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https://calculator.academy/land-to-building-ratio-calculator/	[ "Enter the total square footage of the land parcel and the total square footage of the building to determine the land to building ratio.\n\n## Land to Building Ratio Formula\n\nThe following formula is used to calculate a land to building ratio:\n\nL:B = AL / AB\n\n• Where L:B is the land to building ratio\n• AL is the area of the land (ft^2)\n• AB is the area of the buildling (ft^2)\n\n## What is land to building ratio?\n\nDefinition:\n\nA land to building ratio is defined as the total square footage of a land parcel divided by the total area of the building footprint.\n\nThe land to building ratio is used in real estate to evaluate how much land is taken up by a building to determine if there is room for my development.\n\n## How to calculate a land to building ratio?\n\nExample Problem:\n\nThe following example outlines how to calculate a land to building ratio.\n\nFirst, determine the total square footage taken up by the building. In this case, the total land area is 5000 square feet.\n\nNext, determine the area of the entire lot. In this example, the entire lot of land is 10,000 square feet.\n\nFinally, calculate the land to building ratio using the formula above:\n\nL:B = AL / AB\n\nL:B = 10,000 / 5,000\n\nL:B = 2:1" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90578073,"math_prob":0.9841736,"size":1301,"snap":"2021-43-2021-49","text_gpt3_token_len":301,"char_repetition_ratio":0.21279876,"word_repetition_ratio":0.079166666,"special_character_ratio":0.24058416,"punctuation_ratio":0.10408922,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99935955,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-27T12:05:40Z\",\"WARC-Record-ID\":\"<urn:uuid:aff169a0-e494-4e1f-8dff-79ebba3346e2>\",\"Content-Length\":\"122128\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3b80a6ce-228f-4d44-9bce-a0b4ef6ec45d>\",\"WARC-Concurrent-To\":\"<urn:uuid:ac57f791-603f-42e1-976c-25a1d4b89a0a>\",\"WARC-IP-Address\":\"172.67.69.208\",\"WARC-Target-URI\":\"https://calculator.academy/land-to-building-ratio-calculator/\",\"WARC-Payload-Digest\":\"sha1:2324QUHXCWSPGN6EYHSA44SDG6V3I6SJ\",\"WARC-Block-Digest\":\"sha1:DIUJQ3YMHT25JRNRXTJZ3NRWCVDJPDUW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358180.42_warc_CC-MAIN-20211127103444-20211127133444-00161.warc.gz\"}"}
https://www.advanceexcelforum.com/excel-double-vlookup-iferror-vlookup/	[ "# 03 Best Ways: EXCEL DOUBLE VLOOKUP/ IFERROR VLOOKUP/ NESTED VLOOKUP\n\n0\n\nDOUBLE VLOOKUP is a term in Advance Excel where two VLOOKUP functions nested with the IFERROR function to make a nested formula that sequentially works in two different tables or columns of the same workbook or different workbooks and retrieves the value. Thus, Excel Double VLOOKUP is often known as IFERROR VLOOKUP or NESTED VLOOKUP\n\nBroadly, NESTED VLOOKUP is an advanced level of DOUBLE VLOOKUP where two or more VLOOKUP functions work together with two or more IFERROR functions, making a nested formula that sequentially works in two or more different tables or columns of the same workbook or different workbooks. NESTED VLOOKUP is a kind of IFERROR VLOOKUP in Excel.\n\nThe IFERROR function is a logic function that checks a cell to determine if that cell contains an error or if a formula will result in an error. If no error exists, the function returns the value of the formula. If an error exists, the function returns an error.\n\nThe IFERROR function can’t distinguish the type of error. The error could be #NAME, #N/A, #REF, DIV/0, and so on. What is used as the value_if_error argument will be displayed regardless of the error type.\n\n# A. SIMPLE METHOD OF DOUBLE VLOOKUP / IFERROR VLOOKUP / NESTED VLOOKUP\n\nExcel Double VLOOKUP is a term where two VLOOKUP functions nested with the IFERROR function will make a unique formula, that is able to match the lookup_value in two different columns. It is also known as IFERROR VLOOKUP or NESTED VLOOKUP.\n\nThe logic behind the Excel DOUBLE VLOOKUP formula or IFERROR VLOOKUP formula is that when the first VLOOKUP fails to find the lookup_value in the first column range returns an Error, then the IFERROR function replaces the Error with the second VLOOKUP function which finds the same lookup_value in the second column range and retrieves the value.\n\nAs a result, with a single formula, we can retrieve the value matches with two different columns.\n\n➢ SYNTAX:\n\n➢ STEPS TO START:\n\n=IFERROR(VLOOKUP(\\$G3,\\$B\\$3:\\$D\\$12,3,0), (VLOOKUP(\\$G3,\\$C\\$3:\\$D\\$12,2,0)))\n\n• STEP 1: Select a cell to get the result of Double VLOOKUP (i.e., H3).\n\nIn this cell, place an equality “=” sign to start the formula and just type a few letters ‘=IF….‘ and select the IFERROR function from the Excel provided suggestion list with the help of a down arrow (↓) key, if required.", null, "Then press the ‘Tab’ key, IFERROR syntax appears with the open parenthesis.", null, "The IFERROR function has 2 arguments: value and value_if_error.\n\n• Value is the expression being tested.\n\n• Value_if_error is the text that will be returned if there is an error in the formula (or expression).\n\nWe place the first VLOOKUP in the value position and place the second VLOOKUP in the value_if_error position. Therefore, the new formula works in such a way, if the first VLOOKUP returns any #N/A error, it will be replaced by the value of the second VLOOKUP.\n\n• STEP 2: Then type a few letters of vlookup such as ‘=vlo….’ and select the VLOOKUP function from the given suggestion list.", null, "Then press the ‘Tab’ key, VLOOKUP syntax appears with the open parenthesis within the IFERROR function.", null, "• STEP 3: The first VLOOKUP retrieves the value from a first table_array, likes\n\n=IFERROR( VLOOKUP( \\$G3, \\$B\\$3:\\$D\\$12, 3, 0),\n\n➢ \\$G3 – select the lookup value locates in cell G3 (i.e., CAN-1) and fix the Column address by pressing the F4 key thrice. Thus the range converts from the relative to the mixed cell reference where it indicates the absolute column and relative row.\n\nAs a result, while the formula is copied to the right side or horizontally, the column address does not change but the row address changes accordingly.\n\n➢ \\$B\\$3:\\$D\\$12 lookup_value found in the range is called lookup_array and fixed the range by pressing the F4 key once. Thus the range is converted to absolute from the relative cell reference.\n\nAs a result, the range does not change when the formula is copied to another cell either horizontally or vertically.\n\n➢ 3 – the column_index_num is the count of columns between the lookup value column and the return value column.\n\n➢ 0 – the last argument of the VLOOKUP function is range_lookup. If we are looking for an exact match, place either 0 or FALSE.\n\n■ Note: We had detail explained on Cell Reference in a separate tutorial. Request you read this tutorial: 03 Types of Excel Cell Reference: Relative, Absolute & Mixed\n\n• STEP 4: When the first VLOOKUP formula cannot find a match, it returns the #N/A error (i.e., lookup_value does not find a match in the first column of table_array).\n\nThen the wrapped IFERROR function to replace the #N/A error either with the ‘values’ or ‘suggested texts’. We should place the texts in double quotation marks (” “).\n\nBut here the #N/A error is replaced with the values, those values are getting from the Second VLOOKUP.\n\nWe should place the second VLOOKUP in place of value_if_error, the second argument of the IFERROR function. Then, we write the formula as:\n\n=IFERROR(VLOOKUP(\\$G3, \\$B\\$3:\\$D\\$12, 3, 0), (VLOOKUP(\\$G3, \\$C\\$3:\\$D\\$12, 2, 0)))\n\n➢ \\$G3 – as the same as first Lookup, select the lookup value locates in cell G3 (i.e., CAN-1) and fix the Column address by pressing the F4 key thrice. Thus the range converts from the relative to the mixed cell reference where it indicates the absolute column and relative row.\n\nAs a result, while the formula is copied to the right side or horizontally, the column address does not change but the row address changes accordingly.\n\n\\$C\\$3:\\$D\\$12 – lookup_value found in the range is called lookup_array and in this case, lookup_array should be different from the first lookup_array.\n\nFixed the range by pressing the F4 key once. Thus the range is converted from the relative cell reference to the absolute.\n\nAs a result, the range does not change when the formula is copied to another cell either horizontally or vertically.\n\n➢ 2 – the column_index_num is the count of columns between the lookup value column and the return value column.\n\n0 – the last argument of the VLOOKUP function is range_lookup. If we are looking for an exact match we put either 0 or FALSE.\n\n• STEP 5: Finally, press Enter to accept the formula. Formula ends by default and closes the last parenthesis as well.\n\n• STEP 6: EXTEND THE FORMULA TILL THE END OF THE RANGE (WITHOUT FORMATTING)\n\nCopy (Ctrl+C) the cell with formula Select the range of cells where to copy the formula (Shift+Down Arrow) Press either Alt+E+S+R (press sequentially, Alt, E, S, R) or Alt+Ctrl+V+R (press Alt+Ctrl+V, then R) which will select the “Formulas and number formats” in the Paste Special dialog box Then press Enter or click OK.", null, "• STEP 7: OBSERVATIONS", null, "In the Excel double VLOOKUP two ranges are used for retrieving data: one is the range B3:D12 and the second one is C3:D12.\n\n(01). In the first case, the lookup_value is CAN-1 and the Excel Double VLOOKUP returns the result value CAN-124 as a Number Code (UN). The lookup_value found in the range C3:D12.\n\n(02). In the second case, the lookup_value is IN-91 and the Excel Double VLOOKUP returns the result value IND-356 as a Number Code (UN). The lookup_value found in the different range B3:D12.\n\n(03). In the third case, the lookup_value is USA-840 and the Excel Double VLOOKUP returns the #N/A error. That means the lookup_value does not exist either in range C3:D12 or B3:D12.\n\n(04). In the fourth case, using another IFERROR function to replace the #N/A error with a text likes “Not Found”. The text should be in double quotation. We can keep it blanks ” ” instead of a text.\n\n# B. ADVANCED METHOD OF DOUBLE VLOOKUP / IFERROR VLOOKUP / NESTED VLOOKUP\n\nNested VLOOKUP in Excel is an advanced level of multiple VLOOKUP functions in the IFERROR functions, where individual VLOOKUP function finds the match from the suggested table array and retrieves the value. If the VLOOKUP function fails to match, it returns a #N/A error.\n\nThe IFERROR function in Excel allows replacing the #N/A error with suggested ‘value’ or ‘text’. To make the formula more dynamic, rather than put any value manually replaces it with the second VLOOKUP. Similarly, the second VLOOKUP formula finds the match from another table array and retrieves the value. If it fails to find the match, it returns a #N/A error.\n\nAgain and again, apply the IFERROR function to replace the #N/A error by third, fourth, fifth,…returns by the VLOOKUP functions. The combination of multiple IFERROR and VLOOKUP functions makes a formula that allows to sequential lookup with first table array, second table array, third.., fourth…so on.\n\nThe advanced method of using the IFERROR VLOOKUP in Excel is called the Advanced Double VLOOKUP or Advanced Nested VLOOKUP or Advanced IFERROR VLOOKUP\n\n➢ SYNTAX:", null, "➢ STEPS TO START:\n\n=IFERROR(VLOOKUP(\\$G3,\\$B\\$3:\\$D\\$12,3,0), IFERROR(VLOOKUP(\\$G3,\\$C\\$3:\\$D\\$12,2,0), VLOOKUP(\\$G3,\\$D\\$3:\\$D\\$12,1,0)))\n\n• STEP 1: Select the cell to get the result of Advanced Double VLOOKUP (i.e., H3).\n\nIn this cell, press equality “=” sign to start formula and just type a few letters of IFERROR e.g., =if…. and select the IFERROR function from the Excel provided suggestion list with the help of a down arrow (↓), if required.", null, "Then press the ‘Tab’ key, IFERROR syntax appears with the open parenthesis.", null, "The IFERROR function has two arguments: value and value_if_error.\n\nWe put the first VLOOKUP in place of the value argument and put the second VLOOKUP in place of the value_if_error argument. So, the new formula works in such a way, if the first VLOOKUP returns any #N/A error, it will be replaced by the value of the second VLOOKUP.\n\n• STEP 2: Then type a few letters of the VLOOKUP, for example, =vlo… and select the VLOOKUP function from the Excel provided suggestion list.", null, "Then press the ‘Tab’ key, VLOOKUP syntax appears with the open parenthesis within the IFERROR function.", null, "• STEP 3: The first VLOOKUP retrieves the value from a first table_array. This formula allows retrieving the values against match cases and returns the #N/A errors in non-match cases.\n\nIn this case, we wrap the formula with another IFERROR function to replace the #N/A error. The #N/A error should be replaced with the value, this value comes from the Second VLOOKUP.\n\nIf the second VLOOKUP formula fails to find the match from the range, it returns the #N/A error. This error is replaced by the value of the third VLOOKUP\n\nThe position of the first VLOOKUP in place of the ‘value’ argument of the first IFERROR function and in place of the ‘value_if_error’ argument puts the second IFERROR function. The position of the second VLOOKUP and third VLOOKUP in place of the ‘value’ and ‘value_if_error’ arguments in the second IFERROR function, respectively.\n\nTherefore, now the formula works in such a way, if the first VLOOKUP returns any #N/A error, the second VLOOKUP replaces the error with the value. If the second VLOOKUP returns any #N/A error, it will be replaced by the value of the third VLOOKUP.\n\n=IFERROR(VLOOKUP(\\$G3,\\$B\\$3:\\$D\\$12,3,0), IFERROR(VLOOKUP(\\$G3,\\$C\\$3:\\$D\\$12,2,0), VLOOKUP(\\$G3,\\$D\\$3:\\$D\\$12,1,0)))\n\n➢ \\$G3 – the lookup value locates in cell G3 (i.e., CAN-1) and same for the first, the second and the third VLOOKUP function.\n\nFix the Column address by pressing three times the F4 key. Thus the range converts from the relative to the mixed cell reference where it indicates the absolute column and relative row.\n\nAs a result, while the formula is copied to the right side or horizontally, the column address does not change but the row address changes accordingly.\n\n\\$B\\$3:\\$D\\$12lookup_array for the first VLOOKUP (lookup_array is the range where lookup_value is found).\n\n\\$C\\$3:\\$D\\$12lookup_array for the second VLOOKUP.\n\n\\$D\\$3:\\$D\\$12lookup_array for the third VLOOKUP.\n\nFixed every range by pressing the F4 key once. Thus the range is converted to absolute from the relative cell reference.\n\nAs a result, the range does not change when the formula is copied to another cell either horizontally or vertically.\n\n➢ 3 – column_index_num for the first VLOOKUP (the column_index_num is the count of columns between the lookup value column and the return value column.\n\n2 – column_index_num for the second VLOOKUP.\n\n1 column_index_num for the third VLOOKUP.\n\n➢ 0 – the last argument of the VLOOKUP function is range_lookup. We are looking for an exact match, so we put either 0 or FALSE. So, range_lookup is the same for three VLOOKUP functions.\n\n• STEP 4: Finally, press Enter to accept the formula. Formula ends by default and closes the last parenthesis as well.\n\n• STEP 5: EXTEND THE FORMULA TILL THE END OF THE RANGE (WITHOUT FORMATTING)\n\nCopy (Ctrl+C) the cell with formula Select the range of cells where to copy the formula (Shift+Down Arrow) Press either Alt+E+S+R (press sequentially, Alt, E, S, R) or Alt+Ctrl+V+R (press Alt+Ctrl+V, then R) which will select the “Formulas and number formats” in the Paste Special dialog box Then press Enter or click OK.\n\n# C. DYNAMIC METHOD OF DOUBLE VLOOKUP / IFERROR VLOOKUP / NESTED VLOOKUP\n\nUsing of MATCH function in the Advanced Double VLOOKUP (or Advanced Nested VLOOKUP or Advanced IFERROR VLOOKUP) makes it dynamic formula.\n\nThe MATCH() function returns the position of an item within an array that matches a specific value.\n\nUsing the MATCH() function in place of column_index_num automatically updates the column number that makes the dynamic formula.\n\nThe Syntax of the MATCH function is as follows:", null, "➢ Lookup_value is the item to match. It can be a number, text string, a logical value, or a reference.\n\n➢ Lookup_array is the table or an array containing all of the values to search.\n\n➢ Match_type is a number that specifies how the match will be applied.\n\nA match_type of zero (0) finds the first item in the array that is an exact match with the lookup_value.\n\nTo find the item closest to but less than the lookup_value, use a match_type of -1.\n\nTo find the item closest to but greater than the lookup_value, use a match_type of 1.\n\nIn the last two cases, the values in the lookup_array must be in ascending order for the MATCH function to work correctly.\n\nThe match_type is optional and will default to 1 if omitted from the arguments\n\n➢ SYNTAX:", null, "➢ STEPS TO START:\n\n• STEP 1: Select the cell to get the result of Advanced Double VLOOKUP (i.e., H3).\n\nIn this cell, press equality “=” sign to start formula and just type a few letters of IFERROR e.g., =if…. and select the IFERROR function from the Excel provided suggestion list with the help of a down arrow (↓), if required.", null, "Then press the ‘Tab’ key, the IFERROR syntax appears with the open parenthesis.", null, "The IFERROR function has two arguments: value and value_if_error.\n\nWe put the first VLOOKUP in place of the value argument and put the second VLOOKUP in place of the value_if_error argument. So, the new formula works in such a way, if the first VLOOKUP returns any #N/A error, it will be replaced by the value of the second VLOOKUP.\n\n• STEP 2: Then type a few letters of the VLOOKUP, for example, =vlo… and select the VLOOKUP function from the Excel provided suggestion list.", null, "Then press the ‘Tab’ key, VLOOKUP syntax appears with the open parenthesis within the IFERROR function.", null, "• STEP 3: The position of the first VLOOKUP in place of the ‘value’ argument of the first IFERROR function and in place of the ‘value_if_error‘ argument puts the second IFERROR function. The position of the second VLOOKUP and third VLOOKUP in place of the ‘value‘ and ‘value_if_error‘ arguments in the second IFERROR function, respectively.\n\nSo, the formula works in such a way, if the first VLOOKUP returns any #N/A error, the second VLOOKUP replaces the error with the value. If the second VLOOKUP returns any #N/A error, it will be replaced by the value of the third VLOOKUP.\n\n=IFERROR(VLOOKUP(\\$G3,\\$B\\$3:\\$D\\$12, MATCH(H\\$2,\\$B\\$2:\\$D\\$2,0),0),\n\nIFERROR(VLOOKUP(\\$G3,\\$C\\$3:\\$D\\$12, MATCH(H\\$2,\\$C\\$2:\\$D\\$2,0),0),\n\nVLOOKUP(\\$G3,\\$D\\$3:\\$D\\$12, MATCH(H\\$2,\\$D\\$2:\\$D\\$2,0),0)))\n\n➢ EXPLANATION OF THE FIRST VLOOKUP FUNCTION:\n\nVLOOKUP(\\$G3,\\$B\\$3:\\$D\\$12, MATCH(H\\$2,\\$B\\$2:\\$D\\$2,0),0)\n\n• \\$G3 – the lookup value locates in cell G3 (i.e., CAN-1) and the same for the first, the second and the third VLOOKUP function.\n\nFix the Column address by pressing the F4 key thrice. Thus the range converts from the relative to the mixed cell reference where it indicates the absolute column and relative row.\n\nAs a result, while the formula is copied to the right side or horizontally, the column address does not change but the row address changes accordingly.\n\n• \\$B\\$3:\\$D\\$12 – lookup_array for the first VLOOKUP (lookup_array is the range where lookup_value is found).\n\n• MATCH(H\\$2,\\$B\\$2:\\$D\\$2,0) – we place the MATCH() function in place of column_index_num to update the column number automatically. Please note that the lookup column in both the lookup_array of the VLOOKUP function and the lookup_array of the MATCH function should be the same. Otherwise, the formula returns the #N/A error. In both cases, Column B is the lookup column.\n\n• H\\$2 = lookup_value reference to ‘Number Code (UN)’ and fixed the row address by pressing two times F4 Key. As a result, the lookup_value is converted from the relative cell reference to the mixed column cell reference, where it indicates the relative column and absolute row address.\n\nAs a result, while the formula is copied to the other cells, the row address does not change but the column address changes accordingly.\n\n• \\$B\\$2:\\$D\\$2 = lookup_value found in the range is called lookup_array and fixed the range by pressing a single time F4 key. Thus the range is converted from the relative cell reference to the absolute cell reference.\n\nAs a result, the range does not change when the formula is copied to another cell either horizontally or vertically.\n\n• 0 = for exact match\n\n• 0 – range_lookup the last argument of VLOOKUP function. The value zero (0) or FALSE signifies the exact match.\n\n➢ EXPLANATION OF THE SECOND VLOOKUP FUNCTION:\n\nVLOOKUP(\\$G3, \\$C\\$3:\\$D\\$12, MATCH(H\\$2,\\$C\\$2:\\$D\\$2,0), 0)\n\nWe should follow the same steps described in the first VLOOKUP function.\n\n➢ EXPLANATION OF THE THIRD VLOOKUP FUNCTION:\n\nVLOOKUP(\\$G3,\\$D\\$3:\\$D\\$12, MATCH(H\\$2,\\$D\\$2:\\$D\\$2,0),0)\n\nWe should follow the same steps described in the first VLOOKUP function.\n\n• STEP 4: Finally, press Enter to accept the formula. Formula ends by default and closes the last parenthesis as well.\n\n• STEP 5: EXTEND THE FORMULA TILL THE END OF THE RANGE (WITHOUT FORMATTING)\n\nCopy (Ctrl+C) the cell with formula Select the range of cells where to copy the formula (Shift+Down Arrow) Press either Alt+E+S+R (press sequentially, Alt, E, S, R) or Alt+Ctrl+V+R (press Alt+Ctrl+V, then R) which will select the “Formulas and number formats” in the Paste Special dialog box Then press Enter or click OK.\n\n# D. CONCLUSION\n\n(01). Excel Double VLOOKUP (or IFERROR VLOOKUP or Nested VLOOKUP) is the most important formulas in Advance Excel. The formula helps to reduce the time span in data analysis.\n\n(02). Multiple VLOOKUPS nested with IFERROR function make an advanced formula in Advance Excel that works sequentially in two or more tables or columns in the same workbook or the different workbooks.\n\n(03). Using of MATCH function in Advanced Double VLOOKUP (or Advanced Nested VLOOKUP or Advanced IFERROR VLOOKUP) makes it a dynamic formula.", null, "If you would like to improve your academic and professional career as well, then the below courses help you a lot. Instead of Advanced Excel, a number of best courses suggested you through this platform that boosts your confidence and flies your career high.", null, "Advance Excel Forum", null, "", null, "Microsoft Excel 2019/Office 365 Series\n\nBestseller 4.9/5", null, "", null, "Intermediate Microsoft Access 2019/Office 365\n\nBestseller 4.9/5", null, "", null, "Advanced Microsoft Excel 2019/Office 365 (Self-Paced Tutorial)\n\nBestseller 4.9/5", null, "", null, "Microsoft Excel and Project 2019 Suite\n\nBestseller 4.9/5", null, "", null, "Accounting with MS Excel 2019 Suite\n\nBestseller 4.9/5", null, "", null, "Microsoft Excel 2019 Certification Training\n\nBestseller 4.9/5", null, "", null, "Data Analysis and Visualization\n\nBestseller 4.9/5", null, "", null, "Excel Fundamentals for Data Analysis\n\nBestseller 4.9/5", null, "", null, "Data Visualization in Excel\n\nBestseller 4.9/5", null, "", null, "Bestseller 4.9/5", null, "", null, "Bestseller 4.9/5", null, "", null, "Data Analysis and Reporting in SAS Visual Analytics\n\nBestseller 4.9/5", null, "", null, "Bestseller 4.9/5", null, "", null, "Accounting Analytics\n\nBestseller 4.9/5", null, "", null, "Introduction to Data Engineering\n\nBestseller 4.9/5", null, "", null, "Master Excel Functions in Office 365 – Excel Dynamic Arrays (Learn to Use Excel’s NEW Functions (FILTER, UNIQUE, SORT, XLOOKUP.) to Dramatically Simplify the Work You Do in Excel.)\n\nHigh Rated 4.9/5", null, "", null, "Excel Essentials for the Real World (Complete Excel Course) [Microsoft Excel Beginner to Professional. 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Learn advanced data analysis & earn the Excel Expert Certification (MS Excel 2019/Office 365)\n\nBestseller 4.8/5", null, "", null, "Excel Charts, Graphs & Data Visualization in Excel (Master 20+ Advanced Dynamic Excel Charts and Create Impressive Excel Graphs & Data Visualization in Microsoft Excel)\n\nHigh Rated 4.8/5", null, "12 Examples \|\| How to Use Excel Go To Special?\n\n07 Points Guided You How to Find And Replace in Excel?\n\n05 Points Should Learn How to Freeze Panes in Excel?\n\n12 Things Guided You How to Manage An Excel Workbook\n\n03 Useful Methods : Add Numbers With AutoSum Excel\n\n04 Simple to Advanced Methods: How to Filter in Excel?\n\n05 Best Ways: Create Password Protect Excel & Unprotect it\n\n08 Best Examples: How to Use Excel Conditional Formatting?\n\n04 BEST WAYS: HOW TO TRANSPOSE DATA IN EXCEL\n\nTutorial 01: How to Use Excel SUMIFS Function with Single & Multiple Criteria\n\n0" ]	[ null, 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https://discuss.codechef.com/t/regarding-codechef-problems/376	[ "", null, "# Regarding Codechef Problems\n\nI tried to solve below problems and I have written cod in PHP. Can you tell me why it is giving me wrong answer. I want to know where I am being wrong or what I am doing wrong.\n\nMy Submitted Code for above problem in PHP as below.\n\n``````//Suppose below are my posted array\n\n\\$nTimeMul = array(4,9);\n\\$nToAppear = array(2,3);\nfunction getDezireOutput(\\$nTimeMul = array(), \\$nToAppear = array()) {\n\\$i = 0;\nforeach (\\$nTimeMul as \\$num) {\n\\$tempNum = 1;\nfor (\\$t=0; \\$t < \\$num; \\$t++) {\n\n\\$tempNum = \\$tempNum * \\$num;\n}\necho substr (\\$tempNum, 0, \\$nToAppear[\\$i]).' '.substr (\\$tempNum, -\\$nToAppear[\\$i]).'\n``````\n\n';\n\\$i++;\n}\n}\ngetDezireOutput (\\$nTimeMul, \\$nToAppear);\n\nPlease have look into above problem and let me know the problems with above code so, I can figure it out." ]	[ null, "https://s3.amazonaws.com/discourseproduction/original/3X/2/6/26ff6c8c380a45aa03cbe51d74b4b0f9fd44daa9.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8013211,"math_prob":0.9209762,"size":793,"snap":"2020-24-2020-29","text_gpt3_token_len":238,"char_repetition_ratio":0.13688213,"word_repetition_ratio":0.0,"special_character_ratio":0.31399748,"punctuation_ratio":0.17791411,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98926437,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-11T20:58:24Z\",\"WARC-Record-ID\":\"<urn:uuid:714fd412-c532-4611-b90f-32a710a2ddd7>\",\"Content-Length\":\"14334\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7ac0d553-8352-4151-af8d-c24c2325d830>\",\"WARC-Concurrent-To\":\"<urn:uuid:ba024917-e32c-4598-9804-276d406539ce>\",\"WARC-IP-Address\":\"34.228.136.164\",\"WARC-Target-URI\":\"https://discuss.codechef.com/t/regarding-codechef-problems/376\",\"WARC-Payload-Digest\":\"sha1:4B6IBJRBWYR5E3KIMHCN2TDCRA5QJM2C\",\"WARC-Block-Digest\":\"sha1:XJJDGLH3CG2YLNQZUXPB5CQOPDENINW6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655937797.57_warc_CC-MAIN-20200711192914-20200711222914-00265.warc.gz\"}"}
https://web2.0calc.com/questions/probability_30746	[ "+0\n\n# Probability\n\n0\n33\n1\n\nA positive integer divisor of 11 is chosen at random. What is the probability that this divisor is prime? Express your answer as a common fraction.\n\nJul 12, 2022\n\nIt is $$\\frac{1}{2},$$ since 11 has only two divisors, 1 and 11, and 11 is prime while 1 is not." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.97544366,"math_prob":0.99082696,"size":244,"snap":"2022-27-2022-33","text_gpt3_token_len":67,"char_repetition_ratio":0.1375,"word_repetition_ratio":0.0,"special_character_ratio":0.29918033,"punctuation_ratio":0.12727273,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98267865,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-10T18:16:09Z\",\"WARC-Record-ID\":\"<urn:uuid:9d5a1be2-403d-4332-9493-5a7eb991ea9a>\",\"Content-Length\":\"20878\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dbff4d9b-d622-4da9-860e-6d3a3581c9d6>\",\"WARC-Concurrent-To\":\"<urn:uuid:2f5fc708-50e3-4606-a9ff-35db09fda985>\",\"WARC-IP-Address\":\"49.12.23.161\",\"WARC-Target-URI\":\"https://web2.0calc.com/questions/probability_30746\",\"WARC-Payload-Digest\":\"sha1:3LB3SWAG32E3KH6AHITIUZNEOSDOQG2G\",\"WARC-Block-Digest\":\"sha1:ECSJAVKU3NY2JOFDZISHEG3BZDVROWNZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571198.57_warc_CC-MAIN-20220810161541-20220810191541-00185.warc.gz\"}"}
https://www.teachoo.com/8083/2629/Example-13/category/Addition-of-decimal-numbers/	[ "Chapter 8 Class 6 Decimals\nConcept wise", null, "", null, "Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class\n\n### Transcript\n\nExample 3 Samson travelled 5 km 52 m by bus, 2 km 265 m by car and the rest 1km 30 m he walked. How much distance did he travel in all? Distance travelled by bus = 5 km 52 m Distance travelled by car = 2 km 265 m Distance travelled by foot = 1 km 30 m Total distance travelled = 5 km 52 m + 2 km 265 m + 1 km 30 m = (5 km + 52 m) + (2 km + 265 m) + (1 km + 30 m) = (5 km + 2 km + 1 km) + (52 m + 265 m + 30 m) = 8 km + 347 m = 8 km 347 m = 8 km 347 m Converting to decimals = 8 km + 347 m = 8 km + 347 × 1 m = 8 km + 347 × 1/1000 km = 8 km + 347/1000 km = (8+347/1000) km = ((8 × 1000 + 347)/1000) km = ((8000 + 347)/1000) km = 8347/1000 km = 8.347 km ∴ Total distance = 8.347 km", null, "" ]	[ null, "https://d1avenlh0i1xmr.cloudfront.net/3eeb9bd2-7fbe-410b-a205-b22cc0e3ec59/slide28.jpg", null, "https://d1avenlh0i1xmr.cloudfront.net/affefde0-b93c-4808-951c-2f85de4f9ff9/slide29.jpg", null, "https://www.teachoo.com/static/misc/Davneet_Singh.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7983591,"math_prob":1.0000052,"size":854,"snap":"2023-14-2023-23","text_gpt3_token_len":338,"char_repetition_ratio":0.16705883,"word_repetition_ratio":0.14418605,"special_character_ratio":0.5210773,"punctuation_ratio":0.036458332,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99912244,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,5,null,5,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-01T15:51:49Z\",\"WARC-Record-ID\":\"<urn:uuid:713638bb-04ee-494d-abb9-be17bb8f0752>\",\"Content-Length\":\"153096\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:425e78a1-2e2e-4e0b-95b5-35cb41ce20a4>\",\"WARC-Concurrent-To\":\"<urn:uuid:8067b076-2554-4c58-8fd6-4bc69a88460b>\",\"WARC-IP-Address\":\"104.21.90.10\",\"WARC-Target-URI\":\"https://www.teachoo.com/8083/2629/Example-13/category/Addition-of-decimal-numbers/\",\"WARC-Payload-Digest\":\"sha1:IKEINFUNKK4EDABUMOU26GJUVYNC7WR7\",\"WARC-Block-Digest\":\"sha1:EJYQGMK6NUHROLYAAMPRSAETXLJJHH47\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224647895.20_warc_CC-MAIN-20230601143134-20230601173134-00035.warc.gz\"}"}
https://inches-to-mm.appspot.com/pl/1281-cal-na-milimetr.html	[ "Inches To Mm\n\n# 1281 in to mm1281 Inch to Millimeters\n\nin\n=\nmm\n\n## How to convert 1281 inch to millimeters?\n\n 1281 in * 25.4 mm = 32537.4 mm 1 in\nA common question is How many inch in 1281 millimeter? And the answer is 50.4330708662 in in 1281 mm. Likewise the question how many millimeter in 1281 inch has the answer of 32537.4 mm in 1281 in.\n\n## How much are 1281 inches in millimeters?\n\n1281 inches equal 32537.4 millimeters (1281in = 32537.4mm). Converting 1281 in to mm is easy. Simply use our calculator above, or apply the formula to change the length 1281 in to mm.\n\n## Convert 1281 in to common lengths\n\nUnitUnit of length\nNanometer32537400000.0 nm\nMicrometer32537400.0 µm\nMillimeter32537.4 mm\nCentimeter3253.74 cm\nInch1281.0 in\nFoot106.75 ft\nYard35.5833333334 yd\nMeter32.5374 m\nKilometer0.0325374 km\nMile0.020217803 mi\nNautical mile0.0175687905 nmi\n\n## What is 1281 inches in mm?\n\nTo convert 1281 in to mm multiply the length in inches by 25.4. The 1281 in in mm formula is [mm] = 1281 * 25.4. Thus, for 1281 inches in millimeter we get 32537.4 mm.\n\n## 1281 Inch Conversion Table", null, "## Alternative spelling\n\n1281 Inch to Millimeters, 1281 Inch in Millimeters, 1281 Inches to Millimeter, 1281 Inches in mm, 1281 Inch to Millimeter, 1281 Inch in Millimeter, 1281 in in Millimeter, 1281 in in Millimeters," ]	[ null, "https://inches-to-mm.appspot.com/image/1281.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82373196,"math_prob":0.9107991,"size":914,"snap":"2023-40-2023-50","text_gpt3_token_len":294,"char_repetition_ratio":0.24395604,"word_repetition_ratio":0.0121951215,"special_character_ratio":0.392779,"punctuation_ratio":0.15270936,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9733938,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-10T02:19:57Z\",\"WARC-Record-ID\":\"<urn:uuid:feebb7a2-0039-4c7e-9a4f-aa24ee8a2c8e>\",\"Content-Length\":\"28900\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6a8344ec-bd9b-4ad8-bb17-4d499136ecb1>\",\"WARC-Concurrent-To\":\"<urn:uuid:d8677866-810a-42b1-a74c-34e5f4ee819a>\",\"WARC-IP-Address\":\"172.253.63.153\",\"WARC-Target-URI\":\"https://inches-to-mm.appspot.com/pl/1281-cal-na-milimetr.html\",\"WARC-Payload-Digest\":\"sha1:F77QNPQVIMTPEF7DZYVXIIAI5HA4TVMA\",\"WARC-Block-Digest\":\"sha1:WGCN23T5MMF76HQ7Z3SUR3RP27ALIPIQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100989.75_warc_CC-MAIN-20231209233632-20231210023632-00533.warc.gz\"}"}
