.\n\n``````s = \"It was the best of the times.\"\nprint(s.find(\"the\"))\n> 7``````\n\nSimilarly, `rfind` will behave the same way, but will find the last (highest) index of the given substring:\n\n``````s = \"It was the best of the times.\"\nprint(s.rfind(\"the\"))\n> 19``````\n\nIf you only need to validate that the string contains a given substring rather than return its location, the `in` statement can used to output a boolean value\n\n``````s = \"It was the best of the times.\"\nprint(\"the\" in s)\n> True``````\n\nThis is very handy when searching line by line through a file for a particular string.\n\n### center Method\n\nThe `center` method will center a string by applying a padding string on both sides to achieve a given total length. `center` has the format `[string].find([totallength],[pad])`, where [string] is the given string, [pad] is the substring you want to use for padding, and [totallength] is your desired total width of the final string. The default character for [pad] is a space.\n\nThis Python string operation is useful when you want to print headers with a standard format. Note that [totallength] is the length of the output string, including the original given string.\n\n``````s = \"Introduction\"\nprint(s.center(30, \"-\"))\n> ---------Introduction---------\nprint(t.center(30, \"-\"))\n\n## Python String Formatting\n\nNow that we’ve looked at manipulating string objects, let’s look at how to change the output of variables in print statements and file output.\n\nThere are two ways to format Python output:\n\n1. the old `%` operation, and\n2. the newer `format` method operation.\n\nThe `%` operation for formatting output was used in older releases of Python, and most legacy code and online tutorials will use this method. The `format` method is the preferred option for modern implementations of Python, and indeed some IDEs will flag the `%` method as an error, since it assumes that `%` is only used as the modulo binary operator. The `%` method can still be used in Python Version 3, but the `format` method is preferred unless you are working with older code.\n\nCan't get enough Python?\n\nPython is powerful! Show me more free Python tips\n\n### % Method\n\nThe `%` method works by inserting conversion specifiers into a string, each of which will be substituted by a proceeding variable or object. Let’s look at the details of these specifiers, then we’ll go over some examples.\n\nSpecifiers have the following format of required and optional parameters which must be inserted in the this order:\n\n1. The character `%`\n2. (Optional) Mapping key contained within parentheses: (key)\n3. (Optional) Conversion flags (see below)\n4. (Optional) Minimum field width, or how much character space must be saved for the input variable\n5. (Optional) Precision, or how many decimals are included in floating point numbers\n6. Conversion Type (presented in table below)\n\nAfter a string containing these specifiers is ended, a `%` followed by either a tuple or dictionary is used to indicate the objects to be inserted.\n\nThe following is a table of conversion types that can be used:\n\nSymbol Output Format Note Python Conversion Types \"s\" String Will convert non-string types into strings (e.g. floats and integers) \"d\" Signed (+/-) Integer Decimal \"i\" can also be used \"f\" Floating Point Decimal \"F\" can also be used \"e\" Floating Point in Scientific Notation (e.g. 1.2e10) The symbol \"E\" outputs a capital \"E\" in output (e.g. 1.2E10) \"g\" Floating Point with Variable Scientific Notation Will output a regular floating point, with scientific notation if exponent is less than -4. \"G\" will output a capital \"E\" in exponents \"%\" % Symbol Used as an escape character for literally inserting a % into a string.\n\nThere are additional conversion types available for output in hexadecimal and octal, but those are less commonly used.\n\nThe following is a set of conversion flags for specifying additional operations within the specifier:\n\nSymbol Operation Note Python Conversion Flags \"0\" Zero Padding for Numeric Values If a numeric character is less that the specified output width, \"0\"s will be used as padding \"-\" Left Align Numbers are left-aligned. Overrides \"0\" padding above \" \" (Space) Leading Decimal Space Leave a blank before positive numbers to align with negative numbers starting with a \"-\" \"+\" Force Proceeding Sign Character (+/-) Will require a sign for numeric characters, regardless of negative or positive\n\nOkay, I know we just threw a lot of information out there. Let’s take a look at an example to help clear things up. Suppose we wanted to insert a number into a string:\n\n``````n = 42\ns = \"The secret number is %d\" % n\nprint(s)\n> The secret number is 42``````\n\nLet break down the code from above. Inside of a string, we inserted the specifier `%d`. The specifier started with the required `%`, had no optional modifiers, and ended with the required conversion type `d` for a decimal output format. The string was followed by another `%`, indicating that we wish to format the string with variables, followed by a length 1 tuple of the variable we wanted to insert.\n\nWe can also use more advanced conversion types on the above example:\n\n``````n = 42\nm = -22\ns = \"The secret number is %(is)d, not %(isnot)d\" % {\"is\": n, \"isnot\": m} # Use a dictionary as input\nprint(s)\n> The secret number is 42, not -22\ns = \"The secret number is %d, not %d\" % (n, m) # Use a tuple as input\nprint(s)\n> The secret number is 42, not -22\ns = \"The secret number is %+d, not %+d\" % (n, m) # Force use of signs\nprint(s)\n> The secret number is +42, not -22\n\nn = 42.2222222222222\nm = -22.0\ns = \"The secret number is %07.2f, not %07.2f\" % (n, m) # Pad with 0's, force length of 7, with 2 decimals of precision for floating point\nprint(s)\n> The secret number is 0042.22, not -022.00\ns = \"The secret number is %.2e, not %.2E\" % (n, m) # Use 2 decimals of precision, in two types of scientific notation\nprint(s)\n> The secret number is 4.22e+01, not -2.20E+01``````\n\nFormatting Python strings can be complicated. The best way to get comfortable with all the syntax is to experiment. Create short Python program like the ones above and tinker with the `%` strings using the tables above.\n\n### format Method\n\nThe `format` method uses a method associated with strings rather than an exterior statement like the older `%` method, if that makes any sense.\n\nThe `format` method follows closer to the Python standard of object use, and is preferred for use in new programs. The `format` method uses specifiers within a Python string, followed by a method call that references what values should be replaced by those specifiers.\n\nAt first glance the `format` method can look more confusing than the `%` method, but in practice it really is more straightforward - it just has more options. The nested lists of options will be detailed below. Don’t worry. I’ll follow up with some examples.\n\nLike the `%` format method, there is a required order of symbols used in the specifier:\n\n1. Open curly brace “{“\n2. (Optional) Reference keyword (or integer for tuple of inputs)\n3. (Optional) “!” + Conversion Type. Converts the given variable into another type before formatting. Can only convert to strings with “!s” or “!r”.\n4. (Optional) “:” + Format Specifier. See below.\n5. Close curly brace “}”\n\nThe format specifier (item 4 in the list above) follows an additional set of ordering rules, which closely (but not exactly) mimic the `%` format specifiers. These specifiers must follow the given order:\n\n1. Fill Character. Default is “ “ (space)\n2. Alignment character. See table below.\n3. Sign Option. See table below.\n4. Minimum character width\n5. ”,” to specify commas as thousands separators\n6. ”.” followed by number of decimals of precision\n7. Output Type. See table below.\n\nThe following are the available alignment characters that specify how the string is aligned within the given width:\n\nSymbol Output Format Note Python Format Alignment Characters \"<\" Left Aligned Forces left alignment. Default for all classes except numerics. \">\" Right Aligned Forces right alignment. Default for numeric types. \"=\" Padding After Sign Forces padding after the sign, before a number \"^\" Center Align Forces the field to be centered\n\nThe following is a table of output types that can be used with the `format` method. Note: While there are many similarities between this table and the one for the `%` method, not all entries are identical. The default value for each of these depends on the Python class type of the referenced variable (i.e. int, float, string, etc.). If a type is omitted from the specifier, then the default for the referenced object’s class will be used.\n\nSymbol Output Format Note Python Format Output Types \"s\" String Default for strings \"d\" Signed (+/-) Integer Decimal Default for integer types \"f\" Floating Point Decimal \"F\" can also be used \"e\" Floating Point in Scientific Notation (e.g. 1.2e10) The symbol \"E\" outputs a capital \"E\" in output (e.g. 1.2E10) \"g\" General Format Will output a regular floating point, with scientific notation if exponent is less than a given precision (default is 6). \"G\" will output a capital \"E\" in exponents. Default for floats. \"%\" Percents Multiplies number by 100, and output as \"f\" format\n\nIn most cases, the `format` method follows the same convention as the `%` method, but replaces `%` specifiers with `:` and contains all references within curly braces `{}`.\n\nTo practice, let’s use the same examples we used with the `%` method, but switched to the equivalent `format` method.\n\n``````n = 42\nm = -22\ns = \"The secret number is {0}, not {1}\".format(n, m) # Simple positional reference\nprint(s)\n> The secret number is 42, not -22\ns = \"The secret number is {}, not {}\".format(n, m) # Numbers are not needed for ordered input\nprint(s)\n> The secret number is 42, not -22\ns = \"The secret number is {isthis}, not {isnot}\".format(isthis=n, isnot=m) # Reference by name (no dictionary needed)\nprint(s)\n> The secret number is 42, not -22\ns = \"The secret number is {:+}, not {:+}\".format(n, m) # Force use of signs\nprint(s)\n> The secret number is +42, not -22\n\nn = 42.2222222222222\nm = -22.0\ns = \"The secret number is {:07.2f}, not {:07.2f}\".format(n, m) # Pad with 0's, force length of 7, with 2 decimals of precision for floating point\nprint(s)\n> The secret number is 0042.22, not -022.00\ns = \"The secret number is {:.2e}, not {:.2E}\".format(n, m) # Use 2 decimals of precision, in two types of scientific notation\nprint(s)\n> The secret number is 4.22e+01, not -2.20E+01``````\n\nLooking at these examples, you can see why Python decided to switch from the `%` style to the `format` style. The syntax for the `format` statement is practically identical to the syntax for all the Python string operations we discussed at the top of this tutorial. They all follow a format like `[string].operation([arguments])`. See? Once you get familiar with the syntax, you’ll start recognizing patterns and realize Python isn’t so hard after all!\n\nCan't get enough Python?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7693057,"math_prob":0.88537216,"size":15952,"snap":"2020-45-2020-50","text_gpt3_token_len":3816,"char_repetition_ratio":0.15218209,"word_repetition_ratio":0.13328397,"special_character_ratio":0.26028085,"punctuation_ratio":0.11463495,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9758603,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-20T23:51:25Z\",\"WARC-Record-ID\":\"<urn:uuid:8b45207e-68d7-42e5-bc9c-b56331e9d02f>\",\"Content-Length\":\"66499\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f4e3c44d-9202-494a-91c2-5b7a6be54e3f>\",\"WARC-Concurrent-To\":\"<urn:uuid:ab1af48e-198c-4963-90d2-3b78f62b4ad2>\",\"WARC-IP-Address\":\"216.239.38.21\",\"WARC-Target-URI\":\"https://wellsr.com/python/python-string-operations-and-string-formatting/\",\"WARC-Payload-Digest\":\"sha1:W574EP5C6T3UBO5VAGK5LEFRSC7FF6U3\",\"WARC-Block-Digest\":\"sha1:S3YX2N3KTOGRWPENIV4PVJL5INATMA37\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107874340.10_warc_CC-MAIN-20201020221156-20201021011156-00198.warc.gz\"}"}
https://percent-calc.com/what-is-34-of-181351	[ "# What is 34% of 181351?\n\nA simple way to calculate percentages of X\n\n34% of 181351 is: 61659.34\n\n## Percentage of - Table for 181351\n\nPercentage ofDifference\n1% of 181351 is 1813.51179537.49\n2% of 181351 is 3627.02177723.98\n3% of 181351 is 5440.53175910.47\n4% of 181351 is 7254.04174096.96\n5% of 181351 is 9067.55172283.45\n6% of 181351 is 10881.06170469.94\n7% of 181351 is 12694.57168656.43\n8% of 181351 is 14508.08166842.92\n9% of 181351 is 16321.59165029.41\n10% of 181351 is 18135.1163215.9\n11% of 181351 is 19948.61161402.39\n12% of 181351 is 21762.12159588.88\n13% of 181351 is 23575.63157775.37\n14% of 181351 is 25389.14155961.86\n15% of 181351 is 27202.65154148.35\n16% of 181351 is 29016.16152334.84\n17% of 181351 is 30829.67150521.33\n18% of 181351 is 32643.18148707.82\n19% of 181351 is 34456.69146894.31\n20% of 181351 is 36270.2145080.8\n21% of 181351 is 38083.71143267.29\n22% of 181351 is 39897.22141453.78\n23% of 181351 is 41710.73139640.27\n24% of 181351 is 43524.24137826.76\n25% of 181351 is 45337.75136013.25\n26% of 181351 is 47151.26134199.74\n27% of 181351 is 48964.77132386.23\n28% of 181351 is 50778.28130572.72\n29% of 181351 is 52591.79128759.21\n30% of 181351 is 54405.3126945.7\n31% of 181351 is 56218.81125132.19\n32% of 181351 is 58032.32123318.68\n33% of 181351 is 59845.83121505.17\n34% of 181351 is 61659.34119691.66\n35% of 181351 is 63472.85117878.15\n36% of 181351 is 65286.36116064.64\n37% of 181351 is 67099.87114251.13\n38% of 181351 is 68913.38112437.62\n39% of 181351 is 70726.89110624.11\n40% of 181351 is 72540.4108810.6\n41% of 181351 is 74353.91106997.09\n42% of 181351 is 76167.42105183.58\n43% of 181351 is 77980.93103370.07\n44% of 181351 is 79794.44101556.56\n45% of 181351 is 81607.9599743.05\n46% of 181351 is 83421.4697929.54\n47% of 181351 is 85234.9796116.03\n48% of 181351 is 87048.4894302.52\n49% of 181351 is 88861.9992489.01\n50% of 181351 is 90675.590675.5\n51% of 181351 is 92489.0188861.99\n52% of 181351 is 94302.5287048.48\n53% of 181351 is 96116.0385234.97\n54% of 181351 is 97929.5483421.46\n55% of 181351 is 99743.0581607.95\n56% of 181351 is 101556.5679794.44\n57% of 181351 is 103370.0777980.93\n58% of 181351 is 105183.5876167.42\n59% of 181351 is 106997.0974353.91\n60% of 181351 is 108810.672540.4\n61% of 181351 is 110624.1170726.89\n62% of 181351 is 112437.6268913.38\n63% of 181351 is 114251.1367099.87\n64% of 181351 is 116064.6465286.36\n65% of 181351 is 117878.1563472.85\n66% of 181351 is 119691.6661659.34\n67% of 181351 is 121505.1759845.83\n68% of 181351 is 123318.6858032.32\n69% of 181351 is 125132.1956218.81\n70% of 181351 is 126945.754405.3\n71% of 181351 is 128759.2152591.79\n72% of 181351 is 130572.7250778.28\n73% of 181351 is 132386.2348964.77\n74% of 181351 is 134199.7447151.26\n75% of 181351 is 136013.2545337.75\n76% of 181351 is 137826.7643524.24\n77% of 181351 is 139640.2741710.73\n78% of 181351 is 141453.7839897.22\n79% of 181351 is 143267.2938083.71\n80% of 181351 is 145080.836270.2\n81% of 181351 is 146894.3134456.69\n82% of 181351 is 148707.8232643.18\n83% of 181351 is 150521.3330829.67\n84% of 181351 is 152334.8429016.16\n85% of 181351 is 154148.3527202.65\n86% of 181351 is 155961.8625389.14\n87% of 181351 is 157775.3723575.63\n88% of 181351 is 159588.8821762.12\n89% of 181351 is 161402.3919948.61\n90% of 181351 is 163215.918135.1\n91% of 181351 is 165029.4116321.59\n92% of 181351 is 166842.9214508.08\n93% of 181351 is 168656.4312694.57\n94% of 181351 is 170469.9410881.06\n95% of 181351 is 172283.459067.55\n96% of 181351 is 174096.967254.04\n97% of 181351 is 175910.475440.53\n98% of 181351 is 177723.983627.02\n99% of 181351 is 179537.491813.51\n100% of 181351 is 1813510\n\n## How calculate 34% of 181351\n\nIn the store, the product costs \\$181351, you were given a discount 34% and you want to understand how much you saved.\n\nSolution:\n\nAmount saved = Product price * Percentage Discount/ 100\n\nAmount saved = (34 * 181351) / 100\n\nAmount saved = \\$61659.34\n\nSimply put, when you buy an item for \\$181351 and the discount is 34%, you will pay \\$119691.66 and save \\$61659.34." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84965545,"math_prob":0.9659492,"size":4170,"snap":"2022-40-2023-06","text_gpt3_token_len":2112,"char_repetition_ratio":0.3348536,"word_repetition_ratio":0.0,"special_character_ratio":0.83884895,"punctuation_ratio":0.19277108,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99995375,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-01T02:11:58Z\",\"WARC-Record-ID\":\"<urn:uuid:4dad92cf-dbd1-4a32-ae03-841bf5fbb58f>\",\"Content-Length\":\"31530\",\"Content-Type\":\"application/http; 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https://tex.stackexchange.com/questions/452597/top-alignment-using-longtable-and-multicolumn	[ "# Top alignment using longtable and multicolumn\n\nIn a largetable, I want to obtain the following:\n\n• All cells vertically aligned to the top\n• Left column is horizontally centred, Right column is flushed left\nTo be clear in the picture below:\n• Alpha should be aligned with One\n• Austria should be aligned with 1 I thought that using p instead of m should do the trick but obviously not, though there are many questions about alignment, I did not find the answer to this specific problem.\n\nObviously, long table, multicolumn and the small tabular are used here just to have a working example close to what is my actual document.", null, "\\documentclass[a4paper]{article}\n\\usepackage{longtable}\n\\usepackage{array}\n\\newcolumntype{C}{>{\\centering\\let\\newline\\\\\\arraybackslash\\hspace{0pt}}m{#1}}\n\\newcolumntype{L}{>{\\flushleft\\let\\newline\\\\\\arraybackslash\\hspace{0pt}}p{#1}}\n\\begin{document}\n\\begin{longtable}{\| C{1cm} \| C{1cm} \| L{3cm} \| L{3cm} \|}\n\\hline\n\\multicolumn{2}{\|C{2cm}\|}{\n\\begin{tabular}[t]{c}\nAlpha\\\\\nBeta\\\\\nGamma\\\\\nDelta\n\\end{tabular}\n}&\n\\multicolumn{2}{L{6cm}\|}{\n\\begin{tabular}[t]{c}\nOne\\\\\nTwo\n\\end{tabular}\n}\n\\\\\\hline\n\\multicolumn{2}{\|C{2cm}\|}{\n\\begin{tabular}[t]{c}\nAustria\\\\\nBelgium\n\\end{tabular}\n}&\n\\multicolumn{2}{L{6cm}\|}{\n\\begin{tabular}[t]{c}\n1\\\\\n2\\\\\n3\n\\end{tabular}\n}\n\\\\\\hline\n\\end{longtable}\n\\end{document}\n\n• flushleft is never inteneded to be used in command form it is the \\begin{flushleft} environment and like center it adds vertical space (as you see) the analogue of \\centering is \\raggedright. also you presumably want p not m columns if you want top alignment. – David Carlisle Sep 26 '18 at 15:23\n\nThe problem is that you use \\flushleft instead of \\raggedright. \\flushleft is the begin of the environment flushleft which uses the \\trivlist macro (which is used internally in a lot of places in LaTeX), which happens to introduce some vertical offset.\n\nSo to solve your main issue, just use \\raggedright. The following does fix some more issues and implements a \\mulcol macro, which is \\multicolumn with added code to account the \\tabcolseps in the skipped columns and the vertical rules (adding it to the width). It assumes that next to every included column there are 2\\tabcolsep of space (which is the default) plus a vertical rule. It also assumes a specific pattern in the second argument, being any tokens then a group specifying a width followed by any more tokens.\n\nThe additional issues I fixed are:\n\n• suppress white spaces around the columns in the sub-tabulars using @{}c@{} or similar.\n\n• using p type column for both L and C (this is actually required to get your vertical alignment right)\n\n• use full space available in the columns put together using \\multicolumn (this is done with \\mulcol here, in general you should keep in mind that you also get the space of the stuff in between the columns). This has no big effect on the left aligned columns (you could only see it if you used the full width for the contents), but the centred ones aren't centred compared to the other columns if you forget about this.\n\nComplete code:\n\n\\documentclass[a4paper]{article}\n\\usepackage{longtable}\n\\usepackage{array}\n\\newcolumntype{C}{>{\\centering\\let\\newline\\\\\\arraybackslash}p{#1}}\n\\newcolumntype{L}{>{\\raggedright\\let\\newline\\\\\\arraybackslash}p{#1}}\n\n\\newcommand\\mulcol\n{%\n\\mulcolA{#1}#2\\endmulcolsecarg\n}\n\\newcommand\\mulcolA{}\n\\long\\def\\mulcolA#1#2#{\\mulcolB{#1}{#2}}\n\\newcommand\\mulcolB{}\n\\long\\def\\mulcolB#1#2#3#4\\endmulcolsecarg\n{%\n\\multicolumn\n{#1}\n{%\n#2%\n{%\n\\dimexpr\n#3%\n+#1\\tabcolsep+#1\\tabcolsep-2\\tabcolsep\n+#1\\arrayrulewidth-\\arrayrulewidth\n\\relax\n}%\n#4%\n}%\n}\n\\begin{document}\n\\begin{longtable}{\| C{1cm} \| C{1cm} \| L{3cm} \| L{3cm} \|}\n\\hline\n\\mulcol{2}{\|C{2cm}\|}{\n\\begin{tabular}[t]{@{}c@{}}\nAlpha\\\\\nBeta\\\\\nGamma\\\\\nDelta\n\\end{tabular}\n}&\n\\mulcol{2}{L{6cm}\|}{\n\\begin{tabular}[t]{@{}c@{}}\nOne\\\\\nTwo\n\\end{tabular}\n}\n\\\\\\hline\n\\mulcol{2}{\|C{2cm}\|}{\n\\begin{tabular}[t]{@{}c@{}}\nAustria\\\\\nBelgium\n\\end{tabular}\n}&\n\\mulcol{2}{L{6cm}\|}{\n\\begin{tabular}[t]{@{}c@{}}\n1\\\\\n2\\\\\n3\n\\end{tabular}\n}\n\\\\\\hline\n\\end{longtable}\n\\end{document}", null, "• instead nesting tabular environments inside multicolumn cells i would rather use \\makecell macro from the package of the same name (it is based on tabular environment, but for its use the code is shorter).\n• since your longtable has defined four columns i suspect that the all of them is somewhere used (but not shown in your mwe):\n\nconsidering above mentioned your table can be written as follows:\n\n\\documentclass[a4paper]{article}\n\\usepackage{array, longtable, makecell}\n\\newcolumntype{C}{>{\\centering\\arraybackslash}p{#1}}\n\\newcolumntype{L}{>{\\raggedright\\arraybackslash}p{#1}}\n\n\\begin{document}\n\\begin{longtable}{\| C{1cm} \| C{1cm} \| L{3cm} \| L{3cm} \|}\n\\hline\n\\multicolumn{2}{\|c\|}{\\makecell[tc]{\nAlpha\\\\\nBeta\\\\\nGamma\\\\\nDelta}} &\n\\multicolumn{2}{l\|}{\\makecell[tl]{\nOne\\\\\nTwo}} \\\\\n\\hline\n\\multicolumn{2}{\|c\|}{\\makecell[tc]{\nAustria\\\\\nBelgium}} &\n\\multicolumn{2}{l\|}{\\makecell[tl]{\n1\\\\\n2\\\\\n3}} \\\\\n\\hline\n1 & 2 & 3 & 4 \\\\\n\\hline\n\\end{longtable}\n\\end{document}\n\n\nwhich gives:", null, "spurious vertical align of cells contents is removed on the same way as Skillmon do in his answer.\n\n• \\makecell is in its core just a tabular with some code around it. Also your first column (so \\multicolumn{2}{\|C{2cm}\|}) is not horizontally centred as you left out the two \\tabcolsep and the vertical rule in between those two columns. Also in your picture the contents of the first column seem to wander rightwards. This wouldn't have happened if you used \\let\\newline\\\\ in the column definitions. – Skillmon likes topanswers.xyz Sep 26 '18 at 14:16" ]	[ null, "https://i.stack.imgur.com/ZWddN.png", null, "https://i.stack.imgur.com/kxFQI.png", null, "https://i.stack.imgur.com/maieX.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6936612,"math_prob":0.91245216,"size":1565,"snap":"2020-34-2020-40","text_gpt3_token_len":513,"char_repetition_ratio":0.12107623,"word_repetition_ratio":0.0,"special_character_ratio":0.28242812,"punctuation_ratio":0.04942966,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97253007,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-29T14:30:54Z\",\"WARC-Record-ID\":\"<urn:uuid:1572b179-32fb-4ec6-92d9-7cbc0f093435>\",\"Content-Length\":\"164664\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a2c1b631-e02d-4004-b884-6766618bef76>\",\"WARC-Concurrent-To\":\"<urn:uuid:8e56a648-6cb6-4c40-81ab-1670a7ddbfd5>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://tex.stackexchange.com/questions/452597/top-alignment-using-longtable-and-multicolumn\",\"WARC-Payload-Digest\":\"sha1:R7JXOUKT474K4AEA2JNCFK4DYDKCRP7U\",\"WARC-Block-Digest\":\"sha1:YRV7EJPE4NRSIRI7CA54FKKJZTWMBNMA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600401643509.96_warc_CC-MAIN-20200929123413-20200929153413-00304.warc.gz\"}"}
https://s-space.snu.ac.kr/handle/10371/131272	[ "## Browse\n\nComparison of Robustness between Linear Regression and Logistic Regression in Dichotomous Criterion\n\nCited 0 time in Web of Science Cited 0 time in Scopus\nAuthors\n김용대\nMajor\n자연과학대학 통계학과\nIssue Date\n2013-02\nPublisher\n서울대학교 대학원\nDescription\n학위논문 (석사)-- 서울대학교 대학원 : 통계학과, 2013. 2. 김용대.\nAbstract\nAs \"Big data\" has arisen, simplicity and robustness for analysis have been emphasized. Therefore, we focus on the model which is relatively simpler and more robustness against fluctuation of data.\nAccurate classifcation is one of the most important things to decide a model for decision making in practice. Logistic model is known as an accurate model to estimate the probabilities of dependent categories. However, focusing on the goal of classifcation in practical use, we assume that linear regression can be more efficient in using practical decision making. We start this study to identify the linear regression analysis is better in robustness than the logistic regression analysis.\nIn simulation, by increasing the number of independent variables, we observe the performances of each method. We try diverse data generated by different models to see the robustness of linear regression analysis. Based on the two conjectures, we analyze the prediction errors of each method. By comparing prediction errors, we conclude the linear regression analysis is the most robust method in our simulation. However, because wee only simulate the equal proportioned two classes, the further study is needed.\nLanguage\nEnglish\nURI\nhttps://hdl.handle.net/10371/131272\nFiles in This Item:\nAppears in Collections:\nCollege of Natural Sciences (자연과학대학)Dept. of Statistics (통계학과)Theses (Master's Degree_통계학과)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8856306,"math_prob":0.8185936,"size":1901,"snap":"2022-27-2022-33","text_gpt3_token_len":505,"char_repetition_ratio":0.110173956,"word_repetition_ratio":0.006779661,"special_character_ratio":0.20462914,"punctuation_ratio":0.10443038,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9880894,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-05T19:53:28Z\",\"WARC-Record-ID\":\"<urn:uuid:9b3cfed2-9945-4428-8d54-a39a7be404bd>\",\"Content-Length\":\"21197\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d0863299-d113-4ef3-a4af-be994c34f32c>\",\"WARC-Concurrent-To\":\"<urn:uuid:d138a38b-00f8-4f0c-8fbd-75eeee02cc14>\",\"WARC-IP-Address\":\"147.46.181.99\",\"WARC-Target-URI\":\"https://s-space.snu.ac.kr/handle/10371/131272\",\"WARC-Payload-Digest\":\"sha1:U6JVMCD7TZOSH6Q4YKSZYIJPLXXZLDSK\",\"WARC-Block-Digest\":\"sha1:CDH4X4HV4E4AVVBI3SFWKNSJWD6PCDX3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104597905.85_warc_CC-MAIN-20220705174927-20220705204927-00388.warc.gz\"}"}
https://www.lawdepartmentmanagementblog.com/a_median_of_the/	[ "Published on:\n\n# A median of the minds: to be above average, know what mean means (and some other stats)\n\nBenchmark metrics typically come in three forms: averages, medians and weighted averages. Think of 11 law departments each pooling data on their percentage of certified paralegals. For the average, total the 11 percentages and divide by 11.\n\nFor the median, rank the 11 figures from highest to lowest and pick the middle one (or average the two middle ones if there is an even number of figures).\n\nFor the weighted average, add up all the paralegals of the 11 departments and all the certified paralegals and divide the certified total by the paralegal total.\n\nA variation sometimes seen, which avoids the anomalous outliers that can distort averages, yet draws on more of the data than the median, might be called the average of the middle. After ranking the figures, drop the top and bottom quartile (25%) and calculate the average of the middle fifty percent. So, roughly speaking, drop the top two percentages and the bottom two percentages and average the middle seven.\n\nOne other statistical descriptor is common. The range comes from subtracting the highest and lowest percentage.\n\nPosted in:\nPublished on:\nUpdated:" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8859395,"math_prob":0.9170747,"size":1083,"snap":"2023-40-2023-50","text_gpt3_token_len":223,"char_repetition_ratio":0.16126043,"word_repetition_ratio":0.0,"special_character_ratio":0.19944598,"punctuation_ratio":0.105,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9777942,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-30T12:23:11Z\",\"WARC-Record-ID\":\"<urn:uuid:beddf4df-b9fb-4bb9-afb2-d1c42f40fcc3>\",\"Content-Length\":\"52209\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b67164e3-f82e-41a3-9e82-01da1800cc44>\",\"WARC-Concurrent-To\":\"<urn:uuid:350eccdc-af0a-4474-9147-d2e42565204f>\",\"WARC-IP-Address\":\"3.162.103.8\",\"WARC-Target-URI\":\"https://www.lawdepartmentmanagementblog.com/a_median_of_the/\",\"WARC-Payload-Digest\":\"sha1:JBBL2PBHT3VXP2SFEZQQYJI3YNSHTVJC\",\"WARC-Block-Digest\":\"sha1:6GK5RH2ERG2KHQXH44LUBZUXLXHS3DD6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510676.40_warc_CC-MAIN-20230930113949-20230930143949-00769.warc.gz\"}"}
https://collaborate.princeton.edu/en/publications/the-ugc-hardness-threshold-of-the-lsubpsub-grothendieck-problem	[ "# The UGC hardness threshold of the Lp grothendieck problem\n\nGuy Kindler, Assaf Naor, Gideon Schechtman\n\nResearch output: Contribution to journalArticle\n\n15 Scopus citations\n\n### Abstract\n\nFor p ≥ 2 we consider the problem of, given an n x n matrix A = (a ij) whose diagonal entries vanish, approximating in polynomial time the number Optp(A):= max { σi, j=1n aijxixj: (x1,....,xn) ∈ ℝn ∧ (σi=1n\|x i\|p)1/p≤1}.When p = 2 this is simply the problem of computing the maximum eigenvalue of A, whereas for p = ∞ (actually it suffices to take p≈ log n) it is the Grothendieck problem on the complete graph, which was shown to have a 0(log n) approximation algorithm in Nemirovski et al. [Nemirovski, A., C. Roos, T. Terlaky. 1999. On maximization of quadratic form over intersection of ellipsoids with common center. Math. Program. Ser. A 86(3) 463-473], Megretski [Megretski, A. 2001. Relaxations of quadratic programs in operator theory and system analysis. Systems, Approximation, Singular Integral Operators, and Related Topics (Bordeaux, 2000), Vol. 129. Operator Theory Advances and Applications. Birkhäuser, Basel, 365-392], Charikar and Wirth [Charikar, M., A. Wirth. 2004. Maximizing quadratic programs: Extending Grothendieck's inequality. Proc. 45th Annual Sympos. Foundations Comput. Sci., IEEE Computer Society, 54-60] and was used in the work of Charikar and Wirth noted above, to design the best known algorithm for the problem of computing the maximum correlation in correlation clustering. Thus the problem of approximating Optp(A) interpolates between the spectral (p = 2) case and the correlation clustering (p =∞ ) case. From a physics point of view this problem corresponds to computing the ground states of spin glasses in a hard-wall potential well. We design a polynomial time algorithm which, given p ≥ 2 and an n x n matrix A = (aij) with zeros on the diagonal, computes Optp(A) up to a factor p/e + 30logp. On the other hand, assuming the unique games conjecture (UGC) we show that it is NP-hard to approximate Optp(A) up to a factor smaller than p/e + 14. Hence as p --∞ the UGC-hardness threshold for computing Optp(A) is exactly (p/e)(1 + o(1)).\n\nOriginal language English (US) 267-283 17 Mathematics of Operations Research 35 2 https://doi.org/10.1287/moor.1090.0425 Published - May 1 2010 Yes\n\n### All Science Journal Classification (ASJC) codes\n\n• Mathematics(all)\n• Computer Science Applications\n• Management Science and Operations Research\n\n### Keywords\n\n• Approximation algorithms" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.787688,"math_prob":0.9044324,"size":2435,"snap":"2020-45-2020-50","text_gpt3_token_len":661,"char_repetition_ratio":0.100781575,"word_repetition_ratio":0.03133159,"special_character_ratio":0.26283368,"punctuation_ratio":0.14565217,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9937996,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-19T16:00:41Z\",\"WARC-Record-ID\":\"<urn:uuid:e754beea-6d24-4ff0-8345-54327403c228>\",\"Content-Length\":\"48826\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2d9f5fa4-dc1f-4944-81d8-7f178275fbbe>\",\"WARC-Concurrent-To\":\"<urn:uuid:0ab72b82-189a-4a93-9f3c-5f5f3efa65f3>\",\"WARC-IP-Address\":\"18.210.30.88\",\"WARC-Target-URI\":\"https://collaborate.princeton.edu/en/publications/the-ugc-hardness-threshold-of-the-lsubpsub-grothendieck-problem\",\"WARC-Payload-Digest\":\"sha1:AGZJ5HNO5CYXHQ2G27JZWAVEOFJ57NI2\",\"WARC-Block-Digest\":\"sha1:DKQWSE5S6N2HKVH6YF2BK2WHYM2UMEJ4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107863364.0_warc_CC-MAIN-20201019145901-20201019175901-00170.warc.gz\"}"}
https://miku86.com/js-ds-queue-dequeue/	[ "# JavaScript Data Structures: Queue: Dequeue\n\n## Intro\n\nLast time, we learned to enqueue a node to the end of the Queue.\n\nToday, we learn how to dequeue / remove a new node from the start of the Queue.\n\n## Starter Code ▶️\n\n``````class Node {\nconstructor(value) {\nthis.value = value;\nthis.next = null;\n}\n}\n\nclass Queue {\nconstructor() {\nthis.length = 0;\nthis.start = null;\nthis.end = null;\n}\n\nenqueue(value) {\nconst newNode = new Node(value);\n\nif (!this.length) {\nthis.start = newNode;\nthis.end = newNode;\n} else {\nthis.end.next = newNode;\nthis.end = newNode;\n}\n\nthis.length += 1;\nreturn newNode;\n}\n}\n``````\n\n## Thoughts 💭\n\nFirst, we should think about the constraints and possibilities:\n\nIf the Queue is empty:\n\n• we can't remove a node\n\nIf the Queue has one node:\n\n• set the current start as the node to remove\n• set the node after the start as the new start\n• set the next node of the node to remove to null\n• set the end to null\n• decrease the queue's length by 1\n• return the node to remove\n\nAll remaining cases:\n\n• set the current start as the node to remove\n• set the node after the start as the new start\n• set the next node of the node to remove to null\n• decrease the queue's length by 1\n• return the node to remove\n\nDifferences:\n\n• we only have to change the end of the queue when we start with one node, because then there is no end left, because the queue will be empty\n\n## Example\n\n``````// current queue:\nA (start) ==> B (end)\n\n// desired queue:\nB (start, end)\n``````\n\nSteps:\n\n``````// current queue:\nA (start) ==> B (end)\n\n// set the node after the start as the new start\nA ==> B (start, end)\n\n// set the next node of the node to remove to null\nA B (start, end)\n\n// desired queue:\nB (start, end)\n``````\n\n## Implementation 📝\n\n``````class Node {\nconstructor(value) {\nthis.value = value;\nthis.next = null;\n}\n}\n\nclass Queue {\nconstructor() {\nthis.length = 0;\nthis.start = null;\nthis.end = null;\n}\n\nenqueue(value) {\nconst newNode = new Node(value);\n\nif (!this.length) {\nthis.start = newNode;\nthis.end = newNode;\n} else {\nthis.end.next = newNode;\nthis.end = newNode;\n}\n\nthis.length += 1;\nreturn newNode;\n}\n\ndequeue() {\nif (!this.length) {\nreturn null;\n} else {\n// set the current start as the node to remove\nconst nodeToRemove = this.start;\n\n// set the node after the start as the new start\nthis.start = this.start.next;\n\n// set the next node of the node to remove to null\nnodeToRemove.next = null;\n\n// set the end to null, if the queue will be empty after removing\nif (this.length === 1) {\nthis.end = null;\n}\n\n// decrease the queue's length by 1\nthis.length -= 1;\n\n// return the node to remove\nreturn nodeToRemove;\n}\n}\n}\n``````\n\n## Result\n\nLet's have a look how to use the `dequeue` method and its results.\n\n``````const newQueue = new Queue();\nnewQueue.enqueue(\"new A\");\nnewQueue.enqueue(\"new B\");\n\n// queue with 2 nodes\nconsole.log(newQueue);\n// Queue {\n// length: 2,\n// start: Node { value: 'new A', next: Node { value: 'new B', next: null } },\n// end: Node { value: 'new B', next: null }\n// }\n\n// remove the start, \"new A\"\nconsole.log(newQueue.dequeue());\n// Node { value: 'new A', next: null }\n\n// 1 node should be left, \"new B\"\nconsole.log(newQueue);\n// Queue {\n// length: 1,\n// start: Node { value: 'new B', next: null },\n// end: Node { value: 'new B', next: null }\n// }\n\n// remove the start, \"new B\"\nconsole.log(newQueue.dequeue());\n// Node { value: 'new B', next: null }\n\n// queue should be empty\nconsole.log(newQueue);\n// Queue { length: 0, start: null, end: null }\n``````\n\n## Next Part ➡️\n\nWe will do a small recap of our Queue.\n\nDon't miss interesting stuff, subscribe!\n\n## Questions ❔\n\n• If we would implement the Queue by using an Array, how would we dequeue the start node?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.65483004,"math_prob":0.8491236,"size":3619,"snap":"2021-43-2021-49","text_gpt3_token_len":976,"char_repetition_ratio":0.18063624,"word_repetition_ratio":0.44837758,"special_character_ratio":0.30947775,"punctuation_ratio":0.20505992,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99475265,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-09T12:48:14Z\",\"WARC-Record-ID\":\"<urn:uuid:9574bac8-6bcc-412a-a577-9537868f565c>\",\"Content-Length\":\"8411\",\"Content-Type\":\"application/http; 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https://amsi.org.au/ESA_Senior_Years/SeniorTopic4/4i/4i_5errors_3.html	[ "## Errors in hypothesis testing\n\n### Large $P$-values do not prove the null hypothesis\n\n#### 50 g chocolate bars\n\nA manufacturer wishes to claim that the average weight of chocolate bars is 50 grams. In order to make a check, a random sample of 40 bars is taken and weighed; the mean is 49.5 g, and the standard deviation is 2.5 g. A test of the null hypothesis that $\\mu$, the true mean weight, is 50 g is carried out. The $P$-value is 0.21. How can we interpret this $P$-value?\n\nIs the manufacturer justified in claiming that the large $P$-value proves that the true mean weight is 50 g? Clearly not, but large $P$-values are sometimes interpreted as providing 'proof' that the null hypothesis is correct or true. The data observed are consistent with the null hypothesis, $\\mu = 50 g$, but the data are also consistent with other values of the population parameter. If, for example, the manufacturer had tested the null hypothesis that the true mean weight was 49.8, the $P$-value is 0.45. If the null hypothesis was that the true mean weight was 50.1 g, the $P$-value is 0.13. These are also relatively large $P$-values, and it would be illogical to claim that there was 'proof' that these hypotheses were also true.\n\nAgain, there is value of interpreting the $P$-value with the confidence interval; the 95% confidence interval for the true mean weight was 48.7 g to 50.3 g. The null hypothesis of interest, $\\mu = 50$ g is just one of the plausible values for the true mean that is included in the confidence interval.\n\nNext page - Errors in hypothesis testing - $P$-values do not measure the importance of a result" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.94530815,"math_prob":0.99425864,"size":1628,"snap":"2020-10-2020-16","text_gpt3_token_len":404,"char_repetition_ratio":0.17795567,"word_repetition_ratio":0.021505376,"special_character_ratio":0.26781327,"punctuation_ratio":0.11309524,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9993111,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-24T06:25:36Z\",\"WARC-Record-ID\":\"<urn:uuid:c02a4389-b698-4b5f-94a1-249061c748f6>\",\"Content-Length\":\"4129\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:58c39b18-6fe8-42b9-8052-f89a48f38273>\",\"WARC-Concurrent-To\":\"<urn:uuid:4dece36b-5b0c-4d49-8f8d-9a7931c93271>\",\"WARC-IP-Address\":\"101.0.91.74\",\"WARC-Target-URI\":\"https://amsi.org.au/ESA_Senior_Years/SeniorTopic4/4i/4i_5errors_3.html\",\"WARC-Payload-Digest\":\"sha1:WBTGR2DB7SAVYE3475MYCTMJFKCIFPTX\",\"WARC-Block-Digest\":\"sha1:HLI5GA5VPUGTZF57PVPXQFBXPZQD3PSA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875145897.19_warc_CC-MAIN-20200224040929-20200224070929-00456.warc.gz\"}"}
