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https://www.electro-tech-online.com/threads/pic-running-too-fast.30284/#post-223471	PIC running too fast? Status Not open for further replies. bananasiong New Member Hi, I'm using PIC16F628A for displaying the LCD clock. I've just started and I'm using timer2 interrupt for counting the real time. I've set the interrupt rate to be 40 ms. So I need to count for 25 times to get 1 second accurate. Postscale 10 Prescale 16 4 MHz crystal = 1 us per instruction PR2 249 So, 10 * 16 * 250 * 1 us = 40 ms When I turn on the power, the clock is running, faster than as expected which is almost 2 to 3 times faster. This is the code I have: Code: LIST P=16F628A #include <P16F628A.inc> ERRORLEVEL 0, -302 __config 0x3101 cblock 0x20 count count1 counta countb templcd CountL1 CountL2 count1sec HourTen HourOne MinTen MinOne SecTen SecOne L1C0 ;line1, column 0, and so on.. L1C1 L1C2 L1C3 L1C4 L1C5 L1C6 L1C7 L1C8 L1C9 L1C10 L1C11 L1C12 L1C13 L1C14 L1C15 L2C0 L2C1 L2C2 L2C3 L2C4 L2C5 L2C6 L2C7 L2C8 L2C9 L2C10 L2C11 L2C12 L2C13 L2C14 L2C15 w_temp s_temp p_temp endc LCD_PORT Equ PORTA LCD_TRIS Equ TRISA LCD_RS Equ 0x04 ;LCD handshake lines LCD_E Equ 0x00 org 0x0000 goto Initialize org 0x0004 Interrupt movwf w_temp swapf STATUS, w clrf STATUS movwf s_temp movf PCLATH, w movwf p_temp clrf PCLATH Check1sec decf count1sec, f btfss STATUS, Z ;interrupt happens 25 times? goto Display ;if no display with the same value movlw d'25' ;if yes count for 1 second movwf count1sec StartCount movf SecOne, w xorlw d'9' btfss STATUS, Z goto ISO goto ReSO ISO incf SecOne, f goto CValue ReSO clrf SecOne movf SecTen, w xorlw d'5' btfss STATUS, Z goto IST goto ReST IST incf SecTen, f goto CValue ReST clrf SecTen movf MinOne, w xorlw d'9' btfss STATUS, Z goto IMO goto ReMO IMO incf MinOne, f goto CValue ReMO clrf MinOne movf MinTen, w xorlw d'5' btfss STATUS, Z goto IMT goto ReMT IMT incf MinTen, f goto CValue ReMT clrf MinTen movf HourTen, w btfss STATUS, Z goto CHO movf HourOne, w xorlw d'9' btfss STATUS, Z goto IHO goto ReHO IHO incf HourOne, f goto CValue ReHO clrf HourOne incf HourTen, f goto CValue CHO movf HourOne, w xorlw d'2' btfss STATUS, Z goto IHO clrf HourOne decf HourOne, f goto CValue CValue movf HourTen, w movwf L2C4 movf HourOne, w movwf L2C5 movlw 0x0A movwf L2C6 movf MinTen, w movwf L2C7 movf MinOne, w movwf L2C8 movlw 0x0A movwf L2C9 movf SecTen, w movwf L2C10 movf SecOne, w movwf L2C11 Display movf CountL2, w xorlw d'12' btfsc STATUS, Z call RestoreCL2 movf CountL2, w call LCD_Line2 movf INDF, w call ILCD_Char incf CountL2, f incf FSR call LCD_Line1 goto Int_Re RestoreCL2 movlw d'4' movwf CountL2 movlw L2C4 movwf FSR return ILCD_Cmd movwf templcd swapf templcd, w ;send upper nibble andlw 0x0f ;clear upper 4 bits of W movwf LCD_PORT bcf LCD_PORT, LCD_RS ;RS line to 0 call Pulse_e ;Pulse the E line high movf templcd, w ;send lower nibble andlw 0x0f ;clear upper 4 bits of W movwf LCD_PORT bcf LCD_PORT, LCD_RS ;RS line to 0 call Pulse_e ;Pulse the E line high call Delay5 return ILCD_Char movwf templcd swapf templcd, w ;send upper nibble andlw 0x0f ;clear upper 4 bits of W movwf LCD_PORT bsf LCD_PORT, LCD_RS ;RS line to 1 call Pulse_e ;Pulse the E line high movf templcd, w ;send lower nibble andlw 0x0f ;clear upper 4 bits of W movwf LCD_PORT bsf LCD_PORT, LCD_RS ;RS line to 1 call Pulse_e ;Pulse the E line high call Delay5 return Int_Re movf p_temp, w movwf PCLATH swapf s_temp, w movwf STATUS swapf w_temp, f swapf w_temp, w retfie Initialize movlw 0x07 movwf CMCON bsf STATUS, RP0 ;bank1 movlw 0x00 movwf TRISA movwf TRISB bcf STATUS, RP0 ;bank0 call Delay100 LCD_Init movlw 0x20 ;Set 4 bit mode call LCD_Cmd movlw 0x28 ;Set display shift call LCD_Cmd movlw 0x06 ;Set display character mode call LCD_Cmd movlw 0x0C ;Set display on and cursor off call LCD_Cmd call LCD_Clr ;clear display clrf PORTA movlw d'25' ;interrupt for 25 times = 1 second movwf count1sec movlw d'4' movwf CountL2 movlw HourTen movwf FSR NEXT clrf INDF movf FSR, w xorlw L2C15 btfsc STATUS, Z goto $+3 incf FSR, f goto NEXT movlw L2C4 movwf FSR movlw 0x0A movwf L2C6 movwf L2C9 movlw b'01001111' ;postscale 10, prescale 16, tmr2 on movwf T2CON bsf STATUS, RP0 movlw d'249' ;1 us x 10 x 16 x 250 = 40 ms interrupt rate movwf PR2 bsf PIE1, TMR2IE bcf STATUS, RP0 bsf INTCON, PEIE bsf INTCON, GIE Start ;main program here . . LCD_Line1 movlw 0x80 ;move to 1st row, first column call LCD_Cmd return LCD_Line2 addlw 0xc0 ;move to 2nd row, first column call LCD_Cmd return Delay255 movlw 0xff ;delay 255 mS goto d0 Delay100 movlw d'100' ;delay 100mS goto d0 Delay50 movlw d'50' ;delay 50mS goto d0 Delay20 movlw d'20' ;delay 20mS goto d0 Delay5 movlw 0x05 ;delay 5.000 ms (4 MHz clock) goto d0 Delay4 movlw 0x04 goto d0 d0 movwf count1 d1 movlw 0xC7 ;delay 1mS movwf counta movlw 0x01 movwf countb Delay_0 decfsz counta, f goto$+2 decfsz countb, f goto Delay_0 decfsz count1 ,f goto d1 return Pulse_e bsf PORTB, LCD_E nop bcf PORTB, LCD_E return LCD_Clr movlw 0x01 ;Clear display call LCD_Cmd return end Besides, I've found that the content of the general purpose registers are not zero as default. I thought that they are I have to clear them in initialize. Thanks mcs51mc Member Can you complement an output pin at each interrupt? Place a scope on that pin and check if you really have 40ms. That way you are sure about your timer setting. If you have 40 ms the problem is your code and since I don't know PIC I can't help you further Pommie Well-Known Member You are not clearing the interrupt flag. Try inserting bcf PIR1,TMR2IF in your ISR. Mike. ericgibbs Well-Known Member hi, There seems to be labelling error, LCD_ and ILCD , which is it? If I correct this error and run it thru PIC sim, I get 'stack under flow' error Also in MPLAB IDE the program keeps dropping thru the ISR and keeps on re-initialising, thats probably why its running fast. Also as Mike says, Try inserting bcf PIR1,TMR2IF in your ISR. Pommie Well-Known Member ericgibbs said: hi, There seems to be labelling error, LCD_ and ILCD , which is it? If I correct this error and run it thru PIC sim, I get 'stack under flow' error Also in MPLAB IDE the program keeps dropping thru the ISR and keeps on re-initialising, thats probably why its running fast. Also as Mike says, Try inserting bcf PIR1,TMR2IF in your ISR. Just looked again at the code and the stack underflow is caused by the lack of code where there is a comment saying ";main program here". The OP needs to put a line after the comment something like, Code: hang goto hang I can't see why it would drop through the ISR. It is messy though having subroutines embedded inside the ISR code. Mike. ericgibbs Well-Known Member hi mike, >> I can't see why it would drop through the ISR. It is messy though having subroutines embedded inside the ISR code. I agree its messy, the drop thru 'seemed' to happen in simulation. One point IIRC, he refreshes the display even when there is no change in the data?.. Some of the delays in the LCD write module I'm sure total out to more that his 40mSec Intr??? Hope he reworks the code and re-posts, like to see run OK. Regards Nigel Goodwin Super Moderator It all seems a bit confused?, calling subroutines within an interrupt routine isn't a good idea, and you should really keep ISR's as short and fast as possible - it's a seriously bad idea to have the display routines called from with in an ISR. I've posted timer2 clock code (with alarm as well) in this forum a couple of times in the past - my ISR routine updates the clock registers and NOTHING else. The display is updated in the main routine, this is by far the best way to do it!. Speakerguy Active Member ditto what Nigel said. Keep interrupts as simple as possible, deal with only what you absolutely have to deal with and process everything else in the main program. This is one of the things they really hammer home in embedded systems courses in college. bananasiong New Member Pommie said: You are not clearing the interrupt flag. Try inserting bcf PIR1,TMR2IF in your ISR. Mike. Yes, by referring to the codes of my previous program, I missed out this part. After adding it, it runs as expected. I know that it isn't a good idea especially calling a delay routine in the ISR. I have just started and this is my first program on LCD. If the display routine is not in the same as where the clock register is updated, when there is any changes to the clock register, and at the same time, the program is not running in the display routine (delay routine maybe), then the clock will not be updated correctly. Nigel Goodwin Super Moderator You simply update the clock in the main program, either check to see if the seconds counter has changed, and if so then update the clock. Or, even simpler, set a flag in the interrupt routine every time the seconds are incremented - the main routine checks this flag, and updates when it's set - it then resets the flag ready for the next time. Any other operations (such as adjusting the clock) are also done in the same main loop, something like this: Code: Main: Call Check_Flag ;check flag, update display if required Call Check_Keys ;check if any keys pressed, and service them Call Check_Alarm ;check if alarm should be triggered Call Check_Anythingelse ;whatever you need to do Goto Main bananasiong New Member Oh... seems I'm starting from the wrong direction. I will post up once I finish modifying the program. Thanks all Status Not open for further replies. Replies 4 Views 3K Replies 24 Views 2K Replies 3 Views 2K Replies 0 Views 3K Replies 4 Views 1K	2022-07-07 09:59:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3127012252807617, "perplexity": 5075.8034418357165}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104690785.95/warc/CC-MAIN-20220707093848-20220707123848-00235.warc.gz"}
https://www.nextgurukul.in/wiki/concept/cbse/class-9/science/the-fundamental-unit-of-life/cell-structure/3958037	Notes On Cell Structure - CBSE Class 9 Science The cell is the fundamental and structural unit of life. Eukaryotic cells Eukaryotic cells are the cells which possess a definite nucleus with a distinct nuclear membrane. Let us observe the contents of a eukaryotic cell. The living structures of a cell are called as cell organelles. Cell wall: It is the non-living outermost covering of the cell. The cell wall is composed of cellulose and is permeable. It separates the contents of the cell from the surroundings. It gives shape and protection to the cell. It is present only in plant cells. Functions of cell wall • Cell wall provides rigidity and shape to the cell. • Cell wall is a dead structure but permeable to substances. • Cell wall is made mainly of cellulose. • Cell wall is thick and protective in function. Cell membrane: It is also called as plasma membrane. It is present both in animal and plant cells. It is a thin living membrane made of lipoproteins. It comprises three layers, a lipid layer sandwiched between two protein layers. It is a selectively permeable membrane. Functions of cell membrane • It maintains the shape of the cell. It separates the cellular contents from the external environment. • It helps in absorption of mechanical shocks and protects the cell from injury. • It allows the movement of some substances into and out of the cell. Movement of substances through this semi-permeable membrane can be by the process of diffusion, osmosis etc. • It forms the membrane for various cell organelles. • It keeps the adjacent cells in contact. • Its infolds help in absorption of materials by the process of pinocytosis or phagocytosis. • Its outfolds increase the area for absorption. Difference between diffusion and osmosis DIFFUSION OSMOSIS It is the process by which molecules spread from a region of higher concentration to a region of lower concentration. It is a process by which solvent molecules move from a region of low concentration to theregion of high concentration through a semipermeable membrane. Diffusion occurs in all the mediums. Diffusing particles can belong to either one of the states, solid, liquid or a gas. Osmosis takes place only in liquid medium and only the solvent molecules can move from one place to another. Semipermeable membrane may not be present. Presence of semipermeable membraneseparating the two liquids is necessary. Different types of diffusion may include Brownian motion, facilitated diffusion etc. Forward osmosis and reverse osmosis are the types of osmosis. Nucleus: A true nucleus acts as the control centre of the cell. It is covered by a two layered, perforated nuclear membrane allowing substances to enter and leave the nucleus. Nucleus is filled with semifluid colloidal substance called as nucleoplasm in which float the nucleolus and the chromatin fibres. The nucleus contains chromosomes, which are composed of Deoxyribonucleic acid (DNA) and proteins. Functions of nucleus • Nucleus controls all the activities of the cell. • As the nucleus carries genetic information in the form of DNA, it plays a major role in cell division and cell development. The functional segments of DNA are called genes. • Nucleus plays an important role in protein synthesis and transmission of characters from one generation to another generation. Cytoplasm: It is a jelly like viscous substance occupying entire cell except the nucleus. The cytoplasm supports and protects the cell organelles that perform different metabolic functions. Cell organelles include endoplasmic reticulum, Ribosomes, Golgi apparatus, Mitochondria, Plastids, Lysosomes, and Vacuoles. Inclusions are the non-living substances present in the cytoplasm. CYTOPLASM NUCLEOPLASM It is a jelly like homogeneous substance. It is a transparent fluid present in the nucleus. Protoplasmic part which lies outside the nucleus. Protoplasmic part which is present inside the nucleus. Cell organelles and inclusions are suspended in the cytoplasm. Chromatin fibres and nucleolus are suspended. Both inorganic and organic molecules are present. Composition is rich in nucleotides. Endoplasmic reticulum: It is a network of tubules and flattened sacs perform various activities in the cell. The space inside the endoplasmic reticulum is called as lumen. Endoplasmic reticulum serves as a channel for the transport of proteins between various regions of the cytoplasm. Two types of endoplasmic reticulum are Rough Endoplasmic Reticulum (RER) and Smooth Endoplasmic Reticulum (SER). ROUGH ENDOPLASMIC RETICULUM SMOOTH ENDOPLASMIC RETICULUM These are rough at surface and are associated with ribosomes. These are smooth at surface and are associated with Golgi bodies. These are responsible for the synthesis of proteins and enzymes. These are responsible for the synthesis of glycogen, lipids etc. These provide structural framework for the cell. These take part in the synthesis of fats in adipose cells. These also act as channels for quick transport. These can also detoxify drugs and some poisons. These help in the synthesis of membranes. These help in the formation of rhodopsin from vitamin A. Ribosomes: These are naked granules with no membrane. They are found scattered in the cytoplasm or attached to the outside of the endoplasmic reticulum. They are the biofactories of proteins. Ribosomes can be free ribosomes scattered in the cytoplasm or bound ribosomes seen attached to endoplasmic reticulum. Ribosomes are made up of two subunits, the larger subunit and the smaller subunit. 