http://ixtrieve.fh-koeln.de/birds/litie/document/27790	[ "# Document (#27790)\n\nAuthor\nRinn, R.\nWerner, C.\nTitle\n¬Der Neuerscheinungsdienst Der Deutschen Bibliothek : Eine Bilanz nach eineinhalb Jahren\nSource\nDialog mit Bibliotheken. 16(2004) H.2, S.47-51\nYear\n2004\nAbstract\nMit der Auslieferung der ersten Ausgabe im Januar 2003 löste der Neuerscheinungsdienst (ND) Der Deutschen Bibliothek den CIP-Dienst ab. Über die Gründe für die Neukonzeption der Dienstleistungen bezüglich der Vorankündigungen, über die Ziele und die Funktion des neuen Dienstes sowie über seine ersten Ausgaben wurde bereits im Detail berichtet\". Nur soviel sei in Erinnerung gerufen: Die Deutsche Bibliothek bietet neben ihren anderen Dienstleistungen nun je einen auf die jeweilige Funktion zugeschnittenen Spezialdienst für Erwerbung bzw. Katalogisierung an: Für Erwerbungszwecke ist dies der Neuerscheinungsdienst, der die Verlegermeldungen anzeigt, die parallel dazu auch in das VLB aufgenommen werden. Die Titeldaten der Verleger werden in Der Deutschen Bibliothek mit Sachgruppen versehen, ansonsten aber nicht weiter bearbeitet. Sie enthalten keine hierarchischen Verknüpfungen und ihre Personen- oder Körperschaftseintragungen sind nicht mit den entsprechenden Normdateien PND und GKD verknüpft. Die Daten sind insoweit standardisiert, als dies von dem für die VLB-Meldungen verwendeten ONIX-Format verlangt wird und sie dem Regelwerk des VLB entsprechen, und unter bibliothekarischen Gesichtspunkten teilweise von heterogener Qualität. Für Katalogisierungszwecke werden natürlich weiterhin die Daten der Reihen A, B und C der Deutschen Nationalbibliografie für Eigenkatalogisierung angeboten, deren »autopsierte« Daten nationalbibliografisch autorisiert sind und als endgültige Katalogisate ohne weitere Korrekturen übernommen werden können. Diese beiden Dienste ergänzen sich gegenseitig und sie beide, nicht der Neuerscheinungsdienst allein, sind als funktionale Nachfolger des CIP-Dienstes anzusehen. Um die Nutzbarkeit der beiden Dienste weiter zu verbessern, hat Die Deutsche Bibliothek in den vergangenen eineinhalb Jahren eine ganze Reihe von Maßnahmen ergriffen, auf die hier näher eingegangen werden soll. Dabei sind vor allem auch Anregungen und Kritik berücksichtigt worden, die verschiedene Dienstleistungsbezieher zu den ersten Ausgaben des Neuerscheinungsdienstes dankenswerterweise geäußert haben. Die wichtigsten Einzelpunkte sind im Folgenden vorangestellt. Die statistischen Angaben beziehen sich auf das Jahr 2003, in dem insgesamt ca. 94.000 Titelmeldungen im Neuerscheinungsdienst angezeigt worden sind. - Bereits in der Deutschen Nationalbibliografie angezeigte Titel werden nochmals im ND angezeigt: Durch einen Dublettencheck werden seit März 2003 bereits in der Datenbank Der Deutschen Bibliothek vorhandene Titel erkannt und nicht mehr in den ND übernommen. - Es werden vermehrt Titel ausländischer Verlage ohne deutschen Verlagssitz angezeigt: So genannte NSG-Titel, d. h. Titel, die nicht in das Sammelgebiet Der Deutschen Bibliothek gehören, werden nicht übernommen, soweit sie als solche erkennbar sind. Dazu zählen auch alle Titel von ausländischen Verlagen ohne deutschen Verlagssitz. Zur Information sei erwähnt, dass von den 2003 im ND angezeigten Titeln 96 % eine ISBN mit der Länderkennzeichnung »3-« hatten.\nTheme\nBibliographie\nObject\nDNB-ND\nCIP\nLocation\nD\n\n## Similar documents (author)\n\n1. Rinn, R.: Vom CIP- zum Neuerscheinungsdienst (2003) 2.14\n```2.142108 = sum of:\n2.142108 = product of:\n4.284216 = sum of:\n4.284216 = weight(author_txt:rinn in 992) [ClassicSimilarity], result of:\n4.284216 = score(doc=992,freq=1.0), product of:\n0.7710363 = queryWeight, product of:\n1.1003704 = boost\n8.890302 = idf(docFreq=15, maxDocs=42740)\n0.07881691 = queryNorm\n5.5564384 = fieldWeight in 992, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.890302 = idf(docFreq=15, maxDocs=42740)\n0.625 = fieldNorm(doc=992)\n0.5 = coord(1/2)\n```\n2. Rinn, R.: ¬Die Normdatei für Personennamen : PND (1995) 2.14\n```2.142108 = sum of:\n2.142108 = product of:\n4.284216 = sum of:\n4.284216 = weight(author_txt:rinn in 2194) [ClassicSimilarity], result of:\n4.284216 = score(doc=2194,freq=1.0), product of:\n0.7710363 = queryWeight, product of:\n1.1003704 = boost\n8.890302 = idf(docFreq=15, maxDocs=42740)\n0.07881691 = queryNorm\n5.5564384 = fieldWeight in 2194, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.890302 = idf(docFreq=15, maxDocs=42740)\n0.625 = fieldNorm(doc=2194)\n0.5 = coord(1/2)\n```\n3. Rinn, R.: ¬Die Personennamendatei (PND) (1993) 2.14\n```2.142108 = sum of:\n2.142108 = product of:\n4.284216 = sum of:\n4.284216 = weight(author_txt:rinn in 4356) [ClassicSimilarity], result of:\n4.284216 = score(doc=4356,freq=1.0), product of:\n0.7710363 = queryWeight, product of:\n1.1003704 = boost\n8.890302 = idf(docFreq=15, maxDocs=42740)\n0.07881691 = queryNorm\n5.5564384 = fieldWeight in 4356, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.890302 = idf(docFreq=15, maxDocs=42740)\n0.625 = fieldNorm(doc=4356)\n0.5 = coord(1/2)\n```\n4. Rinn, R.: ¬Das Projekt Personennamendatei : PND-Projekt (1994) 2.14\n```2.142108 = sum of:\n2.142108 = product of:\n4.284216 = sum of:\n4.284216 = weight(author_txt:rinn in 6568) [ClassicSimilarity], result of:\n4.284216 = score(doc=6568,freq=1.0), product of:\n0.7710363 = queryWeight, product of:\n1.1003704 = boost\n8.890302 = idf(docFreq=15, maxDocs=42740)\n0.07881691 = queryNorm\n5.5564384 = fieldWeight in 6568, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.890302 = idf(docFreq=15, maxDocs=42740)\n0.625 = fieldNorm(doc=6568)\n0.5 = coord(1/2)\n```\n5. Rinn, R.: ¬Die überregionale Normdatei für Personennamen (PND) : Bericht zum Projektstand September 1995 (1995) 2.14\n```2.142108 = sum of:\n2.142108 = product of:\n4.284216 = sum of:\n4.284216 = weight(author_txt:rinn in 2771) [ClassicSimilarity], result of:\n4.284216 = score(doc=2771,freq=1.0), product of:\n0.7710363 = queryWeight, product of:\n1.1003704 = boost\n8.890302 = idf(docFreq=15, maxDocs=42740)\n0.07881691 = queryNorm\n5.5564384 = fieldWeight in 2771, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.890302 = idf(docFreq=15, maxDocs=42740)\n0.625 = fieldNorm(doc=2771)\n0.5 = coord(1/2)\n```\n\n## Similar documents (content)\n\n1. 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http://dronesuavreport.com/2019/09/21/how-many-watts-does-a-drone-use/	[ "## How many watts does a drone use?\n\nDrones these days, arrive in a variety of different sizes. Obviously, it is the new technology, and it has been made and modified into so many different models. You can get a drone in whatever size you like. How much power do they use? Here are some examples:\n\n1. A Mavic Air uses 44 Watts/hour.\n\n2. A Mavic 2 uses 59 watts/hour.\n\n3. A DJI Spark uses 17 watts/hour.\n\n4. Holy Stone HS700D 21 watts/hour.\n\nOne thing that is worth noticing is that the flying time is directly proportional to the size of the drone that you have. The bigger the drone, the more is its flying time. However, the smaller drones cannot even last five minutes in the air sometimes.\n\nOn the other hand, the bigger ones are capable of going half an hour in the air, easily. Why do you think, that a larger, and more probably a heavier drone would last longer in the air? To understand this phenomenon, we would have to probe deeper into the science of the drones.\n\nWatt is the unit of power. How do you estimate the power that is needed by a drone to hover in the air? Let’s suppose you have a drone with the rotors that spin. It doesn’t matter how many rotors you have. They could be one, or way more than one as well.\n\nWhat actually matters is that these rotors function by pushing the stationary rotors above them, downward. When the momentum of the air is increased, the rotor will exert a force on the air, and the air will push back on to the rotor. The drone will only be capable of hovering when the weight of the drone equals the weight of the air force.\n\nAirspeed and size of rotors are directly proportional to each other. So, increasing the size means that you would be increasing the power in watts that are being used by the drone.\n\nThis is an important phenomenon when you want to determine the power, which can be expressed by using the following expression.\n\nFrom this expression, we can deduce that the more area a rotor has, the more flying time it has. The power is going to depend upon the area of the rotors of the drone.\n\nPower also depends upon the air pressure. The drone will use less power if its speed is low. On the other hand, if you make it fly at high speed, it will use more power.\n\nPower is defined as the work done in a unit time. It is basically the rate at which a drone will use energy.\n\nThe power will be in watts if you will measure the energy in joules and time interval in seconds.\n\nIn a nutshell, a big drone is going to use more watts, and a smaller drone is going to" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9507389,"math_prob":0.9303913,"size":2469,"snap":"2019-51-2020-05","text_gpt3_token_len":566,"char_repetition_ratio":0.1444219,"word_repetition_ratio":0.0,"special_character_ratio":0.22478737,"punctuation_ratio":0.11111111,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9715049,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-12T06:34:26Z\",\"WARC-Record-ID\":\"<urn:uuid:d73cbce2-02d6-48a7-b989-5232f1cb3942>\",\"Content-Length\":\"27775\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:94794f0d-4f83-460c-891a-68ecc12c224c>\",\"WARC-Concurrent-To\":\"<urn:uuid:f08fe7b1-6680-4b1b-9386-1685099b615e>\",\"WARC-IP-Address\":\"146.66.113.49\",\"WARC-Target-URI\":\"http://dronesuavreport.com/2019/09/21/how-many-watts-does-a-drone-use/\",\"WARC-Payload-Digest\":\"sha1:V2423QYKRADZH6CFXBJWHEPWQ2SSGQE3\",\"WARC-Block-Digest\":\"sha1:GH4HQOIXVLX4ZILYT3L2YT5DAECF5NPT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540537212.96_warc_CC-MAIN-20191212051311-20191212075311-00369.warc.gz\"}"}
https://dev.to/robertobutti/tdd-with-pestphp-13c3	[ "## DEV Community", null, "Roberto B.\n\nPosted on • Updated on\n\n# TDD with PestPHP\n\nTest Driven Development (TDD) is a software development practice.\n\nThe typical flow is:\n\n• writing the test cases starting from the requirements;\n• writing the code that solves the tests;\n• refactoring the code to pass all tests.\n\nWriting tests before the code, allows the developer to:\n\n• focus on requirements;\n• see if some requirements are missing or not detailed;\n• focus about the edge cases (empty params, invalid params etc). This is very useful for thinking outside the \"happy path\".\n\nThis approach helps and supports the developer for creating a \"better and more maintainable code\" because it forces the developer making the code/functions testable from the beginning:\n\n• splitting code in smaller and easier \"pieces\";\n• favoring the single responsibility approach;\n• favoring the code isolation (or encapsulation).\n\nBut, yes, fewer words and more code.\n\nFor our example, we are going to create a class with a static function for calculating the mean (the average).\n\n## Create your project from scratch\n\nCreate a new directory and jump into the new directory:\n\n``````mkdir tdd\ncd tdd\n``````\n\n## Install PestPHP\n\nNow you can install the testing framework PestPHP.\nPestPHP is built on top of the mature PHPUnit, and it aims to provide a nice and smooth interface for the developer that want to write tests.\n\n``````composer require pestphp/pest --dev --with-all-dependencies\nmkdir tests\n./vendor/bin/pest --init\n``````\n\nI'm going to skip the right configuration for PSR-4 autoloading, so I will create a file and include that in the test file via require\n\n## Create Stat class\n\nCreate your \"stat.php\" file with the class \"Stat\":\n\n``````<?php\n\nclass Stat\n{\n}\n``````\n\nYes, I know, the class is empty. Before to write the method and the implementation, you need to create the new test file.\nCreate \"StatTest.php\" file in the \"tests\" directory. The convention to write tests in \"tests\" directory and naming the file with \"Test.php\" suffix, allows PestPHP to automatically find the right tests.\n\n## Create the test file\n\nFile \"tests/StatTest.php\":\n\n``````<?php\nrequire('./stat.php');\n\ntest('test mean', function () {\nexpect(Stat::mean([1,2,3]))->toEqual(2);\n});\n``````\n\nBefore implementing the \"mean()\" method, we are expecting (expect()) that calling \"Stat::mean()\" with an array [1,2,3] as input parameter, it returns a value equal \"2\".\n\nThis is the happy path.\n\nBut I want also to think about \"bad scenarios\" for example if the user will call the method \"Stat::mean()\" with an empty array, I expect to retrieve a \"null\" value\n\n``````test('test mean with empty data', function () {\nexpect(Stat::mean([]))->toBeNull();\n});\n``````\n\nCalling \"Stat::mean()\" with a non array parameter (for example an integer) I expect to have an exception:\n\n``````test('test mean with not array', function () {\nexpect(\nfn () => Stat::mean(42)\n)->toThrow(TypeError::class);\n\n});\n``````\n\nIf you run the tests:\n\n``````./vendor/bin/pest\n``````\n\nYou will receive errors like \"undefined method\" because the class Stat is still empty, and it doesn't implement any methods.\n\n``````Call to undefined method Stat::mean()\n``````\n\nSo now, jump into \"stat.php\" file and implement the method mean() that for now will always return the value 0.0.\n\n``````<?php\n\nclass Stat\n{\npublic static function mean(array \\$data): ?float\n{\nreturn 0.0;\n}\n}\n``````\n\nIf you run the test now, you will still receive errors.", null, "The first one is about the assertion that expect to receive 2 as value. The current implementation is returning 0.0:\n\n`````` • Tests\\StatTest > test mean() with array parameter\nFailed asserting that 0.0 matches expected 2.\n``````\n\nThe second one is about the assertion that expect to receive \"null\" if the input parameter is an empty array. The current implementation is returning 0.0:\n\n`````` • Tests\\StatTest > test mean() with empty array parameter\nFailed asserting that 0.0 is null.\n``````\n\nThe test are failing, so now you need to focus on your implementation, on your code and implement the logic and making the test \"green\".\n\nIn your \"stat.php\" file in the \"mean()\" method, try to implement the mean:\n\n• count the elements in the array;\n• calculate the sum of the elements of the array;\n• divide the sum with the count.\n``````<?php\n\nclass Stat\n{\npublic static function mean(array \\$data): ?float\n{\n\\$count = count(\\$data);\n\\$sum = array_sum(\\$data);\nreturn \\$sum / \\$count;\n}\n}\n``````\n\nIf you execute the tests, the only one that fails is the test that calls \"mean()\" method with empty array and raises a \"Division by zero exception\".", null, "\"Division by zero\" is raised because the count is equal to 0. So, you need to check the length of the array, and if it is 0, you need to return \"null\".\n\n``````<?php\n\nclass Stat\n{\npublic static function mean(array \\$data): ?float\n{\n\\$count = count(\\$data);\nif ( ! \\$count) {\nreturn null;\n}\n\\$sum = array_sum(\\$data);\nreturn \\$sum / \\$count;\n}\n}\n``````\n\nIf you execute the tests ...", null, "All GREEN for us!\n\nSo, we:\n\n• Wrote the test focusing on happy path and \"bad scenarios\";\n• Wrote the code iteratively until all test passes.\n\nAnd you, are you applying TDD practice in your development process?", null, "" ]	[ null, "https://res.cloudinary.com/practicaldev/image/fetch/s--Ya9UksYX--/c_imagga_scale,f_auto,fl_progressive,h_420,q_auto,w_1000/https://dev-to-uploads.s3.amazonaws.com/uploads/articles/77lj8rfwx633k9g8vqft.png", null, "https://res.cloudinary.com/practicaldev/image/fetch/s--qYzPKvV0--/c_limit%2Cf_auto%2Cfl_progressive%2Cq_auto%2Cw_880/https://dev-to-uploads.s3.amazonaws.com/uploads/articles/7git7mxw2uyt8t2f0135.png", null, "https://res.cloudinary.com/practicaldev/image/fetch/s--KzckTukZ--/c_limit%2Cf_auto%2Cfl_progressive%2Cq_auto%2Cw_880/https://dev-to-uploads.s3.amazonaws.com/uploads/articles/eekf5e793w93orgkplfx.png", null, "https://res.cloudinary.com/practicaldev/image/fetch/s--s6tArFuX--/c_limit%2Cf_auto%2Cfl_progressive%2Cq_auto%2Cw_880/https://dev-to-uploads.s3.amazonaws.com/uploads/articles/zt94mj34hc78kbgifroc.png", null, "https://res.cloudinary.com/practicaldev/image/fetch/s--PpzoLqR---/c_fill,f_auto,fl_progressive,h_50,q_auto,w_50/https://dev-to-uploads.s3.amazonaws.com/uploads/user/profile_image/589/jHw75m_5.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8236357,"math_prob":0.6230608,"size":5144,"snap":"2023-14-2023-23","text_gpt3_token_len":1214,"char_repetition_ratio":0.12801556,"word_repetition_ratio":0.08802817,"special_character_ratio":0.2505832,"punctuation_ratio":0.14866205,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96624416,"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\":\"2023-06-05T19:46:18Z\",\"WARC-Record-ID\":\"<urn:uuid:3d7abe35-7d06-42b0-a416-f5c165800fbe>\",\"Content-Length\":\"104474\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:81fb7706-5363-492b-b287-1841651121c2>\",\"WARC-Concurrent-To\":\"<urn:uuid:331a86a7-6a38-48ae-803e-254bde650de8>\",\"WARC-IP-Address\":\"151.101.66.217\",\"WARC-Target-URI\":\"https://dev.to/robertobutti/tdd-with-pestphp-13c3\",\"WARC-Payload-Digest\":\"sha1:UAROKEO2FS4TCYOIHRI34OKGOOXLFODM\",\"WARC-Block-Digest\":\"sha1:MFJWKZ3ZQXSCX35AL5VA5ZGC3J2P5ESW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224652161.52_warc_CC-MAIN-20230605185809-20230605215809-00592.warc.gz\"}"}
https://www.futuristicmath.com/grades.html	[ "# Free Math Activities by Grade For Kids\n\nFree math activities by grade for kids, math quizzes online, math games online, math board games, math flash cards, math puzzles and more. This page contains links to varied math resources which have been arranged by grade.\n\n### Pre-K / Kindergarten Math", null, "Pre-K and Kindergarten math: games, quizzes, worksheets & more.", null, "1st grade math activities: games, quizzes, worksheets & more.", null, "Practice 2nd grade math: games, quizzes, worksheets & more.", null, "Third grade math: games, quizzes, worksheets & more.", null, "4th grade math for kids : games, quizzes, worksheets & more.", null, "5th grade math : games, quizzes, worksheets & more.", null, "", null, "" ]	[ null, "https://www.futuristicmath.com/thumbs/img/kindergarten.png", null, "https://www.futuristicmath.com/thumbs/img/1st-grade.png", null, "https://www.futuristicmath.com/thumbs/img/2nd-grade.png", null, "https://www.futuristicmath.com/thumbs/img/3rd-grade.png", null, "https://www.futuristicmath.com/thumbs/img/4th-grade.png", null, "https://www.futuristicmath.com/thumbs/img/5th-grade.png", null, "https://www.futuristicmath.com/thumbs/img/6th-grade.png", null, "https://www.futuristicmath.com/thumbs/img/7th-grade.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9115201,"math_prob":0.88835275,"size":820,"snap":"2022-27-2022-33","text_gpt3_token_len":180,"char_repetition_ratio":0.23406863,"word_repetition_ratio":0.1627907,"special_character_ratio":0.23048781,"punctuation_ratio":0.25609756,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9851743,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-12T12:01:56Z\",\"WARC-Record-ID\":\"<urn:uuid:fca77937-6237-4bcb-91e7-40dc460418ca>\",\"Content-Length\":\"19309\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:05c63705-dea9-4b10-a8e2-c89e04e8493f>\",\"WARC-Concurrent-To\":\"<urn:uuid:a6ebb389-39f2-4b73-a6ec-6985c0fb3a40>\",\"WARC-IP-Address\":\"69.167.170.47\",\"WARC-Target-URI\":\"https://www.futuristicmath.com/grades.html\",\"WARC-Payload-Digest\":\"sha1:HPBF5QMJM7I4CF6AXUOYWV3KOKXON6ZO\",\"WARC-Block-Digest\":\"sha1:SKWKMX2CEJOSRTICPLG2OO57D7UF3Y2Z\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571692.3_warc_CC-MAIN-20220812105810-20220812135810-00434.warc.gz\"}"}
https://wangdoc.com/es6/destructuring.html	[ "# 变量的解构赋值\n\n## 数组的解构赋值 #\n\n### 基本用法 #\n\nES6 允许按照一定模式，从数组和对象中提取值，对变量进行赋值，这被称为解构（Destructuring）。\n\n``````let a = 1;\nlet b = 2;\nlet c = 3;\n``````\n\nES6 允许写成下面这样。\n\n``````let [a, b, c] = [1, 2, 3];\n``````\n\n``````let [foo, [[bar], baz]] = [1, [, 3]];\nfoo // 1\nbar // 2\nbaz // 3\n\nlet [ , , third] = [\"foo\", \"bar\", \"baz\"];\nthird // \"baz\"\n\nlet [x, , y] = [1, 2, 3];\nx // 1\ny // 3\n\nlet [head, ...tail] = [1, 2, 3, 4];\ntail // [2, 3, 4]\n\nlet [x, y, ...z] = ['a'];\nx // \"a\"\ny // undefined\nz // []\n``````\n\n``````let [foo] = [];\nlet [bar, foo] = ;\n``````\n\n``````let [x, y] = [1, 2, 3];\nx // 1\ny // 2\n\nlet [a, [b], d] = [1, [2, 3], 4];\na // 1\nb // 2\nd // 4\n``````\n\n``````// 报错\nlet [foo] = 1;\nlet [foo] = false;\nlet [foo] = NaN;\nlet [foo] = undefined;\nlet [foo] = null;\nlet [foo] = {};\n``````\n\n``````let [x, y, z] = new Set(['a', 'b', 'c']);\nx // \"a\"\n``````\n\n``````function* fibs() {\nlet a = 0;\nlet b = 1;\nwhile (true) {\nyield a;\n[a, b] = [b, a + b];\n}\n}\n\nlet [first, second, third, fourth, fifth, sixth] = fibs();\nsixth // 5\n``````\n\n### 默认值 #\n\n``````let [foo = true] = [];\nfoo // true\n\nlet [x, y = 'b'] = ['a']; // x='a', y='b'\nlet [x, y = 'b'] = ['a', undefined]; // x='a', y='b'\n``````\n\n``````let [x = 1] = [undefined];\nx // 1\n\nlet [x = 1] = [null];\nx // null\n``````\n\n``````function f() {\nconsole.log('aaa');\n}\n\nlet [x = f()] = ;\n``````\n\n``````let x;\nif ( === undefined) {\nx = f();\n} else {\nx = ;\n}\n``````\n\n``````let [x = 1, y = x] = []; // x=1; y=1\nlet [x = 1, y = x] = ; // x=2; y=2\nlet [x = 1, y = x] = [1, 2]; // x=1; y=2\nlet [x = y, y = 1] = []; // ReferenceError: y is not defined\n``````\n\n## 对象的解构赋值 #\n\n### 简介 #\n\n``````let { foo, bar } = { foo: 'aaa', bar: 'bbb' };\nfoo // \"aaa\"\nbar // \"bbb\"\n``````\n\n``````let { bar, foo } = { foo: 'aaa', bar: 'bbb' };\nfoo // \"aaa\"\nbar // \"bbb\"\n\nlet { baz } = { foo: 'aaa', bar: 'bbb' };\nbaz // undefined\n``````\n\n``````let {foo} = {bar: 'baz'};\nfoo // undefined\n``````\n\n``````// 例一\nlet { log, sin, cos } = Math;\n\n// 例二\nconst { log } = console;\nlog('hello') // hello\n``````\n\n``````let { foo: baz } = { foo: 'aaa', bar: 'bbb' };\nbaz // \"aaa\"\n\nlet obj = { first: 'hello', last: 'world' };\nlet { first: f, last: l } = obj;\nf // 'hello'\nl // 'world'\n``````\n\n``````let { foo: foo, bar: bar } = { foo: 'aaa', bar: 'bbb' };\n``````\n\n``````let { foo: baz } = { foo: 'aaa', bar: 'bbb' };\nbaz // \"aaa\"\nfoo // error: foo is not defined\n``````\n\n``````let obj = {\np: [\n'Hello',\n{ y: 'World' }\n]\n};\n\nlet { p: [x, { y }] } = obj;\nx // \"Hello\"\ny // \"World\"\n``````\n\n``````let obj = {\np: [\n'Hello',\n{ y: 'World' }\n]\n};\n\nlet { p, p: [x, { y }] } = obj;\nx // \"Hello\"\ny // \"World\"\np // [\"Hello\", {y: \"World\"}]\n``````\n\n``````const node = {\nloc: {\nstart: {\nline: 1,\ncolumn: 5\n}\n}\n};\n\nlet { loc, loc: { start }, loc: { start: { line }} } = node;\nline // 1\nloc // Object {start: Object}\nstart // Object {line: 1, column: 5}\n``````\n\n``````let obj = {};\nlet arr = [];\n\n({ foo: obj.prop, bar: arr } = { foo: 123, bar: true });\n\nobj // {prop:123}\narr // [true]\n``````\n\n``````// 报错\nlet {foo: {bar}} = {baz: 'baz'};\n``````\n\n``````const obj1 = {};\nconst obj2 = { foo: 'bar' };\nObject.setPrototypeOf(obj1, obj2);\n\nconst { foo } = obj1;\nfoo // \"bar\"\n``````\n\n### 默认值 #\n\n``````var {x = 3} = {};\nx // 3\n\nvar {x, y = 5} = {x: 1};\nx // 1\ny // 5\n\nvar {x: y = 3} = {};\ny // 3\n\nvar {x: y = 3} = {x: 5};\ny // 5\n\nvar { message: msg = 'Something went wrong' } = {};\nmsg // \"Something went wrong\"\n``````\n\n``````var {x = 3} = {x: undefined};\nx // 3\n\nvar {x = 3} = {x: null};\nx // null\n``````\n\n### 注意点 #\n\n（1）如果要将一个已经声明的变量用于解构赋值，必须非常小心。\n\n``````// 错误的写法\nlet x;\n{x} = {x: 1};\n// SyntaxError: syntax error\n``````\n\n``````// 正确的写法\nlet x;\n({x} = {x: 1});\n``````\n\n（2）解构赋值允许等号左边的模式之中，不放置任何变量名。因此，可以写出非常古怪的赋值表达式。\n\n``````({} = [true, false]);\n({} = 'abc');\n({} = []);\n``````\n\n（3）由于数组本质是特殊的对象，因此可以对数组进行对象属性的解构。\n\n``````let arr = [1, 2, 3];\nlet {0 : first, [arr.length - 1] : last} = arr;\nfirst // 1\nlast // 3\n``````\n\n## 字符串的解构赋值 #\n\n``````const [a, b, c, d, e] = 'hello';\na // \"h\"\nb // \"e\"\nc // \"l\"\nd // \"l\"\ne // \"o\"\n``````\n\n``````let {length : len} = 'hello';\nlen // 5\n``````\n\n## 数值和布尔值的解构赋值 #\n\n``````let {toString: s} = 123;\ns === Number.prototype.toString // true\n\nlet {toString: s} = true;\ns === Boolean.prototype.toString // true\n``````\n\n``````let { prop: x } = undefined; // TypeError\nlet { prop: y } = null; // TypeError\n``````\n\n## 函数参数的解构赋值 #\n\n``````function add([x, y]){\nreturn x + y;\n}\n\n``````\n\n``````[[1, 2], [3, 4]].map(([a, b]) => a + b);\n// [ 3, 7 ]\n``````\n\n``````function move({x = 0, y = 0} = {}) {\nreturn [x, y];\n}\n\nmove({x: 3, y: 8}); // [3, 8]\nmove({x: 3}); // [3, 0]\nmove({}); // [0, 0]\nmove(); // [0, 0]\n``````\n\n``````function move({x, y} = { x: 0, y: 0 }) {\nreturn [x, y];\n}\n\nmove({x: 3, y: 8}); // [3, 8]\nmove({x: 3}); // [3, undefined]\nmove({}); // [undefined, undefined]\nmove(); // [0, 0]\n``````\n\n`undefined`就会触发函数参数的默认值。\n\n``````[1, undefined, 3].map((x = 'yes') => x);\n// [ 1, 'yes', 3 ]\n``````\n\n## 圆括号问题 #\n\n### 不能使用圆括号的情况 #\n\n（1）变量声明语句\n\n``````// 全部报错\nlet [(a)] = ;\n\nlet {x: (c)} = {};\nlet ({x: c}) = {};\nlet {(x: c)} = {};\nlet {(x): c} = {};\n\nlet { o: ({ p: p }) } = { o: { p: 2 } };\n``````\n\n（2）函数参数\n\n``````// 报错\nfunction f([(z)]) { return z; }\n// 报错\nfunction f([z,(x)]) { return x; }\n``````\n\n（3）赋值语句的模式\n\n``````// 全部报错\n({ p: a }) = { p: 42 };\n([a]) = ;\n``````\n\n``````// 报错\n[({ p: a }), { x: c }] = [{}, {}];\n``````\n\n### 可以使用圆括号的情况 #\n\n``````[(b)] = ; // 正确\n({ p: (d) } = {}); // 正确\n[(parseInt.prop)] = ; // 正确\n``````\n\n## 用途 #\n\n（1）交换变量的值\n\n``````let x = 1;\nlet y = 2;\n\n[x, y] = [y, x];\n``````\n\n（2）从函数返回多个值\n\n``````// 返回一个数组\n\nfunction example() {\nreturn [1, 2, 3];\n}\nlet [a, b, c] = example();\n\n// 返回一个对象\n\nfunction example() {\nreturn {\nfoo: 1,\nbar: 2\n};\n}\nlet { foo, bar } = example();\n``````\n\n（3）函数参数的定义\n\n``````// 参数是一组有次序的值\nfunction f([x, y, z]) { ... }\nf([1, 2, 3]);\n\n// 参数是一组无次序的值\nfunction f({x, y, z}) { ... }\nf({z: 3, y: 2, x: 1});\n``````\n\n（4）提取 JSON 数据\n\n``````let jsonData = {\nid: 42,\nstatus: \"OK\",\ndata: [867, 5309]\n};\n\nlet { id, status, data: number } = jsonData;\n\nconsole.log(id, status, number);\n// 42, \"OK\", [867, 5309]\n``````\n\n（5）函数参数的默认值\n\n``````jQuery.ajax = function (url, {\nasync = true,\nbeforeSend = function () {},\ncache = true,\ncomplete = function () {},\ncrossDomain = false,\nglobal = true,\n// ... more config\n} = {}) {\n// ... do stuff\n};\n``````\n\n（6）遍历 Map 结构\n\n``````const map = new Map();\nmap.set('first', 'hello');\nmap.set('second', 'world');\n\nfor (let [key, value] of map) {\nconsole.log(key + \" is \" + value);\n}\n// first is hello\n// second is world\n``````\n\n``````// 获取键名\nfor (let [key] of map) {\n// ...\n}\n\n// 获取键值\nfor (let [,value] of map) {\n// ...\n}\n``````\n\n（7）输入模块的指定方法\n\n``````const { SourceMapConsumer, SourceNode } = require(\"source-map\");\n``````" ]	[ null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.64083236,"math_prob":0.9996126,"size":9344,"snap":"2021-04-2021-17","text_gpt3_token_len":5704,"char_repetition_ratio":0.11884368,"word_repetition_ratio":0.11235217,"special_character_ratio":0.40764126,"punctuation_ratio":0.2438063,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99983627,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-23T07:50:59Z\",\"WARC-Record-ID\":\"<urn:uuid:2d8f6dbc-5014-4ec1-8a2a-cd3a2cc08512>\",\"Content-Length\":\"56774\",\"Content-Type\":\"application/http; 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https://socratic.org/questions/how-do-you-differentiate-y-sin-3x-cot-x-3-8	[ "How do you differentiate y = (sin(3x) + cot(x^3))^8?\n\nJul 31, 2015\n\n${y}^{'} = 24 \\cdot {\\left[\\sin \\left(3 x\\right) + \\cot \\left({x}^{3}\\right)\\right]}^{7} \\cdot \\left[\\cos \\left(3 x\\right) - {x}^{2} \\csc \\left({x}^{3}\\right)\\right]$\n\nExplanation:\n\nNotice that you can write your function $y$ as\n\n$y = {u}^{8}$, with $u = \\sin \\left(3 x\\right) + \\cot \\left({x}^{3}\\right)$\n\nThis means that you can use the power rule and the chain rule to differentiate $y$.\n\nFor a function $y$ that depends on a variable $u$, which in turn depends on a variable $x$, you can determine its derivative by using the chain rule\n\n$\\textcolor{b l u e}{\\frac{d}{\\mathrm{dx}} \\left(y\\right) = \\frac{d}{\\mathrm{du}} \\left(y\\right) \\cdot \\frac{d}{\\mathrm{dx}} \\left(u\\right)}$\n\nIn your case, this would be equivalent to\n\n${y}^{'} = \\frac{d}{\\mathrm{du}} \\left({u}^{8}\\right) \\cdot \\frac{d}{\\mathrm{dx}} \\left(u\\right)$\n\n${y}^{'} = 8 {u}^{7} \\cdot \\frac{d}{\\mathrm{dx}} {\\underbrace{\\left(\\sin \\left(3 x\\right) + \\cot \\left({x}^{3}\\right)\\right)}}_{\\textcolor{red}{a \\left(x\\right)}}$\n\nNow focus on differentiating function $\\textcolor{red}{a \\left(x\\right)}$, which can be written as\n\n$\\frac{d}{\\mathrm{dx}} \\left(a\\right) = \\frac{d}{\\mathrm{dx}} \\left(\\sin \\left(3 x\\right)\\right) + \\frac{d}{\\mathrm{dx}} \\left(\\cot \\left({x}^{3}\\right)\\right)$\n\nOnce again, use the chain rule to differentiate these functions. Remember that\n\n$\\frac{d}{\\mathrm{dx}} \\left(\\sin x\\right) = \\cos x$\n\nand that\n\n$\\frac{d}{\\mathrm{dx}} \\left(\\cot x\\right) = - {\\csc}^{2} x$\n\nThe first one will be\n\n$\\frac{d}{\\mathrm{dx}} \\left(\\sin {u}_{1}\\right) = \\frac{d}{{\\mathrm{du}}_{1}} \\cdot \\left(\\sin {u}_{1}\\right) \\cdot \\frac{d}{\\mathrm{dx}} \\left({u}_{1}\\right)$, where ${u}_{1} = 3 x$\n\n$\\frac{d}{\\mathrm{dx}} \\left(\\sin {u}_{1}\\right) = \\cos {u}_{1} \\cdot \\frac{d}{\\mathrm{dx}} \\left(3 x\\right)$\n\n$\\frac{d}{\\mathrm{dx}} \\left(\\sin \\left(3 x\\right)\\right) = 3 \\cos \\left(3 x\\right)$\n\nThe second one will be\n\n$\\frac{d}{\\mathrm{dx}} \\left(\\cot {u}_{2}\\right) = \\frac{d}{{\\mathrm{du}}_{2}} \\left(\\cot {u}_{2}\\right) \\cdot \\frac{d}{\\mathrm{dx}} \\left({u}_{2}\\right)$, where ${u}_{2} = {x}^{3}$\n\n$\\frac{d}{\\mathrm{dx}} \\left(\\cot {u}_{2}\\right) = - {\\csc}^{2} {u}_{2} \\cdot \\frac{d}{\\mathrm{dx}} \\left({x}^{3}\\right)$\n\n$\\frac{d}{\\mathrm{dx}} \\left(\\cot {u}_{2}\\right) = - {\\csc}^{2} \\left({x}^{3}\\right) \\cdot 3 {x}^{2}$\n\nPlug these back into your tager derivative to get\n\n${y}^{'} = 8 \\cdot {\\left[\\sin \\left(3 x\\right) + \\cot \\left({x}^{3}\\right)\\right]}^{7} \\cdot \\left(3 \\cos \\left(3 x\\right) - {\\csc}^{2} \\left({x}^{3}\\right) \\cdot 3 {x}^{2}\\right)$\n\n${y}^{'} = \\textcolor{g r e e n}{24 \\cdot {\\left[\\sin \\left(3 x\\right) + \\cot \\left({x}^{3}\\right)\\right]}^{7} \\cdot \\left[\\cos \\left(3 x\\right) - {x}^{2} \\csc \\left({x}^{3}\\right)\\right]}$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7450277,"math_prob":1.0000097,"size":1271,"snap":"2019-43-2019-47","text_gpt3_token_len":562,"char_repetition_ratio":0.13496448,"word_repetition_ratio":0.020100502,"special_character_ratio":0.4303698,"punctuation_ratio":0.036231883,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000079,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-15T02:02:13Z\",\"WARC-Record-ID\":\"<urn:uuid:60fe2a42-6171-4267-8011-b55b53990667>\",\"Content-Length\":\"36753\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:eda63b50-ec62-4bb7-9dec-b0fce448170c>\",\"WARC-Concurrent-To\":\"<urn:uuid:11959acc-d5a4-4e44-85cf-610b9cca5d73>\",\"WARC-IP-Address\":\"54.221.217.175\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-you-differentiate-y-sin-3x-cot-x-3-8\",\"WARC-Payload-Digest\":\"sha1:MP257XPNXYMSVJTWM4JP7BM5GQ7X7XY3\",\"WARC-Block-Digest\":\"sha1:YL5BYMZKEPAAHKZI6ZFDCFYDJNZWCDO6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986655735.13_warc_CC-MAIN-20191015005905-20191015033405-00016.warc.gz\"}"}
https://mathematica.stackexchange.com/questions/250503/control-evaluation-for-functional-constraints	[ "# Control evaluation for functional constraints\n\nI'm trying to understand how to use Mathematica to find a solution subject to constraints, where one of the constraints is specified as a predicate function. But I don't know how to control evaluation in order to use the predicate function as a condition.\n\nHere's the problem. I want to find three integers, $$a$$, $$b$$, and $$c$$, subject to these constraints:\n\n1. They sum to 70\n2. Each of the integer is greater than or equal to 15 and less than 30\n3. No digits from 1-9 appears twice if you consider all the digits in the squares of the integers.\n\nSo how do I approach this with Mathematica? I expect I can use FindInstance to find values under a set of constraints.\n\nI can express constraints 1 and 2 by setting variables equal to equations and inequalities:\n\neq = (a + b + c == 70);\ncs = 15 <= a <= 30 && 15 <= b <= 30 && 15 <= c <= 30;\n\n\nIn order to express the third constraint, I have defined this predicate function, which takes a list of numbers and returns true when they do not re-use digits:\n\nUniqueDigitsQ[xss_] := With[\n{nonzeros = Select[Flatten[Map[IntegerDigits,xss]] ,#1!=0&]},\nSortBy[Tally[nonzeros],Last][[-1,-1]] <2];\n\n\nSo I would like to be able to express the third constraint by saying:\n\ncs2 = UniqueDigitsQ[ {a^2,b^2,c^2} ];\n\n\nBut this fails, because the predicate function cannot handle symbolic argument.\n\nIs there a way to fix this problem by defining the function in such a way that it does not evaluate until the arguments are numeric? Or else, what is the right approach?\n\n• Welcome to Mathematica.SE! I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise. Jul 3, 2021 at 4:31\n\nClear[\"Global\"]\n\n\nFind all solutions for the first two criteria then select from those the ones that meet the third criteria.\n\nsol = Select[\nSolve[{a + b + c == 70, 15 <= a < 30, 15 <= b < 30,\n15 <= c < 30}, {a, b, c}, Integers],\n(digits = Flatten[IntegerDigits[{a^2, b^2, c^2}] /. #];\nLength[digits] == Length[Union@digits]) &]\n\n( {{a -> 19, b -> 23, c -> 28}, {a -> 19, b -> 28, c -> 23}, {a -> 23,\nb -> 19, c -> 28}, {a -> 23, b -> 28, c -> 19}, {a -> 28, b -> 19,\nc -> 23}, {a -> 28, b -> 23, c -> 19}} )\n\n\nThese are just permutations of {19, 23, 28}\n\nEDIT: A user-defined function must be used in an equation or an inequality. Consequently, modify your function to return a numeric value.\n\nUniqueDigitsQ[xss_?(VectorQ[#, NumericQ] &)] := With[\n{nonzeros = Select[Flatten[Map[IntegerDigits, xss]], #1 != 0 &]},\nBoole[SortBy[Tally[nonzeros], Last][[-1, -1]] < 2]];\n\nSolve[{a + b + c == 70, 15 <= a < 30, 15 <= b < 30, 15 <= c < 30,\nUniqueDigitsQ[{a^2, b^2, c^2}] == 1}, {a, b, c}, Integers]\n\n( {{a -> 19, b -> 23, c -> 28}, {a -> 19, b -> 28, c -> 23}, {a -> 23, b -> 19,\nc -> 28}, {a -> 23, b -> 28, c -> 19}, {a -> 28, b -> 19,\nc -> 23}, {a -> 28, b -> 23, c -> 19}} *)\n\n• Thanks! Is this a better way because Mathematica’s solvers work only with equations and inequalities, and not user-defined functions? Or is it that it’s dangerous to use substitutions with user-defined functions at all? Functions are obviously a basic form of abstraction, so I’d like to understand how to use them in this sort of case because usually it becomes untenable to express everything in an inline snippet of code. Jul 3, 2021 at 14:58\n• The solvers can work with user-defined functions. I just find the approach that I used simple and straightforward. You could also eliminate the redundancy of the permutations by adding a fourth constraint that the values are ordered, e.g., a < b < c`. Note that the third constraint eliminates any equality. Jul 3, 2021 at 15:07\n• Oh I quite agree your approach is simpler and better! I just remain curious how my kind of approach would work, since sometimes function-defined constraints will be unavoidable, and I wonder if the solver can work with them efficiently or if it can only work analytically with equations and then you have to brute force search over functions. Also, even if the solver could handle UDFs, I think it couldn’t handle mine because mine consumes symbolic arguments eagerly, before substitution rules are applied, unlike IntegerDigits apparently. This is the evaluation behavior I don’t understand. Jul 3, 2021 at 17:12" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8766827,"math_prob":0.9820447,"size":1482,"snap":"2023-40-2023-50","text_gpt3_token_len":374,"char_repetition_ratio":0.11840325,"word_repetition_ratio":0.007633588,"special_character_ratio":0.27530363,"punctuation_ratio":0.117056854,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99621737,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-22T12:34:08Z\",\"WARC-Record-ID\":\"<urn:uuid:dc1e7987-3946-422d-a673-44f3bae27cbb>\",\"Content-Length\":\"160592\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:41e94a66-5254-4052-a873-3eed8f546f08>\",\"WARC-Concurrent-To\":\"<urn:uuid:e27b240b-7b1f-4fa7-bad2-1058ec532c0c>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://mathematica.stackexchange.com/questions/250503/control-evaluation-for-functional-constraints\",\"WARC-Payload-Digest\":\"sha1:6FWW3YRNAJ4DFANWUXJ2N6EW4J3PYQGQ\",\"WARC-Block-Digest\":\"sha1:2LVW4PSK6PJ3CGZA55MW7FKU3435YWWN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506399.24_warc_CC-MAIN-20230922102329-20230922132329-00630.warc.gz\"}"}