https://stackoverflow.com/questions/5453336/plot-correlation-matrix-into-a-graph%22	[ "# Plot correlation matrix into a graph\n\nI have a matrix with some correlation values. Now I want to plot that in a graph that looks more or less like that:", null, "How can I achieve that?\n\nQuick, dirty, and in the ballpark:\n\n``````library(lattice)\n\n#Build the horizontal and vertical axis information\nhor <- c(\"214\", \"215\", \"216\", \"224\", \"211\", \"212\", \"213\", \"223\", \"226\", \"225\")\nver <- paste(\"DM1-\", hor, sep=\"\")\n\n#Build the fake correlation matrix\nnrowcol <- length(ver)\ncor <- matrix(runif(nrowcolnrowcol, min=0.4), nrow=nrowcol, ncol=nrowcol, dimnames = list(hor, ver))\nfor (i in 1:nrowcol) cor[i,i] = 1\n\n#Build the plot\nrgb.palette <- colorRampPalette(c(\"blue\", \"yellow\"), space = \"rgb\")\nlevelplot(cor, main=\"stage 12-14 array correlation matrix\", xlab=\"\", ylab=\"\", col.regions=rgb.palette(120), cuts=100, at=seq(0,1,0.01))\n``````", null, "• It looks very similar to example from OP (fonts, colors, layout). Looks like original was created with lattice too. Great detailed answer, +1. – Marek Mar 28 '11 at 5:16\n• Thank you for the answer. Many people are used to correlation plots in which the diagonal containing 1-s runs from the top left to the bottom right square (see the example figure in the question), rather than from the bottom left to the top right square, as in your solution. Here's how to fix this problem: cor_reversed <- apply(cor, 2, rev); levelplot(t(cor_reversed),...) – skip Jan 11 '16 at 14:47\n• @bill_080 why copy-pasting your code wont print the correlation matrix? – Pavlos Panteliadis Jan 9 '17 at 22:09\n\nRather \"less\" look like, but worth checking (as giving more visual information):\n\nCorrelation matrix ellipses:", null, "Correlation matrix circles:", null, "Please find more examples in the corrplot vignette referenced by @assylias below.\n\n• The site seems to be defunct. Do you have any code or package description for the first plot? – bright-star Mar 25 '14 at 11:13\n• @TrevorAlexander: As far as I remember, the first plot was created by `ellipse:plotcorr`. – daroczig Mar 25 '14 at 12:57\n• I've submitted an edit for link 1 to: improving-visualisation.org/vis/id=250 which provides the same image. – russellpierce May 30 '14 at 14:33\n• Thank you @rpierce, although I see only the image there without the R source. What do I miss here? – daroczig May 30 '14 at 18:45\n• – assylias Sep 7 '15 at 12:50\n\nVery easy with lattice::levelplot:\n\n``````z <- cor(mtcars)\nrequire(lattice)\nlevelplot(z)\n``````", null, "The ggplot2 library can handle this with `geom_tile()`. It looks like there may have been some rescaling done in that plot above as there aren't any negative correlations, so take that into consideration with your data. Using the `mtcars` dataset:\n\n``````library(ggplot2)\nlibrary(reshape)\n\nz <- cor(mtcars)\nz.m <- melt(z)\n\nggplot(z.m, aes(X1, X2, fill = value)) + geom_tile() +\nscale_fill_gradient(low = \"blue\", high = \"yellow\")\n``````", null, "EDIT:\n\n``````ggplot(z.m, aes(X1, X2, fill = value)) + geom_tile() +\nscale_fill_gradient2(low = \"blue\", high = \"yellow\")\n``````", null, "allows to specify the colour of the midpoint and it defaults to white so may be a nice adjustment here. Other options can be found on the ggplot website here and here.\n\n• nice (+1)! Though I would add a manual-break scale (e.g: `c(-1, -0.6, -0.3, 0, 0.3, 0.6, 1)`) with `\"white\"` in the middle to let the colors reflect the symmetry of the correlation efficient. – daroczig Mar 28 '11 at 0:33\n• @Daroczig - Good point. It looks like `scale_fill_gradient2()` achieves the functionality you describe automatically. I didn't know that existed. – Chase Mar 28 '11 at 1:52\n• adding to this: `p <- ggplot(.....) + ... + ....; library(plotly); ggplotly(p)` will make it interactive – schlusie Jul 18 '16 at 15:47\n• To make the diagonal 1's go from top left to bottom right, reversal of factor levels is required for `X1` using: `z.m\\$X1 <- factor(z.m\\$X1, levels = rev(levels( z.m\\$X1 )))` – arun Nov 4 '16 at 4:19\n\nUse the corrplot package:\n\n``````library(corrplot)\ndata(mtcars)\nM <- cor(mtcars)\n## different color series\ncol1 <- colorRampPalette(c(\"#7F0000\",\"red\",\"#FF7F00\",\"yellow\",\"white\",\n\"cyan\", \"#007FFF\", \"blue\",\"#00007F\"))\ncol2 <- colorRampPalette(c(\"#67001F\", \"#B2182B\", \"#D6604D\", \"#F4A582\", \"#FDDBC7\",\n\"#FFFFFF\", \"#D1E5F0\", \"#92C5DE\", \"#4393C3\", \"#2166AC\", \"#053061\"))\ncol3 <- colorRampPalette(c(\"red\", \"white\", \"blue\"))\ncol4 <- colorRampPalette(c(\"#7F0000\",\"red\",\"#FF7F00\",\"yellow\",\"#7FFF7F\",\n\"cyan\", \"#007FFF\", \"blue\",\"#00007F\"))\nwb <- c(\"white\",\"black\")\n\n## different color scale and methods to display corr-matrix\ncorrplot(M, method=\"number\")\ncorrplot(M)\ncorrplot(M, order =\"AOE\")\n\ncorrplot(M, order=\"AOE\", col=col2(200))\n\ncorrplot(M, order=\"AOE\", col=col3(100))\ncorrplot(M, order=\"AOE\", col=col3(10))\n\ncorrplot(M, method=\"color\", col=col1(20), cl.length=21,order = \"AOE\", addCoef.col=\"grey\")\n\nif(TRUE){\n\ncorrplot(M, method=\"square\", col=col2(200),order = \"AOE\")\n\ncorrplot(M, method=\"ellipse\", col=col1(200),order = \"AOE\")\n\ncorrplot(M, method=\"pie\", order = \"AOE\")\n\n## col=wb\ncorrplot(M, col = wb, order=\"AOE\", outline=TRUE, addcolorlabel=\"no\")\n## like Chinese wiqi, suit for either on screen or white-black print.\ncorrplot(M, col = wb, bg=\"gold2\", order=\"AOE\", addcolorlabel=\"no\")\n}\n``````\n\nFor example:", null, "Rather elegant IMO\n\nThat type of graph is called a \"heat map\" among other terms. Once you've got your correlation matrix, plot it using one of the various tutorials out there.\n\n• I'm not sure if calling it a 'heatmap' is a fairly modern invention. It seems to make sense if you are trying to show 'hotspots' by using a red-orange-yellow colour scheme, but in general its just an image plot, or a matrix plot, or a raster plot. I'll be interested to find the oldest reference that calls it a 'heatmap'. tldr; \"[citation needed]\" – Spacedman Mar 28 '11 at 7:04\n• I think you're right that heat map isn't necessarily the earliest name for it. Wikipedia lists a 1957 paper, but I checked that paper and the term \"heat map\" appears nowhere in it (nor do the graphics look exactly like the current form). – Ari B. Friedman Mar 28 '11 at 11:48\n\nI have been working on something similar to the visualization posted by @daroczig, with code posted by @Ulrik using the `plotcorr()` function of the `ellipse` package. I like the use of ellipses to represent correlations, and the use of colors to represent negative and positive correlation. However, I wanted the eye-catching colors to stand out for correlations close to 1 and -1, not for those close to 0.\n\nI created an alternative in which white ellipses are overlaid on colored circles. Each white ellipse is sized so that the proportion of the colored circle visible behind it is equal to the squared correlation. When the correlation is near 1 and -1, the white ellipse is small, and much of the colored circle is visible. When the correlation is near 0, the white ellipse is large, and little of the colored circle is visible.\n\nThe function, `plotcor()`, is available at https://github.com/JVAdams/jvamisc/blob/master/R/plotcor.r.\n\nAn example of the resulting plot using the `mtcars` dataset is shown below.\n\n``````library(plotrix)\nlibrary(seriation)\nlibrary(MASS)\nplotcor(cor(mtcars), mar=c(0.1, 4, 4, 0.1))\n``````", null, "I realise that it's been a while, but new readers might be interested in `rplot()` from the `corrr` package (https://cran.rstudio.com/web/packages/corrr/index.html), which can produce the sorts of plots @daroczig mentions, but design for a data pipeline approach:\n\n``````install.packages(\"corrr\")\nlibrary(corrr)\nmtcars %>% correlate() %>% rplot()\n``````", null, "``````mtcars %>% correlate() %>% rearrange() %>% rplot()\n``````", null, "``````mtcars %>% correlate() %>% rearrange() %>% rplot(shape = 15)\n``````", null, "``````mtcars %>% correlate() %>% rearrange() %>% shave() %>% rplot(shape = 15)\n``````", null, "``````mtcars %>% correlate() %>% rearrange(absolute = FALSE) %>% rplot(shape = 15)\n``````", null, "The corrplot() function from corrplot R package can be also used to plot a correlogram.\n\n``````library(corrplot)\nM<-cor(mtcars) # compute correlation matrix\ncorrplot(M, method=\"circle\")\n``````\n\nseveral articles describing how to compute and visualize correlation matrix are published here:\n\nAnother solution I recently learned about is an interactive heatmap created with the qtlcharts package.\n\n``````install.packages(\"qtlcharts\")\nlibrary(qtlcharts)\niplotCorr(mat=mtcars, group=mtcars\\$cyl, reorder=TRUE)\n``````\n\nBelow is a static image of the resulting plot.", null, "You can see the interactive version on my blog. Hover over the heatmap to see the row, column, and cell values. Click on a cell to see a scatterplot with symbols colored by group (in this example, the number of cylinders, 4 is red, 6 is green, and 8 is blue). Hovering over the points in the scatterplot gives the name of the row (in this case the make of the car).\n\nSince I cannot comment, I have to give my 2c to the answer by daroczig as an anwser...\n\nThe ellipse scatter plot is indeed from the ellipse package and generated with:\n\n``````corr.mtcars <- cor(mtcars)\nord <- order(corr.mtcars[1,])\nxc <- corr.mtcars[ord, ord]\ncolors <- c(\"#A50F15\",\"#DE2D26\",\"#FB6A4A\",\"#FCAE91\",\"#FEE5D9\",\"white\",\n\"#EFF3FF\",\"#BDD7E7\",\"#6BAED6\",\"#3182BD\",\"#08519C\")\nplotcorr(xc, col=colors[5xc + 6])\n``````\n\n(from the man page)\n\nThe corrplot package may also - as suggested - be useful with pretty images found here" ]	[ null, "https://i.stack.imgur.com/I1MUZ.jpg", null, "https://i.stack.imgur.com/9ydmA.jpg", null, "https://i.stack.imgur.com/4aCFY.png", null, "https://i.stack.imgur.com/2Hoq0.png", null, "https://i.stack.imgur.com/bLh1a.png", null, "https://i.stack.imgur.com/IfbRe.png", null, "https://i.stack.imgur.com/dpdSK.png", null, "https://i.stack.imgur.com/B9URo.png", null, "https://i.stack.imgur.com/OBDno.png", null, "https://i.stack.imgur.com/uRmK4.png", null, "https://i.stack.imgur.com/pdBM0.png", null, "https://i.stack.imgur.com/pdWvG.png", null, "https://i.stack.imgur.com/L61rI.png", null, "https://i.stack.imgur.com/zpgAg.png", null, "https://i.stack.imgur.com/hl2kw.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.74503046,"math_prob":0.9315492,"size":7719,"snap":"2019-51-2020-05","text_gpt3_token_len":2274,"char_repetition_ratio":0.14361633,"word_repetition_ratio":0.083333336,"special_character_ratio":0.31286436,"punctuation_ratio":0.18979058,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9904133,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30],"im_url_duplicate_count":[null,6,null,6,null,6,null,6,null,6,null,7,null,6,null,6,null,6,null,6,null,6,null,6,null,6,null,6,null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-23T14:51:22Z\",\"WARC-Record-ID\":\"<urn:uuid:8b23204e-3477-4a93-85fd-583ca14131a0>\",\"Content-Length\":\"234038\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:37118017-058f-4019-a530-bdd5d0e3dc4e>\",\"WARC-Concurrent-To\":\"<urn:uuid:ceefe9bd-3dea-4b36-a5ac-2178dcf16969>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://stackoverflow.com/questions/5453336/plot-correlation-matrix-into-a-graph%22\",\"WARC-Payload-Digest\":\"sha1:J25HKICSAFBNNDEO7D3O6DTPMZEVSWFP\",\"WARC-Block-Digest\":\"sha1:C3IF6SJI5IVIZK25BJHF3EUOOZ6EGEDB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250610919.33_warc_CC-MAIN-20200123131001-20200123160001-00465.warc.gz\"}"}
https://jas.shahroodut.ac.ir/article_858.html	[ "Document Type : Original Manuscript\n\nAuthor\n\nDepartment of Mathematics, Shahrekord University, P.O. Box 115, Shahrekord, Iran.\n\nAbstract\n\nLet R be a commutative ring with identity and M an R-module. In this paper, we associate a graph to M, say\nΓ(RM), such that when M=R, Γ(RM) coincide with the zero-divisor graph of R. Many well-known results by D.F. Anderson and P.S. Livingston have been generalized for Γ(RM). We Will show that Γ(RM) is connected with\ndiam Γ(RM)≤ 3 and if Γ(RM) contains a cycle, then Γ(RM)≤4. We will also show that Γ(RM)=Ø if and only if M is a\nprime module. Among other results, it is shown that for a reduced module M satisfying DCC on cyclic submodules,\ngr (Γ(RM))=∞ if and only if Γ(RM) is a star graph. Finally, we study the zero-divisor graph of free\nR-modules.\n\nKeywords" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7985491,"math_prob":0.89059216,"size":796,"snap":"2022-27-2022-33","text_gpt3_token_len":232,"char_repetition_ratio":0.109848484,"word_repetition_ratio":0.0,"special_character_ratio":0.25376883,"punctuation_ratio":0.14594595,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98896325,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-11T06:11:04Z\",\"WARC-Record-ID\":\"<urn:uuid:ffcd1fc0-2f0d-491c-aa78-6e8087dd1e45>\",\"Content-Length\":\"49446\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f0224191-c9fe-4862-bbde-8dc0c6703ecc>\",\"WARC-Concurrent-To\":\"<urn:uuid:3b3fc958-bcd6-4d10-a0c2-d31329734d56>\",\"WARC-IP-Address\":\"85.185.67.213\",\"WARC-Target-URI\":\"https://jas.shahroodut.ac.ir/article_858.html\",\"WARC-Payload-Digest\":\"sha1:GE2MAVSWFS3ABSOMT3LIJQ35RMQ57AFM\",\"WARC-Block-Digest\":\"sha1:BJXGIJAHQ5UISRSOQ6VHAOXVJ46TVQDX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571234.82_warc_CC-MAIN-20220811042804-20220811072804-00032.warc.gz\"}"}
https://developers.arcgis.com/enterprise-sdk/api-reference/net/INetworkEdge/	[ "# INetworkEdge Interface\n\nProvides access to members that specify the properties of this network edge element.\n\n## Members\n\nName Description", null, "AttributeValue Value of this network element for the given network attribute ID.", null, "AttributeValueAtTime Value of this network element for the given network attribute ID and local time.", null, "AttributeValueByName Value of this network element for the given network attribute name.", null, "CoveredEdgeCount Returns the number of edges covered by this edge.", null, "Direction Direction in which this network edge element is oriented relative to the direction of its source object.", null, "EID Element ID for this network element.", null, "ElementType Type of this network element.", null, "FromAzimuth Direction of travel at the from-end of the network edge element.", null, "HasCoveringHyperedge Indicates if the network edge element has a covering hyperedge.", null, "IsHyperedge Indicates if the network edge element is a hyperedge.", null, "OID Object ID of the object corresponding to this network element.", null, "PartialEdgeAttributeValue Value of this edge element along the given range for the given network attribute ID.", null, "PartialEdgeAttributeValueAtTime Value of this partial edge element for the given network attribute ID and local.", null, "PositionAlongObject Position along the source object at which the specified position along the network edge element lies.", null, "QueryCoveredEdge Used to iterate over the edges covered by this edge.", null, "QueryCoveringHyperedge Queries the covering hyperedge of the network edge element and returns its related positions along the covering hyperedge.", null, "QueryEdgeInOtherDirection Queries the network edge element corresponding to the reverse traversal of this network edge element.", null, "QueryJunctions Queries the network junction elements adjacent to this network edge element.", null, "QueryPositions Queries the positions along the source object at which the from-end and to-end of the network edge element lies.", null, "QueryTurn Queries the index'th network turn element in which this network edge element participates.", null, "SourceID ID of the network dataset source from which this network element was derived.", null, "ToAzimuth Direction of travel at the to-end of the network edge element.", null, "TurnCount Number of network turn elements in which this network edge element participates.", null, "TurnParticipationType Participation of this network edge element within a network turn element.\n\n### INetworkEdge.AttributeValue Property\n\nValue of this network element for the given network attribute ID.\n\n``````Public Function get_AttributeValue ( _\nByVal AttributeID As Integer _\n) As Object\n``````\n``````public object get_AttributeValue (\nint AttributeID\n);\n``````\n\n### INetworkEdge.AttributeValueAtTime Property\n\nValue of this network element for the given network attribute ID and local time.\n\n``````Public Function get_AttributeValueAtTime ( _\nByVal AttributeID As Integer, _\nByVal localTime As DateTime, _\nByVal timeUsage As esriNetworkTimeUsage _\n) As Object\n``````\n``````public object get_AttributeValueAtTime (\nint AttributeID,\nDateTime localTime,\nesriNetworkTimeUsage timeUsage\n);\n``````\n\n### INetworkEdge.AttributeValueByName Property\n\nValue of this network element for the given network attribute name.\n\n``````Public Function get_AttributeValueByName ( _\nByVal AttributeName As String _\n) As Object\n``````\n``````public object get_AttributeValueByName (\nstring AttributeName\n);\n``````\n\n### INetworkEdge.CoveredEdgeCount Property\n\nReturns the number of edges covered by this edge.\n\n``````Public ReadOnly Property CoveredEdgeCount As Integer\n``````\n``````public int CoveredEdgeCount {get;}\n``````\n\n### INetworkEdge.Direction Property\n\nDirection in which this network edge element is oriented relative to the direction of its source object.\n\n``````Public ReadOnly Property Direction As esriNetworkEdgeDirection\n``````\n``````public esriNetworkEdgeDirection Direction {get;}\n``````\n\n#### Remarks\n\nThe Direction property returns the direction of travel along this edge element relative to the digitized direction of the source feature. The direction of travel is either esriNEDAlongDigitized, meaning the direction of travel along the edge element is the same as the feature's direction of digitization, or esriNEDAgainstDigitized, meaning the direction of travel along the edge element is the opposite of the feature's direction of digitization.\n\n### INetworkEdge.EID Property\n\nElement ID for this network element.\n\n``````Public ReadOnly Property EID As Long\n``````\n``````public long EID {get;}\n``````\n\n### INetworkEdge.ElementType Property\n\nType of this network element.\n\n``````Public ReadOnly Property ElementType As esriNetworkElementType\n``````\n``````public esriNetworkElementType ElementType {get;}\n``````\n\n### INetworkEdge.FromAzimuth Property\n\nDirection of travel at the from-end of the network edge element.\n\n``````Public ReadOnly Property FromAzimuth As Double\n``````\n``````public double FromAzimuth {get;}\n``````\n\n#### Remarks\n\nThe FromAzimuth property indicates the direction of travel at the starting end of the edge element. For edge elements generated by an EdgeFeatureSource, the FromAzimuth and ToAzimuth values are measured clockwise in degrees relative to the positive Y-axis direction of the spatial reference of the network dataset.\n\n### INetworkEdge.HasCoveringHyperedge Property\n\nIndicates if the network edge element has a covering hyperedge.\n\n``````Public ReadOnly Property HasCoveringHyperedge As Boolean\n``````\n``````public bool HasCoveringHyperedge {get;}\n``````\n\n### INetworkEdge.IsHyperedge Property\n\nIndicates if the network edge element is a hyperedge.\n\n``````Public ReadOnly Property IsHyperedge As Boolean\n``````\n``````public bool IsHyperedge {get;}\n``````\n\n### INetworkEdge.OID Property\n\nObject ID of the object corresponding to this network element.\n\n``````Public ReadOnly Property OID As Long\n``````\n``````public long OID {get;}\n``````\n\n### INetworkEdge.PartialEdgeAttributeValue Property\n\nValue of this edge element along the given range for the given network attribute ID.\n\n``````Public Function get_PartialEdgeAttributeValue ( _\nByVal fromPosition As Double, _\nByVal toPosition As Double, _\nByVal AttributeID As Integer _\n) As Object\n``````\n``````public object get_PartialEdgeAttributeValue (\ndouble fromPosition,\ndouble toPosition,\nint AttributeID\n);\n``````\n\n### INetworkEdge.PartialEdgeAttributeValueAtTime Property\n\nValue of this partial edge element for the given network attribute ID and local.\n\n``````Public Function get_PartialEdgeAttributeValueAtTime ( _\nByVal fromPosition As Double, _\nByVal toPosition As Double, _\nByVal AttributeID As Integer, _\nByVal localTime As DateTime, _\nByVal timeUsage As esriNetworkTimeUsage _\n) As Object\n``````\n``````public object get_PartialEdgeAttributeValueAtTime (\ndouble fromPosition,\ndouble toPosition,\nint AttributeID,\nDateTime localTime,\nesriNetworkTimeUsage timeUsage\n);\n``````\n\n### INetworkEdge.PositionAlongObject Property\n\nPosition along the source object at which the specified position along the network edge element lies.\n\n``````Public Function get_PositionAlongObject ( _\nByVal positionAlongElement As Double _\n) As Double\n``````\n``````public double get_PositionAlongObject (\ndouble positionAlongElement\n);\n``````\n\n#### Remarks\n\nThe PositionAlongObject property determines, from the given positional value along the edge element, the corresponding positional value along the object from which this edge element was created. The positionAlongElement parameter and the PositionAlongObject value both range from 0.0 to 1.0.\n\nFor example, if the fromPosition and toPosition values from the QueryPositions method is 0.1 and 0.6 respectively, then the PositionAlongObject value for the positionAlongElement = 0.7 is equal to 70% of the way between 0.1 and 0.6, which is 0.45.\n\n### INetworkEdge.QueryCoveredEdge Method\n\nUsed to iterate over the edges covered by this edge.\n\n``````Public Sub QueryCoveredEdge ( _\nByVal Index As Integer, _\nByVal Edge As INetworkEdge _\n)\n``````\n``````public void QueryCoveredEdge (\nint Index,\nINetworkEdge Edge\n);\n``````\n\n### INetworkEdge.QueryCoveringHyperedge Method\n\nQueries the covering hyperedge of the network edge element and returns its related positions along the covering hyperedge.\n\n``````Public Sub QueryCoveringHyperedge ( _\nByVal Edge As INetworkEdge, _\nByRef fromPosition As Double, _\nByRef toPosition As Double _\n)\n``````\n``````public void QueryCoveringHyperedge (\nINetworkEdge Edge,\nref double fromPosition,\nref double toPosition\n);\n``````\n\n### INetworkEdge.QueryEdgeInOtherDirection Method\n\nQueries the network edge element corresponding to the reverse traversal of this network edge element.\n\n``````Public Sub QueryEdgeInOtherDirection ( _\nByVal Edge As INetworkEdge _\n)\n``````\n``````public void QueryEdgeInOtherDirection (\nINetworkEdge Edge\n);\n``````\n\n#### Remarks\n\nThe QueryEdgeInOtherDirection method retrieves the edge element in the opposite direction of travel as this edge.\n\nThis method requires an instantiated NetworkEdge object to be passed in as a parameter. You can create an empty NetworkEdge object by using the INetworkQuery::CreateNetworkElement method.\n\n### INetworkEdge.QueryJunctions Method\n\nQueries the network junction elements adjacent to this network edge element.\n\n``````Public Sub QueryJunctions ( _\nByVal FromJunction As INetworkJunction, _\nByVal ToJunction As INetworkJunction _\n)\n``````\n``````public void QueryJunctions (\nINetworkJunction FromJunction,\nINetworkJunction ToJunction\n);\n``````\n\n#### Remarks\n\nThe QueryJunctions method retrieves the junction elements at the ends of this edge element.\n\nThis method requires two instantiated NetworkJunction objects to be passed in as a parameter. You can create an empty NetworkJunction object by using the INetworkQuery::CreateNetworkElement method.\n\n### INetworkEdge.QueryPositions Method\n\nQueries the positions along the source object at which the from-end and to-end of the network edge element lies.\n\n``````Public Sub QueryPositions ( _\nByRef fromPosition As Double, _\nByRef toPosition As Double _\n)\n``````\n``````public void QueryPositions (\nref double fromPosition,\nref double toPosition\n);\n``````\n\n#### Remarks\n\nThe QueryPositions method queries the positional values along the source feature from which this edge element was created. The position values range from 0.0 to 1.0, where 0.0 is at the from-end of the feature and 1.0 is at the to-end of the feature.\n\nIf the Direction property is esriNEDAlongDigitized, then the fromPosition value will be less than or equal to the toPositionValue. If the Direction property is esriNEDAgainstDigitized, then the fromPosition value will be greater than or equal to the toPositionValue.\n\n### INetworkEdge.QueryTurn Method\n\nQueries the index'th network turn element in which this network edge element participates.\n\n``````Public Sub QueryTurn ( _\nByVal Index As Integer, _\nByVal Turn As INetworkTurn _\n)\n``````\n``````public void QueryTurn (\nint Index,\nINetworkTurn Turn\n);\n``````\n\n#### Remarks\n\nThe QueryTurn method retrieves the turn element that traverses this edge at the specified index. The index values range from 0 to (TurnCount - 1).\n\nThe QueryTurn method requires an instantiated NetworkTurn object to be passed in as a parameter. You can create an empty NetworkTurn object by using the INetworkQuery::CreateNetworkElement method.\n\n### INetworkEdge.SourceID Property\n\nID of the network dataset source from which this network element was derived.\n\n``````Public ReadOnly Property SourceID As Integer\n``````\n``````public int SourceID {get;}\n``````\n\n### INetworkEdge.ToAzimuth Property\n\nDirection of travel at the to-end of the network edge element.\n\n``````Public ReadOnly Property ToAzimuth As Double\n``````\n``````public double ToAzimuth {get;}\n``````\n\n#### Remarks\n\nThe ToAzimuth property indicates the direction of travel at the terminating end of the edge element. For edge elements generated by an EdgeFeatureSource, the FromAzimuth and ToAzimuth values are measured clockwise in degrees relative to the positive Y-axis direction of the spatial reference of the network dataset.\n\n### INetworkEdge.TurnCount Property\n\nNumber of network turn elements in which this network edge element participates.\n\n``````Public ReadOnly Property TurnCount As Integer\n``````\n``````public int TurnCount {get;}\n``````\n\n#### Remarks\n\nThe TurnCount property returns the number of turn elements that traverse this edge.\n\n### INetworkEdge.TurnParticipationType Property\n\nParticipation of this network edge element within a network turn element.\n\n``````Public ReadOnly Property TurnParticipationType As esriNetworkTurnParticipationType\n``````\n``````public esriNetworkTurnParticipationType TurnParticipationType {get;}\n``````\n\n## Classes that implement INetworkEdge\n\nClasses Description\n\n## Remarks\n\nThe INetworkEdge interface is used to access the properties of the network edge element, such as its direction of travel and azimuth values.\n\nYour browser is no longer supported. 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https://math.stackexchange.com/questions/tagged/regularization	[ "# Questions tagged [regularization]\n\nRegularization, in mathematics and statistics and particularly in the fields of machine learning and inverse problems, refers to a process of introducing additional information in order to solve an ill-posed problem or to prevent overfitting. (Def: http://en.wikipedia.org/wiki/Regularization_(mathematics))\n\n356 questions\nFilter by\nSorted by\nTagged with\n19 views\n\n### Penalty function $f(a, b)$ for $a, b \\in \\mathbb{R}^+$ which adds a high positive penalty only when both $a, b$ are very small\n\nI'm doing Bayesian recruitment curve fitting where my curve has two parameters $a$ and $b$. Both $a, b \\in \\mathbb{R}^+$. I have put a Truncated Normal prior on $a$ and a Half Normal prior on $b$. I'm ...\n33 views\n\n29 views\n\n### regularization and punishment method in least square fitting\n\nI have a least square problem: $$\\min \\sum_j^p\| f(x_j)-y_j \|^2 \\\\ \\text{where } f(x) = a x - \\sum_{i}^{n} b_i J_1(c_i x)$$ where $J_1$ is the first order Bessel function. I have to find a set ...\n73 views\n\n### Sobolev regularity required for $\|\\nabla u\|^2 \\in H^1(\\Omega)$\n\nWe have $n=2$ and $\\Omega=\\{x \\in \\mathbb{R}^2\\\| \|\|x\|\|_2<1 \\}$. I need $\|\\nabla u\|^2 \\in H^1(\\Omega)$. So i would choose $u \\in W^{2,4}(\\Omega)$. But since we have $n=2$ I thought that a lower ...\n1 vote\n14 views\n\n### Morozov Discrepancy for PDE-Constrained Optimization With Bound Constraints\n\n$\\textbf{Background and context}$ Let $\\beta:\\Omega\\rightarrow\\mathbb{R}$, be an unknown spatially distributed parameter (over the spatial domain $\\Omega$). Let $\\mathcal{F}$ represent a parameter-to-...\n1 vote\n34 views\n\n### Weighted sum of N images, which minimizes their TV norm\n\nI have $K$ images $I_i, i\\in{1 \\ldots K}$ of the size $M \\times N$. I wish to find weights $w_i$, s.t. $w_i \\in [0,1]$ and $\\sum_1^K w_i = 1$ so that $$\|\\sum_{i=1}^{K} w_i I_i\|_{TV}$$ is minimal. I ...\n85 views\n\n50 views\n\n131 views\n\n### Constant terms of asymptotic expansions of smoothed sums of all prime numbers\n\nConsider the series $$\\sum_{n=1}^\\infty n e^{-n \\varepsilon}$$ For $\\varepsilon \\leq 0$, it diverges. For $\\varepsilon > 0$, it converges and equals $$\\frac{e^\\varepsilon}{(e^\\varepsilon - 1)^2}$$ ...\n45 views\n\n### Convergence of Tikohnov regularization for convex objective\n\nI have the following question: Assume we want to minimize some convex (not strictly convex) and coercive ($f(x) \\to \\infty$ as $\|\|x\|\| \\to \\infty$) function $f \\in C^2(R^n,R)$, which has possibly ...\n146 views\n\n1 vote\n112 views\n\n### Indefinite generalized Tikhonov regularization\n\nSuppose we want to choose $x$ to minimise the following (generalized) Tikhonov regularized least squares objective: $$(Ax-b)^\\top (Ax-b) + \\lambda [(x-c)^\\top W (x-c)],$$ where $W$ is symmetric, but ...\n1 vote\n132 views\n\n### Principal value integral with cosine\n\nWe are doing a practice set for an exam where we have to find the principal value of the integral $$I = P.V. \\int^\\infty_{-\\infty} \\frac{\\cos(x)}{(x-1)(x^2+1)} dx$$ Firstly, I am unsure if this ...\n300 views\n\n### Optimal Transport and Entropic Regularization\n\nWe are working with discrete optimal transport. Let $P$ be a matrix and let $H(P) =- \\sum_{i,j} P_{i,j} (\\log(P_{i,j})-1)$. Let $C$ be the cost matrix. And $\\langle C,P\\rangle$ the Frobenius inner ...\n92 views\n\n### Best optimization technique for solving overdetermined systems with a constraint\n\nI am trying to make a prediction model based on a system of linear equations: $A\\vec{x}=\\vec{b}$, where $\\vec{x}$ ($m\\times1$) is my learning parameters, $A (m\\times n)$ and $\\vec{b}$ $(m\\times1)$ are ...\n1 vote\n138 views\n\n### For $u \\in L^2([0,T])$ there exists $u_m \\in C^\\infty$ such that $u_m \\to u$ in $L^2_{loc}(]0,T[)$.\n\nI am reading Navier-Stokes Equations by Roger Temam and there is a point in a proof I do not understand. Let me explain: We have a function $u: [0,T] \\to H \\in L^2([0,T]; H)$, for $H$ some Hilbert ...\nIn A Generalization of the Cauchy Principal Value, the author presents a way to assign values for hypersingular integrals of the form $$I=\\int_a^b\\frac{f(x)\\,\\mathrm dx}{(x-u)^n},\\quad u\\in(a,b)$$ ..." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.800626,"math_prob":0.9995388,"size":5202,"snap":"2023-40-2023-50","text_gpt3_token_len":1591,"char_repetition_ratio":0.10311659,"word_repetition_ratio":0.00249066,"special_character_ratio":0.3206459,"punctuation_ratio":0.14365153,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999769,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-07T06:38:16Z\",\"WARC-Record-ID\":\"<urn:uuid:7840df0b-ee2d-4e15-99c4-10aa44715725>\",\"Content-Length\":\"347198\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a1505a20-9f3b-4293-93b8-6f5d0c7bf6b0>\",\"WARC-Concurrent-To\":\"<urn:uuid:304edc6e-1f5b-449a-ab51-7ef097a5bc61>\",\"WARC-IP-Address\":\"172.64.144.30\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/tagged/regularization\",\"WARC-Payload-Digest\":\"sha1:SY7K6BV7JGSUVJU6NTCYLWAGWX6PLGKZ\",\"WARC-Block-Digest\":\"sha1:X2JLOMWE5H4KYJA6DK5QL2ZXBCTL5FEQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100650.21_warc_CC-MAIN-20231207054219-20231207084219-00538.warc.gz\"}"}
https://stats.stackexchange.com/questions/164209/relationship-regression-coeffecient-and-line-plot-trendline-slope	[ "Relationship: Regression Coeffecient and Line Plot Trendline Slope\n\nWhy does the slope of my line fit plot not match the coefficient of the same variable in my regression?\n\nFor one of the variables, the coefficient is positive while the line fit plots trendline is negative.\n\nUsing Excel's regression tool... The line best fit plot is for Variable D\n\n• Without a reproducible example there is no way we can tell. You need to be more specific and show at least a snipped of the data so others here can understand what went wrong. – Andy Jul 31 '15 at 22:18\n• Just added screen shots, hopefully that provides enough insight. – Derek Jul 31 '15 at 22:33\n• What is A-H here? I looks like you are looking at only two dimensions X and Y in the plot above, yet your model seems to include an intercept and coefficients A through H. – StatsStudent Jul 31 '15 at 23:27\n• @StatsStudent are you implying that you'd like to see the line fit plots for the other variables? The reason why I am including only variable D's plot is because it has the behavior that I can make the least sense of. From a purely mathematical point of view, I thought that the \"coeffecient in a regression\" is always equal to the slope of the line fit plot's trendline – Derek Aug 3 '15 at 15:50" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.95707643,"math_prob":0.8470524,"size":717,"snap":"2019-43-2019-47","text_gpt3_token_len":149,"char_repetition_ratio":0.11640954,"word_repetition_ratio":0.0,"special_character_ratio":0.20781033,"punctuation_ratio":0.08571429,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99382365,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-21T00:58:55Z\",\"WARC-Record-ID\":\"<urn:uuid:7dc17338-712c-4a9b-9918-6b18ca8bc804>\",\"Content-Length\":\"141529\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6f5db7d3-e066-4b72-af7f-742460f7a6c2>\",\"WARC-Concurrent-To\":\"<urn:uuid:14829d87-db94-4964-8d3a-44929ffd286f>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://stats.stackexchange.com/questions/164209/relationship-regression-coeffecient-and-line-plot-trendline-slope\",\"WARC-Payload-Digest\":\"sha1:IJD3CSEFRLCO2EGXH62WYEXAUS757LPB\",\"WARC-Block-Digest\":\"sha1:ABUSJEGRSAKXR7IX6ZQNB2BOZGYAJ3CM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987750110.78_warc_CC-MAIN-20191020233245-20191021020745-00298.warc.gz\"}"}
https://file.scirp.org/Html/9-1770575_90289.htm	[ " Analysis on Sensitivity of Power System Stability to Generator Parameters\n\nJournal of Power and Energy Engineering\nVol.07 No.01(2019), Article ID:90289,18 pages\n10.4236/jpee.2019.71009\n\nAnalysis on Sensitivity of Power System Stability to Generator Parameters\n\nXiaoming Sun\n\nDepartment of Electrical Engineering, Chongqing Water Resources and Electric Engineering College, Chongqing, China", null, "", null, "", null, "", null, "Received: December 30, 2018; Accepted: January 27, 2019; Published: January 30, 2019\n\nABSTRACT\n\nThe sensitivity of power system stability (including transient and dynamic stabilities) to generator parameters (including parameters of generator model, excitation system and power system stabilizer) is analyzed in depth by simulations. From the tables and plots of the resultant simulated data, a number of useful rules are revealed. These rules can be directly applied to the engineering checking of generator parameters. Because the complex theoretical analyses are circumvented, the checking procedure is greatly simplified, remarkably promoting the working efficiency of electrical engineers on site.\n\nKeywords:\n\nSensitivity Analysis, Power System Stability, Transient Stability, Dynamic Stability, Generator Parameters, Excitation System, Power System Stabilizer", null, "1. Introduction\n\nGenerators are the most important component of power system, and thus the level of power system stability is closely related to generator parameters , which include both the parameters of generator models and the parameters of excitation systems and power system stabilizers (PSS). For this, testing and checking the correctness of generator parameters and the appropriateness of generator parameters’ combinations are very important tasks of electrical engineers prior to further analysis of power system stability. In case generator parameters are mistaken or their combinations are inappropriate, it is very likely to cause decision-making errors or deviations from optimal operating points, bringing unforeseen or unfavorable outcomes. Currently, however, strictly and accurately testing and checking generator parameters are actually impractical and impossible; because in modern bulk power system the number of generators is huge, the field tests on each generator are unavoidably time and cost consuming; what’s more, generator parameters may change with time and operating conditions. Therefore, the electrical engineers at the electric power dispatching center really need a simple and practical method for testing and checking generator parameters quickly and efficiently.\n\nThe main contribution of this paper is based on the actual engineering experiences of the author. From plenty of simulation experiments, the sensitivity of power system stability (including transient stability and dynamic stability) to generator parameters is analyzed and illustrated in detail, and as a result some feasible and fast rules for testing and checking generator parameters are revealed and summarized, which can be directly adopted by the electrical engineers on site. And because the complex theoretical analyses are circumvented, the testing and checking procedure is greatly simplified so that the working efficiency of electrical engineers is greatly promoted.\n\n2. Generator Models and Parameters\n\nThe number and definitions of generator parameters are correlated with specific generator models. To ensure generality, the integral model of generator should be adopted, in which the stator consists of three-phase windings and the rotator consists of salient poles, excitation winding f, d-axis equivalent damping winding D and two q-axis equivalent damping windings g and Q. Based on the Park transform, the per-unit equations of the integral model of generator under dq0 coordinates are as follows (because the magnetic field produced by 0-axis current i0 in stator windings is 0 and has no effect on the electric characteristics of rotator , the equations related to 0-axis component are omitted):\n\n$\\left\\{\\begin{array}{l}{\\upsilon }_{\\text{d}}=-{\\psi }_{\\text{q}}-{R}_{\\text{a}}{i}_{\\text{d}}\\\\ {\\upsilon }_{\\text{q}}={\\psi }_{\\text{d}}-{R}_{\\text{a}}{i}_{\\text{q}}\\\\ {\\upsilon }_{\\text{f}}=\\text{d}{\\psi }_{\\text{f}}/\\text{d}t+{R}_{\\text{f}}{i}_{\\text{f}}\\\\ 0=\\text{d}{\\psi }_{\\text{D}}/\\text{d}t+{R}_{\\text{D}}{i}_{\\text{D}}\\\\ 0=\\text{d}{\\psi }_{\\text{g}}/\\text{d}t+{R}_{\\text{g}}{i}_{\\text{g}}\\\\ 0=\\text{d}{\\psi }_{\\text{Q}}/\\text{d}t+{R}_{\\text{Q}}{i}_{\\text{Q}}\\end{array}\\text{ }\\text{ }\\text{ }\\left\\{\\begin{array}{l}{\\psi }_{\\text{d}}=-{X}_{\\text{d}}{i}_{\\text{d}}+{X}_{\\text{ad}}{i}_{\\text{f}}+{X}_{\\text{ad}}{i}_{\\text{D}}\\\\ {\\psi }_{\\text{f}}=-{X}_{\\text{ad}}{i}_{\\text{d}}+{X}_{\\text{f}}{i}_{\\text{f}}+{X}_{\\text{ad}}{i}_{\\text{D}}\\\\ {\\psi }_{\\text{D}}=-{X}_{\\text{ad}}{i}_{\\text{d}}+{X}_{\\text{ad}}{i}_{\\text{f}}+{X}_{\\text{D}}{i}_{\\text{D}}\\\\ {\\psi }_{\\text{q}}=-{X}_{\\text{q}}{i}_{\\text{q}}+{X}_{\\text{aq}}{i}_{\\text{g}}+{X}_{\\text{aq}}{i}_{\\text{Q}}\\\\ {\\psi }_{\\text{g}}=-{X}_{\\text{aq}}{i}_{\\text{q}}+{X}_{\\text{g}}{i}_{\\text{g}}+{X}_{\\text{aq}}{i}_{\\text{Q}}\\\\ {\\psi }_{\\text{Q}}=-{X}_{\\text{aq}}{i}_{\\text{q}}+{X}_{\\text{aq}}{i}_{\\text{g}}+{X}_{\\text{Q}}{i}_{\\text{Q}}\\end{array}$ (1)\n\nwhere ${\\upsilon }_{\\text{d}}$ , ${\\upsilon }_{\\text{q}}$ and ${\\upsilon }_{\\text{f}}$ are the voltages of d-axis, q-axis and excitation wingding f; id, iq, if, iD, ig and iQ are the currents of d-axis, q-axis, excitation wingding f, equivalent damping windings D, g and Q; Ra, Rf, RD, Rg and RQ are the resistances of one-phase stator winding, excitation winding f, equivalent damping windings D, g and Q; yd, yq, yf, yD, yg and yQ are the total magnetic flux linkages of d-axis and q-axis windings, excitation winding f, equivalent damping windings D, g and Q; Xd, Xq, Xad, Xaq, Xf, XD, Xg and XQ are the synchronous reactances of d-axis and q-axis, the armature reaction reactances of d-axis and q-axis windings, the reactance of excitation winding f, the reactances of equivalent damping windings D, g and Q. And, in Equation (1) two assumptions are made: i) assume dyd/dt ≈ 0 and dyq/dt ≈ 0, that is, the electromagnetic transient process is not considered, or the aperiodic component of the stator current is considered in another way ; ii) assume the per-unit value of the electric angular velocity w ≈ 1, making the equations related to $\\upsilon$ d and $\\upsilon$ q linearized.\n\nApparently, the total magnetic flux linkages yd, yq, yf, yD, yg and yQ in Equation (1) are inconvenient or even impossible to measure in practice, and therefore some measurable variables are introduced to represent these magnetic flux linkages indirectly:\n\n$\\left\\{\\begin{array}{l}{{E}^{\\prime }}_{\\text{d}}=-{X}_{\\text{aq}}{\\psi }_{\\text{g}}/{X}_{\\text{g}}\\\\ {{E}^{\\prime }}_{\\text{q}}={X}_{\\text{ad}}{\\psi }_{\\text{f}}/{X}_{\\text{f}}\\end{array}\\text{ }\\text{ }\\text{ }\\left\\{\\begin{array}{l}{{E}^{″}}_{\\text{d}}=-{X}_{\\text{aq}}\\left({X}_{\\text{σg}}{\\psi }_{\\text{Q}}+{X}_{\\text{σQ}}{\\psi }_{\\text{g}}\\right)/\\left({X}_{\\text{Q}}{X}_{\\text{g}}-{X}_{\\text{aq}}^{2}\\right)\\\\ {{E}^{″}}_{\\text{q}}={X}_{\\text{ad}}\\left({X}_{\\text{σD}}{\\psi }_{\\text{f}}+{X}_{\\text{σf}}{\\psi }_{\\text{D}}\\right)/\\left({X}_{\\text{D}}{X}_{\\text{f}}-{X}_{\\text{ad}}^{2}\\right)\\end{array}$ (2)\n\nwhere ${{E}^{\\prime }}_{\\text{d}}$ , ${{E}^{\\prime }}_{\\text{q}}$ and ${{E}^{″}}_{\\text{d}}$ , ${{E}^{″}}_{\\text{q}}$ are the transient and subtransient electromotive forces of d-axis and q-axis; Xsf, XsD, Xsg and XsQ are the leakage reactances of excitation winding f, equivalent damping windings D, g and Q.\n\nFurther, some measurable parameters are also introduced, not only simplifying the equations but also making the equations’ physical meanings clearer:\n\n$\\left\\{\\begin{array}{l}{{T}^{\\prime }}_{\\text{d}0}={X}_{\\text{f}}/{R}_{\\text{f}}\\\\ {{T}^{\\prime }}_{\\text{q}0}={X}_{\\text{g}}/{R}_{\\text{g}}\\end{array}\\text{ }\\text{ }\\text{ }\\left\\{\\begin{array}{l}{{T}^{″}}_{\\text{d}0}=\\left({X}_{\\text{D}}-{X}_{\\text{ad}}^{2}/{X}_{\\text{f}}\\right)/{R}_{\\text{D}}\\\\ {{T}^{″}}_{\\text{q}0}=\\left({X}_{\\text{Q}}-{X}_{\\text{aq}}^{2}/{X}_{\\text{g}}\\right)/{R}_{\\text{Q}}\\end{array}$ (3)\n\nwhere ${{T}^{\\prime }}_{\\text{d}0}$ , ${{T}^{\\prime }}_{\\text{q}0}$ and ${{T}^{″}}_{\\text{d}0}$ , ${{T}^{″}}_{\\text{q}0}$ are the open circuit transient and subtransient time constants of d-axis and q-axis.