70s ribosomes are present in prokaryotes and 80s ribosomes are present in eukaryotes. Functions of ribosomes • Ribosomes are considered to be the biofactories since they are the sites of protein synthesis. • They are the lodging sites for many enzymes participating in the process of protein synthesis. Golgi apparatus: These cell organelles are named after the biologist, Camillo Golgi, who first described it. The Golgi consists of a stack of membrane-bounded cisternae located between the endoplasmic reticulum and the cell surface. Functions of Golgi apparatus • It synthesises certain biopolymers. • It also consists of some processing enzymes which alter some proteins and phospholipids synthesised by endoplasmic reticulum. Mitochondria: These are cellular organelles termed as ‘power houses of the cells’. These are bounded by a double membrane. The outer membrane is smooth while the inner membrane is thrown into folds called as cristae. The cristae increase the area of cellular respiration. Cristae on their surface possess structures called oxysomes. Oxysomes are rich in ATP synthetase enzyme. Both the membranes are separated by intermembrane space. Mitochondrial matrix is rich in respiratory enzymes. Mitochondria have their own DNA, RNA and ribosomes to synthesise respiratory enzymes. The enzymes in the mitochondria oxidise glucose molecules to produce energy in the form of Adenosine Triphosphate or ATP. Functions of mitochondria • These synthesise energy rich ATP molecules. • These help in the synthesis of fatty acid, amino acids, steroids by providing them with biological intermediates. Lysosomes: Lysosomes are membranous sacs filled with enzymes. The enzymes are hydrolytic enzymes which are capable of digesting cellular macromolecules. When the cell gets damaged, the lysosome may burst and its enzymes may digest thecell itself. Hence, lysosomes are called as ‘suicidal bags’. Functions of lysosomes • Help in killing foreign cells referred to as pathogens. • Help in destroying the diseased cells. Vacuoles: Vacuoles are membrane bound compartments present in both plant and animal cells. These organelles store water, waste products, and substances like amino acids, sugars and proteins. The fluid in the vacuoles is called ‘cell sap’. A vacuole is covered by a living membrane called “tonoplast”. Vacuoles also provide buoyancy to the cell. Functions of vacuoles • They help the cell to maintain its buoyancy and turgidity. • They play an important role in the growth of the cell Plastids: These are major organelles found only in the cells of plants and algae. These are of three types namely, Chloroplasts, Chromoplasts and Leucoplasts. a) Chloroplasts are one kind of plastids present mainly in plant cells. Chloroplasts contain green pigment called as chlorophyll. Chlorophyll absorbs energy from the sunlight necessary for photosynthesis. b) Chromoplasts are the organelles which provide bright colours to the plant structures like buds, flowers etc. c) Leucoplasts are the organelles which store starch, oils and protein granules. Functions of plastids • Plastids are responsible for synthesisof food. • Plastids are responsible for colouration of different parts of the plant. • Plastids are responsible for storage of food molecules. Difference between plant cells and animal cells PLANT CELLS ANIMAL CELLS Plant cells possess cell wall. Animal cells do not possess cell wall. Chloroplasts are present in plant cells. Animal cells do not possess chloroplasts. Plant cells are almost rectangular in shape. Animal cells are round in shape. Plant cells possess large vacuoles. Animal cells have many small vacuoles. Cilia rarely occur in plant cells. Cilia are present in animal cells. Higher plants do not possess centrioles. Animal cells do contain centrioles. Plant cells have no or less number of lysosomes. Animal cells possess many number of vacuoles. Summary The cell is the fundamental and structural unit of life. Eukaryotic cells Eukaryotic cells are the cells which possess a definite nucleus with a distinct nuclear membrane. Let us observe the contents of a eukaryotic cell. The living structures of a cell are called as cell organelles. Cell wall: It is the non-living outermost covering of the cell. The cell wall is composed of cellulose and is permeable. It separates the contents of the cell from the surroundings. It gives shape and protection to the cell. It is present only in plant cells. Functions of cell wall • Cell wall provides rigidity and shape to the cell. • Cell wall is a dead structure but permeable to substances. • Cell wall is made mainly of cellulose. • Cell wall is thick and protective in function. Cell membrane: It is also called as plasma membrane. It is present both in animal and plant cells. It is a thin living membrane made of lipoproteins. It comprises three layers, a lipid layer sandwiched between two protein layers. It is a selectively permeable membrane. Functions of cell membrane • It maintains the shape of the cell. It separates the cellular contents from the external environment. • It helps in absorption of mechanical shocks and protects the cell from injury. • It allows the movement of some substances into and out of the cell. Movement of substances through this semi-permeable membrane can be by the process of diffusion, osmosis etc. • It forms the membrane for various cell organelles. • It keeps the adjacent cells in contact. • Its infolds help in absorption of materials by the process of pinocytosis or phagocytosis. • Its outfolds increase the area for absorption. Difference between diffusion and osmosis DIFFUSION OSMOSIS It is the process by which molecules spread from a region of higher concentration to a region of lower concentration. It is a process by which solvent molecules move from a region of low concentration to theregion of high concentration through a semipermeable membrane. Diffusion occurs in all the mediums. Diffusing particles can belong to either one of the states, solid, liquid or a gas. Osmosis takes place only in liquid medium and only the solvent molecules can move from one place to another. Semipermeable membrane may not be present. Presence of semipermeable membraneseparating the two liquids is necessary. Different types of diffusion may include Brownian motion, facilitated diffusion etc. Forward osmosis and reverse osmosis are the types of osmosis. Nucleus: A true nucleus acts as the control centre of the cell. It is covered by a two layered, perforated nuclear membrane allowing substances to enter and leave the nucleus. Nucleus is filled with semifluid colloidal substance called as nucleoplasm in which float the nucleolus and the chromatin fibres. The nucleus contains chromosomes, which are composed of Deoxyribonucleic acid (DNA) and proteins. Functions of nucleus • Nucleus controls all the activities of the cell. • As the nucleus carries genetic information in the form of DNA, it plays a major role in cell division and cell development. The functional segments of DNA are called genes. • Nucleus plays an important role in protein synthesis and transmission of characters from one generation to another generation. Cytoplasm: It is a jelly like viscous substance occupying entire cell except the nucleus. The cytoplasm supports and protects the cell organelles that perform different metabolic functions. Cell organelles include endoplasmic reticulum, Ribosomes, Golgi apparatus, Mitochondria, Plastids, Lysosomes, and Vacuoles. Inclusions are the non-living substances present in the cytoplasm. CYTOPLASM NUCLEOPLASM It is a jelly like homogeneous substance. It is a transparent fluid present in the nucleus. Protoplasmic part which lies outside the nucleus. Protoplasmic part which is present inside the nucleus. Cell organelles and inclusions are suspended in the cytoplasm. Chromatin fibres and nucleolus are suspended. Both inorganic and organic molecules are present. Composition is rich in nucleotides. Endoplasmic reticulum: It is a network of tubules and flattened sacs perform various activities in the cell. The space inside the endoplasmic reticulum is called as lumen. Endoplasmic reticulum serves as a channel for the transport of proteins between various regions of the cytoplasm. Two types of endoplasmic reticulum are Rough Endoplasmic Reticulum (RER) and Smooth Endoplasmic Reticulum (SER). ROUGH ENDOPLASMIC RETICULUM SMOOTH ENDOPLASMIC RETICULUM These are rough at surface and are associated with ribosomes. These are smooth at surface and are associated with Golgi bodies. These are responsible for the synthesis of proteins and enzymes. These are responsible for the synthesis of glycogen, lipids etc. These provide structural framework for the cell. These take part in the synthesis of fats in adipose cells. These also act as channels for quick transport. These can also detoxify drugs and some poisons. These help in the synthesis of membranes. These help in the formation of rhodopsin from vitamin A. Ribosomes: These are naked granules with no membrane. They are found scattered in the cytoplasm or attached to the outside of the endoplasmic reticulum. They are the biofactories of proteins. Ribosomes can be free ribosomes scattered in the cytoplasm or bound ribosomes seen attached to endoplasmic reticulum. Ribosomes are made up of two subunits, the larger subunit and the smaller subunit. 70s ribosomes are present in prokaryotes and 80s ribosomes are present in eukaryotes. Functions of ribosomes • Ribosomes are considered to be the biofactories since they are the sites of protein synthesis. • They are the lodging sites for many enzymes participating in the process of protein synthesis. Golgi apparatus: These cell organelles are named after the biologist, Camillo Golgi, who first described it. The Golgi consists of a stack of membrane-bounded cisternae located between the endoplasmic reticulum and the cell surface. Functions of Golgi apparatus • It synthesises certain biopolymers. • It also consists of some processing enzymes which alter some proteins and phospholipids synthesised by endoplasmic reticulum. Mitochondria: These are cellular organelles termed as ‘power houses of the cells’. These are bounded by a double membrane. The outer membrane is smooth while the inner membrane is thrown into folds called as cristae. The cristae increase the area of cellular respiration. Cristae on their surface possess structures called oxysomes. Oxysomes are rich in ATP synthetase enzyme. Both the membranes are separated by intermembrane space. Mitochondrial matrix is rich in respiratory enzymes. Mitochondria have their own DNA, RNA and ribosomes to synthesise respiratory enzymes. The enzymes in the mitochondria oxidise glucose molecules to produce energy in the form of Adenosine Triphosphate or ATP. Functions of mitochondria • These synthesise energy rich ATP molecules. • These help in the synthesis of fatty acid, amino acids, steroids by providing them with biological intermediates. Lysosomes: Lysosomes are membranous sacs filled with enzymes. The enzymes are hydrolytic enzymes which are capable of digesting cellular macromolecules. When the cell gets damaged, the lysosome may burst and its enzymes may digest thecell itself. Hence, lysosomes are called as ‘suicidal bags’. Functions of lysosomes • Help in killing foreign cells referred to as pathogens. • Help in destroying the diseased cells. Vacuoles: Vacuoles are membrane bound compartments present in both plant and animal cells. These organelles store water, waste products, and substances like amino acids, sugars and proteins. The fluid in the vacuoles is called ‘cell sap’. A vacuole is covered by a living membrane called “tonoplast”. Vacuoles also provide buoyancy to the cell. Functions of vacuoles • They help the cell to maintain its buoyancy and turgidity. • They play an important role in the growth of the cell Plastids: These are major organelles found only in the cells of plants and algae. These are of three types namely, Chloroplasts, Chromoplasts and Leucoplasts. a) Chloroplasts are one kind of plastids present mainly in plant cells. Chloroplasts contain green pigment called as chlorophyll. Chlorophyll absorbs energy from the sunlight necessary for photosynthesis. b) Chromoplasts are the organelles which provide bright colours to the plant structures like buds, flowers etc. c) Leucoplasts are the organelles which store starch, oils and protein granules. Functions of plastids • Plastids are responsible for synthesisof food. • Plastids are responsible for colouration of different parts of the plant. • Plastids are responsible for storage of food molecules. Difference between plant cells and animal cells PLANT CELLS ANIMAL CELLS Plant cells possess cell wall. Animal cells do not possess cell wall. Chloroplasts are present in plant cells. Animal cells do not possess chloroplasts. Plant cells are almost rectangular in shape. Animal cells are round in shape. Plant cells possess large vacuoles. Animal cells have many small vacuoles. Cilia rarely occur in plant cells. Cilia are present in animal cells. Higher plants do not possess centrioles. Animal cells do contain centrioles. Plant cells have no or less number of lysosomes. Animal cells possess many number of vacuoles. Activities Activity 1 Nobelprize.org has constructed a game for students to gain knowledge about cell structure and its organelles. The template is colourful an made in an interesting way to attract students. Students can follow the instructions by reading the text displayed. Students can jump into organelle sorting game directly by clicking on the button. Organelle sorting game has a Help button to click on which displays the menu. It will allow the user to know the mechanism of centrifuge. It also allows the user to know about the cell organelles individually. Go to Activity Activity 2 Sepuplhs.org has designed a template to explain about ' What cells do?'. User needs to click on the start button. Then click on the 'Next' button. Different types of cells can be constructed. The template allows the student to select the type of cell he wants to construct and later select the organelles to construct that particular type of cells. Template will guide the user by instructing him when he does a mistake. Go to Activity Previous	2021-03-03 00:14:55	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2181989997625351, "perplexity": 9366.601232350402}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178364932.30/warc/CC-MAIN-20210302221633-20210303011633-00229.warc.gz"}
https://rpg.stackexchange.com/questions/128922/by-asking-my-players-to-match-my-roll-am-i-making-their-checks-harder	# By asking my players to match my roll, am I making their checks harder? Sometimes I ask my players to make a roll to see if something random happens (winning a game of chance, for instance, or a wilderness encounter). These rolls won't involve adding any modifiers, they'll almost always be straight up or down rolls. In order to preserve some mystery, I won't tell them beforehand what they have to roll in order to achieve the outcome and I'll assign a random number or range of numbers to the outcome. Primarily this is to try and move away from the 20 = good and 1 = bad mentality that my players tend to have when rolling. It also means that if I make them roll a d20 when their characters are settling down to camp overnight, even though my players might have read the DMG, it won't actually be a roll of 17-20 that results in a random encounter, and instead it'll be a roll of 4-7 instead (for instance). Keeps them on their toes. So what I've been doing is rolling a d20 in secret and then seeing if the player rolls a matching number. If I'm looking for a range I'll just add the size of the range to my rolled number and wrap around. E.g. I roll an 18. For an event with 25% chance of occurrence I'll look for the player to roll an 18, 19, 20, 1 or 2. From a statistics perspective the odds of rolling any particular number on a d20 is 1/20. But to roll two of the same number in a row would be 1/400, which is much less likely. So my question is: by expecting players to match the roll of my dice instead of picking a number at random (or sticking to the suggested ones in the DMG) am I inadvertently making it less likely for them to succeed? • Are you concerned solely with the mathematical probability of passing any particular check? Or would you additionally be concerned about social and psychological issues that this practice may provoke? – user11450 Aug 4, 2018 at 5:27 • @Hurkyl if there are any other techniques that achieve the effect I'm going for then I'd like to hear about them. If I can assure my players that this method is fair (as it seems to be) are there any other social or psychological issues I should be aware of? Aug 4, 2018 at 5:44 • The main worry is that the "bigger = more effective" convention makes the mechanics generally easier to understand, so if you succeed at knocking your players out of that mindset, you've made the game system a little bit more impenetrable overall. Randomizing the success ranges also eliminates feedback that helps the players understand how likely/unlikely some results are. Some social consequences are that you diminish players' excitement of seeing their die roll, and added suspicion of the DM fudging results. – user11450 Aug 6, 2018 at 1:12 # No. From a statistics perspective the odds of rolling any particular number on a d20 is 1/20. But to roll two of the same number in a row would be 1/400, which is much less likely. You are right that the chance of rolling $$\\{20, 20\}\$$ is 1/400, for example. But you actually have 20 possible outcomes that will result in a success for the player, i.e., if you roll 1 and he rolls 1, he succeeds, if you roll 2 and he rolls 2, etc... So, the actual probability of success $$\P(s)\$$ is given by $$P(s) = \sum_{d = 1}^{X} P(d)^2$$ where $$\X\$$ is the dice size (i.e. 20 for a d20) $$\P(d)\$$ is the probability of a specific dice value being rolled (1/X), in the case of the d20, simply 1/20. Since they are uniformly distributed and, by definition of a probability, the sum has to be 1, it goes that $$P(s) = \frac{1}{X} \sum_{d=1}^{X} P(d) = \frac{1}{X}$$ exactly equal to the probability of any dice number being rolled. Another way to see it is a conditional probability: the probability of he rolling a number Y given