http://www.ueda.info.waseda.ac.jp/hydla/index.php?cmd=backup&action=nowdiff&page=Examples&age=8	[ "# Backup diff of Examples vs current(No. 8)\n\n• The added line is THIS COLOR.\n• The deleted line is THIS COLOR.\n#author(\"2017-03-17T11:54:21+09:00\",\"default:Uedalab\",\"Uedalab\")\n#author(\"2017-03-17T17:42:45+09:00\",\"default:Uedalab\",\"Uedalab\")\n#noattach\n* A Bouncing Particle [#p560bc2f]\n>\n* Bouncing particle [#p560bc2f]\n\n// y stands for the height of the ball\nINIT <=> y = 10 & y' = 0. // initial state\nFALL <=> [](y'' = -10). // falling\nBOUNCE <=> [](y- = 0 => y' = -4/5y'-).\n// if the ball reaches the ground, it bounces\nINIT, FALL << BOUNCE. // FALL is weaker than BOUNCE\n<\n\n&ref(./Bouncing_Particle.png,50%);\n----\n A Bouncing Particle with a Parameter [#z5246173]\n>\n* Bouncing particle with a parametric initial height [#z5246173]\n\nINIT <=> 5 < y < 15 & y' = 0. // the initial height is uncertain\nFALL <=> [](y'' = -10).\nBOUNCE <=> [](y- = 0 => y' = -4/5 * y'-).\n\nINIT, FALL << BOUNCE.\n<\n\n&ref(./Bouncing_Particle_with_a_Parameter.png,100%);\n----\n* A Bouncing Paticle on a Curve [#j8167ade]\n>\n* Bouncing particle in a parabolic vase [#j8167ade]\n\n/*\n bouncing particle on a curve of f(x) = (1/2) * x^2\n/\n\nINIT <=> x = 1/2 & y = 10 & x' = 0 & y' = 0 & [](k = 1).\nA <=> [](x'' = 0 & y'' = -98/10).\n/\n sin = f'(x) / (1+f'(x))^(1/2)\n* cos = 1 / (1+f'(x))^(1/2)\n* new x' = (-k * sin^2 + cos^2) * x' + (k+1) * sin * cos * y'\n* new y' = (k+1) * sin * cos * x' + (sin^2 + (-k) * cos^2) * y'\n/\n\nSC <=> [](s = (x-)/(1+(x-)^2)^(1/2)\n& c = 1 /(1+(x-)^2)^(1/2)).\n\nBOUNCE <=> [](y- = (1/2) (x-)^2 =>\nx' = ( (-k) * s^2 + c^2 ) * x'- + ( (k+1) * s * c ) * y'-\n& y' = ( (k+1) * s * c ) * x'- + ( s^2 + (-k) * c^2 ) * y'- ).\n\nINIT, SC, A << BOUNCE.\n<\n\n&ref(./Bouncing_Paticle_on_a_Curve.png,70%);\n----\n* A Bouncing Particle in a Circle [#xbf24e18]\n>\n* Bouncing particle in a circle [#xbf24e18]\n\nINIT <=> x = 0 /\\ 0.5 < y < 0.6 /\\ x' = 2 /\\ y' = 1.\n// the initial position is uncertain\nRUN <=> [](x'' = 0 /\\ y'' = 0).\nBOUNCE <=> []((x-)^2 + (y-)^2 = 1 =>\nx' = x'- - (x- * x'- + y- * y'-) * 2 * (x-)\n/\\ y' = y'- - (x- * x'- + y- * y'-) * 2 * (y-)\n).\nINIT, RUN<<BOUNCE.\n<\n\n&ref(./Bouncing_Particle_in_a_Circle.png,70%);\n----\n* A Hot-Air Balloon with multiple parameters [#tf906f5e]\n>\n* Hot-Air Balloon with multiple parameters [#tf906f5e]\n\n/* A program for a hot-air balloon that repeats rising and falling /\n// The initial condition of the balloon and the timer\n// h: height of the balloon\n// timer: timer variable to repeat rising and falling\nINIT <=> h = 10 /\\ h' = 0 /\\ timer = 0.\n\n// parameters\n// duration: duration of falling\n// riseT: duration of rising\nPARAM<=> 1 < fallT < 4 /\\ 1 < riseT < 3\n/\\ [](riseT' = 0 /\\ fallT' = 0).\n\n// increasing of timer\nTIMER <=> [](timer' = 1).\n\n// rising of the balloon\nRISE <=> [](timer- < riseT =>h'' = 1).\n\n// falling of the balloon\nFALL <=> [](timer- >= riseT => h'' = -2).\n\n// reset the timer to repeat rising and falling\nRESET <=> [](timer- >= riseT + fallT => timer=0).\n\n// assertion for bounded model checking\nASSERT(h > 0).\n\n// constraint hierarchies\nINIT, PARAM, FALL, RISE, TIMER<<RESET.\n<\n\n&ref(./Hot-Air_Balloon.png,70%);\n----\n A Bouncing Particle with Magnetic Force [#a6ea3f5c]\n>\n* Bouncing particle with magnetic force [#a6ea3f5c]\n\nINIT <=> y=10 & y'=0 & mag=0 & timer=0.\nFALL <=> [](y''=-10+mag).\nBOUNCE <=> [](y-=0=>y'=-y'-).\nTRUE <=> [](1=1).\nTIMER <=> [](mag'=0&timer'=1).\nSWITCHON <=> [](timer-=1=>mag=12&timer=0).\n// The magnetic force may be switched on at every one second\nSWITCHOFF <=> [](timer-=1=>mag=0&timer=0).\n// The magnetic force may be switched off at every one second\n\nINIT,TIMER<<(SWITCHOFF,SWITCHON)<<TRUE,FALL<<BOUNCE.\n<\n\n&ref(./Bouncing_Particle_with_Magnetic_Force.png,70%);\n----\n* A Bouncing Particle thrown toward a ceiling [#e682ba1f]\n>\n* Bouncing particle thrown toward a ceiling [#e682ba1f]\n\nINIT <=> 9 < y < 11 & y' = 10.\nFALL <=> [](y'' = -10).\nBOUNCE <=> [](y- = 15 => y' = -(4/5)*y'-).\n\nINIT, FALL << BOUNCE.\n<\n\n#mathjax\n- In this program, the trajectories change qualitatively dependent on the initial height $$y_0$$.\n-- If $$9 < y_0 < 10$$, the ball doesn't reach the ceiling.~\n-- If $$y_0 = 10$$, the ball touches the ceiling, but the velocity remains continuous.~\n-- If $$10 < y_0 < 11$$, the ball bounces on the ceiling.~\n- HyLaGI performs such a case analysis automatically.~\n#norelated\n&ref(./Bouncing_Particle_thrown_toward a_ceiling.png,100%);\n----" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.598027,"math_prob":0.998957,"size":4275,"snap":"2019-13-2019-22","text_gpt3_token_len":1567,"char_repetition_ratio":0.13509716,"word_repetition_ratio":0.100508906,"special_character_ratio":0.47953215,"punctuation_ratio":0.14398943,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9996551,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-24T20:35:59Z\",\"WARC-Record-ID\":\"<urn:uuid:933f4652-8c52-4d1d-bf0f-10f799b66667>\",\"Content-Length\":\"13234\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c1bf0eb8-1234-4da4-9774-f26f94999c38>\",\"WARC-Concurrent-To\":\"<urn:uuid:77a0f10c-d36f-40f5-9121-f0cae567a918>\",\"WARC-IP-Address\":\"133.9.237.10\",\"WARC-Target-URI\":\"http://www.ueda.info.waseda.ac.jp/hydla/index.php?cmd=backup&action=nowdiff&page=Examples&age=8\",\"WARC-Payload-Digest\":\"sha1:LSRBWUTNFFUWKVV2YRG33YNMP25OSRRO\",\"WARC-Block-Digest\":\"sha1:FZFDXVDWLHFXHASVA5AL3S2DI2K4NMTI\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912203491.1_warc_CC-MAIN-20190324190033-20190324212033-00533.warc.gz\"}"}
https://wiki-helper.com/determine-wheter-the-values-given-against-the-quadratic-equation-are-the-root-of-the-equation-or-37400899-28/	[ "# determine wheter the values given against the quadratic equation are the root of the equation or not X2+4x-5=0 ,X=1,-1\n\ndetermine wheter the values given against the quadratic equation are the root of the equation or not X2+4x-5=0 ,X=1,-1\n\n### 1 thought on “determine wheter the values given against the quadratic equation are the root of the equation or not X2+4x-5=0 ,X=1,-1”\n\n1.", null, "" ]	[ null, "https://wiki-helper.com/wp-content/litespeed/avatar/903550a2f70b936bc3f8a4e89e4a4dc9.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7857289,"math_prob":0.99999285,"size":484,"snap":"2023-40-2023-50","text_gpt3_token_len":145,"char_repetition_ratio":0.15625,"word_repetition_ratio":0.5584416,"special_character_ratio":0.2768595,"punctuation_ratio":0.08411215,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9996327,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-30T11:24:47Z\",\"WARC-Record-ID\":\"<urn:uuid:7925b8da-b77d-4cb0-9938-5ce2f3211727>\",\"Content-Length\":\"55128\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0a335761-4de6-4c7c-82ec-1d15776674fa>\",\"WARC-Concurrent-To\":\"<urn:uuid:4a059c0d-53f8-4ab4-ae65-73e3195ad221>\",\"WARC-IP-Address\":\"172.96.191.158\",\"WARC-Target-URI\":\"https://wiki-helper.com/determine-wheter-the-values-given-against-the-quadratic-equation-are-the-root-of-the-equation-or-37400899-28/\",\"WARC-Payload-Digest\":\"sha1:HPJZHJ6WKWREVNLLOJWRZEA7ML3GYXGD\",\"WARC-Block-Digest\":\"sha1:YDYZKUOWU3XCRZA6MCHD3QRIZXIYHGE6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100184.3_warc_CC-MAIN-20231130094531-20231130124531-00459.warc.gz\"}"}
https://www.oreilly.com/library/view/2007-microsoft-office/9780735623248/ch14s02.html	[ "With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, tutorials, and more.\n\nNo credit card required\n\nUsing Functions: A Preview\n\nIn simplest terms, a function is a predefined formula. Many Excel functions are shorthand versions of frequently used formulas. For example, compare the formula =A1+A2+A3+A4+A5+A6+A7+A8+A9+A10 with the formula =SUM(A1:A10). The SUM function makes the formula a lot shorter, easier to read, and easier to create. Some Excel functions perform complex calculations. For example, using the PMT function, you can calculate the payment on a loan at a given interest rate and principal amount.\n\nAll functions consist of a function name followed by a set of arguments enclosed in parentheses. (In the preceding example, A1:A10 is the argument in the SUM function.) If you omit a closing parenthesis when you enter a function, Excel adds ...\n\nWith Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, interactive tutorials, and more.\n\nNo credit card required" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.76730806,"math_prob":0.925256,"size":755,"snap":"2019-43-2019-47","text_gpt3_token_len":165,"char_repetition_ratio":0.15579228,"word_repetition_ratio":0.0,"special_character_ratio":0.22649007,"punctuation_ratio":0.13907285,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99414414,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-14T04:05:57Z\",\"WARC-Record-ID\":\"<urn:uuid:b5d667e0-957f-4b0c-a9b6-74104885abf3>\",\"Content-Length\":\"34267\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4d7ae794-c9cc-4ad8-82b6-b84756d8e972>\",\"WARC-Concurrent-To\":\"<urn:uuid:f68fc27d-d1fc-46d0-96bb-f1e2bcaa8e58>\",\"WARC-IP-Address\":\"104.80.31.164\",\"WARC-Target-URI\":\"https://www.oreilly.com/library/view/2007-microsoft-office/9780735623248/ch14s02.html\",\"WARC-Payload-Digest\":\"sha1:R3KEFOAG4PX4HZAWDBY2WYVLNE2HL7SX\",\"WARC-Block-Digest\":\"sha1:7LFUUYLGFDRIBBQ7SZAOOIRRLUPWZXPF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986649035.4_warc_CC-MAIN-20191014025508-20191014052508-00090.warc.gz\"}"}
http://jarnoralli.com/code/optical-flow-code	[ "# Optical Flow Code\n\nOne complete chapter of my thesis is devoted to explaining how the variational optical flow, disparity and level-set models can be solved. In this chapter I show, step by step, how these models can be solved using different kinds of solvers (Jacobi, Gauss Seidel, Alternating Line Relaxation and Additive Operator Splitting scheme). Related to this chapter, I release the optical flow and disparity codes on this page...these codes are quite similar to the ones that I have used for generating the videos on this site. However, they are not 100% the same...there have been some slight modifications. For anyone interested in the actual code, I suggest first to have a look at the thesis in order to better understand the code itself! The codes are released under LGPL license. If you use the codes, please reference my papers...in my papers, on the other hand, I reference those papers upon which my work is based on. Thanx!!\n\nThe optical flow codes are as follows:\n\n1. Late linerisation optical-flow for large displacements.\n2. Early linearisation optical-flow method for small displacements (basically Horn&Schunck type of formulation).\n3. Late linearisation method for disparity.\n\nPseudo Code\n\nBelow you can see pseudo codes for early- and late-linearization versions of the optical flow. I decided to include these here in order to give a better idea how the different codes actually work. However, all the details are explained in my phd thesis. Therefore, I recommend that you have a look at Chapter 5, Solving Equations.\n\n### Early Linearization\n\n• //----------------------------------------------------------------------------------------------------------------\n• //-COARSE-TO-FINE ALGORITHM FOR CALCULATING OPTICAL FLOW\n• //-Early linearisation (i.e. no warping)\n• //-Quadratic smoothness term (i.e. Tikhonov regulariser)\n• //-Inputs are $$I_0$$ and $$I_1$$, number of scales $$scl$$, and scaling factor $$sclFactor$$\n• //----------------------------------------------------------------------------------------------------------------\n• $$\\mathbf{INPUT:} \\, I_0, \\, I_1, \\, scl, \\, sclFactor$$\n• $$\\mathbf{OUTPUT:} \\, (u,\\, v)$$\n• //Set $$u$$ and $$v$$ to zero\n• $$u=0$$, $$v=0$$\n• //Create image pyramid\n• $$[ Iscl_0\\{\\}, \\, Iscl_1\\{\\} ] = pyramid(I_0, \\, I_1, \\, scl, \\, sclFactor)$$\n• //This is the coarse-to-fine loop\n• WHILE( $$s=scl:-1:1$$ )\n• $$I_0 = Iscl_0\\{s\\}, \\, I_1 = Iscl_1\\{s\\}$$\n• Approximate derivatives for $$I_0$$ and $$I_1$$: $$\\dfrac{\\partial I_k}{\\partial t}$$, $$\\dfrac{\\partial I_{k,0}}{\\partial x}$$, $$\\dfrac{\\partial I_{k,0}}{\\partial y}$$\n• //Solve for new $$u$$ and $$v$$\n• $$[u, \\, v] = SOLVER ( u, \\, v, \\, nLoops, \\, \\dfrac{\\partial I_k}{\\partial t}, \\, \\dfrac{\\partial I_{k,0}}{\\partial x}, \\, \\dfrac{\\partial I_{k,0}}{\\partial y} )$$\n• //Interpolate (prolongate) solution\n• IF ( $$s-1>0$$ )\n• $$[u, \\, v] = prolongate( u, \\, v, \\,sclFactor )$$\n• ENDIF\n• ENDWHILE\n\n### Late Linearization\n\n• //----------------------------------------------------------------------------------------------------------------\n• //-COARSE-TO-FINE ALGORITHM FOR CALCULATING OPTICAL FLOW\n• //-Late linearisation (i.e. uses warping)\n• //-Robust error functions in both the data and the smoothness terms\n• //-Inputs are $$I_0$$ and $$I_1$$, number of scales $$scl$$, and scaling factor $$sclFactor$$\n• //----------------------------------------------------------------------------------------------------------------\n• $$\\mathbf{INPUT:} \\, I_0, \\, I_1, \\, scl, \\, sclFactor$$\n• $$\\mathbf{OUTPUT:} \\, (u,\\, v)$$\n• //Set $$u$$ and $$v$$ to zero\n• $$u=0$$, $$v=0;$$\n• //Create image pyramid\n• $$[ Iscl_0\\{\\} \\, Iscl_1\\{\\} ] = pyramid(I_0, \\, I_1, \\, scl, \\, sclFactor);$$\n• //Coarse-to-fine loop\n• WHILE( $$s=scl:-1:1$$ )\n• $$I_0 =Iscl_0\\{s\\}$$, $$I_1 =Iscl_1\\{s\\};$$\n• //Warping loop\n• WHILE( $$fstLoop$$)\n• //Warp image as per $$u$$ and $$v$$\n• $$I_{k,0}^w = warp( I_{k,0}, u, v );$$\n• Approximate derivatives for $$I_0^w$$ and $$I_1$$: $$\\dfrac{\\partial I_k}{\\partial t} = I_{k,1} - I_{k,0}^w$$, $$\\dfrac{\\partial I_{k,0}^w}{\\partial x}$$, $$\\dfrac{\\partial I_{k,0}^w}{\\partial y}$$\n• //Reset $$du$$ and $$dv$$\n• $$du=0$$, $$dv=0$$\n• //Fixed-point loop due to the robust error functions\n• WHILE( $$sndLoop$$)\n• //Calculate penalizer function values for data\n• $$\\Psi{\\prime} \\Big( (E_k)_D \\Big)$$, where $$\\left( E_k \\right)_D = \\left( \\dfrac{ \\partial I_k }{\\partial t} - \\dfrac{ \\partial I_{k,0}^{w} }{\\partial x}du - \\dfrac{ \\partial I_{k,0}^{w} }{\\partial y}dv \\right)^2 ;$$\n• //Calculate the diffusion weights\n• $$\\Big[ \\Psi{\\prime} \\big( E_R^{l,m} \\big)_W \\, \\Psi{\\prime} \\big( E_R^{l,m} \\big)_N \\, \\Psi{\\prime} \\big( E_R^{l,m} \\big)_E \\, \\Psi{\\prime} \\big( E_R^{l,m} \\big)_S \\Big] = weights(u+du, v+dv);$$\n• //Solve for new $$du$$ and $$dv$$\n• $$\\begin{matrix} [du \\, dv] = SOLVER ( & u, \\, v, \\, du, \\, dv, \\, nLoops, \\, \\dfrac{\\partial I_k}{\\partial t}, \\, \\dfrac{\\partial I_{k,0}^w}{\\partial x}, \\, \\dfrac{\\partial I_{k,0}^w}{\\partial y}, \\, \\Psi{\\prime} \\Big( (E_k)_D \\Big), &\\\\ & \\Psi{\\prime} \\big( E_R^{l,m} \\big)_W, \\, \\Psi{\\prime} \\big( E_R^{l,m} \\big)_N, &\\\\ & \\Psi{\\prime} \\big( E_R^{l,m} \\big)_S, \\, \\Psi{\\prime} \\big( E_R^{l,m} \\big)_E & ); \\end{matrix}$$\n• ENDWHILE\n• //Update $$u$$ and $$v$$\n• $$u=u+du$$, $$v=v+dv;$$\n• ENDWHILE\n• //Interpolate (prolongate) solution\n• IF( $$s-1>0$$ )\n• $$[u \\, v] = prolongate( u, \\, v, \\,sclFactor );$$\n• ENDIF\n• ENDWHILE", null, "Author: Jarno Ralli\nJarno Ralli is a computer vision scientist and a programmer.", null, "", null, "", null, "", null, "", null, "" ]	[ null, "http://jarnoralli.com/images/social/avatars/jarno_avatar_64.png", null, "http://jarnoralli.com/component/jcomments/captcha/97261", null, "http://jarnoralli.com/modules/mod_nice_social_bookmark/icons/facebook_aqu_48.png", null, "http://jarnoralli.com/modules/mod_nice_social_bookmark/icons/twitter_aqu_48.png", null, "http://jarnoralli.com/modules/mod_nice_social_bookmark/icons/google_aqu_48.png", null, "http://jarnoralli.com/modules/mod_nice_social_bookmark/icons/linkedin_aqu_48.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.649838,"math_prob":0.9998425,"size":5426,"snap":"2020-10-2020-16","text_gpt3_token_len":1810,"char_repetition_ratio":0.22150499,"word_repetition_ratio":0.16981132,"special_character_ratio":0.4493181,"punctuation_ratio":0.20042194,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000046,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,null,null,1,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-17T14:31:36Z\",\"WARC-Record-ID\":\"<urn:uuid:46c968ae-1aa7-4e50-bd1d-4f6eea42fc95>\",\"Content-Length\":\"30291\",\"Content-Type\":\"application/http; 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https://calculator.academy/duct-velocity-calculator/	[ "Enter the air CFM (cubic feet per minute) and the duct cross-sectional area into the calculator to determine the duct velocity.\n\n## Duct Velocity Formula\n\nThe following formula is used to calculate the linear air velocity through a duct.\n\nV = CFM / A\n\n• Where V is the duct velocity (FPM)\n• CFM is the volumetric flow rate of air through the duct (typically cubic feet per minute)\n• A is the cross-sectional area (ft^2)\n\n## Duct Velocity Definition\n\nWhat is a duct velocity? A duct velocity is defined as the linear speed at which air is moving through the opening of a duct or air vent. This is the average linear speed of the air since normal airflow is not constant across an entire area.\n\n## Example Problem\n\nHow to calculate duct velocity?\n\n1. First, determine the volumetric flow rate.\n\nThe volumetric flow rate, often denoted CFM (cubic feet per minute) is the total volume of air moving through an area per unit of time. For this problem, the flow rate is 1,500 ft^3/min.\n\n2. Next, determine the cross-sectional area.\n\nFor this example, the cross-sectional area of the duct is 5 ft^2.\n\n3. Finally, calculate the duct velocity.\n\nThe air velocity is then calculated using the formula above. V = 1500 / 5 = 300 ft/min = 5 ft/s." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88339776,"math_prob":0.9960973,"size":1491,"snap":"2021-43-2021-49","text_gpt3_token_len":351,"char_repetition_ratio":0.16677874,"word_repetition_ratio":0.0076045627,"special_character_ratio":0.23272972,"punctuation_ratio":0.08098592,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997532,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-26T19:15:43Z\",\"WARC-Record-ID\":\"<urn:uuid:e2c57162-bd50-4e26-a0e4-1d55e0ff0c46>\",\"Content-Length\":\"75877\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a51cc572-be35-4a43-ae1f-6b7d3d5394be>\",\"WARC-Concurrent-To\":\"<urn:uuid:a62edff6-7940-4383-a626-cffc6b19188b>\",\"WARC-IP-Address\":\"104.26.10.229\",\"WARC-Target-URI\":\"https://calculator.academy/duct-velocity-calculator/\",\"WARC-Payload-Digest\":\"sha1:KWC45VGIRAAAMD76FBTMR63HTZONNKG5\",\"WARC-Block-Digest\":\"sha1:DAFXH74Y5FV2EOJC62H5KH4T6677JNHK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587915.41_warc_CC-MAIN-20211026165817-20211026195817-00665.warc.gz\"}"}
https://miphonedoctorofaustin.com/qa/what-happens-when-irr-is-negative.html	[ "", null, "# What Happens When IRR Is Negative?\n\n## What does negative return mean?\n\nA negative return occurs when a company or business has a financial loss or lackluster returns on an investment during a specific period of time.\n\nSome businesses report a negative return during their early years because of the amount of capital that initially goes into the business to get it off the ground..\n\n## What are the problems with IRR?\n\nThere are some cases in which the cash flow pattern is such that the calculation of IRR actually ends up giving multiple rates. So instead of having one IRR, we would then have multiple IRR’s. Sometimes the IRR number can even go in the negative indicating that the firm is actually losing value.\n\n## Is NPV better than IRR?\n\nBecause the NPV method uses a reinvestment rate close to its current cost of capital, the reinvestment assumptions of the NPV method are more realistic than those associated with the IRR method. … In conclusion, NPV is a better method for evaluating mutually exclusive projects than the IRR method.\n\n## Is a negative IRR bad?\n\nIt is ok to get a negative IRR. Negative IRR indicates that the sum of post-investment cash flows is less than the initial investment; i.e. the non-discounted cash flows add up to a value which is less than the investment.\n\n## What happens when IRR is negative NPV?\n\nIRR of a given series of cash flows is calculated by discounting them at such rate so that the NPV is zero. Therefore higher the IRR (discounting rate), the lower will be the NPV value even falling below zero( both are inversely related). Negative NPV implies a ‘no-go’ investment as expected returns at not delivered.\n\n## Can IRR be negative in Excel?\n\nExcel allows a user to get a negative internal rate of return of an investment using the IRR function." ]	[ null, "https://mc.yandex.ru/watch/66668152", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9298634,"math_prob":0.9174014,"size":1732,"snap":"2020-45-2020-50","text_gpt3_token_len":371,"char_repetition_ratio":0.14930555,"word_repetition_ratio":0.0,"special_character_ratio":0.19745958,"punctuation_ratio":0.08083832,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96996725,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-23T07:46:13Z\",\"WARC-Record-ID\":\"<urn:uuid:6df713eb-ce53-4652-a05a-b94d511a481e>\",\"Content-Length\":\"28728\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8ee76a7a-85bc-4582-b8aa-26873c74f18d>\",\"WARC-Concurrent-To\":\"<urn:uuid:86cce0a2-b3fe-4d57-870d-d7567aec0ccd>\",\"WARC-IP-Address\":\"87.236.16.235\",\"WARC-Target-URI\":\"https://miphonedoctorofaustin.com/qa/what-happens-when-irr-is-negative.html\",\"WARC-Payload-Digest\":\"sha1:RMBWY2N3QSP3GOVK23H4PI7Z4YLMHOUP\",\"WARC-Block-Digest\":\"sha1:7C4344TOAZSRTJSH7535RFPLAOUAJ7Q6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107880878.30_warc_CC-MAIN-20201023073305-20201023103305-00456.warc.gz\"}"}
http://www.tweetnotebook.com/Pennsylvania/normalised-mean-square-error-formula.html	[ "", null, "Address 7314 W Ridge Rd, Fairview, PA 16415 (814) 474-1202 http://microsupportcenter.com\n\n# normalised mean square error formula Cranesville, Pennsylvania\n\nA proposal is then made to obtain the desired results by the use of different indices.Discover the world's research11+ million members100+ million publications100k+ research projectsJoin for free Full-text (PDF)DOI: ·Available from: It is also shown that in certain situations, that have not to be considered as limit cases, the “best” condition to get the lowest value of the NMSE is completely different The final step is to determine if the performance of the competing models is statistically different. Papers of Interest:- 1) V.\n\nnormalization by Co Considering Co/Cp and Cp/Cp, i.e. The same data filtering for FAa calculation is applied for NMSE calculation. This gives a simple relation between NMSE and relative $\\ell^2$ error. Full-text · Article · Feb 2016 Gilberto Fernandes JrLuiz F.\n\nThe RMSD represents the sample standard deviation of the differences between predicted values and observed values. Hanna (1988,1989) has pioneered a significant number of these statistics. Back to Table of Contents Drop in your comments and suggestions to mailto:[email protected] \| The University of Toledo \| \| College of Engineering \| \| Department of Civil Engineering The idea is to find out the quality and reliability of the predictions made by a model when compared to real life data.\n\nKumar, \"Evaluation of the ISC Short Term Model in a Large-Scale Multiple Source Region for Different Stability Classes\", Env. For example, when measuring the average difference between two time series x 1 , t {\\displaystyle x_{1,t}} and x 2 , t {\\displaystyle x_{2,t}} , the formula becomes RMSD = ∑ The quality measurements are the percentage of validation and estimation data unfitness, Akaike's Final Prediction Error (FPE) (Jones, 1975), loss function (Berger, 1985) and mean squared normalized error performance function (MSE) So far, estimation and modeling approaches have enabled a comprehensive understanding of repetitive processes in projects at steady-state.\n\nThough there is no consistent means of normalization in the literature, common choices are the mean or the range (defined as the maximum value minus the minimum value) of the measured The first technique to measure the adjustment degree relative to the current analyzed traffic and DSNSF is the Normalized Mean Square Error (NMSE) (Poli & Cirillo, 1993), which evaluates the difference Similar is the case of Kumar et al (1993) who have used statistical tools to evaluate the prediction of lower flammability distances. Thus, results close to zero indicate excellent traffic characterizations while high values demonstrate that DSNSF is distant from the expected results. \"[Show abstract] [Hide abstract] ABSTRACT: Traffic monitoring and anomaly detection\n\nRetrieved 4 February 2015. ^ \"FAQ: What is the coefficient of variation?\". The bootstrap technique has to be used. To determine the reliability of a model the following criteria suggested by Kumar et al. (1993) could be used. The ideal value for the factor of two should be 1 (100%).\n\nAiming to improve its efficiency, a modification of the Ant Colony Optimization metaheuristic is proposed, which through self-organized agents optimizes the analysis of multidimensional flows attributes and allows it to be So, does anyone know how matlab normalizes the MSE?Many thanks in advance!Hugo 0 Comments Show all comments Tags mseneural networksperformancenormalized Products No products are associated with this question. These approaches underestimate the influence of process repetitiveness, the variation of learning curves and the conservation of processes’ properties. signal-processing share\|cite\|improve this question asked Sep 10 '13 at 0:59 Gummi F 74119 I guess not.\n\nrows or columns)). Pet buying scam Measuring air density - where is my huge error coming from? The system returned: (22) Invalid argument The remote host or network may be down. Generated Fri, 21 Oct 2016 22:00:07 GMT by s_wx1011 (squid/3.5.20)\n\nNote that air quality scientists and engineers do not use all the performance measures mentioned below. Maximal number of regions obtained by joining n points around a circle by straight lines Questions about convolving/deconvolving with a PSF A crime has been committed! ...so here is a riddle On the other hand, high NMSE values do not necessarily mean that a model is completely wrong. These requirements cause the calibration of models to be a very expensive and often time-consuming study.\n\nNevertheless, most estimation, planning, and scheduling approaches overlook the dynamics of project-based systems in construction. Scott Armstrong & Fred Collopy (1992). \"Error Measures For Generalizing About Forecasting Methods: Empirical Comparisons\" (PDF). By using this site, you agree to the Terms of Use and Privacy Policy. An Error Occurred Unable to complete the action because of changes made to the page.\n\nAn internet search however only shows strange definitions like $$\\frac{ \\sum_i (x_i-y_i)^2}{N\\sum_i (x_i)^2} \\quad\\text{or} \\quad \\frac{N \\sum_i (x_i-y_i)^2}{\\sum_i x_i \\sum_i y_i}$$ Is my interpretation not the standard definition? When an 'NA' value is found at the i-th position in obs OR sim, the i-th value of obs AND sim are removed before the computation. It is written in symbolic form as: iii) Normalized Mean Square Error This statistic emphasizes the scatter in the entire data set and is known as Normalized Mean Square Error Not the answer you're looking for?\n\nBesides, there is the possibility to calculate the same MSE normalized setting 'standard' or 'percent'.I have looked for the algorithm to calculate both of them with no success. Old literature in the fields of science and engineering is full of such examples. Later on correlation coefficient between the observed and predicted values became a popular way of looking at the performance of a model. See all ›40 CitationsSee all ›7 ReferencesShare Facebook Twitter Google+ LinkedIn Reddit Download Full-text PDF On the use of the normalized mean square error in evaluating dispersion model performanceArticle (PDF Available) in Atmospheric\n\nBased on your location, we recommend that you select: . Difference measures represent a quantitative estimate of the size of the differences between observed and predicted values. The resulting detection system was tested with real and simulated data, achieving high detection rates while the false alarm rate remains low.Article · Jan 2016 Luiz Fernando CarvalhoSylvio BarbonLeonardo de Souza Full-text · Conference Paper · Jul 2015 · Expert Systems with ApplicationsRicardo Magno Santos AntunesVicente a.\n\nOn the other hand, high NMSE values do not necessarily mean that a model is completely wrong. Apply Today MATLAB Academy New to MATLAB? Ret_type is a switch to select the return output (1= RMSD (default), 2= NRMSD, 3= CV(RMSD)). Aiming an automated management to detect and prevent potential problems, we present and compare two novel anomaly detection mechanisms based on statistical procedure Principal Component Analysis and the Ant Colony Optimization\n\nRMSD is a good measure of accuracy, but only to compare forecasting errors of different models for a particular variable and not between variables, as it is scale-dependent. Contents 1 Formula In equation form it is represented as: v) Geometric Mean Bias The geometric mean bias ( MG ) ig given by: vi) Geometric Mean Variance The geometric mean variance These are NOT standard definitions for regression/curve-fitting or classification/pattern-recognition." ]	[ null, "http://www.tweetnotebook.com/maps/Cranesville_PA.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88555896,"math_prob":0.8724282,"size":7897,"snap":"2019-43-2019-47","text_gpt3_token_len":1691,"char_repetition_ratio":0.09400735,"word_repetition_ratio":0.02138158,"special_character_ratio":0.2114727,"punctuation_ratio":0.08872727,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96337956,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-12T06:24:12Z\",\"WARC-Record-ID\":\"<urn:uuid:aebfc941-88a7-445c-877f-a83a3454db3e>\",\"Content-Length\":\"25679\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:02297d36-cfd1-43b1-8b1a-29eb8f6347f5>\",\"WARC-Concurrent-To\":\"<urn:uuid:1d09baf8-68a6-4a98-a7d7-aaf64a912010>\",\"WARC-IP-Address\":\"104.27.157.221\",\"WARC-Target-URI\":\"http://www.tweetnotebook.com/Pennsylvania/normalised-mean-square-error-formula.html\",\"WARC-Payload-Digest\":\"sha1:ZWBBFRC64OZ6F3CGMA57MU7YR4SATVKG\",\"WARC-Block-Digest\":\"sha1:PIP24MY54JSFMJZXIA3ZFFGRVFSLLJRD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496664752.70_warc_CC-MAIN-20191112051214-20191112075214-00096.warc.gz\"}"}