\n\nBy substituting Equations (2) and (3) into Equation (1), the 6th-order practical model of generator can be derived:\n\n$\\left\\{\\begin{array}{l}\\left\\{\\begin{array}{l}{\\upsilon }_{\\text{d}}={{E}^{″}}_{\\text{d}}+{{X}^{″}}_{\\text{q}}{i}_{\\text{q}}-{R}_{\\text{a}}{i}_{\\text{d}}\\\\ {\\upsilon }_{\\text{q}}={{E}^{″}}_{\\text{q}}-{{X}^{″}}_{\\text{d}}{i}_{\\text{d}}-{R}_{\\text{a}}{i}_{\\text{q}}\\end{array}\\\\ {{T}^{\\prime }}_{\\text{d}0}\\frac{\\text{d}{{E}^{\\prime }}_{\\text{q}}}{\\text{d}t}=\\frac{{X}_{\\text{ad}}{\\upsilon }_{\\text{f}}}{{R}_{\\text{f}}}-\\frac{{X}_{\\text{d}}-{X}_{\\text{σa}}}{{{X}^{\\prime }}_{\\text{d}}-{X}_{\\text{σa}}}{{E}^{\\prime }}_{\\text{q}}+\\frac{{X}_{\\text{d}}-{{X}^{\\prime }}_{\\text{d}}}{{{X}^{\\prime }}_{\\text{d}}-{X}_{\\text{σa}}}{{E}^{″}}_{\\text{q}}-\\frac{\\left({X}_{\\text{d}}-{{X}^{\\prime }}_{\\text{d}}\\right)\\left({{X}^{″}}_{\\text{d}}-{X}_{\\text{σa}}\\right)}{{{X}^{\\prime }}_{\\text{d}}-{X}_{\\text{σa}}}{i}_{\\text{d}}\\\\ {{T}^{″}}_{\\text{d}0}\\frac{\\text{d}{{E}^{″}}_{\\text{q}}}{\\text{d}t}=\\frac{{{X}^{″}}_{\\text{d}}-{X}_{\\text{σa}}}{{{X}^{\\prime }}_{\\text{d}}-{X}_{\\text{σa}}}{{T}^{″}}_{\\text{d}0}\\frac{\\text{d}{{E}^{\\prime }}_{\\text{q}}}{\\text{d}t}-{{E}^{″}}_{\\text{q}}+{{E}^{\\prime }}_{\\text{q}}+\\left({{X}^{\\prime }}_{\\text{d}}-{{X}^{″}}_{\\text{d}}\\right){i}_{\\text{d}}\\\\ {{T}^{\\prime }}_{\\text{q0}}\\frac{\\text{d}{{E}^{\\prime }}_{\\text{d}}}{\\text{d}t}=-\\frac{{X}_{\\text{q}}-{X}_{\\text{σa}}}{{{X}^{\\prime }}_{\\text{q}}-{X}_{\\text{σa}}}{{E}^{\\prime }}_{\\text{d}}+\\frac{{X}_{\\text{q}}-{{X}^{\\prime }}_{\\text{q}}}{{{X}^{\\prime }}_{\\text{q}}-{X}_{\\text{σa}}}{{E}^{″}}_{\\text{d}}+\\frac{\\left({X}_{\\text{q}}-{{X}^{\\prime }}_{\\text{q}}\\right)\\left({{X}^{″}}_{\\text{q}}-{X}_{\\text{σa}}\\right)}{{{X}^{\\prime }}_{\\text{q}}-{X}_{\\text{σa}}}{i}_{\\text{q}}\\\\ {{T}^{″}}_{\\text{q0}}\\frac{\\text{d}{{E}^{″}}_{\\text{d}}}{\\text{d}t}=\\frac{{{X}^{″}}_{\\text{q}}-{X}_{\\text{σa}}}{{{X}^{\\prime }}_{\\text{q}}-{X}_{\\text{σa}}}{{T}^{″}}_{\\text{q0}}\\frac{\\text{d}{{E}^{\\prime }}_{\\text{d}}}{\\text{d}t}-{{E}^{″}}_{\\text{d}}+{{E}^{\\prime }}_{\\text{d}}+\\left({{X}^{\\prime }}_{\\text{q}}-{{X}^{″}}_{\\text{q}}\\right){i}_{\\text{q}}\\\\ \\frac{\\text{d}\\delta }{\\text{d}t}=\\omega -1\\end{array}$ (4)\n\nwhere Xsa, ${{X}^{\\prime }}_{\\text{d}}$ , ${{X}^{\\prime }}_{\\text{q}}$ and ${{X}^{″}}_{\\text{d}}$ , ${{X}^{″}}_{\\text{d}}$ are the leakage reactance of stator winding, the transient and subtransient reactances of d-axis and q-axis. And the 6 equations are the voltage equations of stator, excitation winding f, equivalent damping windings D, g and Q, and the motion equation of stator, successively. From Equation (4), the specific position of each generator parameter in the formula is determined clearly.\n\n3. Fast Rules for Testing and Checking Generator Parameters in Models of Different Orders\n\nBy reducing Equation (4) to different extent, i.e., neglecting a certain number of windings or introducing new assumptions, the 5th-order, 4th-order, 3rd-order and 2nd-order models of generator can be obtained , and they are not listed out in this paper for brevity. In these models, only one or two parameters have slight discrepancies, and other parameters are the same. It is because of these slight discrepancies that some evident incorrectness of generator parameters can be tested and checked rapidly in terms of the following 6 rules.\n\n1) Rules for the 6th-order model (in this model d-axis, q-axis windings, excitation winding f and equivalent damping windings D, g and Q are all considered, and it is the detailed model of solid steam turbine or non-salient pole machine): ${{T}^{\\prime }}_{\\text{q0}}>0$ , ${X}_{\\text{q}}\\ne {{X}^{\\prime }}_{\\text{q}}$ .\n\n2) Rules for the 5th-order model (in this model equivalent damping winding g is neglected, and it is the detailed model of hydraulic turbine or salient pole machine): ${{T}^{\\prime }}_{\\text{q0}}=0$ , ${X}_{\\text{q}}\\ne {{X}^{\\prime }}_{\\text{q}}$ .\n\n3) Rules for the 4th-order model (in this model only d-axis, q-axis windings, excitation winding f and equivalent damping winding g are considered, and it is suitable for describing the solid steam turbine): ${{T}^{\\prime }}_{\\text{q0}}>0$ , ${X}_{\\text{q}}\\ne {{X}^{\\prime }}_{\\text{q}}$ .\n\n4) Rules for the 3rd-order model (in this model only d-axis, q-axis windings and excitation winding f are considered, and it is suitable for describing the salient pole machine when high computational accuracy is not required): ${{T}^{\\prime }}_{\\text{q0}}=0$ , ${X}_{\\text{q}}={{X}^{\\prime }}_{\\text{q}}$ .\n\n5) Rules for the 2nd-order model (in this model it is assumed that the excitation system is so strong that it can maintain the constancy of ${{E}^{\\prime }}_{\\text{d}}$ and ${{E}^{\\prime }}_{\\text{q}}$ ): ${{T}^{\\prime }}_{\\text{d}0}$ = a very big value.\n\n6) Rules for both the salient and non-salient pole machines: ${X}_{\\text{q}}\\ne {{X}^{\\prime }}_{\\text{d}}$ .\n\nThe rules above are merely the qualitative rules, and a number of quantitative testing and checking rules (variation ranges) derived from engineering experiences are listed in Table 1. From Table 1, two complementary rules can be summarized: a) ${X}_{\\text{d}}\\ge {X}_{\\text{q}}\\ge {{X}^{\\prime }}_{\\text{q}}>{{X}^{\\prime }}_{\\text{d}}>{{X}^{″}}_{\\text{q}}\\ge {{X}^{″}}_{\\text{d}}>{X}_{\\text{σa}}$ ; b) ${T}_{\\text{J}}>{{T}^{\\prime }}_{\\text{d0}}>{{T}^{\\prime }}_{\\text{q0}}>{{T}^{″}}_{\\text{d0}}\\ge {{T}^{″}}_{\\text{q0}}$ .\n\nProvided that the generator parameters are prominently deviating from the aforementioned 6 requirements, the above 2 rules and the reference ranges of Table 1, it is justified in doubting that the parameters are incorrect, and more careful testing means should be taken. However, it is a simple and fast method to test the generator parameters and is very suitable for the preliminary test of the newly obtained parameters.\n\n4. Sensitivity of Power System Stability to Generator Parameters\n\nThe testing and checking rules of generator parameters presented in Section 3 are necessary conditions but not sufficient conditions to guarantee power system\n\nTable 1. Quantitative testing and checking rules (variation ranges) of generator parameters.\n\nstability. Whether power system can maintain stable or not depends also on generator parameters’ combinations. This section is aimed at analyzing and illustrating the sensitivity of power system stability (including transient stability and dynamic stability) to generator parameters’ combinations by plenty of simulation experiments. To make it possible for readers to reproduce the simulation results, the IEEE 9-node test system is selected as the simulation model. And the simulation experiments are carried out on one of the most famous simulation software platforms of power system, the Chinese edition BPA, i.e. PSD-BPA (Power System Department―Bonneville Power Administration) .\n\n4.1. Overview of IEEE 9-Node Test System\n\nThe geographically interconnected diagram of the IEEE 9-node test system is shown in Figure 1, which displays the power flow distribution under normal operation condition. For brevity, this subsection lists out only generator parameters in Table 2, and other detailed parameters, i.e. the steady-state and transient parameters of loads, buses, transmission lines and transformers can be found in . The units of the parameters in Table 2 are the same as those in Table 1, and because Ra ≈ 0, Ra is omitted from Table 2.\n\nBy comparing the generator parameters in Table 1 and Table 2, it can be inferred that GEN2 is a hydraulic turbine and GEN3 is a steam turbine. However, GEN1 and GEN2’s ${{X}^{″}}_{\\text{d}}$ and ${{X}^{″}}_{\\text{d}}$ (indicated by the shadings in Table 2) do not strictly comply with the variation range in Table 1, meaning that Table 1 should only be used to judge the prominent incorrectness of generator parameters but\n\nFigure 1. The geographically interconnected diagram of the IEEE 9-node test system. The units of node voltages, active power and reactive power are kV, MW and MVar, respectively; “G” denotes the output energy of generator and “L” denotes the load of station.\n\nTable 2. Generator parameters in IEEE 9-node test system.\n\nnot be obeyed inflexibly, and generator parameters’ combinations should also be taken into consideration.\n\n4.2. Sensitivity of Transient Stability of Power System to Generator Parameters\n\nFrom a good many simulation experiments, the author finds that the transient stability of power system is very sensitive to the values of ${{X}^{″}}_{\\text{d}}/{{X}^{\\prime }}_{\\text{d}}$ and ${{X}^{″}}_{\\text{q}}/{{X}^{\\prime }}_{\\text{q}}$ . If one of these two values is greater than 0.95, under large-disturbance the power angle curve of generator tends to oscillate greatly and damp slowly, or even diverges, implying generator is out of transient stability. Simulation results in Figure 2 and Figure 3 have illustrated this fact, and therefore the appropriateness of the combinations of ${{X}^{\\prime }}_{\\text{d}}$ , ${{X}^{″}}_{\\text{d}}$ , ${{X}^{\\prime }}_{\\text{q}}$ and ${{X}^{″}}_{\\text{q}}$ can be tested and checked by transient stability simulation experiments.\n\nFigure 2 shows the power angle curves of GEN3 under two different conditions, i.e. ${{X}^{″}}_{\\text{d}}/{{X}^{\\prime }}_{\\text{d}}<0.95$ and ${{X}^{″}}_{\\text{d}}/{{X}^{\\prime }}_{\\text{d}}>0.95$ , respectively. The large-disturbance set in the simulation is a three-phase permanent fault on the 220 kV transmission line\n\nFigure 2. The sensitivity of transient stability of power system to ${{X}^{″}}_{\\text{d}}/{{X}^{\\prime }}_{\\text{d}}$ .\n\nFigure 3. The sensitivity of transient stability of power system to ${{X}^{″}}_{\\text{q}}/{{X}^{\\prime }}_{\\text{q}}$ .\n\nBUS1-STATION B: at 0 s, a three-phase short-circuit fault occurs on the side of STATION B; at 0.2 s, the breakers on both sides trip to clear the fault but do not reclose. From the comparison of the power angle curves in Figure 2(a) and Figure 2(b), it is seen that the former returns to a smooth line quickly, while the latter oscillates ceaselessly and cannot damp to a straight line for a very long time.\n\nThe large-disturbance set in the simulation of Figure 3 is the same as that in Figure 2. Because the power angle curve of GEN3 with ${{X}^{″}}_{\\text{q}}/{{X}^{\\prime }}_{\\text{q}}<0.95$ is similar to Figure 2(a), the simulation of Figure 3 considers only the situation of ${{X}^{″}}_{\\text{q}}/{{X}^{\\prime }}_{\\text{q}}>0.95$ . Again, because GEN3 in this situation has been out of transient stability, the power angle curve oscillates so great that it has exceeded the drawing ranges of PSD-BPA and thus can not be displayed properly. Therefore, the power angle curve here is replaced by the system frequency curve shown in Figure 3, which is recorded by the simulation process supervisor and presents the status of power system indirectly. In Figure 3, the system frequency curve oscillates severely and has already been impossible to return to a straight line, illustrating also that GEN3 is out of transient stability.\n\n4.3. Sensitivity of Dynamic Stability of Power System to Generator Parameters\n\nThe common steps to analyze the dynamic stability of power system are as follows.\n\n1) Use the small-disturbance analysis method, i.e. the frequency-domain method, to decompose the oscillation modes, and then calculate the real part, imaginary part, frequency and damping ratio of each oscillation mode, together with the electromechanical circuit correlation ratio, the modulus and phase angle of the right eigenvector, and the participation factors of the generators participating in the oscillation.\n\n2) Use Prony method, a famous time-domain method, to analyze the active powers of some important interconnection transmission lines under large-disturbance and decompose them into a number of oscillation modes as well.\n\n3) Try to find out the oscillation modes under large-disturbance, which are consistent with those found by small-disturbance analysis. If such oscillation modes are found, it can be concluded that the analytic results in frequency domain and time domain are consistent with each other, and that the dynamic stability analysis above is valid.\n\nAccording to the foregoing steps, firstly, apply the small-disturbance analysis to the IEEE 9-node test system. From the small-disturbance analysis, 6 oscillation modes are obtained, wherein only 1 oscillation mode has an electromechanical circuit correlation ratio that is greater than 1.0, meaning that it is the dominant oscillation mode (DOM). Due to this, for brevity, the tables and figures in this subsection show only the details of this DOM. And in Tables 3-7 and Figures 4-8, the sensitivities of the frequency and damping ratio of the DOM to GEN3’s parameters ${{T}^{\\prime }}_{\\text{d0}}$ , ${{T}^{\\prime }}_{\\text{q0}}$ , ${{T}^{″}}_{\\text{d0}}$ , ${{T}^{″}}_{\\text{q0}}$ and TJ are illustrated respectively, where the variation ranges of GEN3’s parameters are conform to Table 1.\n\nTable 3 and Figure 4 show that under small disturbance, with ${{T}^{\\prime }}_{\\text{d0}}$ increasing, the frequency and damping ratio of the DOM are gradually decreasing. Table 4 and Figure 5 show that under small disturbance, with ${{T}^{\\prime }}_{\\text{q0}}$ increasing, the frequency and damping ratio of the DOM are slightly increasing; however, on the whole they are not sensitive to ${{T}^{\\prime }}_{\\text{q0}}$ . Table 5 and Figure 6 show that under small disturbance, with ${{T}^{″}}_{\\text{d0}}$ increasing, the frequency of the DOM is almost unchanged, while the damping ratio is gradually decreasing. Table 6 and Figure 7 show that under small disturbance, with ${{T}^{″}}_{\\text{q0}}$ increasing, the frequency of the DOM is slightly decreasing and not sensitive to ${{T}^{″}}_{\\text{q0}}$ on the whole, while the\n\nTable 3. The sensitivity of DOM’s characteristics to ${{T}^{\\prime }}_{\\text{d0}}$ under small disturbance.\n\nFigure 4. The curves of the sensitivity of DOM’s characteristics to ${{T}^{\\prime }}_{\\text{d0}}$ under small disturbance.\n\nTable 4. The sensitivity of DOM’s characteristics to ${{T}^{\\prime }}_{\\text{q0}}$ under small disturbance.\n\nFigure 5. The curves of the sensitivity of DOM’s characteristics to ${{T}^{\\prime }}_{\\text{q0}}$ under small disturbance.\n\nTable 5. The sensitivity of DOM’s characteristics to ${{T}^{″}}_{\\text{d0}}$ under small disturbance.\n\nFigure 6. The curves of the sensitivity of DOM’s characteristics to ${{T}^{″}}_{\\text{d0}}$ under small disturbance.\n\nTable 6. The sensitivity of DOM’s characteristics to ${{T}^{″}}_{\\text{q0}}$ under small disturbance.\n\nFigure 7. The curves of the sensitivity of DOM’s characteristics to ${{T}^{″}}_{\\text{q0}}$ under small disturbance.\n\ndamping ratio has a small tendency of increasing. Table 7 and Figure 8 show that under small disturbance, only when TJ is smaller than a specific value, e.g. 6.5s as shown in Figure 8, are the frequency and damping ratio of the DOM sensitive to TJ.\n\nTable 7. The sensitivity of DOM’s characteristics to TJ under small disturbance.\n\nFigure 8. The curves of the sensitivity of DOM’s characteristics to TJ under small disturbance.\n\nSecondly, apply the large-disturbance analysis to the IEEE 9-node test system to testify the simulation results derived from the preceding small-disturbance analysis. The large disturbance set in the simulation here is the same three-phase permanent fault on 220 kV transmission line BUS1-STATION B as the one in Subsection 4.2. And the resultant active power of the 220 kV transmission line BUS2-STATION A is analyzed by Prony method. Likewise, in Tables 8-12 and Figures 9-13, the sensitivities of the frequency and damping ratio of the DOM to GEN3’s parameters ${{T}^{\\prime }}_{\\text{d0}}$ , ${{T}^{\\prime }}_{\\text{q0}}$ , ${{T}^{″}}_{\\text{d0}}$ , ${{T}^{″}}_{\\text{q0}}$ and TJ are illustrated respectively.\n\nTable 8 and Figure 9 show that under large disturbance, with ${{T}^{\\prime }}_{\\text{d0}}$ increasing, the frequency of the DOM is gradually decreasing, while the damping ratio is first increasing and then (at about 6 s) decreasing. Table 9 and Figure 10 show that under large disturbance, with ${{T}^{\\prime }}_{\\text{q0}}$ increasing, the frequency and damping ratio of the DOM are slightly oscillating across a straight line and their mean values are almost unchanged. Table 10 and Figure 11 show that under large disturbance, with ${{T}^{″}}_{\\text{d0}}$ increasing, the frequency of the DOM is not sensitive to ${{T}^{″}}_{\\text{d0}}$ and is almost unchanged, while the damping ratio is first increasing and then (at about 0.035 s) decreasing. Table 11 and Figure 12 show that under large disturbance, with ${{T}^{″}}_{\\text{q0}}$ increasing, the frequency of the DOM is slightly decreasing, while the damping ratio has a tendency of increasing. Table 12 and Figure 13 show that under large disturbance, only when TJ is smaller than a specific value, e.g. 6.5 s as shown in Figure 13, is the frequency of the DOM sensitive to TJ, while the damping ratio is first increasing and then (at about 4.7 s) decreasing.\n\nNow, by successively comparing the simulation results obtained from the time-domain method under large disturbance with those obtained from the frequency-domain method under small disturbance, we can see that the DOM’s frequency changes with generator parameters in both the time-domain analysis\n\nTable 8. The sensitivity of DOM’s characteristics to ${{T}^{\\prime }}_{\\text{d0}}$ under large disturbance.\n\nFigure 9. The curves of the sensitivity of DOM’s characteristics to ${{T}^{\\prime }}_{\\text{d0}}$ under large disturbance.\n\nTable 9. The sensitivity of DOM’s characteristics to ${{T}^{\\prime }}_{\\text{q0}}$ under large disturbance.\n\nFigure 10. The curves of the sensitivity of DOM’s characteristics to ${{T}^{\\prime }}_{\\text{q0}}$ under large disturbance.\n\nTable 10. The sensitivity of DOM’s characteristics to ${{T}^{″}}_{\\text{d0}}$ under large disturbance.\n\nFigure 11. The curves of the sensitivity of DOM’s characteristics to ${{T}^{″}}_{\\text{d0}}$ under large disturbance.\n\nTable 11. The sensitivity of DOM’s characteristics to ${{T}^{″}}_{\\text{q0}}$ under large disturbance.\n\nFigure 12. The curves of the sensitivity of DOM’s characteristics to ${{T}^{″}}_{\\text{q0}}$ under large disturbance.\n\nand frequency-domain analysis are consistent with each other, while the DOM’s damping ratio changes with generator parameters, except ${{T}^{\\prime }}_{\\text{q0}}$ and ${{T}^{″}}_{\\text{q0}}$ , are different from each other. It is because the frequency of an oscillation mode is determined by the inherent structural characteristic of power system, and thus is irrelevant to the operation mode and disturbance type of power system. And it is\n\nTable 12. The sensitivity of DOM’s characteristics to TJ under large disturbance.\n\nFigure 13. The curves of the sensitivity of DOM’s characteristics to TJ under large disturbance.\n\nlike the natural vibration frequency of a mechanical system. However, the damping ratio is not determined by the inherent structural characteristic of power system, and therefore it is sensitive to the operation mode and disturbance types of power system.\n\nAt the end of this section, two perspectives should be pointed out.\n\n1) The non-dominant oscillation modes’ frequency and damping ratio changes with generator parameters are similar to those of the DOM’s, and therefore the conclusions derived from the DOM are applicable to the non-dominant oscillation modes.\n\n2) If the damping ratio of certain oscillation mode is used to test and check generator parameters, the results obtained under small disturbance and large disturbance should be analyzed separately.\n\n5. Testing and Checking Excitation Systems and PSS Parameters\n\nExcept generator parameters per se, the adjustment of the parameters of excitation systems and PSS also plays an important role in ensuring the stability of generators and power system, and therefore the parameters of excitation systems and PSS are always treated as a requisite component of generator parameters. For this, this section proposes the engineering methods for testing and checking the parameters of excitation systems and PSS parameters.\n\n5.1. Testing and Checking Excitation System Parameters\n\nThe engineering method for testing and checking excitation systems parameters includes 4 steps.\n\n1) Stop the operation of the PSS of the tested generator.\n\n2) Isolate the tested generator from the electric network.\n\n3) Adjust the reference voltage of excitation system according to a specific function, e.g. step function or ramp function.\n\n4) Investigate whether the output voltage of generator is able to track the reference voltage effectively, and here the “effective track” means that the rising edge is steep, the overshoot is small and the steady area has no great oscillations.\n\nIt should be pointed out that a practical and effective way to finish step 4) is to compare the output voltage curve of the tested generator with that of a reference generator which has the correct parameters of both generator and excitation system. Here, the output voltage curve of the reference generator is named as classic curve. If the differences of the two curves are fairly small, it then can be inferred that the excitation system parameters are appropriate from the perspective of application; otherwise, it can be concluded that the excitation system parameters are inappropriate or ineffective, and should be adjusted again or measured directly from field test.\n\nFigure 14 shows the output voltage curve of GEN3 in the IEEE 9-node test system, where for comparison a classic curve is superposed on the figure. From Figure 14(a) and Figure 14(b), it is seen that the differences between the output voltage curve of GEN3 and the classic curve with different excitation system parameters are both acceptably small, meaning that the excitation system parameters\n\nFigure 14. The simulation testing and checking of excitation system parameters.\n\nof GEN3 has good robustness. It should be noted that excitation system parameters depend mainly on the dynamic amplification coefficient of its automatic voltage regulator (AVR) . In Figure 14(a), the dynamic amplification coefficient is relatively big, and it is seen that the response of the output voltage of GEN3 is fast but the overshoot is large; in Figure 14(b), the dynamic amplification coefficient is relatively small, and it is seen that the output voltage of GEN3 has no overshoot but the response is slow. Therefore, on the premise of guaranteeing stability, the dynamic amplification coefficient can be properly adjusted to meet the specific practical requirement.\n\n5.2. Testing and Checking PSS Parameters\n\nThe engineering method for testing and checking PSS parameters also includes 4 steps.\n\n1) Set up the “double machines and double lines” simulation system as shown in Figure 15(a), where GEN3’s parameters are the same as those in the IEEE 9-node test system, and other component’s parameters are labeled in the figure.\n\n2) Set a three-phase permanent fault on transformer’s high-voltage side bus: at 0 s, a three-phase short-circuit fault occurs; at 0.1 s, the breakers on both sides of the transmission lines trip to clear the fault and do not reclose.\n\n3) Switch on and off the PSS of GEN3, respectively.\n\n4) Compare the damping speeds of the oscillations of GEN3’s output power under the two conditions above. If the damping speed of the oscillation of the output power is much faster when PSS is switched on than that when PSS is switched off, it can be concluded that PSS parameters are appropriate; otherwise, PSS parameters should be readjusted.\n\nFigure 15. The simulation testing and checking of PSS parameters.\n\n6. Conclusion\n\nThis paper analyzes the sensitivity of power system stability to generator parameters, including the parameters of generator, excitation system and PSS from plenty of simulation experiments. Because the simulation processes are closely related to the engineering practice and do not involve any complex theoretical analysis, the summarized rules and conclusions are evident and practical, and thus can be directly referenced or applied by electrical engineers. Based on these rules and conclusions, some feasible methods are proposed for testing and checking generator parameters quickly, which may greatly promote the working efficiencies of electrical engineers. Although the simulation examples used in this paper is very simple, the resultant conclusions are of generality and can be easily testified by readers in their researching and working activities.\n\nAcknowledgements\n\nThis research is supported by the Science and Technology Research Project of Chongqing Educational Committee (KJ1603605) and the Natural Science Fund Project of Yongchuan District Science and Technology Committee (Ycstc, 2016nc3001).\n\nConflicts of Interest\n\nThe author declares no conflicts of interest regarding the publication of this paper.\n\nCite this paper\n\nSun, X.M. (2019) Analysis on Sensitivity of Power System Stability to Generator Parameters. Journal of Power and Energy Engineering, 7, 165-182. https://doi.org/10.4236/jpee.2019.71009\n\nReferences\n\n1. 1. Kim, D.J., Moon, Y.H., Lee, J.J., Ryu, H.S. and Kim, T.H. (2018) A New Method of Recording Generator Dynamics and Its Application to the Derivation of Synchronous Machine Parameters for Power System Stability Studies. IEEE Transactions on Energy Conversion, 33, 605-616. https://doi.org/10.1109/TEC.2017.2772234\n\n2. 2. Hasan, K.N., Preece, R. and Milanović, J. (2018) Application of Game Theoretic Approaches for Identification of Critical Parameters Affecting Power System Small-Disturbance Stability. International Journal of Electrical Power and Energy Systems, 97, 344-352. https://doi.org/10.1016/j.ijepes.2017.11.027\n\n3. 3. Izena, A., Kihara, H., Shimojo, T., Hirayama, K. and Furukawa, N. (2008) Generator Voltage Building-Up Field Test for 500 kV Transformer Energization for Black-Start Power System. IEEJ Transactions on Power and Energy, 128, 641-646. https://doi.org/10.1541/ieejpes.128.641\n\n4. 4. Hasegawa, K. and Imai, Y. (2002) Field Test of 70 MW Class Superconducting Generator. Cryogenics, 42, 191-197. https://doi.org/10.1016/S0011-2275(02)00034-6\n\n5. 5. Leonard, L.G. (2012) Power System Stability and Control. 3rd Edition, JCRC Press Inc., Taylor & Francis Group, Boca Raton.\n\n6. 6. IEEE (2007) IEEE Guide for Synchronous Generator Modeling Practices and Applications in Power System Stability Analyses. Revision of IEEE Std., 1110-1991. https://ieeexplore.ieee.org/document/1251520\n\n7. 7. Guo, Y.Z., Huang, R.M., Zhu, M. and Cai, X. (2018) Reexamination of New-Generation General Wind Turbine Models in PSD-BPA Transient Stability Simulation Program. Proceedings of the 13th IEEE Conference on Industrial Electronics and Applications, ICIEA 2018, Wuhan, 31 May-2 June 2018, 588-593. https://doi.org/10.1109/ICIEA.2018.8397784\n\n8. 8. Yan, C.Y., Wang, M. and Lu, J.J. (2013) The Contrast between the Models’ in PSD-BPA and those in PSASP. Advanced Materials Research, 614-615, 1055-1064. https://doi.org/10.4028/http://www.scientific.net/AMR.614-615.1055\n\n9. 9. Han, S., Rong, N., Sun, T. and Peng, X.J. (2013) Study on Conversion between the Common Models of PSD-BPA and PSS/E. Proceedings of 2013 IEEE 11th International Conference on Electronic Measurement and Instruments, ICEMI 2013, Harbin, 16-19 August 2013, 64-69. https://doi.org/10.1109/ICEMI.2013.6743040\n\n10. 10. Power System Department of China (2007) PSD Software Program Training Manual. https://wenku.baidu.com/view/0282520a79563c1ec5da713f.html\n\n11. 11. Yegireddy, N.K., Panda, S., Papinaidu, T. and Yadav, K.P.K. (2018) Multi-Objective Non-Dominated Sorting Genetic Algorithm-II Optimized PID Controller for Automatic Voltage Regulator Systems. Journal of Intelligent and Fuzzy Systems, 35, 4971-4975. https://doi.org/10.3233/JIFS-169781\n\n12. 12. Hoque, M.M. (2014) Design, Implementation and Performance Study of Programmable Automatic Voltage Regulator. Journal of Electrical Systems, 10, 472-483." ]	[ null, "https://html.scirp.org/file/9-1770575x1.png", null, "https://html.scirp.org/file/9-2500537x3.png", null, "https://html.scirp.org/file/9-2500537x2.png", null, "https://html.scirp.org/file/18-1760983x4.png", null, "https://html.scirp.org/file/2-1410169x6.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86796343,"math_prob":0.9914802,"size":29444,"snap":"2023-14-2023-23","text_gpt3_token_len":6410,"char_repetition_ratio":0.17252038,"word_repetition_ratio":0.13731144,"special_character_ratio":0.21641082,"punctuation_ratio":0.13612375,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9984918,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,4,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-08T16:23:50Z\",\"WARC-Record-ID\":\"<urn:uuid:1dadd79a-994b-481a-a6e2-a4720238bca4>\",\"Content-Length\":\"124298\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:248edbb8-3b46-4211-839f-b90a8e4be757>\",\"WARC-Concurrent-To\":\"<urn:uuid:50dd0030-1ce3-4dd9-8bff-9b5fcb51f677>\",\"WARC-IP-Address\":\"107.191.112.46\",\"WARC-Target-URI\":\"https://file.scirp.org/Html/9-1770575_90289.htm\",\"WARC-Payload-Digest\":\"sha1:BJ52I7U5MU4VZ6LIM3OD2EMXU2JGABP2\",\"WARC-Block-Digest\":\"sha1:UHWYL4QTNO5P5L676N2OZFYNEL7OSGW5\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224655027.51_warc_CC-MAIN-20230608135911-20230608165911-00496.warc.gz\"}"}
https://www.watchawear.com/help/tips-tutorials-tricks/40-smooth-day-rotation.html	[ "### Login to WatchAwear\n\n9\n\n#### smooth day rotation\n\nI've tried to make a smooth day rotation and here is my (quick & dirty) solution:\nphotoshop: draw a circle path and place numbers (01 - 31) [text ring layer doesn't work] - save it as an image; load it into watchmaker\n\n-- lua testscript for smooth day rotation\nstart_rotation = -3 -- in this example, our real 0 position is -3\nday_degrees = 11,86 -- degrees until next day (360/31 = 11,61)\nvar_s_daterotation = 0 -- this is our rotation\n\n-- * we need some global vars to test \n-- date vars\ndd = {dd} -- day\nddim = {ddim} -- day of month\n-- hour vars\ndh24 = {dh24} -- hour\ndm = {dm} -- minute\nds = {ds} -- second\n\nfunction get_start_rotation(dd, dh, dm, ds)\nr = day_degrees (dd - 1) -- r = start rotation\nr = r * -1 -- we rotate counterclockwise, so multiply with -1\nr = r + start_rotation\n\n-- r is now our rotation from (today, 00:00:00)\n\ndp = day_degrees / 86400 -- each second, we will rotate x percent more\ns = (dh * 3600) + (dm * 60) + ds -- our actually seconds\n\nreturn r + ((s * dp) * -1) -- give our real rotation back\nend\n\n-- * init our rotation \nvar_s_daterotation = get_start_rotation(dd, dh24, dm, ds)\n\n-- just testing, but nessessary *\nx = 51 -- in real, change to 1 (I've add 51 to each second & minute for testing)\n\nfunction on_second()\nds = ds + x\nif ds > 59 then\nds = 0\ndm = dm + x\nif dm > 59 then\ndm = 0\ndh24 = dh24 + 1\nif dh24 > 23 then\ndh24 = 0\ndd = dd + 1\nif dd > ddim then\ndd = 1\nddim = {ddim}\nend\nend\nend\nend\n\nvar_s_daterotation = get_start_rotation(dd, dh24, dm, ds)\nend\n\nPublished by Sabine Mayerhagen on 17 March 2017WatchMaker Tips & Tricks Posted\n\nDownload the\n'WatchAwear - Resources for WatchMaker' App", null, "Download The Best Watch Designing App\nfor Wear OS, and Tizen:", null, "No smartwatch required!!\nBest clock wallpaper for Android!", null, "" ]	[ null, "https://www.watchawear.com/images/app.png", null, "https://www.watchawear.com/images/watchmaker-app.png", null, "https://www.watchawear.com/images/watchmaker-live-wallpaper-app.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6067799,"math_prob":0.98808783,"size":1521,"snap":"2021-04-2021-17","text_gpt3_token_len":504,"char_repetition_ratio":0.14106789,"word_repetition_ratio":0.012618297,"special_character_ratio":0.39250493,"punctuation_ratio":0.08058608,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9806441,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-17T12:17:04Z\",\"WARC-Record-ID\":\"<urn:uuid:5953afbd-b451-4632-9b75-9d35296f6f11>\",\"Content-Length\":\"22813\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3dc9e3f9-07c6-4f23-bec2-db51dbe00148>\",\"WARC-Concurrent-To\":\"<urn:uuid:0555cbbb-57c6-4d27-835b-f59edd60aac9>\",\"WARC-IP-Address\":\"198.12.251.148\",\"WARC-Target-URI\":\"https://www.watchawear.com/help/tips-tutorials-tricks/40-smooth-day-rotation.html\",\"WARC-Payload-Digest\":\"sha1:BYO3URGOB5BYOQH3FHO6RVTUPCAV6ILJ\",\"WARC-Block-Digest\":\"sha1:JYXL7GCAPDLOSCU7KSOCEM3GSLIOSSQY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038119532.50_warc_CC-MAIN-20210417102129-20210417132129-00144.warc.gz\"}"}
https://fatareforms.org/3335/how-do-you-find-the-class-boundaries-in-a-frequency-distribution/	[ "# How do you find the class boundaries in a frequency distribution?\n\nThe lower boundary of each class is calculated by subtracting half of the gap value 12=0.5 1 2 = 0.5 from the class lower limit. On the other hand, the upper boundary of each class is calculated by adding half of the gap value 12=0.5 1 2 = 0.5 to the class upper limit. Simplify the lower and upper boundaries columns.\n\nThe lower class boundary is found by subtracting 0.5 units from the lower class limit and the upper class boundary is found by adding 0.5 units to the upper class limit. The difference between the upper and lower boundaries of any class.\n\ndoes the frequency distribution appear to have a normal distribution? No, the distribution does not appear to be normal. Does the frequency distribution appear to have a normal? distribution? Explain. Yes, because the frequencies start? low, proceed to one or two high? frequencies, then decrease to a low? frequency, and the distribution is approximately symmetric.\n\nAlso, how do you find the frequency distribution?\n\nSteps to Making Your Frequency Distribution\n\n1. Step 1: Calculate the range of the data set.\n2. Step 2: Divide the range by the number of groups you want and then round up.\n3. Step 3: Use the class width to create your groups.\n4. Step 4: Find the frequency for each group.\n\nHow do you find the class frequency?\n\nCount the tally marks to determine the frequency of each class. The relative frequency of a data class is the percentage of data elements in that class. The relative frequency can be calculated using the formula fi=fn f i = f n , where f is the absolute frequency and n is the sum of all frequencies.\n\n### How do you find the class width in a frequency table?\n\nCalculating Class Width in a Frequency Distribution Table Calculate the range of the entire data set by subtracting the lowest point from the highest, Divide it by the number of classes. Round this number up (usually, to the nearest whole number).\n\n### How do you calculate upper class boundaries?\n\nThe lower boundary of each class is calculated by subtracting half of the gap value 12=0.5 1 2 = 0.5 from the class lower limit. On the other hand, the upper boundary of each class is calculated by adding half of the gap value 12=0.5 1 2 = 0.5 to the class upper limit. Simplify the lower and upper boundaries columns.\n\n### How do you find the upper class limit?\n\nTo find the upper limit of the first class, subtract one from the lower limit of the second class. Then continue to add the class width to this upper limit to find the rest of the upper limits. Find the boundaries by subtracting 0.5 units from the lower limits and adding 0.5 units from the upper limits.\n\n### What do you mean by frequency distribution?\n\nFrequency distribution is a representation, either in a graphical or tabular format, that displays the number of observations within a given interval. Frequency distributions are typically used within a statistical context.\n\n### How do you find the cumulative frequency?\n\nThe cumulative frequency is calculated by adding each frequency from a frequency distribution table to the sum of its predecessors. The last value will always be equal to the total for all observations, since all frequencies will already have been added to the previous total.\n\n### What is class mark?\n\nThe class midpoint (or class mark) is a specific point in the center of the bins (categories) in a frequency distribution table; It’s also the center of a bar in a histogram. It is defined as the average of the upper and lower class limits.\n\n### What is class boundary?\n\nClass boundary is the midpoint of the upper class limit of one class and the lower class limit of the subsequent class. Each class thus has an upper and a lower class boundary.\n\n### How do you find lower frequency and relative class limit?\n\nMidpoint = Lower class limit + Upper class limit 2 . The “relative frequency” of each class is the proportion of the data that falls in that class. It can be calculated for a data set of size n by: Relative frequency = Class frequency Sample size = f n .\n\n### What is the difference between relative frequency and cumulative frequency?\n\nWhat is the difference between relative frequency and cumulative? frequency? Relative frequency of a class is the percentage of the data that falls in that? class, while cumulative frequency of a class is the sum of the frequencies of that class and all previous classes.\n\n### How do you find upper and lower limits?\n\nAdd three times the standard deviation to the average to get the upper control limit. Subtract three times the standard deviation from the average to get the lower control limit.\n\n### What is class limits in statistics?\n\nClass limits. The lower class limit of a class is the smallest data value that can go into the class. The upper class limit of a class is the largest data value that can go into the class. Class limits have the same accuracy as the data values; the same number of decimal places as the data values." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89243674,"math_prob":0.9811587,"size":5339,"snap":"2022-05-2022-21","text_gpt3_token_len":1134,"char_repetition_ratio":0.20431115,"word_repetition_ratio":0.18335089,"special_character_ratio":0.21408503,"punctuation_ratio":0.103415556,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99868995,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-18T06:12:21Z\",\"WARC-Record-ID\":\"<urn:uuid:aebc2840-9093-46f9-a50f-9c7c4634246f>\",\"Content-Length\":\"40694\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:75827223-96cf-4ea9-8cd1-83669bb04245>\",\"WARC-Concurrent-To\":\"<urn:uuid:30b2428c-055e-44c6-8e02-573f0eedfbf6>\",\"WARC-IP-Address\":\"104.21.2.45\",\"WARC-Target-URI\":\"https://fatareforms.org/3335/how-do-you-find-the-class-boundaries-in-a-frequency-distribution/\",\"WARC-Payload-Digest\":\"sha1:P5GFIHA2CI3PC675ITVIVQL6NRR6IX37\",\"WARC-Block-Digest\":\"sha1:YZLSYVZCYFKAI77O7CBZOE7GJ655YYSL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662521152.22_warc_CC-MAIN-20220518052503-20220518082503-00193.warc.gz\"}"}
https://www.it.uu.se/edu/course/homepage/ai/vt05/AI-theorem.html	[ "# Theorem proving, resolution\n\nLuger: 2.3, 13, (15)\n\n## Why theorem proving in an AI course?\n\n• proving theorems is considered to require high intelligence\n• if knowledge is represented by logic, theorem proving is reasoning\n• theorem proving uses AI techniques, such as (heuristic) search\n• (study how people prove theorems. Differently!)\n\n## What is theorem proving?\n\nReasoning by theorem proving is a weak method, compared to experts systems, because it does not make use of domain knowledge. This, on the other hand, may be a strength, if no domain heuristics are available (reasoning from first principles). Theorem proving is usually limited to sound reasoning.\n\nDifferentiate between\n\n• theorem provers: fully automatic\n• proof assistants: require steps as input, take care of bookkeeping and sometimes 'easy' proofs.\nTheorem proving requires\n• a logic (syntax)\n• a set of axioms and inference rules\n• a strategy on when how to search through the possible applications of the axioms and rules\nExamples of axioms\np -> (q->p)\n(p->(q->r)) -> ((p->q) ->(p->r))\np \\/ ~p\np->(~p->q)\nNotation: I use ~ for \"not\", since it's on my keyboard.\n\nExamples of inference rules\n\n name from derive modus ponens p, p->q q modus tollens p->q, ~q ~p and elimination p/\\q p and introduction p, q p/\\q or introduction p p\\/q instantiation for all X p(X) p(a) rename for all X phi(X) for all Y phi(Y) exists-introduction p(a) exists X p(X) substitution phi(p) phi(psi) replacement p->q ~p\\/q implication assume p ... ,q p->q contradiction assume ~p ...,false p resolution p\\/phi, ~p\\/psi phi \\/ psi (special case) p, ~p\\/psi psi (more special case) p, ~p false\n\nStrategies\nforwards - start from axioms, apply rules\nbackwards - start from the theorem (in general: a set of goals), work backwards to the axioms\n\nwhen to apply which rule\n\ngeneral questions:\nare the rules correct (sound)?\nis there a proof for every logical consequence (complete)?\ncan we remove rules (redundant)?\nHaving redundant rules may allow shorter proofs, but a larger search space.\n\n## Resolution and refutation\n\nWe want to prove theory -> goal.\nThe theory is usually a set of facts and rules, that can be treated as axioms (by the rule called implication above). A theory is usually quite stable, and used to prove various goals.\n• we use contradiction, add ~goal to the axioms, and try to prove false.\n• the theory and ~goal are put in clausal form - a set (conjunction) of clauses\n[see Luger p 558-560]\n• use resolution and unification to derive false\nClauses\na clause is a universially quantified disjunction of literals\na literal is an atomic formula or its negation\nexamples of clauses: p \\/ q, p(X) \\/ q(Y), ~p(X) \\/ q(X).\nA special case is Horn-clauses: they have at most one positive (not negated) literal.\nThree subclasses:\nfacts (1 pos, 0 neg): p(a,b).\nrules (1 pos, > 0 neg): ~p \\/ ~q \\/ r - often written as: p /\\ q -> r\ngoals(0 pos): ~p \\/ ~q - if we want to prove p/\\q from the theory, we add this clause - can be written as p /\\ q -> false\n\nSkolemization (p. 559)\nHow to remove existential quantification? Use function symbols!\nExample: every person has a mother: (for all X) person(X) -> (exists Y) mother(X,Y)\nGive a name to Y: the mother of X. (for all X) person(X) -> mother(X,mother_of(X))\n\nThis allows us to define datastructures in the logic.\nExample: if X is an element and Y a list, then there is a list with head X and tail Y.\n(for all X,Y) elt(X) /\\ list(Y) -> (exists Z) list(Z) /\\ head(Z,X) /\\ tail(Z,Y)\nfirst we name Z the cons of X and Y, then we get 3 rules:\nelt(X) /\\ list(Y) -> list(cons(X,Y)).\nelt(X) /\\ list(Y) -> head(cons(X,Y), X).\nelt(X) /\\ list(Y) -> tail(cons(X,Y), Y).\n\n### Unification\n\nUnification (2.3.2) is used to perform instantiation before (during) resolution.\n\nSimple case:\nResolve list(nil) with ~elt(X) \\/ ~list(Y) \\/ list(cons(X,Y)).\nFirst we must instantiate Y to nil, then we can derive ~elt(X) \\/ list(cons(X,nil)).\n\nHard case:\nI want to prove that there is a list that has the head mother_of(john).\nSo I take as a goal its negation ~(exists X)(list(X) /\\ head(X,mother_of(john)))\nThis gives the clause ~list(X) \\/ ~head(X,mother_of(john))\nWe can resolve with ~elt(X) \\/ ~list(Y) \\/ head(cons(X,Y),X).\nProblem 1: the X's in both clauses \"accidentally\" have the same name.\nSolution 1: rename to ~elt(Z) \\/ ~list(Y) \\/ head(cons(Z,Y),Z).\nThe unifier is: {Z/mother_of(john), X/cons(mother_of(john),Y)}\nThe resolvent (resulting clause) is\n(~list(X) \\/ ~elt(Z) \\/ ~list(Y)){Z/mother_of(john), X/cons(mother_of(john),Y)} =\n~list(cons(mother_of(john),Y)) \\/ ~elt(mother_of(john)) \\/ ~list(Y)\n\nExercise: use the first list-rule and the facts elt(mother_of(john)) and list(nil)to complete the proof.\n\nUnification as rewriting\nthe algorithm maintains a set of equalities. Each time, remove an equality and apply the applicable rule.\n\n• Var1 = Var1 - nothing (just remove the equality)\n• Var1 = Expr - in all other equalities, replace Var1 by Expr, keep Var1 = Expr\nexception: if Var1 occurs in Expr, then FAIL\n• f(A1,..,An) = f(B1,..,Bn) - add {A1 = B1, .., An = Bn}\n• f(A1,..,An) = g(B1,..,Bm) - FAIL\nExample (the equality selected is underlined)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.77942294,"math_prob":0.9807572,"size":6228,"snap":"2020-10-2020-16","text_gpt3_token_len":1811,"char_repetition_ratio":0.123071976,"word_repetition_ratio":0.005934718,"special_character_ratio":0.29174694,"punctuation_ratio":0.14771849,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9985334,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-27T22:53:36Z\",\"WARC-Record-ID\":\"<urn:uuid:c89c3be5-264a-4efb-95de-673982eea241>\",\"Content-Length\":\"9072\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f4928132-8f53-4685-91db-92ce687f3bac>\",\"WARC-Concurrent-To\":\"<urn:uuid:20637e11-4c60-4787-b4dc-f2f75ed65907>\",\"WARC-IP-Address\":\"130.238.12.100\",\"WARC-Target-URI\":\"https://www.it.uu.se/edu/course/homepage/ai/vt05/AI-theorem.html\",\"WARC-Payload-Digest\":\"sha1:JKRE7HDVFHQAQCCB2L76O64HN6ZDRSCW\",\"WARC-Block-Digest\":\"sha1:BYSVMDGO6ZYA2AVPTA3ILVS43YCMQDD6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875146907.86_warc_CC-MAIN-20200227221724-20200228011724-00220.warc.gz\"}"}