that you already rolled Y is 1/X. Or, skipping the math and answering it: no, you don't change the probability at all. It stays exactly and perfectly the same. ## No, but maybe yes. HellSaint is right about the math. As long as the number of possible outcomes that are a "success" is the same, it doesn't matter which numbers they are. However, in any situation where they would get to add their characters' stats or bonuses to the roll, this method would negate the benefit of doing things the character is good at. If you want to leave it up to pure chance that's fine, but then you might as well just roll the die yourself. Take the chance of a wilderness encounter. Is that really pure luck, or does a seasoned explorer have some advantage? Is there anything they can do--post more guards, don't light a fire, get a watchdog--to be less likely to get a hostile encounter or more likely to get a helpful one? With a straight high roll, you can give bonuses, which rewards your players for having good ideas and for playing to their party's strengths. With the method you're using, you could expand the range of success numbers, but the players don't see it happening, so they won't feel rewarded. In general you shouldn't make your players roll for success or failure on a task that they have no control over. Even if their characters are in a gambling den and the die roll represents a literal die roll, why have them gamble and not give them the option to cheat? • What about Death Saving Throws? They are completely by luck, with no modifier whatsoever, but the players roll it, not the DM, even if the players (or their characters) have no control over it. Aug 4, 2018 at 1:10 • @HellSaint death saving throws are special because they are still Saving Throws and though they get no ability/proficiency they are still explicitly boosted by things that generically improve saving throws: "... aided only by spells and features that improve your chances of succeeding on a saving throw." So the players can have some influence over how likely they (or at least their allies) are to succeed at such a roll. Aug 4, 2018 at 9:07 • I think this is a fair suggestion - to get the same outcome (keeping the mystery) without the overhead of worrying about whether the suggested approach is fair, it's easier to do the rolls myself and not worry about it. Aug 4, 2018 at 9:17 You said: From a statistics perspective the odds of rolling any particular number on a d20 is 1/20. But to roll two of the same number in a row would be 1/400, which is much less likely. This is an incorrect analogy. While the chances of rolling a particular (e.g. "3") twice in a row is indeed 1/400 for a d20, that is not what you're asking them to do. You're asking them to roll the same number you rolled. One way of looking at this is to realize that of the 400 possible combinations that come up, 20 of them are valid hits - if you roll 1 and they roll 1; if you roll 2 and they roll 2; and so forth. If 20 of the 400 are hits, then the chances of a hit is 1/20. The other way of looking at this is that your die roll is GUARANTEED to come up with a number; any number is valid. After you have rolled, there is a 1/20 chance their roll comes up with the same number. Of course, if both of you roll the same die, and it isn't balanced, or if your two dice are unbalanced in the same way, this will increase their chances (of course, if they are unbalanced in different ways, it will work against them).	2022-05-28 17:53:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.47779205441474915, "perplexity": 653.29174998178}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652663016949.77/warc/CC-MAIN-20220528154416-20220528184416-00330.warc.gz"}
https://gmatclub.com/forum/a-number-when-divided-by-105-leaves-99-as-remainder-173670.html	Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st. It is currently 17 Jul 2019, 01:36 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # A number when divided by 105 leaves 99 as remainder Author Message TAGS: ### Hide Tags Intern Joined: 21 Jun 2014 Posts: 3 A number when divided by 105 leaves 99 as remainder [#permalink] ### Show Tags Updated on: 01 Jul 2014, 22:57 3 3 00:00 Difficulty: 5% (low) Question Stats: 90% (01:15) correct 10% (01:30) wrong based on 231 sessions ### HideShow timer Statistics A number when divided by 105 leaves 99 as remainder. What will be the remainder if the number is divided by 21? A. 9 B. 14 C. 20 D. 6 E. 15 Originally posted by Gmatkarma101 on 01 Jul 2014, 22:54. Last edited by Gnpth on 01 Jul 2014, 22:57, edited 1 time in total. Tweak Senior Manager Joined: 13 Jun 2013 Posts: 271 Re: A number when divided by 105 leaves 99 as remainder [#permalink] ### Show Tags 02 Jul 2014, 00:52 8 1 Gmatkarma101 wrote: A number when divided by 105 leaves 99 as remainder. What will be the remainder if the number is divided by 21? A. 9 B. 14 C. 20 D. 6 E. 15 let the number be x, then we have x=105k+99 , where k=0,1,2,3... now remainder when x is divided by 21, will be (105k+99)/21 here 105 is a multiple of 21. therefore will leave zero remainder. whereas 99 when divided by 21 will leave 15 as a remainder. therefore overall remainder is 15 ##### General Discussion SVP Status: The Best Or Nothing Joined: 27 Dec 2012 Posts: 1787 Location: India Concentration: General Management, Technology WE: Information Technology (Computer Software) Re: A number when divided by 105 leaves 99 as remainder [#permalink] ### Show Tags 02 Jul 2014, 01:07 2 1 Let the number be 105+99 = 204 204 when divided by 105, give a remainder of 99 Dividing 204 by 21 Remainder = 204 - 189 = 15 _________________ Kindly press "+1 Kudos" to appreciate Manager Joined: 25 Apr 2014 Posts: 104 Re: A number when divided by 105 leaves 99 as remainder [#permalink] ### Show Tags 08 Jul 2014, 18:10 Hi PareshGmat Is this method shown by you going to work for all of such type of questions? Or you just did it this way for this particular question? SVP Status: The Best Or Nothing Joined: 27 Dec 2012 Posts: 1787 Location: India Concentration: General Management, Technology WE: Information Technology (Computer Software) Re: A number when divided by 105 leaves 99 as remainder [#permalink] ### Show Tags 08 Jul 2014, 18:42 1 maggie27 wrote: Hi PareshGmat Is this method shown by you going to work for all of such type of questions? Or you just did it this way for this particular question? It would work with all such type of questions _________________ Kindly press "+1 Kudos" to appreciate Senior Manager Joined: 24 Jun 2016 Posts: 340 GMAT 1: 770 Q60 V60 GPA: 4 Re: A number when divided by 105 leaves 99 as remainder [#permalink] ### Show Tags 26 Jun 2016, 02:07 A = 105k + 99 = 215k + 214 + 15, so A/21 remainder 15. _________________ Stuck in the 600's and want to score 700+ on the GMAT? If this describes you, we should talk. I specialize in getting motivated students into the 700's. \$75/hour as of July 2019. I am not accepting any more students for the Fall 2019 application cycle, but if you are planning to apply in 2020, feel free to reach out! HanoiGMATTutor@gmail.com Manager Joined: 12 Jun 2016 Posts: 212 Location: India WE: Sales (Telecommunications) Re: A number when divided by 105 leaves 99 as remainder [#permalink] ### Show Tags 07 Aug 2017, 00:32 Hi Moderators/Math experts, Let N be the number. So, we have $$N = 105q + 99$$ (where q is an integer). Dividing throughout by 21 we get $$\frac{N}{21} = 3q + \frac{99}{21}$$ From the responses given by fellow club members, I do see that we should not reduce $$\frac{99}{21}$$ to its lowest form $$\frac{33}{7}$$ for finding the reaminder. My question is - why should we not do so and then find the remainder? I wanted to know this because the answer will be different in both cases and worry that there may be a Trap answer in real GMAT TIA for the help! _________________ My Best is yet to come! Intern Joined: 01 Aug 2017 Posts: 8 Re: A number when divided by 105 leaves 99 as remainder [#permalink] ### Show Tags 07 Aug 2017, 06:47 1 susheelh wrote: Hi Moderators/Math experts, Let N be the number. So, we have $$N = 105q + 99$$ (where q is an integer). Dividing throughout by 21 we get $$\frac{N}{21} = 3q + \frac{99}{21}$$ From the responses given by fellow club members, I do see that we should not reduce $$\frac{99}{21}$$ to its lowest form $$\frac{33}{7}$$ for finding the reaminder. My question is - why should we not do so and then find the remainder? I wanted to know this because the answer will be different in both cases and worry that there may be a Trap answer in real GMAT TIA for the help! The question is asking you to divide by 21 not by 7 .. the remainder for divisor 21 will vary from 1 - 20 and for 7 will vary from 1-6 . Always better to stick to the original divisor Senior Manager Joined: 06 Jul 2016 Posts: 360 Location: Singapore Concentration: Strategy, Finance Re: A number when divided by 105 leaves 99 as remainder [#permalink] ### Show Tags 07 Aug 2017, 11:05 N = 105Q + 99 Let Q = 1 => N = 204 204 divided by 21 = 219 + 15 Sent from my iPhone using GMAT Club Forum _________________ Put in the work, and that dream score is yours! CEO Joined: 12 Sep 2015 Posts: 3847 Re: A number when divided by 105 leaves 99 as remainder [#permalink] ### Show Tags 07 Aug 2017, 11:12 2 Top Contributor Gmatkarma101 wrote: A number when divided by 105 leaves 99 as remainder. What will be the remainder if the number is divided by 21? A. 9 B. 14 C. 20 D. 6 E. 15 When it comes to remainders, we have a nice rule that says: If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc. For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc. A number when divided by 105 leaves 99 as remainder. So, the possible values of the number are: 99, 99+105, 99+(2)(105), 99+(3)(105), 99+(4)(105), etc. Let's see what happens when test out the smallest possible value: 99 What will be the remainder if the number is divided by 21? 99 divided by 21 equals 4 with remainder 15 RELATED VIDEO _________________ Test confidently with gmatprepnow.com Director Joined: 04 Dec 2015 Posts: 745 Location: India Concentration: Technology, Strategy WE: Information Technology (Consulting) Re: A number when divided by 105 leaves 99 as remainder [#permalink] ### Show Tags 07 Aug 2017, 22:06 Gmatkarma101 wrote: A number when divided by 105 leaves 99 as remainder. What will be the remainder if the number is divided by 21? A. 9 B. 14 C. 20 D. 6 E. 15 Let the number be $$= n$$ Number $$n$$ is in form ; $$n = 105q + 99$$ Dividing $$n$$ by $$21$$, we get; $$\frac{n}{21} => \frac{105q + 99}{21} => \frac{105q}{21} + \frac{99}{21}$$ $$105$$ is divisible by $$21$$ leaving remainder $$0$$. ($$215 = 105$$) $$\frac{99}{21}$$ gives ; $$21* 4 = 84$$ $$99-84 = 15$$ Remainder $$= 15$$ _________________ Please Press "+1 Kudos" to appreciate. Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 6923 Location: United States (CA) Re: A number when divided by 105 leaves 99 as remainder [#permalink] ### Show Tags 10 Aug 2017, 10:31 Gmatkarma101 wrote: A number when divided by 105 leaves 99 as remainder. What will be the remainder if the number is divided by 21? A. 9 B. 14 C. 20 D. 6 E. 15 n = 105Q + 99 Thus, when that number is divided by 21, we have: (105Q + 99)/21 5Q + 99/21 Since 5Q is an integer, the remainder will come from 99 divided by 21: 99/21 = 4 R 15 We see that the remainder is 15. _________________ # Scott Woodbury-Stewart Founder and CEO Scott@TargetTestPrep.com 122 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Target Test Prep Representative Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2823 Re: A number when divided by 105 leaves 99 as remainder [#permalink] ### Show Tags 18 Aug 2018, 20:11 Gmatkarma101 wrote: A number when divided by 105 leaves 99 as remainder. What will be the remainder if the number is divided by 21? A. 9 B. 14 C. 20 D. 6 E. 15 The number could be 105 + 99 = 204 since 204/105 = 1 R 99. Since 204/21 = 9 R 15, the remainder is 15. _________________ # Jeffrey Miller Jeff@TargetTestPrep.com 122 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Re: A number when divided by 105 leaves 99 as remainder [#permalink] 18 Aug 2018, 20:11 Display posts from previous: Sort by	2019-07-17 08:36:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7002187967300415, "perplexity": 3515.271832977812}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195525133.20/warc/CC-MAIN-20190717081450-20190717103450-00101.warc.gz"}
http://eprints.iisc.ernet.in/907/	# Field Theories of Frustrated Heisenberg Antiferromagnets Rao, Sumathi and Sen, Diptiman (1995) Field Theories of Frustrated Heisenberg Antiferromagnets. [Preprint] Preview PDF 9506145.pdf We study the Heisenberg antiferromagnetic chain with both dimerization and frustration. The classical ground state has three phases: a Neel phase, a spiral phase and a colinear phase. In each phase, we discuss a non-linear sigma model field theory governing the low energy excitations. We study the theory in the spiral phase in detail using the renormalization group. The field theory, based on an $SO(3)$ matrix-valued field, becomes $SO(3) \times SO(3)$ and Lorentz invariant at long distances where the elementary excitation is analytically known to be a massive spin-$1/2$ doublet. The field theory supports $Z_2 ~$ solitons which lead to a double degeneracy in the spectrum for half-integer spins (when there is no dimerization).	2015-07-08 02:37:11	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7202597856521606, "perplexity": 1273.9208528912031}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375635604.22/warc/CC-MAIN-20150627032715-00287-ip-10-179-60-89.ec2.internal.warc.gz"}
https://solvedlib.com/n/2-10-pts-let-f-be-a-function-whose-derivative-r-2-i6r-quot,19063504	# 2. (10 pts) Let f ()be a function whose derivative (r)=2 I6r" 30r. Use second derivative test t0 find all ###### Question: 2. (10 pts) Let f ()be a function whose derivative (r)=2 I6r" 30r. Use second derivative test t0 find all _ values which f (x) has local maximum and or mimnimumi Local Maximum at Local minimum J( #### Similar Solved Questions ##### Student.athesting.com Reading-Sthudy Materials-My AT Grammany TEAS SmartPrep: Reading ati OCLOSE End of Lesson Quiz: 1.3 The... student.athesting.com Reading-Sthudy Materials-My AT Grammany TEAS SmartPrep: Reading ati OCLOSE End of Lesson Quiz: 1.3 The house was situated in a neighborhood that had rarely experienced any crime. In fact, no one could remember the last time anyone's home had been broken into. The community ... ##### Kimmel, Accounting, 7e Help System Announcements Do It! Review 10-04 Sandhill Co. provides you with the... 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Current assets Long-term assets Total assets $14,360 31,800$46,160 Current liabilities Long-term liabilities Stockholders' equity ... ##### Select the functional groups that are present in the following molecule (select 3 best answers)OHH;cAlkeneAlkyneAromaticAlkyl HalideAlcoholThiolAldehydeKetoneEtherCarboxylic AcidEsterAcid ChlorideAmineAmideNitrile Select the functional groups that are present in the following molecule (select 3 best answers) OH H;c Alkene Alkyne Aromatic Alkyl Halide Alcohol Thiol Aldehyde Ketone Ether Carboxylic Acid Ester Acid Chloride Amine Amide Nitrile... ##### Estimate Anunknow populalion mean Use the given information find the minimum sample size required ntemtor 5129 confidence level: 9990, 5507 Margin C) 103 D) 59 A) 52 the critical value za/z (6) find the critical value talz (c) state that Do one of the following aPpropriate: (2) Find neither the normal nor the distribution applies_ appean mnminat distribuled. 6) 999;n = 17; is unknown; population C)ta 2.898 D) tz 2921 2.567 B) za/2 583 A) zan Use the given . confidence level and sample data t0 fi estimate Anunknow populalion mean Use the given information find the minimum sample size required ntemtor 5129 confidence level: 9990, 5507 Margin C) 103 D) 59 A) 52 the critical value za/z (6) find the critical value talz (c) state that Do one of the following aPpropriate: (2) Find neither the norm... ##### [-/1 Points] DETAILS LARLINALG8 4.3.003. 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The ciphertext is c = 3 When will they meet?... ##### Xy' - y = Vx? +y? xy' - y = Vx? +y?... ##### Kitchen Supply, Inc. (KSI), manufactures three types of flatware: Institutional, standard, and silver. It applies all... Kitchen Supply, Inc. (KSI), manufactures three types of flatware: Institutional, standard, and silver. It applies all Indirect costs according to a predetermined rate based on direct labor-hours. A consultant recently suggested that the company switch to an activity-based costing system and prepared... ##### C 4 H 9 , 114.21 g/mol Express your answer as a chemical formula. C 4 H 9 , 114.21 g/mol Express your answer as a chemical formula.... ##### Biologist is studying a certain type of beetle. We know that the Cactus Eating Beetle spends 25% of its time eating; 15% of its time burrowing, 30% of its time sleeping, and 30% of its time sunning itself. The biologist has found a bug with unusual markings, and wants to see if this beetle acts like a CEB. The biologist watches the beetle for 100 hours, and observes eating for 30 hours, burrowing for 20 hours, sleeping for 22 hours, and sunning for 28 hours _ Can the biologist be 95% sure that t biologist is studying a certain type of beetle. We know that the Cactus Eating Beetle spends 25% of its time eating; 15% of its time burrowing, 30% of its time sleeping, and 30% of its time sunning itself. 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How many ounces of medication are required for a person who weights 175 lbs.? 1.5 1.75 2.5... ##### Find all critical points of the function f(s,y) =2? 4r+y 6y + 5 on the closed triangular region with vertices 0,0}, (4.01. and (4,2}. (Multiple answers possible} (Multiple answers possible means that for each correct item you select; you will get a point; and for each incorrect itemn you select you will lose point)0(o}0 (2301200iook (401(21 04,n) P# I3WV C on Find all critical points of the function f(s,y) =2? 4r+y 6y + 5 on the closed triangular region with vertices 0,0}, (4.01. and (4,2}. (Multiple answers possible} (Multiple answers possible means that for each correct item you select; you will get a point; and for each incorrect itemn you select you ...	2022-11-27 07:59:31	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5902740359306335, "perplexity": 7039.198357213784}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710218.49/warc/CC-MAIN-20221127073607-20221127103607-00252.warc.gz"}