https://stackoverflow.com/questions/9909038/formatting-hexadecimal-number-to-short-uuid-in-javascript	[ "# Formatting hexadecimal Number to short UUID in JavaScript\n\nI have a hexadecimal number in javascript. For display purposes, I would like to format the string as:\n\n``````ffffffff-ffff-ffff\n00000000-0000-01ff\n``````\n\n`(8 digits)-(4 digits)-(4 digits)` with padded zeros on the front\n\nI've been trying to write my own loop to format an arbitrary hexadecimal number into this format, but this seems like something that should be available in JavaScript already.\n\nIs there a built-in way to format a hexadecimal number in JavaScript?\n\n• There is unfortunately no way to format such hexadecimal numbers in JavaScript. But check out this link, maybe it helps.\n– Neq\nMar 28, 2012 at 14:05\n\nI would do a two-step process:\n\n1) convert number to 16 digit hex with leading zeros:\n\n``````var i = 12345; // your number\nvar h = (\"000000000000000\" + i.toString(16)).substr(-16);\n``````\n\n``````var result = h.substr(0, 8)+'-'+h.substr(8,4)+'-'+h.substr(12,4);\n``````\n\nIf your number is really a full 64 bits long you should be aware that javascript has only doubles, which top out at around 53 bits of precision. E.g.\n\n``````var i = 0x89abcdef01234567; // a 64-bit constant\nvar h = (\"000000000000000\" + i.toString(16)).substr(-16); // \"89abcdef01234800\"\n``````\n\nSo you probably want to split this into two 32-bit numbers, and format them 8 digits at a time. Then the second caveat strikes: javascript performs bitwise ops on signed 32-bit integers, and this formatting code can't handle negative numbers.\n\n``````var i = 0xffd2 << 16; // actually negative\nvar h = (\"0000000\" + i.toString(16)).substr(-8); // \"0-2e0000\"\n``````\n\nSince it's fairly likely that numbers you want formatted in hexadecimal are the result of bitwise manipulations, the code can be tweaked to print in two's complement instead:\n\n``````var i = 0xffd2 << 16; // actually negative\nvar h = (\"0000000\" + ((i\|0)+4294967296).toString(16)).substr(-8); // \"ffd20000\"\n``````\n\nThis produces the hex representation of the bottom 32 bits of the integral part of arbitrary positive and negative numbers. This is probably what you want (it's approximately `printf(\"%08x\")`). Some more corner cases:\n\n``````var i = 1.5; // non-integers are rounded\nvar h = (\"0000000\" + ((i\|0)+4294967296).toString(16)).substr(-8); // \"00000001\"\n\nvar i = -1.5; // rounding is towards zero\nvar h = (\"0000000\" + ((i\|0)+4294967296).toString(16)).substr(-8); // \"ffffffff\"\n\nvar i = NaN; // not actually a number\nvar h = (\"0000000\" + ((i\|0)+4294967296).toString(16)).substr(-8); // \"00000000\"\n``````\n• While `(i\|0)+4294967296` will adequately force the number to be in range and unsigned, `i >>> 0` achieves this purpose as well, and it's a lot clearer too. Oct 2, 2020 at 17:15\n\nES6 Version\n\n``````function toPaddedHexString(num, len) {\nstr = num.toString(16);\nreturn \"0\".repeat(len - str.length) + str;\n}\n\n``````\n• This is simplified further by padStart in EMCAScript 2017. `num.toString(16).padStart(len, \"0\")` Dec 4, 2018 at 17:36\n\nThere are two methods for `String` called `padStart` and `padEnd`\n\n``````const num = 15;\n``````\n``````const num = 15;\n``````\n\n## The shortest version as conclusion\n\nFor ID like `00000000-0000-01ff`\n\n``````function createID_8_4_4(s)\n{\nreturn(1e16+s).slice(-16).replace(/^.{8}\|.{4}(?!\\$)/g,'\\$&-')\n/\nOR the version for easy understanding:\nreturn(1e16+s).slice(-16).match(/..../g).join('-').replace('-','')\nreplace on the end replaces only first '-'\n/\n}\n\nconsole.log(createID_8_4_4('01ff')); //00000000-0000-01ff``````\n\nFor ID like `0000-0000-0000-01ff`\n\n``````function createID_4_4_4_4(s)\n{\nreturn(1e16+s).slice(-16).match(/..../g).join('-')\n}\n\nconsole.log(createID_4_4_4_4('01ff')); //0000-0000-0000-01ff``````\n\ni don't think there is anything related to that in pure javascript, but frameworks provide this method, in ExtJS 3 it is implemented this way\n\n`````` /*\n Pads the left side of a string with a specified character. This is especially useful\n* for normalizing number and date strings. Example usage:\n* <pre><code>\nvar s = String.leftPad('123', 5, '0');\n// s now contains the string: '00123'\n* </code></pre>\n* @param {String} string The original string\n* @param {Number} size The total length of the output string\n* @param {String} char (optional) The character with which to pad the original string (defaults to empty string \" \")\n* @return {String} The padded string\n* @static\n*/\nleftPad : function (val, size, ch) {\nvar result = String(val);\nif(!ch) {\nch = \" \";\n}\nwhile (result.length < size) {\nresult = ch + result;\n}\nreturn result;\n}\n``````\n• This is just a padding function, what about the hex formatting and the dashes? Mar 28, 2012 at 14:06\n\nI assume the number is already in the form of a string due to the limitations described in hexwab's answer. If it isn't, start the process by getting it into a hex string of arbitrary length (called `str`) using whatever mechanism is appropriate for your situation. Then:\n\n``````function format(str) {\nreturn (1e15+str).slice(-16).match(/^.{8}\|.{4}/g).join('-');\n}\n\nconsole.log(format('723e9f4911'))``````\n\n• Very nice and short solution! (+) I wrote a little bit shorted version from your answer. See my answer. Apr 21, 2020 at 15:59\n• @Bharata I love golf! Let's do this. Amended my answer. Apr 21, 2020 at 18:55" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7066172,"math_prob":0.9595059,"size":1530,"snap":"2023-40-2023-50","text_gpt3_token_len":445,"char_repetition_ratio":0.15072083,"word_repetition_ratio":0.13478261,"special_character_ratio":0.40261438,"punctuation_ratio":0.14436619,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98563695,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-02T09:35:05Z\",\"WARC-Record-ID\":\"<urn:uuid:c5d539c8-2ba4-4e4b-a0f5-c837bbd77003>\",\"Content-Length\":\"235572\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9eccbdfb-98e3-4a7a-b865-65ce5ad2a880>\",\"WARC-Concurrent-To\":\"<urn:uuid:de2cda38-3877-4402-baaf-2550d81acdb4>\",\"WARC-IP-Address\":\"104.18.23.201\",\"WARC-Target-URI\":\"https://stackoverflow.com/questions/9909038/formatting-hexadecimal-number-to-short-uuid-in-javascript\",\"WARC-Payload-Digest\":\"sha1:C2GPVMSSDYE5A45SFOORZGHHESIGRSUQ\",\"WARC-Block-Digest\":\"sha1:KFLIFCOR3F34GO3KRR2ZMWVKLDNVIKUL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510983.45_warc_CC-MAIN-20231002064957-20231002094957-00397.warc.gz\"}"}
http://www.webconversiononline.com/radiation-exposure-conversion.aspx	[ "", null, "LengthConversion WeightConversion TemperatureConversion Date & TimeConversion VolumeConversion AreaConversion SpeedConversion ScientificConversion HealthCalculators OtherCalculators Wizard &Dictionary\n Radiation", null, "Radiation Conversion", null, "Radioactivity conversion", null, "Radiation Exposure", null, "Absorbed Dose", null, "Half Life Period", null, "Isotopes of Element", null, "Atomic Symbol", null, "Atomic Number", null, "Atomic Weight", null, "Energy Conversion", null, "Frequency to Wavelength Calculator Magnetism", null, "Magnetic Field Strength", null, "Magnetic Flux", null, "Magnetic Flux Density", null, "Magnetomotive Force Chemistry", null, "Atomic Number", null, "Atomic Symbol", null, "Atomic Weight", null, "Boiling Point", null, "Chemical Formula Calc", null, "Color of Elements", null, "Compounds of Element", null, "Flame Color Test", null, "Freezing Point", null, "Half Life Period", null, "Isotopes of Element", null, "Melting Point", null, "Molecular Weight Calc", null, "Properties of Element", null, "Volume to Weight Calc Health", null, "Blood Sugar conversion", null, "BMI Calculator", null, "BMR Calculator", null, "Body Fat Calculator", null, "Calories Burned Calc", null, "Calories in Foods", null, "Cloth Size Conversion", null, "Cooking Conversion", null, "Ideal Weight Calculator", null, "Shoe Size Conversion", null, "Volume to Weight of Cooking Ingredient", null, "Waist to Hip Ratio Calc", null, "Weight Management Fluid & Viscosity", null, "Density Conversion", null, "Dynamic Viscosity", null, "Flow Conversion", null, "Fluid Concentration", null, "Kinematic Viscosity", null, "Mass Flow Conversion", null, "Mass Flux Density", null, "Molar Concentration", null, "Molar Flow Conversion", null, "Permeability conversion", null, "Pressure Conversion", null, "Surface Tension", null, "Surface Tension Conversion", null, "Thermal Expansion Coefficient", null, "Viscosity of Fluids Technology", null, "Computer Unit Conversion", null, "Data Rate Conversion", null, "Display Resolution", null, "Frequency Conversion", null, "HTML Color codes", null, "HTML Symbols", null, "Image Resolution Conversion", null, "IP Lookup", null, "My Device Information", null, "Number Conversion", null, "Prefix Conversion", null, "Sound Conversion", null, "Speed of Device", null, "Wavelength Calculator", null, "What is my IP", null, "What is my User Agent Temperature", null, "Temperature Conversion", null, "Boiling Point", null, "Freezing Point", null, "Melting Point", null, "Energy Density", null, "Energy Mass", null, "Heat Capacity", null, "Heat Density", null, "Heat Flux Density", null, "Heat Transfer Coefficient Conversion", null, "Thermal Conductivity", null, "Thermal Expansion", null, "Thermal Expansion Coefficient", null, "Thermal Resistance Pregnancy Banking & Finance More Topics\n Home Tools Topics Mobile Version", null, "Convert Radiation Exposure between Coulomb/Kg, Millicoulomb/Kg, Microcoulomb/Kg, Roentgen, Parker, Rep and other radiation exposure measurement units\n Enter radiation exposure quantity, select units and click 'Convert'\n Quantity From To Coulomb/KgMillicoulomb/KgMicrocoulomb/KgRoentgenParkerRep = ? Coulomb/KgMillicoulomb/KgMicrocoulomb/KgRoentgenParkerRep\n Use Radiation Exposure Conversion Wizard Instead? How to use this calculator...Use current calculator (page) to convert Radiation Exposure from Parker to Coulomb/Kg. Simply enter Radiation Exposure quantity and click ‘Convert’. Both Parker and Coulomb/Kg are Radiation Exposure measurement units.For conversion to different Radiation Exposure units, select required units from the dropdown list (combo), enter quantity and click convertFor very large or very small quantity, enter number in scientific notation, Accepted format are 3.142E12 or 3.142E-12 or 3.142x10*12 or 3.142x10^12 or 3.14210*12 or 3.14210^12 and like wise\n\n Related ... » Radiation Conversion » Radiation Activity Conversion » Radiation Absorbed Dose Conversion » Isotopes of different elements » Half-life period of different element » Properties of different element » Atomic symbols of different element » Trigonometry Calculator » Logarithm Calculator » Distance between Cities (or Town) » Molecular Weight Calculator » World Time - Find time at different places » BMI Calculator (Body Mass Index) » Body Fat Calculator » BMR Calculator (Find daily calorie requirement) » Waist to Hip Ratio Calculator » Calculate Calories Burned in different activities » Calculate Your Ideal Weight » Weight Management » Due Date Calculator (Pregnancy) » When is Ovulation (Ovulation Calculator) » Pregnancy Calculator » Safe Period Calculator » Birth Control Methods » Find Definition of different Units » List of different measurement Units\n\n Supported Conversion Types ... Acceleration Angle Angular Acceleration Angular Velocity Area Blood Sugar Clothing Size Computer Storage Unit Cooking Volume Cooking Weight Data Transfer Rate Date Density Dynamic Viscosity Electric Capacitance Electric Charge Electric Conductance Electric Conductivity Electric Current Electric Field Strength Electric Potential Electric Resistance Electric Resistivity Energy Energy Density Energy Mass Euro Currency Fluid Concentration Fluid Flow Rate Fluid Mass Flow Rate Force Frequency Fuel Economy Heat Capacity Heat Density Heat Flux Density Heat Transfer Coefficient Illumination Image Resolution Inductance Kinematic Viscosity Length Luminance (Light) Light Intensity Linear Charge Density Linear Current Density Magnetic Field Strength Magnetic Flux Magnetic Flux Density Magnetomotive Force Mass Flux Density Molar Concentration Molar Flow Rate Moment of Inertia Number Permeability Power Prefix Pressure Radiation Radiation Absorbed Radiation Exposure Radioactivity Shoe Size Sound Specific Volume Speed Surface Charge Density Surface Current Density Surface Tension Temperature Thermal Conductivity Thermal Expansion Thermal Resistance Time Torque Volume Volume Charge Density Water Oil Viscosity Weight\n\n Topic of Interest ... 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https://answers.everydaycalculation.com/add-fractions/15-42-plus-8-20	[ "Solutions by everydaycalculation.com\n\n15/42 + 8/20 is 53/70.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 42 and 20 is 420\n2. For the 1st fraction, since 42 × 10 = 420,\n15/42 = 15 × 10/42 × 10 = 150/420\n3. Likewise, for the 2nd fraction, since 20 × 21 = 420,\n8/20 = 8 × 21/20 × 21 = 168/420", null, "Download our mobile app and learn to work with fractions in your own time:" ]	[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.77688074,"math_prob":0.99467903,"size":330,"snap":"2020-10-2020-16","text_gpt3_token_len":133,"char_repetition_ratio":0.21472393,"word_repetition_ratio":0.0,"special_character_ratio":0.5030303,"punctuation_ratio":0.064935066,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99656934,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-22T22:55:08Z\",\"WARC-Record-ID\":\"<urn:uuid:92161493-332f-4115-ae94-3fd93238c5ab>\",\"Content-Length\":\"7887\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:80ed49c4-20ff-4e7b-b9f3-ca488b954f0b>\",\"WARC-Concurrent-To\":\"<urn:uuid:2413a270-7af0-4046-a2fc-d37fa8867bda>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/add-fractions/15-42-plus-8-20\",\"WARC-Payload-Digest\":\"sha1:44TL52YLERY6PLEIOKHVODCKY6XOCCL4\",\"WARC-Block-Digest\":\"sha1:YHX4FT5WFLBQRSEKJLCPNBYNFC4HBRPX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875145729.69_warc_CC-MAIN-20200222211056-20200223001056-00505.warc.gz\"}"}
https://support.bioconductor.org/p/62039/	[ "exporting limma log2 data\n1\n0\nEntering edit mode\nJahn Davik ▴ 110\n@jahn-davik-5747\nLast seen 8.0 years ago\n\nDear all,\n\nI am quite new to limma and need to ask how to export the normalized data. I want to use them in graphical data exploration and presenatation. So, I do something like this (following the Mattick Lab post):\n\nlibrary(limma)\n\nx <- read.maimages(targets, path = 'mydirectory', source = 'agilent', green.only =T)\n\ny <- backgroundCorrection(x,method = 'normexp')\n\ny <- normalizeBetweenArrays(y, method = 'quantile')\n\nSo I guess 'y' now contains the log2 expression data.\n\nQuestion: how can I get them out of R and into a, say, tab-delimited text file?\n\nI would be much obliged.\n\nThank you.\n\njd\n\nlimma export log2 data agilent • 2.5k views\n2\nEntering edit mode\n@james-w-macdonald-5106\nLast seen 32 minutes ago\nUnited States\n\nThe object you are calling 'y' is an EList, and y$E contains the normalized data. So you could hypothetically use write.table() to output to a file. But I would recommend that you resist this temptation. There are any number of Bioconductor packages (one of which is limma, the package you are currently using) that are useful for exploration and presentation of your data. In fact I would argue that the easiest and most productive way for you to proceed from here is to keep your data in R and finish the analysis using limma. I can't imagine what you would do with those data that you couldn't do (probably more efficiently) with Bioconductor. If you say what exactly you are trying to do, I am sure we can come up with some pointers. ADD COMMENT 0 Entering edit mode Dear MacDonald, Thanks for the reply. I wanted to extract the data in order to do boxplots of probes of my interest. But if that is possible within limma, advice is greatly appreciated. jd ADD REPLY 0 Entering edit mode I am not sure exactly what you mean by 'boxplots of probes'. I could interpret that to mean that you want to make boxplots for individual probes, across (some or all) samples. Or I could interpret that to mean that you want to make boxplots for each sample, restricted to a set of probes. There isn't any functionality in limma that I know of to make boxplots, but that's because it would be duplication of existing functionality elsewhere in R. Since I don't know exactly what you want, here are some suggestions. In these examples, I will use 'y' as the EList input object. Boxplot of individual probes, all samples: probevec <- <character vector of probe IDs of interest> boxplot(t(y$E[probevec,]), xaxt = \"none\")\naxis(1, seq_len(length(probevec)), probevec, las = 2)\n\nBoxplot of individual probes, separated by groups of samples:\n\nlibrary(lattice)\ngroupfac <- <factor of groups>\n## this can be any vector that has the same value for each sample of a given type. As an example:\ngroupfac <- factor(c(\"Control\",\"Control\",\"Treated\",\"Control\",\"Treated\",\"Treated\"))\nprobes <- t(y$E[probevec,]) forlattice <- data.frame(groupfac = rep(groupfac, length(probevec), probes = colnames(probes)[col(probes)], vals = as.vector(probes)) bwplot(vals~probes\|groupfac, forlattice) Boxplot of individual samples, subsetted by a set of probes: boxplot(y$E[probevec,])\n\nDoes that help?\n\n0\nEntering edit mode\n\nIt helps. Thank you so much for the feedback.\n\njd" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9124097,"math_prob":0.77833676,"size":1622,"snap":"2022-40-2023-06","text_gpt3_token_len":383,"char_repetition_ratio":0.111248456,"word_repetition_ratio":0.066176474,"special_character_ratio":0.2268804,"punctuation_ratio":0.13931888,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9642407,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-28T16:59:07Z\",\"WARC-Record-ID\":\"<urn:uuid:9c7ecfc0-0904-4349-b1dd-4697857ed227>\",\"Content-Length\":\"24752\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:82b9b76b-abe9-48eb-b46e-26c48483043b>\",\"WARC-Concurrent-To\":\"<urn:uuid:0c111267-db5c-4e8d-9812-9d57eac439fe>\",\"WARC-IP-Address\":\"54.164.10.157\",\"WARC-Target-URI\":\"https://support.bioconductor.org/p/62039/\",\"WARC-Payload-Digest\":\"sha1:HHYGW4SCM5HN7VPQ46U7L3NN3GB4R2TU\",\"WARC-Block-Digest\":\"sha1:RUP4BDU66NN4S3UGWY4SV6R3ETMPCFKN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335257.60_warc_CC-MAIN-20220928145118-20220928175118-00159.warc.gz\"}"}
https://www.aimsciences.org/article/doi/10.3934/dcds.2017187	[ "", null, "", null, "", null, "", null, "August 2017, 37(8): 4379-4390. doi: 10.3934/dcds.2017187\n\n## Livšic theorem for banach rings\n\n 1 Dept. of Math & Computer Science, St. John's University, Queens, NY, USA 2 Deptartment of Mathematics, The Pennsilvania State University, University Park, PA, USA\n\nReceived November 2016 Revised May 2017 Published April 2017\n\nThe Livšic Theorem for Hölder continuous cocycles with values in Banach rings is proved. We consider a transitive homeomorphism ${\\sigma :X\\to X}$ that satisfies the Anosov Closing Lemma and a Hölder continuous map ${a:X\\to B^\\times}$ from a compact metric space $X$ to the set of invertible elements of some Banach ring $B$. The map $a(x)$ is a coboundary with a Hölder continuous transition function if and only if $a(\\sigma^{n-1}p)\\ldots a(\\sigma p)a(p)$ is the identity for each periodic point $p=\\sigma^n p$.\n\nCitation: Genady Ya. Grabarnik, Misha Guysinsky. Livšic theorem for banach rings. Discrete & Continuous Dynamical Systems - A, 2017, 37 (8) : 4379-4390. doi: 10.3934/dcds.2017187\n##### References:\n H. Bercovici and V. Nitica, A Banach algebra version of the Livšic theorem, Discr. Contin. Dyn. Syst., 4 (1998), 523-534. doi: 10.3934/dcds.1998.4.523.", null, "", null, "Google Scholar H. Federer, Geometric Measure Theory, Springer, New York, 1969.", null, "Google Scholar H. Furstenberg and H. Kesten, Products of random matrices, The Annals of Mathematical Statistics, 31 (1960), 457-469. doi: 10.1214/aoms/1177705909.", null, "", null, "Google Scholar M. Guysinsky, Livšic Theorem for cocycles with values in the group of diffeomorphisms, preprint, 2013. Google Scholar B. Kalinin, Livšic theorem for matrix cocycles, Annals of Mathematics, 173 (2011), 1025-1042. doi: 10.4007/annals.2011.173.2.11.", null, "", null, "Google Scholar B. Kalinin and V. 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https://encyclopediaofmath.org/wiki/Dickman_function	[ "# Dickman function\n\nThe function", null, "defined on", null, "by the initial condition", null, "for", null, "and by the differential-delay equation", null, "for", null, ". Interest attaches to this function because of its connection to \"smooth\" numbers, i.e. numbers that are the product of many small prime numbers. Let", null, "denote the number of positive integers less than or equal to", null, "and free of prime divisors greater than", null, ". When", null, "is much larger than", null, ", it is a simple matter of inclusion-and-exclusion counting (cf. also Inclusion-exclusion formula) to show that", null, ", where", null, "denotes a prime number. But the error terms grow rapidly, and the \"main\" term gives the wrong answer in the ranges of greatest interest, including the case when", null, "is comparable to a fixed fractional power of", null, ". For this case, K. Dickman found that", null, ". If, in place of the restriction", null, "for", null, "one takes the condition", null, ", the resulting", null, "is approximated by", null, ", where", null, "is the Buchstab function, defined by", null, ",", null, ", and", null, ",", null, ", where for", null, "the right-hand derivative has to be taken, [a1]. Unlike", null, ",", null, "oscillates and tends to a positive limit, equal to", null, ".\n\nThere are two combinatorial identities linking the Dickman function to", null, ". Early work is based on the Buchstab identity: With", null, "denoting a prime number, for", null, ",", null, "The usual heuristic device of replacing a sum over prime numbers by an integral with \"prime density\"", null, "and replacing", null, "with", null, "leads to an identity which, when", null, "and", null, ", simplifies to an integral equivalent to the definition of", null, ". N.G. de Bruijn carried this idea to its limits in [a2], but accuracy suffers when large and comparable estimated quantities must be subtracted. The more recent Hildebrand identity involves only additions and has the further advantage that the second input is the same", null, "throughout:", null, "Applications require estimates uniform in", null, "; the best known estimate along these lines, due to A. Hildebrand and based on the identity above, is", null, "uniformly in", null, "and", null, ". There are similar results for algebraic integers, [a3].\n\nThere are also results concerning the number of smooth integers in an interval, and concerning the distribution of smooth integers into congruence classes [a4], [a5].\n\nThe Riemann hypothesis (cf. Riemann hypotheses) implies", null, "[a8].\n\nThe analytical properties of", null, "are reasonably well understood; calculus, analysis of the Laplace transform, and the saddle-point method are the key tools. The first extensive analysis is due to De Bruijn, and the function", null, "is sometimes termed the Dickman–De Bruijn function. One has [a10]:\n\na)", null, "for", null, "(so that", null, "is positive for all positive", null, ", and hence, from the definition, decreasing);\n\nb)", null, "is log-concave, that is,", null, "for", null, "and", null, "(K. Alladi);\n\nc) The Laplace transform", null, "", null, "d)", null, "as", null, ".\n\nThe Dickman function is one of a parameterized family of related functions", null, ", [a12], and a wider class of similar delay-differential equations has been studied in [a7]. A quick and simple bit of Mathematica code suffices to calculate", null, "to reasonable accuracy in the interval", null, "using step-size", null, ". This code calculates a table of values of", null, "at intervals of length", null, ", working back recursively into", null, ":\n\n<tbody> </tbody>", null, "", null, ",", null, ";", null, ";", null, "", null, "How to Cite This Entry:\nDickman function. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Dickman_function&oldid=41921\nThis article was adapted from an original article by D. Hensley (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. 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https://metanumbers.com/16649	[ "## 16649\n\n16,649 (sixteen thousand six hundred forty-nine) is an odd five-digits prime number following 16648 and preceding 16650. In scientific notation, it is written as 1.6649 × 104. The sum of its digits is 26. It has a total of 1 prime factor and 2 positive divisors. There are 16,648 positive integers (up to 16649) that are relatively prime to 16649.\n\n## Basic properties\n\n• Is Prime? Yes\n• Number parity Odd\n• Number length 5\n• Sum of Digits 26\n• Digital Root 8\n\n## Name\n\nShort name 16 thousand 649 sixteen thousand six hundred forty-nine\n\n## Notation\n\nScientific notation 1.6649 × 104 16.649 × 103\n\n## Prime Factorization of 16649\n\nPrime Factorization 16649\n\nPrime number\nDistinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 16649 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 9.72011 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 16,649 is 16649. Since it has a total of 1 prime factor, 16,649 is a prime number.\n\n## Divisors of 16649\n\n2 divisors\n\n Even divisors 0 2 2 0\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 16650 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 8325 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 129.031 Returns the nth root of the product of n divisors H(n) 1.99988 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 16,649 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 16,649) is 16,650, the average is 8,325.\n\n## Other Arithmetic Functions (n = 16649)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 16648 Total number of positive integers not greater than n that are coprime to n λ(n) 16648 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1927 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares\n\nThere are 16,648 positive integers (less than 16,649) that are coprime with 16,649. And there are approximately 1,927 prime numbers less than or equal to 16,649.\n\n## Divisibility of 16649\n\n m n mod m 2 3 4 5 6 7 8 9 1 2 1 4 5 3 1 8\n\n16,649 is not divisible by any number less than or equal to 9.\n\n• Arithmetic\n• Prime\n• Deficient\n\n• Polite\n\n• Prime Power\n• Square Free\n\n## Base conversion (16649)\n\nBase System Value\n2 Binary 100000100001001\n3 Ternary 211211122\n4 Quaternary 10010021\n5 Quinary 1013044\n6 Senary 205025\n8 Octal 40411\n10 Decimal 16649\n12 Duodecimal 9775\n20 Vigesimal 21c9\n36 Base36 cuh\n\n## Basic calculations (n = 16649)\n\n### Multiplication\n\nn×i\n n×2 33298 49947 66596 83245\n\n### Division\n\nni\n n⁄2 8324.5 5549.67 4162.25 3329.8\n\n### Exponentiation\n\nni\n n2 277189201 4614923007449 76833853151018401 1279206821111305358249\n\n### Nth Root\n\ni√n\n 2√n 129.031 25.5346 11.3592 6.98679\n\n## 16649 as geometric shapes\n\n### Circle\n\n Diameter 33298 104609 8.70816e+08\n\n### Sphere\n\n Volume 1.93309e+13 3.48326e+09 104609\n\n### Square\n\nLength = n\n Perimeter 66596 2.77189e+08 23545.2\n\n### Cube\n\nLength = n\n Surface area 1.66314e+09 4.61492e+12 28836.9\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 49947 1.20026e+08 14418.5\n\n### Triangular Pyramid\n\nLength = n\n Surface area 4.80106e+08 5.43874e+11 13593.9\n\n## Cryptographic Hash Functions\n\nmd5 bd896f3dbc16b0042625fbf0a8ab8b3a 74e0468cab3eff3280ec3bd2316121e29878476a b856aac2f8db97177fc96cdb7c7df3c1913ac76a9ee41240a1df79d40951512b f2d58d9e44c936d1dfeaa34b7abb3e3506bd267484748d09545c794d1a84d84b3195c24d9d50afdfed9bd5fcbd1d55ad3111a3ebf4f8df3fd5afa837bf86dd8f e80beee40e81aae44105cb9c7df9817a813b929a" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.63873214,"math_prob":0.9795389,"size":4495,"snap":"2020-34-2020-40","text_gpt3_token_len":1589,"char_repetition_ratio":0.12113115,"word_repetition_ratio":0.029717682,"special_character_ratio":0.44716352,"punctuation_ratio":0.07622739,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99632317,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-22T08:09:44Z\",\"WARC-Record-ID\":\"<urn:uuid:ab87f419-9bfc-481e-944b-bd3c0fe343bf>\",\"Content-Length\":\"47763\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9a9eae3d-db96-41e5-9977-b07f8e6ee469>\",\"WARC-Concurrent-To\":\"<urn:uuid:8d28ba8c-2716-48ed-bd14-d999d4692acc>\",\"WARC-IP-Address\":\"46.105.53.190\",\"WARC-Target-URI\":\"https://metanumbers.com/16649\",\"WARC-Payload-Digest\":\"sha1:DREYKBXR27UYDPNMMZ6MTFADAHUBCORU\",\"WARC-Block-Digest\":\"sha1:BBGWQJN5SA6OIMBA7LQENGJZA2QEFEUT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400204410.37_warc_CC-MAIN-20200922063158-20200922093158-00303.warc.gz\"}"}
https://www.dml.cz/handle/10338.dmlcz/703068	[ "# Article\n\nKeywords:\ngeneralized Newtonian fluids; Navier-Stokes equations; OpenFOAM; bypass\nSummary:\nThe aim of this work is to present numerical results of non-Newtonian fluid flow in a model of bypass. Different angle of a connection between narrowed channel and the bypass graft is considered. Several rheology viscosity models were used for the non-Newtonian fluid, namely the modified Cross model and the Carreau-Yasuda model. The results of non-Newtonian fluid flow are compared to the results of Newtonian fluid. The fundamental system of equations is the generalized system of Navier-Stokes equations for incompressible laminar flow. Generalized Newtonian fluids flow in the bypass is numerically simulated by using an open source CFD tool, OpenFOAM." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9089126,"math_prob":0.90651417,"size":756,"snap":"2021-43-2021-49","text_gpt3_token_len":159,"char_repetition_ratio":0.14893617,"word_repetition_ratio":0.018867925,"special_character_ratio":0.16931216,"punctuation_ratio":0.1007752,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9903952,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-05T23:23:33Z\",\"WARC-Record-ID\":\"<urn:uuid:f8eea92e-c51b-418b-ad6c-6132767efa16>\",\"Content-Length\":\"12564\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3ccc60b6-bb4a-4dc9-bade-c1f0cd88edb2>\",\"WARC-Concurrent-To\":\"<urn:uuid:04ef700d-31fe-4550-b480-82d870180eda>\",\"WARC-IP-Address\":\"147.251.6.150\",\"WARC-Target-URI\":\"https://www.dml.cz/handle/10338.dmlcz/703068\",\"WARC-Payload-Digest\":\"sha1:YQLDBXY62JT4VEAGAA2TECU7GWJQPB5X\",\"WARC-Block-Digest\":\"sha1:TWWM67J2JVHMSDEWBZOF5HKZACETCMZ6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363226.68_warc_CC-MAIN-20211205221915-20211206011915-00035.warc.gz\"}"}
http://mathcentral.uregina.ca/QandQ/topics/pick%20a%20number	[ "", null, "", null, "Math Central - mathcentral.uregina.ca", null, "", null, "Quandaries & Queries", null, "", null, "", null, "", null, "Q & Q", null, "", null, "", null, "", null, "Topic:", null, "pick a number", null, "", null, "", null, "start over\n\nOne item is filed under this topic.", null, "", null, "", null, "", null, "Page1/1", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Pick a number greater than 1 2004-06-25", null, "From A student:I understand that when you pick a number greater than 1 and less than 10; multiply it by 7 and add 23, then add the digits of that number until you get a one digit number. Then multiply that number by 9, add the digits of that number until you get a one digit number, subtract 3 from that number and divide the difference by 3; that this process will always give you the result of 2. Does this have a name or theory for it as to why the answer will always be 2?Answered by Penny Nom.", null, "", null, "", null, "", null, "", null, "", null, "Page1/1", null, "", null, "", null, "", null, "Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.", null, "", null, "", null, "", null, "about math central :: site map :: links :: notre site français" ]	[ null, "http://mathcentral.uregina.ca/lid/images/pixels/transparent.gif", null, "http://mathcentral.uregina.ca/lid/images/pixels/transparent.gif", null, "http://mathcentral.uregina.ca/lid/images/pixels/transparent.gif", null, "http://mathcentral.uregina.ca/lid/images/pixels/transparent.gif", null, "http://mathcentral.uregina.ca/lid/images/pixels/transparent.gif", null, "http://mathcentral.uregina.ca/lid/images/pixels/transparent.gif", null, "http://mathcentral.uregina.ca/lid/images/pixels/transparent.gif", null, 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https://ncatlab.org/schreiber/show/Rational+parameterized+stable+homotopy+theory	[ "# Schreiber Rational parameterized stable homotopy theory\n\nReferences\n\nAn article that we are writing:\n\n• Rational parameterized stable homotopy theory\n\nAbstract We combine Quillen-Sullivan’s classical rational homotopy theory with Schwede-Shipley’s stable homotopy theory of HR-module spectra to prove that rational parameterized spectra over a rational parameter space are equivalently modeled by the homotopy theory of dg(co-)modules over the Quillen-Sullivan dg(co-)algebras that models the parameter spaces. This implies also that rational parameterized spectra form a full subcategory of the slice homotopy theory of unbounded $L_\\infty$-algebras, thereby extending Hinich‘s embedding of rational homotopy theory in $L_\\infty$-algebras from non-negative to undounded degrees. Minimal models of augmented Sulllivan algebras regarded just as minimal dg-modules over the base algebra have been studied before by A. Roig, and our result shows that these are models for the fiberwise suspension spectra of the corresponding rational fibration.\n\nAn interesting example is the fiberwise suspension spectrum over the 3-sphere of the homotopy quotient of the 4-sphere $S^4 \\simeq S(\\mathbb{R} \\oplus \\mathbb{H})$ by the circle action: this turns out to contain a copy of twisted K-theory, rationally: $\\Sigma^\\infty_{S^3} (S^4/S^1) \\underset{\\mathbb{Q}}{\\simeq} ku/BU(1) \\oplus \\cdots$.\n\n$\\,$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.851294,"math_prob":0.98472875,"size":1308,"snap":"2022-05-2022-21","text_gpt3_token_len":293,"char_repetition_ratio":0.15260737,"word_repetition_ratio":0.0,"special_character_ratio":0.18425076,"punctuation_ratio":0.075555556,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9881253,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-24T19:46:12Z\",\"WARC-Record-ID\":\"<urn:uuid:142a28f4-225c-4359-a2f2-8725f01203a3>\",\"Content-Length\":\"17233\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ac9e679d-6018-4f2d-88a9-bd1adf1c6d1a>\",\"WARC-Concurrent-To\":\"<urn:uuid:7b509c9a-9ddb-4811-a468-e10bd23ac68b>\",\"WARC-IP-Address\":\"172.67.177.13\",\"WARC-Target-URI\":\"https://ncatlab.org/schreiber/show/Rational+parameterized+stable+homotopy+theory\",\"WARC-Payload-Digest\":\"sha1:ZERPXC5YM4X5FLBOVQIU5DSJH7GXBHEO\",\"WARC-Block-Digest\":\"sha1:4RQWRISZZBAHRPSKFVEBTG63Q7HOZ77Z\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662573189.78_warc_CC-MAIN-20220524173011-20220524203011-00146.warc.gz\"}"}