https://math.stackexchange.com/questions/1332807/hermitian-matrices-with-repeated-eigenvalues-has-codimension-3	[ "# Hermitian Matrices with Repeated Eigenvalues has Codimension 3?\n\nIt is sometimes claimed that\n\nthe space of $n\\times n$ Hermitian matrices with at least one repeated eigenvalue has codimension 3.\n\nThe proof of this in dimension two is very simple: by diagonalizing a $2\\times 2$ Hermitian matrix $H$, we easily conclude that if $H$ has the same eigenvalue twice, then $H$ must be a constant multiple of the $2\\times 2$ identity matrix, and this space has dimension 1, versus 4 for the space of $2\\times 2$ Hermitian matrices.\n\nHowever, I'm having difficulties with (what I believe to be) the standard demonstration of this in higher dimensions (provided in the paper \"On the Behaviour of Eigenvalues in Adiabatic Processes\" by Von Neumann and Wigner). From what I can tell, the demonstration (outline) goes goes as follows:\n\n1. An $n\\times n$ Hermitian matrix $H$ can be written as $UDU^$, where $U$ is unitary and $D$ is diagonal and contains the eigenvalues, say in increasing order.\n2. The decomposition $UDU^$ is not unique, but if we have $WDW^$ for another unitary matrix $W$, then we can write $W=UV$, where $V$ commutes with $D$, and so one could argue we can fix $U$ and $D$, and then write $H=(UV)D(UV)^$, where $V$ can be any unitary matrix that commutes with $D$.\n3. Thus, the number of real parameters one needs to specify the matrix $H$ would be $n^2+f-v$, where $n^2$ is the number of real parameters to specify a fixed unitary matrix $U$, $f$ is the number of parameters to specify the real eigenvalues of $D$, and $v$ is the number of parameters to specify $V$.\n4. We can show that, if $f$ is the number of eigenvalues of $H$, and $g_1,\\ldots,g_f$ are the multiplicities of the eigenvalues (i.e., $g_1+\\cdots+g_f=n$), then $v=g_1^2+g_2^2+\\cdots+g_f^2$.\n5. Therefore, if there is a repeated eigenvalue ($g_i\\geq 2$ for at least one $i$), the number of parameters will be at most $n^2-3$, hence the space of Hermitian matrices with repeated eigenvalues has codimension 3.\n\nMy problems with this are the following:\n\n• In the case of $2\\times 2$ matrices, it is clear that the space of Hermitian matrices with repeated eigenvalues is a vector space, as it consists of the multiples of the identity matrix. However, it is not at all clear to me that this is the case in higher dimensions.\n• In step 3. above, I don't understand how one just removes the parameters used to specify $V$ from the total amount of parameters. If it were argued that we have double counted some parameters and that we must now remove them, this would be fine, but it seems that we start with $n^2+f$ parameters and that we then somehow remove parameters that are completely unrelated to the previous ones.\n• Finally, I'm not sure I understand the connection between parameters and dimension. While it is true that Unitary matrices can be specified with $n^2$ numbers, we can't say that the space of Unitary matrices has dimension $n^2$ because unitary matrices do not form a linear subspace. So, I'm not sure how this business of counting parameters rigorously coincides with dimension of subspaces.\n• In the statement \"the space of $n\\times n$ Hermitian matrices with at least one repeated eigenvalue has codimension 3\", the word dimension does not refer to dimensions of linear subspaces, but instead to dimensions of (nonlinear) submanifolds and varieties. The argument you give is using fairly standard techniques to determine the dimension of this variety. Jun 24, 2015 at 14:38\n• @JimBelk Could you perhaps briefly describe how this parameter counting can be made into a rigorous determination of the codimension? Apr 3, 2016 at 20:18\n\nLet $K_n$ be the set of hermitian matrices of dimension $n$. Note that $E_n=\\{H\\in K_n\|card(spectrum(H))\\leq n-1\\}$ is a REAL algebraic set and $dim(E_n)$ is the greatest dimension of its components. A component of maximal dimension is obtained by considering $F_n=\\{H\\in K_n\|card(spectrum(H))= n-1\\}$. We consider the function $f:(U,D)\\in U(n)\\times DR\\rightarrow UDU^*\\in F_n$ where $DR=\\{diag((\\lambda_i)_i)\|\\lambda_i\\in\\mathbb{R},\\lambda_1=\\lambda_2\\}$. The stabilizer, $R=\\{V\\in U(n)\|VD=DV\\}=\\{V\\in U(n)\|V=diag(T_2,\\mu_3,\\cdots,\\mu_n)\\}\\approx U(2)\\times (S_1)^{n-2}$, has real dimension $2^2+1+\\cdots+1=n+2$; thus we obtain the real dimension $dim(F_n)=dim(U(n)\\times DR)-dim(R)=n^2+(n-1)-(n+2)=n^2-3$.\nRemark. $E_n=\\{H\\in K_n\|discrim(\\det(H-xI_n),x)=0\\}$. Note that $p(H)=discrim(\\det(H-xI_n),x)\\in\\mathbb{Z}[(u_{ij})_{ij}]$ where the $u_{ij}$ are the $n^2$ real parameters that define $K_n$. The previous result says that $p=0$ can be decomposed in $3$ independent real algebraic relations." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8700116,"math_prob":0.9998267,"size":3040,"snap":"2023-14-2023-23","text_gpt3_token_len":805,"char_repetition_ratio":0.14262187,"word_repetition_ratio":0.034951456,"special_character_ratio":0.2542763,"punctuation_ratio":0.09618574,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99998033,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-07T15:56:24Z\",\"WARC-Record-ID\":\"<urn:uuid:366065bb-65ee-49be-9ef6-c96642477e8f>\",\"Content-Length\":\"136560\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:092139fc-6a5e-4ec1-b35d-3d0a56fb9efe>\",\"WARC-Concurrent-To\":\"<urn:uuid:a8f90fa5-5fc1-4e54-a79d-1115ffebe091>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/1332807/hermitian-matrices-with-repeated-eigenvalues-has-codimension-3\",\"WARC-Payload-Digest\":\"sha1:HLI6NYBYQXWG3HOQ2W5ZUOI6ABHXY7BO\",\"WARC-Block-Digest\":\"sha1:HOGPXHGUA5TDFHYKZN5APVFTSUJ3JAXD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224653930.47_warc_CC-MAIN-20230607143116-20230607173116-00227.warc.gz\"}"}
https://www.colorhexa.com/32da18	[ "# #32da18 Color Information\n\nIn a RGB color space, hex #32da18 is composed of 19.6% red, 85.5% green and 9.4% blue. Whereas in a CMYK color space, it is composed of 77.1% cyan, 0% magenta, 89% yellow and 14.5% black. It has a hue angle of 112 degrees, a saturation of 80.2% and a lightness of 47.5%. #32da18 color hex could be obtained by blending #64ff30 with #00b500. Closest websafe color is: #33cc00.\n\n• R 20\n• G 85\n• B 9\nRGB color chart\n• C 77\n• M 0\n• Y 89\n• K 15\nCMYK color chart\n\n#32da18 color description : Vivid lime green.\n\n# #32da18 Color Conversion\n\nThe hexadecimal color #32da18 has RGB values of R:50, G:218, B:24 and CMYK values of C:0.77, M:0, Y:0.89, K:0.15. Its decimal value is 3332632.\n\nHex triplet RGB Decimal 32da18 `#32da18` 50, 218, 24 `rgb(50,218,24)` 19.6, 85.5, 9.4 `rgb(19.6%,85.5%,9.4%)` 77, 0, 89, 15 112°, 80.2, 47.5 `hsl(112,80.2%,47.5%)` 112°, 89, 85.5 33cc00 `#33cc00`\nCIE-LAB 76.609, -72.327, 71.634 26.55, 50.884, 9.286 0.306, 0.587, 50.884 76.609, 101.797, 135.276 76.609, -67.68, 91.364 71.333, -58.395, 42.215 00110010, 11011010, 00011000\n\n# Color Schemes with #32da18\n\n• #32da18\n``#32da18` `rgb(50,218,24)``\n• #c018da\n``#c018da` `rgb(192,24,218)``\nComplementary Color\n• #93da18\n``#93da18` `rgb(147,218,24)``\n• #32da18\n``#32da18` `rgb(50,218,24)``\n• #18da5f\n``#18da5f` `rgb(24,218,95)``\nAnalogous Color\n• #da1893\n``#da1893` `rgb(218,24,147)``\n• #32da18\n``#32da18` `rgb(50,218,24)``\n• #5f18da\n``#5f18da` `rgb(95,24,218)``\nSplit Complementary Color\n• #da1832\n``#da1832` `rgb(218,24,50)``\n• #32da18\n``#32da18` `rgb(50,218,24)``\n• #1832da\n``#1832da` `rgb(24,50,218)``\n• #dac018\n``#dac018` `rgb(218,192,24)``\n• #32da18\n``#32da18` `rgb(50,218,24)``\n• #1832da\n``#1832da` `rgb(24,50,218)``\n• #c018da\n``#c018da` `rgb(192,24,218)``\n• #229510\n``#229510` `rgb(34,149,16)``\n• #27ac13\n``#27ac13` `rgb(39,172,19)``\n• #2dc315\n``#2dc315` `rgb(45,195,21)``\n• #32da18\n``#32da18` `rgb(50,218,24)``\n• #3fe725\n``#3fe725` `rgb(63,231,37)``\n• #53e93c\n``#53e93c` `rgb(83,233,60)``\n• #67ec52\n``#67ec52` `rgb(103,236,82)``\nMonochromatic Color\n\n# Alternatives to #32da18\n\nBelow, you can see some colors close to #32da18. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #62da18\n``#62da18` `rgb(98,218,24)``\n• #52da18\n``#52da18` `rgb(82,218,24)``\n• #42da18\n``#42da18` `rgb(66,218,24)``\n• #32da18\n``#32da18` `rgb(50,218,24)``\n• #22da18\n``#22da18` `rgb(34,218,24)``\n• #18da1e\n``#18da1e` `rgb(24,218,30)``\n• #18da2f\n``#18da2f` `rgb(24,218,47)``\nSimilar Colors\n\n# #32da18 Preview\n\nThis text has a font color of #32da18.\n\n``<span style=\"color:#32da18;\">Text here</span>``\n#32da18 background color\n\nThis paragraph has a background color of #32da18.\n\n``<p style=\"background-color:#32da18;\">Content here</p>``\n#32da18 border color\n\nThis element has a border color of #32da18.\n\n``<div style=\"border:1px solid #32da18;\">Content here</div>``\nCSS codes\n``.text {color:#32da18;}``\n``.background {background-color:#32da18;}``\n``.border {border:1px solid #32da18;}``\n\n# Shades and Tints of #32da18\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #010601 is the darkest color, while #f5fef3 is the lightest one.\n\n• #010601\n``#010601` `rgb(1,6,1)``\n• #051803\n``#051803` `rgb(5,24,3)``\n• #092905\n``#092905` `rgb(9,41,5)``\n• #0e3b06\n``#0e3b06` `rgb(14,59,6)``\n• #124d08\n``#124d08` `rgb(18,77,8)``\n• #165e0a\n``#165e0a` `rgb(22,94,10)``\n• #1a700c\n``#1a700c` `rgb(26,112,12)``\n• #1e820e\n``#1e820e` `rgb(30,130,14)``\n• #229310\n``#229310` `rgb(34,147,16)``\n• #26a512\n``#26a512` `rgb(38,165,18)``\n• #2ab714\n``#2ab714` `rgb(42,183,20)``\n• #2ec816\n``#2ec816` `rgb(46,200,22)``\n• #32da18\n``#32da18` `rgb(50,218,24)``\n• #3ae61f\n``#3ae61f` `rgb(58,230,31)``\n• #49e831\n``#49e831` `rgb(73,232,49)``\n• #59ea43\n``#59ea43` `rgb(89,234,67)``\n• #69ec54\n``#69ec54` `rgb(105,236,84)``\n• #78ee66\n``#78ee66` `rgb(120,238,102)``\n• #88f078\n``#88f078` `rgb(136,240,120)``\n• #97f289\n``#97f289` `rgb(151,242,137)``\n• #a7f49b\n``#a7f49b` `rgb(167,244,155)``\n``#b6f6ad` `rgb(182,246,173)``\n• #c6f8be\n``#c6f8be` `rgb(198,248,190)``\n``#d6fad0` `rgb(214,250,208)``\n• #e5fce2\n``#e5fce2` `rgb(229,252,226)``\n• #f5fef3\n``#f5fef3` `rgb(245,254,243)``\nTint Color Variation\n\n# Tones of #32da18\n\nA tone is produced by adding gray to any pure hue. In this case, #767d75 is the less saturated color, while #24ed05 is the most saturated one.\n\n• #767d75\n``#767d75` `rgb(118,125,117)``\n• #6f866c\n``#6f866c` `rgb(111,134,108)``\n• #699062\n``#699062` `rgb(105,144,98)``\n• #629959\n``#629959` `rgb(98,153,89)``\n• #5ba250\n``#5ba250` `rgb(91,162,80)``\n• #54ab47\n``#54ab47` `rgb(84,171,71)``\n• #4db53d\n``#4db53d` `rgb(77,181,61)``\n• #46be34\n``#46be34` `rgb(70,190,52)``\n• #40c72b\n``#40c72b` `rgb(64,199,43)``\n• #39d121\n``#39d121` `rgb(57,209,33)``\n• #32da18\n``#32da18` `rgb(50,218,24)``\n• #2be30f\n``#2be30f` `rgb(43,227,15)``\n• #24ed05\n``#24ed05` `rgb(36,237,5)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #32da18 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5701744,"math_prob":0.7579417,"size":3676,"snap":"2020-34-2020-40","text_gpt3_token_len":1608,"char_repetition_ratio":0.12037037,"word_repetition_ratio":0.011090573,"special_character_ratio":0.5609358,"punctuation_ratio":0.23430493,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.985896,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-09T00:52:06Z\",\"WARC-Record-ID\":\"<urn:uuid:a26bff1c-3946-4815-8b33-9e4613d2d48e>\",\"Content-Length\":\"36258\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fdf62c46-3a47-4c52-ba7e-cf2df7f3fc32>\",\"WARC-Concurrent-To\":\"<urn:uuid:53236185-d3ad-4a7e-9f31-923853f46621>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/32da18\",\"WARC-Payload-Digest\":\"sha1:U2CBOB7Y4RX7UUTBCUJLWPZIVBUPSVYI\",\"WARC-Block-Digest\":\"sha1:3VLN6F7H6YZC4VQOYZSZUXMSX5VMMURF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738366.27_warc_CC-MAIN-20200808224308-20200809014308-00277.warc.gz\"}"}
https://www.griddlers.net/pages/gexample2	[ "# Griddlers Tutorial - Solving Puzzles\n\n## Example 2: Solving a Triangle Puzzle\n\nTriangle puzzles are solved in basically the same manner as a color puzzle. Triangles can be placed right beside solid-colored squares and differently rotated triangles. No background square is required between them.\n\nLet's look at this puzzle.", null, "Note the first row (marked with red). The first two triangles are different, therefore they can be beside each other in the puzzle. However, the 2nd and 3rd triangle are the same. They must be separated by at least one background square. The 3rd and 4th triangles are different. They can be side-by-side also. In the last row (marked with green), the triangles are the same. They must be separated by at least one background square.", null, "Let's begin. We can immediately fill in row 3. There are 5 squares available and the clues add up to 5.", null, "Fill in 3 squares of clue 4 in the fourth row using overlapping counting.", null, "Since the third column is complete, fill in the background color.", null, "Look at the triangle in the first column (circled in red). It matches this triangle clue. Therefore this triangle is the last clue and anything below it is background color.", null, "The first column can now be completed as there are only 2 clue blocks and 2 spaces remaining.", null, "Fill in the second and fifth columns in the same way.", null, "Fill in with background the remaining squares in the second and fifth rows since they have all the necessary blocks. Look at the remaining unplaced clues. We must place that triangle in the last empty square.", null, "Insert the triangle and the image is complete!" ]	[ null, "https://www.griddlers.net/images/ex2_1.gif", null, "https://www.griddlers.net/images/ex2_2.gif", null, "https://www.griddlers.net/images/ex2_3.gif", null, "https://www.griddlers.net/images/ex2_4.gif", null, "https://www.griddlers.net/images/ex2_5.gif", null, "https://www.griddlers.net/images/ex2_6.gif", null, "https://www.griddlers.net/images/ex2_7.gif", null, "https://www.griddlers.net/images/ex2_8.gif", null, "https://www.griddlers.net/images/ex2_9.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.93640375,"math_prob":0.92612696,"size":1581,"snap":"2023-14-2023-23","text_gpt3_token_len":348,"char_repetition_ratio":0.15852885,"word_repetition_ratio":0.07434944,"special_character_ratio":0.21252371,"punctuation_ratio":0.10130719,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9584491,"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,5,null,5,null,5,null,5,null,5,null,5,null,5,null,5,null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-05T18:24:08Z\",\"WARC-Record-ID\":\"<urn:uuid:834ac18c-cea0-4bf3-8a88-cc976ac5a565>\",\"Content-Length\":\"44388\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c52568aa-4ed4-40a3-9a09-428f51aba4f5>\",\"WARC-Concurrent-To\":\"<urn:uuid:a76ae877-65e6-4c6c-86b2-a2389df38268>\",\"WARC-IP-Address\":\"207.180.236.117\",\"WARC-Target-URI\":\"https://www.griddlers.net/pages/gexample2\",\"WARC-Payload-Digest\":\"sha1:5SVE2HL76NWI6BWRKGRAMN4NCN7QYL66\",\"WARC-Block-Digest\":\"sha1:A536SY47JNLCKVRZWNVWG7N3OWH2EDQA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224652149.61_warc_CC-MAIN-20230605153700-20230605183700-00757.warc.gz\"}"}
https://www.wisdomjobs.com/e-university/financial-management-tutorial-289/applications-of-the-general-dividend-valuation-model-6600.html	[ "# Applications of the General Dividend Valuation Model - Financial Management\n\n## The Generalized Dividend Valuation Model\n\nThe general dividend valuation model can be simplified if a firm’s dividend payments over time are expected to follow one of several different patterns, including zero growth, constant growth, and nonconstant growth.\n\nZero Growth Dividend Valuation Model\n\nIf a firm’s future dividend payments are expected to remain constant forever, then Dt in Equation, the general dividend valuation model, can be replaced by a constant value D to yield the following:", null, "This equation represents the value of a perpetuity and is analogous to those used for valuing a perpetual bond and a preferred stock developed in the previous chapter. It can be simplified to obtain\n\nP0 = D/ke\n\nThis model is valid only when a firm’s dividend payments are expected to remain constant forever. Although few common stocks strictly satisfy these conditions, the model can still be used to approximate the value of a stock for which dividend payments are expected to remain constant for a relatively long period into the future. To illustrate the zero growth dividend valuation model, assume that the Mountaineer Railroad common stock pays an annual dividend of $1.50 per share, which is expected to remain constant for the foreseeable future.What is the value of the stock to an investor who requires a 12 percent rate of return? Substituting$1.50 for D and 12 percent (0.12) for ke in Equation yields the following:\n\nP0 = $1.50/0.12 =$12.50\n\nRelationship Between Financial Decisions and Shareholder Wealth", null, "Dividend Growth Patterns\n\nConstant Growth Dividend Valuation Model\n\nIf a firm’s future dividend payments per share are expected to grow at a constant rate, g, per period forever, then the dividend at any future time period t can be forecast as follows:\n\nDt = D0(1 + g)t\n\nwhere D0 is the dividend in the current period (t = 0). The expected dividend in period 1 is D1 = D0(1 + g)1, the expected dividend in period 2 is D2 = D0(1 + g)2, and so on. The constant-growth curve in Figure illustrates such a dividend pattern.\n\nSubstituting Equation for Dt in the general dividend valuation model (Equation ) yields the following:", null, "Assuming that the required rate of return, ke, is greater than the dividend growth rate,17 g, Equation can be transformed algebraically to obtain the following simplified common stock valuation model:\n\nP0 = D1/(ke-g)\n\nNote that in the constant growth valuation model, the dividend value in the numerator is D1, that is, the dividend expected to be received one year from now. The model assumes that D0, the current dividend, has just been paid and does not enter the (forward -looking) valuation process.\nThe constant growth valuation model assumes that a firm’s earnings, dividends and stock price are expected to grow at a constant rate, g, into the future. Hence, to apply this model to a specific common stock, it is necessary to estimate the expected future growth rate, g. Considerable research evidence indicates that (1) the most accurate estimates of future growth are those provided by security analysts, and (2) consensus analyst forecasts of growth are an excellent proxy for growth expectations of investors. Sources of analyst growth rate forecasts include\n\n1. Value Line Investment Survey. Value Line reports, even though they represent only one analyst’s forecast for each company, are readily available at most public and university libraries and have been shown to be reasonably accurate and closely related to investor expectations.\n2. Zacks Earnings Estimates. It provides summaries of long-term (five years) and shortterm earnings growth expectations from more than 2,100 analysts covering more than 3,500 companies. Forecasts from Zacks are available through the Internet.\n3. Thomson Financial/First Call. This service is similar to that of Zacks and can be accessed through the Internet.\n\nThe constant growth dividend valuation model can be used to illustrate the two forms of returns an investor can expect to receive from holding a common stock. Solving Equation for ke yields the following:\n\nke = (D1/P0)+g\n\nThe investor’s required rate of return is equal to the expected dividend yield, D1/P0, plus the price appreciation yield, g—the expected increase in dividends and, ultimately, in the price of the stock.\n\nTo illustrate the application of the constant growth valuation model, consider Eaton Corporation. Dividends for Eaton are expected to be $1.76 per share next year. According to estimates from Value Line and First Call, earnings and dividends are expected to grow at about 6.5 percent annually. To determine the value of a share of this stock to an investor who requires a 12 percent rate of return, substituting$1.76 for D1, 6.5 percent(0.065) for g, and 12 percent(0.12) for ke in Equation 8.12 yields the following value for a share of Eaton’s common stock:\n\nP0 = $1.76/(0.12-0.065) =$32.00\n\nThus, the investor’s 12 percent required return consists of a 5.5 percent dividend yield (D1/P0 = $1.76/$32.00) plus a growth return of 6.5 percent annually.\n\nNonconstant Growth Dividend Valuation Model\n\nMany firms experience growth rates in sales, earnings, and dividends that are not constant. Typically, many firms experience a period of above -normal growth as they exploit new technologies, new markets, or both; this generally occurs relatively early in a firm’s life cycle. Following this period of rapid growth, earnings and dividends tend to stabilize or grow at a more normal rate comparable to the overall average rate of growth in the economy. The reduction in the growth rate occurs as the firm reaches maturity and has fewer sizable growth opportunities. The upper curve in Figure illustrates a nonconstant growth pattern.\n\nENTREPRENEURIAL ISSUES\n\nValuation of Closely Held Firms\n\nThe ownership of many small firms is closely held. An active market for shares of closely held companies normally does not exist. As a result ,entrepreneurs occasionally need to have the value of their enterprises estimated by independent appraisers.The reasons for these valuations include mergers and acquisitions, divestitures and liquidations, initial public offerings, estate and gift tax returns, leveraged buyouts, recapitalizations, employee stock ownership plans, divorce settlements, estate valuation, and various other litigation matters.\n\nThe principles of valuation developed in this chapter and applied to large publicly traded firms also apply to the valuation of small firms. Small firm valuation poses several unique challenges.When valuing the shares of a closely held corporation, many factors are considered, including the nature and history of the business, the general economic outlook and the condition and outlook of the firm’s industry, earnings capacity, dividend paying capacity, the book value of the company, the company’s financial condition, whether the shares represent a majority or minority (less than 50 percent) interest, and whether the stock is voting or nonvoting.\n\nSpecifically, however, in valuing a company that sells products and services, earnings capacity is usually the most important factor to be considered. Typically, the company as a whole is valued by determining a normal earnings level and multiplying that figure by an appropriate price —earnings multiple.This approach is known as the “capitalization of earnings” approach and results in a “going concern value.” If the shares represent a minority interest in the corporation, a discount is taken for the lack of marketability of these shares.\n\nDetermination of Normal Earnings\n\nThe determination of an average, or “normal,” earnings figure usually involves either a simple average or some type of weighted average of roughly the last five years of operations. For example, if the earnings of the company have been growing, some type of weighted average figure that places greater emphasis on more recent results is often used.\n\nIn some cases, the reported earnings of the company may not be an appropriate figure for valuation purposes. For example, suppose a portion of the salary paid to the president (and principal stockholder) really constitutes dividends paid as salary to avoid the payment of income taxes. In this situation, it is appropriate to adjust the reported earnings to account for these dividends.\n\nDetermination of an Appropriate Price–Earnings Multiple\n\nThe next step in the valuation process is to determine an appropriate rate at which to capitalize the normal earnings level.This is equivalent to multiplying the earnings by a price –earnings multiple. Had there been recent armslength transactions in the stock, the price –earnings multiple would be known and observable in the financial marketplace. However, this situation rarely exists for closely held corporations.As a consequence, the analyst must examine the price–earnings multiple for widely held companies in the same industry as the firm being valued in an attempt to find firms that are similar to the firm of interest.The price –earnings multiple for these comparable firms is used to capitalize the normal earnings level.\n\nMinority Interest Discount\n\nThe owner of a minority interest in a closely held corporation has an investment that lacks control and often marketability.There is usually either no market for the shares or the only buyer is either the other owners or the corporation itself. In addition, the owner of minority interest shares generally receives little, if any, dividends. The minority interest shareholder lacks control and is not able to change his or her inferior position.As a result of these problems, it is a widely practiced and accepted principle of valuation that the value of minority interest shares should be discounted.\n\nThe usual procedure in valuing minority interest shares is to value the corporation as a whole. Next, this value is divided by the number of shares outstanding to obtain a per-share value. Finally, a discount is applied to this per-share value to obtain the minority interest share value.These discounts have ranged from a low of 6 percent to more than 50 percent.\n\nIn conclusion, the basic valuation concepts are the same for small and large firms. However, the lack of marketability of shares for many small firms and the problem of minority interest positions pose special problems for the analyst.\n\nNonconstant growth models can also be applied to the valuation of firms that are experiencing temporary periods of poor performance, after which a normal pattern of growth is expected to emerge. (The lower nonconstant growth line in Figure illustrates this type of pattern.) For example, during 1992 IBM paid a dividend of $1.21 per share. This dividend was cut to$0.40 in 1993 and $0.25 in 1994. As IBM’s restructuring efforts progressed, earnings and dividend growth resumed. Dividends in 2000 grew to$0.51. Value Line projects year 2004 dividends of $0.70 per share and$1.25 per share by 2007. An investor attempting to value IBM’s stock at the beginning of 1993 should have reflected the declining, then increasing pattern of dividend growth for the firm.\n\nThere is no single model or equation that can be applied when nonconstant growth is anticipated. In general the value of a stock that is expected to experience a nonconstant growth rate pattern of dividends is equal to the present value of expected yearly dividends during the period of nonconstant growth plus the present value of the expected stock price at the end of the nonconstant growth period, or: Present value of expected Present value of the expected\n\nP0= dividends during period + stock price at the end of the of nonconstant growth nonconstant growth period\n\nThe stock price at the end of a nonconstant growth rate period can be estimated in a number of ways:\n\n1. Security analysts such as Value Line provide an estimated (five -year) future price range for the stocks they follow.\n2. Value Line, Zacks, and Thomson Financial/First Call all provide earnings growth rate estimates for five years into the future. These growth rate estimates can be used to derive earnings per share (EPS) forecasts for five years in the future. The EPS forecasts can be multiplied by the expected price -to-earnings (P/E) multiple, estimated by looking at current P/E multiples for similar firms, to get an expected price in five years.\n3. At the end of the period of nonconstant growth, an estimate of the value of the stock can be derived by applying the constant growth rate valuation model. For example, consider a firm expected to experience dividend growth at a nonconstant rate for m periods. Beginning in period m + 1, dividends are expected to grow at rate g2 forever. The value of the stock, Pm, at the end of period m is equal to\nPm = (Dm + 1)/(ke – g2)\n\nTo demonstrate how this model works, suppose an investor expects the earnings and common stock dividends of HILO Electronics to grow at a rate of 12 percent per annum for the next five years. Following the period of above -normal growth, dividends are expected to grow at the slower rate of 6 percent for the foreseeable future. The firm currently pays a dividend, D0, of $2 per share.What is the value of HILO common stock to an investor who requires a 15 percent rate of return? Value of HILO Electronics Common Stock", null, "Table illustrates the step -by -step solution to this problem. First, compute the sum of the present values of the dividends received during the nonconstant growth period (years 1 through 5 in this problem). This equals$9.25. Second, use the constant growth model to determine the value of HILO common stock at the end of year 5; P5 equals $41.56. Next, determine the present value of P5, which is$20.66. Finally, add the present value of the dividends received during the first five years ($9.25) to the present value of P5 ($20.66) to obtain the total value of the common stock of \\$29.91 per share. A timeline showing the expected cash flows from the purchase of HILO common stock is given in Figure.", null, "Financial Management Topics" ]	[ null, "https://www.wisdomjobs.com/tutorials/zero-growth-dividend-valuation-model.png", null, "https://www.wisdomjobs.com/tutorials/relationship-between-financial-decisions-and-shareholder-wealth.png", null, "https://www.wisdomjobs.com/tutorials/constant-growth-dividend-valuation-model.png", null, "https://www.wisdomjobs.com/tutorials/value-of-hilo-electronics-common-stock.png", null, "https://www.wisdomjobs.com/tutorials/cash-flows-from-hilo-electronics-common-stock.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9074488,"math_prob":0.9589133,"size":14034,"snap":"2019-13-2019-22","text_gpt3_token_len":2833,"char_repetition_ratio":0.15773343,"word_repetition_ratio":0.039251484,"special_character_ratio":0.1975203,"punctuation_ratio":0.09338678,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9805348,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-22T03:17:17Z\",\"WARC-Record-ID\":\"<urn:uuid:53dfb711-c535-4ff2-8837-6f0650c0bd14>\",\"Content-Length\":\"324843\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2c8e4362-108f-4a39-9130-acd7718a33f5>\",\"WARC-Concurrent-To\":\"<urn:uuid:e2ad81aa-335e-44eb-9e91-69b3fa04b586>\",\"WARC-IP-Address\":\"139.59.43.240\",\"WARC-Target-URI\":\"https://www.wisdomjobs.com/e-university/financial-management-tutorial-289/applications-of-the-general-dividend-valuation-model-6600.html\",\"WARC-Payload-Digest\":\"sha1:A6ZODUMVR25BOHO4NMF7EG6P4LHC2JTV\",\"WARC-Block-Digest\":\"sha1:7LR7Q2M477EF2OW4LGBXFUBOVNSB23AR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232256724.28_warc_CC-MAIN-20190522022933-20190522044933-00472.warc.gz\"}"}
http://mathcentral.uregina.ca/QQ/database/QQ.09.04/kate1.html	[ "", null, "Quandaries and Queries Two circles, C1 and C2, touch each other externally; and the line l is a common tangent. The line m is parallel to l and touches the two circls C1 and C3. The three circles are mutually tangent. If the radius of C2 is 9 and if the radius of C3 is 4, what is the radius of C1? Hi Kate, I drew a diagram and labeled some points. P, R and T are the centers of the circles. PQ and ST are parallel to the common tangent lines m and l.", null, "I then redrew part of the diagram, larger so that I can include some dimensions. I let r be the radius of C1.", null, "Since PQ and ST are parallel to the tangent lines. PQR and RST are right triangles and hence Pythagoras' Theorem can be used to find the lengths of PQ ans ST. Now draw a line trough T, parallel to SQ and meeting PQ at W.", null, "PWT is a right triangle and hence Pythagoras' Theorem applied to this triangle gives a quadratic you can solve for r. This is one configuration for the circles you describe in your problem but there is at least one other configuration. Chris Go to Math Central" ]	[ null, "http://mathcentral.uregina.ca/QQ/database/QQ.09.04/MCIcon.gif", null, "http://mathcentral.uregina.ca/QQ/database/QQ.09.04/kate1.1.gif", null, "http://mathcentral.uregina.ca/QQ/database/QQ.09.04/kate1.2.gif", null, "http://mathcentral.uregina.ca/QQ/database/QQ.09.04/kate1.3.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9269815,"math_prob":0.93845344,"size":1033,"snap":"2020-24-2020-29","text_gpt3_token_len":259,"char_repetition_ratio":0.1292517,"word_repetition_ratio":0.029268293,"special_character_ratio":0.23814134,"punctuation_ratio":0.0969163,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9985044,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-07T03:33:49Z\",\"WARC-Record-ID\":\"<urn:uuid:539573c1-7c09-4b19-8341-996ce1c34d02>\",\"Content-Length\":\"3877\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7e3bc5b6-fe06-4327-bffa-4e5255db37ac>\",\"WARC-Concurrent-To\":\"<urn:uuid:d2853ab4-7a77-4a67-b802-bb396427425b>\",\"WARC-IP-Address\":\"142.3.156.43\",\"WARC-Target-URI\":\"http://mathcentral.uregina.ca/QQ/database/QQ.09.04/kate1.html\",\"WARC-Payload-Digest\":\"sha1:E66T35MTXKUAZMEWI35BL2YNHRETNREM\",\"WARC-Block-Digest\":\"sha1:CQXXEWEWDSMVNPF5URBNKQHKDUXW3JSU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655891640.22_warc_CC-MAIN-20200707013816-20200707043816-00024.warc.gz\"}"}
https://www.colorhexa.com/00e192	[ "# #00e192 Color Information\n\nIn a RGB color space, hex #00e192 is composed of 0% red, 88.2% green and 57.3% blue. Whereas in a CMYK color space, it is composed of 100% cyan, 0% magenta, 35.1% yellow and 11.8% black. It has a hue angle of 158.9 degrees, a saturation of 100% and a lightness of 44.1%. #00e192 color hex could be obtained by blending #00ffff with #00c325. Closest websafe color is: #00cc99.\n\n• R 0\n• G 88\n• B 57\nRGB color chart\n• C 100\n• M 0\n• Y 35\n• K 12\nCMYK color chart\n\n#00e192 color description : Pure (or mostly pure) cyan - lime green.\n\n# #00e192 Color Conversion\n\nThe hexadecimal color #00e192 has RGB values of R:0, G:225, B:146 and CMYK values of C:1, M:0, Y:0.35, K:0.12. Its decimal value is 57746.\n\nHex triplet RGB Decimal 00e192 `#00e192` 0, 225, 146 `rgb(0,225,146)` 0, 88.2, 57.3 `rgb(0%,88.2%,57.3%)` 100, 0, 35, 12 158.9°, 100, 44.1 `hsl(158.9,100%,44.1%)` 158.9°, 100, 88.2 00cc99 `#00cc99`\nCIE-LAB 79.57, -63.701, 26.103 32.111, 55.922, 36.294 0.258, 0.45, 55.922 79.57, 68.841, 157.717 79.57, -69.048, 46.885 74.781, -54.219, 23.571 00000000, 11100001, 10010010\n\n# Color Schemes with #00e192\n\n• #00e192\n``#00e192` `rgb(0,225,146)``\n• #e1004f\n``#e1004f` `rgb(225,0,79)``\nComplementary Color\n• #00e122\n``#00e122` `rgb(0,225,34)``\n• #00e192\n``#00e192` `rgb(0,225,146)``\n• #00c0e1\n``#00c0e1` `rgb(0,192,225)``\nAnalogous Color\n• #e12200\n``#e12200` `rgb(225,34,0)``\n• #00e192\n``#00e192` `rgb(0,225,146)``\n• #e100c0\n``#e100c0` `rgb(225,0,192)``\nSplit Complementary Color\n• #e19200\n``#e19200` `rgb(225,146,0)``\n• #00e192\n``#00e192` `rgb(0,225,146)``\n• #9200e1\n``#9200e1` `rgb(146,0,225)``\n• #4fe100\n``#4fe100` `rgb(79,225,0)``\n• #00e192\n``#00e192` `rgb(0,225,146)``\n• #9200e1\n``#9200e1` `rgb(146,0,225)``\n• #e1004f\n``#e1004f` `rgb(225,0,79)``\n• #009560\n``#009560` `rgb(0,149,96)``\n• #00ae71\n``#00ae71` `rgb(0,174,113)``\n• #00c881\n``#00c881` `rgb(0,200,129)``\n• #00e192\n``#00e192` `rgb(0,225,146)``\n• #00fba3\n``#00fba3` `rgb(0,251,163)``\n``#15ffad` `rgb(21,255,173)``\n• #2fffb6\n``#2fffb6` `rgb(47,255,182)``\nMonochromatic Color\n\n# Alternatives to #00e192\n\nBelow, you can see some colors close to #00e192. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #00e15a\n``#00e15a` `rgb(0,225,90)``\n• #00e16d\n``#00e16d` `rgb(0,225,109)``\n• #00e17f\n``#00e17f` `rgb(0,225,127)``\n• #00e192\n``#00e192` `rgb(0,225,146)``\n• #00e1a5\n``#00e1a5` `rgb(0,225,165)``\n• #00e1b8\n``#00e1b8` `rgb(0,225,184)``\n• #00e1ca\n``#00e1ca` `rgb(0,225,202)``\nSimilar Colors\n\n# #00e192 Preview\n\nThis text has a font color of #00e192.\n\n``<span style=\"color:#00e192;\">Text here</span>``\n#00e192 background color\n\nThis paragraph has a background color of #00e192.\n\n``<p style=\"background-color:#00e192;\">Content here</p>``\n#00e192 border color\n\nThis element has a border color of #00e192.\n\n``<div style=\"border:1px solid #00e192;\">Content here</div>``\nCSS codes\n``.text {color:#00e192;}``\n``.background {background-color:#00e192;}``\n``.border {border:1px solid #00e192;}``\n\n# Shades and Tints of #00e192\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000906 is the darkest color, while #f5fffb is the lightest one.\n\n• #000906\n``#000906` `rgb(0,9,6)``\n• #001d13\n``#001d13` `rgb(0,29,19)``\n• #00301f\n``#00301f` `rgb(0,48,31)``\n• #00442c\n``#00442c` `rgb(0,68,44)``\n• #005839\n``#005839` `rgb(0,88,57)``\n• #006b46\n``#006b46` `rgb(0,107,70)``\n• #007f52\n``#007f52` `rgb(0,127,82)``\n• #00935f\n``#00935f` `rgb(0,147,95)``\n• #00a66c\n``#00a66c` `rgb(0,166,108)``\n• #00ba79\n``#00ba79` `rgb(0,186,121)``\n• #00cd85\n``#00cd85` `rgb(0,205,133)``\n• #00e192\n``#00e192` `rgb(0,225,146)``\n• #00f59f\n``#00f59f` `rgb(0,245,159)``\n• #09ffa9\n``#09ffa9` `rgb(9,255,169)``\n• #1dffb0\n``#1dffb0` `rgb(29,255,176)``\n• #30ffb6\n``#30ffb6` `rgb(48,255,182)``\n• #44ffbd\n``#44ffbd` `rgb(68,255,189)``\n• #58ffc4\n``#58ffc4` `rgb(88,255,196)``\n• #6bffcb\n``#6bffcb` `rgb(107,255,203)``\n• #7fffd2\n``#7fffd2` `rgb(127,255,210)``\n• #93ffd9\n``#93ffd9` `rgb(147,255,217)``\n• #a6ffe0\n``#a6ffe0` `rgb(166,255,224)``\n• #baffe7\n``#baffe7` `rgb(186,255,231)``\n• #cdffee\n``#cdffee` `rgb(205,255,238)``\n• #e1fff4\n``#e1fff4` `rgb(225,255,244)``\n• #f5fffb\n``#f5fffb` `rgb(245,255,251)``\nTint Color Variation\n\n# Tones of #00e192\n\nA tone is produced by adding gray to any pure hue. In this case, #687973 is the less saturated color, while #00e192 is the most saturated one.\n\n• #687973\n``#687973` `rgb(104,121,115)``\n• #5f8276\n``#5f8276` `rgb(95,130,118)``\n• #578a78\n``#578a78` `rgb(87,138,120)``\n• #4e937b\n``#4e937b` `rgb(78,147,123)``\n• #459c7d\n``#459c7d` `rgb(69,156,125)``\n• #3da480\n``#3da480` `rgb(61,164,128)``\n``#34ad83` `rgb(52,173,131)``\n• #2bb685\n``#2bb685` `rgb(43,182,133)``\n• #23be88\n``#23be88` `rgb(35,190,136)``\n• #1ac78a\n``#1ac78a` `rgb(26,199,138)``\n• #11d08d\n``#11d08d` `rgb(17,208,141)``\n• #09d88f\n``#09d88f` `rgb(9,216,143)``\n• #00e192\n``#00e192` `rgb(0,225,146)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #00e192 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.511334,"math_prob":0.8402999,"size":3704,"snap":"2019-51-2020-05","text_gpt3_token_len":1599,"char_repetition_ratio":0.13378379,"word_repetition_ratio":0.010989011,"special_character_ratio":0.5572354,"punctuation_ratio":0.23146068,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99201447,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-28T04:20:44Z\",\"WARC-Record-ID\":\"<urn:uuid:45a7d582-b7df-4e49-8780-2fb80f5b932f>\",\"Content-Length\":\"36276\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6101a2e9-fad1-42c5-a223-de910f2db6ba>\",\"WARC-Concurrent-To\":\"<urn:uuid:8f027f1b-e95f-4882-9533-5adf01732241>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/00e192\",\"WARC-Payload-Digest\":\"sha1:F5GFHEVIGVOGKXFOTVH3SP6FDLTVA3ZU\",\"WARC-Block-Digest\":\"sha1:7FIESRYETSBIWPX57BOATIUNLPU6OGLV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251773463.72_warc_CC-MAIN-20200128030221-20200128060221-00287.warc.gz\"}"}