https://docs.babylon.finance/babl/mining	BABL Mining Program BABL Mining Program was the first governance proposal. It was approved unanimously by the token holders. Rewards started on October 16th 2021. 500,000 tokens will be allocated to the BABL Mining Program, representing 50% of the total token supply over ten years. You can read the initial Medium announcement here. # TLDR • The Mining Program lasts for aprox. 10 years. • Only active strategies earn BABL. • Members can claim the BABL only after the strategy has completed. • The longer the strategy lasts, the more BABL it earns. • The more capital the strategy has, the more BABL it earns. • The more returns the strategy has, the more BABL it earns. • Staked prophets in gardens give garden members additional bonuses in the mining program. • Strategist receives 10% of the BABL of a strategy (if it returns profits). • Voters receive 10% of the BABL of a strategy. • Garden members receive 80% of the BABL of the strategy weighted by their ownership of the garden and how long were they in the garden relative to the duration of the strategy. # Supply Curve Let’s start with the supply curve. The supply curve is designed to optimize the long-term sustainability of the protocol. The rewards are front-loaded, slowly decreasing quarter by quarter. Participation rewards will run for at least 8 years. We are targeting a specific number of rewards per quarter but the protocol may under-allocate or over-allocate depending on the market conditions. Here is how we will calculate the rewards target for a given quarter E: $RewardsQuarter(E)=R_q(E) = RP -RP(1/1.12)^E - RpSpent(E)$ $RP = BABL for Mining Program = 500,000$ $RpSpent(E) = Rewards SpentUntilE$ This creates the following supply curve for the Participation Rewards Program: # Rewards per epoch and per block Given the supply curve, we can calculate how many rewards we can distribute per second of a specific quarter. Rq(E) for a given epoch E. An epoch is three months or a quarter. Then we calculate the max number of rewards to distribute in a second dividing by the number of seconds in the epoch. Rs (Rewards per second) is then calculated as follows: $Rs(E) = Rq(E) /(90 days/Epoch 24 h/day * 60 min/h * 60 secs/min)$ # Strategy Rewards (baseline) It is important to make sure that the incentives of all the participants are aligned. To that end, we only reward participants involved in active Strategies. Active Strategies are the ones that received enough votes to attract capital from the Garden. The rewards of a Strategy are based both on the reserve asset (ETH) allocated (“Principal”) and the number of seconds that the Strategy was active. These rewards will also be based on the amount of principal allocated to other Strategies in the protocol during that same time period. In short, the max rewards of a Strategy can be modeled as follows: $StrategyPower = StrategyPrincipal * StrategyDuration$ $ProtocolPower (E)= Σ_i^n (StrategyPrincipal (strategy_i) (ETH) * StrategyDuration (strategy_i) (secs))$ $StrategyRewards(E)(strategy_i)= StrategyPower(strategy_i) / ProtocolPower(E)Rq(E)$ ## Example As an example, if a Strategy "i" received 1ETH and lasted 1 hour: StrategyPower(strategy_i) = 1 6060 = 3600 weight If we also assume that there were other 10 strategies in addition to our strategy "i" with a total of 50 ETH (5 ETH each) during that same time period and that we are in Q1, where rewards are 53,571 BABL: ProtocolPower(E) = 10 (5 ETH* 3600 secs) + (1 ETH * 3600 secs)= 183,6K weight StrategyRewards(Q1)(strategy_i) = 3600 / 1.836 * 10^5 = 0.02 * 53,571 = 1,050.41 BABL When a Strategy duration is longer than a quarter, it will get proportional BABL tokens per quarter. In that case, the strategy "i" will get: $StrategyRewards (strategy_i) = Σ_j^N StrategyRewards(E_j)(strategy_i)$ $Σ_j^N ((StrategyPower(E_j)(strategy_i)/ProtocolPower(E_j))Rq(E_j))$ The Strategy will get the sum of the corresponding rewards from each epoch (each of them will depend on each protocol power and the available number of tokens per quarter/epoch). This calculation provides the strategy rewards "baseline". In the following section we will discuss about the new weights introduced by the decentralized governance. $StrategyRewardsBaseline (strategy_i)$ ## Adjusting based on performance vs. principal As part of the decentralized governance proposal bip#1, and later in bip#7, the community decided to introduce new weights to penalize strategies that are farming BABL without generating wealth. The performance (returns) must have higher impact than the principal (capital allocated) in the final rewards received by each strategy. These are the weights: • Principal Weight (5%): represents the principal weight to apply to the baseline. • Profit Weight (95%): represents the weight of profits to apply to the baseline. Therefore, the profits of a strategy represent the most important factor to determine BABL rewards earned. We want to encourage strategists, stewards an garden creators to create, vote and execute the best performing strategies as possible. Note that the profit can be positive or negative, which means that we can get more BABL than provided by the baseline but negative profit strategies will get less than if they were only based on principal. ## Negative, break-even and positive strategies In order to carry out the goal to reward strategies, BIP-7 divides the strategies into three different buckets based on its profits. Bad strategies: all strategies that are under -20% returns annualized belong to this segment. These strategies will get only 15% of the rewards (85% slashed). Break-even strategies: all strategies which are above -20% but are still under 4% returns annualized belong to this segment. These strategies will get only 60% of the rewards (40% slashed). Profitable strategies: all strategies which are equal or above 4% annualized returns belong to this segment. These strategies will get an additional bonus of 15% rewards i.e they will receive 115%. After BIP-7 there is no max cap for the strategies, so higher returns mean higher BABL without any limit. Each Strategy will get then its BABL rewards as follows: $StrategyRewards(E)(strategy_i)= StrategyRewardsBaseline(E)(strategy_i)(PrincipalWeight + Profit(strategy_i)ProfitWeightBenchmarkRatioForSegment)$ where "BenchmarkRatioForSegment" is 15% for segment 1, 60% for segment 2 and 115% for segment 3. ## Heart strategies In case of the strategy belongs to the Heart of Babylon, it will always get a bonus of 50% (i.e. they will receive 150%). # Rewards per participant At this point, we know how much BABL to allocate to a finalized Strategy. However, we still need to distribute this BABL among all the participants for that Strategy. In addition to BABL rewards (which are based mostly on participation), the protocol will also distribute the Profit Rewards (if any) among the different participants depending on their active role. 5% of the profits are allocated to the treasury of the protocol, the other 95% is allocated to the participants following the garden default or customized distribution. With regard to BABL rewards, LPs get 80%, the Strategist gets 10% and the Voters get 10% of Strategy BABL rewards. To summarize, we outline the profit-sharing below but we will focus on the BABL rewards. You can read more about each participant and the profit split in the litepaper. Now, we are going to cover the reward baseline for each type of participant. ## 🧠 Strategists 10% in BABL Rewards The creator of the Strategy will receive 10% of BABL rewards if the Strategy is successful. In order to align incentives, Strategies that result in losses for the Garden will not result in any BABL rewards for their Strategists. e.g. If 100 BABL rewards are delivered to the Strategy"i" in total, 10 goes to the Strategist in case it is a profitable Strategy. The Strategist will also receive 10% (or the customized % value per each garden, if any) of the Strategy profits (if any) if the Strategy was successful. On the other hand, if the Strategy is not successful, the protocol will slash the Strategist's stake. e.g. If 100 ETH are returned as profits, the Strategist will get 10 ETH. ## 📰 Voters 10% in BABL Rewards All the Voters that upvoted the strategy will receive their proportional share of the rewards based on the number of votes cast by each member. e.g. If 100 BABL tokens are delivered to the Strategy" i" in total; 10 will be split between Voters according to the votes cast by each Voter. Voters will also receive a total performance fee of 5% (or the customized % defined at the garden level, if any) of all the Strategy Profits in ETH. You can see a table with the specifics in the litepaper. e.g. If 100 ETH are returned as profits, Voters will get in total 5 ETH that will be split according to the votes cast by each Voter. Adjustments: as there are multiple cases, we recommend to read the section "Adjusting based on performance and expected returns". Anyway, here we provide a summary of the different adjustments: • if the strategy got no profits at all and all users voted for the strategy, there will be no rewards for stewards as a mechanism to protect the protocol from bad strategies. • If some users voted against its execution and it finally had no profits, the 10% will be distributed only between those who tried to protect the protocol voting against it. • On the other hand, stewards voting against a strategy returning profits above its expected return will get no rewards as stewards, as they were voting against the execution of a good strategy. In this case, the 10% will be distributed among the users who voted for the strategy. ## 🧧 Liquidity Providers 80% in BABL Rewards All Garden LPs that deposit in gardens with active strategies receive 80% of the BABL rewards split based on the amount deposit by each member. The amount depends also on the time the user has been a member of the garden relative to the duration of the strategy e.g. If 100 BABL rewards are delivered to the Strategy" i" in total, 80 goes to LPs, proportional to their ETH deposited during the Strategy execution period. LPs will also receive a total of 80% (or the garden customized value, if any) of the Strategy profits proportional to their ETH deposited during the Strategy execution period. e.g. If 100 ETH are returned as Strategy profits, LPs will get 80 ETH split between them based on the amount they had deposited in the Garden at the moment of strategy execution. Contributor power will consider when contributor deposits took place and the size of each deposit out of all garden deposits. ## 🌱 Gardeners 1.10x Multiplier in BABL Rewards The Gardener is also an active member of the Garden. As the Gardener of the community, they gets a special bonus for all the other activities that they perform as a member. # 🛡️Prophet Bonuses Prophets that are staked in a garden gives the user extra bonuses in the mining rewards of that garden. Prophets offer boosts to the participants based on the bonus type. • 🧧LP Bonus. The LPs get in aggregate 80% of the BABL of the strategy. If you stake a prophet that has a lp bonus, you will receive a bonus on your proportional share of the lp BABL rewards. e.g. If 100 BABL rewards are delivered to the Strategy" i" in total, 80 goes to LPs, proportional to their assets deposited during the Strategy execution period. If you have 10% of the assets and a 3% prophet staked bonus, you would receive 8.24 BABL instead of 8. • 📰 Voter Bonus. The voters get in aggregate 10% of the BABL of the strategy. If you stake a prophet that has a voter bonus, you will receive a bonus on your proportional share of the voting BABL rewards. e.g. If 100 BABL tokens are delivered to the Strategy" i" in total; 10 will be split between Voters according to the votes cast by each Voter. If you have 20% of the voting power and a 2% prophet staked bonus, you would receive 2.04 BABL instead of 2. • 🧠Strategist Bonus. If you stake a prophet that has a strategist bonus, it will be added to the 10% mentioned above. e.g. if your great prophet gives you a 1% bonus, your strategist BABL rewards will be 11% of the strategy instead of 10%. • 🌱 Gardener Bonus. If you stake a prophet that has a gardener bonus, it will be added to the 1.10x multiplier above. e.g. If your great prophet gives you a 3% bonus, your multiplier will be 1.13x instead of 1.10. ## Summary The total sum of BABL rewards will then be divided into the different profiles: $StrategyRewards(strategy_i)= StrategyRewardsStrategists(strategy_i) + StrategyRewardsVoters(strategy_i) + StrategyRewardsLPs(strategy_i)$ where: $StrategyRewardsStrategist (strategy_i)= StrategyRewards(strategy_i) * StrategistWeight$ $StrategyRewardsVoters(strategy_i) = StrategyRewards (strategy_i) * VotersWeight$ $StrategyRewardsLPs(strategy_i) = StrategyRewards(strategy_i) * LPsWeight$ Note: the Gardener will get the 1.10x multiplier on all the other roles taken in this Garden. Strategist and Voters will be able to get additional tokens (limited by a 2x cap) depending on the expected return as described below. # Adjusting based on performance and expected returns The BABL rewards explained above are refined based on performance to ensure that Babylon rewards Strategies that return capital and respect the opportunity cost associated with executing them. ## Expected return The "expected return" concept is a key element in Babylon. It is set by the Strategist based on market conditions, knowledge, and experience. Voters review the Strategy details, including the expected return, set by the Strategist to decide whether or not it is a good use of the Garden's capital. Voters need to stake their Garden Tokens to vote so Voters need to pick the Strategies they want to vote on wisely. The protocol tries to ensure that the dominant tactic for each participant is aligned with the goals of Babylon. • Strategies are penalized if they inflate their expected returns. Their BABL rewards are tuned based on the accuracy of their prediction. • Voters are disincentivized from upvoting or downvoting without enough information. They are locking their BABL tokens and will only get rewarded when they provide an accurate signal. ## Negative Strategies Strategies that result in losses will only provide BABL rewards to Voters that downvoted the strategy. The Strategist and the Voters that upvoted it do not receive any rewards. The Strategist's stake gets also slashed with a quadratic penalty of -1.75x the % of losses. e.g. If a Strategy loses 5% of capital (capital returned / capital allocated = 95%), the Strategist will get a quadratic penalty of 1.75x 5% = 8.75% of the Strategist stake will then be burned by the protocol. ## Real Return vs Expected Return Adjustments to the baseline model are done based on the distance between the expected return and the real return. ### 🧠 Strategist Bonus Strategist receives a bonus (capped to a max of 2x) of BABL rewards based on the accuracy of their prediction. The more accurate it is, the more BABL rewards the Strategist receives. On the other hand, a negative strategy (with negative profits) will not provide BABL rewards to Strategists and the Strategist's stake will be slashed as defined above. ### 📰 Voter Bonus Voters voting in favor of a profitable Strategy will get proportional BABL rewards depending on their stake (vote) out of the total positive votes. Users voting in favor of bad strategies with negative results will get 0 BABL rewards as they contributed to draining the capital of the Garden. Voters voting against a Strategy will get BABL rewards only if the returns are below the expected return. The bonus has a max cap of 2x. It will also be proportional to the stake used for the vote out of the total negative votes.	2022-06-30 10:55:43	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 15, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5037379860877991, "perplexity": 2907.6396832411033}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103671290.43/warc/CC-MAIN-20220630092604-20220630122604-00645.warc.gz"}