https://www.colorhexa.com/b2220c	[ "#b2220c Color Information\n\nIn a RGB color space, hex #b2220c is composed of 69.8% red, 13.3% green and 4.7% blue. Whereas in a CMYK color space, it is composed of 0% cyan, 80.9% magenta, 93.3% yellow and 30.2% black. It has a hue angle of 8 degrees, a saturation of 87.4% and a lightness of 37.3%. #b2220c color hex could be obtained by blending #ff4418 with #650000. Closest websafe color is: #993300.\n\n• R 70\n• G 13\n• B 5\nRGB color chart\n• C 0\n• M 81\n• Y 93\n• K 30\nCMYK color chart\n\n#b2220c color description : Dark red.\n\n#b2220c Color Conversion\n\nThe hexadecimal color #b2220c has RGB values of R:178, G:34, B:12 and CMYK values of C:0, M:0.81, Y:0.93, K:0.3. Its decimal value is 11674124.\n\nHex triplet RGB Decimal b2220c `#b2220c` 178, 34, 12 `rgb(178,34,12)` 69.8, 13.3, 4.7 `rgb(69.8%,13.3%,4.7%)` 0, 81, 93, 30 8°, 87.4, 37.3 `hsl(8,87.4%,37.3%)` 8°, 93.3, 69.8 993300 `#993300`\nCIE-LAB 38.964, 55.437, 47.902 19, 10.638, 1.401 0.612, 0.343, 10.638 38.964, 73.266, 40.829 38.964, 110.408, 28.111 32.616, 46.902, 20.285 10110010, 00100010, 00001100\n\nColor Schemes with #b2220c\n\n• #b2220c\n``#b2220c` `rgb(178,34,12)``\n• #0c9cb2\n``#0c9cb2` `rgb(12,156,178)``\nComplementary Color\n• #b20c49\n``#b20c49` `rgb(178,12,73)``\n• #b2220c\n``#b2220c` `rgb(178,34,12)``\n• #b2750c\n``#b2750c` `rgb(178,117,12)``\nAnalogous Color\n• #0c49b2\n``#0c49b2` `rgb(12,73,178)``\n• #b2220c\n``#b2220c` `rgb(178,34,12)``\n• #0cb275\n``#0cb275` `rgb(12,178,117)``\nSplit Complementary Color\n• #220cb2\n``#220cb2` `rgb(34,12,178)``\n• #b2220c\n``#b2220c` `rgb(178,34,12)``\n• #0cb222\n``#0cb222` `rgb(12,178,34)``\n• #b20c9c\n``#b20c9c` `rgb(178,12,156)``\n• #b2220c\n``#b2220c` `rgb(178,34,12)``\n• #0cb222\n``#0cb222` `rgb(12,178,34)``\n• #0c9cb2\n``#0c9cb2` `rgb(12,156,178)``\n• #6a1407\n``#6a1407` `rgb(106,20,7)``\n• #821909\n``#821909` `rgb(130,25,9)``\n• #9a1d0a\n``#9a1d0a` `rgb(154,29,10)``\n• #b2220c\n``#b2220c` `rgb(178,34,12)``\n• #ca270e\n``#ca270e` `rgb(202,39,14)``\n• #e22b0f\n``#e22b0f` `rgb(226,43,15)``\n• #f0371b\n``#f0371b` `rgb(240,55,27)``\nMonochromatic Color\n\nAlternatives to #b2220c\n\nBelow, you can see some colors close to #b2220c. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #b20c1f\n``#b20c1f` `rgb(178,12,31)``\n• #b20c12\n``#b20c12` `rgb(178,12,18)``\n• #b2140c\n``#b2140c` `rgb(178,20,12)``\n• #b2220c\n``#b2220c` `rgb(178,34,12)``\n• #b2300c\n``#b2300c` `rgb(178,48,12)``\n• #b23e0c\n``#b23e0c` `rgb(178,62,12)``\n• #b24c0c\n``#b24c0c` `rgb(178,76,12)``\nSimilar Colors\n\n#b2220c Preview\n\nThis text has a font color of #b2220c.\n\n``<span style=\"color:#b2220c;\">Text here</span>``\n#b2220c background color\n\nThis paragraph has a background color of #b2220c.\n\n``<p style=\"background-color:#b2220c;\">Content here</p>``\n#b2220c border color\n\nThis element has a border color of #b2220c.\n\n``<div style=\"border:1px solid #b2220c;\">Content here</div>``\nCSS codes\n``.text {color:#b2220c;}``\n``.background {background-color:#b2220c;}``\n``.border {border:1px solid #b2220c;}``\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, #0d0201 is the darkest color, while #fffaf9 is the lightest one.\n\n• #0d0201\n``#0d0201` `rgb(13,2,1)``\n• #1f0602\n``#1f0602` `rgb(31,6,2)``\n• #310903\n``#310903` `rgb(49,9,3)``\n• #440d05\n``#440d05` `rgb(68,13,5)``\n• #561006\n``#561006` `rgb(86,16,6)``\n• #681407\n``#681407` `rgb(104,20,7)``\n• #7b1708\n``#7b1708` `rgb(123,23,8)``\n• #8d1b0a\n``#8d1b0a` `rgb(141,27,10)``\n• #a01e0b\n``#a01e0b` `rgb(160,30,11)``\n• #b2220c\n``#b2220c` `rgb(178,34,12)``\n• #c4260d\n``#c4260d` `rgb(196,38,13)``\n• #d7290e\n``#d7290e` `rgb(215,41,14)``\n• #e92d10\n``#e92d10` `rgb(233,45,16)``\n• #f0391d\n``#f0391d` `rgb(240,57,29)``\n• #f1492f\n``#f1492f` `rgb(241,73,47)``\n• #f25941\n``#f25941` `rgb(242,89,65)``\n• #f36954\n``#f36954` `rgb(243,105,84)``\n• #f57966\n``#f57966` `rgb(245,121,102)``\n• #f68979\n``#f68979` `rgb(246,137,121)``\n• #f7998b\n``#f7998b` `rgb(247,153,139)``\n• #f8a99d\n``#f8a99d` `rgb(248,169,157)``\n• #fabab0\n``#fabab0` `rgb(250,186,176)``\n• #fbcac2\n``#fbcac2` `rgb(251,202,194)``\n``#fcdad4` `rgb(252,218,212)``\n• #fdeae7\n``#fdeae7` `rgb(253,234,231)``\n• #fffaf9\n``#fffaf9` `rgb(255,250,249)``\nTint Color Variation\n\nTones of #b2220c\n\nA tone is produced by adding gray to any pure hue. In this case, #625d5c is the less saturated color, while #b91d05 is the most saturated one.\n\n• #625d5c\n``#625d5c` `rgb(98,93,92)``\n• #695855\n``#695855` `rgb(105,88,85)``\n• #70524e\n``#70524e` `rgb(112,82,78)``\n• #784d46\n``#784d46` `rgb(120,77,70)``\n• #7f483f\n``#7f483f` `rgb(127,72,63)``\n• #864238\n``#864238` `rgb(134,66,56)``\n• #8d3d31\n``#8d3d31` `rgb(141,61,49)``\n• #953729\n``#953729` `rgb(149,55,41)``\n• #9c3222\n``#9c3222` `rgb(156,50,34)``\n• #a32d1b\n``#a32d1b` `rgb(163,45,27)``\n• #ab2713\n``#ab2713` `rgb(171,39,19)``\n• #b2220c\n``#b2220c` `rgb(178,34,12)``\n• #b91d05\n``#b91d05` `rgb(185,29,5)``\nTone Color Variation\n\nColor Blindness Simulator\n\nBelow, you can see how #b2220c 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":0.51380366,"math_prob":0.706406,"size":3652,"snap":"2019-26-2019-30","text_gpt3_token_len":1647,"char_repetition_ratio":0.12746711,"word_repetition_ratio":0.011111111,"special_character_ratio":0.5605148,"punctuation_ratio":0.23516238,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9867699,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-25T20:43:16Z\",\"WARC-Record-ID\":\"<urn:uuid:efb469bc-7db6-4fd5-bc85-6d8cbf932957>\",\"Content-Length\":\"36366\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b7c933fb-c95b-4dc3-a210-5d8e5e7964a6>\",\"WARC-Concurrent-To\":\"<urn:uuid:631661b8-f51f-45a0-939d-b2716636eea3>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/b2220c\",\"WARC-Payload-Digest\":\"sha1:W5T2OGYMJ2TT7KJWZ6SWD667YQ4PDUFK\",\"WARC-Block-Digest\":\"sha1:ZQB3PN37LTKU4DJ5WOXZVROXD7HVRH7D\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999946.25_warc_CC-MAIN-20190625192953-20190625214953-00109.warc.gz\"}"}
https://www.tradingmasters.io/cfd-course/moving-average	[ "Another function that can be used in charts for online trading is the moving average, which is widely used for the analysis of story series and especially for technical analysis. On the trading platforms (we recommend this one) you can set this type of graphic function and usually find three types: simple, exponential, and weighted.\n\nIn this lesson we will detail what these three types of moving averages are and how they differ, without going too far into technical details that you can explore personally later depending on your needs.\n\n## Simple Moving Average\n\nThe simple moving average (SMA) is also called arithmetic average and is in fact the most widely used by analysts and by those who trade online. The data of a given period is used and the average is calculated simply by adding it together and dividing the total by the number of values (e.g., 3 + 5 + 6 + 8 and divided by 4). However, the simple moving average is subject to criticism because it assigns the same importance (or weight) to each value. In practice, if we had a 10-period moving average, the same weight would be given to every single value (each of the 10 periods is 10%).\n\n## Weighted Moving Average\n\nThe weighted moving average (WMA) is used to address the previously highlighted problem with the simple moving average. Greater weight is given to the last values of a series. As for the final calculation, this remains unchanged: all the values of the same series are added together. For example, if we had 10 periods we would associate “weight” 1 to the first value, “weight” 2 to the second, and so on. These “weights” must be multiplied by the value of the data. For example, if the first period has a value of 3, we will multiply it by weight 1 and it will ultimately have a value of 3. If the fifth period is 7, we will multiply it by 5 and obtain a last value of 35. In the end, we add up the final values and divide them by the sum of the weights used. Example with three periods (5-6-7): the result will be [(5 × 1) + (6 × 2) + (7 × 3) / (1 + 2 + 3)] or 38/6 = 6.3333.\n\nEven this is subject to criticism because it fails to instantaneously provide an idea of what is happening on the market.\n\n## Exponential Moving Average\n\nThe exponential moving average (EMA) is much more complex than the averages described above and is therefore used by more experienced users. As in the weighted average, a different weight is assigned to the various prices (or values) being considered, greater for the most recent and lesser for the oldest, with the major difference being that it takes into account much more data (e.g., many “older” values).\n\nBeing a very difficult calculation medium, it is practically impossible to generate except through the use of a computer. By the way, we can get a perfect EMA with a single mouse clic, on the trading platform.\n\nHowever, to calculate the exponential moving average two values are taken into consideration:\n\n• The arithmetic mean (seen previously)\n• The Alpha coefficient, which will make the average more reactive.\n\nTo calculate Alpha, use this formula:\n\nAlpha = 2/(n + 1) where n is the number of periods.\n\nThe exponential moving average is obtained by:\n\nEMA (exp. M.) = (Close – previous EMA) * Alpha + previous EMA\n\nWith respect to the weighted average, the weighting coefficients are assigned with an exponential and non-linear progressive value. The difference in “weight” between the different values, therefore, varies exponentially.\n\nGo to lesson 15 – Alligator indicator" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9408227,"math_prob":0.99077666,"size":3480,"snap":"2019-35-2019-39","text_gpt3_token_len":774,"char_repetition_ratio":0.13636364,"word_repetition_ratio":0.009803922,"special_character_ratio":0.23103449,"punctuation_ratio":0.09985097,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9977535,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-21T11:51:11Z\",\"WARC-Record-ID\":\"<urn:uuid:934b67ae-5e6c-408a-a70a-119cc4b06879>\",\"Content-Length\":\"91094\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:247b1d12-d1ed-4ca3-9ea2-dee5085b47a0>\",\"WARC-Concurrent-To\":\"<urn:uuid:47331dac-d4de-40ff-89ed-a388a7f8b372>\",\"WARC-IP-Address\":\"104.18.55.123\",\"WARC-Target-URI\":\"https://www.tradingmasters.io/cfd-course/moving-average\",\"WARC-Payload-Digest\":\"sha1:DPBWSHWUEEU5YEDBFS72DFF52AOHSNVS\",\"WARC-Block-Digest\":\"sha1:4Q7TBVDQ3DWFETKDUUSGRBBYKFADAEGS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514574409.16_warc_CC-MAIN-20190921104758-20190921130758-00068.warc.gz\"}"}
https://www.cuemath.com/ncert-solutions/direct-and-inverse-proportions-class-8-maths/	[ "# NCERT Class 8 Maths Direct and Inverse Proportions\n\nThe chapter 13 begins with an introduction to Direct and Inverse Proportions by citing situations in our day-to-day life, where variation in one quantity bringing in variation in the other to delve into the subject of interest.After this , Direct Proportion is explained by stating some real-life scenario and there are some related examples to dig deep into it.Similarly, Inverse Proportion i.e if one quantity increases, the other quantity decreases and vice versa, is also discussed in the same manner with some real-life examples and both of them have some exercise problems to work on.\n\nDownload FREE PDF of Chapter-13 Direct and Inverse Proportions\n\n## Chapter 13 Ex.13.1 Question 1\n\nFollowing are the car parking charges near a railway station up to:\n\n$$4$$ hours ₹$$60$$\n\n$$8$$ hours ₹$$100$$\n\n$$12$$ hours ₹$$140$$\n\n$$24$$ hours ₹$$180$$\n\nCheck if the parking charges are in direct proportion to the parking time.\n\n### Solution\n\nWhat is Known?\n\nParking charges for different hours.\n\nWhat is Unknown?\n\nParking charges are direct proportion to the parking time or not.\n\nReasoning:\n\nIf two quantities are related in such a way that an increase in one lead to a corresponding proportional increase in the other, then such a variation is called direct variation.\n\nSteps:\n\nThe parking charge for $$1$$ hour in all the four cases then the variation is direct.\n\nWe have:\n\n\\begin{align}\\frac{{60}}{4}&=\\frac{{15}}{1}\\\\&=15\\\\ \\\\ \\frac{{100}}{8}&=\\frac{{25}}{2}\\\\ &= 12.5\\\\ \\\\ \\frac{{140}}{{12}} &=\\frac{{35}}{3}\\\\&=11.67 \\\\ \\\\\\frac{{180}}{{24}}&=\\frac{{15}}{2}\\\\&= 7.50\\end{align}\n\nSince all the values are not the same, the parking charges are not in direct proportion to parking times.\n\n## Chapter 13 Ex.13.1 Question 2\n\nA mixture of paint is prepared by mixing $$1$$ part of red pigments with $$8$$ parts of base. In the following table find the parts of base that needed to be added.\n\n Parts of red pigment 1 4 7 12 20 Parts of base 8 ... ... ... ...\n\n### Solution\n\nWhat is Known?\n\nParts of red pigments used.\n\nWhat is Unknown?\n\nParts of base used.\n\nReasoning:\n\nTwo numbers $$x$$ and $$y$$ are said to vary in direct proportion if\n\n$\\frac{x}{y} = k,x = yk$\n\nWhere, $$k$$ is a constant.\n\nSteps:\n\nLet the parts of red pigments used be $$x$$ and parts of base used be $$y.$$\n\n\\begin{align}\\therefore \\frac{{{x_1}}}{{{y_1}}} = \\frac{{{x_2}}}{{{y_2}}}\\end{align}\n\nHere\n\n\\begin{align} \\quad{x_1} &= 1, \\qquad {x_2} = 4 \\\\ {y_1} &= 8,\\qquad {y_2} = ?\\\\ \\frac{{{x_1}}}{{{y_1}}} &= \\frac{{{x_2}}}{{{y_2}}} \\\\ \\frac{1}{8} &= \\frac{4}{{y2}}\\\\ \\,{y_2} &= 8 \\times 4 \\\\ &= 32{\\text{ parts }}\\end{align}\n\n$$32$$ parts of base is needed for $$4$$ parts of red pigment.\n\nHere\n\n\\begin{align} {{x}_{1}}&=1,\\quad \\ {{x}_{2}}=7 \\\\ {{y}_{1}}&=8,\\quad{{y}_{2}}=? \\\\ \\frac{{{x}_{1}}}{{{y}_{1}}}&=\\frac{{{x}_{2}}}{{{y}_{2}}} \\\\ \\,\\,\\frac{1}{8}&=\\frac{7}{{{y}^{2}}} \\\\ {{y}_{2}}&=8\\times 7 \\\\ \\,\\,\\,\\,\\,\\,&=56\\,\\ \\text{parts} \\end{align}\n\n$$56$$ parts of base is needed for $$7$$ parts of red pigment.\n\nHere\n\n\\begin{align} \\,\\,\\,\\,\\,\\,\\,\\,{{x}_{1}}&=1,\\quad{{x}_{2}}=12 \\\\ {{y}_{1}}&=8,\\quad{{y}_{2}}=? \\\\ \\frac{{{x}_{1}}}{{{y}_{1}}}&=\\frac{{{x}_{2}}}{{{y}_{2}}} \\\\ \\frac{1}{8}&=\\frac{12}{{{y}_{2}}} \\\\ {{y}_{2}}&=8\\times 12\\quad \\\\ &=96 \\end{align}\n\n$$96$$ parts of base is needed for $$12$$ parts of red pigment.\n\nHere\n\n\\begin{align} {{x}_{1}}&=1,\\quad{{x}_{2}}=20 \\\\ {{y}_{1}}&=8,\\quad{{y}_{2}}=? \\\\ \\frac{{{x}_{1}}}{{{y}_{1}}}&=\\frac{{{x}_{2}}}{{{y}_{2}}} \\\\ \\frac{1}{8}&=\\frac{20}{{{y}_{2}}} \\\\ {{y}_{2}}&=8\\times 20 \\\\ &=160 \\end{align}\n\n$$160$$ parts of base are needed for $$20$$ parts of red pigment.\n\n## Chapter 13 Ex.13.1 Question 3\n\nIn question ($$2$$), above if $$1$$ part of red pigment requires $$75\\,\\rm{ml}$$ of base, how much red pigment should we mix with $$1800\\,\\rm{ml}$$ of base?\n\n### Solution\n\nWhat is Known?\n\n$$1$$ part of red pigment requires $$75\\,\\rm{ml}$$ of base.\n\nWhat is Unknown?\n\n$$1800$$ mL of base needed how much red pigment?\n\nReasoning:\n\nTwo numbers $$x$$ and $$y$$ are said to vary in direct proportion if,\n\n\\begin{align}\\frac{x}{y} = k, \\qquad x = y\\,k\\end{align}\n\nWhere $$k$$ is a constant.\n\nSteps:\n\nLet the number of parts of red pigment be $$x$$\n\nAs the number of parts of red pigment increases, amount of base also increases in the same ratio. So it is a case of direct proportion.\n\nHere,\n\n\\begin{align} \\frac{{{x}_{1}}}{{{y}_{1}}}&=\\frac{{{x}_{2}}}{{{y}_{2}}} \\\\ {{x}_{1}}&=1,\\qquad {{x}_{2}}=? \\\\ {{y}_{1}}&=8,\\qquad{{y}_{2}}=1800 \\\\ \\frac{1}{75}&=\\frac{{{x}_{2}}}{1800} \\\\ {{x}_{2}}&=\\frac{1\\ \\times \\ 1800}{75} \\\\ {{x}_{2}}&=24 \\end{align}\n\n$$24$$ parts of red pigment should be mixed with $$1800\\,\\rm{ml}$$ of base.\n\n## Chapter 13 Ex.13.1 Question 4\n\nA machine in a soft drink factory fills $$840$$ bottles in $$6$$ hours. How many bottles it will fill in five hours?\n\n### Solution\n\nWhat is Known?\n\n$$840$$ bottles can be filled in $$6$$ hours.\n\nWhat is Unknown?\n\nBottles filled in $$5$$ hours.\n\nReasoning:\n\nTwo numbers $$x$$ and $$y$$ are said to be in direct proportion, if\n\n\\begin{align}\\frac{x}{y}=k,\\quad x=y\\,k\\end{align}\n\nWhere $$k$$ is a constant.\n\nSteps:\n\n No of bottles Time in hours $${x}_{1}=840$$ $${y}_{1}=6$$ $${x}_{2}=?$$ $${y}_{2}=5$$\n\nSo, the number of bottles filled, and numbers of hours are directly proportional to each other.\n\n\\begin{align} \\frac{{{x}_{1}}}{{{y}_{1}}}&=\\frac{{{x}_{2}}}{{{y}_{2}}} \\\\ \\frac{840}{6}&=\\frac{{{x}_{2}}}{5} \\\\ 6{{x}_{2}}&=840\\,\\times \\,5 \\\\ {{x}_{2}}&=\\frac{840\\times 5}{6} \\\\ {{x}_{2}}&=700 \\end{align}\n\n$$700$$ bottles will be filled in $$5$$ hours.\n\n## Chapter 13 Ex.13.1 Question 5\n\nA photograph of a bacteria is enlarged $$50,000$$ times attains a length of $$5 \\rm{cm}$$ as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged $$20,000$$ times only, what would be its enlarged length?\n\n### Solution\n\nWhat is Known?\n\nBacteria enlarged $$50,000$$ times attain a length of $$5 \\,\\rm{cm.}$$\n\nWhat is Unknown?\n\nActual length of the bacteria\n\nif $$20,000$$ times enlarged what will be the length of the bacteria?\n\nReasoning:\n\nTwo numbers $$x$$ and $$y$$ are said in direct proportion if,\n\n\\begin{align}\\frac{x}{y}=k,\\quad x=y\\,k\\end{align}\n\nWhere $$k$$ is a constant.\n\nSteps:\n\n\\begin{align} \\text{Actual length,}\\ l&=\\frac{{{y}_{1}}}{{{x}_{1}}} \\\\ l&=\\frac{5}{50000} \\\\ l&=0.0001\\ \\rm{cm} \\\\ \\end{align}\n\n Number of times enlarged Length attained ${{x_1} = {\\rm{50,000}}}$ ${{y_{\\rm{1}}} = {\\rm{5}}}$ ${{x_2} = {\\rm{20,000}}}$ ${{y_{\\rm{2}}} = {\\rm{?}}}$\n\nThe number of times enlarged is directly proportional to the length attained.\n\nActual length $$= 0.0001 \\,\\rm{cm}$$\n\nEnlarged length will be $$2 \\,\\rm{cm.}$$\n\n## Chapter 13 Ex.13.1 Question 6\n\nIn a model of a ship, the mast is $$9 \\,\\rm{cm}$$ high, while the mast of the actual ship is $$12\\, \\rm{cm}$$ high. If the length of the actual ship is $$28\\;\\rm{m}$$, how long is the model ship?\n\n### Solution\n\nWhat is Known?\n\nThe mast is $$9 \\,\\rm{cm}$$ high while the mast of actual ships is $$12\\, \\rm{cm}$$ high.\n\nWhat is Unknown?\n\nIf the length of the ship is $$28\\;\\rm{m}$$ How long is the model ship?\n\nReasoning:\n\nTwo numbers $$x$$ and $$y$$ are said in direct proportion if,\n\n\\begin{align}\\frac{x}{y} = k,\\quad x = y\\,k\\end{align}\n\nWhere $$k$$ is a constant.\n\nSteps:\n\n Actual ship Model ship $${y}_{1}=12\\;\\rm{m}$$ $${y}_{2}=9\\;\\rm{cm}$$ $${x}_{1}=28\\rm{m}$$ $${x}_{2}=?$$\n\nMore the length of the ship more would be the length of its mast. Hence, this is a direct proportion.\n\n\\begin{align}\\frac{{{x_1}}}{{{y_1}}} &= \\frac{{{x_2}}}{{{y_2}}}\\\\\\frac{{28}}{{12}} &= \\frac{{{x_2}}}{9}\\\\12 \\times {x_2} &= 28 \\times 9\\\\{x_2} &= \\frac{{28 \\times 9}}{{12}}\\\\{x_2} &= 21\\;{\\rm{m}}\\end{align}\n\nLength of the model ship is $$21\\;\\rm{ m.}$$\n\n## Chapter 13 Ex.13.1 Question 7\n\nSuppose $$2\\,\\rm{kg}$$ of sugar contains $$9 \\times {10^6}$$ crystals. How many sugar crystals are there in\n\n(1) $$5\\,\\rm{kg}$$ of sugar?\n\n(2) $$1.2\\,\\rm{kg}$$ of sugar?\n\n### Solution\n\n(i) How many crystals are there in $$5\\, \\rm{kg}$$ of crystals?\n\nWhat is Known?\n\n$$2\\,\\rm{kg}$$ of sugar contains $$9 × 10^6$$ crystals.\n\nWhat is Unknown?\n\n(i) $$5\\,\\rm{kg}$$ of sugar contains how many crystals?\n\nReasoning:\n\nTwo numbers $$x$$ and $$y$$ are said in direct proportion if,\n\n\\begin{align}\\frac{x}{y} = k,\\quad\\;\\; x = yk\\,\\end{align}\n\nWhere $$k$$ is a constant.\n\nSteps (i):\n\n Amount of sugar No. of crystals $$2$$ $${9 \\times 10^6}$$ $$5$$ ?\n\nMore the amount of sugar more will be the number of crystals. Hence this is a direct proportion.\n\n\\begin{align}\\frac{{{x_1}}}{{{y_1}}} &= \\frac{{{x_2}}}{{{y_2}}}\\\\\\frac{2}{{9 \\times {{10}^6}}}&= \\frac{5}{{{y_2}}}\\\\2 \\times y_2 &= 9 \\times {10^6} \\times 5\\\\ y_2 &= \\frac{{9 \\times {{10}^6} \\times 5}}{2}\\\\{y_2} &= 22.5 \\times {10^6}\\\\{y_2} &= 2.25 \\times {10^7}\\end{align}\n\nHence there are $$2.25 \\times {10^7}$$ crystals.\n\n(ii) How many crystals are there in $$1.2\\,\\rm{kg}$$ of crystals?\n\nWhat is Known?\n\n$$2\\,\\rm{kg}$$ of sugar contains $$9 × 10^6$$ crystals.\n\nWhat is UnKnown?\n\nHow many crystals are there in $$1.2\\,\\rm{kg}$$ of crystals?\n\nSteps (ii):\n\n\\begin{align}\\frac{{{x_1}}}{{{y_1}}} &= \\frac{{{x_2}}}{{{y_2}}}\\\\\\frac{2}{{9 \\times {{10}^6}}} &= \\frac{{1.2}}{{{y_2}}}\\\\2 \\times y_2 &= 9 \\times {10^6} \\times 1.2\\\\ y_2 &=\\frac{{9 \\times {{10}^6} \\times 5}}{2}\\\\{y_2} &= 5.4 \\times {10^6}\\end{align}\n\nHence there are $$5.4 \\times {10^6}$$ crystals.\n\n## Chapter 13 Ex.13.1 Question 8\n\nRashmi has a road map with a scale of $$1 \\;\\rm{cm}$$ representing $$18 \\,\\rm{km}$$. She drives on a road for $$72 \\,\\rm{km}$$. What would be her distance covered in the map?\n\n### Solution\n\nWhat is Known:\n\nThe scale of representing $$1\\; \\rm{cm}$$ $$=$$ $$18 \\,\\rm{km}$$\n\nWhat is Unknown:\n\nThe distance covered in map when the distance on road is $$72 \\,\\rm{km.}$$\n\nReasoning:\n\nTwo numbers $$x$$ and $$y$$ are said in direct proportion if\n\n\\begin{align}\\frac{x}{y} = k,\\, \\qquad x = y\\,k\\end{align}\n\nWhere $$k$$ is a constant.\n\nThe map is a representation of very large region. The scale shows the representation of the actual length and the length represented in map.\n\nSteps:\n\n$$1\\,\\rm{cm}$$ on map represents $$18 \\,\\rm{km}$$ of actual distance then $$2\\, \\rm{cm}$$ on the map represents $$36 \\,\\rm{km}$$. Hence the scale is based on the concept of direct proportion.\n\n\\begin{align}1:18 &= x:72\\\\\\frac{{1}}{18} &= \\frac{x}{{72}}\\\\18 \\times x &= 72 \\times 1\\\\x &= \\frac{{72}}{{18}}\\\\x &= 4\\end{align}\n\nThe distance covered in the map would be $$4 \\,\\rm{cm}$$.\n\n## Chapter 13 Ex.13.1 Question 9\n\nA $$5\\,\\rm{m}$$ $$60\\,\\rm {cm}$$ high vertical pole casts a shadow $$3\\;\\rm{m}$$ $$20 \\;\\rm{cm}$$ long. Find at the same time\n\n(i) Length of the shadow cast by another pole $$10 \\;\\rm{m}$$ $$50 \\;\\rm{cm}$$ high.\n\n(ii) The height of the pole which casts a shadow of $$5\\;\\rm{m}$$ long.\n\n### Solution\n\n(i) Length of the shadow cast by another pole $$10\\; \\rm{m}$$ $$50 \\;\\rm{cm}$$ high.\n\nWaht is Known?\n\n$$5.6\\, \\rm{m}$$ vertical pole casts a shadow of $$3.2\\; \\rm{m}$$ long.\n\nWaht is Unknown?\n\nThe length of a shadow cast by a pole $$10.5\\; \\rm{m}$$ high.\n\nReasoning:\n\nTwo numbers $$x$$ and $$y$$ are said in direct proportion if\n\n\\begin{align}\\frac{x}{y} = k,\\quad x= y\\,k\\end{align}\n\nWhere $$k$$ is a constant.\n\nSteps:\n\n Height of the pole Length of the shadow $${5.6{\\rm{m}}}$$ $${3.2{\\rm{m}}}$$ $${10.5{\\rm{m}}}$$ $$?$$\n\nAs the height of the pole increases the length of the shadow also increases. So, it is a direct proportion.\n\n\\begin{align}\\frac{{{x_1}}}{{{y_1}}}&= \\frac{{{x_2}}}{{{y_2}}}\\\\\\frac{{5.6}}{{3.2}}&= \\frac{{10.5}}{{{y_2}}}\\\\\\,\\,5.6\\,\\, \\times &= 10.5 \\times 3.2\\\\{y_2}&= \\frac{{10.5 \\times 3.2}}{{5.6}}\\\\ y_2 & = 6\\end{align}\n\nIf the height of the pole is $$10.5\\; \\rm{m}$$, then length of the shadow is $$6\\;\\rm{m}$$.\n\n(ii) The height of the pole which casts a shadow of $$5\\,\\rm{m}$$ long.\n\nWaht is Known?\n\n$$5.6 \\,\\rm{m}$$ vertical pole casts a shadow of $$3.2\\,\\rm{m}$$ long.\n\nWaht is Unknown?\n\nThe height of the pole when the length of the shadow is $$5\\,\\rm{m}$$ long.\n\nSteps:\n\n\\begin{align}\\frac{{{x_1}}}{{{y_1}}}&= \\frac{{{x_2}}}{{{y_2}}}\\\\\\frac{{5.6}}{{3.2}}&= \\frac{{{x_2}}}{5}\\\\3.2x &= 5 \\times 5.6\\\\{x_2}&= \\frac{{5 \\times 5.6}}{{3.2}}\\\\{x_2}&= 8.75\\end{align}\n\nIf the height of the pole is $$5\\,\\rm{m}$$, then length of the shadow is $$8.75\\, \\rm{m.}$$\n\n## Chapter 13 Ex.13.1 Question 10\n\nA loaded truck travels $$14\\,\\rm{km}$$ in $$25$$ minutes. If the speed remains the same, how far it travels in $$5$$ hours?\n\n### Solution\n\nWhat is Known?\n\nTruck travels $$14\\,\\rm{km}$$ in $$25$$ minutes.\n\nWhat is Unknown?\n\nDistance travelled in $$5$$ hours.\n\nReasoning:\n\nTwo numbers $$x$$ and $$y$$ are said in direct proportion if,\n\n\\begin{align}\\frac{x}{y} = k,\\quad x = k\\,y\\end{align}\n\nWhere $$k$$ is a constant.\n\nSteps:\n\nIn $$25$$ minutes, it travels $$14 \\,\\rm{km}$$. In $$5$$ hours, it will travel more distance. So, it is a case of direct proportion.\n\n Distance Time in minutes $${{\\rm{14}}}$$ $${{\\rm{25}}}$$ $${\\,{\\rm{?}}}$$ $$5 \\times 60$$ ($$1$$ hour $$=$$ $$60$$ minutes)\n\n[For comparison the unit should be same]\n\n\\begin{align}\\frac{{{x_1}}}{{{y_1}}} &= \\frac{{{x_2}}}{{{y_2}}}\\\\\\frac{{14}}{{25}} &= \\frac{{{x_2}}}{{5 \\times 60}}\\\\25\\,{x_2} &= 5 \\times 60 \\times 14\\\\{x_2} &= \\frac{{5 \\times 60 \\times 14}}{{25}}\\\\{x_2} &= 168\\;{\\rm{km}}\\end{align}\n\nHence the truck can travel $$168 \\,\\rm{km}$$ in $$5$$ hours.\n\nDownload FREE PDF of Chapter-13 Direct and Inverse Proportions\nDirect and Inverse Proportions \| NCERT Solutions\nLearn from the best math teachers and top your exams\n\n• Live one on one classroom and doubt clearing\n• Practice worksheets in and after class for conceptual clarity\n• Personalized curriculum to keep up with school" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.69488686,"math_prob":1.0000049,"size":13436,"snap":"2021-21-2021-25","text_gpt3_token_len":4757,"char_repetition_ratio":0.15589637,"word_repetition_ratio":0.14330874,"special_character_ratio":0.44805002,"punctuation_ratio":0.1370007,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99997616,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-13T06:17:23Z\",\"WARC-Record-ID\":\"<urn:uuid:cc1440f5-dc46-4ba1-ac3b-c147fdde16fe>\",\"Content-Length\":\"147189\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cc9bb148-3938-45c2-ada9-f637a887ac09>\",\"WARC-Concurrent-To\":\"<urn:uuid:24eae51f-66fa-4ed7-a3d5-d1b9c253f862>\",\"WARC-IP-Address\":\"104.18.18.168\",\"WARC-Target-URI\":\"https://www.cuemath.com/ncert-solutions/direct-and-inverse-proportions-class-8-maths/\",\"WARC-Payload-Digest\":\"sha1:REUXFKNONWFUDM5KGZQCXL4V4BO7VEB3\",\"WARC-Block-Digest\":\"sha1:2CTF554F4OKXHYMILY7EQ3WJ476NDYOE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487600396.21_warc_CC-MAIN-20210613041713-20210613071713-00193.warc.gz\"}"}
https://financetrainingcourse.com/education/tag/var-calculations/	[ "", null, "# VaR calculations", null, "## Calculating Value at Risk Excel\n\n2 mins read time Calculating Value at Risk Excel Our second course on Risk […]\n\n## Calculating Value at Risk – Approach Specific Steps\n\n5 mins read time Calculating Variance-Covariance (VCV) Value at Risk (VaR) This method assumes […]\n\n## Calculating Value at Risk – Data & Return Series\n\n1 Comment\n\n4 mins read time Methodology Setting the Scene Sample Portfolio Our sample portfolio that […]\n\n## Calculating Value at Risk – Introduction\n\n1 Comment\n\n2 mins read time One of the most pertinent questions in risk management has […]" ]	[ null, "data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==", null, "data:image/svg+xml,%3Csvg%20xmlns=%22http://www.w3.org/2000/svg%22%20viewBox=%220%200%20480%20220%22%3E%3C/svg%3E", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.80267364,"math_prob":0.9315951,"size":462,"snap":"2019-43-2019-47","text_gpt3_token_len":100,"char_repetition_ratio":0.17685589,"word_repetition_ratio":0.32894737,"special_character_ratio":0.22510822,"punctuation_ratio":0.0,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9833866,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-17T01:53:43Z\",\"WARC-Record-ID\":\"<urn:uuid:1cb18706-2274-4620-947e-ce9005d67478>\",\"Content-Length\":\"30552\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:262ae6ea-ad18-4024-8437-d35d8b14df43>\",\"WARC-Concurrent-To\":\"<urn:uuid:be3cf0a9-f2be-4df8-9af0-b0abb26d593c>\",\"WARC-IP-Address\":\"104.27.158.78\",\"WARC-Target-URI\":\"https://financetrainingcourse.com/education/tag/var-calculations/\",\"WARC-Payload-Digest\":\"sha1:RZOYP4I5BZ3NA5G7FQSGI7GNEQ7F2YBZ\",\"WARC-Block-Digest\":\"sha1:EIRBUDCTQAZW4TXBNMJZC3ABUQSU3HZG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668782.15_warc_CC-MAIN-20191117014405-20191117042405-00214.warc.gz\"}"}