https://essay.ua-referat.com/Ratio_Of_Charge_To_Mass_For_The	[ "", null, "# Ratio Of Charge To Mass For The\n\nскачати\n\nRatio Of Charge To Mass For The Electron Essay, Research Paper\n\nIntroduction: The object of this lab was to determine the measure of the ratio of an electron to its mass. This is done by accelerating a stream of electrons through a measured potential difference. The stream of electrons moves through a uniform magnetic field. It is perpendicular to the velocity of the electrons. The path of the electrons is circular because of this fact. The ratio of e/m can be found by the relationships between the measured accelerating potential difference, the diameter of the circular path described by the electron, and the magnetic flux density.\n\nTheory: British scientist Sir J.J. Thompson (1856-1940) first discovered that the electron was a discrete particle of electricity. From his discoveries came the accepted value for e/m which is 1.7589010^11 coulombs/kg. With this information we could then accurately determine the mass of the electron.\n\nThe force F acting upon a charge that is moving with a velocity v perpendicular to the magnetic field B is\n\nThis force is centripetal. These forces cause the electron to move in a circular path. The centrifugal force of reaction of the electron is equal in magnitude to the force on the electron by the magnetic field. Therefore the following equation is valid for this experiment. R is the radius of the path of the electrons.\n\nThrough a potential difference, the kinetic energy acquired by the falling electron is:\n\nFrom these last two equations, we can make a third equation involving all of the variables.\n\nWith this apparatus for this experiment, we can determine values for V, B, and r. With these values we can determine the ratio for e/m. The current in these two Hemholtz coils produces a magnetic field which bends the beam. Since the coils are vertical, the beam is horizontal. This is because the beam and the magnetic field are perpendicular. In the experiment, since the distance between the coils is equal of the radius of both of the coils, a nearly uniform magnetic field is produced at the midway point. The currents in the coils must yield fields of the coils that are in the same direction as their common axis.\n\nAt a central point, the magnitude of the flux density B is:\n\nN is the number of turns per coil. In this experiment N=129. I is the current in the coils. This is always changing. R is the coil radius. For the experiment R=10.75 is the permeablilty of free space. =4 10^-7 weber/ampm . Later we will come to find that when the specified units are used, e/m is expressed in coulombs/kilogram.\n\nData: Connect the apparatus as shown in the following diagram. The electron stream should have a diameter of about 2mm.\n\nSet the rheostat for high resistance close to the circuit field coils and then vary the current to (3-5) amp until the electron beam bends into a complete semicircle. Arrange the plate potential so that the accelerating potential difference can very and change the field current until the beam falls on the marked circles. The plate potential, the field current, and the radius of the described circle were recorded as follows.\n\nBecause these values for e/m were so different than the predicted value of e/m (1.7589010^11 coulombs/kg) the experiment was run a second time, but with a higher voltage and current. The data was recorded as follows:\n\nConclusion: From this data, and use of the working equation for e/m, it can be concluded that our experiment was consistent, but off from the given value of e/m by a fairly constant amount. This can be caused by several sources of error. One might be that we did not take the magnetic field of the earth into account. Another could be a magnetic field created by a nearby object. Still a third source of error might have been that we could not determine which ring on the plate that the electron beam curve actually hit. This experiment does show that given a Magnetic field, a V, an I, and an R, we can calculate e/m and compare this value to our given value for e/m.\n\nSpecial Thanks to Dr. Tarvin and the Samford University Department of Physics for direction and extreme patience with this project\n\nДодати в блог або на сайт\n\nЦей текст може містити помилки.\n\nA Free essays \| Essay\n\nRelated works:\nThe Golden Ratio\nDetermining The Ratio Of Circumference To Diameter\nThe Charge\nElectric Charge\nDulce Et Decorum Est Vs. The Charge\nComparing Dulce Et Decorum Est The Charge\nMass 2\nMass\nMass Spectrometer" ]	[ null, "https://essay.ua-referat.com/ua-referat.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9390837,"math_prob":0.9764974,"size":4126,"snap":"2020-24-2020-29","text_gpt3_token_len":866,"char_repetition_ratio":0.15041243,"word_repetition_ratio":0.0,"special_character_ratio":0.20940378,"punctuation_ratio":0.09547123,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9962194,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-26T09:02:38Z\",\"WARC-Record-ID\":\"<urn:uuid:cdbbfd6d-ec5b-496f-b29f-cfccb8a10eef>\",\"Content-Length\":\"19823\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:51b2364c-c662-4faf-b0cd-bd3d6c38ff5f>\",\"WARC-Concurrent-To\":\"<urn:uuid:828eab01-07e8-4bb8-88a8-96414f111ede>\",\"WARC-IP-Address\":\"176.9.102.205\",\"WARC-Target-URI\":\"https://essay.ua-referat.com/Ratio_Of_Charge_To_Mass_For_The\",\"WARC-Payload-Digest\":\"sha1:K7LZZBPIBIVP4LHT633CYN3IVBCF3NGI\",\"WARC-Block-Digest\":\"sha1:UX7HF2Z424K2AB67TZWLMCTFMQSLJKRG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347390755.1_warc_CC-MAIN-20200526081547-20200526111547-00492.warc.gz\"}"}
https://socratic.org/questions/590cd8797c014915525a4c01	[ "# Question a4c01\n\nMay 5, 2017\n\n$1.7 \\cdot {10}^{3} \\text{ppm}$\n\n#### Explanation:\n\nThe thing to remember about a mixture's concentration in parts per million, or ppm, is that it is supposed to tell you the number of parts of a given component of the mixture present for every\n\n${10}^{6} = 1 , 000 , 000$\n\nparts of mixture. In your case, you know that the sample has a total mass of $\\text{70 g}$. This sample is said to contain $\\text{0.119 g}$ of iron.\n\nYour goal here is to figure out how many grams of iron you would get for ${10}^{6}$ $\\text{g}$ of sample by using its known composition.\n\n10^6 color(red)(cancel(color(black)(\"g sample\"))) * \"0.119 g Fe\"/(70color(red)(cancel(color(black)(\"g sample\")))) = 1.7 * 10^3# $\\text{g Fe}$\n\nThis means that the sample's concentration in parts per million is equal to\n\n$\\textcolor{\\mathrm{da} r k g r e e n}{\\underline{\\textcolor{b l a c k}{\\text{concentration Fe\" = 1.7 * 10^3color(white)(.)\"ppm}}}}$\n\nI'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the mass of the sample." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9010058,"math_prob":0.9964203,"size":886,"snap":"2023-14-2023-23","text_gpt3_token_len":241,"char_repetition_ratio":0.117913835,"word_repetition_ratio":0.0,"special_character_ratio":0.29119638,"punctuation_ratio":0.07386363,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9972782,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-28T00:15:44Z\",\"WARC-Record-ID\":\"<urn:uuid:1561b4ec-c3c3-4cda-aebc-1568d52e3b49>\",\"Content-Length\":\"34944\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0a7245e7-ecea-4a66-99b0-19079ead7b2d>\",\"WARC-Concurrent-To\":\"<urn:uuid:f88ce3cb-d984-4d81-8f63-bb38d15f0718>\",\"WARC-IP-Address\":\"216.239.36.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/590cd8797c014915525a4c01\",\"WARC-Payload-Digest\":\"sha1:GKQI4ISBUJBZK2DC2LYJ7OE3JZZOX334\",\"WARC-Block-Digest\":\"sha1:QNSK55QOVGIOBH347EM37UHPU7DNJHXS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296948708.2_warc_CC-MAIN-20230327220742-20230328010742-00335.warc.gz\"}"}
http://dotnumerics.com/NumericalLibraries/LinearAlgebra/CSharpCodeFiles/dggqrf.aspx	[ "", null, "", null, "` 1: #region Translated by Jose Antonio De Santiago-Castillo.`\n` 2: `\n` 3: //Translated by Jose Antonio De Santiago-Castillo. `\n` 4: //E-mail:[email protected]`\n` 5: //Web: www.DotNumerics.com`\n` 6: //`\n` 7: //Fortran to C# Translation.`\n` 8: //Translated by:`\n` 9: //F2CSharp Version 0.71 (November 10, 2009)`\n` 10: //Code Optimizations: None`\n` 11: //`\n` 12: #endregion`\n` 13: `\n` 14: using System;`\n` 15: using DotNumerics.FortranLibrary;`\n` 16: `\n` 17: namespace DotNumerics.CSLapack`\n` 18: {`\n` 19: /// <summary>`\n` 20: /// -- LAPACK routine (version 3.1) --`\n` 21: /// Univ. of Tennessee, Univ. of California Berkeley and NAG Ltd..`\n` 22: /// November 2006`\n` 23: /// Purpose`\n` 24: /// =======`\n` 25: /// `\n` 26: /// DGGQRF computes a generalized QR factorization of an N-by-M matrix A`\n` 27: /// and an N-by-P matrix B:`\n` 28: /// `\n` 29: /// A = QR, B = QTZ,`\n` 30: /// `\n` 31: /// where Q is an N-by-N orthogonal matrix, Z is a P-by-P orthogonal`\n` 32: /// matrix, and R and T assume one of the forms:`\n` 33: /// `\n` 34: /// if N .GE. M, R = ( R11 ) M , or if N .LT. M, R = ( R11 R12 ) N,`\n` 35: /// ( 0 ) N-M N M-N`\n` 36: /// M`\n` 37: /// `\n` 38: /// where R11 is upper triangular, and`\n` 39: /// `\n` 40: /// if N .LE. P, T = ( 0 T12 ) N, or if N .GT. P, T = ( T11 ) N-P,`\n` 41: /// P-N N ( T21 ) P`\n` 42: /// P`\n` 43: /// `\n` 44: /// where T12 or T21 is upper triangular.`\n` 45: /// `\n` 46: /// In particular, if B is square and nonsingular, the GQR factorization`\n` 47: /// of A and B implicitly gives the QR factorization of inv(B)A:`\n` 48: /// `\n` 49: /// inv(B)A = Z'(inv(T)R)`\n` 50: /// `\n` 51: /// where inv(B) denotes the inverse of the matrix B, and Z' denotes the`\n` 52: /// transpose of the matrix Z.`\n` 53: /// `\n` 54: ///</summary>`\n` 55: public class DGGQRF`\n` 56: {`\n` 57: `\n` 58: `\n` 59: #region Dependencies`\n` 60: `\n` 61: DGEQRF _dgeqrf; DGERQF _dgerqf; DORMQR _dormqr; XERBLA _xerbla; ILAENV _ilaenv; `\n` 62: `\n` 63: #endregion`\n` 64: `\n` 65: `\n` 66: #region Fields`\n` 67: `\n` 68: bool LQUERY = false; int LOPT = 0; int LWKOPT = 0; int NB = 0; int NB1 = 0; int NB2 = 0; int NB3 = 0; `\n` 69: `\n` 70: #endregion`\n` 71: `\n` 72: public DGGQRF(DGEQRF dgeqrf, DGERQF dgerqf, DORMQR dormqr, XERBLA xerbla, ILAENV ilaenv)`\n` 73: {`\n` 74: `\n` 75: `\n` 76: #region Set Dependencies`\n` 77: `\n` 78: this._dgeqrf = dgeqrf; this._dgerqf = dgerqf; this._dormqr = dormqr; this._xerbla = xerbla; this._ilaenv = ilaenv; `\n` 79: `\n` 80: #endregion`\n` 81: `\n` 82: }`\n` 83: `\n` 84: public DGGQRF()`\n` 85: {`\n` 86: `\n` 87: `\n` 88: #region Dependencies (Initialization)`\n` 89: `\n` 90: LSAME lsame = new LSAME();`\n` 91: XERBLA xerbla = new XERBLA();`\n` 92: DLAMC3 dlamc3 = new DLAMC3();`\n` 93: DLAPY2 dlapy2 = new DLAPY2();`\n` 94: DNRM2 dnrm2 = new DNRM2();`\n` 95: DSCAL dscal = new DSCAL();`\n` 96: DCOPY dcopy = new DCOPY();`\n` 97: IEEECK ieeeck = new IEEECK();`\n` 98: IPARMQ iparmq = new IPARMQ();`\n` 99: DGEMV dgemv = new DGEMV(lsame, xerbla);`\n` 100: DGER dger = new DGER(xerbla);`\n` 101: DLARF dlarf = new DLARF(dgemv, dger, lsame);`\n` 102: DLAMC1 dlamc1 = new DLAMC1(dlamc3);`\n` 103: DLAMC4 dlamc4 = new DLAMC4(dlamc3);`\n` 104: DLAMC5 dlamc5 = new DLAMC5(dlamc3);`\n` 105: DLAMC2 dlamc2 = new DLAMC2(dlamc3, dlamc1, dlamc4, dlamc5);`\n` 106: DLAMCH dlamch = new DLAMCH(lsame, dlamc2);`\n` 107: DLARFG dlarfg = new DLARFG(dlamch, dlapy2, dnrm2, dscal);`\n` 108: DGEQR2 dgeqr2 = new DGEQR2(dlarf, dlarfg, xerbla);`\n` 109: DGEMM dgemm = new DGEMM(lsame, xerbla);`\n` 110: DTRMM dtrmm = new DTRMM(lsame, xerbla);`\n` 111: DLARFB dlarfb = new DLARFB(lsame, dcopy, dgemm, dtrmm);`\n` 112: DTRMV dtrmv = new DTRMV(lsame, xerbla);`\n` 113: DLARFT dlarft = new DLARFT(dgemv, dtrmv, lsame);`\n` 114: ILAENV ilaenv = new ILAENV(ieeeck, iparmq);`\n` 115: DGEQRF dgeqrf = new DGEQRF(dgeqr2, dlarfb, dlarft, xerbla, ilaenv);`\n` 116: DGERQ2 dgerq2 = new DGERQ2(dlarf, dlarfg, xerbla);`\n` 117: DGERQF dgerqf = new DGERQF(dgerq2, dlarfb, dlarft, xerbla, ilaenv);`\n` 118: DORM2R dorm2r = new DORM2R(lsame, dlarf, xerbla);`\n` 119: DORMQR dormqr = new DORMQR(lsame, ilaenv, dlarfb, dlarft, dorm2r, xerbla);`\n` 120: `\n` 121: #endregion`\n` 122: `\n` 123: `\n` 124: #region Set Dependencies`\n` 125: `\n` 126: this._dgeqrf = dgeqrf; this._dgerqf = dgerqf; this._dormqr = dormqr; this._xerbla = xerbla; this._ilaenv = ilaenv; `\n` 127: `\n` 128: #endregion`\n` 129: `\n` 130: }`\n` 131: /// <summary>`\n` 132: /// Purpose`\n` 133: /// =======`\n` 134: /// `\n` 135: /// DGGQRF computes a generalized QR factorization of an N-by-M matrix A`\n` 136: /// and an N-by-P matrix B:`\n` 137: /// `\n` 138: /// A = QR, B = QTZ,`\n` 139: /// `\n` 140: /// where Q is an N-by-N orthogonal matrix, Z is a P-by-P orthogonal`\n` 141: /// matrix, and R and T assume one of the forms:`\n` 142: /// `\n` 143: /// if N .GE. M, R = ( R11 ) M , or if N .LT. M, R = ( R11 R12 ) N,`\n` 144: /// ( 0 ) N-M N M-N`\n` 145: /// M`\n` 146: /// `\n` 147: /// where R11 is upper triangular, and`\n` 148: /// `\n` 149: /// if N .LE. P, T = ( 0 T12 ) N, or if N .GT. P, T = ( T11 ) N-P,`\n` 150: /// P-N N ( T21 ) P`\n` 151: /// P`\n` 152: /// `\n` 153: /// where T12 or T21 is upper triangular.`\n` 154: /// `\n` 155: /// In particular, if B is square and nonsingular, the GQR factorization`\n` 156: /// of A and B implicitly gives the QR factorization of inv(B)A:`\n` 157: /// `\n` 158: /// inv(B)A = Z'(inv(T)R)`\n` 159: /// `\n` 160: /// where inv(B) denotes the inverse of the matrix B, and Z' denotes the`\n` 161: /// transpose of the matrix Z.`\n` 162: /// `\n` 163: ///</summary>`\n` 164: /// <param name=\"N\">`\n` 165: /// (input) INTEGER`\n` 166: /// The number of rows of the matrices A and B. N .GE. 0.`\n` 167: ///</param>`\n` 168: /// <param name=\"M\">`\n` 169: /// (input) INTEGER`\n` 170: /// The number of columns of the matrix A. M .GE. 0.`\n` 171: ///</param>`\n` 172: /// <param name=\"P\">`\n` 173: /// (input) INTEGER`\n` 174: /// The number of columns of the matrix B. P .GE. 0.`\n` 175: ///</param>`\n` 176: /// <param name=\"A\">`\n` 177: /// = QR, B = QTZ,`\n` 178: ///</param>`\n` 179: /// <param name=\"LDA\">`\n` 180: /// (input) INTEGER`\n` 181: /// The leading dimension of the array A. LDA .GE. max(1,N).`\n` 182: ///</param>`\n` 183: /// <param name=\"TAUA\">`\n` 184: /// (output) DOUBLE PRECISION array, dimension (min(N,M))`\n` 185: /// The scalar factors of the elementary reflectors which`\n` 186: /// represent the orthogonal matrix Q (see Further Details).`\n` 187: ///</param>`\n` 188: /// <param name=\"B\">`\n` 189: /// (input/output) DOUBLE PRECISION array, dimension (LDB,P)`\n` 190: /// On entry, the N-by-P matrix B.`\n` 191: /// On exit, if N .LE. P, the upper triangle of the subarray`\n` 192: /// B(1:N,P-N+1:P) contains the N-by-N upper triangular matrix T;`\n` 193: /// if N .GT. P, the elements on and above the (N-P)-th subdiagonal`\n` 194: /// contain the N-by-P upper trapezoidal matrix T; the remaining`\n` 195: /// elements, with the array TAUB, represent the orthogonal`\n` 196: /// matrix Z as a product of elementary reflectors (see Further`\n` 197: /// Details).`\n` 198: ///</param>`\n` 199: /// <param name=\"LDB\">`\n` 200: /// (input) INTEGER`\n` 201: /// The leading dimension of the array B. LDB .GE. max(1,N).`\n` 202: ///</param>`\n` 203: /// <param name=\"TAUB\">`\n` 204: /// (output) DOUBLE PRECISION array, dimension (min(N,P))`\n` 205: /// The scalar factors of the elementary reflectors which`\n` 206: /// represent the orthogonal matrix Z (see Further Details).`\n` 207: ///</param>`\n` 208: /// <param name=\"WORK\">`\n` 209: /// (workspace/output) DOUBLE PRECISION array, dimension (MAX(1,LWORK))`\n` 210: /// On exit, if INFO = 0, WORK(1) returns the optimal LWORK.`\n` 211: ///</param>`\n` 212: /// <param name=\"LWORK\">`\n` 213: /// (input) INTEGER`\n` 214: /// The dimension of the array WORK. LWORK .GE. max(1,N,M,P).`\n` 215: /// For optimum performance LWORK .GE. max(N,M,P)max(NB1,NB2,NB3),`\n` 216: /// where NB1 is the optimal blocksize for the QR factorization`\n` 217: /// of an N-by-M matrix, NB2 is the optimal blocksize for the`\n` 218: /// RQ factorization of an N-by-P matrix, and NB3 is the optimal`\n` 219: /// blocksize for a call of DORMQR.`\n` 220: /// `\n` 221: /// If LWORK = -1, then a workspace query is assumed; the routine`\n` 222: /// only calculates the optimal size of the WORK array, returns`\n` 223: /// this value as the first entry of the WORK array, and no error`\n` 224: /// message related to LWORK is issued by XERBLA.`\n` 225: ///</param>`\n` 226: /// <param name=\"INFO\">`\n` 227: /// (output) INTEGER`\n` 228: /// = 0: successful exit`\n` 229: /// .LT. 0: if INFO = -i, the i-th argument had an illegal value.`\n` 230: ///</param>`\n` 231: public void Run(int N, int M, int P, ref double[] A, int offset_a, int LDA, ref double[] TAUA, int offset_taua`\n` 232: , ref double[] B, int offset_b, int LDB, ref double[] TAUB, int offset_taub, ref double[] WORK, int offset_work, int LWORK, ref int INFO)`\n` 233: {`\n` 234: `\n` 235: #region Array Index Correction`\n` 236: `\n` 237: int o_a = -1 - LDA + offset_a; int o_taua = -1 + offset_taua; int o_b = -1 - LDB + offset_b; `\n` 238: int o_taub = -1 + offset_taub; int o_work = -1 + offset_work; `\n` 239: `\n` 240: #endregion`\n` 241: `\n` 242: `\n` 243: #region Prolog`\n` 244: `\n` 245: // `\n` 246: // -- LAPACK routine (version 3.1) --`\n` 247: // * Univ. of Tennessee, Univ. of California Berkeley and NAG Ltd..`\n` 248: // * November 2006`\n` 249: // `\n` 250: // .. Scalar Arguments ..`\n` 251: // * ..`\n` 252: // * .. Array Arguments ..`\n` 253: // * ..`\n` 254: // `\n` 255: // Purpose`\n` 256: // * =======`\n` 257: // `\n` 258: // DGGQRF computes a generalized QR factorization of an N-by-M matrix A`\n` 259: // * and an N-by-P matrix B:`\n` 260: // `\n` 261: // A = QR, B = QTZ,`\n` 262: // `\n` 263: // * where Q is an N-by-N orthogonal matrix, Z is a P-by-P orthogonal`\n` 264: // * matrix, and R and T assume one of the forms:`\n` 265: // `\n` 266: // if N >= M, R = ( R11 ) M , or if N < M, R = ( R11 R12 ) N,`\n` 267: // * ( 0 ) N-M N M-N`\n` 268: // * M`\n` 269: // `\n` 270: // where R11 is upper triangular, and`\n` 271: // `\n` 272: // if N <= P, T = ( 0 T12 ) N, or if N > P, T = ( T11 ) N-P,`\n` 273: // * P-N N ( T21 ) P`\n` 274: // * P`\n` 275: // `\n` 276: // where T12 or T21 is upper triangular.`\n` 277: // `\n` 278: // In particular, if B is square and nonsingular, the GQR factorization`\n` 279: // * of A and B implicitly gives the QR factorization of inv(B)A:`\n` 280: // `\n` 281: // * inv(B)A = Z'(inv(T)R)`\n` 282: // `\n` 283: // * where inv(B) denotes the inverse of the matrix B, and Z' denotes the`\n` 284: // * transpose of the matrix Z.`\n` 285: // `\n` 286: // Arguments`\n` 287: // * =========`\n` 288: // `\n` 289: // N (input) INTEGER`\n` 290: // * The number of rows of the matrices A and B. N >= 0.`\n` 291: // `\n` 292: // M (input) INTEGER`\n` 293: // * The number of columns of the matrix A. M >= 0.`\n` 294: // `\n` 295: // P (input) INTEGER`\n` 296: // * The number of columns of the matrix B. P >= 0.`\n` 297: // `\n` 298: // A (input/output) DOUBLE PRECISION array, dimension (LDA,M)`\n` 299: // * On entry, the N-by-M matrix A.`\n` 300: // * On exit, the elements on and above the diagonal of the array`\n` 301: // * contain the min(N,M)-by-M upper trapezoidal matrix R (R is`\n` 302: // * upper triangular if N >= M); the elements below the diagonal,`\n` 303: // * with the array TAUA, represent the orthogonal matrix Q as a`\n` 304: // * product of min(N,M) elementary reflectors (see Further`\n` 305: // * Details).`\n` 306: // `\n` 307: // LDA (input) INTEGER`\n` 308: // * The leading dimension of the array A. LDA >= max(1,N).`\n` 309: // `\n` 310: // TAUA (output) DOUBLE PRECISION array, dimension (min(N,M))`\n` 311: // * The scalar factors of the elementary reflectors which`\n` 312: // * represent the orthogonal matrix Q (see Further Details).`\n` 313: // `\n` 314: // B (input/output) DOUBLE PRECISION array, dimension (LDB,P)`\n` 315: // * On entry, the N-by-P matrix B.`\n` 316: // * On exit, if N <= P, the upper triangle of the subarray`\n` 317: // * B(1:N,P-N+1:P) contains the N-by-N upper triangular matrix T;`\n` 318: // * if N > P, the elements on and above the (N-P)-th subdiagonal`\n` 319: // * contain the N-by-P upper trapezoidal matrix T; the remaining`\n` 320: // * elements, with the array TAUB, represent the orthogonal`\n` 321: // * matrix Z as a product of elementary reflectors (see Further`\n` 322: // * Details).`\n` 323: // `\n` 324: // LDB (input) INTEGER`\n` 325: // * The leading dimension of the array B. LDB >= max(1,N).`\n` 326: // `\n` 327: // TAUB (output) DOUBLE PRECISION array, dimension (min(N,P))`\n` 328: // * The scalar factors of the elementary reflectors which`\n` 329: // * represent the orthogonal matrix Z (see Further Details).`\n` 330: // `\n` 331: // WORK (workspace/output) DOUBLE PRECISION array, dimension (MAX(1,LWORK))`\n` 332: // * On exit, if INFO = 0, WORK(1) returns the optimal LWORK.`\n` 333: // `\n` 334: // LWORK (input) INTEGER`\n` 335: // * The dimension of the array WORK. LWORK >= max(1,N,M,P).`\n` 336: // * For optimum performance LWORK >= max(N,M,P)max(NB1,NB2,NB3),`\n` 337: // where NB1 is the optimal blocksize for the QR factorization`\n` 338: // * of an N-by-M matrix, NB2 is the optimal blocksize for the`\n` 339: // * RQ factorization of an N-by-P matrix, and NB3 is the optimal`\n` 340: // * blocksize for a call of DORMQR.`\n` 341: // `\n` 342: // If LWORK = -1, then a workspace query is assumed; the routine`\n` 343: // * only calculates the optimal size of the WORK array, returns`\n` 344: // * this value as the first entry of the WORK array, and no error`\n` 345: // * message related to LWORK is issued by XERBLA.`\n` 346: // `\n` 347: // INFO (output) INTEGER`\n` 348: // * = 0: successful exit`\n` 349: // * < 0: if INFO = -i, the i-th argument had an illegal value.`\n` 350: // `\n` 351: // Further Details`\n` 352: // * ===============`\n` 353: // `\n` 354: // The matrix Q is represented as a product of elementary reflectors`\n` 355: // `\n` 356: // Q = H(1) H(2) . . . H(k), where k = min(n,m).`\n` 357: // `\n` 358: // Each H(i) has the form`\n` 359: // `\n` 360: // H(i) = I - taua * v * v'`\n` 361: // `\n` 362: // where taua is a real scalar, and v is a real vector with`\n` 363: // * v(1:i-1) = 0 and v(i) = 1; v(i+1:n) is stored on exit in A(i+1:n,i),`\n` 364: // * and taua in TAUA(i).`\n` 365: // * To form Q explicitly, use LAPACK subroutine DORGQR.`\n` 366: // * To use Q to update another matrix, use LAPACK subroutine DORMQR.`\n` 367: // `\n` 368: // The matrix Z is represented as a product of elementary reflectors`\n` 369: // `\n` 370: // Z = H(1) H(2) . . . H(k), where k = min(n,p).`\n` 371: // `\n` 372: // Each H(i) has the form`\n` 373: // `\n` 374: // H(i) = I - taub * v * v'`\n` 375: // `\n` 376: // where taub is a real scalar, and v is a real vector with`\n` 377: // * v(p-k+i+1:p) = 0 and v(p-k+i) = 1; v(1:p-k+i-1) is stored on exit in`\n` 378: // * B(n-k+i,1:p-k+i-1), and taub in TAUB(i).`\n` 379: // * To form Z explicitly, use LAPACK subroutine DORGRQ.`\n` 380: // * To use Z to update another matrix, use LAPACK subroutine DORMRQ.`\n` 381: // `\n` 382: // =====================================================================`\n` 383: // `\n` 384: // .. Local Scalars ..`\n` 385: // * ..`\n` 386: // * .. External Subroutines ..`\n` 387: // * ..`\n` 388: // * .. External Functions ..`\n` 389: // * ..`\n` 390: // * .. Intrinsic Functions ..`\n` 391: // INTRINSIC INT, MAX, MIN;`\n` 392: // * ..`\n` 393: // * .. Executable Statements ..`\n` 394: // `\n` 395: // Test the input parameters`\n` 396: // `\n` 397: `\n` 398: #endregion`\n` 399: `\n` 400: `\n` 401: #region Body`\n` 402: `\n` 403: INFO = 0;`\n` 404: NB1 = this._ilaenv.Run(1, \"DGEQRF\", \" \", N, M, - 1, - 1);`\n` 405: NB2 = this._ilaenv.Run(1, \"DGERQF\", \" \", N, P, - 1, - 1);`\n` 406: NB3 = this._ilaenv.Run(1, \"DORMQR\", \" \", N, M, P, - 1);`\n` 407: NB = Math.Max(NB1, Math.Max(NB2, NB3));`\n` 408: LWKOPT = Math.Max(N, Math.Max(M, P)) NB;`\n` 409: WORK[1 + o_work] = LWKOPT;`\n` 410: LQUERY = (LWORK == - 1);`\n` 411: if (N < 0)`\n` 412: {`\n` 413: INFO = - 1;`\n` 414: }`\n` 415: else`\n` 416: {`\n` 417: if (M < 0)`\n` 418: {`\n` 419: INFO = - 2;`\n` 420: }`\n` 421: else`\n` 422: {`\n` 423: if (P < 0)`\n` 424: {`\n` 425: INFO = - 3;`\n` 426: }`\n` 427: else`\n` 428: {`\n` 429: if (LDA < Math.Max(1, N))`\n` 430: {`\n` 431: INFO = - 5;`\n` 432: }`\n` 433: else`\n` 434: {`\n` 435: if (LDB < Math.Max(1, N))`\n` 436: {`\n` 437: INFO = - 8;`\n` 438: }`\n` 439: else`\n` 440: {`\n` 441: if (LWORK < Math.Max(1, Math.Max(N, Math.Max(M, P))) && !LQUERY)`\n` 442: {`\n` 443: INFO = - 11;`\n` 444: }`\n` 445: }`\n` 446: }`\n` 447: }`\n` 448: }`\n` 449: }`\n` 450: if (INFO != 0)`\n` 451: {`\n` 452: this._xerbla.Run(\"DGGQRF\", - INFO);`\n` 453: return;`\n` 454: }`\n` 455: else`\n` 456: {`\n` 457: if (LQUERY)`\n` 458: {`\n` 459: return;`\n` 460: }`\n` 461: }`\n` 462: // `\n` 463: // QR factorization of N-by-M matrix A: A = QR`\n` 464: // `\n` 465: this._dgeqrf.Run(N, M, ref A, offset_a, LDA, ref TAUA, offset_taua, ref WORK, offset_work`\n` 466: , LWORK, ref INFO);`\n` 467: LOPT = (int)WORK[1 + o_work];`\n` 468: // `\n` 469: // Update B := Q'B.`\n` 470: // `\n` 471: this._dormqr.Run(\"Left\", \"Transpose\", N, P, Math.Min(N, M), ref A, offset_a`\n` 472: , LDA, TAUA, offset_taua, ref B, offset_b, LDB, ref WORK, offset_work, LWORK`\n` 473: , ref INFO);`\n` 474: LOPT = Math.Max(LOPT, Convert.ToInt32(Math.Truncate(WORK[1 + o_work])));`\n` 475: // `\n` 476: // RQ factorization of N-by-P matrix B: B = TZ.`\n` 477: // `\n` 478: this._dgerqf.Run(N, P, ref B, offset_b, LDB, ref TAUB, offset_taub, ref WORK, offset_work`\n` 479: , LWORK, ref INFO);`\n` 480: WORK[1 + o_work] = Math.Max(LOPT, Convert.ToInt32(Math.Truncate(WORK[1 + o_work])));`\n` 481: // `\n` 482: return;`\n` 483: // `\n` 484: // * End of DGGQRF`\n` 485: // *`\n` 486: `\n` 487: #endregion`\n` 488: `\n` 489: }`\n` 490: }`\n` 491: }`" ]	[ null, "http://dotnumerics.com/WebResource.axd", null, "http://dotnumerics.com/WebResource.axd", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5025309,"math_prob":0.99014914,"size":16311,"snap":"2023-40-2023-50","text_gpt3_token_len":5668,"char_repetition_ratio":0.1588275,"word_repetition_ratio":0.23084322,"special_character_ratio":0.47857276,"punctuation_ratio":0.2848501,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99892604,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-08T10:04:02Z\",\"WARC-Record-ID\":\"<urn:uuid:b1865ce4-cba6-4303-875e-ea5a630c871e>\",\"Content-Length\":\"69683\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0816c1d4-acd5-4eb7-a8ed-4326b7ff431a>\",\"WARC-Concurrent-To\":\"<urn:uuid:8f9e032c-fdbf-43e7-841f-e0321aad0b15>\",\"WARC-IP-Address\":\"45.58.159.42\",\"WARC-Target-URI\":\"http://dotnumerics.com/NumericalLibraries/LinearAlgebra/CSharpCodeFiles/dggqrf.aspx\",\"WARC-Payload-Digest\":\"sha1:GHFVIO775SBJDOPSBOPRTSB6TKKX6ORZ\",\"WARC-Block-Digest\":\"sha1:D56CLLPJSX4XEGQ6ARC5I4YWPOR4D5YL\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100739.50_warc_CC-MAIN-20231208081124-20231208111124-00219.warc.gz\"}"}
https://answers.everydaycalculation.com/multiply-fractions/15-9-times-21-70	[ "Solutions by everydaycalculation.com\n\n## Multiply 15/9 with 21/70\n\n1st number: 1 6/9, 2nd number: 21/70\n\nThis multiplication involving fractions can also be rephrased as \"What is 15/9 of 21/70?\"\n\n15/9 × 21/70 is 1/2.\n\n#### Steps for multiplying fractions\n\n1. Simply multiply the numerators and denominators separately:\n2. 15/9 × 21/70 = 15 × 21/9 × 70 = 315/630\n3. After reducing the fraction, the answer is 1/2\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:\nAndroid and iPhone/ iPad\n\n×" ]	[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89403224,"math_prob":0.9656605,"size":447,"snap":"2022-40-2023-06","text_gpt3_token_len":175,"char_repetition_ratio":0.16478555,"word_repetition_ratio":0.0,"special_character_ratio":0.4541387,"punctuation_ratio":0.08571429,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9680753,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-27T01:26:04Z\",\"WARC-Record-ID\":\"<urn:uuid:0a2609b4-3149-4500-87f0-ea6b03ac2901>\",\"Content-Length\":\"7223\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9b689599-34b4-4630-ae72-6ec3fbc91e78>\",\"WARC-Concurrent-To\":\"<urn:uuid:02a4c926-4e5d-4a1c-a254-88619ac3df0d>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/multiply-fractions/15-9-times-21-70\",\"WARC-Payload-Digest\":\"sha1:2URLPMYOHH7A5A6SZZWA4ZUJ3MHPBSHX\",\"WARC-Block-Digest\":\"sha1:BKBZETLXUKQWNYIHJK3CHDIISRRMWNVX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030334974.57_warc_CC-MAIN-20220927002241-20220927032241-00781.warc.gz\"}"}
https://www.classcentral.com/course/youtube-calculus-2-full-length-videos-53097	[ "Class Central is learner-supported. When you buy through links on our site, we may earn an affiliate commission.\n\n# Calculus 2\n\n## Syllabus\n\nCalculus 2 Lecture 6.1: The Natural Log Function.\nCalculus 2 Lecture 6.2: Derivatives of Inverse Functions.\nCalculus 2 Lecture 6.3: Derivatives and Integrals of Exponential Functions.\nCalculus 2 Lecture 6.4: Derivatives and Integrals of General Exponential Functions.\nCalculus 2 Lecture 6.5: Calculus of Inverse Trigonometric Functions.\nCalculus 2 Lecture 6.6: A Discussion of Hyperbolic Functions.\nCalculus 2 Lecture 6.7: Evaluating Limits of Indeterminate Forms.\nCalculus 2 Lecture 7.1: Integration By Parts.\nCalculus 2 Lecture 7.2: Techniques For Trigonometric Integrals.\nCalculus 2 Lecture 7.3: Integrals By Trigonometric Substitution.\nCalculus 2 Lecture 7.4: Integration By Partial Fractions.\nCalculus 2 Lecture 7.6: Improper Integrals.\nCalculus 2 Lecture 8.1: Solving First Order Differential Equations By Separation of Variables.\nCalculus 2 Lecture 9.1: Convergence and Divergence of Sequences.\nCalculus 2 Lecture 9.2: Series, Geometric Series, Harmonic Series, and Divergence Test.\nCalculus 2 Lecture 9.3: Using the Integral Test for Convergence/Divergence of Series, P-Series.\nCalculus 2 Lecture 9.4: The Comparison Test for Series and The Limit Comparison Test.\nCalculus 2 Lecture 9.5: Showing Convergence With the Alternating Series Test, Finding Error of Sums.\nCalculus 2 Lecture 9.6: Absolute Convergence, Ratio Test and Root Test For Series.\nCalculus 2 Lecture 9.7: Power Series, Calculus of Power Series, Ratio Test for Int. of Convergence.\nCalculus 2 Lecture 9.8: Representation of Functions by Taylor Series and Maclauren Series.\nCalculus 2 Lecture 9.9: Approximation of Functions by Taylor Polynomials.\nCalculus 2 Lecture 10.2: Introduction to Parametric Equations.\nCalculus 2 Lecture 10.3: Calculus of Parametric Equations.\nCalculus 2 Lecture 10.4: Using Polar Coordinates and Polar Equations.\nCalculus 2 Lecture 10.5: Calculus of Polar Equations.\nNumerical Integration With Trapezoidal and Simpson's Rule.\n\n### Taught by\n\nProfessor Leonard\n\n## Reviews\n\nStart your review of Calculus 2\n\n###", null, "Never Stop Learning!\n\nGet personalized course recommendations, track subjects and courses with reminders, and more." ]	[ null, "https://ccweb.imgix.net/https%3A%2F%2Fwww.classcentral.com%2Fimages%2Fsignup-illus-mobile.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.60302764,"math_prob":0.98485625,"size":1985,"snap":"2021-43-2021-49","text_gpt3_token_len":547,"char_repetition_ratio":0.30893487,"word_repetition_ratio":0.014545455,"special_character_ratio":0.2277078,"punctuation_ratio":0.22278482,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9911575,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-23T01:39:17Z\",\"WARC-Record-ID\":\"<urn:uuid:9b6d75b9-9d76-4475-9ef1-fdb920e61a49>\",\"Content-Length\":\"581916\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6e5f5616-5c11-4ef2-9c91-36a3a6b36721>\",\"WARC-Concurrent-To\":\"<urn:uuid:7288bac0-4a1a-40b4-acc7-05227e455af7>\",\"WARC-IP-Address\":\"34.68.4.21\",\"WARC-Target-URI\":\"https://www.classcentral.com/course/youtube-calculus-2-full-length-videos-53097\",\"WARC-Payload-Digest\":\"sha1:ZCITBIYPPTTDPUXQZD4Y7YGVUVNVY2DW\",\"WARC-Block-Digest\":\"sha1:QMGGZYU7LW7MJIH4MH5MTHAJSSIFU5XU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585537.28_warc_CC-MAIN-20211023002852-20211023032852-00220.warc.gz\"}"}
https://studylib.net/doc/10916513/sharp-l--estimates-for-the-wave-equation-on-heisenberg-ty..	[ "# SHARP L -ESTIMATES FOR THE WAVE EQUATION ON HEISENBERG TYPE GROUPS", null, "```SHARP Lp -ESTIMATES FOR THE WAVE EQUATION\nON HEISENBERG TYPE GROUPS\nLECTURE NOTES ORLEANS, APRIL 2008\nHTTP://ANALYSIS.MATH.UNI-KIEL.DE/MUELLER/\nDETLEF MÜLLER\nAbstract. In these lectures, which are based on recent joint work\nwith A. Seeger , I shall present sharp analogues of classical estimates by Peral and Miyachi for solutions of the standard wave\nequation on Euclidean space in the context of the wave equation\nassociated to the sub-Laplacian on a Heisenberg type group. Some\nrelated questions, such as spectral multipliers for the sub-Laplacian\nor Strichartz-estimates, will be briefly addressed. Our results improve on earlier joint work of mine with E.M. Stein. The new\nmore clearly the connections of the problem with the underlying\nsub-Riemannian geometry.\nDate: April 8, 2008.\n1\n2\nDETLEF MÜLLER\nContents\n1. Introduction\n2\n1.1. Connections with spectral multipliers and further facts\n7\n2. The sub-Riemannian geometry of a Heisenberg type group\n10\n3. The Schrödinger and the wave propagators on a Heisenberg\ntype group\n14\n3.1. Twisted convolution and the metaplectic group\n16\n3.2. A subordination formula\n18\n4. Estimation of Aλ,k,l,± if k ≥ 1\n22\n√\n5. Estimation of (1 + L)−(d−1)/4 ei\nL\nδ\n25\n5.1. Anisotropic re-scaling for k fixed\n27\n6. L2 -estimates for components of the wave propagator\n27\n7. Estimation for p = 1\n28\n8. Proof of Theorem 1.1\n31\n9. Appendix: The Fourier transform on a group of Heisenberg\ntype\n34\nReferences\n38\n1. Introduction\nLet\ng = g1 ⊕ g2 ,\nwith dim g1 = 2m and dim g2 = n, be a Lie algebra of Heisenberg type,\nwhere\n[g, g] ⊂ g2 ⊂ z(g) ,\nTHE WAVE EQUATION ON GROUPS OF HEISENBERG TYPE\n3\nz(g) being the center of g. This means that g is endowed with an inner\nproduct h , i such that g1 and g2 or orthogonal subspaces and the\nfollowing holds true:\nIf we define for µ ∈ g∗2 \\ {0} the skew form ωµ on g1 by\nωµ (V, W ) := µ [V, W ] ,\nthen there is a unique skew-symmetric linear endomorphism Jµ of g1\nsuch that\nωµ (V, W ) = hµ, [V, W ]i = hJµ (V ), W i\n(here, we also used the natural identification of g∗2 with g2 via the inner\nproduct). Then\nJµ2 = −\|µ\|2 I\n(1.1)\nfor every µ ∈ g∗2 \\ {0}.\nNote that this implies in particular that [g1 , g1 ] = g2 .\nAs the corresponding connected, simply connected Heisenberg type\nLie group G we shall then choose the linear manifold g, endowed with\nthe Baker-Campbell-Hausdorff product\n(V1 , U1 ) (V2 , U2 ) := (V1 + V2 , U1 + U2 + 12 [V1 , V2 ])\nand identity element e = 0.\nNote that the nilpotent part in the Iwasawa decomposition of a simple Lie group of real rank one is always of Heisenberg type or Euclidean.\nAs usual, we shall identify X ∈ g with the corresponding leftinvariant vector field on G given by the Lie-derivative\nXf (g) :=\nd\nf (g exp(tX))\|t=0 ,\ndt\nwhere exp : g → G denotes the exponential mapping, which agrees\nwith the identity mapping in our case.\nLet us next fix an orthonormal basis X1 , . . . , X2m of g1 , and let us\ndefine the non-elliptic sub-Laplacian\nL := −\n2m\nX\nj=1\nXj2\n4\nDETLEF MÜLLER\non G. Since the vector fields Xj together with their commutators span\nthe tangent space to G at every point, L is still hypoelliptic and provides an example of a non-elliptic “sum of squares operator ” in the\nsense of Hörmander (). Moreover, L takes over in many respects of\nanalysis on G the role which the Laplacian plays on Euclidean space.\nTo simplify the notation, we shall also fix an orthonormal basis\nU1 , . . . , Un of g2 , and shall in the sequel identify g = g1 + g2 and G\nwith R2m × Rn by means of the basis X1 , . . . , X2m , U1 , . . . , Un of g.\nThen our inner product\non g will agree with the canonical Euclidean\nP\n2m+n\nproduct z · w = 2m+n\nz\n, and Jµ will be identified with\nj wj on R\nj=1\na skew-symmetric 2m × 2m matrix. Moreover, the Lebesgue measure\ndx du on R2m+n is a bi-invariant Haar measure on G. By\nd := 2m + n\nwe shall denote the topological dimension of G. We also introduce the\nautomorphic dilations\nδr (x, u) := (rx, r 2 u),\nr > 0,\non G, and the Koranyi norm\nk(x, u)kK := (\|x\|4 + \|4u\|2)1/4 .\nNotice that this is a homogeneous norm with respect to the dilations δr ,\nand that L is homogeneous of degree 2 with respect to these dilations.\nMoreover, if we denote the corresponding balls by\nQr (x, u) := {(y, v) ∈ G : k(y, v)−1(x, u)kK < r}, (x, u) ∈ G, r > 0,\nthen the volume \|Qr (x, u)\| is given by\nwhere\n\|Qr (x, u)\| = \|Q1 (0, 0)\| r D ,\nD := 2m + 2n\nis the homogeneous dimension of G. We shall also have to work with\nthe Euclidean balls\nBr (x, u) := {(y, v) ∈ G : \|(y − x, v − u)\| < r}, (x, u) ∈ G, r > 0,\nwith respect to the Euclidean norm\n\|(x, u)\| := (\|x\|2 + \|u\|2)1/2 .\nIn the special case n = 1 we may assume that Jµ = µJ, µ ∈ R, where\n0\nIm\nJ :=\n−Im 0\nTHE WAVE EQUATION ON GROUPS OF HEISENBERG TYPE\n5\nThis is the case of the Heisenberg group Hm , which is R2m ×R, endowed\nwith the product\n(z, u) · (z ′ , u′) = (z + z ′ , u + u′ + 21 hJz, z ′ i).\nA basis of the Lie algebra hm of Hm is then given by X1 , . . . , X2m , U,\n∂\nis central.\nwhere U := ∂u\nConsider the following Cauchy problem for the wave equation on\nG × R associated to L:\n∂2u\n∂u (1.2)\n−\n(−L)u\n=\n0,\nu\n=\nf,\n= g,\n∂t2\n∂t t=0\nt=0\nwhere t ∈ R denotes time.\nThe solution to this problem is formally given by\n!\n√\n√\nsin(t L)\n√\ng (x) + (cos(t L)f )(x),\nu(x, t) =\nL\n(x, t) ∈ G × R.\nIn fact, the above expression for u makes perfect sense at least for\nf, g ∈ L2 (G), if one defines the functions of L involved by the spectral\ntheorem (noticeR that L is essentially selfadjoint on C0∞ (G) ) as\n∞\nfollows: If L = 0 s dEs denotes the spectral resolution of L on L2 (G),\nthen every Borel function m on R+ gives rise to a (possibly unbounded)\noperator\nZ\n∞\nm(L) :=\nm(s) dEs\n0\non L2 (G). Let us also note that if m is a bounded spectral multiplier, then, by left–invariance and the Schwartz kernel theorem, it is\neasy\nto see that m(L) is a convolution operator m(L)f = f ∗ Km =\nR\nf (y)Km(y −1·) dy, where a priori the convolution kernel Km is a tem-\nG\npered distribution. We also write Km = m(L)δ.\nIf one decides to measure smoothness properties of the solution\nu(x, t) for fixed time t in terms of Sobolev norms of the form\nkf kLpα := k(1 + L)α/2 f kLp ,\none is √naturally led√to study the mapping properties of operators such\nsin(t L)\neit L\n√\nas (1+L)\nas operators on Lp (G) into itself.\nα/2 or\nL(1+L)α/2\n6\nDETLEF MÜLLER\nFor the classical wave equation on Euclidian space, sharp estimates\nfor the corresponding operators have been established by Peral and\nMiyachi .\n√\nd\n−α/2 it −∆\nIn particular, if ∆ denotes the Laplacian\non\ne\nR , then (1−∆)\n1 1\np\nd\nis bounded on L (R ), if α ≥ (d − 1) p − 2 , for 1 < p < ∞. Moreover,\n(d−1)/2\n(1 − ∆)− 2\ninto L1 (Rd ).\n√\neit\n−∆\nis bounded from the real Hardy space H 1 (Rd )\nLocal analogues of these results hold true for solutions to strictly\nhyperbolic differential equations (see e.g. ,, ).\nIndeed, as has been shown in and , the estimates in and\n locally hold true more generally for large classes of Fourier integral operators, and solutions to strictly hyperbolic equations can be\nexpressed in terms of such operators.\nThe problem in studying the wave equation associated to the subLaplacian on the Heisenberg group is the lack of strict hyperbolicity,\nsince L is degenerate-elliptic, and classical Fourier integral operator\ntechnics do not seem to be available any more. However, as we shall\nsee, it is still possible to represent solutions to (1.2) as infinite sums of\nFourier integral operators with degenerate phases and amplitudes.\nInteresting information about solutions to (1.2) have been obtained\nby Nachman . Among other things, Nachman showed that the\nwave operator on Hm admits a fundamental solution supported in a\n“forward light cone”, whose singularities lie along the cone Γ formed\nby the characteristics through the origin. Moreover, he computed the\nasymptotic behaviour of the fundamental solution as one approaches\na generic singular point on Γ. His method does, however, not provide\nuniform estimates on these singularities, so that it cannot be used to\nprove Lp -estimates for solutions to (1.2). What his results do reveal,\nhowever, is that Γ is by far more complex for Hm than the corresponding cone in the Euclidian case. This is related to the underlying, more\ncomplex sub-Riemannian geometry, which we shall briefly discuss.\nNevertheless, the following theorem is true, which improves on earlier\nwork in on the Heisenberg group, by also including the endpoint\nregularity, and which is a direct analogue of the results by Peral and\nMiyachi in the Euclidean setting.\nTHE WAVE EQUATION ON GROUPS OF HEISENBERG TYPE\n7\n√\ni L\ne\np\nTheorem 1.1. (1+L)\nα/2 extends to a bounded operator on L (G), for\n.\n1 < p < ∞, when α ≥ (d − 1)\| p1 − 12 \|, and for p = 1, if α > d−1\n2\nThe proof will be based on some “Hardy space type result” for the\nendpoint p = 1, which will become evident later. The main difference compared to the Euclidean setting seems, however, that there is\nnot just one single Hardy space, but a whole sequence of local Hardy\nspaces associated with the problem, whose definitions are based on\nsome interesting interplay between two different homogeneities, namely\nthose coming from the isotropic dilations of the underlying Euclidean\nstructure of G and the one coming from the non-isotropic automorphic\ndilations on G (compare Remark 7.3).\nRemark 1.2. The same result holds for\ntors (1 + L)−α/2 (respectively\n√ −(α−1) (1 + L)\n(respectively (1 + L)\n).