https://tex.stackexchange.com/questions/594196/node-contents-when-placing-a-node-on-path	# Node contents when placing a node on path In TikZ one can specify a node by a command like \path (0,0) node[node contents=A]; This is very useful if one wishes to apply a command to the text in the node, say, to avoid writing \node at (0,0) {\contour{red}{A}}; and simply set \path (0,0) node[node contents={\contour{red}{A}}]; (which comes very handy when used repeatedly!) However, there are some restrictions in the syntax, because the parsing of the command stops after node contents is found. In particular, this seems to prevent one from placing such a node on a path by means of decorations.markings library. Below is the minimal (non-)working example: \documentclass{standalone} \usepackage{tikz} \usetikzlibrary{decorations.markings} \begin{document} \begin{tikzpicture} \path (.5,0) node[node contents={A}]; \draw[postaction=decorate,decoration={markings,mark=at position 0.5 with \node {B};}] (0,-1) to (1,-1); \draw[postaction=decorate,decoration={markings,mark=at position 0.5 with \node[node contents=C];}] (0,-2) to (1,-2); \end{tikzpicture} \end{document} The last command raises an error "Paragraph ended before \tikz@fig@scan@options was complete. ...node[node contents=C];}] (0,-2) to (1,-2);" Is there a workaround? While I agree that node contents has flaws, in this case you are just lacking braces. In general you can often use execute at begin node to add something to the node contents, this key has, according to what I find, less flaws. \documentclass{standalone} \usepackage{tikz} \usetikzlibrary{decorations.markings} \begin{document} \begin{tikzpicture} \path (.5,0) node[node contents={A}]; \draw[postaction=decorate,decoration={markings,mark=at position 0.5 with \node {B};}] (0,-1) to (1,-1); \draw[postaction=decorate,decoration={markings,mark=at position 0.5 with {\node[node contents=C];}}] (0,-2) to (1,-2); \draw[postaction=decorate,decoration={markings,mark=at position 0.5 with {\node[execute at begin node=D]{};}}] (0,-3) to (1,-3); \end{tikzpicture} \end{document} Just put the code between braces. with {\node {B};} \documentclass{standalone} \usepackage{tikz} \usetikzlibrary{decorations.markings} \begin{document} \begin{tikzpicture} \path (.5,0) node[node contents={A}]; \draw[postaction=decorate,decoration={markings,mark=at position 0.5 with {\node {B};}}] (0,-1) to (1,-1); \draw[postaction=decorate,decoration={markings,mark=at position 0.5 with {\node[node contents=C];}}] (0,-2) to (1,-2); \end{tikzpicture} \end{document}	2021-10-16 18:54:14	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8825374841690063, "perplexity": 8117.120838568817}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323584913.24/warc/CC-MAIN-20211016170013-20211016200013-00465.warc.gz"}
https://mathematica.stackexchange.com/questions/152625/integral-involving-besselj-functions	# Integral involving BesselJ functions In this paper, page 252, there is an equality (D.69), the numerical results integrals on the left is of interest: $$\int_{-\infty}^{\infty}\dfrac{x\cdot J_0(x\cdot r)}{x^2-k^2}dx=\left\{ \begin{array}{cc} \dfrac{i \cdot\pi \cdot H_0^{(1)}(k\cdot r)}{2}&\text{Im}(k)>0 \\ \\ \dfrac{i \cdot\pi \cdot H_0^{(2)}(k\cdot r)}{2} &\text{Im}(k)<0\\ \end{array} \right. \tag{D.69}$$ Where $J_0$ is the BesselJ, $H_0$'s are the Hankel functions respectively. I don't know how to prove the integral equality or whether it holds only for the non-real $k$'s . But I am interested in the numerical (or analytical when possible) result when $k$ is a real positive number. But how to obtain a numerical solution of such integrals? e.g., I tried to let $k=10$ and $r=6$, but found most of the available integral algorithms in NIntegrate didnot lead to easy convergence probably due to the existence of a discontinuity point $x=k$ of the integrand and the improper upper limit. e.g.. NIntegrate[(x BesselJ[0, 6 x])/(x^2 - 100), {x, 0, Infinity}, Method -> {"OscillatorySelection", "TermwiseOscillatory" -> True, Method -> {"GlobalAdaptive", "MaxErrorIncreases" -> 10000}}, MaxRecursion -> 100, PrecisionGoal -> 8, AccuracyGoal -> 5] gives 0.155652 and warning messages: NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small. NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in x near {x} = {10.0009}. NIntegrate obtained 0.16591870616051807and 0.2705956661691797 for the integral and error estimates. • The integration doesn't converge: Integrate[(x BesselJ[0, 6 x])/(x^2 - 100), {x, -Infinity, Infinity}] Jul 30, 2017 at 5:22 Integrate[(x BesselJ[0, r x])/(x^2 - k^2), {x, 0, Infinity}, • @J.M. OP's actual code shows the integral from zero to Infinity, so I presume that is what he meant. Furthermore, the integrand is antisymmetric, so the integral from -Infinity to Infinity is zero. Jul 30, 2017 at 6:37 • Indeed, so we'll have to wait for OP to clarify. The only problem now is $k=0$, where the integral is genuinely divergent. Jul 30, 2017 at 6:40	2023-03-28 08:21:55	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9154738783836365, "perplexity": 1341.2118757191959}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948817.15/warc/CC-MAIN-20230328073515-20230328103515-00209.warc.gz"}
https://docs.opengosim.com/reference/linear_solver/	# LINEAR_SOLVER¶ Specifies linear solver parameters associated with the simulation. A typical general purpose example is as follows: LINEAR_SOLVER FLOW KSP_TYPE FGMRES PC_TYPE CPR END An unlikely example demonstrating all possible options is: LINEAR_SOLVER FLOW KSP_TYPE FGMRES PC_TYPE CPR ATOL 1.d-6 RTOL 1.d-6 DTOL 1.d-6 MAXIT 50 GMRES_RESTART 200 GMRES_MODIFIED_GS CPR_OPTIONS CPR_MANUAL_AMG_CONFIG CPR_AMG_REPORT END END The meanings of the keywords above are given in the following sections. The preconditioner CPR is the implementation of the compressed pressure residual preconditioner that OGS has developed. This is a two stage preconditioner which uses AMG (algebraic multigrid) as the first stage; currently block Jacobi is the only available second stage. CPR is a very effective modern preconditioner and should provide a significant increase in efficiency and decrease in linear iterations in almost all nontrivial situations. In addition to the options described above there is a sub-card CPR_OPTIONS which collects options specifically for the CPR preconditioner. These are listed in Advanced AMG/CPR configuration section. ## LINEAR_SOLVER Type¶ The word after LINEAR_SOLVER at the start of this card can be either FLOW or TRANSPORT. For TOIL_IMS, GAS_WATER, BLACK_OIL,TODD_LONGSTAFF and SOLVENT_TL, this must be FLOW. ## KSP_TYPE¶ Specifies solver type, where options include: DIRECT, ITERATIVE, GMRES, BCGS, IBCGS. DIRECT uses LU and ITERATIVE employs Bi-CGStab (BCGS) and block Jacobi preconditioning with ILU[0] in each block. ## PC_TYPE¶ Specifies preconditioner type, where options include: NONE, ILU, LU, BJACOBI, ADDITIVE_SCHWARZ or ASM, HYPRE, or CPR. CPR is the compressed pressure residual preconditioner which OGS has implemented for PFLOTRAN. This is the AMG linear solver option, more on the CPR and its options are discussed above. ## ATOL¶ absolute tolerance. Absolute size of 2-norm of residual. ## RTOL¶ Relative tolerance. Relative decrease in size of 2-norm of residual. ## DTOL¶ Divergence tolerance. Relative increase in residual norm. ## MAXIT¶ Maximum number of linear solver iterations before reporting failed convergence. ## GMRES_RESTART¶ How many linear iterations to allow before restarting the Krylov subspace, if using GMRES or FGMRES as the linear solver. ## GMRES_MODIFIED_GS¶ Use modified instead of classical Gram-Schmidt orthogonalization in the construction of the Krylov subspace if using GMRES or FGMRES as the linear solver.	2021-05-18 06:52:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4430030286312103, "perplexity": 10238.176590071469}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989756.81/warc/CC-MAIN-20210518063944-20210518093944-00416.warc.gz"}
https://socratic.org/questions/how-do-you-solve-2-5n-5-2-75	# How do you solve 2^(5n-5)+2=75? Sep 12, 2016 ${2}^{5 n - 5} = 73$ $\log \left({2}^{5 n - 5}\right) = \log 73$ $\left(5 n - 5\right) \log 2 = \log 73$ $5 n \log 2 - 5 \log 2 = \log 73$ $5 n \log 2 = \log 73 + \log 32$ $n = \log \frac{2336}{\log} 32$ The decimal approximation will be $n \cong 2.24$. Hopefully this helps!	2019-07-19 20:38:14	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4183264672756195, "perplexity": 2723.818222010945}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195526359.16/warc/CC-MAIN-20190719202605-20190719224605-00156.warc.gz"}
https://codereview.stackexchange.com/questions/115869/saturated-signed-addition	So this a exercise from the book COMPUTERS SYSTEMS A PROGRAMMERS PERSPECTIVE I need to add two signed numbers and in case it overflows or underflows return TMAX (011..1b) or TMIN (100..0b) respectively. In this case we can assumed two's complement representation. The book imposes a set of rules that the solution must follow : Forbidden: 1. Conditionals, loops, function calls and macros. 2. Division, modulus and multiplication. 3. Relative comparison operators (<, >, <= and >=). Allowed operations: 1. All bit level and logic operations. 2. Left and right shifts, but only with shift amounts between 0 and w - 1 4. Equality (==) and inequality (!=) tests. 5. Casting between int and unsigned. My code int saturating_add(int x , int y) { int sum = x + y; int w = (sizeof(int) << 3) -1; int mask = (~(x ^ y) & (y ^ sum) & (x ^ sum)) >> w; int max_min = (1 << w) ^ (sum >> w); } Compiled code in my machine Note: I used the following command -> gcc -O2 -S sat_add.c. leal (%rdi,%rsi), %edx movl %edx, %eax movl %edx, %ecx xorl %esi, %eax xorl %edi, %esi xorl %edx, %edi notl %esi sarl $31, %ecx andl %esi, %eax addl$-2147483648, %ecx andl %edi, %eax sarl \$31, %eax movl %eax, %esi andl %ecx, %eax notl %esi andl %edx, %esi leal (%rsi,%rax), %eax ret So I want to know if there is a better solution in terms of elegance and performance. Also if there is a solution that compiles to a single instruction in x86_64 (Maybe PADDS although this instruction may not be the one I am looking for). Also any other kind of feedback is welcome. • The line int sum = x + y; will overflow (and invoke undefined behavior) if x + y is greater than INT_MAX. But you might be able to salvage something by casting x and y to unsigned int first, and then using unsigned int throughout. Care to try again? :) – Quuxplusone Jan 5 '16 at 5:33 • hmmm yes it will overflow , thats the point when it overflows I need to return INT_MAX. Also forgot to mention that two's complement representation is assumed. So casting to unsinged int will result in the same bit level representation. – MAG Jan 5 '16 at 5:39 • C doesn't assume two's complement, though. If you get to assume two's complement, then the problem becomes pretty much trivial. Are you looking specifically for the best answer in (inline) x86-64 assembly? – Quuxplusone Jan 5 '16 at 5:54 • Yes I know C doesn't assumed two's complement but the exercise stated it and i forgot to mention it (Already edit it). No Im not looking for a inline x86-64 assembly solution. Just wanted to know if there is a more concise way of doing it and If there is a solution that can be compiled to a single instruction in x86-64 with a certain level of optimization (0g, 01, 02.. etc) – MAG Jan 5 '16 at 6:01 ### Undefined behavior from signed overflow Technically, your first line causes undefined behavior: int sum = x + y; It should be written instead as: int sum = (unsigned int) x + y; In C, signed integer overflow is undefined behavior but unsigned integer overflow is not. Your compiler probably will treat the two lines above identically, but to be safe you should use the unsigned add. ### Save a couple of instructions This line here could be optimized: int mask = (~(x ^ y) & (y ^ sum) & (x ^ sum)) >> w; to this: int mask = (~(x ^ y) & (x ^ sum)) >> w; If x and y have the same sign, then you only need to check one of them against sum instead of both of them. This saves 2 assembly instructions when you compile it. • Ah I can't believe I missed that second one, thanks – MAG Jan 5 '16 at 20:53	2020-01-22 03:45:53	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3801582157611847, "perplexity": 3231.528817542194}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250606269.37/warc/CC-MAIN-20200122012204-20200122041204-00296.warc.gz"}
https://www.physicsforums.com/threads/lagrangian-equations-of-motion.539604/	# Lagrangian equations of motion ## Homework Statement Find equations of motion (eom) of a particle moving in a D-dimensional flat space with the following Lagrangian L = (1/2)mv2i - k/ra, r = root(x2i), m,k,a are constants ## The Attempt at a Solution The equations of motion are given by d/dt(∂L/∂vi) - ∂L/∂xi = 0 So, when I work all this out I get ma = ka/xia+1 I have a feeling this isn;t correct though. Am I doing the partials wrong? ## Answers and Replies vela Staff Emeritus Homework Helper Yes, you're taking the partial with respect to xi incorrectly. Try writing out the potential term in terms of the xi's and differentiating. Also why is a multiplying m? How did the exponent of r get over there? Yes, you're taking the partial with respect to xi incorrectly. Try writing out the potential term in terms of the xi's and differentiating. Also why is a multiplying m? How did the exponent of r get over there? U = k/ra = k/root(x2i)a = k/xai = kx-ai Differentiating this with respect to xi gives -akx-a-1 As for the last part; sorry, when I wrote a on the LHS I was referring to the second derivative of xi w.r.t. time vela Staff Emeritus Homework Helper U = k/ra = k/root(x2i)a = k/xai = kx-ai Differentiating this with respect to xi gives -akx-a-1 How are you going from k/root(x2i)a to k/xai? Are you saying, for instance, that $\sqrt{x_1^2+x_2^2} = x_1+x_2$? EDIT: Oh, I see why we're getting different answers. I think there's an implied summation: $$r=\sqrt{x_i^2} = \sqrt{x_i x_i} = \sqrt{\sum_i x_i^2}$$	2021-06-16 18:12:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8646434545516968, "perplexity": 2097.8251004277226}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487625967.33/warc/CC-MAIN-20210616155529-20210616185529-00455.warc.gz"}
https://tex.stackexchange.com/questions/124838/remove-the-ugly-borders-and-set-different-colors-to-hyperlinks-using-elsarticle	# Remove the ugly borders and set different colors to hyperlinks using elsarticle template I am using the elsarticle template to write something. I want to remove the ugly border and use different colors instead. My setting is as follows: \usepackage[colorlinks=true,linkcolor=red,citecolor=green]{hyperref} The results is that ugly borders disappear but the all of linkcolor and citecolor are blue despite of no blue setting. Any suggestion is greatly appreciated. Class elsarticle also sets the colors: \AtBeginDocument{\@ifpackageloaded{hyperref} \def\@anchorcolor{blue} \def\@citecolor{blue} \def\@filecolor{blue} \def\@urlcolor{blue} \def\@pagecolor{blue} Therefore your colors should be set at a later time: \documentclass{elsarticle} • @user22986: The "first component" is just the code snipset of elsarticle.cls, proving that the class is responsible for the blue color and shows the time, when the colors are set (\AtBeginDocument). – Heiko Oberdiek Jul 20 '13 at 21:57	2019-10-13 22:21:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4611947238445282, "perplexity": 3028.7598382531514}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986648343.8/warc/CC-MAIN-20191013221144-20191014004144-00319.warc.gz"}