https://www.hindawi.com/journals/amp/2019/4364108/	[ "/ / Article\n\nResearch Article \| Open Access\n\nVolume 2019 \|Article ID 4364108 \| 6 pages \| https://doi.org/10.1155/2019/4364108\n\n# Symmetry, Pulson Solution, and Conservation Laws of the Holm-Hone Equation\n\nRevised06 Jan 2019\nAccepted13 Jan 2019\nPublished03 Feb 2019\n\n#### Abstract\n\nIn this paper, we focus on the Holm-Hone equation which is a fifth-order generalization of the Camassa-Holm equation. It was shown that this equation is not integrable due to the nonexistence of a suitable Lagrangian or bi-Hamiltonian structure and negative results from Painlevé analysis and the Wahlquist-Estabrook method. We mainly study its symmetry properties, travelling wave solutions, and conservation laws. The symmetry group and its one-dimensional optimal system are given. Furthermore, preliminary classifications of its symmetry reductions are investigated. Also we derive some solitary pattern solutions and nonanalytic first-order pulson solution via the ansatz-based method. Finally, some conservation laws for the fifth-order equation are presented.\n\n#### 1. Introduction\n\nIn the study of shallow water waves, Camassa and Holm derived a nonlinear dispersive shallow water wave equation which is called Camassa-Holm (CH) equation now. The here denotes the fluid velocity at time in the direction . Eq. (1) admits bi-Hamiltonian structure and it is completely integrable [3, 4]. What is more, the equation has infinite conservation laws as well as a spike solitary wave solution where is an arbitrary constant. The solitary wave curve of the solution has a cusp at the peak, and the first derivative of the cusp is not continuous. Accordingly, it is called a peakon [6, 7].\n\nWith the further researches on the CH equation, a lot of findings about the equation have been obtained and it is impossible to give a comprehensive overview here. For example, Eq. (1) represents the equation for geodesics on the Bott-Virasoro group and owns the geometric interpretation . The CH equation possesses both global solutions and solutions developing singularities in finite time and the blow-up happens in a way which resembles wave breaking to some extent [9, 10]. The well-developed inverse scattering theory can also be used to integrate the CH flow . In , it was shown that the well-known CH equation is included in the negative order CH hierarchy and a class of new algebro-geometric solutions of the CH equation was presented. Moreover, the time evolution of traveling-wave solutions and the interaction of peaked and cusped waves were numerically studied .\n\nRecently, one of the latest trends is that researchers are trying to generalize the CH equation to higher order, which is also the subject of this paper. In fact, some higher-order CH equations are well considered. For example, Wazwaz studied the nonlinear fourth-order dispersive variants of the generalized CH equation by using sine-cosine method . The existence of global weak solutions was established for a higher-order CH equation describing exponential curves of the manifold of smooth orientation-preserving diffeomorphisms of the unit circle in the plane . Nonsmooth travelling wave solutions of a generalized fourth-order nonlinear CH equation were studied in . In , the CH model was extended to fifth order and some interesting solutions were obtained including explicit single pseudo-peakons, two-peakon, and -peakon solutions.\n\nEspecially, Holm and Hone discussed a fifth-order partial differential equation (PDE)which is a generalization of the integrable CH equation . We will call it the Holm-Hone equation in the following. This fifth-order PDE admits exact solutions in terms of an arbitrary number of superposed pulsons. The pulsons of Eq. (3) are weak solutions with discontinuous second derivatives at isolated points. Numerical simulations show that the pulsons are stable and they dominate the initial value problem and scatter elastically . These characteristics are reminiscent of solitons in integrable systems. However, after demonstrating the nonexistence of a suitable Lagrangian or bi-Hamiltonian structure and obtaining negative results from Painlevé analysis and the Wahlquist-Estabrook method , they asserted that the Holm-Hone equation is not integrable. Consequently, most effective methods for integrable systems are not applicable to this equation. However, the Lie symmetry approach originated from Norwegian mathematician Sophus Lie systematically unifies and extends well-known ad hoc techniques to construct explicit solutions for differential equations, no matter the equations are integrable or not.\n\nActually, there are several important applications of symmetry analysis in the investigation of differential equations . Since some solutions of partial differential equations asymptotically tend to solutions of lower-dimensional equations obtained by symmetry reduction, some of these special solutions illustrate important physical phenomena. In particular, the exact solutions arising from symmetry methods can often be used effectively to study properties such as asymptotics and blow-up. Besides, the explicit solutions found by symmetry methods can play important roles in the design and testing of numerical integrators, and these solutions provide an important practical check on the accuracy and reliability of such integrators.\n\nThis paper is organized as follows. We first present the common form of the infinitesimal generators of the Holm-Hone equation. Then we establish the optimal system of one-dimensional subalgebras and consider the similarity reductions of the equation. Furthermore, the travelling wave solutions are investigated. At last, some conservation laws of the equation are also derived.\n\n#### 2. Lie Symmetries for the Holm-Hone Equation\n\nIn this section, we investigate the Lie symmetries and similarity reductions of the Holm-Hone equation through the classical methods. The infinitesimal generators corresponding to the one-parameter transformation group are presented. Furthermore, the one-dimensional optimal system of the group is derived with the help of the adjoint representation among the vector fields. Finally, the reduced equations are given through the similarity transformation.\n\n##### 2.1. Infinitesimal Generators\n\nWe introduce the infinitesimal form of the single parameter transformation groupwhere is the infinitesimal group parameter and , , are the infinitesimals of the transformation for the independent and dependent variables, respectively. The vector field associated with the above group of transformations can be written as Using the software package GeM , we obtainwhere , , are arbitrary constants. The infinitesimal generators of the corresponding Lie algebra are given by\n\nIn order to obtain the group transformation which is generated by the infinitesimal generators for , we need to solve the first-order ordinary differential equations Exponentiating the infinitesimal symmetries we get the one-parameter groups generated by for as Accordingly, if is a solution of the Holm-Hone equation, so are the functions\n\n##### 2.2. One-Dimensional Optimal System\n\nIn general, the Lie group has infinite subgroups; it is not usually feasible to list all possible group invariant solutions to the system. We need an effective systematic approach to classify these solutions, so the optimal 1-dimensional subalgebras of group invariant solutions can be derived. In this section, the optimal system is obtained by computing the adjoint representation of the vector fields . We use the Lie series where is the commutator for the Lie algebra, is a parameter, and . The commutator table of the Lie point symmetries for Eq. (3) and the adjoint representation of the symmetry group on its Lie algebra are presented in Tables 1 and 2, respectively.\n\n 0 0 0 0 0 0 0\n\nGiven a nonzero vector our task is to simplify as many of the coefficients as possible through judicious applications of adjoint maps to .\n\nFirstly, we suppose that . Scaling if necessary, we can assume that . Referring to Table 2, if we act on such a by , we can make the coefficient of vanish and the coefficients of cannot be eliminated further, so we can make the coefficient of either , , or . Thus any one-dimensional subalgebra spanned by with is equivalent to one spanned by either , , or .\n\nThe remaining one-dimensional subalgebras are spanned by vectors of the above form with . If , we scale to make , and then no coefficient vanishes by any action on , so that is equivalent to .\n\nThe remaining case, , is equivalent to . And it is impossible to make further simplification.\n\nUntil now, we have found the optimal system of one-dimensional subalgebras spanned by\n\nThe list can be reduced slightly if we admit the discrete symmetry , which maps to , and thereby the number of inequivalent subalgebras is reduced to four.\n\n##### 2.3. Similarity Reductions\n\nThe Holm-Hone equation is expressed in the coordinates , so we try to reduce this equation in order to search for its form in specific coordinates which can be constructed by solving the characteristic equationBased on the optimal system presented before, we obtain the following four kinds of reductions:\n\nReduction 1. For , we get the reduction\n\nReduction 2. For , we get the traveling wave reduction , , where satisfies\n\nReduction 3. For , we get the reduction , where satisfies\n\nReduction 4. For , we get the reduction , where satisfies\n\n#### 3. Travelling Wave Solutions of the Holm-Hone Equation\n\nThe appearance of nonanalytic peakon-type solutions has increased the menagerie of solutions appearing in nonlinear partial differential equations, both integrable and nonintegrable. The pulson solutions for Eq. (3) have a finite jump in second derivative of the solutions. In a series of papers, Wazwaz proposed some schemes to determine new sets of soliton solutions, in addition to the peakon solutions obtained before, for the family of CH equations [29, 30]. The method rests mainly on some ansatzes that use one hyperbolic function or combine two hyperbolic functions as follows:\n\n(1) A sinh-cosh Ansatz I\n\n(2) A sinh-cosh Ansatz II\n\n(3) A tanh Ansatz or a coth Ansatz or\n\n(4) The Exponential Peakon Ansatz where , and are parameters that will be determined. By this method, solitons, solitary patterns solutions, periodic solutions, compactons, and peakons solutions for a family of CH equations with distinct parameters are obtained. However, we should modify the first ansatz and combine it into the second one since and . In what follows, we try to describe some specific travelling wave solutions for the Holm-Hone equation via this modified method.\n\nFirstly, we obtain the travelling wave solutions with or and are arbitrary constants. Moreover, it can be easily verified that there is no effective result for the third ansatz for the Holm-Hone equation. And in order to take the last exponential peakon ansatz, we are advised to rewrite Eq. (3) aswith\n\nSupposing that , then Eq. (24) becomeswhereLet us find all possible nonconstant solutions satisfying and . Since , the corresponding characteristic equation is . Then the characteristic values are and , which yield It is easy to see that if and only if Nevertheless, when Then the solution can be reduced to Considering the continuity of solution at , we look for a solution of the following form:where , are arbitrary constants. Next we explore the relationship between the coefficients and . Substituting Eq. (32) into Eq. (27) and using the property we get the travelling wave solution of the original equation This solution is called the first-order pulson solution which is different from the peakons of CH equation. Its first derivative is continuous and the second derivative of the cusp is not continuous.\n\n#### 4. Conservation Laws for the Holm-Hone Equation\n\nConservation laws are widely applied in the analysis of PDEs, particularly in the study of existence, uniqueness, and stability of solutions. The concept of conservation laws and the relationship between symmetries and conservation laws arise in a wide variety of applications and contexts [25, 26].\n\nA conservation law for (3) is of the form where and denote the total derivatives as and the subscripts denote partial derivatives. The vector is a conserved vector for the partial differential equation. It was shown that Eq. (3) has the conservation law where and it does not own bi-Hamiltonian structure . In this section, we mainly solve the conservation laws of the Holm-Hone equation by multiplier method .\n\nA multiplier has the property thatfor all solutions . Generally speaking, each multiplier is a function as , where denotes all th order derivatives of with respect to all independent variables . Here we only consider multipliers of the form , although multipliers which depend on the first-order and higher-order partial derivatives of could also be considered, but the calculations become more complicated and we fail to find any result.\n\nThe right-hand side of (38) is a divergence expression which leads to the determining equation for the multiplier aswhere is the standard Euler operator.\n\nFrom the system (39), we can obtain the solutionwhere and are arbitrary constants. Therefore we get that any conserved vector of the Holm-Hone equation with multiplier of the form is a linear combination of the two conserved vectors\n\n#### 5. Conclusions\n\nIn summary, we have found the most general Lie point symmetries group for the nonintegrable Holm-Hone equation which is a fifth-order generalization of the CH equation. Meanwhile, we constructed the optimal system of one-dimensional subalgebras. Afterwards, we created the preliminary classifications of similarity reductions. The Lie invariants and similarity reduced equations corresponding to infinitesimal symmetries have been obtained. In order to obtain the traveling wave solutions of the equation, we adopted the method of ansatz. We also found some conservation laws from the multiplier method.\n\n#### Data Availability\n\nNo data were used to support this study.\n\n#### Conflicts of Interest\n\nThe authors declare that they have no conflicts of interest.\n\n#### Acknowledgments\n\nThis work is supported by the 13th Five-Year National Key Research and Development Program of China with Grant No. 2016YFC0401407.\n\n1. R. Camassa and D. D. Holm, “An integrable shallow water equation with peaked solitons,” Physical Review Letters, vol. 71, no. 11, pp. 1661–1664, 1993. View at: Publisher Site \| Google Scholar \| MathSciNet\n2. R. Camassa, D. D. Holm, and J. M. Hyman, “A new integrable shallow water equation,” Advances in Applied Mechanics, vol. 31, pp. 1–33, 1994. 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View at: Publisher Site \| Google Scholar \| MathSciNet\n33. S. C. Anco, “Generalization of Noether’s theorem in modern form to non-variational partial differential equations,” in Recent progress and Modern Challenges in Applied Mathematics, Modeling and Computational Science, vol. 79, pp. 119–182, Fields Institute Communications, 2017. View at: Google Scholar\n\n#### More related articles\n\nWe are committed to sharing findings related to COVID-19 as quickly and safely as possible. Any author submitting a COVID-19 paper should notify us at [email protected] to ensure their research is fast-tracked and made available on a preprint server as soon as possible. We will be providing unlimited waivers of publication charges for accepted articles related to COVID-19. Sign up here as a reviewer to help fast-track new submissions." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86563927,"math_prob":0.9282456,"size":21648,"snap":"2020-24-2020-29","text_gpt3_token_len":5088,"char_repetition_ratio":0.16253927,"word_repetition_ratio":0.09238152,"special_character_ratio":0.22833519,"punctuation_ratio":0.17619517,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99261266,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-03T15:45:28Z\",\"WARC-Record-ID\":\"<urn:uuid:bea7ec7a-92ba-4b6a-9c51-5681b36fcb20>\",\"Content-Length\":\"1049254\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:216127d9-7b64-4f97-ac5f-ef8bbec03874>\",\"WARC-Concurrent-To\":\"<urn:uuid:9e6a0d7f-c3b6-4f70-8238-9951ad423d6d>\",\"WARC-IP-Address\":\"99.84.191.103\",\"WARC-Target-URI\":\"https://www.hindawi.com/journals/amp/2019/4364108/\",\"WARC-Payload-Digest\":\"sha1:N6ZACL2C4D6KNAU3HZ3TFIBRPIKGCR46\",\"WARC-Block-Digest\":\"sha1:Q64VNX3YLYJLLAFERCUUGA6UE7CCNIDH\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347435238.60_warc_CC-MAIN-20200603144014-20200603174014-00597.warc.gz\"}"}
https://jwcn-eurasipjournals.springeropen.com/articles/10.1186/s13638-018-1149-7	[ "# Random forests for resource allocation in 5G cloud radio access networks based on position information\n\n## Abstract\n\nNext generation 5G cellular networks are envisioned to accommodate an unprecedented massive amount of Internet of things (IoT) and user devices while providing high aggregate multi-user sum rates and low latencies. To this end, cloud radio access networks (CRAN), which operate at short radio frames and coordinate dense sets of spatially distributed radio heads, have been proposed. However, coordination of spatially and temporally denser resources for larger sets of user population implies considerable resource allocation complexity and significant system signalling overhead when associated with channel state information (CSI)-based resource allocation (RA) schemes. In this paper, we propose a novel solution that utilizes random forests as supervised machine learning approach to determine the resource allocation in multi-antenna CRAN systems based primarily on the position information of user terminals. Our simulation studies show that the proposed learning based RA scheme performs comparably to a CSI-based scheme in terms of spectral efficiency and is a promising approach to master the complexity in future cellular networks. When taking the system overhead into account, the proposed learning-based RA scheme, which utilizes position information, outperforms legacy CSI-based scheme by up to 100%. The most important factor influencing the performance of the proposed learning-based RA scheme is antenna orientation randomness and position inaccuracies. While the proposed random forests scheme is robust against position inaccuracies and changes in the propagation scenario, we complement our scheme with three approaches that restore most of the original performance when facing random antenna orientations of the user terminal.\n\n## Introduction\n\nOne of the key challenges with the fifth generation (5G) cellular networking technology is to ensure high data rate provision to all users, irrespective of their location and time of network access. Typically, the service requirements on 5G systems include a 1000 × increase in system capacity compared to Long Term Evolution-Advanced (LTE-A) systems , an end-to-end latency reduction of at least 10 × compared to LTE-A systems , and support for medium- to high-mobility users, with high throughput and always-on connectivity requirements .\n\nDensification of the radio access network (RAN) toward network deployments of high access node density has been suggested to massively increase the system capacity of mobile radio networks. However, a massive densification of the radio access network resources implies high coordination requirements that the existing LTE system architecture cannot meet. For this reason, new network architectures had been proposed, among which the cloud radio access network (CRAN) architecture constitutes a promising solution for implementing dense networks that can achieve fast coordination at relatively moderate costs. In CRAN, the radio access units, which are formed from distributed antenna systems, are separated from the central processing units, that handle all the baseband processing. The central processing units, essentially being small cloud-like data processing units, are also connected to each other through a backbone supporting fast coordination among them. A single unit of the distributed antenna systems is called remote radio head (RRH), which when densely placed with other RRHs in an area of interest forms an antenna domain, and this set up is formally known as ultra-dense network (UDN) deployment . Such UDN deployments allow for tight interference coordination between RRHs, and consequently result in higher system capacity, and thus can achieve the aforementioned targets for 5G communication systems.\n\nServing an unprecedented massive amount of terminals in 5G, comprising Internet-of-things (IoT) and user devices, with five times smaller frame duration than in LTE requires an extensive overhead for the channel state information (CSI) acquisition. Comparing with traditional resource allocation (RA) schemes [7, 8], this overhead becomes excessive for moderate to high speeds of the user terminals, where the channel states fluctuate intensely . On the other hand, the densification implies more terminal connections in line-of-sight (LOS) positions, leading to lower statistical channel variability. This motivates the consideration of alternative non-CSI based RA approaches that utilize pure position information of the terminals in the antenna domain. In , it is shown that RA based on CSI is much more expensive in terms of system overhead compared to location-based RA in the context of device-to-device (D2D) communications. Although position estimation based on, e.g., uplink (UL) pilot reference signals, or beacons, requires a much lower overhead than CSI acquisition, it remains an open research question how such schemes perform in terms of network key performance indicators (KPIs) when benchmarked with CSI-based schemes. Furthermore, it remains to be studied how accurate and complex a realization of a position-based RA is, and how robust it is to the changes in the propagation scenario or to terminal position information inaccuracies.\n\nIn order to address these research issues, we propose the usage of a supervised machine learning approach based on the random forests algorithm. The random forests algorithm is well known for its inherent robustness and superiority over other known supervised learning techniques in case of missing data values [11, 12]. In this article, we model and devise a learning-based RA scheme as centralized solution for resource allocation in 5G systems by using random forests as multi-class classifier. Our solution essentially aims at predicting the modulation and coding scheme (MCS) to be used for a given terminal position. Through numerical evaluations, we study the basic efficiency of the approach which achieves a spectral efficiency performance comparable to CSI-based schemes. By considering the corresponding overhead, we also show that the learning-based approach outperforms CSI-based approaches significantly. We demonstrate the robustness of the proposed scheme with respect to different variations of users’ position accuracy, showing that even for quite large variations the learning-based approach can still provide good performance. In addition we show the increasing accuracy on the system performance that can be achieved by an increasing number of training and test samples. Finally, we study the impact of a random orientation of the user antennas, which may lead to significant performance deterioration, and discuss several compensation schemes, which successfully addresses this challenge.\n\nThe remaining paper is structured in the following manner: Section 2 presents relevant prior art research, while Section 3 presents the system model and the detailed problem statement. Some background information on machine learning and random forests algorithm is presented in Section 4, along with the details for the design of learning-based RA scheme. The performance evaluation of the proposed scheme is then elaborated in Section 5. Finally, Section 6 concludes the article accompanied by a discussion of the future work.\n\n## Related work\n\nOptimizing resource allocation based on terminals’ CSI for multi-antenna systems has been extensively used and studied in the prior art literature where both centralized and distributed solutions have been suggested . To this end, CRAN defines a suitable platform for resource allocation in 5G that may allow for both centralized and distributed control of a common pool of resources belonging to multiple operators or multiple service providers . Despite the recent CRAN advances, related work with respect to machine learning for resource allocation within the context of CRAN is still quite sparse. In , the authors propose a resource allocation scheme based on linearisation of mixed integer non-linear program for mobile users present in 5G CRAN systems, where they formulate the problem as maximization of network throughput, with a constraint on maximum network capacity. The authors in have shown that resource allocation based on CSI is much expensive in terms of system overhead compared to location-based resource allocation scheme in the context of device-to-device (D2D) communications, which are an integral part of 5G system design. In such case, when perfect CSI is used for resource allocation, the system overhead can be as large as about 25% of the system capacity. The authors in use the reinforcement learning, a machine learning technique, for adaptive modulation and coding in orthogonal frequency division multiplex-multiple input, multiple output (OFDM-MIMO) based 5G systems. In our previous work , we used the random forests algorithm as a binary classifier for allocating resources to users present in CRAN-based 5G system. In that case, the random forests classifier was coupled with a system scheduler, which validated the prediction provided by the random forests, and then the appropriate resources were allocated to serve the given set of users in the system. Though we evaluated the robustness of the RA scheme based on the binary random forests classifier for different system parametrization, but the scope of such investigations was quite limited. We thus conclude that all of our above addressed challenges are still open, some of which are investigated in this work.\n\n## System model\n\n### General system components and assumptions\n\nFigure 1 presents the CRAN system, consisting of the following essential components:\n\n• N terminals spread out across the entire CRAN system;\n\n• A number of R remote radio heads (RRHs) serving the terminals; and\n\n• A baseband unit (denoted by ‘BBU’ in Fig. 1) that performs baseband processing in the entire system, and to which all RRHs are connected to by a fast back-haul.\n\nFor simplicity, we consider the case where each RRH is serving only one terminal in the downlink (DL) in a given time frame. It is further assumed that the CRAN system operates in time division duplex (TDD), where time frames of duration Tf are used for a communication link. Each time frame consists of a number of L sub-frames, each of duration Tsub. Orthogonal frequency-division multiplexing (OFDM) waveform is considered, and therefore, each sub-frame consists of a number of symbols Stotal and fsc OFDM sub-carriers. The operating frequency of the system is fc, with a system bandwidth W. Dense deployment of RRHs is considered within the given CRAN system; an example could be placing distributed antenna systems on top of street lights . Terminals are roaming freely within the area with varying velocities and directions. Each RRH and terminal is equipped with ATx and ARx antennas, respectively.\n\nFinally, the following notations are followed in the description of the 5G CRAN system model and throughout the rest of this paper. Bold-faced capital letter, e.g. A, denotes a matrix, whereas bold-faced small letter, e.g., a, denotes a vector. $$\\mathbb {A}$$ denotes a set, whereas A or a denote scalars. (.) denotes the Hermitian of a vector or a matrix, while the transpose of a vector or a matrix is denoted by (.)T.\n\n### Resource allocation and channel model\n\nIn the considered CRAN system, the baseband unit is the central RA unit, which allocates resources on a per-frame basis. First, it performs the assignment of each RRH to the terminals present in the system. Let Xt denote the binary assignment matrix, where each element $$x_{r,n}^{t}$$ denotes the assignment of nth terminals to rth RRH at time t. The assignment of the terminals to RRH, is in the literature referred to as user assignment. Once the user assignment is determined, the baseband unit decides on the selection of a transmit beam $$\\pmb {v}_{r,n}^{t}$$ to be used by the assigned RRH, and also the receive filter $$\\pmb {u}_{r,n}^{t}$$ at the terminal. The transmit beams and receive filters belong to the pre-defined sets of beams, i.e., $$\\pmb {v}_{r,n}^{t}\\in \\mathbb {V}$$ and $$\\pmb {u}_{r,n}^{t}\\in \\mathbb {U}$$, that are available at each RRH and terminal, respectively. Finally, the baseband unit selects a modulation and coding scheme (MCS), $$m_{r,n}^{t}$$ from the set $$\\mathbb {M}$$, for data transmission between each RRH-terminal link. The resource allocation is done by the baseband unit on per-frame basis, where the objective is to maximize the system goodput.\n\nThe propagation scenario is interference-limited, having densely deployed RRHs within the area of interest. Given a certain allocation of terminals to RRHs in the system, combined with the selection of transmit beams and receive filters for each allocation, the signal-to-interference-and-noise ratio (SINR) of a terminal n for a given time t is given by\n\n$$\\begin{array}{{20}l} \\gamma_{n}^{t} = \\frac{P_{r, n}^{t}}{ \\sigma^{2} + \\sum\\limits_{\\substack{q=1 \\\\ q \\neq r} }^{R} P_{q, n}^{t}}, \\end{array}$$\n(1)\n\nwhere, $$P_{r, n}^{t}$$ is the signal power received by terminal n, from the rth RRH, at time t, $$P_{q, n}^{t}$$ is the signal power received by terminal n, from the qth RRH (other than the rth RRH), at the same time frame t, and σ2 is the noise power. The received signal power $$P_{r, n}^{t}$$ is given by\n\n$$\\begin{array}{{20}l} P_{r, n}^{t} = \\ P_{\\text{Tx}} \\cdot \\ \| (\\pmb{u}_{r,n}^{t})^{\\dagger} \\pmb{H}_{r, n}^{t} \\pmb{v}_{r,n}^{t} \|^{2}. \\end{array}$$\n(2)\n\nIn Eq. (2), PTx denotes the transmit power allocated per RRH, and $$\\pmb {H}_{r, n}^{t}$$ is the channel matrix for the time frame t between RRH r and terminal n. Throughout the duration of a frame we assume that the SINR remains constant in both time and frequency.\n\nEach element of the channel matrix $$\\pmb {H}_{r, n}^{t}$$ represents the complex polarimetric channel impulse response between each transmitter antenna element aTx and receiver antenna element aRx, and is denoted by $$H_{a_{\\text {Rx}}, a_{\\text {Tx}}} (t, \\tau)$$. In reality, this channel response is the combination of different path components, i.e., reflection, diffraction, and scattering, which can be modeled as k different multipath components. The channel impulse response is the sum of the impulse responses from k different multipath components, between aTx and aRx antenna elements and is given by\n\n$$\\begin{array}{{20}l} H_{a_{\\text{Rx}}, a_{\\text{Tx}}} (t, \\tau) = & \\sum\\limits_{k = 1}^{K} \\tilde{\\pmb{h}}_{k, a_{\\text{Rx}}, a_{\\text{Tx}}}(t) \\cdot \\\\ & \\cdot e^{\\frac {j 2 \\pi d_{k} (t)}{\\lambda}} \\delta \\left(\\tau - \\tau_{k, a_{\\text{Rx}}, a_{\\text{Tx}}}(t)\\right). \\end{array}$$\n(3)\n\nHere, K is the total number of multipath components. $$\\tilde {\\pmb {h}}_{k, a_{\\text {Rx}}, a_{\\text {Tx}}}(t)$$ is the impulse response of kth multipath, including the relevant pathloss. λ denotes the wavelength, and d k is the total distance for multipath k at time t. $$\\delta (\\tau - \\tau _{k, a_{\\text {Rx}}, a_{\\text {Tx}}}(t))$$ is the delta function representing the evolution of channel impulse response with respect to different multipath delays $$\\tau _{k, a_{\\text {Rx}}, a_{\\text {Tx}}}$$.\n\nGiven the SINR per time frame, the system utilizes the MCS $$m_{r,n}^{t}$$ chosen by the resource allocation unit, i.e., the baseband unit, to convey backlogged information to the corresponding terminal. This results in a certain spectral efficiency combined with a block error rate $$e_{m_{r,n}^{t}} (\\gamma _{n}^{t})$$. Thus, assuming full-buffer at the baseband unit, the choice of MCS determines a certain payload size $$b_{m_{r,n}^{t}}$$ that can be sent over the channel, and depending on the resulting block error rate, the goodput for the corresponding link at time t can be calculated as\n\n$$\\begin{array}{{20}l} G_{r,n}^{t} = \\frac{(1 - e_{m_{r,n}^{t}} (\\gamma_{n}^{t})) \\cdot b_{m_{r,n}^{t}}}{T_{\\mathrm{f}}}. \\end{array}$$\n(4)\n\nFor the determination of user assignments and resource allocations, the baseband unit utilizes either the position estimates of the terminals present in the system, or their CSI. Acquisition of either of these comes at a certain signalling expense, which we model as a system overhead. We assume a time frame structure, as shown in Fig. 2, where the frame duration Tf is 1 ms, and it comprises of L = 5 sub-frames, each of duration Tsub. The first few symbols of each sub-frame are used for acquiring terminals’ position estimates or their CSI estimates. Since TDD-based system operation is assumed, channel reciprocity holds and, therefore, the UL pilots can be used for terminals’ positions or CSI estimates in the DL subframe.\n\nThe first symbol of each sub-frame is used for position acquisition, i.e., narrow-band pilots are sufficient for acquiring the terminals’ position estimates. The next four consecutive symbols in each sub-frame are the full-band pilots to be used for CSI acquisition. The number of pilots needed for position or CSI estimation depend on the number of terminals present in the system; the greater the number of terminals, the greater the number of full-band pilots needed for CSI estimation. Typically, the number of narrow-band pilots spans a few time symbols within a frame Tf. To avoid inter-carrier interference, the adjacent CSI-sensing pilots are scheduled based on the cyclic-prefix compensation distance, as explained in . Also, if any of the pilots in the sub-frames within a frame are not used for position or CSI sensing, they can be used for data transmission in downlink.\n\nBased on these parameters, the percentage overhead for position acquisition per frame can be calculated as\n\n$$\\begin{array}{{20}l} OH_{\\text{pos}} = \\frac{S_{\\text{pos}} \\cdot f_{\\mathrm{sc,pos}} }{S_{\\text{total}} \\cdot f_{\\mathrm{sc,total}}}, \\end{array}$$\n(5)\n\nwhere Spos is the number of OFDM symbols used for position estimation of terminals in the system, fsc,pos denotes the number of sub-carriers used in the positioning beacon, and Stotal and fsc,total are the total number of OFDM symbols and sub-carriers available in the time frame Tf, respectively.\n\nSimilarly, for CSI acquisition per frame, the percentage overhead can be computed as\n\n$$\\begin{array}{{20}l} OH_{\\text{CSI}} = \\frac{S_{\\text{CSI}} \\cdot f_{\\mathrm{sc,CSI}} }{S_{\\text{total}} \\cdot f_{\\mathrm{sc,total}}}, \\end{array}$$\n(6)\n\nwhere SCSI and fsc,CSI denote the number of OFDM symbols and the number of sub-carriers used for CSI acquisition of terminals present in the system in a transmitted frame, respectively.\n\n### Problem statement\n\nThe objective of this research work is to design an efficient resource allocation scheme that maximizes the sum-goodput of the CRAN based 5G system having medium- to high-speed mobility terminals. The traditional approach utilizes the terminals’ CSI for doing RA in a centralized or distributed fashion for multi-antenna systems [7, 8, 17]. Certainly, the same approach can be applied for a 5G system and combined with the advantage of coordination between RRHs provided by the CRAN architecture; however, the resulting system overhead can lead to deteriorating performance for an increased number of terminals in the system. In addition, the CSI-based RA approach leads to increased computational complexity when the number of terminals requiring service becomes large.\n\nTo alleviate these scalability problems in 5G CRAN system, we propose the usage of acquired terminals’ position estimates as an input to the RA scheme. Specifically, we design a RA scheme using machine learning algorithm to correlate the acquired terminals’ position information with different system parameters and to compare the resulting sum-goodput with that obtained from the traditional CSI-based RA scheme. Our proposed scheme is a centralized solution for resource allocation and is considerably less complex, but leads to other potential issues relevant to the context of using machine learning algorithms, namely, (a) the number of training samples for learning model needed to achieve a comparable system performance, and (b) the robustness of the learned structure to changes in the live environment, which we address in this research study. In the following, we will first discuss a learning-based approach which utilizes the random forests algorithm, and afterwards present a thorough performance evaluation.\n\n## Learning-based resource allocation scheme\n\nMachine learning is a tool used for making a computer program or a machine “develop new knowledge or skill from the existing and non-existing data samples to optimize some performance criteria” . In this work, we use the random forests algorithm , a supervised machine learning technique, for designing the learning-based RA scheme. For a basic understanding of the learning approach and its application to CRAN resource allocation, a brief introduction of the random forests algorithm, followed by the details of the proposed resource allocation scheme, are presented in the remainder of this section.\n\n### Random forests algorithm\n\nAs the name suggests, the algorithm uses a combination of multiple ‘random’ binary decision trees, which make up the forest, for predicting one (or a set of) outcome(s). Being a supervised learning technique, random forests algorithm relies on provision of a training dataset to generate the decision trees. The training dataset D consists of two parts: a set of data characteristics or features F, and a set of output variables Y. Each instance d i of the training dataset is called an input feature vector. The algorithm then constructs Ω t binary random trees, each with a depth Ω d , using the different features (selected randomly) in the training dataset. Each tree typically consists of a root node, one or more interior nodes and terminates at leaf nodes, as shown in the sample tree in Fig. 3. The leaf nodes store the output variable(s), technically called a ‘vote’, and the output variable predicted by the algorithm is the mode of those votes from all trees in the random forest. It is worthy to mention here that the algorithm strives to learn the input-output correlation so as to maximize the overall accuracy of prediction, irrespective of the distribution of individual values of the output variable y. Therefore, care has to be taken that the set of output variables Y is not ‘biased’ toward some particular values in the training data. Once a forest has been trained (during the operational phase), the input features of a new (and potentially unknown) instance is presented to the decision forest, leading to a prediction (through the voting of the trees) of the output variable. More detailed descriptions about the working of the random forests algorithm can be found in .