\n√\nsin\n√\nL\n, or with the fac√\n) replaced by (1 + L)−α ,\nL(1+L)\n−(α−1)/2\nα−1\n2\nNotice that the restriction to time t = 1 in our theorem is inessential,\nsince L is homogeneous of degree 2 with respect to the automorphic\ndilations (z, t) 7→ (rz, r 2 t), r > 0.\n1.1. Connections with spectral multipliers and further facts\nabout the wave equation. If m is a bounded spectral multiplier,\nthen clearly the operator m(L) is bounded on L2 (G). An important\nquestion is then under which additional conditions on the spectral multiplier m the operator m(L) extends from L2 ∩Lp (M) to an Lp -bounded\noperator, for a given p 6= 2. If so, m is called an Lp -multiplier for L.\nFix a non-trivial cut-off function χ ∈ C0∞ (R) supported in the interval [1, 2], and define for α > 0\nkmksloc,α := sup kχ m(r·)kH α ,\nr>0\nwhere H α = H α (R) denotes the classical Sobolev-space of order α.\nThus, kmksloc,α < ∞, if m is locally in H α , uniformly on every scale.\nNotice also that, up to equivalence, k·ksloc,α is independent of the choice\nof the cut-off function χ. Then the following analogue of the classical\nMihlin-Hörmander multiplier theorem holds true (see M. Christ ,\nand also Mauceri-Meda ):\n8\nDETLEF MÜLLER\nTheorem 1.3. If G is a stratified Lie group of homogeneous dimension\nD, and if kmksloc,α < ∞ for some α > D/2, then m(L) is bounded on\nLp (G) for 1 < p < ∞, and of weak type (1,1).\nObserve that, in comparizon to the classical case G = Rd , the homogeneous dimension D takes over the role of the Euclidian dimension\nd. Because of this fact, which is an outgrowth of the homogeneity of\nL with respect to the automorphic dilations, the condition α > D/2\nin Theorem 1.3 appeared natural and was expected to be sharp for\na while. The following result, which was found in joint work with\nE.M. Stein , and independently also by W. Hebisch a bit later,\ncame therefore as a surprise:\nTheorem 1.4. For the sub-Laplacian L on a Heisenberg type group G,\nthe statement in Theorem 1.3 remains valid under the weaker condition\nα > d/2 instead of α > D/2.\nTheorem 1.1 can be used to give a new, short proof of this (sharp)\nmultiplier theorem, based on the following simple Subordination Principle, respectively variants of it:\n√\ni L\ne\nProposition 1.5. Assume that (1+L)\nα/2 extends to a bounded operator\n1\non L (G), and let β > α + 1/2. Then there is a constant C > 0, such\nthat for any multiplier ϕ ∈ H β (R) supported in [1, 2], the corresponding\nconvolution kernel Kϕ = ϕ(L)δ is in L1 (G), and\nkKϕ kL1 (G) ≤ CkϕkH β (R) .\nProof. Observe first that (1 + L)−ε δ ∈ L1 (G) for any ε > 0. This\nfollows from the formula\nZ ∞\nZ ∞\n1\n1\n−ε\nε−1 −t(1+L)\n(1 + L) δ =\nt e\nδ dt =\ntε−1 e−t pt dt\nΓ(ε) 0\nΓ(ε) 0\nand the fact that the heat kernel pt := e−tL δ is a probability measure\non G. Write\n√\nϕ(s) = ψ( s)(1 + s)−γ , with γ > α/2,\nand put g := (1 + L)−γ δ ∈ L1 (G). Then \|\|ψ\|\|H β ≃ \|\|ϕ\|\|H β , and\nZ ∞\n√\n√\n√\n−γ\nψ̂(t)eit L g dt.\nKϕ = ψ( L)((1 + L) δ) = ψ( L)g = c\n−∞\nTHE WAVE EQUATION ON GROUPS OF HEISENBERG TYPE\n9\nAnd, our assumption in combination with some easy multiplier estimates (cf. ) and the homogeneity of L yields\n√\nkeit\nL\ngk1 . k(1 + t2 L)α/2 gk1 . k(1 + t2 )α/2 (1 + L)α/2 gk1,\nhence\nkKϕ k1 .\nZ\n∞\n−∞\n\|ψ̂(t)\|(1 + \|t\|)α k(1 + L)α/2 gk1dt.\nBut, (1 + L)α/2 g = (1 + L)α/2−γ δ ∈ L1 , hence, by Hölder’s inequality\nand Plancherel’s theorem,\nZ ∞\n1/2 Z ∞\n1/2\nβ 2\nkKϕ \|\|1 .\n\|ψ̂(t)(1+\|t\|) \| dt\n·\n(1+\|t\|)2(α−β) dt\n. kψkH β .\n−∞\n−∞\nQ.E.D.\n+ 12 = d2 in this subordination prinIndeed, we may choose β > d−1\n2\nciple. This is just the required regularity of the multiplier in Theorem\n1.4, and one can in fact deduce this theorem from Theorem 1.1 by\nmeans of a refinement of the above subordination principle and standard arguments from Calderón-Zygmund theory.\nIn contrast to the close analogy between the wave equation on G and\non Euclidean space expressed in Theorem 1.1, there are other aspects\nwhere the solutions behave quite differently compared to the classical\nsetting.\nOne instance of this is the “almost” failure of dispersive estimates\nfor solutions to (1.2) on Heisenberg groups. Indeed, in , Bahouri,\nGérard and Xu prove that solutions to (1.2) on Hm with sufficiently\nsmooth initial data in L1 satisfy\nku(t)k∞ ≤ C\|t\|−1/2 ,\nand that the exponent −1/2 is optimal, whereas for the Laplacian on\nRd a dispersive estimate holds with exponent −(d − 1)/2. Correspondingly, the Strichartz estimates proven in are much weaker than the\nEuclidean analogues would suggest.\nA related result is the “almost” failure of (spectral) restriction theorems for Hm proven in .\n10\nDETLEF MÜLLER\n2. The sub-Riemannian geometry of a Heisenberg type\ngroup\nLet M be a smooth manifold, and let D ⊂ T M be a smooth distribution, i.e., a smooth subbundel of T M. A sub-Riemannian structure\n(D, g) on M is given by a smooth distribution D and a Riemannian metric g on D. For p ∈ M and X, Y ∈ D(p) we write hX, Y ig := gp (X, Y )\n1/2\nand kXk = hX, Xig . M, endowed with such a structure, is called a\nsub-Riemannian manifold.\nAn absolutely continuous curve γ : [0, T ] → M in M is then called\nhorizontal, if\nd\nγ̇(t) = γ(t) ∈ D(γ(t))\ndt\nfor almost every t ∈ [0, T ]. As in Riemannian geometry, the length of a\nhorizontal curve γ is then defined by\nZ T\nL(γ) :=\nkγ̇(t)kg dt.\n0\nThe Carnot–Carathéodory distance (or sub-Riemannian distance) on\n(M, D, g) is then defined by\ndCC (p, q) := inf{L(γ) : γ is a horinzontal curve joining p with q},\np, q ∈ M, where we use the convention that inf ∅ := ∞.\nWe shall here be interested in the special situation where D =\nspan {X1 , . . . , Xk }, and where X1 , . . . , Xk are smooth vector fields on\nM which form an orthonormal system with respect to g at every point.\nThen a curve γ is horizontal iff there are functions aj (t) such that γ̇(t) =\n1/2\nRT P\nPk\n2\na\n(t)X\n(γ(t))\nfor\na.e.\nt\n∈\n[0,\nT\n],\nand\nL(γ)\n=\na\n(t)\ndt.\nj\nj\nj\nj=1\nj\n0\nMoreover, if these vector fields are bracket generating, i.e., if they\nsatisfy Hörmander’s condition, then by the Chow-Rashevskii theorem\ndCC (p, q) < ∞ for every p, q ∈ M, and dCC is a metric on M that\ninduces the topology of M.\nLet us now consider the sub-Riemannian structure on a group G of\nHeisenberg type, where D = span {X1 , . . . , X2m }. Here, the Carnot–\nCarathéodory distance is left-invariant, i.e.,\ndCC (x, y) = kx−1 ykCC ,\nTHE WAVE EQUATION ON GROUPS OF HEISENBERG TYPE\n11\nwhere kxkCC := dCC (0, x) is again a homogeneous norm on G. Observe\nthat there exists a constant C ≥ 1 such that\nC −1 kxkCC ≤ kxkK ≤ CkxkCC ,\n(2.1)\nsince any two homogeneous norms are equivalent . In analogy with\nwave equations on Riemannian manifolds, one expects that singularities\nof solutions propagate along “geodesics” of the underlying geometry.\nHowever, it turns out that the notion of geodesic of a sub-Riemannian\nmanifold is not as clear as one might expect. The perhaps best approach to the problem of determining length minimizing curves in this\ncontext is by means of control theory, as it is outlined in the article\n by Liu and Sussman, which I warmly recommend.\nOne type of geodesic can be constructed following classical geometrical optics ideas:\nIf x ∈ M, we endow the dual space D(x)∗ with the dual metric gx∗\nto gx on D(x) ⊂ T ∗ M. This allows to define the length of a cotangent\nvector (x, ξ) ∈ T ∗ M as kξkg := kξ\|D(x)kgx∗ . The function H : T ∗ M → R\ngiven by\nH(x, ξ) := 21 kξk2g\nis called the Hamiltonian of the sub-Riemannian manifold M. Observe\nthat this is, apart from the factor 21 , just the principal symbol of the\nassociated sub-Laplacian. Since T ∗ M is a symplectic manifold\nin a\nP\ncanonical way with respect to the symplectic form ωM := j dξj ∧ dxj\n(in canonical coordinates), we may associate to H the Hamiltonian vector field H, i.e., ωM (X, H) = dH(X) for every X ∈ T M. The integral\ncurves (x(t), ξ(t)) of H are called bicharacteristics. They satisfy the\nHamilton-Jacobi equations\nẋ(t) =\n∂H\n(x(t), ξ(t)),\n∂ξ\n∂H\nξ˙ = −\n(x(t), ξ(t)),\n∂x\nand H is constant along every bicharacteristic. A normal geodesic of\nthe sub-Riemannian manifold M will then be the space projection x(t)\nof a bicharacteristic (x(t), ξ(t)) along which the Hamiltonian H does\nnot vanish. This notion is in accordance with the idea that the wave\nfront set of a solution to our wave equation should evolve along the\nbicharacteristics of the wave equation.\nIt is known that every length minimizing curve in a sub-Riemannian\nmanifold M is either such a normal geodesic, or an “abnormal geodesic” . I won’t give a definition of the latter ones here, since\n12\nDETLEF MÜLLER\nit is known that groups of Heisenberg type (and more generally,\nMetivier groups) do not admit non-trivial abnormal geodesics.\nHowever, it should be warned here that in general abnormal geodesics\ndo exist, even in nilpotent Lie groups (see for an example of a group\nof step 4, and for an example of step 2). This fact has been ignored\nin some cases in the literature, which had led to some confusion.\nOn a Heisenberg type group G, one easily finds that the Hamiltonian\nis given by\nH = 12 \|ξ − 21 Jµ x\|2 ,\nif we write coordinates for T ∗ G as ((x, u), (ξ, µ)) ∈ (R2m ×Rn )×(R2m ×\nRn ). The Hamilton-Jacobi equations are then given by\n∂H\n∂H\n,\nu̇ =\n∂ξ\n∂µ\n∂H\n∂H\nξ˙ = −\n,\nµ̇ = −\n.\n∂x\n∂u\nBy left-invariance, it suffices to condsider bicharacteristics starting at\nthe origin, i.e., x(0) = 0, u(0) = 0. We also put η := ξ − 12 Jµ x. Then\nH = 12 hη, ηi, so that \|η\| is constant along the bicharacteristic. Also\nµ = const., and\nẋ =\nẋ = η, ξ˙ = − 21 Jµ η\nη̇ = ξ˙ − 12 Jµ ẋ = −Jµ η,\nhence\nη(t) = e−tJµ α,\nif we put ξ(0) = η(0) := α. This implies x(t) =\nJµ2 = −\|µ\|2 I, we have\n(2.2)\n1−e−tJµ\nα,\nJµ\nand since\nsin( 2t \|µ\|) − t Jµ\ne 2 α.\nx(t) = 2\n\|µ\|\nFor simplicity, let us now assume that n = 1, i.e., that we are dealing\nwith a Heisenberg group. Then Jµ = µJ, with µ ∈ R, and we have\nsin( 2t \|µ\|) −tJµ\nt\nhe\nα, Je− 2 Jµ αi\nu̇ =\n=−\n\|µ\|\nt\nsin( 2 \|µ\|)\nt\nhα, Je 2 Jµ αi.\n= −\n\|µ\|\nhη, − 21 Jxi\nTHE WAVE EQUATION ON GROUPS OF HEISENBERG TYPE\n13\nt\nLooking at the Taylor series of e 2 Jµ and using the fact that J is skew\nand satisfies J 2 = −I, this yields\nµ\nt\nu̇ =\n\|α\|2 sin2 ( \|µ\|),\n2\n\|µ\|\n2\nif µ 6= 0. Consequently,\n(\nsin( 2t \|µ\|) cos( 2t \|µ\|)\n) \|α\|2 \|µ\|µ2 ,\n( 2t −\n\|µ\|\n(2.3)\nu(t) =\n0, if µ = 0.\nif µ 6= 0\nThe same formula remains valid on an arbitrary Heisenberg type group\n. Observe that the curve γ(t) := (x(t), u(t)) is a normal geodesic\niff α 6=P\n0, and it has constant speed kγ̇kg = \|ẋ(t)\| = \|η\| = \|α\|, since\nγ̇(t) = j ẋj (t)Xj (γ(t)).\nDenote by Σ1 the set of endpoints of all geodesics of length 1 (see\nfigure 1). These endpoints are given by (2.2),(2.3), with t = 1, where\nα can be chosen arbitrarily within the unit sphere in R2m and µ in Rn .\nTherefore,\n(2.4)\nsin s s − sin s cos s Σ1 = {(x, u) ∈ G : \|x\| = , 4\|u\| = for some s ∈ R.}\n2\ns\ns\nWe then expect that Σ1 is exactly the singular support of any of the\nwave propagators\n√\n√\n√\nsin(t L)\n√\nδ, cos(t L)δ or eit L δ,\nL\nfor time t = 1, and this turns out to be true.\nSingular support of a wave emenating from a single point in the Heisenberg group\nFigure 1. Σ1\n14\nDETLEF MÜLLER\nThe “outer hull” of the set Σ1 is just the unit sphere with respect to\nthe Carnot-Carathéodory distance.\nLet us finally mention the well-known fact that solutions to our wave\nequation have finite propagation speed (see , or for an elementary√proof for Lie groups), i.e., if Pt denotes any of the wave propagators\n√\nsin(t L)\n√\nL)δ, then, for t ≥ 0,\nδ,\ncos(t\nL\n(2.5)\nsupp Pt ⊂ {(x, u) ∈ G : k(x, u)kCC ≤ t}\n3. The Schrödinger and the wave propagators on a\nHeisenberg type group\nOne principal problem in proving estimates for solutions to wave\nequations consists in finding sufficiently explicit expressions for the\nassociated wave propagators. For strictly hyperbolic equations, such\nexpressions can be given by means of Fourier integral operators, at least\nlocally in space and time. These technics, however, do not apply to\nwave equations associated to non-elliptic Laplacians, and no sufficiently\nexplicit expressions for the wave propagators are known to date in these\nmore general situations, except for the case of Heisenberg type groups\nand a few related examples. What allows us to give such expressions\non Heisenberg type groups is the existence of explicit formulas for the\nSchrödinger propagators eitL δ on these groups.\nThere are several possible approaches to these formulas. One would\nconsist in appealing to the Gaveau-Hulanicki formula , for the\nheat kernels e−tL δ, which can be derived by means of representation\ntheory and Mehler’s formula, and then trying to apply physicists approach and replacing time t by complex time −it.\nWe shall here sketch another approach, which is more far-reaching\nand has turned out to be useful also in the study of questions of solvability of general second order left-invariant differential operators on\n2-step nilpotent Lie groups (see , also for further references to this\ntopic).\nLet G be a group of Heisenberg type as before. Given f ∈ L1 (G) and\nµ ∈ g∗2 = R2m , we define the partial Fourier transform f µ of f along\nTHE WAVE EQUATION ON GROUPS OF HEISENBERG TYPE\nthe center by\nµ\nf (x) :=\nZ\nRn\n15\nf (x, u)e−2πiµ·u du (x ∈ R2m ).\nMoreover, if we define the µ–twisted convolution of two suitable functions ϕ and ψ on g1 = R2m by\nZ\n(ϕ ∗µ ψ)(x) :=\nϕ(x − y)ψ(y)e−iπωµ(x,y) dy,\nR2m\nthen\n(f ∗ g)µ = f µ ∗µ g µ ,\nwhere f ∗ g denotes the convolution product of the two functions f, g ∈\nL1 (G). Accordingly, the vector fields Xj are transformed into the µtwisted first order differential operators Xjµ such that (Xj f )µ = Xjµ f µ ,\nand the sub-Laplacian is transformed into the µ-twisted Laplacian Lµ ,\ni.e.,\n2m\nX\n(Lf )µ = Lµ f µ = −\n(Xjµ )2\nj=1\n(say for f ∈ S(G)). An easy computation shows that explicitly\nXjµ =\n∂\n+ iπωµ (x, Xj ).\n∂xj\nWhat turns out to be important for us is that the one-parameter\nµ\nSchrödinger group eitL , t ∈ R, generated by Lµ can be computed\nexplicitly:\nProposition 3.1. For f ∈ S(G),\nµ\neitL f = f ∗µ γtµ ,\nt ∈ R,\nwhere γtµ ∈ S ′ (R2m ) is a tempered distribution given by the imaginary\nGaussian\nm\nπ\n\|µ\|\n2\nµ\n−m\nγt (x) = 2\ne−i 2 \|µ\| cot(2πt\|µ\|)\|x\| ,\n(sin(2πt\|µ\|)\nwhenever sin(2πt\|µ\|) 6= 0. Moreover, by well-known formulas for the\ninverse Fourier transform of a generalized Gaussian , its Fourier\ntransform is\n2π\n1\ntan(2πt\|µ\|)\|ξ\|2\ni \|µ\|\nµ\nγc\ne\nt (ξ) =\n(cos(2πt\|µ\|))m\nwhenever cos(2πt\|µ\|) 6= 0.\n16\nDETLEF MÜLLER\n3.1. Twisted convolution and the metaplectic group. It is not\ndifficult to reduce the proof of Proposition 3.1 to the case of the Heisenberg group G = Hm , which we shall consider now. For any real, symmetric matrix A := (ajk )j,k=1,...,2m let us put\n(3.1)\nS := −AJ,\n0\nIm\nis as in the definition of the product in Hm .\n−Im 0\nIt is easy to see that the mapping A 7→ S is a vector space isomorphism\nof the space of all real, symmetric 2m×2m matrices onto the Lie algebra\nwhere J :=\nsp(m, R) := {S : tSJ + JS = 0}.\nLet us define\n∆S :=\n2m\nX\najk Xj Xk ,\nj,k=1\nS ∈ sp(m, R),\nwhere A is related to S by (3.1). Then one verifies easily that for any\nS1 , S2 ∈ sp(m, R), we have\n(3.2)\n[∆S1 , ∆S2 ] = −2U∆[S1 ,S2 ] ,\nwhere U denotes again the central basis element of hm . Applying a\npartial Fourier transform along the (here one-dimensional) center, this\nleads to the following commutation relations among the µ-twisted convolution operators ∆µS :\n[∆µS1 , ∆µS2 ] = −4πiµ ∆µ[S1 ,S2 ] .\nSince ∆µS is formally self-adjoint, this means that the mapping\n(3.3)\nS 7→\ni\n∆µ\n4πµ S\nis a representation of sp(m, R) by (formally) skew-adjoint operators on\nL2 (R2m ).\nLet us consider the case µ = 1 (it is not difficult to reduce considerations to this case). In this case, we just speak of the twisted convolution,\nand write ϕ × ψ in place of ϕ ∗1 ψ.\nIn , R. Howe has proved for this case that the map (3.3) can\nbe exponentiated to a unitary representation of the metaplectic group\nMp (m, R), a two-fold covering of the symplectic group Sp(m, R). The\nTHE WAVE EQUATION ON GROUPS OF HEISENBERG TYPE\n17\ngroup Mp (m, R) can in fact be represented by twisted convolution operators of the form f 7→ f × γ, where the γ’s are suitable measures\nwhich, generically, are multiples of purely imaginary Gaussians\neA (z) := e−iπ\ntz·A·z\n,\nwith real, symmetric 2m × 2m matrices A. In particular, one has\nt\n1\nei 4π ∆S f = f × γt,S ,\n(3.4)\nt ∈ R.\nThe distributions γt,S have been determined explicitly in . To\nindicate how this can be accomplished, let us argue on a completely\nformal basis:\nIf eA , eB are two Gaussians as before such that det(A + B) 6= 0, one\ncomputes that\neA × eB = [det(A + B)]−1/2 eA−(A−J/2)(A+B)−1 (A+J/2) ,\nwhere a suitable determination of the root has to be chosen. Choosing\nA = 21 JS1 , B = 21 JS2 , with S1 , S2 ∈ sp(m, R), and assuming that S1\nand S2 commute, one finds that\ne 1 JS1 × e 1 JS2 = 2n (det(S1 + S2 ))−1/2 e 1 J[S1 S2 +I)(S1 +S2 )−1 ] .\n2\n2\n2\nThis reminds of the addition law for the hyperbolic cotangent, namely\ncoth x coth y + 1\n.\ncoth x + coth y\nWe are thus led to define, for non-singular S,\n1\n(3.5)\nA(t) := J coth(tS/2),\n2\nwhich is well-defined at least for \|t\| > 0 small.\ncoth(x + y) =\nThen\neA(t1 ) × eA(t2 ) = 2n (det(A(t1 ) + A(t2 )))−1/2 eA(t1 +t2 ) .\nAnd, from\nsinh(x + y)\n,\nsinh x sinh y\nwe obtain (ignoring again the determination of roots)\ncoth x + coth y =\n[det sinh((t1 + t2 )S/2)]1/2 [det(A(t1 ) + A(t2 ))]−1/2\n= [det sinh(t1 S/2)]1/2 [det sinh(t2 S/2)]1/2 .\n18\nDETLEF MÜLLER\nTogether this shows that\n(3.6)\nγt,S := 2−n [det sinh(tS/2)]−1/2 eA(t)\nforms a (local) 1-parameter group under twisted convolution, and it is\nnot hard to check that its infinitesimal generator is 4πi ∆1S .\nFinally, if we choose S := J, so that A = −I and ∆S = L, we\nobtain Proposition 3.1, at least for \|µ\| = 1; the general case follows by\na scaling argument.\nRemark 3.2. It is important to note that Howe’s construction of the\nmetaplectic group by means of twisted convolution operators can be extended to the so-called oscillator semigroup. If we denote by sp+ (C, m)\nthe cone of all elements S ∈ sp(C, m) such that the associated complex, symmetric matrix A = SJ has positive semi-definite real part,\nthen\naccordingly explicit formulas for the one-parameter semigroups\nt∆µ\nS\ne , t ≥ 0, have been derived in .\n3.2. A subordination formula. Fix a bump function χ0 ∈ C0∞ (R)\nsupported in the interval [1/2, 2]. If a ∈ S ν is a symbol of order ν ∈ R\non T ∗ R = R2 , i.e., if\n\|∂sα ∂σβ a(s, σ)\| ≤ Cα,β (1 + \|σ\|)ν−β ,\n(s, σ) ∈ R2 ,\nfor every α, β ∈ N, then we say that a is supported over the subset\nI ⊂ R if supp a(·, σ) ⊂ I for every σ ∈ R.\nProposition 3.3. There exist a symbol a0 ∈ S 0 supported over the\ninterval [1/16, 4] and a Schwartz function ρ ∈ S(R2 ) such that, for\nx ≥ 0,\n√x √\ntx it x\ne\nχ0\n= ρ\n, λt\nλ\nλZ\n√\nλt\nts\n+\n(3.7)\nλt a0 (s, λt) ei 4s ei λ x ds,\nfor every λ ≥ 1 and t > 0 such that λt ≥ 1.\nProof. This can easily be obtained by the method of stationary phase.\nµ\nThe explicit formulas in Proposition 3.1 for eitL in combination\nwith Proposition 3.3 allow√ us to describe spectrally localized parts of\nthe wave propagators e±it L δ. Of course, there is no loss of generality\nin assuming that t ≥ 0 and that the sign in the exponent is positive.\nTHE WAVE EQUATION ON GROUPS OF HEISENBERG TYPE\n19\nMoreover, since L is homogeneous of degree 1 with respect to the automorphic dilations δr , we may reduce ourselves to considering the case\nwhere t = 1.\nLet us recall here that it is well-known that if ψ ∈ S(R), then the\nconvolution kernel ψ(L)δ will also be a Schwartz function (on G), whose\nL1 (G)-norm will be controlled by some Schwartz norm on S(R) (see,\n; also for a connection with finite propagation speed).\nProposition 3.4. Let χ0 ∈ C0∞ (R) be a bump function supported in\nthe interval [1/2, 2], and assume that λ ≥ 1. Then there exist Schwartz\nfunctions ψλ ∈ S(G) and Kλ ∈ S(G) such that\n√L √\nχ0\nei L δ = ψλ + Kλ ,\n(3.8)\nλ\nand so that the following hold true:\n(i) For every N ∈ N there is constant CN ≥ 0 so that\n\|\|ψλ \|\|1 ≤ CN λ−N\nfor every λ ≥ 1.\n(ii) The partial Fourier transform Kλµ ∈ S(R2m ), (µ 6= 0), of Kλ is\ngiven by an oscillatory integral of the form\n√ Z\nλ\nµ\nµ\n, ϕi dt, ϕ ∈ S(R2m ),\nhKλ , ϕi = λ a0 (t, λ) ei 4t hγt/λ\nwhere a0 ∈ S 0 is a symbol of order 0 supported over the interval\n[1/16, 4] and where γtµ is given by Proposition 3.1.\nProof. Apart from (i), all the claims follow easily from the subordination\nformula (3.7) in Proposition 3.3, the spectral resolution\nR\nL = R+ s dEs and Fubini’s theorem in combination with Proposition\n3.1.\nTo prove (i), notice that we may choose ψλ := ρ Lλ , λ δ, with a\nSchwartz\nρ ∈ S(R2 ). In particular, every Schwartz norm of\nfunction\nx 7→ ρ λx , λ decays like = O(λ−N ) for every N ∈ N. By the previous remark, it is then also clear\nthat\nthe convolution kernel ψλ of the\nspectral multiplier operator ρ Lλ , λ will satisfy the estimates in (i).\n20\nDETLEF MÜLLER\nIn view of this proposition, it will in the sequel suffice to estimate\nthe convolution operators defined by the kernels Kλ . Moreover, for\nsimplicity, we may and shall assume that\n√ Z\nλ\nµ\nµ\n(3.9)\nhKλ , ϕi = λ χ0 (t) ei 4t hγt/λ\n, ϕi dt, ϕ ∈ S(R2m ),\nwhere χ0 ∈ C0∞ (R) is supported in the interval [1/16, 4]. Indeed, this\nwill be justified since our estimates will only depend on some C k -norm\nof χ0 .\n, 3π\n] and\nη0 ∈ C0∞ (R) such that supp η0 ⊂ [− 3π\n8\n8\nPWe next choose\nπ\nη\n(s\n+\nk\n)\n=\n1\nfor\nevery\ns\n∈\nR,\nand\nput\n0\nk∈Z\n2\nZ\nλ\nµ\nµ\n1/2\n, ϕi dt,\nχ0 (t) η0 (2πt\|µ\|/λ − kπ) ei 4t hγt/λ\nhAλ,k , ϕi := λ\nZ\nλ\nµ\nµ\n1/2\n, ϕi dt.\nχ0 (t) η0 (2πt\|µ\|/λ − kπ − π2 ) ei 4t hγt/λ\nhBλ,k , ϕi := λ\nBy Aλ,k and Bλ,k we shall denote the tempered distributions defined\non G whose partial Fourier transforms along the center are given by\nµ\nAµλ,k and Bλ,k\n, respectively. Then\n(3.10)\nKλ =\n∞\nX\n(Aλ,k + Bλ,k )\nk=0\nin the sense of distributions.\nFor the sake of this exposition, we shall restrict ourselves to considering the Aλ,k . Since\nµ\nµ\nhγt/λ\n, ϕi = hγd\nb\nt/λ , ϕi,\nµ\nµ\nwhere γd\nt/λ is given by Proposition 3.1, it is clear that Aλ,k is a wellµ\ndefined tempered distribution. However, since the integral kernel γt/λ\nhas a singularity in t where 2πt\|µ\|/λ = kπ if k 6= 0, we shall here perform a priori a dyadic decomposition with respect to t. More precisely,\nif k ≥ 1, let us fix another bump\nχ1 ∈ C0∞ (R) supported in\nP∞ function\nthe interval [1/2, 2] so that l=0 χ1 (2l s) = 1 for every s ∈]0, 3π\n], and\n8\ndecompose the distribution Aµλ,k as\n(3.11)\nhAµλ,k , ϕi\n=\n∞\nX\nl=0\n(hAµλ,k,l,+, ϕi + hAµλ,k,l,−, ϕi),\nTHE WAVE EQUATION ON GROUPS OF HEISENBERG TYPE\n21\nwhere Aλ,k,l,± is defined to be the smooth function\nR\nAµλ,k,l,±(x) := λ1/2 χ0 (t)χ1 ± 2l (2πt\|µ\|/λ − kπ) η0 (2πt\|µ\|/λ − kπ)\nm λ\nπ\n2\n\|µ\|\n2−m (sin(2πt\|µ\|/λ)\nei 4t 2−m e−i 2 \|µ\| cot(2πt\|µ\|/λ)\|x\| dt\nBy Fourier inversion in the central variable u, we have\nZ\nAλ,k,l,± (x, u) =\nAµλ,k,l,± (x)e2πiu·µ dµ,\nRn \\{0}\nIn the integration w.r. to µ, we introduce polar coordinates µ =\nrω, r > 0, ω ∈ S n−1 . Then dµ = r n−1 dσn−1 (ω), where dσn−1 denotes\nthe surface measure on the sphere S n−1 . It is well-known by the\nmethod of stationary phase that\nZ\neiu·ω dσn−1 (ω) = ei\|u\| a+ (\|u\|) + e−2πi\|u\| a− (\|u\|), u ∈ Rn ,\nS n−1\nwhere a+ , a− ∈ S −(n−1)/2 are symbols of order −(n − 1)/2.\nSince the integral defining Aµλ,k,l,± is absolutely convergent in t, applying this and suitable changes of coordinates, we then find that\n(3.12)\nAλ,k,l,± (x, u) = Aλ,k,l,±,a+ (x, \|u\|) + Aλ,k,l,±,a− (x, −\|u\|),\nwhere for any symbol a ∈ S −(n−1)/2 the function Aλ,k,l,±,a(x, v) on R2m ×\nR is given by\nRR\nAλ,k,l,±,a(x, v) = λm+n+1/2\nχ(t) χ1 ± 2l (s − kπ) η(s − kπ)sm+n−1\n1\nsin(s)−m eiλts( s −cot(s)\|x\|\n2 +4v)\na(λst4\|v\|) dtds.\nSince k ≥ 1, this can be re-written in the form\nR R t χ1 (±2l s)\nAλ,k,l,±,a(x, v) = λm+n+1/2 k m+n−1\nχ s/k+π sinm (s) η(s)\n(3.13)\niλkt\ne\n1\n−cot(s)\|x\|2 +4v\ns+kπ\na(λkt4\|v\|) dsdt,\nwhere the functions χ and η have similar properties as χ1 and η0 .\nA similar expressions can be given for k = 0.\n22\nDETLEF MÜLLER\n4. Estimation of Aλ,k,l,± if k ≥ 1\nWe next need to establish estimates for the kernels Aλ,k,l,±, into which\nthe kernels Aλ,k decompose, and their derivatives. In some regions, the\nestimates that can be obtained for these kernels will be useful only if\nλ & k. However, this problem can be fixed in an easy way, and we\nshall not go into details here but prefer to give a somewhat simplified\naccount of the actual estimates that we can obtain, for the sake of\nclarity of exposition.\nWe shall only consider Aλ,k,l,+,a, since Aλ,k,l,−,a, can be estimated very\nmuch in the same way. Let us use the short-hand notation Fl (x, v) :=\nAλ,k,l,+,a(x, v), so that, by (3.13), for k ≥ 1 we have\nZ Z t\nm+n+1/2 m+n−1 (m−1)l\nχ −l −1\nFl (x, v) = λ\nk\n2\nχ(l) (r)\n2 k r+π\niλkt ψ(2−l r)+4v\nη(2−l r)e\na(λkt4\|v\|) drdt,\nwhere the function ψ(s) is given by\n1\n− \|x\|2 cot(s),\nψ(s) :=\ns + kπ\nand where\nχ1 (r)\nχ(l) (r) := l\n(2 sin(2−l r))m\nis again supported where r ∼ 1. Moreover, if 2−l r ∈ supp η and k ≥ 1,\nthen \|2−l r\| ≤ 3π\n≤ kπ\n, so that\n8\n2\n1\n∼ k −1\n+ kπ\nfor every k ≥ 1, l ≥ 0 (the case of the negative sign will actually become\nrelevant for the Aλ,k,l,−,a only.)\n±2−l r\nNotice that all derivatives of χ(l) (r) are uniformly bounded in l. We\ncan therefore re-write\nFl (x, v) = λm+n+1/2 k m+n−1 2(m−1)l\nZ Z\niλkt ψ(2−l r)+4v\n(4.1)\n·\nχk,l (r, t)e\na(λkt4\|v\|) drdt,\nwhere χk,l (r, t) is a smooth function supported where\nr ∼ 1, t ∼ 1,\nTHE WAVE EQUATION ON GROUPS OF HEISENBERG TYPE\n23\nsuch that\n\|∂rα ∂tβ χk,l (r, t)\| ≤ Cα,β for every α, β ∈ N,\n(4.2)\nuniformly in k and l.\nSince\n(4.3)\n1\n+ \|x\|2 sin(s)−2 ,\n(s + kπ)2\n2\nψ ′′ (s) =\n− 2\|x\|2 cos(s) sin(s)−3 ,\n(s + kπ)3\nψ ′ (s) = −\nwe see that the critical points s = sc of ψ are given by the equation\nsin(sc ) = ±\|x\|(sc + kπ), hence those of ψ(2−l ·) by r = 2l sc . If r is\nsupposed to lie in our domain of integration, then \|x\| ∼ 2−l k −1 . For\nsuch x, there is then in fact exactly one critical point sc = sc,k (\|x\|) ∼ 2−l\nsufficiently close to our domain of integration, given by\nsin(sc ) = \|x\|(sc + kπ).\n(4.4)\nFor later use, let us also put\nγk (\|x\|) := ψ(sc,k (\|x\|)) =\n1\n− \|x\|2 cot(sc,k (\|x\|)).\nsc,k (\|x\|) + kπ\nObserve that, by (4.4), this can be re-written as\n(4.5)\nγk (\|x\|) =\nsk (\|x\|) − sin sk (\|x\|) cos sk (\|x\|)\n,\nsk (\|x\|)2\nif we put sk (\|x\|) := sc,k (\|x\|) + kπ.\nAssume that λ & 2l k.\nIn this case, one gains from the r-integration in Fl , which we perform\nfirst. To this end, recall that ψ(s) has now a unique critical point\nsc = sc,k (\|x\|) ∼ 2−l if \|x\| ∼ 2−l k −1 .\nWe may thus apply the method of stationary phase to the integration in r. Noticing that the critical point of the phase ψl is given by\n2l sc,k (\|x\|), one eventually finds that\nFl (x, v) = λ\nm+n+1/2 m+n−1 (m−1)l\nk\n2\nZ\nχ(t)a(λkt4\|v\|)\niλkt ψ(sc,k (\|x\|))+4v\nb(k 2 22l \|x\|2 , λk −1 2−l t)e\ndrdt,\n24\nDETLEF MÜLLER\nwhere χ(t) is again a smooth cut-off function supported where t ∼ 1,\nand where b ∈ S −1/2 is a symbol of order −1/2 on T ∗ R, which may in\nfact depend also on k and l, but which lie in a bounded set of S −1/2\nwith respect to the natural Fréchet topology on S −1/2 .\nConsequently,\nAλ,k,l,+,a(x, ±\|u\|) = λ\n(4.6)\nm+n+1/2 m+n−1 (m−1)l\nk\n2\nZ\nχ(t)a(λkt4\|u\|)\niλkt γk (\|x\|)±4\|u\|\nb(k 2 22l \|x\|2 , λk −1 2−l t)e\ndt,\nwhere γk (\|x\|) := ψ(sc,k (\|x\|)) is explicitly given by (4.5).\nNow, integrations by parts in (4.6) easily yield that, for every N ∈ N,\n1\n1\n\|Aλ,k,l,+,a(x, ±\|u\|)\| ≤ CN λm+n k m+n− 2 2(m− 2 )l\n−N\nn−1\n(4.7)\n.\n·(1 + λk\|u\|)− 2 1 + λk\|γk (\|x\|) ± 4\|u\|\|\nThe actual details of these arguments are more involved, and a complete estimate of Aλ,k,l,+,a requires a careful case analysis and domain\ndecompositions.\nRemarks 4.1. (a) Observe that (4.7) shows that the singularities of\nAλ,k,l,+(x, u) are located where \|u\| = γk (\|x\|). In view of (4.4) and (4.5),\nthese are thus located where the pair of norms (\|x\|, 4\|u\|) lies on the\npiece of the parametric curve\nsin s s − sin s cos s γ : s 7→ , ,\ns\ns2\ncorresponding to the range of parameters where s − kπ ∼ 2−l .\nThis corresponds exactly to the set Σ1 , in view of (2.4).\n(b) Analogous estimates and results hold true for Aλ,k,l,− (x, u). Here,\nwe have sc (\|x\|) < 0, and the singularities correspond to the range of\nparameters where s − kπ ∼ −2−l .\nTHE WAVE EQUATION ON GROUPS OF HEISENBERG TYPE\n25\n1/k\n1/(k+1)\n1/k\nFigure 2. curve γ\n√\n5. Estimation of (1 + L)−(d−1)/4 ei\nLet us put\n1 1\nα(p) := (d − 1) − ,\np 2\nL\nδ\n1 ≤ p < ∞.\n√\nWe need to estimate (1 + L)−α(p)/2 ei L δ on Lp (G) in Theorem 1.1.\nBy means of a dyadic decomposition, this can easily be reduced to\nestimating the operator\n∞\n√L √\nX\n−α(p)j\nei L ,\n(5.1)\n2\nχ̃\nj\n2\nj=0\nfor a suitable cut-off function χ̃ supported in [1/16, 4]. In view of (3.10),\nthis means that we have to estimate the convolution operators on Lp (G)\ngiven by the integral kernels\np,±\nEk,J\n(x, u)\n:=\nJ X\n∞\nX\n2−α(p)j A2j ,k,l,±(x, u)\nj=0 l=0\nand the corresponding kernels, with Aλ,k,l,± replaced by Bλ,k,l,±. These\nkernels will converge, in the sense of distributions, to distributions Ekp,±\nas J → ∞, and we shall have to show that the estimates sum in k.\n26\nDETLEF MÜLLER\nOur estimates will follow from some rather elementary estimate for\nthe case p = 2, and a deeper estimate for p = 1 on a suitable local\nHardy space h1k , whose definition will depend on k, by means of complex\ninterpolation. The corresponding interpolation argument is standard\n(cf. ,).\nLet’s here concentrate on the case p = 1, and the kernels\n+\nEk,J\n:=\n1,+\nEk,J\n=\nJ X\n∞\nX\n2−\n(d−1)\nj\n2\nA2j ,k,l,+.\nj=0 l=0\nWhat can be proved in the end for these kernels is a result, which,\n±\nProposition 5.1. Let k ≥ 1. There exist kernel functions Kλ,k,l\n∈\nl\n1\nS(G), for λ ≥ 2 k, and Rk,J ∈ L (G) so that\n(5.2)\n+\nEk,J\n=\n∞\nX\nJ\nX\n(K2+j ,k,l + K2−j ,k,l) + Rk,J ,\nl=0 j=l+log2 k\nand such that the following hold true:\n±\n(a) The kernel Kλ,k,l\nis supported in the set\n{(x, u) ∈ G : \|x\| ∼ 2−l k −1 and λk\|u\| & 1},\nand can be estimated, for every α ∈ N2m , β ∈ Nn , by\nn+1\n(5.3)\n1\n1\n±\n\|∂xα ∂uβ Kλ,k,l\n(x, u)\| ≤ CN,α,β λ\|α\| (λk)\|β\| λ 2 k m+n− 2 2(m− 2 )l\n−N\nn−1\n(1 + λk\|u\|)− 2 1 + λk\|γk (\|x\|) ± 4\|u\|\|\n,\nfor every N ∈ N, with constants CN,α,β which independent of\nλ, k and l, and where γk (\|x\|) is given by (4.5). In particular,\n(5.4)\n1\n±\nk∂xα ∂uβ Kλ,k,l\nk1 . λ\|α\| (λk)\|β\| (2l k)−m− 2 .\n(b) The sequence {Rk,J }J converges for J → ∞ in L1 (G) towards\na function Rk ∈ L1 (G), and\n(5.5)\n1\nkRk k1 ≤ Ck −m− 2 ,\nwith a constant C not depending on k.\nTHE WAVE EQUATION ON GROUPS OF HEISENBERG TYPE\n27\n5.1. Anisotropic re-scaling for k fixed. Let k ≥ 1, and denote by\nF̃ (x, u) := k −D F (k −1 x, k −2 u)\nthe L1 -norm preserving scaling of a function F on G by δk−1 . Let us\nalso put\nk\n−Γk (v) := k 2 γk (k −1 v) − .\nπ\nWe can then make the interesting observation that, by (5.3), (5.4), the\n±\n]\nnon-isotropically re-scaled kernels K\nare supported in the set\nλ,k,l\n{(x, u) ∈ G : \|x\| ∼ 2−l and\nλ\n\|u\| & 1},\nk\nand satisfy the isotropic estimates\nλ \|α\|+\|β\| n+1\n1\n1\nα β ]\n±\n\|∂x ∂u Kλ,k,l (x, u)\| ≤ CN,α,β\nλ 2 k −(m+n+ 2 ) 2(m− 2 )l\nk\n−N\nλ k\nλ\n− n−1\n2\n(5.6)\n1 + − Γk (\|x\|) ± 4\|u\|\n(1 + \|u\|)\nk\nk π\nand\nλ \|α\|+\|β\|\n1\n±\n]\n(5.7)\nk∂xα ∂uβ K\nk\n.\n(2l k)−m− 2 .\n1\nλ,k,l\nk\nMoreover, one easily verifies that\n1\np\n−2\n−1\n2\n.\n(5.8)\nΓk (v) = π arcsin(πv) + π v 1 − (πv) + O\nk\n6. L2 -estimates for components of the wave propagator\nDenote by Aλ,k,l,± the convolution operator\nAλ,k,l,±f := f ∗ Aλ,k,l,±\non L2 (G). It follows from Proposition 3.1 that\nR\nAµλ,k,l,± = λ1/2\nχ0 (t)χ1 (±2l (2πt\|µ\|/λ − kπ)) η0 (2πt\|µ\|/λ − kπ)\nλ\nµ\nei 4t eit/λ L dt δ.\nComparing with (9.13) and (9.12) in the Appendix, which obtained by\nmeans of Plancherel’s formula on G, we then find that the operator\nnorm of Aλ,k,l,± on L2 (G) is given by\n(6.1)\nkAλ,k,l,±k = λ1/2 sup{\|αλ,k,l,±(ν, q)\| : ν ≥ 0, q ∈ N},\n28\nDETLEF MÜLLER\nwhere\nαλ,k,l,±(ν, q)\nZ\nν\nλ\n:=\nχ0 (t)χ1 (±2l (t λν − kπ)) η0 (t λν − kπ)ei{ 4t +t λ (m+2q)} dt.\nThe method of stationary phase then implies that\nkAλ,k,l,±k ≤ C,\nuniformly in λ, k and l. By the same method, we also get, for any α ≥ 0,\n∞ X\nX\n2−αj A2j ,k,l,±k ≤ C 2−αj0 ,\n(6.2)\nk\nl=0 j≥j0\nuniformly in l and j0 .\n7. Estimation for p = 1\nWe introduce a non-standard atomic local Hardy space h1 on G as\nfollows:\nAn atom centered at the origin is an L2 -function a supported in a\nEuclidean ball Br (0) of radius r ≤ 1 satisfying the normalizing condition\n(7.1)\nkak2 ≤ r −d/2\nand, if r < 1, in addition the moment condition\nZ\n(7.2)\na(x) dx = 0.\nG\nNotice that in particular kak1 . 1. An atom centered at z ∈ G is\nthe left-translate by z of an atom centered at the origin. The local\nHardy space h1 (G) consists of all L1 -functions which admit an atomic\ndecomposition\nX\n(7.3)\nf=\nλj aj ,\nj\nP\nwith atoms aj and coefficients λj such that j \|λj \| < ∞. We define the\nP\nnorm kf kh1 as the infimum of the sums j \|λj \| < ∞ over all possible\natomic decompositions (7.3) of f.\nIn comparison, let us denote by h1E := h1 (R2m × Rn ) the classical\nEuclidean local Hardy space introduced by Goldberg . Since atoms\nTHE WAVE EQUATION ON GROUPS OF HEISENBERG TYPE\n29\nin this space which are supported in a ball of radius r ≥ 1 need not\nsatisfy (7.2), such atoms can be decomposed into atoms supported in\nballs of radius 1, so that one can give the same atomic definition for\nh1E as for h1 , only with the non-commutative group translation in G\nreplaced by Euclidean translation.\nConsider the cylindrical sets Z(y) := {(x, u) ∈ G : \|x − y\| ≤ 10}, for\ny ∈ R2m . Note the following easy, but important observations:\nLemma 7.1. a) Any atom a supported in Z(0) is indeed also a Euclidean atom in h1E (possibly up to a fixed factor depending only on G),\nand vice versa.\nb) If HE1 := H 1 (R2m × Rn ) denote the classical Euclidean Hardy\nspace, and if η ∈ C0∞ (R2m ), then multiplication with η ⊗ 1 maps HE1\ncontinuously into h1E .\nb) is known for compactly supported smooth multipliers, and the\nproof carries over easily.\nSince G can be covered by sets Z(yj ) with bounded overlap, if we\nchoose a suitable ε-separated set of points {yj }j for a suitable constant\nε > 0, one can use a) and b) in combination with classical arguments\n to show that our non-standard Hardy space h1 (G) interpolates with\nL2 (G) by the complex method.\nWe shall not do this here, but indicate the main idea in the interpolation argument explained later.\nProposition 7.2. There is a constant C > 0, such that\nfor every f ∈ h1 (G).\n1,±\n−m\nkf ∗ Eg\nkf kh1\nk k1 ≤ Ck\nProof. We again consider only Ek1,+ = Ek+ . In view of the atomic\ndecomposition of f and because of left-invariance, one can show that\nit suffices to assume that f = a is an atom centered at the origin and\nsupported in a ball Br (0). We shall also assume that r < 1, so that a\nhas vanishing integral, since the case r = 1 is somewhat easier. In view\nof Proposition 5.1, we have to understand in particular the functions\n±\n^\na∗K\n.\n2j ,k,l\n30\nDETLEF MÜLLER\nTo simplify the argument a bit, assume that n = 1, so that \|u\| = ±u.\n±\n]\nLet’s also assume that we consider the part of K\nwhere u > 0, and\nλ,k,l\n±\n]\nthat K\nλ,k,l were not only essentially, but actually supported in the set\nwhere\nk\nk\n− Γk (\|x\|) + 4u\| ≤ .\nπ\nλ\nTranslating these kernels along the center by πk , in view of (5.6) we can\nthen make the (slightly oversimplified) assumption that\nk\n±\n−l\n1\n]\nu\n−\nΓ\n(\|x\|)\n(7.4) supp K\n⊂\n{(x,\nu)\n∈\nG\n:\n\|x\|\n∼\n2\nand\n≤ },\nλ,k,l\n4 k\nλ\nand that\nλ \|α\|+\|β\|\n1\n1\n±\n]\n(7.5)\n\|∂xα ∂uβ K\n(x,\nu)\|\n≤\nC\nλ k −(m+n+ 2 ) 2(m− 2 )l\nN,α,β\nλ,k,l\nk\nDefine for ρ > 0 the set\n1\nΩρ := {(x, u) ∈ G : \|x\| . 1 and u − 4 Γk (\|x\|) ≤ Cρ}\nwhere C is a sufficiently large constant, and choose j0 so that\nk2−j0 = r.\nObrserve that with this choice of j0\nXX\n±\n^\nK\n) ⊂ Ωr ,\n(7.6)\nsupp a ∗ (\n2j ,k,l\nj≥j0\nl\nsince supp a ⊂ Br (0). We therefore split\nEk1,± = Ek,r + Fk,r ,\nwhere\nEk,r :=\nFk,r :=\nP∞ P\nl=0\nP∞\nl=0\nj≥j0\nP\nj<j0\n2−α(1)j A2j ,k,l,±,\n2−α(1)j A2j ,k,l,± .\ng\nWe estimate a ∗ E\nk,r separately on the set Ωr and it’s complement. By\nHölder’s inequality and (6.2), we get\n(d−1)\n1/2\ng\nka ∗ E\nkak2 2−α(1)j0 . r 1/2 r −(m+1/2) 2− 2 j0 ≤ k −m .\nk,r kL1 (Ωr ) . \|Ωr \|\nP P\n±\n^\ng\nMoreover, since by Proposition 5.1 Ek,r = l j≥j0 K\n2j ,k,l +Rk,r , where\n1\nkRk,r k1 . k −m− 2 , we obtain from (7.6) that\n1\n−m−\ng\ng\n2,\nka ∗ E\nk,r kL1 (G\\Ωr ) ≤ ka ∗ Rk,r kL1 (G) . k\nTHE WAVE EQUATION ON GROUPS OF HEISENBERG TYPE\n31\nso that\n−m\ng\nka ∗ E\n.\nk,r k1 . k\n(7.7)\ng\ng\nConsider next a∗ F\nk,r . Again, by Proposition 5.1, we can write Fk,r =\n1\nP P\n−m−\n±\n′\n^\n′\ng\n2 . And, for j < j0 , we\nl\nj<j0 K2j ,k,l + Rk,r , where kRk,r k1 . k\nhave, by (7.4),\nX\n±\n^\nK\n) ⊂ Ω −j .\nsupp a ∗ (\nk2\n2j ,k,l\nl\nR\nMoreover, making use of the cancellation a dx = 0 and (7.5) in es±\n^\ntimating the convolution product a ∗ K\n, we gain a factor of order\nO( k2r−j ) compared to (5.7) and obtain\nka ∗\nX\nl\n±\n^\nK\n2j ,k,l k1 .\nSumming in j, this gives\n(7.8)\n∞\nX\nl=0\n2j ,k,l\n1\n1\nr\nr\n(2l k)−m− 2 . 2j k −m− 2\n−j\nk2\nk\n−m− 12\ng\n.\nka ∗ F\nk,r k1 . k\nThe estimates (7.7) and (7.8) imply the proposition.\nRemark 7.3. Let us denote by h1k (G) the re-scaled local Hardy space\nconsisting of all function\nf (x, u) := k D f˜(kx, k 2 u), f˜ ∈ h1 (G).\nThen Proposition 7.1 can be re-stated as\nkf ∗ Ek1,± k1 ≤ Ck −m kf kh1k\nfor every f ∈ h1k (G). Thus, h1k (G) is a natural space associated with\nthe k-th piece Ek1,± of the wave propagator whose definition depends\non the parameter k.\n8. Proof of Theorem 1.1\nObserve that by (6.2) convolution with the kernel\n∞ X\n∞\nX\n2,±\nEk :=\nA2j ,k,l,±\nj=0 l=0\nis bounded on L2 (G), and\n(8.1)\nkf ∗ Ek2,± k2 ≤ Ckf k2 ,\n32\nDETLEF MÜLLER\nwhere C is independent of k. Consider now the analytic family of operators\n∞ X\n∞\nX\n2−αj A^\n0 ≤ Re α ≤ (d−1)\nTα : f 7→ f ∗\n.\n2j ,k,l,± ,\n2\nj=0 l=0\nBy ignoring some error terms, we may assume that all A^\n2j ,k,l,± are\nsupported in Z(0) (even a smaller cylindrical set). This implies that if\nf is supported in a cylindrical set Z(y), then Tα f is supported in the\n“cylindrical double” 2Z(y) := {(x, u) ∈ G : \|x − y\| ≤ 20}, for every\ny ∈ R2m , which allows to reduce to functions f supported in a set Z(y),\nand we may even assume y = 0, by translation invariance.\nChoosing a suitable cut-off function η ∈ C0∞ (R2m ) so that η ⊗ 1\nlocalizes to Z(0), we therefore consider the localized operators\nTα0 f := Tα ((η ⊗ 1)f ).\n0\nBy Lemma 7.1 and Proposition 7.3, T d−1\n: HE1 → L1 continuously, and\n2\nsince the proof of Proposition 7.3 remains valid for T d−1 +iβ , we have\n2\n0\nkT d−1 +iβ f k1 ≤ Ck\n−m\n2\nkf kHE1\n∀β ∈ R.\nSimilarly, by (8.1), we have\nkTiβ0 f k2 ≤ Ckf k2\n∀β ∈ R.\nWe can thus apply the complex interpolation result by C. Fefferman\nand Stein in to obtain\n2\n0\nkTα(p)\nf kp ≤ Ck −( p −1)m kf kp ,\n1 < p ≤ 2.\nIn view of the afore-mentioned support property of the convolution\nkernels involved, this immediately implies\nα(p),±\nkf ∗ Ek\n2\nkp ≤ Ck −( p −1)m kf kp ,\n1 < p ≤ 2.\nIf m ≥ 2, these estimates sum in k for p sufficiently close to 1, so that\n(8.2)\n√\nk(1 + L)−α(p)/2 ei\nL\nf kp ≤ Ckf kp\nfor these p. Interpolating these estimates with the corresponding trivial\nestimate for p = 2 we then obtain Theorem 1.1 for 1 < p ≤ 2, and the\ncase p > 2 follows by duality.\nFor m = 1, the result can be obtained by means of a minor refinement\nof the arguments.\nTHE WAVE EQUATION ON GROUPS OF HEISENBERG TYPE\n33\nThe L1 -estimate in Theorem 1.1 follows a lot more easily from Propod−1\nsition 5.1, since for α > d−1\nwe have an additional factor λ−(α− 2 )\n2\nin the corresponding estimate (5.4), so that the corresponding norms\nkK2±j ,k,l k1 simply sum in j, k and l.\n34\nDETLEF MÜLLER\n9. Appendix: The Fourier transform on a group of\nHeisenberg type\nLet us first briefly recall some facts about the unitary representation\ntheory of a Heisenberg type group G (compare ,). For the convenience of the reader, we shall show how this representation theory\ncan easily be reduced to the well-known case of the Heisenberg group\nHm = R2m × R. In slightly modified notation, the product in Hm is\ngiven by\n1\n(z, t) · (z ′ , t′ ) = (z + z ′ , t + t′ + ω(z, z ′ )),\n2\nwhere ω denotes the canonical symplectic form\n0 −Im\n(9.1)\nω(z, w) := hJz, wi,\nJ :=\n,\nIm\n0\non R2m . Let us split coordinates z = (x, y) ∈ Rm × Rm in R2m , and\nconsider the associated natural basis of left-invariant vector fields\n∂\n∂\n∂\n∂\n∂\nX̃j :=\n− 12 yj , Ỹj :=\n+ 12 xj , j = 1, . . . , m, and T := ,\n∂xj\n∂t\n∂yj\n∂t\n∂t\nof the Lie algebra of Hm .\nFor τ ∈ R× := R \\ {0}, the Schrödinger representation ρτ of Hm acts\non the Hilbert space L2 (Rm ) as follows:\n1\n[ρτ (x, y, t)h](ξ) := e2πiτ (t+y·ξ+ 2 y·x) h(ξ + x),\nh ∈ L2 (Rm ).\nThis is an irreducible, unitary representation, and every irreducible\nunitary representation of Hm which acts non-trivially on the center is\nin fact unitarily equivalent to exactly one of these, by the Stone-von\nNeumann theorem (a good reference to these and related results is for\nNext, if π is any unitary representation, say, of a Heisenberg type\ngroup G, we denote by\nZ\nπ(f ) :=\nf (g)π(g) dg, f ∈ L1 (G),\nG\nthe associated representation of the group algebra L1 (G). Going back\nto the Heisenberg group,\nR if f ∈ S(Hm ), then it is well-known and\neasily seen that ρτ (f ) = R2m f −τ (z)ρτ (z, 0) dz is a trace class operator\non L2 (Rm ), and its trace is given by\nZ\n−m\nf (0, 0, t)e2πiτ t dt = \|τ \|−m f −τ (0, 0),\n(9.2)\ntr(ρτ (f )) = \|τ \|\nR\nTHE WAVE EQUATION ON GROUPS OF HEISENBERG TYPE\n35\nfor every τ ∈ R× .