https://www.math.uni-potsdam.de/institut/veranstaltungen/archiv/details-archiv/veranstaltungsdetails/continuum-limits-of-tree-valued-stochastic-processes	Continuum limits of tree-valued stochastic processes 15.07.2019, 12:00 – Golm Haus 9 Raum 0.13 Forschungsseminar Wahrscheinlichkeitstheorie Wolfgang Löhr (Universität Duisburg-Essen) Markov chains on sets of (finite, graph-theoretic) trees arise in applications, e.g., as evolving genealogical trees in population models or in Markov chain Monte Carlo methods for the reconstruction of phylogenetic trees. In order to find useful limit processes as the trees become large, we need an appropriate state space including continuum trees''. I will compare the more classical approach with metric measure trees and a new approach with algebraic measure trees. A particular example will be the Aldous chain on cladograms which converges in the new state space to an ergodic diffusion. (Joint work with Leonid Mytnik and Anita Winter) zu den Veranstaltungen	2023-01-28 01:03:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.43618014454841614, "perplexity": 3887.6950946078305}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499468.22/warc/CC-MAIN-20230127231443-20230128021443-00132.warc.gz"}
http://openstudy.com/updates/4f1a4d0ee4b04992dd22a62e	## kcbrosell Group Title find an equation for the line with the given properties Question: Perpendicular to the line x − 2y = −5; containing the point (0, 4) 2 years ago 2 years ago 1. myininaya Group Title so put that equation in this form: y=mx+b 2. myininaya Group Title so we can find the slope of the perpendicular line after we do that 3. myininaya Group Title Do you need help putting in that form? 4. MrYantho Group Title 1. Convert equation to y=mx+b, to find slope 2. Our second slope is the negative reciprocal of the given line's slope 3. Substitute the new slope, and the values of x and y into y=mx+b to solve for b (the y-int) 5. kcbrosell Group Title I am not understanding 6. myininaya Group Title Do you know how to write the equation given in this form: y=mx+b? 7. kcbrosell Group Title no i do not 8. myininaya Group Title You are just trying to solve for y: $x-2y=-5$ $\text{ add } 2y \text{ on both sides }$ $x-2y+2y=-5+2y$ $x=-5+2y$ Now add 5 on both sides $x+5=2y$ Now divide both sides by 2 $\frac{x+5}{2}=y =>y=\frac{x}{2}+\frac{5}{2} =>\frac{1}{2}x+\frac{5}{2}$ so what is the slope here? 9. kcbrosell Group Title 1/5=5 & 2/2=1, is that right 10. myininaya Group Title y=mx+b where is m is slope so what is in front of x up there? 1/2 right? so what is the slope of a perpendicular line to this one? in other words what satisfies this equation 1/2 * ? = -1 11. kcbrosell Group Title -1/2 12. myininaya Group Title close -2 would be the slope of the perpendicular line 13. myininaya Group Title so we have all lines are in this form y=mx+b => this line has this form: y=-2x+b 14. myininaya Group Title now we know a point on the line (0,4) 15. myininaya Group Title where x=0 and y=4 so we have enough info to find b replace x with 0 and y with 4 4=-2(0)+b how do you solve this for b? 16. kcbrosell Group Title wouldnt you multiply each side by -4 17. myininaya Group Title 4=0+b since -2(0)=0 4=b since 0+b=b 18. myininaya Group Title so the equation us y=-2x+4	2014-09-01 08:15:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6782571077346802, "perplexity": 2163.4668267467155}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1409535917463.1/warc/CC-MAIN-20140901014517-00279-ip-10-180-136-8.ec2.internal.warc.gz"}
http://mathoverflow.net/questions/111470/a-degree-of-an-arbitrary-polynomial-knot	# A Degree of an Arbitrary Polynomial Knot Here a degree of a polynomial knot is a minimal degree which can define a long knot. I would like to find out how this degree can be bounded below, according to the number of crossing points, for instance, or maybe some other properties. Do you know some articles or literature concerning to this problem? All materials, I found, are about particular examples of polynomial reprsentations of some knots, or just says that this is the interesting problem. After the first answer given, I need to add some information. I know the bound given in the answer, this is natural. I would like to come up with more precise bounds (both lower and upper ones) and, ideally, to derive a formula which gives a degree according to a given knot. This problem seems to be difficult, that's why, first of all, I try to construct more narrow bounds for a degree and I would like to know what research has been done already. If you know any materials concerning to that, write it, please. Thank you. - Suppose one has a polynomial long knot given as the image of the function $P(t)=(p_1(t),p_2(t),p_3(t)), t\in \mathbb{R}$, where $degree(p_i)\leq d, 1\leq i\leq 3$. Then the crossing number will be bounded by the number of double points of the projection $(p_1,p_2):\mathbb{R}\to \mathbb{R}^2$, assuming this is generic. This projection is a plane curve of degree bounded by $d$ (the degree of a plane curve is the number of intersections with a generic line $ax+by=c$, and is bounded by $d$ since the number of solutions of $ap_1(t)+bp_2(t)=c$ will be bounded by $d$ by the fundamental theorem of algebra). By Bezout's theorem, the number of double points will be bounded by $d^2$. So one obtains a bound on the crossing number $c(K)\leq d^2$, or $d\geq \sqrt{c(K)}$ since you requested a lower bound on $d$.	2014-03-10 06:27:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8382748961448669, "perplexity": 116.21577856525388}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394010672371/warc/CC-MAIN-20140305091112-00075-ip-10-183-142-35.ec2.internal.warc.gz"}
https://www.gap-system.org/Manuals/pkg/homalg-2017.10.26/doc/chap6.html	Goto Chapter: Top 1 2 3 4 5 6 7 8 9 10 11 12 A B C D E F Bib Ind ### 6 Complexes #### 6.1 Complexes: Category and Representations ##### 6.1-1 IsHomalgComplex ‣ IsHomalgComplex( C ) ( category ) Returns: true or false The GAP category of homalg (co)complexes. (It is a subcategory of the GAP category IsHomalgObject.) ##### 6.1-2 IsComplexOfFinitelyPresentedObjectsRep ‣ IsComplexOfFinitelyPresentedObjectsRep( C ) ( representation ) Returns: true or false The GAP representation of complexes of finitley presented homalg objects. (It is a representation of the GAP category IsHomalgComplex (6.1-1), which is a subrepresentation of the GAP representation IsFinitelyPresentedObjectRep.) ##### 6.1-3 IsCocomplexOfFinitelyPresentedObjectsRep ‣ IsCocomplexOfFinitelyPresentedObjectsRep( C ) ( representation ) Returns: true or false The GAP representation of cocomplexes of finitley presented homalg objects. (It is a representation of the GAP category IsHomalgComplex (6.1-1), which is a subrepresentation of the GAP representation IsFinitelyPresentedObjectRep.) #### 6.2 Complexes: Constructors ##### 6.2-1 HomalgComplex ‣ HomalgComplex( M[, d] ) ( function ) ‣ HomalgComplex( phi[, d] ) ( function ) ‣ HomalgComplex( C[, d] ) ( function ) ‣ HomalgComplex( cm[, d] ) ( function ) Returns: a homalg complex The first syntax creates a complex (i.e. chain complex) with the single homalg object M at (homological) degree d. The second syntax creates a complex with the single homalg morphism phi, its source placed at (homological) degree d (and its target at d-1). The third syntax creates a complex (i.e. chain complex) with the single homalg (co)complex C at (homological) degree d. The fourth syntax creates a complex with the single homalg (co)chain morphism cm (--> HomalgChainMorphism (7.2-1)), its source placed at (homological) degree d (and its target at d-1). If d is not provided it defaults to zero in all cases. To add a morphism (resp. (co)chain morphism) to a complex use Add (6.5-1). gap> ZZ := HomalgRingOfIntegers( ); Z gap> M := HomalgMatrix( "[ 2, 3, 4, 5, 6, 7 ]", 2, 3, ZZ ); <A 2 x 3 matrix over an internal ring> gap> M := LeftPresentation( M ); <A non-torsion left module presented by 2 relations for 3 generators> gap> N := HomalgMatrix( "[ 2, 3, 4, 5, 6, 7, 8, 9 ]", 2, 4, ZZ ); <A 2 x 4 matrix over an internal ring> gap> N := LeftPresentation( N ); <A non-torsion left module presented by 2 relations for 4 generators> gap> mat := HomalgMatrix( "[ \ > 0, 3, 6, 9, \ > 0, 2, 4, 6, \ > 0, 3, 6, 9 \ > ]", 3, 4, ZZ ); <A 3 x 4 matrix over an internal ring> gap> phi := HomalgMap( mat, M, N ); <A "homomorphism" of left modules> gap> IsMorphism( phi ); true gap> phi; <A homomorphism of left modules> The first possibility: <A homomorphism of left modules> gap> C := HomalgComplex( N ); <A non-zero graded homology object consisting of a single left module at degre\ e 0> gap> C; <A complex containing a single morphism of left modules at degrees [ 0 .. 1 ]> The second possibility: gap> C := HomalgComplex( phi ); <A non-zero acyclic complex containing a single morphism of left modules at de\ grees [ 0 .. 1 ]> ##### 6.2-2 HomalgCocomplex ‣ HomalgCocomplex( M[, d] ) ( function ) ‣ HomalgCocomplex( phi[, d] ) ( function ) ‣ HomalgCocomplex( C[, d] ) ( function ) ‣ HomalgCocomplex( cm[, d] ) ( function ) Returns: a homalg complex The first syntax creates a cocomplex (i.e. cochain complex) with the single homalg object M at (cohomological) degree d. The second syntax creates a cocomplex with the single homalg morphism phi, its source placed at (cohomological) degree d (and its target at d+1). The third syntax creates a cocomplex (i.e. cochain complex) with the single homalg cocomplex C at (cohomological) degree d. The fourth syntax creates a cocomplex with the single homalg (co)chain morphism cm (--> HomalgChainMorphism (7.2-1)), its source placed at (cohomological) degree d (and its target at d+1). If d is not provided it defaults to zero in all cases. To add a morphism (resp. (co)chain morphism) to a cocomplex use Add (6.5-1). gap> ZZ := HomalgRingOfIntegers( ); Z gap> M := HomalgMatrix( "[ 2, 3, 4, 5, 6, 7 ]", 2, 3, ZZ ); <A 2 x 3 matrix over an internal ring> gap> M := RightPresentation( Involution( M ) ); <A non-torsion right module on 3 generators satisfying 2 relations> gap> N := HomalgMatrix( "[ 2, 3, 4, 5, 6, 7, 8, 9 ]", 2, 4, ZZ ); <A 2 x 4 matrix over an internal ring> gap> N := RightPresentation( Involution( N ) ); <A non-torsion right module on 4 generators satisfying 2 relations> gap> mat := HomalgMatrix( "[ \ > 0, 3, 6, 9, \ > 0, 2, 4, 6, \ > 0, 3, 6, 9 \ > ]", 3, 4, ZZ ); <A 3 x 4 matrix over an internal ring> gap> phi := HomalgMap( Involution( mat ), M, N ); <A "homomorphism" of right modules> gap> IsMorphism( phi ); true gap> phi; <A homomorphism of right modules> The first possibility: <A homomorphism of right modules> gap> C := HomalgCocomplex( M ); <A non-zero graded cohomology object consisting of a single right module at de\ gree 0> gap> C; <A cocomplex containing a single morphism of right modules at degrees [ 0 .. 1 ]> The second possibility: gap> C := HomalgCocomplex( phi ); <A non-zero acyclic cocomplex containing a single morphism of right modules at\ degrees [ 0 .. 1 ]> #### 6.3 Complexes: Properties ##### 6.3-1 IsSequence ‣ IsSequence( C ) ( property ) Returns: true or false Check if all maps in C are well-defined. ##### 6.3-2 IsComplex ‣ IsComplex( C ) ( property ) Returns: true or false Check if C is complex. ##### 6.3-3 IsAcyclic ‣ IsAcyclic( C ) ( property ) Returns: true or false Check if the homalg complex C is acyclic, i.e. exact except at its boundaries. ##### 6.3-4 IsRightAcyclic ‣ IsRightAcyclic( C ) ( property ) Returns: true or false Check if the homalg complex C is acyclic, i.e. exact except at its left boundary. ##### 6.3-5 IsLeftAcyclic ‣ IsLeftAcyclic( C ) ( property ) Returns: true or false Check if the homalg complex C is acyclic, i.e. exact except at its right boundary. ‣ IsGradedObject( C ) ( property ) Returns: true or false Check if the homalg complex C is a graded object, i.e. if all maps between the objects in C vanish. ##### 6.3-7 IsExactSequence ‣ IsExactSequence( C ) ( property ) Returns: true or false Check if the homalg complex C is exact. ##### 6.3-8 IsShortExactSequence ‣ IsShortExactSequence( C ) ( property ) Returns: true or false Check if the homalg complex C is a short exact sequence. ##### 6.3-9 IsSplitShortExactSequence ‣ IsSplitShortExactSequence( C ) ( property ) Returns: true or false Check if the homalg complex C is a split short exact sequence. ##### 6.3-10 IsTriangle ‣ IsTriangle( C ) ( property ) Returns: true or false Set to true if the homalg complex C is a triangle. ##### 6.3-11 IsExactTriangle ‣ IsExactTriangle( C ) ( property ) Returns: true or false Check if the homalg complex C is an exact triangle. #### 6.4 Complexes: Attributes ##### 6.4-1 BettiTable ‣ BettiTable( C ) ( attribute ) Returns: a homalg diagram The Betti diagram of the homalg complex C of graded modules. ##### 6.4-2 FiltrationByShortExactSequence ‣ FiltrationByShortExactSequence( C ) ( attribute ) Returns: a homalg diagram The filtration induced by the short exact sequence C on its middle object. #### 6.5 Complexes: Operations and Functions ‣ Add( C, phi ) ( operation ) ‣ Add( C, mat ) ( operation ) Returns: a homalg complex In the first syntax the morphism phi is added to the (co)chain complex C (--> 6.2) as the new highest degree morphism and the altered argument C is returned. In case C is a chain complex, the highest degree object in C and the target of phi must be identical. In case C is a cochain complex, the highest degree object in C and the source of phi must be identical. In the second syntax the matrix mat is interpreted as the matrix of the new highest degree morphism psi, created according to the following rules: In case C is a chain complex, the highest degree left (resp. right) object C_d in C is declared as the target of psi, while its source is taken to be a free left (resp. right) object of rank equal to NrRows(mat) (resp. NrColumns(mat)). For this NrColumns(mat) (resp. NrRows(mat)) must coincide with the NrGenerators(C_d). In case C is a cochain complex, the highest degree left (resp. right) object C^d in C is declared as the source of psi, while its target is taken to be a free left (resp. right) object of rank equal to NrColumns(mat) (resp. NrRows(mat)). For this NrRows(mat) (resp. Columns(mat)) must coincide with the NrGenerators(C^d). gap> ZZ := HomalgRingOfIntegers( ); Z gap> mat := HomalgMatrix( "[ 0, 1, 0, 0 ]", 2, 2, ZZ ); <A 2 x 2 matrix over an internal ring> gap> phi := HomalgMap( mat ); <A homomorphism of left modules> gap> C := HomalgComplex( phi ); <A non-zero acyclic complex containing a single morphism of left modules at de\ grees [ 0 .. 1 ]> gap> C; <A sequence containing 2 morphisms of left modules at degrees [ 0 .. 2 ]> gap> Display( C ); ------------------------- at homology degree: 2 Z^(1 x 2) ------------------------- [ [ 0, 1 ], [ 0, 0 ] ] the map is currently represented by the above 2 x 2 matrix ------------v------------ at homology degree: 1 Z^(1 x 2) ------------------------- [ [ 0, 1 ], [ 0, 0 ] ] the map is currently represented by the above 2 x 2 matrix ------------v------------ at homology degree: 0 Z^(1 x 2) ------------------------- gap> IsComplex( C ); true gap> IsAcyclic( C ); true gap> IsExactSequence( C ); false gap> C; <A non-zero acyclic complex containing 2 morphisms of left modules at degrees [ 0 .. 2 ]> ##### 6.5-2 ByASmallerPresentation ‣ ByASmallerPresentation( C ) ( method ) Returns: a homalg complex It invokes ByASmallerPresentation for homalg (static) objects. InstallMethod( ByASmallerPresentation, "for homalg complexes", [ IsHomalgComplex ], function( C ) List( ObjectsOfComplex( C ), ByASmallerPresentation ); if Length( ObjectDegreesOfComplex( C ) ) > 1 then List( MorphismsOfComplex( C ), DecideZero ); fi; IsZero( C ); return C; end ); This method performs side effects on its argument C and returns it. gap> ZZ := HomalgRingOfIntegers( ); Z gap> M := HomalgMatrix( "[ 2, 3, 4, 5, 6, 7 ]", 2, 3, ZZ ); <A 2 x 3 matrix over an internal ring> gap> M := LeftPresentation( M ); <A non-torsion left module presented by 2 relations for 3 generators> gap> N := HomalgMatrix( "[ 2, 3, 4, 5, 6, 7, 8, 9 ]", 2, 4, ZZ ); <A 2 x 4 matrix over an internal ring> gap> N := LeftPresentation( N ); <A non-torsion left module presented by 2 relations for 4 generators> gap> mat := HomalgMatrix( "[ \ > 0, 3, 6, 9, \ > 0, 2, 4, 6, \ > 0, 3, 6, 9 \ > ]", 3, 4, ZZ ); <A 3 x 4 matrix over an internal ring> gap> phi := HomalgMap( mat, M, N ); <A "homomorphism" of left modules> gap> IsMorphism( phi ); true gap> phi; <A homomorphism of left modules> gap> C := HomalgComplex( phi ); <A non-zero acyclic complex containing a single morphism of left modules at de\ grees [ 0 .. 1 ]> gap> Display( C ); ------------------------- at homology degree: 1 [ [ 2, 3, 4 ], [ 5, 6, 7 ] ] Cokernel of the map Z^(1x2) --> Z^(1x3), currently represented by the above matrix ------------------------- [ [ 0, 3, 6, 9 ], [ 0, 2, 4, 6 ], [ 0, 3, 6, 9 ] ] the map is currently represented by the above 3 x 4 matrix ------------v------------ at homology degree: 0 [ [ 2, 3, 4, 5 ], [ 6, 7, 8, 9 ] ] Cokernel of the map Z^(1x2) --> Z^(1x4), currently represented by the above matrix ------------------------- And now: gap> ByASmallerPresentation( C ); <A non-zero acyclic complex containing a single morphism of left modules at de\ grees [ 0 .. 1 ]> gap> Display( C ); ------------------------- at homology degree: 1 Z/< 3 > + Z^(1 x 1) ------------------------- [ [ 0, 0, 0 ], [ 2, 0, 0 ] ] the map is currently represented by the above 2 x 3 matrix ------------v------------ at homology degree: 0 Z/< 4 > + Z^(1 x 2) ------------------------- Goto Chapter: Top 1 2 3 4 5 6 7 8 9 10 11 12 A B C D E F Bib Ind generated by GAPDoc2HTML	2019-05-23 23:19:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6352237462997437, "perplexity": 9138.68801566644}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232257432.60/warc/CC-MAIN-20190523224154-20190524010154-00190.warc.gz"}