\n\nIn general, random forests are known to be easy-to-use but robust machine learning data structures for noisy data. Furthermore, they can be used to calculate the ‘input variable importance,’ which signifies the influence of an input feature on prediction of the output variable. In this study, we consider system-level simulations modeled according to realistic scenarios, so the collected data may sometimes be noisy which will influence the overall performance of the learning algorithm. Keeping this in view, the aforementioned properties motivated our choice for using random forests algorithm in the design of the proposed RA scheme.\n\n### The proposed scheme\n\nAs already mentioned, the main motivation for designing the proposed approach is to use the acquired terminals’ position estimates for allocating the resources efficiently. The resources include transmit beam, $$\\pmb {v}_{r,n}^{t}$$, per RRH-terminal link, receive filter, $$\\pmb {u}_{r,n}^{t}$$, per terminal, as well as the appropriate MCSs’ selection, $$m_{r,n}^{t}$$, for each RRH-terminal user assignment. We consider the case of fixed user assignments, where each RRH serves only one terminal throughout the operational phase. We apply a simplistic approach for designing the dataset to use for the training of random forests algorithm: we use the acquired terminals’ position estimates in combination with some system resources as input features, and train the algorithm for predicting the MCS as an output variable. Specifically, the input parameters, which constitute the input feature vector, are the following:\n\n• Terminal position estimate $$\\mathcal {P}_{n}^{t}$$ of the dedicated terminal. In reality, this is acquired with certain precision using an extended Kalman filter, along with the direction of arrival and time of arrival estimates of those terminals .\n\n• Transmit beam $$\\pmb {v}_{r,n}^{t}$$ used for serving the dedicated terminal. Essentially, they belong to a set of fixed beams $$\\mathbb {V}$$ based on geometric beamforming, with a certain angular separation.\n\n• Receive filter $$\\pmb {u}_{r,n}^{t}$$ used by the terminal to receive transmitted data in a specific direction. These filters are also geometric beams, with an angular separation dependent on the number of antennas at the terminal.\n\n• Interfering transmit beams $$\\pmb {v}_{q,n}^{t}, q\\neq r$$ used by the interfering RRHs for nth terminal.\n\nRandom forests algorithm is trained on these ‘input features’ to predict the MCS, $$m_{r,n}^{t}$$, such that the sum-goodput is maximized. Essentially, we deploy the random forests as multi-class classifier, where the output variable $$m_{r,n}^{t}$$ has multiple values, or classes.\n\nFigure 3 illustrates an example of a binary decision tree of depth Ω d = 3 from a set of trees comprising a random forest. The root node comprises the input variable corresponding to the user position along the x-axis, while the input features corresponding to transmit beam, receive filter, interference beam for the first interfering terminal, and user position along the y-axis constitute the interior nodes of the tree. A binary decision tree organizes the classification in a set of binary choices about each input in steps, starting at the root node of the tree and progressing down to the leaf nodes which represent the classification decision. At each step of the tree construction, the input feature and its threshold value providing the highest information gain are selected. An input feature and its corresponding threshold value show the highest information gain if they divide the sample training dataset in two (equally large) subsets. Table 1 shows the details of the input variables used for constructing the input feature vector for training the random forests algorithm. Here, ‘Tx_IF1’ refers to the interfering transmit beam for the first interfering terminal, ‘Tx_IF2’ is the interfering transmit beam for the second interferer, and so on. The leaf nodes, which corresponds to the output variable values, comprises the classification decision, i.e., the MCS indices of which only a limited number is shown in Fig. 3 for illustration simplicity (c.f. Table 2).\n\nIt has to be noted that the organization of each trained decision tree of the random forest is determined by the bootstrap sample subset that is extracted from the training dataset during the training phase. For constructing the training dataset, we use ‘exhaustive search’ to determine the optimal allocation of resources, i.e., $$\\pmb {v}_{r,n}^{t}$$, $$\\pmb {u}_{r,n}^{t}$$, and $$m_{r,n}^{t}$$ for a terminal location $$\\mathcal {P}_{n}^{t}$$, to be served by a pre-allocated RRH. In a realistic system, a heuristic approach can be used to determine the optimal resource allocation for the acquired terminal position estimates and construct the training dataset. Upon completion of the training phase, random forests is used as a ‘scheduler’ for predicting the resource variable, MCS $$m_{r,n}^{t}$$, for a newly acquired terminal position estimate and testing its performance. This is done by constructing a test dataset, where the newly acquired position information is compared to the known position estimates used in training dataset. The input features for the closest matching terminal position are then combined with the new position estimate to construct the test data sample, for which a prediction of MCS is obtained from the trained random forests’ data structure. Using the predicted MCS, the sum-goodput is computed by combining all RRH-terminal links’ goodput, calculated using Eq. 4, for evaluating the overall system performance.\n\nDue to the inherent property of the random forests algorithm, this learning-based RA scheme can be expected to be robust to noisy data. Since in reality the acquired terminal position estimates can be inaccurate for some instances, the robustness property of the random forests suggests that this will not have much impact on the overall system performance. Nevertheless, in general, this robustness only holds up to a certain limit. In the performance evaluation section that follows, we determine these limits and discuss remedies.\n\n## Performance evaluation\n\nIn this research study, the performance of the random forests algorithm has been evaluated by means of simulation experiments. In this section, the simulation assumptions and scenarios of our evaluation methodology are first presented. This is followed by a performance analysis of the random forests algorithm providing a basis for the analysis of the performance for system level simulations. Next, a thorough analysis of various sets of performance results of the learning-based RA scheme is given in comparison to a set of benchmark schemes. Finally, various aspects on the robustness and performance limitation of the proposed learning-based RA scheme are discussed in further detail.\n\n### Evaluation methodology\n\nThe performance evaluation of the proposed learning-based RA scheme is done by performing simulations using the discrete event simulator Horizon . Figure 4 shows the simulation scenario, comprising 4 RRHs, each serving a single mobile terminal (MT). This represents a simpler multi-RRH, multi-terminal scenario for 5G CRAN system where only inter-RRH interference exists. The terminal position estimates can either be accurate or can be erroneous, where the error in position is modeled using normal distribution with zero mean and a given standard deviation. A fixed set of transmit beams is designed using geometric beamforming, with an angular separation of 3°. The receive filters are designed in the same way as the transmit beams, but the angular separation is set to 12°. Other parameter settings for the simulation set up are given in Table 3. Assuming DL communication, channel coefficients for each RRH-terminal link are based on TDD-based downlink and are extracted by using the map-based METIS channel model for Madrid grid . A ray-tracer-based channel model was implemented for this purpose, the details for which can be found in . Note that in Table 3, hTx refers to the height of the RRH antennas, while hRx refers to the terminal antenna height from the ground. Finally, v Rx denotes the terminals’ velocity.\n\nDepending on the investigation scenario (c.f., Section 3), the training dataset is constructed next by using the procedure outlined in Section 4. These training datasets are used to construct the multi-class Random forests model, where we rely on the implementation provided by OpenCV . A total of 100 terminal positions, per terminal, are selected randomly from a set of 1000 terminal positions generated by Horizon, per terminal, to create training datasets of 0.25 million samples for each investigation scenario. The output from the random forests model is used to compute the goodput at the terminal, referred to as user goodput, using Eq. (4). The system goodput is computed by taking the sum of the user goodput for each time instance, and its average over all considered 100 terminal positions is used for system-based performance evaluation.\n\n### Performance analysis of random forests algorithm\n\nBefore a forest-based data structure can be applied, it first needs to be learned off-line based on training data. For this, a considerable amount of instances needs to be collected, for instance, from an optimal CSI-based RA scheme. It is important to analyze the collected dataset itself, before constructing the random forest. Table 2 shows the distribution of the output variable, i.e., the considered 8 MCSs, in the training dataset. We notice that the output variables are concentrated more toward lower or higher MCS selection values; this is because the channel gain for a particular terminal is severely influenced by the interference present in the system. Based on this observation, we need to consider an appropriate structure for constructing the random forest, such that the prediction of the output variable is not ‘biased’ toward the output variable value present in the majority of the training data instances in the given dataset.\n\nIn addition to the details of the input variables used for constructing the input feature vector for training the random forests algorithm, Table 1 also shows the variable importance of each input feature obtained by applying the random forests’ property, to observe its dependence on the output variable in the training dataset. From the table, it can be observed that the index of the selected receive filter beam is the most important variable that influences the value of the output variable. This is because the output variable, the MCS selection, is tied to the SINR value of the given terminal n for a given time instance t, which is highly dependent on the receive filter selected to serve the user n. The terminal position is the next most important variable, which itself indicates the expected interference level seen by terminal n when served by a selected transmit beam and receive filter combination by an RRH. The interfering beam indices also considerably influence the output variable, since they are also important in deciding the level of interference seen by terminal n at time t. The assigned terminal ID is the least important parameter, since it only indicates the terminal considered for a given sample instance, and is loosely tied to the output variable, i.e., the MCS selected for serving the indicated terminal.\n\nAfter analyzing the data characteristics, the next step is to construct and train the random forests algorithm. The training of the algorithm essentially optimizes the forest data structure for accuracy. Here, an important aspect relates to the dimensioning of the forest itself, as it impacts the training and test accuracy. Dimensioning relates to the depth of the trees as well as the number of trees to be used in the forest. The number of random features selected for creating a node split in each tree is chosen to be $$\\sqrt {I}$$ according to the analysis study of the random forests algorithm presented in , where I is the number of input features.\n\nThe training accuracy is obtained by using a subset of training data for validation of the constructed random forests model. Once a sufficiently dimensioned random forest structure has been found, a test dataset is then used to compute the test accuracy of the model by passing each instance of the test dataset through each of the random trees in the model. The minimum number of training samples needed to learn a given function (e.g., the resource allocation function in this case) is known as the sample complexity in statistical learning theory. As the problem is open for many functions, determining the number of training samples is usually done empirically. In this work, the number of test samples is the same as that of the training samples, i.e., 0.25 million, with 100 terminal positions drawn randomly from among 1000 terminal positions’ data. In terms of performance evaluation, the accuracy of the data structure built by the random forests algorithm is an important metric; the higher the number of correctly predicted output variables by the model (whether for the validation dataset or the test dataset), the higher the accuracy. However, having a very high training accuracy is not an indicator of an appropriate learned structure. It could be the case that the random forests structure works perfectly for the training dataset, but shows a low accuracy for test dataset. Such a structure is then an over-fit to the training data. For building a robust random forests structure, we need to vary the number of trees Ω t in the forest, as well as the depth of the trees Ω d , in such a way that the model achieves a fairly high training accuracy while it shows good test accuracy for any test dataset with similar input feature vector composition. Hence, for some data collected from a first system set-up (see the next sub-section for details), we study in Table 4 the training and test accuracy obtained for different parametrization of the random forests structure. Based on these investigations, we used the best possible random forests model for the design of the learning-based RA scheme, with 100 trees, each with a maximum depth of 10.\n\nOnce the learned random forests structure achieves an optimal training accuracy, based on different parametrization of the algorithm, it is available for predicting the output variable for the test dataset generated at run-time of the considered CRAN system.\n\n### Evaluation results for the proposed learning-based RA scheme\n\nWe initially start with benchmarking the raw goodput for different schemes based on perfect system status knowledge, i.e., position or CSI. In detail, we consider the following schemes:\n\n• The proposed learning-based RA scheme; where the multi-class random forests algorithm is used for allocating appropriate resources.\n\n• A random MCS allocation scheme; which uses the same input features as used for the learning-based scheme, but assigns a randomly selected MCS to serve a given terminal. This scheme serves as a benchmark to directly determine the value of learning the MCS for a given input feature vector.\n\n• A geometric-based RA scheme; where the terminal position information is used for allocating the transmit beams and receive filters for serving a given terminal, while again selecting the MCS randomly. This scheme benchmarks, in addition, the value of the pre-processing.\n\n• A legacy CSI-based scheme; where for simplicity, we consider a scheme that determines the optimal transmit beam and receive filters based on the given CSI. This serves as an upper bound on the system performance.\n\nTable 5 presents the average system goodput for all the abovementioned comparison schemes. Note that in this investigation, we do not consider the impact from the overhead model. We initially recognize that the learning-based RA scheme achieves a performance quite close to the CSI-based scheme. In contrast, the scheme based on random MCS assignment performs a lot worse than the learning-based RA scheme, because of the spatial selectivity of the system. This also signifies the importance of learning the correlation between different system parameters. The geometric-based RA scheme shows the lowest system goodput compared to the goodput obtained from the CSI-based scheme; the reason being a severely interference-limited system considered for the given case. In general, the transmit beam and receive filter selected purely on the basis of user terminal position are strictly non-optimal in an interference-limited system. Also, since the selection of MCS for serving a given terminal with known position is done at random, therefore, the system goodput degrades even further. From this point onwards, we will provide a comparison of results for the proposed learning-based scheme with the CSI-based RA scheme only, since the random MCS allocation scheme as well as the geometric-based RA scheme reap off very low system goodput.\n\nWe next turn to the evaluation of the different approaches taking the system overhead into account. Since the transmitted frame duration is set to 1 ms, we assume L=5 TDD-frames to be used for position, or CSI, acquisition and data transmission for all terminals present in the system. This serves as basic parametrization for the overhead calculations presented in Eqs. 5 and 6. Our goal is to study the impact of overhead on performance of the learning-based and CSI-based RA schemes as the number of terminals in the system (for which the state information needs to be collected) grows. Note that we consider at this step still all state information to be perfectly accurate (i.e., the position information as well as the CSI).\n\nFigure 5 shows the results of the average system goodput obtained using accurate terminal position information at all RRHs for the learning-based and CSI-based RA schemes. The colored bars show the effective average system goodput, i.e. the system goodput obtained after taking into account the effect of system overhead due to position beaconing or CSI sensing, while the underlying gray bars represent the system performance without taking the overhead into account. Overall, the proposed learning-based RA scheme achieves about 96% of the system goodput achieved by the CSI-based scheme, without considering any overhead. However, if the system overhead is accounted for, we observe that the proposed scheme is either at par or better in performance compared to the CSI-based scheme for all possible number of terminals present in the system. In particular, as we increase the number of terminals in the system, the number of narrow-band beacons for acquiring terminals’ position estimates increases gradually per TTI, and thus the overhead scales up only marginally for the learning-based RA scheme. In contrast, the overhead for the CSI-based scheme grows much stronger with the increase in the number of terminals present in the system, reaching up to 48% of the frame time, showing that effective system performance degrades severely if CSI-based scheme is used for resource allocation in a system with high terminal density.\n\nThese two initial results are quite striking: firstly, with respect to pure spectral efficiency, a learning-based RA scheme using position information can achieve quite a good performance already in comparison to a CSI-based scheme. This holds at least for the considered system scenario, which nevertheless has been designed carefully and contains a typical level of detail for a system-level simulation of a 5G network. Second, if the overhead or the state acquisition is factored in, due to the high cost of the CSI acquisition, the learning-based RA scheme can significantly outperform CSI-based approaches (up to 100% performance improvement).\n\n### Robustness of learning-based RA scheme\n\nThis performance advantage motivates a more thorough study on the robustness of our learning-based RA scheme. We start with considering the most obvious potential source of inaccuracy influencing the learning-based scheme, namely, the accuracy of the position information.\n\nFigure 6 shows the results for the average system goodput obtained when a random error is involved in the position estimation for the terminals being served by RRHs. It can be seen that the classifier trained on perfect terminal position information is enough to guarantee good system performance up to a certain degree of error involved in the position estimation. However, if the error margin in the terminal position estimates exceeds 2 m, the learning-based RA scheme trained on perfect terminal position estimates fails to provide satisfactory system performance. Better system goodput can be obtained by using the learning-based RA scheme trained on inaccurate position estimates, but the traditional CSI-based RA scheme provides still about 10% better effective system performance.\n\nThis shows the robustness of the proposed scheme for small degrees of error involved in acquired terminal position information. However, when the error margin becomes excessively large, the CSI-based RA scheme provides better effective system performance, when the best-case terminal density scenario is considered, i.e., the scenario where the number of terminals present in the system is equal to the number of RRHs serving them.\n\nWe next turn to the question how sensitive the learning-based RA scheme is to a change in the propagation scenario in contrast to the one from which the training data has been acquired. Figure 7 shows the results for changing obstacle/scatterer density when the random forests model is trained only for a fixed system parameterization. We observe that the average system goodput varies only marginally with varying scatterers’ density ranging between 0.01 and 0.2/m2. Overall, the proposed scheme experiences only 7% loss compared to the traditional CSI-based RA scheme in terms of effective system goodput obtained for all considered scenarios.\n\n### Performance accuracy\n\nAn important factor for the design of such a scheme is the quantitative analysis of the dataset, i.e., how many samples are needed to achieve a comparable system performance between the learning-based RA scheme and the traditional CSI-based RA scheme. For this purpose, the system-based performance with an increasing number of total samples has been computed and presented in Fig. 8. More specifically, Fig. 8 represents the system goodput in comparison to the CSI-based scheme for different number of samples used for training and testing the random forests model, without considering any overhead. These results are for the case when perfect position estimates of the users are available at each RRH, and the random forests model is trained (and tested) with the given number of samples. For each case, the learning model has 100 trees, each with a maximum depth of 10.\n\nAs shown in Fig. 8, the performance of the learning-based RA scheme improves with greater number of samples used for training and testing; with more samples comes more information about correlation between the input data features and the output variable, and thus the learning of RF model improves with increasing number of training instances with improved test accuracy. However, the performance saturates after a certain point; this happens for a total of 250,000 instances used for training and testing, in the case shown in Fig. 8. Therefore, we used a total of 250,000 instances for all the abovementioned performance results’ evaluation, which is good enough to guarantee a comparable performance of the learning-based RA scheme to the CSI-based RA scheme.\n\n### Sensitivity to random antenna orientation\n\nIn the learning-based RA scheme, one of the allocated resources includes the receive filter, which is based on beamforming in the direction closest to the direction of the received signal. For this to work perfectly, it is necessary to have the knowledge of the terminal’s antenna orientation at the RRH serving the related terminal. The antenna orientation of the terminal defines the radiation pattern of the receiving antenna, which dictates the selection of the receive filter. However, the terminal antenna orientation is typically random and can therefore be defined in the local coordinate system (LCS), whereas the terminal antenna orientation known at RRH is defined in the global coordinate system (GCS). In order to compute the correct direction of receive filter, the following transformation between GCS and LCS has to be used (based on the discussion given in the METIS channel model documentation ):\n\n$$F_{\\text{GCS}} (\\theta, \\phi) = \\left[ \\begin{array}{ll} cos \\varphi & -sin \\varphi \\\\ sin \\varphi & cos \\varphi\\\\ \\end{array} \\right] \\left[ \\begin{array}{l} F_{\\theta, {\\text{LCS}}} (\\theta ', \\phi ') \\\\ F_{\\phi, {\\text{LCS}}} (\\theta ', \\phi ') \\\\ \\end{array} \\right].$$\n(7)\n\nHere, FGCS represents the antenna radiation pattern in GCS, θ and ϕ are the elevation and azimuth angles in LCS, and Fθ,LCS and Fϕ,LCS denote the radiation patterns of terminal antenna in elevation and azimuth planes, respectively. cosφ and sinφ are the system transformation variables, given by :\n\n$$\\begin{array}{{20}l} cos \\varphi = \\pmb{e}_{\\theta, {\\text{GCS}}} (\\theta, \\phi)^{T} \\pmb{R}\\: \\pmb{e}_{\\theta, {\\text{LCS}}} (\\theta ', \\phi '), \\end{array}$$\n(8)\n$$\\begin{array}{{20}l} sin \\varphi = \\pmb{e}_{\\phi, {\\text{GCS}}} (\\theta, \\phi)^{T} \\pmb{R}\\: \\pmb{e}_{\\theta, {\\text{LCS}}} (\\theta ', \\phi '), \\end{array}$$\n(9)\n\nwhere, eθ,GCS and eϕ,GCS are the basis vectors in GCS for elevation and azimuth planes, respectively. R is the rotation matrix applied for correcting the angular orientation in GCS based on the orientation in LCS, and eθ,LCS and eϕ,LCS are the basis vectors in LCS for elevation and azimuth planes, respectively. The derivation of the rotation matrix is given in the Appendix.\n\nIf not known a priori at the serving RRH, the antenna orientation of the user terminal is expected to affect the average system goodput. We therefore study next the impact of such a random orientation of the terminal antenna on the performance of the learning-based RA scheme. Figure 9 shows the effect of misalignment in terminal antenna orientation information in the training and test datasets for the learning-based RA scheme. It can be seen that the average system goodput is adversely affected by the misalignment in antenna orientation at the receiver, with system goodput only being about 27% of that for the traditional CSI-based scheme for resource allocation. This is by far the biggest impact on the performance of the learning-based RA scheme found in our research work. Thus, it is important to investigate approaches to mitigate the performance degradation from random antenna orientation caused by users’ hand movements. One way to mitigate the effect of the misalignment in antenna orientation is to train the classification model for the learning-based approach using random terminal antenna orientation information, and then test it for dataset with random terminal antenna orientation information embedded within. This case is shown as ‘solution 1’ in Fig. 9. In this case, the random terminal antenna orientation helps the classifier learn the correlation between different resources and terminal-related system parameters effectively, thus resulting in the performance gap of only 6% from the system goodput for the CSI-based RA scheme.\n\nAnother option to mitigate the effect of terminal antenna orientation is to apply a rotation matrix to adjust the predicted receive filter settings according to realistic terminal antenna orientation. The mathematical analysis for applying this solution is based on the derivation above. The performance result for this method is shown in Fig. 9 by the bar labeled ‘solution 2.’ In this case, we achieve almost 85% of the average system goodput compared to the CSI-based scheme, which is fairly good but worse than the performance seen for ‘solution 1.’ A possible reason for this performance loss is the interference present in the system, which makes the solution of only rotating the predicted receive filter for good reception at terminal a sub-optimal approach.\n\nYet another solution can be applied to mitigate the effect of unknown terminal antenna orientation, i.e., by making the terminal antenna orientation a part of the input feature vector, exclusively. Note that this would require some additional signalling from the terminal to the baseband unit, for which we do not here account for the overhead. The performance of this approach is shown as ‘solution 3’ in Fig. 9, where we observe that the performance of the proposed technique reaps off almost the same average system goodput as in case of ‘solution 1.’ Since we use the same number of features for random selection in building the decision trees in random forests model, the randomization of trees in the model results in the variation of the obtained system goodput, for the case when terminal antenna orientation is embedded or is exclusively incorporated as an input feature for training the random forests model, i.e., for ‘solution 1’ and ‘solution 3,’ respectively. We conclude with the remarkable observation that the random antenna orientation can basically deteriorate performance strongly; however, especially by including this effect in the training data, more robust MCS selections can be trained to compensate for this randomness.\n\n### Change in channel statistics\n\nA special case for testing the performance of the proposed scheme is when the LOS links are no more existent between the RRHs and the relevant terminals in the system. In this case, the specular component is totally neglected when computing the channel matrix for a given RRH-terminal link, thus resulting in a non-LOS (NLOS) scenario. Figure 10 shows the average system goodput obtained from the proposed learning-based, as well as the traditional CSI-based RA schemes, for different inaccuracy ranges involved in the acquired terminal position estimates. We kept the range of terminal position inaccuracy fairly small in this case, since NLOS consideration is already enough to result in performance degradation using only terminal-position estimates for resource allocation in the system. Overall, for perfect terminal position information availability, the proposed scheme still performs fairly well, reaping off almost 90% of the system goodput obtained using CSI-based RA scheme. The effects on average system performance for different variation in terminal position inaccuracies do not show a specific trend, because of the changing channel statistics in NLOS scenario. However, training on inaccurate terminal position estimates proves to be beneficial in improving the system goodput, in contrast to the effect seen in LOS case, where the learning-based scheme trained only on perfect terminal position information is enough to guarantee a system performance comparable to the traditional CSI-based RA scheme, when lower accuracy is involved in the terminals’ position estimates.\n\n## Conclusions\n\nWe presented the design of a learning-based RA scheme which has much lower system overhead, as well as lower complexity, than the traditionally used CSI-based RA scheme, because of its dependence on only the acquired terminal position estimates. Random forests algorithm is used for designing learning-based RA scheme, that works as a scheduler for appropriate resource allocation in 5G CRAN system, serving the different terminals using only their position information. A comparison analysis was done for the RA scheme based on random forests model and the CSI-based RA scheme, in different contexts. The proposed scheme shows either comparable or significantly better effective system performance compared to the CSI-based RA scheme for different terminal densities in the system. In terms of the design parameter variations, the proposed scheme is fairly robust to the inaccuracy involved in the terminal position estimation. Training the random forests model on the dataset involving variation in either the system metrics or the design parameters for the learning-based scheme is very beneficial in case when the same model trained on fixed system parametrization shows degraded system performance. In general, for LOS or NLOS cases, the proposed scheme is robust to small error margin involved in the acquired terminal position information, as well as to the variation in system characterization (such as changing scatterers’ density). The change in terminal antenna orientation affects the performance of the proposed scheme most severely, but the effect can be mitigated by training on top of the terminal antenna orientation information, either embedded or provided explicitly in the training data for constructing the random forest. The performance limitations of the learning-based scheme for extreme channel characterization variation is still an open question, which will be a part of the future work. Also, the scope of the proposed scheme is limited to a centralized solution for resource allocation in CRAN-based 5G systems. Designing a similar learning-based RA scheme as a distributed solution for CRAN architecture, with the interference coordination between the baseband units, is a potential topic for future research.