\nFrom these facts, it is easy to derive the Plancherel formula for our\nHeisenberg type group G. Given µ ∈ g∗2 = Rn , µ 6= 0, consider the\nmatrix Jµ introduced in Section 3. If \|µ\| = 1, then Jµ2 = −I, because\nof (1.1). In particular, Jµ has only eigenvalues ±i, and since it is\northogonal, it is easy to see that there exists an orthonormal basis\nXµ,1 , . . . , Xµ,m , Yµ,1, . . . , Yµ,m\nof g1 = R2m which is symplectic with respect to the form ωµ , i.e., ωµ\nis represented by the standard symplectic matrix J in (9.1).\nThis means that, for every µ ∈ Rn \\ {0}, there is an orthogonal\nµ ∈ O(2m, R) such that\nmatrix Rµ = R \|µ\|\n(9.3)\nJµ = \|µ\|Rµ J tRµ .\nNotice also that tRµ = Rµ−1 , and that condition (9.3) is in fact equivalent to G being of Heisenberg type.\nWe remark that (9.3) easily implies that the subalgebra L1r (G) of\nL1 (G), consisting of all radial functions f (x, u) in the sense that they\ndepend only on \|x\| and u, is commutative. This fact is well-known and\neasy to prove for Heisenberg groups (,). And, by means of (9.3),\none easily checks that\n(ϕ ∗µ ψ) ◦ Rµ = (ϕ ◦ Rµ ) ×\|µ\| (ψ ◦ Rµ )\nfor all functions ϕ, ψ ∈ L1 (R2m ). Therefore, if f, g ∈ L1r (G), then f µ ◦\nRµ , g µ ◦Rµ are radial on R2m , and thus f µ ◦Rµ ×\|µ\| g µ ◦Rµ = g µ ◦Rµ ×\|µ\|\nf µ ◦ Rµ , which is again radial. Consequently, f µ ∗µ g µ = g µ ∗µ f µ , for\nevery µ, and thus\n(9.4)\nf ∗g = g∗f\nfor every f, g ∈ L1r (G).\nThe following lemma is easy to check and establishes a useful link\nbetween representations of G and those of Hm .\nLemma 9.1. The mapping αµ : G → Hm , given by\nαµ (z, u) := ( tRµ z, µ·u\n),\n\|µ\|\n(z, u) ∈ R2m × Rn ,\nis an epimorphism of Lie groups. In particular, G/ ker αµ is isomorphic\nto Hm , where ker αµ = µ⊥ is just the orthogonal complement of µ in\nthe center Rn of G.\n36\nDETLEF MÜLLER\nGiven µ ∈ Rn \\ {0}, we can now define an irreducible unitary representation πµ of G on L2 (Rm ) by simply putting\nπµ := ρ\|µ\| ◦ αµ .\nObserve that then πµ (0, u) = e2πiµ·u I. In fact, any irreducible representation of G with central character e2πiµ·u factors through the kernel of\nαµ and hence, by the Stone-von Neumann theorem, must be equivalent\nto πµ .\nOne then computes that, for f ∈ S(G),\nZ\nπµ (f ) =\nf −µ (Rµ z)ρ\|µ\| (z, 0) dz,\nR2m\nso that the trace formula (9.2) yields the analogous trace formula\ntr πµ (f ) = \|µ\|−m f −µ (0)\nn\nRon G. The Fourier minversion formula in R then leads to f (0, 0) =\ntr πµ (f ) \|µ\| dµ. When applied to δg−1 ∗ f, we arrive at the\nµ∈Rn \\{0}\nFourier inversion formula\nZ\n(9.5)\nf (g) =\ntr (πµ (g)∗πµ (f )) \|µ\|m dµ, g ∈ G.\nµ∈Rn \\{0}\nApplying this to f ∗ ∗ f at g = 0, where f ∗ (g) := f (g −1), we obtain the\nPlancherel formula\nZ\n2\n(9.6)\nkf k2 =\nkπµ (f )k2HS \|µ\|m dµ,\nµ∈Rn \\{0}\nwhere kT k2HS = tr (T ∗ T ) denotes the Hilbert-Schmidt norm.\nLet us next consider the group Fourier transform of our sub-Laplacian\nL on G.\nWe first observe that dαµ (X) = tRµ X for every X ∈ g1 = R2m , if\nwe view, for the time being, elements of the Lie algebra as tangential\nvectors at the identity element. Moreover, by (9.3), we see that\nt\nRµ Xµ,1 , . . . , tRµ Xµ,m , tRµ Yµ,1 , . . . , tRµ Yµ,m\nforms a symplectic basis with respect to the canonical symplectic form\nω on R2m . We may thus assume without loss of generality that this\nbasis agrees with our basis X̃1 , . . . , X̃m , Ỹ1 , . . . , Ỹm of R2m , so that\ndαµ (Xµ,j ) = X̃j , dαµ (Yµ,j ) = Ỹj ,\nj = 1, . . . , m.\nTHE WAVE EQUATION ON GROUPS OF HEISENBERG TYPE\n37\nBy our construction of the representation πµ , we thus obtain for the\nderived representation dπµ of g that\n(9.7)\ndπµ (Xµ,j ) = dρ\|µ\| (X̃j ), dπµ (Yµ,j ) = dρ\|µ\| (Ỹj ),\nj = 1, . . . , m.\nLet us define the sub-Laplacians Lµ on G and L̃ on Hm by\nLµ := −\nm\nX\n2\n(Xµ,j\n+\n2\nYµ,j\n),\nj=1\nm\nX\nL̃ := −\n(X̃j2 + Ỹj2 ),\nj=1\nwhere from now on we consider elements of the Lie algebra again as\nleft-invariant differential operators. Then, by (9.6),\ndπµ (Lµ ) = dρ\|µ\| (L̃).\nMoreover, since the basis Xµ,1 , . . . , Xµ,m , Yµ,1 , . . . , Yµ,m and our original\nbases X1 , . . . , X2m of g1 are both orthonormal bases, it is easy to verify\nthat the distributions Lδ0 and Lµ δ0 do agree. Since Af = f ∗ (Aδ0 )\nfor every left-invariant differential operator A, we thus have L = Lµ ,\nhence\n(9.8)\ndπµ (L) = dρ\|µ\| (L̃).\nBut, it is well-known that dρ\|µ\| (L̃) = ∆ξ −(2π\|µ\|)2\|ξ\|2 is just a re-scaled\nHermite operator, and an orthonormal basis of L2 (Rm ) is given by the\ntensor products\n\|µ\|\n\|µ\|\nh\|µ\|\nα := hα1 ⊗ · · · ⊗ hαm ,\nα ∈ Nm ,\nwhere hµk (x) := (2π\|µ\|)1/4hk ((2π\|µ\|)1/2 x). Here,\ndk −x2\ne\ndxk\ndenotes the L2 -normalized Hermite function of order k on R. Consequently,\nhk (x) = ck (−1)k ex\n(9.9)\n2 /2\n\|µ\|\ndπµ (L)h\|µ\|\nα = 2π\|µ\|(m + 2\|α\|)hα ,\nIt is also easy to see that\n(9.10)\ndπµ (Uj ) = 2πiµj I,\nα ∈ Nm .\nj = 1, . . . , n.\nNow, the operators L, −iU1 , . . . , −iUn form a commuting set of selfadjoint operators, with joint core S(G), so that they admit a joint\nspectral resolution, and we can thus give meaning to expressions like\nϕ(L, −iU1 , . . . , −iUn ) for each continuous function ϕ defined on the\ncorresponding joint spectrum. For simplicity of notation we write\nU := (−iU1 , . . . , −iUn ).\n38\nDETLEF MÜLLER\nIf ϕ is bounded, then ϕ(L, U) is a bounded, left invariant operator on\nL2 (G), so that it is a convolution operator\nϕ(L, U)f = f ∗ Kϕ ,\nf ∈ S(G),\nwith a convolution kernel Kϕ ∈ S ′ (G) which will also be denoted by\nϕ(L, U)δ. Moreover, if ϕ ∈ S(R × Rn ), then ϕ(L, U)δ ∈ S(G) (compare , ). Since functional calculus is compatible with unitary\nrepresentation theory, we obtain in this case from (9.9), (9.10) that\n(9.11)\n\|µ\|\nπµ (ϕ(L, U)δ)h\|µ\|\nα = ϕ(2π\|µ\|(m + 2\|α\|), 2πµ)hα\n(this identity in combination with the Fourier inversion formula could\nin fact be taken as the definition of ϕ(L, U)δ). In particular, the\nPlancherel theorem implies then that the operator norm on L2 (G) is\ngiven by\n(9.12)\nkϕ(L, U)k = sup{\|ϕ(\|µ\|(m + 2q), µ)\| : µ ∈ Rn , q ∈ N}.\nFinally, observe that\n(9.13)\nKφµ = ϕ(Lµ , 2πµ)δ;\nthis follows for instance by applying the unitary representation induced\nfrom the character e2πiµ·u on the center of G to Kϕ .\nReferences\n Hajer Bahouri, Patrick Gérard, and Chao-Jiang Xu. Espaces de Besov et estimations de Strichartz généralisées sur le groupe de Heisenberg. J. Anal. Math.,\n82:93–118, 2000.\n R. Michael Beals. Lp boundedness of Fourier integral operators. Mem. Amer.\nMath. Soc., 38(264):viii+57, 1982.\n Michael Christ. Lp bounds for spectral multipliers on nilpotent groups. Trans.\nAmer. Math. Soc., 328(1):73–81, 1991.\n Y. Colin de Verdière and M. Frisch. Régularité lipschitzienne et solutions de\nl’équation des ondes sur une variété riemannienne compacte. Ann. Sci. École\nNorm. Sup. (4), 9(4):539–565, 1976.\n Ewa Damek and Fulvio Ricci. Harmonic analysis on solvable extensions of\nH-type groups. J. Geom. Anal., 2(3):213–248, 1992.\n C. Fefferman and E. M. Stein. H p spaces of several variables. Acta Math.,\n129(3-4):137–193, 1972.\n G. B. Folland and Elias M. Stein. Hardy spaces on homogeneous groups, volume 28 of Mathematical Notes. Princeton University Press, Princeton, N.J.,\n1982.\n Gerald B. Folland. Harmonic analysis in phase space, volume 122 of Annals of\nMathematics Studies. Princeton University Press, Princeton, NJ, 1989.\nTHE WAVE EQUATION ON GROUPS OF HEISENBERG TYPE\n39\n Bernard Gaveau. Principe de moindre action, propagation de la chaleur et\nestimées sous elliptiques sur certains groupes nilpotents. Acta Math., 139(12):95–153, 1977.\n David Goldberg. A local version of real Hardy spaces. Duke Math. J., 46(1):27–\n42, 1979.\n Chr. Golé and R. Karidi. A note on Carnot geodesics in nilpotent Lie groups.\nJ. Dynam. Control Systems, 1(4):535–549, 1995.\n Waldemar Hebisch. Multiplier theorem on generalized Heisenberg groups. Colloq. Math., 65(2):231–239, 1993.\n Lars Hörmander. Hypoelliptic second order differential equations. Acta Math.,\n119:147–171, 1967.\n Lars Hörmander. The analysis of linear partial differential operators. I, volume\n256 of Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences]. Springer-Verlag, Berlin, 1983. Distribution\ntheory and Fourier analysis.\n Roger Howe. The oscillator semigroup. In The mathematical heritage of Hermann Weyl (Durham, NC, 1987), volume 48 of Proc. Sympos. Pure Math.,\npages 61–132. Amer. Math. Soc., Providence, RI, 1988.\n A. Hulanicki. The distribution of energy in the Brownian motion in the Gaussian field and analytic-hypoellipticity of certain subelliptic operators on the\nHeisenberg group. Studia Math., 56(2):165–173, 1976.\n Andrzej Hulanicki. A functional calculus for Rockland operators on nilpotent\nLie groups. Studia Math., 78(3):253–266, 1984.\n Wensheng Liu and Héctor J. Sussman. Shortest paths for sub-Riemannian\nmetrics on rank-two distributions. Mem. Amer. Math. Soc., 118(564):x+104,\n1995.\n Giancarlo Mauceri and Stefano Meda. Vector-valued multipliers on stratified\ngroups. Rev. Mat. Iberoamericana, 6(3-4):141–154, 1990.\n Richard Melrose. Propagation for the wave group of a positive subelliptic\nsecond-order differential operator. In Hyperbolic equations and related topics\n(Katata/Kyoto, 1984), pages 181–192. Academic Press, Boston, MA, 1986.\n Akihiko Miyachi. On some estimates for the wave equation in Lp and H p . J.\nFac. Sci. Univ. Tokyo Sect. IA Math., 27(2):331–354, 1980.\n D. Müller and F. Ricci. Analysis of second order differential operators on\nHeisenberg groups. I. Invent. Math., 101(3):545–582, 1990.\n D. Müller and A. Seeger. Sharp Lp -estimates for the wave equation on Heisenberg type groups. in preparation, 2008.\n D. Müller and E. M. Stein. On spectral multipliers for Heisenberg and related\ngroups. J. Math. Pures Appl. (9), 73(4):413–440, 1994.\n Detlef Müller. A restriction theorem for the Heisenberg group. Ann. of Math.\n(2), 131(3):567–587, 1990.\n Detlef Müller. Analysis of invariant PDO’s on the Heisenberg group.\nLecture notes ICMS-Instructional conference, Edinburgh, April 1999,\nhttp://analysis.math.uni-kiel.de/mueller/, 1999.\n Detlef Müller. Marcinkiewicz multipliers and multi-parameter structure on\nHeisenberg groups. Lecture notes of a minicorso at Padova, June 2004,\nhttp://analysis.math.uni-kiel.de/mueller/, 2004.\n40\nDETLEF MÜLLER\n Detlef Müller. Local solvability of linear differential operators with double characteristics. II. Sufficient conditions for left-invariant differential operators of\norder two on the Heisenberg group. J. Reine Angew. Math., 607:1–46, 2007.\n Detlef Müller, Fulvio Ricci, and Elias M. Stein. Marcinkiewicz multipliers\nand multi-parameter structure on Heisenberg (-type) groups. I. Invent. Math.,\n119(2):199–233, 1995.\n Detlef Müller, Fulvio Ricci, and Elias M. Stein. Marcinkiewicz multipliers\nand multi-parameter structure on Heisenberg (-type) groups. II. Math. Z.,\n221(2):267–291, 1996.\n Detlef Müller and Elias M. Stein. Lp -estimates for the wave equation on the\nHeisenberg group. Rev. Mat. Iberoamericana, 15(2):297–334, 1999.\n Adrian I. Nachman. The wave equation on the Heisenberg group. Comm. Partial Differential Equations, 7(6):675–714, 1982.\n Edward Nelson and W. Forrest Stinespring. Representation of elliptic operators\nin an enveloping algebra. Amer. J. Math., 81:547–560, 1959.\n Martin Paulat. Sub-Riemannian geometry and heat kernel estimates. Dissertation, C.A.University Kiel, 2008.\n Juan C. Peral. Lp estimates for the wave equation. J. Funct. Anal., 36(1):114–\n145, 1980.\n Fulvio Ricci. Harmonic analysis on generalized Heisenberg groups. unpublished\npreprint.\n Andreas Seeger, Christopher D. Sogge, and Elias M. Stein. Regularity properties of Fourier integral operators. Ann. of Math. (2), 134(2):231–251, 1991.\nD. Müller, Mathematisches Seminar, C.A.-Universität Kiel, LudewigMeyn-Str.4, D-24098 Kiel, Germany" ]	[ null, "https://s2.studylib.net/store/data/010916513_1-1cd38b438237763de2ce0e78a255c362.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8189958,"math_prob":0.99134636,"size":62400,"snap":"2020-10-2020-16","text_gpt3_token_len":21880,"char_repetition_ratio":0.12045644,"word_repetition_ratio":0.049160767,"special_character_ratio":0.3423077,"punctuation_ratio":0.15772913,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9983257,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-25T22:08:16Z\",\"WARC-Record-ID\":\"<urn:uuid:541603d4-3af6-44d1-95d9-ed3f8203a5a7>\",\"Content-Length\":\"131680\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:54908f04-14cc-4623-ab32-0a376e4b87a8>\",\"WARC-Concurrent-To\":\"<urn:uuid:b7a038f5-c792-43f1-93e0-017fef084a54>\",\"WARC-IP-Address\":\"104.24.124.188\",\"WARC-Target-URI\":\"https://studylib.net/doc/10916513/sharp-l--estimates-for-the-wave-equation-on-heisenberg-ty..\",\"WARC-Payload-Digest\":\"sha1:BHQJTRXQWO632T26MJGCEYVGJQMN4XFJ\",\"WARC-Block-Digest\":\"sha1:Z7RNFKVE43GTWFT6M7I2SIIJUQQ4T333\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875146160.21_warc_CC-MAIN-20200225202625-20200225232625-00296.warc.gz\"}"}
https://terraria.fandom.com/zh/wiki/%E6%B0%B8%E5%A4%9C%E5%88%83	[ "•", null, "", null, "", null, "类型 \\l!--\\n --\\g\\{\\{#vardefine:_typepage\\!\\{\\{#or:\\{\\{l10n\\!item_infobox\\!typepage:@@@@\\}\\}\\!\\{\\{tr\\!\\{\\{ucfirst:@@@@\\}\\}\\!link\\ey\\}\\}\\}\\}\\}\\}\\l!--\\_\\n --\\g\\{\\{#vardefine:_typetext\\!\\{\\{#or:\\{\\{l10n\\!item_infobox\\!type:@@@@\\}\\}\\!\\{\\{tr\\!\\{\\{ucfirst:@@@@\\}\\}\\}\\}\\}\\}\\}\\}\\l!--\\_\\n --\\g\\{\\{nowrap\\!class\\etag\\!\\{\\{ifexists\\!\\{\\{#var:_typepage\\}\\}\\!\\(\\(\\{\\{#var:_typepage\\}\\}\\!\\{\\{#var:_typetext\\}\\}\\)\\)\\!\\{\\{#var:_typetext\\}\\}\\}\\}\\}\\}\\l!--\\n --\\g\\l!--\\n --\\g\\{\\{#vardefine:_typepage\\!\\{\\{#or:\\{\\{l10n\\!item_infobox\\!typepage:@@@@\\}\\}\\!\\{\\{tr\\!\\{\\{ucfirst:@@@@\\}\\}\\!link\\ey\\}\\}\\}\\}\\}\\}\\l!--\\_\\n --\\g\\{\\{#vardefine:_typetext\\!\\{\\{#or:\\{\\{l10n\\!item_infobox\\!type:@@@@\\}\\}\\!\\{\\{tr\\!\\{\\{ucfirst:@@@@\\}\\}\\}\\}\\}\\}\\}\\}\\l!--\\_\\n --\\g\\{\\{nowrap\\!class\\etag\\!\\{\\{ifexists\\!\\{\\{#var:_typepage\\}\\}\\!\\(\\(\\{\\{#var:_typepage\\}\\}\\!\\{\\{#var:_typetext\\}\\}\\)\\)\\!\\{\\{#var:_typetext\\}\\}\\}\\}\\}\\}\\l!--\\n --\\g 42 (近战) 4.5（普通击退力） 4% 21（快速度） / 27（普通速度）", null, "4 需要 1 份\n\n使用 \\{\\{#lstmap:\\$soundfiles\\$\\!,\\!%%%%\\!\\l!--\\_print\\_sounds\\n --\\g\\lspan\\_style\\e\"margin-left:2px;\"\\g\\{\\{sound\\!\\!\\{\\{trim\\!%%%%\\}\\}\\}\\}\\l/span\\g\\l!--\\n --\\g\\!\\}\\}\n\n!!Error: （请输入有效的值。参见Template:Modifier。）\n\n## 制作\n\ntotal: 2 row(s)\n\n### 制作树\n\n × 20(", null, "+", null, ") × 60( 地狱 ) × 30 40 (", null, "", null, ") × 30 40 (", null, "", null, ") × 12 × 12 15 (", null, "", null, ")", null, "", null, "", null, "", null, "", null, "(", null, ") × 10 × 10", null, "/", null, "", null, "/", null, "", null, "/", null, "", null, "/", null, "或", null, "/", null, "total: 2 row(s)\n\n## 备注\n\n• 因为永夜刃需要来自地牢村正，因此它只在骷髅王被打败之后才可获得。\n• 在中，几乎总是会在地牢守卫生成的深度以上生成一个锁住的金箱，并且其中总会有一把村正。因此，有可能在打败骷髅王之前就得到永夜刃。\n\n## 花絮\n\n• 1.1 更新以前，永夜刃是游戏中最强大的剑，尽管此地位现在已被推翻了。尽管如此，它仍然是最强的困难模式之前的剑。\\{\\{comma/item\\!\\l!--\\n --\\g\\{\\{#var:_s_bi_start\\}\\}\\l!--\\_bold/italic\\_opening\\_tags\\n --\\g\\{\\{@@@@\\_version\\!\\{\\{#var:_s_mode\\}\\}\\!short\\e\\{\\{#var:_s_short\\}\\}\\!nl\\e\\{\\{#var:_s_nl\\}\\}\\!small\\e\\{\\{#var:_s_small\\}\\}\\}\\}\\l!--\\n --\\g\\{\\{#var:_s_bi_end\\}\\}\\l!--\\_bold/italic\\_closing\\_tags\\n --\\g\\}\\}上除外，在其中晶光刃在困难模式之前已经可用。\n• 永夜刃是唯一一种在祭坛上制作又不是 Boss 召唤物品的物品。\n• 用于制作它的四把剑各自代表了困难模式之前的一种主要的困难生物群落：草剑代表的是丛林，村正是地牢，炽焰巨剑是地狱，魔光剑/血腥屠刀代表的是邪恶生物群落\n• 永夜刃在现实中也有相应的玩具，由 Jazwares 制造。\n• 它在《地牢守护者 2》中作为 Phantom's Edge 出现。\n• 无论用的是哪个制作配方，因为永夜刃散发出来的颗粒的关系，它都像是腐化主题的剑。\n• 此剑的微粒包括了制作此剑所用的全部剑：村正（正十字）、炽焰巨剑（斜十字）、草剑（小泡泡）、和魔光剑/血腥屠刀（大泡泡）。\n\n## 历史\n\n• 电脑版 1.4.1\n• 使用时间由 27 降低至 21。\n• 现在能自动挥舞了。\n• 主机版 1.02\n• 现在用于原版永夜刃的制作。\n• 血腥屠刀现在可以用来在制作配方中替代魔光剑。\n• 出售价值由 54 提升至 18\n• 主机正式版：伴随着电脑版 1.0.6 的相应外观和机制引入。\n• 移动版 1.2.11212\n• 现在用于原版永夜刃的制作。\n• 血腥屠刀现在可以用来在制作配方中替代魔光剑。\n• 出售价值由 54 提升至 18\n• 移动正式版：伴随着电脑版 1.0.6 的相应外观和机制引入。\n• 3DS正式版：和机制、配方、外观一道由电脑版 1.2 引入。\n\n## 参考\n\n1. Terraria-Based Toys Now Available! 2014 年 10 月 30 日" ]	[ null, "https://static.wikia.nocookie.net/terraria_gamepedia/images/9/98/Night%27s_Edge.png/revision/latest", null, "https://static.wikia.nocookie.net/terraria_gamepedia/images/c/ca/Night%27s_Edge_%28old%29.png/revision/latest", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null, "data:image/gif;base64,R0lGODlhAQABAIABAAAAAP///yH5BAEAAAEALAAAAAABAAEAQAICTAEAOw%3D%3D", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.52381647,"math_prob":0.9910934,"size":4679,"snap":"2022-27-2022-33","text_gpt3_token_len":3109,"char_repetition_ratio":0.26395723,"word_repetition_ratio":0.12612613,"special_character_ratio":0.49369523,"punctuation_ratio":0.25925925,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.983442,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56],"im_url_duplicate_count":[null,1,null,1,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,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\":\"2022-07-06T03:18:13Z\",\"WARC-Record-ID\":\"<urn:uuid:5753fcb3-55c1-47d1-8dce-c78f77aada2f>\",\"Content-Length\":\"555636\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c3a51a8d-914d-4de3-b856-3b2c7539e300>\",\"WARC-Concurrent-To\":\"<urn:uuid:623f1c3c-23bf-4243-a810-867726ed67a4>\",\"WARC-IP-Address\":\"151.101.0.194\",\"WARC-Target-URI\":\"https://terraria.fandom.com/zh/wiki/%E6%B0%B8%E5%A4%9C%E5%88%83\",\"WARC-Payload-Digest\":\"sha1:3QNVXHZ4IUHZA67QTNKOVSCSTL55UW33\",\"WARC-Block-Digest\":\"sha1:B3IGCKM2XYLZOJUGS2F47AGYFRHERMVN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104660626.98_warc_CC-MAIN-20220706030209-20220706060209-00731.warc.gz\"}"}
https://resources.altium.com/p/s-parameter-measurements-and-errors-power-integrity	[ "# S-Parameter Measurements and Errors in Power Integrity", null, "Zachariah Peterson\n\| Created: February 11, 2021 \| Updated: February 12, 2021", null, "No matter where you look, it seems S-parameters never go away! They are mandatory tools for understanding some systems, such as an interconnect or antenna, while other network parameters can sometimes give a better conceptual understanding of electrical behavior. These parameters are normally reserved for signal integrity among electronics engineers, but if you look hard enough, you’ll find that S-parameters are also used for power integrity. This should make sense intuitively simply from a power flow perspective: Kurokawa’s original formulation of S-parameters was in terms of power carried by a signal, so why not use this for power integrity?\n\nIn PDN design, particularly for high speed digital components, we care about designing to a low PDN impedance. Low PDN impedance leads to low voltage disturbances measured between power rails for a given transient current draw. Network parameters can be used to characterize the PDN and determine its impedance, but the use of S-parameters requires the use of the appropriate reference (port) impedance for an accurate PDN impedance calculation. Let’s look at exactly how errors in S-parameter measurements propagate into Z-parameter measurements in a simple case to gain some intuition, then I’ll discuss general N-port PDNs and how errors in the S-parameter matrix create errors in the impedance matrix.\n\n## S-Parameters and Power Integrity\n\nWhen measuring S-parameters, every measurement will be bandlimited and discretely sampled. This leads to errors in the measurement that are unavoidable. In other words, the S-parameters you measure are not the true S-parameters, leading to problems with causality. Since S-parameters can be used to calculate other network parameters (including Z-parameters), how does the S-parameter error influence the Z-parameter error? Let’s look at this for a 2-port PDN, then an N-port PDN.\n\n### Errors in 2-Port PDNs with Large S11\n\nFirst, let’s look at errors in a 2-port PDN as this is an easy problem we can solve to get some insight. To get started, we can use a basic conversion to relate the S-parameters in our PDN back to the Z-parameters, then calculate the Z-parameters in the presence of some error.\n\nIn the following equation, I’ve defined my PDN self-impedance in terms of an S-parameter matrix for the PDN in the presence of 2 errors. The e term is my S11/S22 error, and the f term is my S21/S12 error. Assuming reciprocity holds (Sij = Sji), we have:\n\nJust to focus on the critical aspects of self-impedance, let's assume the PDN is reciprocal and lossless. In this case, the S-parameters are S21 = S12 = 0 and S11 = S22, and the above equation reduces to the familiar conversion between S11 and self-impedance. We can get a good approximation on the Z11 error if we take the difference between the high and low error as defined above, and by setting and squared error terms to zero (i.e., e2 << e). This gives the following simple expression for errors in the PDN impedance due to measurement errors in S11:", null, "Self-impedance in terms of S-parameter error for the special case of a lossless reciprocal PDN.\n\nIn this example, let’s suppose that S11 = -0.9 in our hypothetical lossless reciprocal PDN. In this case, a 1% error in my S-parameter measurement translates into a 10.5% error in Z11. That’s a 10-fold amplification in error!\n\nThat may seem like a major mistake, but it matches with remarks from other experts in this area. In particular, note the remark on page 8 of this study from Keysight, where a 1-2% error in S-parameter measurements leads to a PDN impedance measurement of 300 to 400 mOhms. Just for a sanity check, let’s bring this back into our example. When the true impedance is ~10 mOhms and the standard VNA port impedance of 50 Ohms is used, we have S11 = -0.9996 and Z11 measurement error of 250%. Such large impedance mismatches at the input port are highly undesirable when we’re trying to use S-parameters to determine PDN impedance.\n\n### Errors in 2-Port PDNs with Small S11\n\nNow let’s suppose my reference impedance is brought much closer to my PDN impedance so that S11 = 0.1 with up to 1% error. The error in the Z11 is now only 2.02%. When we have very close matching to the true PDN impedance, we have a reduction in error in the calculated Z11 value. As it turns out, the critical S11 value in this example where your PDN impedance error will perfectly match your S-parameter measurement error is S11 = 0.268.\n\nThis should show how a large impedance mismatch amplifies S-parameter error measurements when calculating the impedance parameters for a 2-port PDN. Note that this is frequency dependent, but the process applies at each frequency measurement; you might have very accurate impedance at some frequencies, and your results might be very inaccurate at others. This can then be extended to N-port networks using the general S-to-Z-parameter conversion.\n\n### Errors in N-Port PDNs\n\nN-port problems are much more difficult to deal with analytically; this requires using the general Z-parameter matrix for an N-port network (including self-impedances and transfer impedances). In general, you’ll need to perform the same process as detailed above, but with a general S-to-Z-parameter conversion matrix for an N-port network:\n\nThis takes a ton of algebra just to derive an expression relating S-parameter errors to Z-parameter errors. Therefore, this problem of solving such a matrix for a set of S-parameter measurements and port impedances is best solved with Matlab or Mathematica. The point here is, the error is going to be inversely proportional to products of squared (1 - S) terms. Therefore, we’re going to have a situation where an N output port PDN will need to have its reference impedance approximately reduced by a factor N to ensure the error will be low.\n\n## What Reference Impedance Should be Used?\n\nFrom the above discussion, it seems you would want as small of an impedance mismatch as possible if you were using a VNA to measure the S-parameters for an N-port PDN, and then use these measurements to determine the impedance matrix. This would then give you the smallest possible error magnitude in your Z-parameters for a given S-parameter measurement error. Since the PDN impedance is at mOhm levels, your reference impedance should also be around mOhm levels, not at the 50 Ohm level normally set in commercial VNAs.\n\nFinally, if you apply renormalization to drop down the S-parameter reference to be closer to the PDN impedance (maybe down to 50 mOhms), the error term also propagates nonlinearly because normalization involves an S-parameter multiplication operation. In other words, there may be some S-parameter values squared, which may amplify the error in the Z-parameter values you calculate. I’ll leave this up to the reader, just apply the following equation and calculate the Z-parameter values with the processes I’ve described above.", null, "" ]	[ null, "https://resources.altium.com/sites/default/files/styles/medium/public/user_pictures/headshot-small-350.jpg", null, "https://resources.altium.com/sites/default/files/styles/max_width_1300/public/blogs/S-Parameter%20Measurements%20and%20Errors%20in%20Power%20Integrity-72453.jpg", null, "data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==", null, "https://resources.altium.com/sites/default/files/styles/medium/public/user_pictures/headshot-small-350.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.875579,"math_prob":0.92160994,"size":7474,"snap":"2022-05-2022-21","text_gpt3_token_len":1628,"char_repetition_ratio":0.17443106,"word_repetition_ratio":0.0016353229,"special_character_ratio":0.21019535,"punctuation_ratio":0.08391608,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99126637,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,3,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-23T07:49:55Z\",\"WARC-Record-ID\":\"<urn:uuid:3b253488-a79a-45d0-836c-e5d23c3905e3>\",\"Content-Length\":\"72127\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9e2e293f-4c75-4a60-ab60-4c4293138ce1>\",\"WARC-Concurrent-To\":\"<urn:uuid:da05045d-2559-4e83-8664-1cb1fc67d8fa>\",\"WARC-IP-Address\":\"54.158.179.14\",\"WARC-Target-URI\":\"https://resources.altium.com/p/s-parameter-measurements-and-errors-power-integrity\",\"WARC-Payload-Digest\":\"sha1:KMHCRRBBPRTABP3E6CXBVEYCMCQQBKMB\",\"WARC-Block-Digest\":\"sha1:YGJZR2YNIIOVDZF736SBNZBDBDOKFPBW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662556725.76_warc_CC-MAIN-20220523071517-20220523101517-00231.warc.gz\"}"}
http://www.nutils.org/en/v3.1/library/numeric/	[ "# Numeric¶\n\nThe numeric module provides methods that are lacking from the numpy module.\n\nnutils.numeric.overlapping(arr, axis=-1, n=2)[source]\n\nreinterpret data with overlaps\n\nnutils.numeric.normdim(ndim, n)[source]\n\ncheck bounds and make positive\n\nnutils.numeric.get(arr, axis, item)[source]\n\ntake single item from array axis\n\nnutils.numeric.contract(A, B, axis=-1)[source]\nnutils.numeric.dot(A, B, axis=-1)[source]\n\nTransform axis of A by contraction with first axis of B and inserting remaining axes. Note: with default axis=-1 this leads to multiplication of vectors and matrices following linear algebra conventions.\n\nnutils.numeric.meshgrid(*args)[source]\n\nmulti-dimensional meshgrid generalisation\n\nnutils.numeric.normalize(A, axis=-1)[source]\n\ndevide by normal\n\nnutils.numeric.diagonalize(arg, axis=-1, newaxis=-1)[source]\n\ninsert newaxis, place axis on diagonal of axis and newaxis\n\nnutils.numeric.ix(args)[source]\n\nversion of numpy.ix_() that allows for scalars\n\nnutils.numeric.ext(A)[source]\n\nExterior For array of shape (n,n-1) return n-vector ex such that ex.array = 0 and det(arr;ex) = ex.ex\n\nclass nutils.numeric.const[source]" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.67438537,"math_prob":0.81410056,"size":681,"snap":"2021-43-2021-49","text_gpt3_token_len":183,"char_repetition_ratio":0.14771049,"word_repetition_ratio":0.0,"special_character_ratio":0.24229075,"punctuation_ratio":0.16296296,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9970185,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-07T14:20:12Z\",\"WARC-Record-ID\":\"<urn:uuid:8d7389ed-6509-4a44-89d2-0963b56cebfe>\",\"Content-Length\":\"16492\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5eda3710-39e2-44ef-854f-3f5bc4448412>\",\"WARC-Concurrent-To\":\"<urn:uuid:33df818e-7f8c-4e3d-8d1c-0b18f14c83a6>\",\"WARC-IP-Address\":\"104.17.33.82\",\"WARC-Target-URI\":\"http://www.nutils.org/en/v3.1/library/numeric/\",\"WARC-Payload-Digest\":\"sha1:TJBGUMYVUND5XT34QOARJ5UEQI5EBSIG\",\"WARC-Block-Digest\":\"sha1:GGBPZEDI3Q37AVULZKVF2ILP2AIPZPVA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363400.19_warc_CC-MAIN-20211207140255-20211207170255-00589.warc.gz\"}"}
https://www.paddlepaddle.org.cn/documentation/docs/zh/beginners_guide/basics/label_semantic_roles/index.html	[ "# 语义角色标注¶\n\n## 说明¶\n\n1. 本教程可支持在 CPU/GPU 环境下运行\n\n2. Docker镜像支持的CUDA/cuDNN版本\n\n如果使用了Docker运行Book，请注意：这里所提供的默认镜像的GPU环境为 CUDA 8/cuDNN 5，对于NVIDIA Tesla V100等要求CUDA 9的 GPU，使用该镜像可能会运行失败;\n\n3. 文档和脚本中代码的一致性问题\n\n请注意：为使本文更加易读易用，我们拆分、调整了train.py的代码并放入本文。本文中代码与train.py的运行结果一致，可直接运行train.py进行验证。\n\n## 背景介绍¶", null, "1. 构建一棵句法分析树，例如，图1是对上面例子进行依存句法分析得到的一棵句法树。\n2. 从句法树上识别出给定谓词的候选论元。\n3. 候选论元剪除；一个句子中的候选论元可能很多，候选论元剪除就是从大量的候选项中剪除那些最不可能成为论元的候选项。\n4. 论元识别：这个过程是从上一步剪除之后的候选中判断哪些是真正的论元，通常当做一个二分类问题来解决。\n5. 对第4步的结果，通过多分类得到论元的语义角色标签。可以看到，句法分析是基础，并且后续步骤常常会构造的一些人工特征，这些特征往往也来自句法分析。", null, "", null, "## 模型概览¶\n\n### 栈式循环神经网络（Stacked Recurrent Neural Network）¶", null, "### 双向循环神经网络（Bidirectional Recurrent Neural Network）¶", null, "### 条件随机场 (Conditional Random Field)¶\n\nCRF是一种概率化结构模型，可以看作是一个概率无向图模型，结点表示随机变量，边表示随机变量之间的概率依赖关系。简单来讲，CRF学习条件概率$P(X\|Y)$，其中 $X = (x_1, x_2, ... , x_n)$ 是输入序列，$Y = (y_1, y_2, ... , y_n)$ 是标记序列；解码过程是给定 $X$序列求解令$P(Y\|X)$最大的$Y$序列，即$Y^* = \\mbox{arg max}_{Y} P(Y \| X)$。", null, "", null, "", null, "$\\omega$是特征函数对应的权值，是CRF模型要学习的参数。训练时，对于给定的输入序列和对应的标记序列集合$D = \\left[(X_1, Y_1), (X_2 , Y_2) , ... , (X_N, Y_N)\\right]$ ，通过正则化的极大似然估计，求解如下优化目标：", null, "### 深度双向LSTM（DB-LSTM）SRL模型¶\n\n1. 构造输入；\n• 输入1是谓词，输入2是句子\n• 将输入1扩展成和输入2一样长的序列，用one-hot方式表示；\n1. one-hot方式的谓词序列和句子序列通过词表，转换为实向量表示的词向量序列；\n2. 将步骤2中的2个词向量序列作为双向LSTM的输入，学习输入序列的特征表示；\n3. CRF以步骤3中模型学习到的特征为输入，以标记序列为监督信号，实现序列标注；\n\n• 谓词上下文：上面的方法中，只用到了谓词的词向量表达谓词相关的所有信息，这种方法始终是非常弱的，特别是如果谓词在句子中出现多次，有可能引起一定的歧义。从经验出发，谓词前后若干个词的一个小片段，能够提供更丰富的信息，帮助消解歧义。于是，我们把这样的经验也添加到模型中，为每个谓词同时抽取一个“谓词上下文” 片段，也就是从这个谓词前后各取$n$个词构成的一个窗口片段；\n• 谓词上下文区域标记：为句子中的每一个词引入一个0-1二值变量，表示它们是否在“谓词上下文”片段中；\n\n1. 构造输入\n• 输入1是句子序列，输入2是谓词序列，输入3是谓词上下文，从句子中抽取这个谓词前后各$n$个词，构成谓词上下文，用one-hot方式表示，输入4是谓词上下文区域标记，标记了句子中每一个词是否在谓词上下文中；\n• 将输入2~3均扩展为和输入1一样长的序列；\n1. 输入1~4均通过词表取词向量转换为实向量表示的词向量序列；其中输入1、3共享同一个词表，输入2和4各自独有词表；\n2. 第2步的4个词向量序列作为双向LSTM模型的输入；LSTM模型学习输入序列的特征表示，得到新的特性表示序列；\n3. CRF以第3步中LSTM学习到的特征为输入，以标记序列为监督信号，完成序列标注；", null, "## 数据介绍¶\n\nconll05st-release/\n└── test.wsj\n├── props # 标注结果\n└── words # 输入文本序列\n\n\n1. 将文本序列和标记序列其合并到一条记录中；\n2. 一个句子如果含有$n$个谓词，这个句子会被处理$n$次，变成$n$条独立的训练样本，每个样本一个不同的谓词；\n3. 抽取谓词上下文和构造谓词上下文区域标记；\n4. 构造以BIO法表示的标记；\n5. 依据词典获取词对应的整数索引。\n\nA set n't been set . × 0 B-A1\nrecord set n't been set . × 0 I-A1\ndate set n't been set . × 0 I-A1\nhas set n't been set . × 0 O\nn't set n't been set . × 1 B-AM-NEG\nbeen set n't been set . × 1 O\nset set n't been set . × 1 B-V\n. set n't been set . × 1 O\n\nword_dict 输入句子的词典，共计44068个词\nlabel_dict 标记的词典，共计106个标记\npredicate_dict 谓词的词典，共计3162个词\nemb 一个训练好的词表，32维\n\nfrom __future__ import print_function\n\nimport math, os\nimport numpy as np\nimport six\nimport time\n\nwith_gpu = os.getenv('WITH_GPU', '0') != '0'\n\nword_dict, verb_dict, label_dict = conll05.get_dict()\nword_dict_len = len(word_dict)\nlabel_dict_len = len(label_dict)\npred_dict_len = len(verb_dict)\n\nprint('word_dict_len: ', word_dict_len)\nprint('label_dict_len: ', label_dict_len)\nprint('pred_dict_len: ', pred_dict_len)\n\n\n## 模型配置说明¶\n\n• 定义输入数据维度及模型超参数。\nmark_dict_len = 2 # 谓上下文区域标志的维度，是一个0-1 2值特征，因此维度为2\nword_dim = 32 # 词向量维度\nmark_dim = 5 # 谓词上下文区域通过词表被映射为一个实向量，这个是相邻的维度\nhidden_dim = 512 # LSTM隐层向量的维度： 512 / 4\ndepth = 8 # 栈式LSTM的深度\nmix_hidden_lr = 1e-3 # linear_chain_crf层的基础学习率\n\nIS_SPARSE = True # 是否以稀疏方式更新embedding\nPASS_NUM = 10 # 训练轮数\nBATCH_SIZE = 10 # batch size 大小\n\nembedding_name = 'emb'\n\n\n• 如上文提到，我们用基于英文维基百科训练好的词向量来初始化序列输入、谓词上下文总共6个特征的embedding层参数，在训练中不更新。\n# 这里加载PaddlePaddle保存的二进制参数\nwith open(file_name, 'rb') as f:\nreturn np.fromfile(f, dtype=np.float32).reshape(h, w)\n\n\n## 训练模型¶\n\n• 我们根据网络拓扑结构和模型参数来进行训练，在构造时还需指定优化方法，这里使用最基本的SGD方法(momentum设置为0)，同时设定了学习率、正则等。\n\nuse_cuda = False #在cpu上执行训练\nsave_dirname = \"label_semantic_roles.inference.model\" #训练得到的模型参数保存在文件中\nis_local = True\n\n\n### 数据输入层定义¶\n\n# 句子序列\nword = fluid.data(\nname='word_data', shape=[None, 1], dtype='int64', lod_level=1)\n\n# 谓词\npredicate = fluid.data(\nname='verb_data', shape=[None, 1], dtype='int64', lod_level=1)\n\n# 谓词上下文5个特征\nctx_n2 = fluid.data(\nname='ctx_n2_data', shape=[None, 1], dtype='int64', lod_level=1)\nctx_n1 = fluid.data(\nname='ctx_n1_data', shape=[None, 1], dtype='int64', lod_level=1)\nctx_0 = fluid.data(\nname='ctx_0_data', shape=[None, 1], dtype='int64', lod_level=1)\nctx_p1 = fluid.data(\nname='ctx_p1_data', shape=[None, 1], dtype='int64', lod_level=1)\nctx_p2 = fluid.data(\nname='ctx_p2_data', shape=[None, 1], dtype='int64', lod_level=1)\n\n# 谓词上下区域标志\nmark = fluid.data(\nname='mark_data', shape=[None, 1], dtype='int64', lod_level=1)\n\n\n### 定义网络结构¶\n\n#预训练谓词和谓词上下区域标志\npredicate_embedding = fluid.embedding(\ninput=predicate,\nsize=[pred_dict_len, word_dim],\ndtype='float32',\nis_sparse=IS_SPARSE,\nparam_attr='vemb')\n\nmark_embedding = fluid.embedding(\ninput=mark,\nsize=[mark_dict_len, mark_dim],\ndtype='float32',\nis_sparse=IS_SPARSE)\n\n#句子序列和谓词上下文5个特征并预训练\nword_input = [word, ctx_n2, ctx_n1, ctx_0, ctx_p1, ctx_p2]\n# 因词向量是预训练好的，这里不再训练embedding表，\n# 参数属性trainable设置成False阻止了embedding表在训练过程中被更新\nemb_layers = [\nfluid.embedding(\nsize=[word_dict_len, word_dim],\ninput=x,\nparam_attr=fluid.ParamAttr(\nname=embedding_name, trainable=False)) for x in word_input\n]\n#加入谓词和谓词上下区域标志的预训练结果\nemb_layers.append(predicate_embedding)\nemb_layers.append(mark_embedding)\n\n\n# 共有8个LSTM单元被训练，每个单元的方向为从左到右或从右到左，\n# 由参数is_reverse确定\n# 第一层栈结构\nhidden_0_layers = [\nfluid.layers.fc(input=emb, size=hidden_dim, act='tanh')\nfor emb in emb_layers\n]\n\nhidden_0 = fluid.layers.sums(input=hidden_0_layers)\n\nlstm_0 = fluid.layers.dynamic_lstm(\ninput=hidden_0,\nsize=hidden_dim,\ncandidate_activation='relu',\ngate_activation='sigmoid',\ncell_activation='sigmoid')\n\n# 用直连的边来堆叠L-LSTM、R-LSTM\ninput_tmp = [hidden_0, lstm_0]\n\n# 其余的栈结构\nfor i in range(1, depth):\nmix_hidden = fluid.layers.sums(input=[\nfluid.layers.fc(input=input_tmp, size=hidden_dim, act='tanh'),\nfluid.layers.fc(input=input_tmp, size=hidden_dim, act='tanh')\n])\n\nlstm = fluid.layers.dynamic_lstm(\ninput=mix_hidden,\nsize=hidden_dim,\ncandidate_activation='relu',\ngate_activation='sigmoid',\ncell_activation='sigmoid',\nis_reverse=((i % 2) == 1))\n\ninput_tmp = [mix_hidden, lstm]\n\n# 取最后一个栈式LSTM的输出和这个LSTM单元的输入到隐层映射，\n# 经过一个全连接层映射到标记字典的维度，来学习 CRF 的状态特征\nfeature_out = fluid.layers.sums(input=[\nfluid.layers.fc(input=input_tmp, size=label_dict_len, act='tanh'),\nfluid.layers.fc(input=input_tmp, size=label_dict_len, act='tanh')\n])\n\n# 标注序列\ntarget = fluid.data(\nname='target', shape=[None, 1], dtype='int64', lod_level=1)\n\n# 学习 CRF 的转移特征\ncrf_cost = fluid.layers.linear_chain_crf(\ninput=feature_out,\nlabel=target,\nparam_attr=fluid.ParamAttr(\nname='crfw', learning_rate=mix_hidden_lr))\n\navg_cost = fluid.layers.mean(crf_cost)\n\n# 使用最基本的SGD优化方法(momentum设置为0)\nsgd_optimizer = fluid.optimizer.SGD(\nlearning_rate=fluid.layers.exponential_decay(\nlearning_rate=0.01,\ndecay_steps=100000,\ndecay_rate=0.5,\nstaircase=True))\n\nsgd_optimizer.minimize(avg_cost)\n\n\ncrf_decode = fluid.layers.crf_decoding(\ninput=feature_out, param_attr=fluid.ParamAttr(name='crfw'))\n\nbatch_size=BATCH_SIZE)\n\nplace = fluid.CUDAPlace(0) if use_cuda else fluid.CPUPlace()\n\n\nfeeder = fluid.DataFeeder(\nfeed_list=[\nword, ctx_n2, ctx_n1, ctx_0, ctx_p1, ctx_p2, predicate, mark, target\n],\nplace=place)\nexe = fluid.Executor(place)\n\n\nmain_program = fluid.default_main_program()\n\nexe.run(fluid.default_startup_program())\nembedding_param = fluid.global_scope().find_var(\nembedding_name).get_tensor()\nembedding_param.set(\nplace)\n\nstart_time = time.time()\nbatch_id = 0\nfor pass_id in six.moves.xrange(PASS_NUM):\nfor data in train_data():\ncost = exe.run(main_program,\nfeed=feeder.feed(data),\nfetch_list=[avg_cost])\ncost = cost\n\nif batch_id % 10 == 0:\nprint(\"avg_cost: \" + str(cost))\nif batch_id != 0:\nprint(\"second per batch: \" + str((time.time(\n) - start_time) / batch_id))\n# Set the threshold low to speed up the CI test\nif float(cost) < 60.0:\nif save_dirname is not None:\nfluid.io.save_inference_model(save_dirname, [\n'word_data', 'verb_data', 'ctx_n2_data',\n'ctx_n1_data', 'ctx_0_data', 'ctx_p1_data',\n'ctx_p2_data', 'mark_data'\n], [feature_out], exe)\nbreak\n\nbatch_id = batch_id + 1\n\n\n## 应用模型¶\n\nuse_cuda = False #在cpu上进行预测\nsave_dirname = \"label_semantic_roles.inference.model\" #调用训练好的模型进行预测\n\nplace = fluid.CUDAPlace(0) if use_cuda else fluid.CPUPlace()\nexe = fluid.Executor(place)\n\n\nlod = [[3, 4, 2]]\nbase_shape = \n\n# 构造假数据作为输入，整数随机数的范围是[low, high]\nword = fluid.create_random_int_lodtensor(\nlod, base_shape, place, low=0, high=word_dict_len - 1)\npred = fluid.create_random_int_lodtensor(\nlod, base_shape, place, low=0, high=pred_dict_len - 1)\nctx_n2 = fluid.create_random_int_lodtensor(\nlod, base_shape, place, low=0, high=word_dict_len - 1)\nctx_n1 = fluid.create_random_int_lodtensor(\nlod, base_shape, place, low=0, high=word_dict_len - 1)\nctx_0 = fluid.create_random_int_lodtensor(\nlod, base_shape, place, low=0, high=word_dict_len - 1)\nctx_p1 = fluid.create_random_int_lodtensor(\nlod, base_shape, place, low=0, high=word_dict_len - 1)\nctx_p2 = fluid.create_random_int_lodtensor(\nlod, base_shape, place, low=0, high=word_dict_len - 1)\nmark = fluid.create_random_int_lodtensor(\nlod, base_shape, place, low=0, high=mark_dict_len - 1)\n\n\n[inference_program, feed_target_names,\n\n\nassert feed_target_names == 'word_data'\nassert feed_target_names == 'verb_data'\nassert feed_target_names == 'ctx_n2_data'\nassert feed_target_names == 'ctx_n1_data'\nassert feed_target_names == 'ctx_0_data'\nassert feed_target_names == 'ctx_p1_data'\nassert feed_target_names == 'ctx_p2_data'\nassert feed_target_names == 'mark_data'\n\n\nresults = exe.run(inference_program,\nfeed={\nfeed_target_names: word,\nfeed_target_names: pred,\nfeed_target_names: ctx_n2,\nfeed_target_names: ctx_n1,\nfeed_target_names: ctx_0,\nfeed_target_names: ctx_p1,\nfeed_target_names: ctx_p2,\nfeed_target_names: mark\n},\nfetch_list=fetch_targets,\nreturn_numpy=False)\n\n\nprint(results.lod())\nnp_data = np.array(results)\nprint(\"Inference Shape: \", np_data.shape)\n\n\n## 参考文献¶\n\n1. Sun W, Sui Z, Wang M, et al. Chinese semantic role labeling with shallow parsing[C]//Proceedings of the 2009 Conference on Empirical Methods in Natural Language Processing: Volume 3-Volume 3. Association for Computational Linguistics, 2009: 1475-1483.\n2. Pascanu R, Gulcehre C, Cho K, et al. How to construct deep recurrent neural networks[J]. arXiv preprint arXiv:1312.6026, 2013.\n3. Cho K, Van Merriënboer B, Gulcehre C, et al. Learning phrase representations using RNN encoder-decoder for statistical machine translation[J]. arXiv preprint arXiv:1406.1078, 2014.\n4. Bahdanau D, Cho K, Bengio Y. Neural machine translation by jointly learning to align and translate[J]. arXiv preprint arXiv:1409.0473, 2014.\n5. Lafferty J, McCallum A, Pereira F. Conditional random fields: Probabilistic models for segmenting and labeling sequence data[C]//Proceedings of the eighteenth international conference on machine learning, ICML. 2001, 1: 282-289.\n6. 李航. 统计学习方法[J]. 清华大学出版社, 北京, 2012.\n7. Marcus M P, Marcinkiewicz M A, Santorini B. Building a large annotated corpus of English: The Penn Treebank[J]. Computational linguistics, 1993, 19(2): 313-330.\n8. Palmer M, Gildea D, Kingsbury P. The proposition bank: An annotated corpus of semantic roles[J]. Computational linguistics, 2005, 31(1): 71-106.\n9. Carreras X, Màrquez L. Introduction to the CoNLL-2005 shared task: Semantic role labeling[C]//Proceedings of the Ninth Conference on Computational Natural Language Learning. Association for Computational Linguistics, 2005: 152-164.\n10. Zhou J, Xu W. End-to-end learning of semantic role labeling using recurrent neural networks[C]//Proceedings of the Annual Meeting of the Association for Computational Linguistics. 2015.", null, "" ]	[ null, "https://githubraw.cdn.bcebos.com/PaddlePaddle/book/develop/07.label_semantic_roles/image/Eqn1.png", null, "https://githubraw.cdn.bcebos.com/PaddlePaddle/book/develop/07.label_semantic_roles/image/dependency_parsing.png", null, "https://githubraw.cdn.bcebos.com/PaddlePaddle/book/develop/07.label_semantic_roles/image/bio_example.png", null, "https://githubraw.cdn.bcebos.com/PaddlePaddle/book/develop/07.label_semantic_roles/image/stacked_lstm.png", null, "https://githubraw.cdn.bcebos.com/PaddlePaddle/book/develop/07.label_semantic_roles/image/bidirectional_stacked_lstm.png", null, "https://githubraw.cdn.bcebos.com/PaddlePaddle/book/develop/07.label_semantic_roles/image/linear_chain_crf.png", null, "https://githubraw.cdn.bcebos.com/PaddlePaddle/book/develop/07.label_semantic_roles/image/Eqn2.gif", null, "https://githubraw.cdn.bcebos.com/PaddlePaddle/book/develop/07.label_semantic_roles/image/Eqn3.gif", null, "https://githubraw.cdn.bcebos.com/PaddlePaddle/book/develop/07.label_semantic_roles/image/Eqn4.png", null, "https://githubraw.cdn.bcebos.com/PaddlePaddle/book/develop/07.label_semantic_roles/image/db_lstm_network.png", null, "https://paddlepaddleimage.cdn.bcebos.com/bookimage/camo.png", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.6934632,"math_prob":0.95232785,"size":17784,"snap":"2020-10-2020-16","text_gpt3_token_len":10603,"char_repetition_ratio":0.1003937,"word_repetition_ratio":0.05496183,"special_character_ratio":0.25179937,"punctuation_ratio":0.17972536,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98295295,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22],"im_url_duplicate_count":[null,5,null,10,null,10,null,10,null,10,null,10,null,5,null,5,null,5,null,10,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-17T08:01:46Z\",\"WARC-Record-ID\":\"<urn:uuid:7d93af28-6c45-4a51-9261-89a516404534>\",\"Content-Length\":\"177515\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7cf4ecb9-1c9a-4910-bfdd-b2e9619f12a1>\",\"WARC-Concurrent-To\":\"<urn:uuid:fb2e97bd-c763-462e-89cd-f23ed72ee819>\",\"WARC-IP-Address\":\"180.76.58.84\",\"WARC-Target-URI\":\"https://www.paddlepaddle.org.cn/documentation/docs/zh/beginners_guide/basics/label_semantic_roles/index.html\",\"WARC-Payload-Digest\":\"sha1:YCTDJLFGYLZDUS26RAF6JPLSW4BFPMTN\",\"WARC-Block-Digest\":\"sha1:UNKCMNOXOTIVKQWAV3E5Q3Y2GH57B6ZR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875141749.3_warc_CC-MAIN-20200217055517-20200217085517-00503.warc.gz\"}"}