https://math.stackexchange.com/questions/3148643/example-of-a-non-negative-martingale-satisfying-certain-conditions	# Example of a Non-negative Martingale Satisfying Certain Conditions Question The question is to find an example of a non-negative martingale $$(X_n)$$ with $$EX_n=1$$ for all $$n$$ such that $$X_n$$ converges almost surely to a random variable $$X$$ where $$EX\neq 1$$ and $$\text{Var}(X)>0$$. My attempt An example of a martingale that I thought could fit the bill was the product martingale with $$X_n=\prod_{i=1}^n Y_i$$ where $$(Y_{i})$$ are i.i.d non-negative random variables with mean $$1$$ and $$P(Y_i=1)<1$$. Unfortunately $$X_n\to 0$$ a.s and hence the limit is degenerate. Other examples, I tried to cook up (e.g. branching process with one individual) all had degenerate limits. I am having trouble coming up with an example that does not have a degenerate limit. Modifying your example a little, by letting $$Y_0$$ be any non-negative random variable independent of $$\{Y_n\}_{n\ge 1}$$ with $$E(Y_0)=1$$, $$M_0=\frac12Y_0+\frac 12,\ \ M_n=\frac12\left(Y_0+\prod_{i=1}^nY_i\right),\ \ n\ge 1$$ is a martingale converging to $$\frac 12 Y_0$$. ### Hint Convex combinations of martingales are martingale. Your example converges to $$0$$. modify if to get an example converging to something else, mix the two. Let $$X_n$$ be the martingale you described; $$X_n=\prod_{i=1}^n Y_i$$, where $$Y_i\stackrel{\text{iid}}\sim \operatorname{Unif}\{1/2,3/2\}$$. Let $$(X_n')_{n\ge 1}$$ be an independent copy of $$(X_n)_{n\ge 1}$$. Finally, let $$\xi$$ be Bernoulli$$(1/3)$$, independent of the previous variables. Then $$X_n\cdot{\bf 1} ({\xi=1})+(2-X_n')\cdot {\bf 1}(\xi=0)$$ is a martingale whose expectation is always one, and in the limit equals $$0$$ with probability $$1/3$$ and equals $$2$$ with probability $$2/3$$.	2019-05-26 23:25:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 30, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9754514098167419, "perplexity": 148.9299198526763}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232260161.91/warc/CC-MAIN-20190526225545-20190527011545-00048.warc.gz"}
https://en.wikibooks.org/wiki/Space_Transport_and_Engineering_Methods/Physics	# Section 1.1 - Basic Sciences: Physics ## Physics as a Subject Physics is the study of how the Universe behaves in its component parts: matter, energy, forces, motion, space, time, and so on. Our understanding of physics is based on experiment and observation, and ideas developed to explain what we see. There are many ideas, but only one reality. So the Scientific Method uses experiment and observation to determine which ideas best match reality. Ideas are loosely graded in quality; as hypotheses, theories, principles, and laws. These terms are applied based on how firmly and widely they have been tested. In the sciences, no idea is considered final or absolute truth. Rather, they are always subject to revision or replacement when confronted by new observations and experiments. Each new observation increases our degree of confidence for the ideas that it supports, and many ideas have been tested for so long, and in so many ways, that we can rely on them routinely, even in engineering projects whose failure would be catastrophic and expensive. Some ideas in theoretical physics explore aspects of reality that are not easily observed. They require devices like larger and more expensive telescopes or particle accelerators, or nobody has figured out a way to test the idea yet. Theoretical ideas can be developed with nothing more than mathematics, a pencil, and pad of paper. However, most of physics is empirical - based things we can see, test, and use. For example, quantum mechanics is illustrated and tested by the double-slit experiment, and special and general relativity by observing bodies in motion through the solar system. These also have engineering consequences in some applications, such as integrated circuit design and the timing of satellite signals (such as GPS). Conversely, some observations, like rotation curves and gravitational lensing of galaxies, don't yet have a good explanation. We call what causes the observations Dark Matter, but we don't yet know what dark matter is. So physics and the other sciences are unfinished works in progress. Many parts are well understood and settled, but around the edges are parts still being worked on. [Move to page 3 Astronomy: In some cases, it is possible to test something by looking for logical consequences which would not be true in competing scenarios. For example, measurements of the cosmological constant has helped narrow down the theoretical possibilities regarding the origins of the universe by excluding those which predict higher or lower energy levels.] ### Scope of this section In this section we discuss key physical principles that relate to space projects. These are only a subset of physics as a whole. We first look at them in ideal terms for the purpose of understanding them individually. However, realistic design work has to consider less than ideal conditions, such as friction or perturbing additional forces, and also must account for uncertainties in how well we know any measured property. This difference represents the difference between physical laws and practical engineering. Physical principles are usually expressed mathematically as algebraic formulas and geometric relationships, with supporting explanations to provide meaning and context. When known numerical values with proper units are inserted into these formulas, you can solve for an unknown value you wish to know. The ability to calculate unknown values is enormously useful when designing or operating space projects. As noted in the introduction to Part 1, the reader should have an understanding of mathematics if they want to use these formulas themselves. It is not our purpose here to include an entire physics textbook, but rather a summary of the most important relationships that apply to space systems. For more detail on physics in general, you can refer to one of the following sources: ## Units and Coordinates ### Units In order to obtain the correct results from a formula, a consistent set of units and method of measuring physical quantities such as position in those units is necessary. For example, adding two feet to three meters to get five of something does not produce a meaningful result because the units are different. The International System of Units (PDF file), abbreviated from the French to SI, is the preferred system of units for engineering and scientific work. It is also known as the Metric System because the base unit of length is the "Metre" (or meter in English). For historical reasons some values in space systems design are reported in US customary units, but these should be converted to SI values. There are also units of convenience, such as gravities being a multiple of Earth's surface gravity. It is convenient to express acceleration effects on humans in this way, relative to the Earth normal value. It should always be recognized as a convenience, and converted to SI units when doing calculations by it's standard ratio of 1 gravity = 9.80665 m/s2. Physical quantities include both the numerical value and the units, and units must be carried through properly when doing calculations. The base SI units are not defined in terms of other units, but rather by a description of how to measure them from nature. In one case, the kilogram, the base is a physical artifact, but that may soon be replaced by measurements. The base units are currently the Meter for length, Kilogram for mass, Second for time, Ampere for electric current, Kelvin for thermodynamic temperature, mole for amount of substance, and candela for luminous intensity. Efforts are underway to define these base units in terms of constants of nature, but they are not complete. Derived SI units are the products of powers of the base units. For example, the unit of force, called the Newton, is 1 kilogram-meter times seconds-2. Many of the derived units are named after famous physicists, but these named units are identical to the form expressed in base units. Multiples and sub-multiples of units are indicated by prefixes which indicate integer powers of ten ranging from -24 to +24. Among the more common are kilo, indicating 103 or 1000, and milli- indicating 10-3 or 0.001. ### Position In modern physics there is no absolute or preferred reference frame in the universe. Therefore position is measured relative to a starting point known as the Origin, which is given a value of zero. Altitude on Earth is measured relative to sea level in the direction opposite the local gravity direction. On bodies without an ocean to serve as a reference, an average ellipsoid based on the shape of the planet or satellite is defined as zero altitude. On gas giants, which do not have a visible solid or liquid surface, altitude is based on pressure. On near-spherical bodies, Latitude and Longitude are measured relative to the points where the surface meets the axis of rotation, which are called the Poles, and a point assigned values of zero in both coordinates called the Zero Point. Units of Degrees, which are 1/360th of a circle measure the location relative to the zero point. Objects in space are generally moving in relation to each other in paths defined by gravitational forces. When the paths are purely the result of gravity they are called Orbits and are measured by six parameters called Orbital Elements from which you can calculate position at a given time. When absolute position is more useful, it can be measured in three dimensions relative to an origin, such as the center of the Sun, with axes defined relative to the stellar background. Alternately a radial distance and two angles relative to a reference plane can be used. Most often the reference plane is that which contains the Earth's orbit around the Sun, known as the Ecliptic. In a Universe of three physical dimensions, it takes three values to define a position uniquely, either distance in three axes, or a radius and two angles. These values are known as the object's Coordinates. ## Motion Displacement is the change in position. It has both amount and direction, such as "three kilometers North". Velocity is the rate of change in position per unit time. When stated without a direction and purely as an amount it has a single value in units of meters per second. Where x is position in the direction of motion, t is time, and the Greek letter delta (which looks like a triangle) indicates change in those values, then velocity v is given by the following formula: ${\displaystyle {\bar {v}}={\frac {\Delta x}{\Delta t}}}$ Acceleration is the rate of change in velocity, or the second derivative of position, with respect to time. Thus acceleration a is: ${\displaystyle {\bar {a}}={\frac {\Delta v_{x}}{\Delta t}}={\frac {d^{2}x}{dt^{2}}}}$ The horizontal bar over v and a in the respective formulas indicates velocity is directional. A value such as this when stated with both a magnitude and direction is called a Vector, while a value without a direction is called a Scalar. The direction can be given in terms of two angles, or the velocity can be expressed as components in the three (x, y, z) axes of a reference system, but either way a total of three values are required to state a velocity vector. Vector algebra is a method for doing calculations with vectors. It is somewhat different and more complex than simple algebra. In accelerated motion, the velocity at any given instant is changing. We can define an Instantaneous Velocity at a particular time, and an Average Velocity over an interval. In a closed orbit the moving object returns to its starting point. Thus for a full orbit, the net change in position is zero, and as a vector the average velocity is also zero. If you measure the total length of the orbital path and divide by the time one orbit takes, you can obtain an average orbital speed as a positive scalar value. This illustrates how different vector and scalar values can be. Acceleration can also change with time. For example, the accelerating force due to gravity changes as the inverse square of distance. Thus a falling object will increase in acceleration as it gets closer. Under constant acceleration in a straight line we can determine change in position or distance d from: ${\displaystyle d=\Delta x=1/2\times at^{2}}$ In circular motion, where v is the velocity and r is the radius, we can find the acceleration a from the following formula: ${\displaystyle a={\frac {v^{2}}{r}}}$ One use for this formula is finding a required velocity for a circular orbit from the acceleration of gravity (see under Forces below) and the radial distance from the center of the body. Note there is no centrifugal force. Gravity or a rotating structure provide an inwardly directed force to maintain circular motion of the object, but there is no outward one. ## Forces The 20th century theories of General Relativity and Quantum Mechanics are more accurate predictors of how objects behave in the realms of the fast and the small, but in many cases the simpler formulas of Classical Mechanics are sufficiently accurate to use. Examples where classical mechanics is not accurate enough include long term changes in the orbit of Mercury, which being the closest planet to the Sun, moves the fastest, and GPS navigation, which relies on extreme accuracy of the orbits of the satellites, and the effect of gravity on their signals, to determine user position. Isaac Newton formulated many of the basic ideas of classical mechanics in his Principia, published in 1687. These include his three laws of motion, conservation of momentum and angular momentum, and the law of universal gravitation. ### Newton's Three Laws These are laws in the mathematical sense, which Newton deduced from experiments performed by others before him. They involve two opposing concepts: Forces which tend to create motion, and Mass which tends to oppose it via the property of Inertia. The relationship of forces to the motion they create is known as Dynamics. Forces are vectors, having magnitude and direction, and multiple forces act as the vector sum of the component forces. This also means single forces can be decomposed into components, such as vertical and horizontal components relative to an axis system, or perpendicular (normal) and parallel components relative to a surface. Decomposition is done when it is useful in solving a problem. The three laws are: First: Inertia - A body acted on by no net force moves with constant velocity (which may be zero) and zero acceleration: ${\displaystyle \sum {\vec {F}}_{i}=0\Rightarrow a=0}$ This is contrary to common earthly experience where friction acts to stop objects in motion. Objects moving freely in the vacuum of space demonstrate this Law more clearly since it is a frictionless environment. An airplane in level flight has multiple forces acting on it (gravity, lift, thrust, and drag), but if the vector sum of all the forces is zero, it will continue moving at the same altitude and velocity in the same direction. From the fact that buildings typically are not accelerating we can deduce there is no net force acting on them. To put it another way, the sum of the forces is zero. Since gravity acts to pull the building down, there must be an equal force from the bedrock acting to hold it up. Applying this idea to every structural component of the building is a powerful way to determine the necessary design of those components - at each point where components connect, the forces must sum to zero, therefore you can calculate the forces which a particular component must withstand. Second: Force - When a force does act on a body of mass m, the acceleration a is related to the magnitude of the force F by the formula ${\displaystyle {\vec {F}}=m{\vec {a}}}$ The arrows above the symbols indicate they are vectors, meaning a quantity in a particular direction. Thus a force in a given direction produces an acceleration in the same direction. Manipulating this simple formula has very wide ranging use in space systems. Given any two of the values, we can find the third. Summing