\n\n## Appendix\n\nThis appendix presents a derivation for the rotation matrix. The basic rotation matrices for rotating the vectors by an angle in x-, y-, or z-axes using the right-hand rule are given as follows :\n\n$$\\pmb{R}_{x} (\\theta) = \\left[ \\begin{array}{lll} 1 & 0 & 0 \\\\ 0 & cos \\theta & -sin \\theta\\\\ 0 & sin \\theta & cos \\theta \\end{array} \\right],$$\n(10)\n$$\\pmb{R}_{y} (\\theta) = \\left[ \\begin{array}{lll} cos \\theta & 0 & sin \\theta\\\\ 0 & 1 & 0\\\\ -sin \\theta & 0 & cos \\theta \\end{array} \\right],$$\n(11)\n$$\\pmb{R}_{z} (\\theta) = \\left[ \\begin{array}{lll} cos \\theta & -sin \\theta & 0\\\\ sin \\theta & cos \\theta & 0\\\\ 0 & 0 & 1 \\end{array} \\right],$$\n(12)\n\nFor the pre-defined orientation angles of the receive filter for a terminal, the rotation matrix can be computed as:\n\n$$\\begin{array}{{20}l} \\pmb{R} = \\pmb{R}_{z}(\\phi_{0}) \\pmb{R}_{y}(\\theta_{0}). \\end{array}$$\n(13)\n\nExpanding the above expression, we get the following form:\n\n$$\\pmb{R} = \\left[ \\begin{array}{lll} cos \\phi_{0} & -sin \\phi_{0} & 0\\\\ sin \\phi_{0} & cos \\phi_{0} & 0\\\\ 0 & 0 & 1 \\end{array} \\right] \\left[ \\begin{array}{lll} cos \\theta_{0} & 0 & sin \\theta_{0}\\\\ 0 & 1 & 0\\\\ -sin \\theta_{0} & 0 & cos \\theta_{0} \\end{array} \\right]$$\n(14)\n$$\\pmb{R} = \\left[ \\begin{array}{lll} cos \\phi_{0}\\:cos \\theta_{0} & -sin \\phi_{0} & cos \\phi_{0} \\:sin \\theta_{0}\\\\ sin \\phi_{0} \\:cos \\theta_{0} & cos \\phi_{0} & sin \\phi_{0} \\:sin \\theta_{0}\\\\ -sin \\theta_{0} & 0 & cos \\theta_{0} \\end{array} \\right]$$\n(15)\n\nInserting this rotation matrix expression in Eq. 8 results in the following expressions for cosφ and sinφ:\n\n$${}cos \\varphi\\! =\\! \\left[ \\begin{array}{lll} \\!cos \\theta\\:cos \\phi & \\!cos \\theta\\: sin \\phi & -sin\\theta \\end{array} \\right] \\pmb{R} \\left[\\! \\begin{array}{lll} cos \\theta'\\:cos \\phi'\\\\ cos \\theta'\\:sin \\phi'\\\\ -sin \\theta' \\end{array} \\!\\right],$$\n(16)\n$$sin \\varphi = \\left[ \\begin{array}{lll} -sin \\phi & cos \\phi & 0 \\end{array} \\right] \\pmb{R} \\left[ \\begin{array}{lll} cos \\theta'\\:cos \\phi'\\\\ cos \\theta'\\:sin \\phi'\\\\ -sin \\theta' \\end{array} \\right].$$\n(17)\n\nSimplifying these matrix multiplications gives\n\n$$\\begin{array}{{20}l} cos \\varphi = & \\: (cos\\theta \\: cos \\theta_{0} \\: cos\\phi'\\: +\\:sin \\theta \\:sin \\theta_{0}) cos \\theta'\\:cos \\phi'\\: \\\\ & + cos\\theta\\:sin \\phi'\\:cos\\theta'\\:sin\\phi' \\\\ & - sin\\theta' (cos\\theta\\:sin\\theta_{0}\\:cos\\phi'\\: - sin\\theta\\:cos\\theta_{0}), \\end{array}$$\n(18)\n\nand\n\n$$\\begin{array}{*{20}l} sin \\varphi = & -cos\\theta_{0}\\, \\: sin \\phi'\\: cos\\theta'\\:cos \\phi' +\\:cos \\phi' \\:cos \\theta'\\:sin\\phi' \\\\ & -\\:sin\\theta_{0}\\, \\:sin\\phi'\\:sin\\theta' . \\end{array}$$\n(19)\n\n## Abbreviations\n\n5G:\n\n5th generation\n\nBBU:\n\nBaseband unit\n\nCRAN:\n\nCSI:\n\nChannel state information\n\nD2D:\n\nDevice-to-device\n\nDL:\n\nGCS:\n\nGlobal coordinate system\n\nIoT:\n\nInternet of things\n\nKPI:\n\nKey performance indicator\n\nLCS:\n\nLocal coordinate system\n\nLOS:\n\nLine-of-sight\n\nLTE:\n\nLong-term evolution\n\nLTE-A:\n\nMIMO:\n\nMultiple-input multiple-output\n\nMCS:\n\nModulation and coding scheme\n\nMT:\n\nMobile terminal\n\nNLOS:\n\nNon-line-of-sight\n\nOFDM:\n\nOrthogonal frequency-division multiplexing\n\nRA:\n\nResource allocation\n\nRAN:\n\nRRH:\n\nSINR:\n\nSignal-to-interference-and-noise ratio\n\nTDD:\n\nTime division duplex\n\nTTI:\n\nTime transmission interval\n\nUDN:\n\nUltra-dense network\n\nUL:\n\n## References\n\n1. 1\n\nQC Li, H Niu, AT Papathanassiou, G Wu, 5G Network capacity: key elements and technologies. 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ACM. 55(6), 61–69 (2012). http://doi.acm.org/10.1145/2184319.2184337.\n\n24. 24\n\nEW Swokowski, Calculus With Analytic Geometry (Prindle, Weber, and Schmidt, Boston, 1979).\n\n## Acknowledgements\n\nThe authors would like to thank Mr. Christer Qvarfordt and Dr. Kari Leppänen both with Huawei Technologies for valuable discussions.\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to Georgios P. Koudouridis.\n\n## Ethics declarations\n\n### Competing interests\n\nThe authors declare that they have no competing interests.\n\n### Publisher’s Note\n\nSpringer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.\n\n## Rights and permissions", null, "" ]	[ null, "https://jwcn-eurasipjournals.springeropen.com/track/article/10.1186/s13638-018-1149-7", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8991858,"math_prob":0.9269262,"size":62211,"snap":"2021-43-2021-49","text_gpt3_token_len":12988,"char_repetition_ratio":0.18071921,"word_repetition_ratio":0.034838438,"special_character_ratio":0.20793751,"punctuation_ratio":0.11467686,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98892504,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-04T13:13:15Z\",\"WARC-Record-ID\":\"<urn:uuid:64429482-f7c9-401b-8328-4bb3deb1b41f>\",\"Content-Length\":\"307783\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4c47b4c9-bb95-4a53-8877-882948022b58>\",\"WARC-Concurrent-To\":\"<urn:uuid:0d8935b4-126e-4cf1-8e37-2ccfa3a2a4b1>\",\"WARC-IP-Address\":\"146.75.28.95\",\"WARC-Target-URI\":\"https://jwcn-eurasipjournals.springeropen.com/articles/10.1186/s13638-018-1149-7\",\"WARC-Payload-Digest\":\"sha1:YAA5RKFBZMJLD4PHX5ZD7EUR3AMUCIZH\",\"WARC-Block-Digest\":\"sha1:AJOCFVI73W6VH3SVBHXSAF4ZS5UJNJB5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964362992.98_warc_CC-MAIN-20211204124328-20211204154328-00372.warc.gz\"}"}
https://physics.stackexchange.com/questions/242437/throwing-a-football-is-it-truly-parabolic	[ "Throwing a Football. Is it truly parabolic?\n\nDriving into work, I started thinking about the arc of something being thrown and was puzzled about how gravity's affect is squared per second for falling bodies. Intuitively that implies the shape would be \"like\" a parabolic shape. But I'm curious if it truly is parabolic. Doing some googling, I've found some rudimentary explanations that remind me of explanations I received in algebra as a kid (such as http://entertainment.howstuffworks.com/physics-of-football1.htm)\n\nMy confusion is around the idea that gravity's effect on falling bodies (reference http://en.wikipedia.org/wiki/Equations_for_a_falling_body) is increased over time. If this is the case, then wouldn't this affect the arc in such a way that it's longer on the release end and curves downward faster at the end? This might not be as noticeable for short distances, but intuitively it could play an important role in longer distances.\n\nIn other words, my question is pretty simple, as soon as something is thrown (shot, projected, etc) is it considered a falling body. Either way, is the arc or path truly parabolic, or is the path elongated on the throwing end and curve downward faster at the end? If it is truly parabolic, can you please give a clean explanation as to why the effect of gravity over time doesn't apply? If my intuition is correct about it being elongated, can you please share a useful reference as well?\n\nA couple of assumptions:\n\n1. What is being thrown is small and close to a large body. Like throwing a football 1,000 miles across to the earth's surface.\n2. Ignore air resistance or other factors for simplicity sake.", null, "Update:\n\nThe Newton cannon illustration in @HariPrasad's answer shows us the flight path is elliptical not parabolic. It shows how modifying the initial vector's magnitude, when the angle is tangent to the earth effects the ellipse. It however does not show how changes to the initial vector's angle affects the ellipse.\n\nCan we formulate an equation for the path (Reference: https://en.wikipedia.org/wiki/Ellipse)? I'm hoping for an answer that explains how the foci positions, and sum of red and blue line (need editor to give technical name) change in relation to changes in the direction and magnitude of the initial vector.", null, "• Given your assumptions, yes, it is parabolic. Gravity's effect isn't increased over time, it is a constant force $\\vec{f} = m \\vec{g}$. – Dimitri Mar 9 '16 at 13:59\n• Doing some more googling, it does look like things drop faster at the end for bullets/artillery (en.wikipedia.org/wiki/External_ballistics) however, what you are saying is this is due to air/wind resistance or other factors? – VenomFangs Mar 9 '16 at 14:03\n• Okay, I see what you mean. The trajectory of any object (pointlike or of finite size) will always be parabolic in a constant force field if there is no friction. In the case of a football some shape effects coupled to air resistance might indeed modify the trajectory. – Dimitri Mar 9 '16 at 14:10\n• For something like a football, you can not neglect air resistance and that's why, indeed, the path is elongated on the throwing end and curved downward faster at the end. You also need to consider the Magnus effect, which allows football players to “bend” a shot in different ways. – leftaroundabout Mar 9 '16 at 15:00\n• The faster drop at the end of a bullet's trajectory is caused by the bullet beginning to tumble instead of pointing directly forward. This is indeed caused by air resistance and loss of spin imparted by the barrel of the gun. A bullet fired in a friction-less environment would not have these issues. – Darrel Hoffman Mar 9 '16 at 19:52\n\nMany sources may tell you, that the path of a trajectory is a parabola. Indeed, all of the mathematical formula and calculations dealing with trajectories of objects falling, thrown or propelled support that interpretation. But when dealing with earth satellites and ballistic missiles, the truth is that their orbits are portions of ellipses.\n\nThe Newton's canon on a Mountain\n\nNewton's cannonball was a thought experiment by Isaac Newton used to hypothesize that the force of gravity was universal, and it was the key force for planetary motion.", null, "Here is an interactive version of it: Newton's Cannon on a Mountain\n\nIf an object has less than escape velocity (For earth it is 11.2 km/s), its path is an ellipse. If the object has velocity equal to escape velocity, it has a parabolic trajectory. If it is greater than escape velocity, it is hyperbolic.\n\nNormally when we throw an object the actual path of the object is a part of a larger ellipse as the below image shows but since the velocity is not enough the object hits the ground before completing a full elliptical path which seems to be a parabola", null, "The parabolic paths become flatter and flatter as the cannon is fired faster. Newton imagined that the mountain was so high that air resistance could be ignored, and the canon was sufficiently powerful.\n\nPS: Newton's mountain was impossibly high but he realized that the moon's circular path around the earth could be caused by the same gravitational force that pulls cannonball in its orbit, in other words, the same force that causes objects to fall.\n\nThe answer to your question is that the path of football is truly \"elliptical\" since its velocity is way less than escape velocity. But to us, we \"approximate its path to a parabola\".\n\nWe can use equations of projectile motion as follows.", null, "Equation for the trajectory of a projectile motion:\n\n$\\displaystyle y= x\\tan\\theta -{\\frac{g}{2u^2\\cos^2\\theta}}x^2$\n\n(yes it is an equation of parabola but I have mentioned earlier that the mathematical formula and calculations dealing with trajectories of object are approximated to parabola)\n\nNow from your question we can have to situations:\n\nCASE-1: When the object is thrown inclined at an angle $\\theta_1$ with a velocity $u$\n\nThen the maximum height the object will reach is given by:\n\n$\\displaystyle h=\\frac{u^2\\sin^2\\theta_1}{2g}$\n\nNow if $\\theta_1= 30^\\circ$ and initial velocity $u= 100\\ \\mathrm{m/s}$ (just for consideration)\n\nThen the maximum Height the object will reach is equal to:\n\n$h= 127.55$ meters\n\nNow using the same angle and velocity, if we calculate the maximum distance traveled(called the range of projectile) we have\n\n$\\displaystyle R_\\text{max}=\\frac{u^2\\sin2\\theta_1}{g}$\n\nNow by plugging in the values, we have $R=883.69$ meters\n\nCASE-2: When the object is thrown at a higher angle than before but with same velocity.\n\nNow say the angle $\\theta_2=60^\\circ$ (Higher angle than before) and $u=100\\ \\mathrm{m/s}$\n\nThen by using the same equation used before we have\n\n$h= 382.65$ meters and $R= 441.83$ meters\n\nRESULT:\n\nWe can clearly see that the maximum height in case-1 is less than that of case-2 and the maximum range in case-1 is higher than that of case-2\n\nWhich means the path in case-1 is less high and more far. But the path in case-2 is higher and less far. See the below image for more clarification.", null, "(Sorry for the funky colors :P)\n\n• @VenomFangs \"The velocity is more important than the angle considering the type of path(parabolic, elliptical or hyperbolic)\". But since you asked about on earth phenomenon i would say that if the cannon was angled up slightly, the portion of the ellipse will look like \"elongated at the start and curving down faster at the end\" – hxri Mar 9 '16 at 14:49\n• This is a lovely answer. The only thing you could add (if you felt like it) is the calculation of the difference between the elliptical curve, and the parabolic approximation. This is something that would best be done numerically; I expect the difference is tiny (certainly much smaller than the difference due to ignoring effects of air drag and, in the case of a football, lift). – Floris Mar 9 '16 at 16:51\n• On the \"parabola vs. ellipse\" bit, if the Earth were flat, the football would, in fact, follow a parabolic trajectory. Since, on the scale of kicked footballs, the Earth looks flat, we can use the easier-to-calculate parabolic approximation rather than the exact elliptical solution. – Mark Mar 9 '16 at 21:36\n• 'The answer to your question is that the path of football is truly \"elliptical\" since its velocity is way less than escape velocity.' This is itself a weak gravity approximation. The real path of the object is to very high orders elliptical, but even neglecting air resistance etc., it is still not exact. – Zorawar Mar 9 '16 at 22:00\n• Absolutely right! Not many people aware of this fact. And we are all brainwashed in Physics 101 to believe the misconception of a parabolic flight path. Great presentation of the answer BTW. – docscience Mar 10 '16 at 15:35\n\nTo build up a little bit on our comments, do you mean that you expect the trajectory to be of different shape at the end than at the beginning ? This is again only possible with friction. If there is no friction, your Newton's equation of motion reads\n\n$m \\frac{d^2 \\vec{r}}{dt^2} = m \\vec{g}$\n\nand is invariant under the transformation $t \\mapsto -t$ (it is said to have time reversal symmetry). The practical consequence is that your trajectory must have some axis of symmetry : the football cannot move differently flying upwards than falling downwards because each of these movements map eachother through the time-reversal transformation.\n\nHowever, if you add a friction term like\n\n$m \\frac{d^2 \\vec{r}}{dt^2} = m \\vec{g} - k \\frac{d \\vec{r}}{dt}$\n\n(with $k >0$)\n\nthen you break time-reversal symmetry because $\\vec{v}$ becomes $- \\vec{v}$ under this transformation. Thus, as @leftaroundabout pointed out, you need a friction term to get the trajectory distorsion you described in your question." ]	[ null, "https://i.stack.imgur.com/i6LTwl.gif", null, "https://i.stack.imgur.com/2xMixm.gif", null, "https://i.stack.imgur.com/K1Et3.jpg", null, "https://i.stack.imgur.com/BLl31.gif", null, "https://i.stack.imgur.com/mIBMt.jpg", null, "https://i.stack.imgur.com/QcSE0.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9533185,"math_prob":0.96690404,"size":2234,"snap":"2019-43-2019-47","text_gpt3_token_len":476,"char_repetition_ratio":0.09372197,"word_repetition_ratio":0.005586592,"special_character_ratio":0.20411818,"punctuation_ratio":0.108490564,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9955253,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,5,null,5,null,5,null,5,null,5,null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-17T00:08:11Z\",\"WARC-Record-ID\":\"<urn:uuid:f7809b8d-54d8-4e50-a7f6-fca6066c53f3>\",\"Content-Length\":\"162057\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:32f7c5ee-14ec-42d8-ab5c-368c25a7f456>\",\"WARC-Concurrent-To\":\"<urn:uuid:bbea5f70-9824-4f5c-9bee-060c3c3a41eb>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/questions/242437/throwing-a-football-is-it-truly-parabolic\",\"WARC-Payload-Digest\":\"sha1:WL7DFK5E4IUSMHLJTLVSPKK6OB5TH7RH\",\"WARC-Block-Digest\":\"sha1:OENVNPDF6SJMW4MDJHLZD72TU7JEVJ4J\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986672431.45_warc_CC-MAIN-20191016235542-20191017023042-00097.warc.gz\"}"}
https://www.800score.com/gmat-prep/prep-guide/independent-events/	[ "## Independent Events\n\nTwo outcomes are independent if the occurrence of one outcome has no effect on the occurrence of the other. For example, if a coin is tossed twice, the first outcome (H or T) has no effect on the second outcome (H or T).\n\nThere are two ways multiple events can take place in a single probability problem: either they can each occur separately (A or B) or they must occur together (A and B).\n\n## “Or”\n\nIn some scenarios, the events do not have to occur together to have the desired result. One of the events is enough.\n\nI will be happy today if I win the lottery OR have email.\n\n“Or” means that either outcome is desired. With more possible desired outcomes, the probability is greater than for one event alone.\n\nFor independent events with “or,” add the probabilities of the events.\n\nA fair six-sided die is rolled once. What is the probability of rolling a 5 or a 6?\n\n### Solution\n\nFirst, notice the word “or” in the question. There are 6 possible outcomes, and either a 5 or a 6 is wanted. There is a 1 in 6 probability of a 5 and a 1 in 6 probability of a 6.\n\nBottom: 6 outcomes are possible\nTop: 2 possible desired outcomes\nProbability: \\dfrac{2}{6} = \\dfrac{1}{3}\n\nA deck of 52 cards contains cards numbered 1 through 13 in each of 4 different colors. John will win $100 if he pulls a card numbered either 7 or 9 from the deck. What is the probability that John will win$100?\n\n### Solution\n\nJohn can win by pulling out either a 7 or a 9. His chance of doing this is higher than if he could only win by pulling out a 7. In that case, he’d only have 4 cards that would win, in this case there are 8 winning cards.\n\nMethod 1\n\nTo find the total probability, figure out the probability of each event and then add probabilities.\n\nPulling a 7\nBottom: 52 outcomes are possible\nTop: 4 possible desired outcomes\nProbability: \\dfrac{4}{52} = \\dfrac{1}{13}\n\nPulling a 9\nBottom: 52 outcomes are possible\nTop: 4 possible desired outcomes\nProbability: \\dfrac{4}{52} = \\dfrac{1}{13}\n\nProbability of pulling a 7 or a 9:\n\\dfrac{1}{13} + \\dfrac{1}{13} = \\dfrac{2}{13}\n\nMethod 2\n\nBecause the total possible outcomes are the same for both events, there is no need to calculate their probabilities separately.\nBottom: 52 outcomes are possible\nTop: 8 possible desired outcomes\nProbability: \\dfrac{8}{52} = \\dfrac{2}{13}\n\nA fair coin is flipped 3 times. What is the probability that heads will come up only once?\n\n### Solution\n\nThis is an “or” question, even though the word “or” isn’t written. The probability is independent. Think about what it means to have heads come up only once. Be careful to recognize that the desired outcome is heads only once, not at least once. Write the possible desired scenarios.\n\nH T T or T H T or T T H\n\nThere are three outcomes that would achieve the desired result.\n\nProbability of H T T = \\dfrac{1}{2} × \\dfrac{1}{2} × \\dfrac{1}{2} = \\dfrac{1}{8}\n\nProbability of T H T = \\dfrac{1}{2} × \\dfrac{1}{2} × \\dfrac{1}{2} = \\dfrac{1}{8}\n\nProbability of T T H = \\dfrac{1}{2} × \\dfrac{1}{2} × \\dfrac{1}{2} = \\dfrac{1}{8}\n\nSince these are independent events, add.\n\n\\dfrac{1}{8} × \\dfrac{1}{8} × \\dfrac{1}{8} = \\dfrac{3}{8}\nA fair coin is flipped once. A fair six-sided die is rolled once. What is the probability of getting a heads or rolling a 3?\n\n### Solution\n\nThe probabilities are independent. Each event has a different number of possible outcomes. The only way to solve this type of problem, where the events have a different number of possible outcomes, is to add the separate probabilities.\n\nBottom: 2 outcomes are possible\nTop: 1 possible desired outcome\nProbability: \\dfrac{1}{2}\n\nRolling a 3\nBottom: 6 outcomes are possible\nTop: 1 possible desired outcome\nProbability: \\dfrac{1}{6}\n\nProbability of getting a heads or rolling a 3: \\dfrac{1}{2} + \\dfrac{1}{6} = \\dfrac{3}{6} + \\dfrac{1}{6}\n= \\dfrac{4}{6} = \\dfrac{2}{3}\n\n## “And”\n\nIn some scenarios, all the events have to occur in order to attain the desired result.\n\nI will be happy today if I win the lottery AND have email.\n\n“And” means that both outcomes are needed. With more outcomes required, the probability is less.\n\nFor independent events with “and,” multiply the probabilities of the events.\n\n#### 800score Tip\n\nIn general, the probability of many events occurring is less than just one event occurring.\n\nProbability is expressed as a fraction between 0 and 1. When you multiply these fractions, you get smaller fractions, as in “AND” scenarios. When you add these fractions you get greater fractions, as in “OR” scenarios.\n\nIf a coin is tossed twice, what is the probability that the first toss is heads and the second toss is tails?\n\n### Solution\n\nFirst, notice the word “and” in the question. That means both events must occur. So the desired coin toss is H T.\n\nThe probability that the coin will land on heads is \\dfrac{1}{2}. The probability that the coin will land on tails is also \\dfrac{1}{2}.\n\nSince both must happen, multiply the individual probabilities.\n\n\\dfrac{1}{2} × \\dfrac{1}{2} = \\dfrac{1}{4}\n\nAs expected, the probability of both events occurring is less than the probabilities of either individual event occurring.\n\nNot all “and” questions will include the word “and.” It may say “then” or “both” or it may just be implied.\n\nTwo fair six-sided die are rolled once. What is the probability of rolling two fives?\n\n### Solution\n\nIn this case, the first die must be a 5 and the second die must also be a 5. This is an implied “and.”\n\nProbability of a 5 on the first die is \\dfrac{1}{6}\n\nProbability of a 5 on the second die is \\dfrac{1}{6}\n\nProbability both are 5 is \\dfrac{1}{6} × \\dfrac{1}{6}\n= \\dfrac{1}{36}\n\n## With Replacement\n\nSometimes independent event questions use the term “with replacement.” This means that the first card or marble that is chosen is put back or replaced, so it could be chosen again and not affect the probability of subsequent events (i.e., all the events are independent).\n\nTwo tokens are pulled from a bag that contains 52 tokens, an equal number of which are red, blue, green and white. After the first token is pulled it is placed back in the bag, then the second token is pulled. What is the probability that both tokens are red?\n\n### Solution\n\nSince both tokens must be red, this is an “and” question. Since the events are the same and the first token is replaced, they will each have the same probability. Calculate the probability then multiply.\n\nBottom: 52 outcomes are possible\nTop: 13 tokens are red\nProbability: \\dfrac{13}{52} = \\dfrac{1}{4}\n\nProbability of pulling 2 reds:\n\\dfrac{1}{4} × \\dfrac{1}{4} = \\dfrac{1}{16}\n\nA bag contains 3 white marbles, 2 blue marbles and 5 red marbles. What is the probability of drawing a red marble then a blue marble if the first marble is replaced after it is pulled?\n\n### Solution\n\nThe “then” in the question implies “and.” Since both outcomes must occur, and there is replacement, these are independent events. Multiply the probabilities.\n\nProbability of a red marble\nBottom: 10 outcomes are possible\nTop: 5 marbles are red\nProbability: \\dfrac{5}{10} = \\dfrac{1}{2}\n\nProbability of a blue marble\nBottom: 10 outcomes are possible\nTop: 2 marbles are blue\nProbability: \\dfrac{2}{10} = \\dfrac{1}{5}\n\nProbability of red then blue:\n\\dfrac{1}{2} × \\dfrac{1}{5} = \\dfrac{1}{10}\n\n#### Independent Events\n\nBest viewed in landscape mode\n\n2 questions with video explanations\n\n100 seconds per question", null, "" ]	[ null, "https://i.ytimg.com/vi/mEf0ebVfHFg/maxresdefault.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9051925,"math_prob":0.99041766,"size":6129,"snap":"2021-31-2021-39","text_gpt3_token_len":1675,"char_repetition_ratio":0.2324898,"word_repetition_ratio":0.10433071,"special_character_ratio":0.2788383,"punctuation_ratio":0.099673204,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9986788,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-25T09:12:37Z\",\"WARC-Record-ID\":\"<urn:uuid:026222ef-312a-40f8-9961-d5c0f3cfcdac>\",\"Content-Length\":\"355410\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:17131780-5b04-414a-bf23-5fa49e0db1ba>\",\"WARC-Concurrent-To\":\"<urn:uuid:51286ab6-6ba2-43d0-a8af-9f1db6a0030e>\",\"WARC-IP-Address\":\"67.227.198.168\",\"WARC-Target-URI\":\"https://www.800score.com/gmat-prep/prep-guide/independent-events/\",\"WARC-Payload-Digest\":\"sha1:23XJ7Q4NNQ7YSIHXJKXCT7EGELKQZHI7\",\"WARC-Block-Digest\":\"sha1:FS3TK4MP66BYO6NXUQQA7NSHCZCSFRNX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057615.3_warc_CC-MAIN-20210925082018-20210925112018-00057.warc.gz\"}"}
https://www.nagwa.com/en/videos/369162172531/	[ "Question Video: Dividing Two-Digit Numbers by Eight Mathematics • 3rd Grade\n\nThe robot divides any number by 8 and gives a result. In this case, the number 16 gives a result of 2 as 16 ÷ 8 = 2. What will the result be in the following case? 40 ÷ 8 = ＿ What will the result be in the following case? 64 ÷ 8 = ＿\n\n03:15\n\nVideo Transcript\n\nThis robot divides any number by eight and gives a result. In this case, the number 16 gives a result of two as 16 divided by eight equals two. What will the result be in the following case? 40 divided by eight equals what. And what will the result be in the following case? 64 divided by eight equals what.\n\nIn the first part of this question, we have to help the robot calculate 40 divided by eight. And in the second part of the question, we have to help the robot calculate 64 divided by eight. Did you notice that in all of the calculations we’re dividing by eight? How could we calculate 40 divided by eight? We could use an array and our knowledge of the eight times table to help.\n\nWhen we’re dividing by eight, we can think of this as dividing a number into equal groups of eight. Here’s one group of eight, which gives us a total of eight. Two eights make 16. Three eights make 24. Eight, 16, 24. Four eights are 32, and five eights make 40. Eight, 16, 24, 32, 40. So, if the number 40 goes in and the robot divides it by eight, the result will be five. There are five eights in 40. 40 divided by eight equals five.\n\nNow, we just need to answer the second part of the question. What is 64 divided by eight? How many eights are there in 64? Let’s keep on counting in eights. We already know that five groups of eight makes 40, so let’s just keep counting forward in eights. Five eights are 40, six eights are 48, seven eights are 56, and eight eights are 64. There are eight eights in 64. If we put the number 64 into the robot and he divides it by eight, the result will be eight because 64 divided by eight equals eight.\n\n40 divided by eight equals five and 64 divided by eight equals eight. We found the results using arrays and our knowledge of the eight times table." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9534017,"math_prob":0.9992485,"size":1793,"snap":"2022-05-2022-21","text_gpt3_token_len":472,"char_repetition_ratio":0.20793739,"word_repetition_ratio":0.12209302,"special_character_ratio":0.26045734,"punctuation_ratio":0.12623763,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-21T02:53:43Z\",\"WARC-Record-ID\":\"<urn:uuid:b310e465-e5c8-4fbf-bb0b-67250eced646>\",\"Content-Length\":\"93301\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:373639e8-aaef-4058-9109-38322a431823>\",\"WARC-Concurrent-To\":\"<urn:uuid:126e7fd1-e0bb-43e2-ac68-86680c0158df>\",\"WARC-IP-Address\":\"76.223.114.27\",\"WARC-Target-URI\":\"https://www.nagwa.com/en/videos/369162172531/\",\"WARC-Payload-Digest\":\"sha1:7VCKVTZLLWYVE4KXZ7LT2PEKTTU6JKZS\",\"WARC-Block-Digest\":\"sha1:GMJUBUV277RIMC34KUTXEK4ZAJGD77XL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320302715.38_warc_CC-MAIN-20220121010736-20220121040736-00667.warc.gz\"}"}
https://math.stackexchange.com/questions/793754/let-f-n-n-in-mathbbn-rightarrow-f-on-0-infty-true-or-false-lim/793792	[ "# Let $(f_n)_{n\\in\\mathbb{N}} \\rightarrow f$ on $[0,\\infty)$. True or false: $\\lim_{n\\to\\infty}\\int_0^{\\infty}f_n(x) \\ dx = \\int_0^{\\infty}f(x) \\ dx.$\n\nThe Assignment:\n\nLet $(f_n)_{n\\in\\mathbb{N}}$ converge uniformly to $f$ on $[0,\\infty)$ and let the improper integrals of $f_n$ and $f$ exist on $[0,\\infty)$. True or false: $$\\lim_{n\\to\\infty}\\int_0^{\\infty}f_n(x) \\ dx = \\int_0^{\\infty}f(x) \\ dx$$ Prove it.\n\nSince I know that under certain circumstances (if $f_n$ is integrable $\\forall n≥1$) on the closed interval $[a,b]$: $$\\lim_{n\\to\\infty} \\int_a^bf_n(x) \\ dx = \\int_a^bf(x) \\ dx$$ I have the tendency to say that the statement above is false, since these \"safe\" conditions (compact interval) are substituted by others.\n\nAny help would be appreciated\n\n## 3 Answers\n\nHint:\n\nFill a rubber square aquarium with water until it's completely filled. Now, take the right edge of the aquarium and stretch it to the right. The amount of water never changes, even though the height of the water decreases. What happens when you keep stretching it farther and farther to the right, without restriction?\n\nEdit:\n\nSince I answered your question with a riddle, I'll add a useful tip that might seem more math-like:\n\nWhenever you encounter the question\n\nDoes condition XYZ guarantee $\\displaystyle \\lim_{n\\to\\infty} \\int f_n = \\int \\lim_{n\\to\\infty} f_n$ ?\n\nit is helpful to make the following observation:\n\nIf it is true in the special case that $f_n \\to 0$, then it is true anyway.\n\nThis observation helps you in that you only really need to prove this case, or to find a counterexample of this sort.\n\nProof: Suppose it is true in the special case $f_n \\to 0$. Consider another case, where $f_n \\to f \\ne 0$. Define the sequence $g_n = f_n - f$. Then $g_n \\to 0$, and condition XYZ probably holds for $g_n$ as well as it does for $f_n$, so you can get $\\lim \\int (f_n - f) = 0$, which is exactly what you want.\n\n• I'm still trying to figure out what you're trying to say. Basically you're comparing a definite integral from $0$ to $b$ with water in an rubber square aquarium. Stretching the right edge farther to the right without restriction lets $b \\to \\infty$. While it makes sense in your scenario, why would the integral \"even out\" when one bound is $\\infty$? Edit: I'll read through your edit now. – Nhat May 13 '14 at 22:30\n• The integral is the amount of water. Since the amount of water doesn't change, the integral of $f_n$ is always $1$. However, the integral of $f$ is $0$. (What are $f_n$ and $f$?) – Yoni Rozenshein May 13 '14 at 22:33\n• Firstly, I liked that riddle, haha. Also it hasn't come to my mind to only prove / disprove the special case for $f_n \\rightarrow f = 0$. ... So I'm looking for an $f_n$ such that $\\forall n \\in \\mathbb{N}$ $\\int_0^{\\infty}f_n \\ dx = 1$ and $f_n \\rightarrow 0$ uniformly? – Nhat May 13 '14 at 22:44\n• Yes, exactly. – Yoni Rozenshein May 13 '14 at 22:48\n• Can you help me or give me a hint? – Nhat May 13 '14 at 23:33\n\nConsider $f_n:=\\frac{1}{n}\\chi_{[n,2n]}$ where $\\chi$ denotes the indicator function. Obviously $\|\|f_n\|\|_\\infty\\to 0$, but $\\int_0^\\infty f_n=1~\\forall n\\in\\mathbb{N}$.\n\nConsider the flattening mountain of constant integral\n\n$\\begin{array}{ccccc} f_n & : & R^+ & \\to & R^+\\\\ & & x & \\mapsto &\\frac{x}{n^2}-1+\\frac{1}{n} \\; \\text{when} \\; n^2-n<x\\leq n^2\\\\ && x & \\mapsto & \\frac{-x}{n^2}+1+\\frac{1}{n} \\; \\text{when} \\; n^2<x<n^2 +n\\\\ && x & \\mapsto & 0\\text{when} \\; n^2-n>x \\; \\text{or} \\; x>n^2 +n \\\\ \\end{array}$\n\nEach $f_n$ has integral $1$ but the sequence converges uniformly to $0$." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.76077694,"math_prob":0.99967253,"size":604,"snap":"2021-04-2021-17","text_gpt3_token_len":206,"char_repetition_ratio":0.115,"word_repetition_ratio":0.0,"special_character_ratio":0.33278146,"punctuation_ratio":0.08547009,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99984837,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-11T10:10:29Z\",\"WARC-Record-ID\":\"<urn:uuid:1effcebd-68b9-4888-9be0-8493a64069f2>\",\"Content-Length\":\"185481\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d22f487d-f424-4ea9-8a52-f4754d20e1de>\",\"WARC-Concurrent-To\":\"<urn:uuid:67bc636b-bea1-4539-875a-0b9e7f97a2b0>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/793754/let-f-n-n-in-mathbbn-rightarrow-f-on-0-infty-true-or-false-lim/793792\",\"WARC-Payload-Digest\":\"sha1:3GDYLRTOBTCOEAVCCL2S4GFTXJD4Q5M6\",\"WARC-Block-Digest\":\"sha1:HQKYQ7GQUJNTPUOL24JVIGQRZ452XBMB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038061820.19_warc_CC-MAIN-20210411085610-20210411115610-00245.warc.gz\"}"}