http://documentation.statsoft.com/STATISTICAHelp.aspx?path=sepath/Sepath/Dialogs/AnalysisParameters	[ "", null, "Analysis Parameters\n\nClick the Set parameters button on the Advanced tab of the Structural Equation Modeling Startup Panel to display the Analysis Parameters dialog box. Note that you can also access this dialog box by clicking the Parameters button on the Path Construction Tool or the Set Parameters button on the Model tab of the Monte Carlo Analysis dialog box.\n\nData to analyze. Use the options under Data to analyze to select which type of data to analyze. For standard path analyses or structural equation models, you choose either Covariances or Correlations, for models specifying intercepts, or structured means, you must specify the Moments option.\n\nCovariances. Select the Covariances option button to analyze the covariance matrix of the input variables, regardless of what kind of data are input. So, for example, if you input a correlation matrix file with standard deviations, STATISTICA will calculate the covariance matrix and analyze it. If your input file is a covariance matrix file, STATISTICA will analyze it directly. If your file contains raw data, STATISTICA will analyze it and calculate the covariance matrix for you. See also, Covariance Matrices vs. Correlation Matrices for information on how STATISTICA analyzes these two types of matrices.\n\nCorrelations. Select the Correlations option button to calculate the correlation matrix from the input data and analyze it. STATISTICA analyzes the correlation matrix correctly, using constrained estimation theory developed by Michael Browne (see Browne, 1982; Mels, 1989; Browne & Mels, 1992), and implemented first in the computer program RAMONA. As a result, STATISTICA gives the correct standard errors, estimates, and test statistics when a correlation matrix is analyzed directly. When combined with the Standardization - New option STATISTICA can estimate a completely standardized path model, where all variables are standardized to have unit variance, and standard errors can be estimated as well.\n\nMoments. Select the Moments option button to analyze the augmented product-moment matrix instead of the covariance matrix. This option is selected if you are analyzing models involving intercepts or structured means. STATISTICA will also add, to the input variable list, an additional variable called CONSTANT that always takes on the value 1. CONSTANT is a dummy variable used to represent means in structured means models, i.e., those with an intercept variable.\n\nOutput options. Use the options in the Output options frame for preliminary control over the appearance of your output. You can preselect the number of decimal places to be printed in report editor, and you can select whether or not to print estimated standard errors for model coefficients.\n\nNo. of decimal places. Enter a number between 1 and 6 in the No. of decimal places field. The number is number of decimal places displayed by default in the results spreadsheets.\n\nStandard errors. Select the Standard errors check box to display estimated standard errors for all parameters in the PATH1 text output, and Model Summary Spreadsheet. Note that if OLS estimation is performed, or the \"old\" standardization method is employed, standard errors will not be available.\n\nConvergence criteria. Use the options in the Convergence criteria frame to adjust constants that can directly affect when the program decides convergence has occurred during iteration. The default values produce desirable results for a wide variety of problems, and you will seldom need to adjust these criteria.\n\nMax. residual cosine. Enter a value for the maximum residual cosine in the Max. residual cosine field. This criterion becomes small when parameter values have stabilized. You can alter this criterion in the input field. The default value of .0001 works well across a wide variety of situations. For a technical definition of this criterion and a discussion of its merits, see the discussion of Unconstrained Minimization Techniques.\n\nRelative funct. change. Enter a value for the relative function change in the Relative funct. change field. This criterion becomes small when the discrepancy function being minimized (see the section on Statistical Estimation Theory for a discussion) is no longer changing. On occasion, especially when boundary cases are encountered, it will not be possible for the program to iterate to a point where a local or global minimum is obtained. In that case, the Maximum Residual Cosine criterion may not fall below, but the discrepancy function will not change. At that point, this criterion will stop iteration. In general, it should be kept very low, or it may cause iteration to terminate prematurely. For a technical definition of the criterion see the discussion of Unconstrained Minimization Techniques.\n\nDiscrepancy function. Use the options in the Discrepancy function frame to specify which discrepancy function or functions will be minimized to yield parameter estimates. Consult the section on Statistical Estimation for an in-depth discussion of these methods.\n\nMaximum Likelihood (ML). Select the Maximum Likelihood (ML) button to perform Maximum Wishart Likelihood estimation if Correlations or Covariances are analyzed, and Maximum Normal Likelihood estimation if Moments are analyzed.\n\nGeneralized Least Squares (GLS). Select the Generalized Least Squares (GLS) button to perform Generalized Least Squares estimation.\n\nGLS --> ML. Select the GLS--> ML (GLS followed by ML) option button to perform five iterations using the Generalized Least Squares estimation procedure followed by Maximum Likelihood Estimation. If selected, this option disregards the current settings in the Maximum No. of Iterations field.\n\nOrdinary Least Squares (OLS). Select the Ordinary Least Squares (OLS) option button to perform Ordinary Least Squares estimation.\n\nADFG. Select the ADFG option button to perform Asymptotically Distribution Free (Gramian) estimation, which does not require the assumption of multivariate normality. In this variant, the weight matrix is guaranteed to be Gramian. As a preliminary to ADFG estimation, STATISTICA performs GLS estimation.\n\nADFU. Select the Asymptotically Distribution Free Unbiased (ADFU) option button to perform Asymptotically Distribution Free (Unbiased) estimation, which does not require the assumption of multivariate normality. In this variant, the weight matrix is an unbiased estimate of the true weight matrix. However, this estimated matrix may not be Gramian (i.e., invertible) in all cases. If it is not, STATISTICA gives an error message. At that point, you should restart the estimation using the ADFG option. As a preliminary to ADFU estimation, STATISTICA performs GLS estimation.\n\nGlobal iteration parameters. Use the options under Global iteration parameters to enter parameters that control the basic iterative process. For a technical description of the iteration process, see Unconstrained Minimization Techniques.\n\nMaximum no. of iterations. Enter the maximum number of iterations allowed in the Maximum no. of iterations field. The default number is 30. When this number of iterations is reached, the iterative process will automatically terminate, and a message will be issued to indicate that the maximum number of iterations was exceeded. Minimum is zero (in which case the discrepancy function and estimated covariance matrix will be computed, and control will be returned to the user), maximum is 1000.\n\nMaximum step length. Enter the maximum length of the step vector that will be allowed in the Maximum step length field. See Unconstrained Minimization Techniques for a discussion of this parameter. See the Solving Iteration Problems for suggestions on how to use this parameter to solve some problems encountered during iteration.\n\nSteepest descent iterations. Enter a number of steepest descent iterations to proceed the standard iterations in the Steepest descent iterations field. See Solving Iteration Problems for suggestions on how to use this parameter to solve some problems encountered during iteration.\n\nStep tolerance. Enter a tolerance value in the Step tolerance field. STATISTICA uses this value for when a parameter is temporarily eliminated from the iterative process. The tolerance value is basically one minus the squared multiple correlation of a parameter with the other parameters. If a parameter becomes highly redundant with other parameters during iteration, the approximate Hessian employed during Gauss-Newton iteration becomes unstable. This parameter controls when a parameter is temporarily removed from the iterative process. Very low values of this parameter mean that a parameter will never be removed. This can cause iteration to \"blow up\" if the approximate Hessian becomes nearly singular. High Values mean that parameters will not be varied while they are moderately correlated with other parameters. This can cause iteration to fail if parameters are fairly highly correlated at the solution point. Change this parameter seldom, and in small increments. The default value seems to work well on a wide range of problems.\n\nInitial values. Use the options in the Initial values frame to select the method employed to find initial values for free parameters. The default method uses .5 for all free parameters, except variances and covariances (or correlations) of manifest exogenous variables. These parameters are initialized at the values obtained from the sample data.\n\nDefault. Select the Default option button to use .5 for all free parameters, except variances and covariances (or correlations) of manifest exogenous variables. These parameters are initialized at the values obtained from the sample data.\n\nAutomatic. Select the Automatic option button to obtain the initial values via a method that is a minor adaptation of the technique described by McDonald and Hartmann (1992).\n\nStandardization. Use the options under Standardization to choose to generate either a standardized solution (where latent variables all have unit variance) by one of two methods, or an unstandardized solution.\n\nNote that the New and Old options will cause the program to compute standardized solutions based on the (standardized) endogenous latent variables (this is the common practice in most programs of this kind); if your model does not contain latent endogenous variables (e.g., only contains manifest exogenous variables), these methods of standardization may not be appropriate, and you may want to use a correlation matrix as input instead, and choose the New method to obtain estimates that are standardized in the manifest exogenous variables. See also, New Method for Standardizing Endogenous Latent Variables in the Technical Aspects of SEPATH section for additional details.\n\nNew. Check the New option button to estimate a standardized solution via constrained estimation. This approach produces a solution where all latent variables, both independent and dependent, have variances of 1. Unlike the old method, however, standard errors are available with this option. Combining this option with the Data to Analyze - Correlations option (see above) allows one to estimate a completely standardized path model, where all variables, manifest and latent, have unit variances, and standard errors can be estimated for all parameters.\n\nOld. Check the Old option button to estimate a standardized solution using older methods. Older programs generate a standardized solution after iteration is complete, then perform a calculation after the fact to transform the solution to a standardized form. This method gets a solution faster than the New option described above, because it does not need to use constrained estimation. However, standard errors cannot be computed.\n\nNone. Check the None option button to calculate an unstandardized solution.\n\nManifest exogenous. Use the options under Manifest exogenous to account for exogenous manifest variables. Most models do not have manifest variables that are exogenous. If a model does have such variables, their variances and covariances must be accounted for. In most cases (but not all), the variances and covariances estimated under one of the standard estimation techniques will be identical to the observed variances and covariances, and consequently these parameters are not of much interpretive interest. STATISTICA provides two approaches to calculating these parameters automatically and keeping them out of view (the Fixed and Free options below) and also allows them to be declared explicitly.\n\nFree. Check the Free option button to treat the variances and covariances among the manifest exogenous variables as free parameters and add them to the model, although their values are not shown. Start values are the observed variances and covariances. If you select this option and attempt (in the PATH1 input) to specify the variance or covariance of a manifest exogenous variable, you will generate an error message.\n\nFixed. Check the Fixed option button to treat the variances and covariances for manifest exogenous variables as fixed (at the value of the observed variances and covariances) during iteration. After iteration, they are treated as if they were free parameters. This approach eliminates, in the case of several manifest exogenous variables, the addition of a number of extra free parameters, which can slow down iteration and make it less reliable. The user should take the output obtained by this approach, and resubmit it using the Free option described below to guarantee correct estimates. If you select this option and attempt (in the PATH1 input) to specify the variance or covariance of a manifest exogenous variable, you will generate an error message.\n\nUser. Check the User option button if you want to specify the variances and covariances of all manifest exogenous variables, using standard PATH1 syntax.\n\nLine search method. Use the options under Line search method to choose a basic line search method. Once the step direction has been chosen, the minimization problem is basically reduced from a problem in n unknowns to a problem in 1 unknown, i.e., the length of the step. There are three methods for choosing the length of the step. Their technical aspects are discussed in the section on Unconstrained Minimization Techniques.\n\nCubic interpolation. Check the Cubic interpolation option button to use cubic interpolation, a method that is reasonably fast and rather robust. It works well in the vast majority of circumstances.\n\nGolden section. Check the Golden section option button to use golden section, a method that tries to solve the one dimensional minimization problem exactly on each iteration. It often converges in slightly fewer iterations than cubic interpolation, but takes longer, because it requires more function evaluations on each iteration.\n\nSimple stephalving. Check the Simple stephalving option button to use simple stephalving. Although this is the fastest method, it will fail to converge for a fair number of problems that the other two, more sophisticated methods succeed on.\n\nLine search parameters. Use the options under Line search parameters to choose numerical parameters that control the performance of the line search method you have chosen. Only a subset of the parameters are relevant to each line search method. Only relevant parameters will be enabled for the currently selected line search method.\n\nMax. no. of stephalves. Enter a value in the Max. no. of stephalves field to set the maximum number of stephalves allowed on a single iteration if the Simple Stephalving line search method is used.\n\nStephalve fraction. Enter a value in the Stephalve fraction field to set the fraction the current step is multiplied by when Simple Stephalving is used as the line search method.\n\nCubic LS alpha. Enter a value in the Cubic LS alpha field to control how large a reduction in the discrepancy function has to be made before a step is considered acceptable when the Cubic Interpolation line search method is used. The default value, .0001, allows virtually any improvement to be considered acceptable.\n\nGolden search tau. Enter a value in the Golden search aau field to control how wide a range to which the Golden Section line search is limited.\n\nGolden srch precision. Enter a value in the Golden srch precision field to control the precision of estimation in a Golden Section line search." ]	[ null, "http://documentation.statsoft.com/STATISTICAHelp/sepath/Images/M_SEM.GIF", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7579328,"math_prob":0.9220952,"size":16337,"snap":"2019-26-2019-30","text_gpt3_token_len":3400,"char_repetition_ratio":0.15692157,"word_repetition_ratio":0.09061876,"special_character_ratio":0.17849055,"punctuation_ratio":0.10573158,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9602031,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-24T09:05:33Z\",\"WARC-Record-ID\":\"<urn:uuid:2365a3b6-2809-4ebc-b002-ccba34ada4e4>\",\"Content-Length\":\"77653\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e6c4fc73-d02a-45f6-bd5f-03cf3cbe7e85>\",\"WARC-Concurrent-To\":\"<urn:uuid:23f8d0bb-474a-4794-9884-97eb048fcc6f>\",\"WARC-IP-Address\":\"99.157.40.5\",\"WARC-Target-URI\":\"http://documentation.statsoft.com/STATISTICAHelp.aspx?path=sepath/Sepath/Dialogs/AnalysisParameters\",\"WARC-Payload-Digest\":\"sha1:BLUYUAPKXGVCMQNCOZDCU6ZPYQQDTIKX\",\"WARC-Block-Digest\":\"sha1:XACU2XUY27VFONSAJHQW6A7NVV5DVP3X\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999298.86_warc_CC-MAIN-20190624084256-20190624110256-00016.warc.gz\"}"}
https://www.questaal.org/tutorial/lmf/bandedge/	[ "# Extremal points and effective mass\n\nThis tutorial demonstrates how to find extremal points (maxima, minima, and saddle points) in the Brillouin zone, and calculate effective masses using the band-edge utility. LDA silicon was chosen for simplicity, though it is a trivial example as its extremal points are found on high-symmetry lines. band-edge is particularly useful when searching for multiple extremal points, and/or points distinct from those of high symmetry.\n\nAt the end of this tutorial the band-edge manual documents its usage in more detail.\n\n### Preliminaries\n\nThis tutorial uses a number of Questaal executables and scripts, e.g. band-edge, blm, lmfa, lmf, pfit, fmain, plbnds, and fplot. They are assumed to be in your path.\n\n### Tutorial\n\n#### Self-consistent density and energy bands\n\nThe starting point is a self-consistent LDA density, you may want to review the DFT tutorial for silicon. Cut and paste the following lines into file init.si:\n\nLATTICE\nALAT=10.26\nPLAT= 0.00000000 0.50000000 0.50000000\n0.50000000 0.00000000 0.50000000\n0.50000000 0.50000000 0.00000000\n# pos means cartesian coordinates, units of alat\nSITE\nATOM=Si POS= 0.00000000 0.00000000 0.00000000\nATOM=Si POS= 0.25000000 0.25000000 0.25000000\n\n\nRun the following command to obtain a self-consistent density:\n\n$blm --express si --nk=4 --gmax=5$ cp actrl.si ctrl.si\n$lmfa si$ cp basp0.si basp.si\n$lmf si It will be helpful to have a band structure to refer to when finding the extremal points, you may want to review the silicon band plotting tutorial. Create a symmetry file syml.si joining L, Γ, X, W, and Γ again. lmchk si --syml~n=41~lbl=LGXWG~q=.5,.5,.5,0,0,0,1,0,0,1,.5,0,0,0,0 Generate the energy bands (file bnds.si) $ lmf si --rs=1,0 -vnit=1 --band~fn=syml\n\n\nThe following two commands will plot and display the bands:\n\n$echo -6,6,10,15 \| plbnds -fplot -ef=0 -scl=13.6 -lbl=L,G,X,W,G bnds.si$ fplot -f plot.plbnds && gs fplot.ps\n\n\nTake a look at the band structure plot. The valence band maximum falls at the $\\Gamma$ point while the conduction band minimum lies between $\\Gamma$ and X, at about 0.85 of the distance to X.\n\n#### Find the conduction band effective mass\n\nWe will now use the band-edge script to accurately locate the position of the conduction band minimum and to calculate the effective mass. This is done in three steps. First do a rough search by ‘floating’ to a point near the minimum. Next, do a more refined search by carrying out a minimization until the gradient is negligibly small. Lastly, you calculate the effective mass around this point.\n\n##### 1. Float to low-energy point\n\nband-edge script has a ‘float’ option that is useful for doing a quick search to find a low-energy (or high-energy) region of k-space. You specify a starting point, then the script creates a cluster of points around it and checks what is the lowest-energy point. It then uses the lowest-energy point as the next central point, creates a new set of points around it and again moves to the lowest-energy one. This process is repeated until the central point is the lowest-energy point.\n\nLet’s pick a random point (0.1,0.2,0.3) and let band-edge float downhill. Run the following command:\n\n$band-edge -floatmn -maxit=20 -r=0.1 -band=5 -q0=0.1,0.2,0.3 si ‘−floatmn’ tells band-edge to seek a minimum-energy point (see additional exercises for a maximum-energy point example). ‘−maxit’ switch limits the number iterations (number of times a set of points is created) in case convergence is not reach before then, ‘−r=’ sets a range that defines how far from the centre the points are generated, ‘−band’ is for the band considered (here conduction band is 5 since 4 electrons and spin degenerate) and lastly −q0 is the starting k-point. To see what switches band-edge has, invoke it without any arguments, or see documentation below. You should get an output similar to the following: check that \"lmf si\" reads input file without error ... ok start iteration 1 of 20 lmf si --band~lst=5~box=0.1,n=12+o,q0=0.1,0.2,0.3 > out.lmf qnow, E : 0.100000 0.200000 0.300000 0.3728323 gradient : -0.154879 0.182604 -0.061755 0.247276 q* : 0.160109 0.113958 0.240887 use : 0.027639 0.147427 0.344721 ... start iteration 8 of 20 lmf si --band~lst=5~box=0.1,n=12+o,q0=-0.017083,0.009789,0.834163 > out.lmf qnow, E : -0.017083 0.009789 0.834163 0.2202239 gradient : -0.045667 0.028215 -0.015231 0.055799 q* : -0.004562 0.001050 0.845041 cluster center is extremal point ... exiting loop q @ minimum gradient : -0.017083 0.009789 0.834163 Final estimate for q : -0.017083 0.009789 0.834163 Take a look at the first line beginning with lmf. band-edge tells you what command it uses for the energy of each point in the cluster around your starting point. In each iteration qnow gives the current central k-point and its energy in Rydbergs. ‘use’ prints the lowest-energy k-point in the cluster of points around the middle point; this will then be used as the central point in the next iteration. The cluster of 13 k-points and their energies are printed to bnds.si. Take a look and you will see that (-0.017083,0.009789,0.834163) is indeed the lowest-energy point in the cluster. As the iterations proceed, note that the energies at qnow are going down as we float to a low-energy region. After 8 iterations, the following is printed: ‘cluster center is extremal point … exiting loop’. The central q-point is the lowest-energy point and the float routine is finished. ##### 2. Gradient minimization Starting from the low-energy point we floated to, the next step is to do a more refined search using a gradient minimization approach. band-edge creates a new cluster of points, does a quadratic fit and then traces the gradients to a minimum point. Run the following command: $ band-edge -edge2 -maxit=20 -r=.04 -band=5 -gtol=.0001 -q0=\"-0.017083 0.009789 0.834163\" si\n\n\n‘edge2’ part specifies what gradient minimization algorithm to use. All the switches are explained in the documentation below\n\nThe output is similar to before but now we will pay attention to the gradient line which prints the x, y and z gradient components and the last column gives the magnitude. Note how the gradient magnitude is decreasing with each step until it falls below the specified tolerance (gtol). At this point, the gradients minimization is converged and the following is printed ‘gradient converged to tolerance gtol = .0001’. It found the minimum gradient to be (0,0,0.846).\n\n##### 3. Calculate effective mass\n\nNow that we have accurately determined the conduction band minimum, we can calculate the effective mass. This is done by fitting a quadratic form to a set of points around the conduction band minimum. Run the following command:\n\n\\$ band-edge -mass -alat=10.26 --bin -r=.0005,.001,.002,.003 -band=5 -q0=0,0,0.846 si\n\n\nmass tells band-edge to estimate the effective mass tensor. alat is the lattice constant (found in various places such as the lmf output, init or site file). The lattice constant is needed to convert to atomic units since the code reports k-points in units of $2\\pi/\\mathrm{alat}$. r gives the radii of the four clusters of points around the central point; each radius has 32 points (points and faces of an icosohedron). The extra points improve the accuracy of the quadratic fitting.\n\nThe last line of the output prints the three effective mass components in atomic units. So for silicon, the effective mass is anisotropic with lighter masses in two directions and a heavier effective mass in the third direction.\n\n 0.918580 0.185604 0.185532\n\n\nThe masses are fairly close to experimental values for Si.\n\nTo see the principal axes (eigenvectors of the mass tensor) do the following:\n\necho princ \| pfit 2 bndsa.si\n\n\nThe three columns below eval are the three principal axes (eigenvectors of the quadratic form). In this case they are just the Cartesian axes.\n\n### The band-edge manual\n\nband-edge is shell script that calls lmf (or any Questaal executable that can generate energy bands through the --band switch.\n\nIts purpose is to find band extrema and it operates in four distinct modes. All modes start from a reference q point, and compute energy bands in a cluster of points around the reference. One band in particular is selected out. The cluster of points are arranged in some regular polyhedron with 12, 20, or 32 points in a shell. You specify the radius of the polyhedron (or radii of multiple shells if desired). Energy bands are generated by invoking lmf or similar code with the --band switch, which is invoked in a special-purpose box mode that makes bands for clusters of q points centered around a reference.\n\nThe four modes are:\n\n1. (-floatmx or -floatmn) Update the current q point by indentifying whichever point in the cluster has the maximum value (-floatmx) or minimum value (-floatmn). If the central point is the extremal point, band-edge exits. Otherwise, the extremal point becomes the new central point and the cycle is repeated.\n2. (-edge) The selected energy band at a given cluster of points is subject to a least-squares fit and the gradient in q extracted. The q point and gradient are fed to a Broyden minimization algorithm built into the fmin utility. fmin returns a new estimate for the extremal point. The process is iterated until the gradient falls below a specified tolerance.\n3. (-edge2) The selected energy band at a given cluster of points is fit to a quadratic polynomial in qx, qy, qz and the extremal point is estimated from the extremal point of the parabola. This new point becomes the reference point; the process is iterated until the gradient falls below a specified tolerance.\n4. (-mass) The selected energy band at a given cluster of points is fit to a quadratic polynomial. The normal matrix is diagonalized; its eigenvalues are the tensor effective masses and eigenvectors are the principal axes of the mass tensor.\n\nUsage : band-edge switches ext\n\next is the extension in ctrl.ext, the input file the band code uses to generate $E_n(\\mathbf{q})$. By default band-edge uses lmf (but see -cmd below).\n\nChoose between one of the following modes :\n\n• -edge\nSeek extremal q by calculating $\\nabla_q E_n$ and minimizing it using a Broyden algorithm\n\n• -edge2 \| -edge2=fac\nSeek extremal q using 1st and 2nd derivatives of $\\nabla_q E_n$, from a polynomial fit to $E_n$ on the cluster.\nCalculates qnew = (1-fac)qold + facqmin where qmin is extremal point estimated from quadratic form.\nfac is used if it is specified; otherwise fac=1. Independently of fac, the change in q might be limited by dqmx (below)\n\n• -floatmx\nUpdate reference q with largest $E_n$ in cluster. Iterate until the reference point (cluster center) is the maximum point.\n\n• -floatmn\nUpdate reference q with smallest $E_n$ in cluster. Iterate until the reference point (cluster center) is the minimum point.\n\n• -mass\nFit bands around q0 to estimate effective mass tensor.\n\nRequired switches:\n\n• -q0=#,#,#\nStarting value of reference q\n\n• -band=n\nSpecify band index n to optimize\n\n• -r=#[,#2,#3,…]\n\n• -alat=a\nLattice constant a (a.u.). It is not needed except in the calculation of effective mass (required there because q has units 2πa).\n\nOptional switches:\n\n• -h \| --h \| --help\nshow this message\n\n• -cmd=strn\nuse strn in place of default command to generate bands, e.g.\ncmd=\"mpirun -n 16 lmf\"\nDefault cmd is lmf.\n\n• -n=12 \| -n=20 \| -n=32\nSpecify number of points in shell of radius r\n\n• -dqmx=#\nMaximum allowed change n any component of q for one iteration\n\n• -gtol=#\nConvergence tolerance in the gradient (applies to -edge and -edge2)\n\n• –maxit=#\nMaximum number of iterations to attempt\n\n• –spin1\nreplace --band with -band~spin1\n\n• –spin2\nreplace --band with -band~spin2\n\n• –noexec\nShow what cmd will be executed, without executing it\n\n• –bin’ Tell the bands maker to write bands in binary format\n\nAlthough we know the valence band maximum for silicon is at the $\\Gamma$ point, we will use it as an example for finding a maximum point. Try starting from the point (0.1,0.1,0.1). In the float step, change the floatmn (for minimum) to floatmx (for maximum) and change the band number to 4 (for valence band). Notice the energies are going up. Now do a gradient minimization from the point you floated to, remember to change the band number to 4. The valence band maximum is in a flatter region so try using a smaller excursion radius, say 0.001. Then calculate the effective mass. Your values should be around (-0.34, -0.56, -0.88). The negative signs indicate a maximum point, a saddle point would have at least one positive sign." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7998922,"math_prob":0.988384,"size":13057,"snap":"2021-43-2021-49","text_gpt3_token_len":3559,"char_repetition_ratio":0.14042749,"word_repetition_ratio":0.06132303,"special_character_ratio":0.28329632,"punctuation_ratio":0.15037327,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9961853,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-26T21:25:23Z\",\"WARC-Record-ID\":\"<urn:uuid:c0338c28-42f7-4f99-ad96-dd49211fed9a>\",\"Content-Length\":\"47801\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b32ec373-e5b6-49ec-9bad-24c35b8e67a5>\",\"WARC-Concurrent-To\":\"<urn:uuid:646ab34b-e3bb-4b70-97c4-589ff15919d0>\",\"WARC-IP-Address\":\"173.212.228.104\",\"WARC-Target-URI\":\"https://www.questaal.org/tutorial/lmf/bandedge/\",\"WARC-Payload-Digest\":\"sha1:3MWEQ3UOZZSHWRBM2ZUG5DH4TIMQTFWL\",\"WARC-Block-Digest\":\"sha1:U5XWD56BCF7NZGHI5AUTLF5UEOTKVLG6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587926.9_warc_CC-MAIN-20211026200738-20211026230738-00682.warc.gz\"}"}
http://www.baojianyuan.cn/gs/69zz83/	[ "# 面相极贵的男人", null, "2021-12-01\n\n### 什么面相的男人注定富贵?", null, "=x1^3-x2^3\n\n=(x1-x2)(x1^2+x2^2+x1x2)\n\n=(x1-x2)[(x1+x2/2)^2+3(x2^2)/4]\n\n∵ x1<x2\n\n∴ x1-x2<0 ,(x1+x2/2)^2+3(x2^2)/4>0\n\n∴ (x1-x2)[(x1+x2/2)^2+3(x2^2)/4]<0\n\n∴ y2-y1<0\n\n### 日月角骨很明显的人面相\n\n1、性平的食物一年四季都可食用。\n2、性温的食物除夏季适当少食用外，其它季节都可食用。\n3、性凉的食物夏季可经常食用，其它季节如要食用须配合性温的食物一起吃。\n4、性寒的食物尽量少吃，如要食用必须加辣椒、花椒、生姜等性温热的食物一起吃。\n\n### 日后注定富贵男人面相有哪些特点\n\n• 1.宿世缘分是什么\n• 2.属马人2021年运势运程每月运程男\n• 3.老黄历姓名测吉凶查询\n• 4.辟邪神兽的本领\n• 5.女性打喷嚏吉凶\n• 6.算命说感情不顺晚婚\n• 7.六爻六爻\n• 8.五行属木有什么字最好取名" ]	[ null, "http://www.baojianyuan.cn/pic/w40xrs1e4aw.jpg", null, "https://iknow-pic.cdn.bcebos.com/bba1cd11728b47103479e6e3cfcec3fdfc03234e", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.94877845,"math_prob":0.9959005,"size":5358,"snap":"2021-43-2021-49","text_gpt3_token_len":7323,"char_repetition_ratio":0.046507284,"word_repetition_ratio":0.0,"special_character_ratio":0.24710713,"punctuation_ratio":0.02929293,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96603364,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-01T09:31:21Z\",\"WARC-Record-ID\":\"<urn:uuid:36654790-8886-4110-998f-fabc286a0a8a>\",\"Content-Length\":\"24279\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a93c4759-7ffa-4a4c-9317-a839622fb8eb>\",\"WARC-Concurrent-To\":\"<urn:uuid:0509c4d3-76af-4cb6-86c1-2265679e5a73>\",\"WARC-IP-Address\":\"103.86.47.220\",\"WARC-Target-URI\":\"http://www.baojianyuan.cn/gs/69zz83/\",\"WARC-Payload-Digest\":\"sha1:NF7KP2A6L2GSNAIRHPDAPQ2OYSZMSJC4\",\"WARC-Block-Digest\":\"sha1:K6U2GOVYGIUZCCQA762YUTJNGGFC52G7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964359976.94_warc_CC-MAIN-20211201083001-20211201113001-00353.warc.gz\"}"}
https://discusstest.codechef.com/t/maze_shortest-path/1698	[ "", null, "# maze_shortest path\n\nhttp://www.codechef.com/BTFR2013/problems/NITA07\n\nHow to solve this??\n\n1 Like\n\nBfs can be used to solve it.\n\n1 Like\n\nI didn’t solve this one yet, but it seems as simple shortest path problem, so simply use one of the well known algorithm f.e.:\n\n2 Likes\n\nthanks!!!\n\n1 Like\n\nthanks!!!\n\n@hariprasath\n\nThis is a very known problem called as Rat in a Maze.\n\nYou can use BFS or Backtracking to solve this problem.\n\n@betlista i think you do not need to use Dijkstra algorithm for this problem.\n\n3 Likes\n\ny dijkstra??\n\n1 Like\n\nthis function can solve the question recursive and easy", null, "int minCost(int cost[][C],int m, int n)\n{\nif (n < 0 \|\| m < 0)\nreturn INT_MAX;\nelse if (m == 0 && n == 0)\nreturn cost[m][n];\nelse\nreturn cost[m][n] + min( minCost(cost,m-1, n-1),\nminCost(cost,m-1, n),\nminCost(cost,m, n-1) );\n}\n\n5 Likes\n//" ]	[ null, "https://s3.amazonaws.com/discoursestaging/original/2X/4/45bf0c3f75fc1a2cdf5d9042041a80fa6dd3106b.png", null, "https://discusstest.codechef.com/images/emoji/apple/slight_smile.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7886716,"math_prob":0.90124094,"size":795,"snap":"2021-31-2021-39","text_gpt3_token_len":238,"char_repetition_ratio":0.13400759,"word_repetition_ratio":0.015748031,"special_character_ratio":0.30943397,"punctuation_ratio":0.20833333,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98931015,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-25T05:42:42Z\",\"WARC-Record-ID\":\"<urn:uuid:3d8867e8-da65-41c2-a135-6cbd01698b8b>\",\"Content-Length\":\"25342\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ace1c348-7d42-4320-9fea-999600841a23>\",\"WARC-Concurrent-To\":\"<urn:uuid:03aa1b6a-6c7b-4399-8592-c1f5ee1431ff>\",\"WARC-IP-Address\":\"54.165.214.21\",\"WARC-Target-URI\":\"https://discusstest.codechef.com/t/maze_shortest-path/1698\",\"WARC-Payload-Digest\":\"sha1:6BOIWWHS5TNWBZ5BTIX2NI575SJ2T3RR\",\"WARC-Block-Digest\":\"sha1:ZM4PC4FE3INT2VFIJOBVLUJK53YY5V7S\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046151638.93_warc_CC-MAIN-20210725045638-20210725075638-00707.warc.gz\"}"}
https://studylib.net/doc/11133721/homework-%23-1--due-10-feb-%3D-1-magnitude-leads-to	[ "# Homework # 1, due 10 Feb = 1 magnitude leads to", null, "```Homework # 1, due 10 Feb\n1. Show that an extinction of Aλ = 1 magnitude leads to\na flux Fλ decrease to 40% of its original value.\n2. Assuming dust grains are 0.1µ m in radius, that the\ngas density in the ISM is nH = 1 cm−3, and the number density of dust grains is nd = 10−12 cm−3 and\nuniformly distribued, find the column density NH of\nhydrogen to be expected along a line of sight in which\na source is observed to have an extinction AV = 1 magnitude. What is the mean free path of light along this\nline of sight? What is the distance to the source?\n3. Assume a star cluster has 200 F5 stars at the main\nsequence turnoff and 20 K0III giant stars. Determine\nthe quantities MV and B − V for each type of star\nsingly and also for the cluster. Suppose the cluster is\nat a distance of 2 kpc. What is mV for the cluster?\n4. In a galaxy at a distance of d Mpc, what would be\nmB⊙? In this galaxy, what length does 1 arcsec correspond to? If the surface brightness of this galaxy is\nIB = 27 mag arsec−2, show that the brightness of this\ngalaxy is equivalent to 0.99 LB⊙ pc−2. Why doesn’t\nthis quantity depend on the distance?\n```" ]	[ null, "https://s2.studylib.net/store/data/011133721_1-5c2c55095584e5f868918db54b294c7c-768x994.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9326764,"math_prob":0.9703055,"size":1117,"snap":"2021-31-2021-39","text_gpt3_token_len":305,"char_repetition_ratio":0.10961366,"word_repetition_ratio":0.009049774,"special_character_ratio":0.2676813,"punctuation_ratio":0.0996016,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9830845,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-20T16:49:04Z\",\"WARC-Record-ID\":\"<urn:uuid:fe037eb4-d4cd-43ca-afe2-c4c16d2a0584>\",\"Content-Length\":\"48313\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ccac37cb-08e8-46e5-94f6-dcfa6fd85956>\",\"WARC-Concurrent-To\":\"<urn:uuid:933fa405-111c-400b-b8c3-67a0bb05c768>\",\"WARC-IP-Address\":\"104.21.72.58\",\"WARC-Target-URI\":\"https://studylib.net/doc/11133721/homework-%23-1--due-10-feb-%3D-1-magnitude-leads-to\",\"WARC-Payload-Digest\":\"sha1:BW7TJDAXTIG4P2QDILWONDHOI4WP22U3\",\"WARC-Block-Digest\":\"sha1:RKDMJXSBVWAK4HEWBNCDIZM3OC6SO3YP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057083.64_warc_CC-MAIN-20210920161518-20210920191518-00324.warc.gz\"}"}
https://jp.maplesoft.com/support/help/Maple/view.aspx?path=LinearAlgebra/Modular/IntegerDeterminant	[ "", null, "IntegerDeterminant - Maple Help\n\nLinearAlgebra[Modular]\n\n IntegerDeterminant\n determinant of an integer matrix using modular methods", null, "Calling Sequence IntegerDeterminant(M)", null, "Parameters\n\n M - Square Matrix with integer entries", null, "Description\n\n • The IntegerDeterminant function computes the determinant of the integer matrix M. This is a programmer level function, it does not perform argument checking. Thus, argument checking must be handled external to this function.\n • Note: The IntegerDeterminant command uses a probabilistic approach that achieves great gains for structured systems. Information on controlling the probabilistic behavior can be found in EnvProbabilistic.\n • This function is used by the Determinant function in the LinearAlgebra package when a Matrix is determined to contain only integer entries.\n • This command is part of the LinearAlgebra[Modular] package, so it can be used in the form IntegerDeterminant(..) only after executing the command with(LinearAlgebra[Modular]). However, it can always be used in the form LinearAlgebra[Modular][IntegerDeterminant](..).", null, "Examples\n\nA 3x3 matrix\n\n > $\\mathrm{with}\\left(\\mathrm{LinearAlgebra}\\left[\\mathrm{Modular}\\right]\\right):$\n > $M≔\\mathrm{Matrix}\\left(\\left[\\left[2,1,3\\right],\\left[4,3,1\\right],\\left[-2,1,-3\\right]\\right]\\right)$\n ${M}{≔}\\left[\\begin{array}{ccc}{2}& {1}& {3}\\\\ {4}& {3}& {1}\\\\ {-2}& {1}& {-3}\\end{array}\\right]$ (1)\n > $\\mathrm{IntegerDeterminant}\\left(M\\right)$\n ${20}$ (2)\n\nA 100x100 matrix\n\n > $M≔\\mathrm{LinearAlgebra}\\left[\\mathrm{RandomMatrix}\\right]\\left(100\\right):$\n > $\\mathrm{tt}≔\\mathrm{time}\\left(\\right):$\n > $\\mathrm{IntegerDeterminant}\\left(M\\right)$\n ${38562295347802366242417909657285032281105091485000162871067163275296273582728190925949289361981964881806516849833008824879568403928373759144147382030798909099402726531205056808283212790472544339698767179236612577117605985054960334148934541347201762137455}$ (3)\n > $\\mathrm{time}\\left(\\right)-\\mathrm{tt}$\n ${0.042}$ (4)" ]	[ null, "https://bat.bing.com/action/0", null, "https://jp.maplesoft.com/support/help/Maple/arrow_down.gif", null, "https://jp.maplesoft.com/support/help/Maple/arrow_down.gif", null, "https://jp.maplesoft.com/support/help/Maple/arrow_down.gif", null, "https://jp.maplesoft.com/support/help/Maple/arrow_down.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6001329,"math_prob":0.9944781,"size":1645,"snap":"2022-05-2022-21","text_gpt3_token_len":435,"char_repetition_ratio":0.1773309,"word_repetition_ratio":0.011976048,"special_character_ratio":0.31610942,"punctuation_ratio":0.13043478,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9984615,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-28T14:53:25Z\",\"WARC-Record-ID\":\"<urn:uuid:0f449728-a2c0-490b-9344-b12815b464d0>\",\"Content-Length\":\"165808\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7ba7860f-d625-4be0-b21d-73f17cb7182e>\",\"WARC-Concurrent-To\":\"<urn:uuid:3bb9de13-81d3-4d70-a4e1-e3190193d5f3>\",\"WARC-IP-Address\":\"199.71.183.28\",\"WARC-Target-URI\":\"https://jp.maplesoft.com/support/help/Maple/view.aspx?path=LinearAlgebra/Modular/IntegerDeterminant\",\"WARC-Payload-Digest\":\"sha1:6VYGPX32B7A4ZUPQX5FEPMXLD7LSZQUY\",\"WARC-Block-Digest\":\"sha1:U6SN4SOQTI456BE6DS7NYCUATWJNG7FI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652663016853.88_warc_CC-MAIN-20220528123744-20220528153744-00465.warc.gz\"}"}