across time, we can find total velocity change. Since mass has units of kilograms, and acceleration has units of meters per second squared, then by the above formula force has units of kilogram-meter per second squared. We call this unit a Newton (abbreviated "N") in the SI system of units, and is named after the scientist. The force which the Earth exerts on a falling apple is coincidentally, given stories about the scientist and falling apples, about 1 Newton. The force of gravity on an object is referred to as Weight. Most humans live where acceleration of gravity is within 0.2% of the same value, so we often confuse weight with mass. They are proportional, but they have different units. On another planet, the same object would have a different weight. Weight does not disappear when in orbit - aboard the Space Station the force of the Earth's gravity is only 10% less than on the ground. So-called "zero gravity" is more properly described as Free Fall. The astronauts inside the Station and the Station itself are both affected by the same acceleration of gravity. So the difference between them is zero, and the astronauts do not feel their bodies pressed against anything. On Earth what you feel is parts of your body pressed against the ground or furniture, and internally pressing against other parts of your body. This pressure is what you experience as "weight". The product of mass times velocity is called momentum. It is given the symbol p since mass already uses the letter m. Force also equals the change of momentum with respect to time, since acceleration is the change in velocity with respect to time and we are just adding the multiplier of mass: ${\displaystyle {\vec {F}}={\frac {d{\vec {p}}}{dt}}}$. Third: Reaction - Single forces do not act in isolation. At the most fundamental level the particles which carry the four forces of nature act on both the emitter and absorber of the particles. At the macroscopic level we live in, where forces are the combined action of many particles, we observe the dual action as for every force there is an equal and opposite reaction force. Where the subscripts indicate the force of object A on object B and object B on object A: ${\displaystyle {\vec {F}}_{\mathrm {a} b}=-{\vec {F}}_{\mathrm {b} a}}$ Therefore a body can never move itself by applying forces only to itself, because the reaction force would cancel it out. You cannot lift yourself above the ground no matter how hard you try by applying forces to your own body. A pole vaulter, however, can raise their body a considerable distance by applying force to the ground. The reaction force of the ground through the pole then acts to raise their body. Of great interest for space systems is that a rocket engine applies a great deal of force to expel gases in one direction, and the gas applies a reaction force in the opposite direction, which moves the vehicle. Multiplying both sides of the above formula by units of time, and subtracting the left side from the right side, we find that sum of momentum (mass times velocity) changes is always zero. This is known as the Law of Conservation of Momentum. It is referred to as a physical law because it has never been observed to be violated, and conservation in the physics sense means a value which does not change. It is found to be conserved both for linear and rotational motion. The latter is referred to as Angular Momentum. Thus the Earth would continue to rotate forever unless acted on by outside forces. Such forces do in fact act, mainly tidal forces from the Moon. So the Earth's rotation is slowing down measurably - the day is getting longer by 23 microseconds per year. But since angular momentum is conserved, slowing the Earth's rotation means the Moon increases its angular momentum. This increases the size of its orbit by a measurable amount (3.8 cm/year) ### The Forces of Nature There are only four fundamental forces that we know of, responsible for all motion in the Universe. These are the gravitational, electromagnetic, weak nuclear, and strong nuclear forces. These forces interact via gravitons, photons, W and Z bosons, and gluons respectively. For more detail see Fundamental Interaction. The latter two are short range forces which mostly occur within atomic nuclei, so the two that concern space projects the most are gravity and electromagnetism. #### Gravity Gravitons are the hypothetical particles which should carry the gravitational force. They are hypothetical because they have not yet been observed. Because gravitons never decay, their range is infinite, and the gravitational field of any object in the Universe affects every other object in the Universe. As a result, the total gravitational field surrounding an object remains the same at any distance. The area of a sphere surrounding an object is 4 × pi × the radius squared. So the gravitational field per unit area decreases with the square of the radius r. The rates of graviton production and absorption are both proportional to the mass of an object. Between any two objects the total gravitational force depends on the product of the two masses, since the first object emits a number depending on it's mass M, and then the second object absorbs some of them according to it's second mass m. The rate of graviton production and absorption is measured as a universal constant G which applies to every object in the Universe, as far as we know: ${\displaystyle G=6.67\times 10^{-11}Nm^{2}/kg^{2}}$ Gravity always acts to attract two objects to each other, in other words reduce the distance, therefore the force is given a negative value. As a practical matter, since the field falls as the square of distance, objects sufficiently far away can be ignored to the extent you need to accurately calculate the total gravitational force on an object. The force acts on a line between each pair of objects and the total force is simply the sum of the individual forces accounting for the direction of each, and each is found by the formula ${\displaystyle F=-{\frac {GMm}{r^{2}}}}$ Since force is also mass times acceleration, we can equate them and remove mass m from both sides of the equation, giving the acceleration due to gravity of an object with mass M as ${\displaystyle {\vec {a}}=-{\frac {GM}{r^{2}}}}$ When restrained from accelerating, such as when you stand on the surface of the Earth, you experience the downward force as weight. Your mass does not change according to what object you are standing on, but the acceleration does, due to the object's different mass and radius. Therefore your weight will be different on other planets and satellites. When unrestrained from accelerating, also known as free fall, then the parts of your body, for instance if riding in vehicle, all will accelerate at the same rate. They do not have any acceleration relative to each other, sometimes called Zero gravity, but incorrectly since nowhere in the Universe is there truly no gravitational field. So in the case of unrestrained acceleration "free fall" is the more correct term, and "zero apparent gravity" will provide a better indication as to express how it appears to a human relative to their surroundings. The local acceleration at the Earth's surface has a standard value of 9.80665 meters per second squared, but actually varies about 2% depending on location. The standard value is given the symbol g, and accelerations are sometimes stated as multiples of standard Earth gravity to give an impression of how humans would experience it, but for calculation purposes meters/second squared should be used to avoid unit errors. Similarly weight should not be confused with mass. Weight is in reference to the local gravity field, while mass is the more correct unit to use at any location. #### Electromagnetism Photons, the particles which carry the electromagnetic force, behave similarly to gravitons in that they do not decay as they travel, and obey an inverse square field law. Where gravity is the result of mass, electromagnetic force is the result of electric charge. Unlike gravity, charge comes in two types which we call positive and negative. The names are arbitrary, positive charges are not larger or higher than negative ones, but they have the property that like charges repel each other, and unlike charges attract. The electromagnetic force is found by the formula ${\displaystyle F=k_{\mathrm {e} }{\frac {q_{1}q_{2}}{r^{2}}}}$ Where F is the force, k(e) is fixed value called Coulomb's constant, q1 and q2 are the electric charges, and r is the distance between them. Note the form of this equation is similar to the one for gravitational force. When both of the charges are positive, or both negative, their product is positive, and so is the force. Positive forces act to increase the distance between charges. When the charges are unlike, one positive and one negative, the product is negative, and thus the force acts to decrease distance. Coulomb's constant, where C is charge in units of 1 Coulomb = 1 mole of elementary charges is ${\displaystyle k_{\mathrm {e} }=8.987\times 10^{9}\ \mathrm {N\cdot m^{2}/C^{2}} }$ Elementary charges are those on a single electron or proton, and are always observed as integer multiples of those charges, never fractions. Charges are additive by simple arithmetic, with negative charges canceling the fields of positive charges. Since unlike charges attract each other, they tend to annihilate if they are antiparticles or form neutral atoms if they are protons within atomic nuclei and electrons. So large quantities of matter tend to have low net charge. Since mass is always positive, large quantities of matter always have large amounts of gravity. Moving electric charges create a magnetic field, including the imputed spin of the charge from elementary particles. Materials with aligned atomic spins can have a static magnetic field. A steady flow of electric charges is called a Current, and also creates a field. Magnetic fields in turn affect the motion of electric charges creating a force F, where I is the current, l is the length of the wire, and B is the strength of the magnetic field: ${\displaystyle \mathbf {F} =I{\boldsymbol {\ell }}\times \mathbf {B} \,\!}$ The bold face symbols indicate these are vector values, having directions. The force is perpendicular to both the direction of the wire/current and the magnetic field. Natural magnetic fields, such as the Earth's, are assumed to be caused by electric currents within the metal core.	2016-07-27 11:49:55	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 14, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7013684511184692, "perplexity": 309.0061465928298}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257826773.17/warc/CC-MAIN-20160723071026-00296-ip-10-185-27-174.ec2.internal.warc.gz"}
https://socratic.org/questions/how-do-you-find-the-critical-points-for-y-x-2-3-x-2-16	# How do you find the critical points for y = x^(2/3)(x^2-16) ? Mar 27, 2015 For $f \left(x\right) = y = {x}^{\frac{2}{3}} \left({x}^{2} - 16\right)$, the critical points are $0 , 2 , - 2$ Solution $c$ is a critcal point for $f$ if $c$ is in the domain of $f$ and either $f ' \left(c\right) = 0$ or $f ' \left(c\right)$ does not exist. Find $f ' \left(x\right)$ We have choices for finding $f ' \left(x\right)$. We could leave the function as written and use the product rule, or we could distribute the ${x}^{\frac{2}{3}}$ and avoid the product rule. Using the product rule $\left(\text{I use:} \left(F S\right) ' = F ' S + F S '\right)$ $f ' \left(x\right) = \frac{2}{3} {x}^{- \frac{1}{3}} \left({x}^{2} - 16\right) + {x}^{\frac{2}{3}} \left(2 x\right)$ $= \frac{2 \left({x}^{2} - 16\right)}{3 \sqrt[3]{x}} + \frac{2 x \cdot {x}^{\frac{2}{3}}}{1}$ $= \frac{2 \left({x}^{2} - 16\right)}{3 \sqrt[3]{x}} + \frac{6 {x}^{2}}{3 \sqrt[3]{x}} = \frac{8 {x}^{2} - 32}{3 \sqrt[3]{x}}$ .Re-writing $f$ before differentiating $f \left(x\right) = {x}^{\frac{8}{3}} - 16 {x}^{\frac{2}{3}}$ $f ' \left(x\right) = \frac{8}{3} {x}^{\frac{5}{3}} - 16 \cdot \frac{2}{3} {x}^{- \frac{1}{3}} = \frac{8 {x}^{\frac{5}{3}}}{3} - \frac{32}{3 \sqrt[3]{x}}$ $= \frac{8 {x}^{\frac{5}{3}} \cdot {x}^{\frac{1}{3}}}{3 \sqrt[3]{x}} - \frac{32}{3 \sqrt[3]{x}} = \frac{8 {x}^{2} - 32}{3 \sqrt[3]{x}}$ . Using either method, we get $f ' \left(x\right) = \frac{8 {x}^{2} - 32}{3 \sqrt[3]{x}}$ Critical points: $c$ is a critcal point for $f$ if $c$ is in the domain of $f$ and either $f ' \left(c\right) = 0$ or $f ' \left(c\right)$ does not exist. $f \left(x\right) = y = {x}^{\frac{2}{3}} \left({x}^{2} - 16\right)$ For this function, the domain is $\left(- \infty , \infty\right)$ So the critical points for this $f$ will be the zeros of $f '$. and all zeros of the denominator of $f '$ $f ' \left(x\right) = \frac{8 {x}^{2} - 32}{3 \sqrt[3]{x}} = 0$ when $8 \left({x}^{2} - 4\right) = 0$ at $x = \pm 2$, and $f ' \left(x\right)$ fails to exist at $x = 0$ Because these are all in the domain of $f$, these are all critical points for this $f$.	2021-06-13 22:52:03	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 38, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8788031935691833, "perplexity": 192.47988077548234}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487611089.19/warc/CC-MAIN-20210613222907-20210614012907-00291.warc.gz"}
https://www.gamedev.net/forums/topic/350898-2d-stairs/	# 2D Stairs This topic is 4691 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic. ## Recommended Posts Anyone played one of the latter level by level Castlevanias (Castlevania 4, Dracula X)? How could I implement a stair system like that? Ive tried a few things but nothing is getting acceptable results. ##### Share on other sites I assume you mean, like when you press up at the foot of a stair case you go up? I assume the just have a flag in the tile that identifies it as 'the foot of a staircase'. If the user is on it, and presses up, they go up. What have you tried so far? What isn't working? Matt Hughson ##### Share on other sites Well the main problem Ive had so far is not with the starting to climb the stairs but when landing into them after a jump. Ideally, the player would latch onto the stairs at the same spot on the x axis corresponding to the stair they are colliding with if they are pressing up.The problem is that the point of collision will change and hence the coordinates wont always correspond. ##### Share on other sites Not that I think it's an ideal solution, but didn't Castlevania just let you fall right through the stairs when jumping onto them? Maybe they fixed that by the SNES version though, not sure. Matt Hughson ##### Share on other sites I've never played Castlevania so I'm not sure on the specific stair-case you're referring to, so I'll try to explain as best I can. If you're using a 2D engine you could create another sprite set of your character(s) climbing a stair-case, and get those sprites to be used when you hit a certain block set to a "stair-case" type. E.G.: while ( map.block[1] == staircase ){ LoadSprites(player_climb.bmp); UseSprites(player_climb.bmp);} That's a very basic example, assuming you've defined everything where needed. ##### Share on other sites Never mind guys, I dont think I explained myself too well. Nevertheless, I'm pretty sure I thought up a decent solution to my problem. ##### Share on other sites In Castlevania: Symphony of the Night the collision is tested against a slope and the stairs are simply a graphical affectation (you can see this if you use a cheat to enter the test screen). The stair graphic protrudes above the collision slope, obscuring the characters feet, thus hiding the fact that he is not standing on any particular stair. Perhaps if you used a similar system your problem would become less complex. ##### Share on other sites Define a line that indicates the walkable portion of the stairs. Then if the character's feet intersect that line while up is pressed then the character is on the stairs. That should work for both jumping, walking, or falling. 1. 1 2. 2 3. 3 4. 4 frob 15 5. 5 • 16 • 12 • 20 • 12 • 16 • ### Forum Statistics • Total Topics 632155 • Total Posts 3004484 ×	2018-08-18 10:08:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.25128117203712463, "perplexity": 2210.0567984876297}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221213540.33/warc/CC-MAIN-20180818095337-20180818115337-00485.warc.gz"}
https://www.scireslit.com/PublicHealth/AJEPH-ID37.php	Research Article # Latrine Utilization and Factors Associated among Rural Community in Chire Woreda at Sidama National Regional state, Ethiopia: A Community-based Cross-sectional Study ### Demelash Wachamo Wako* and Yadessa Tegene Woldie Department of Public Health, Paradise Valley College, West Arsi Zone, Oromia, Ethiopia *Address for Correspondence: Demelash Wachamo Wako, Department of Public Health, Paradise Valley College, West Arsi Zone, Oromia, Ethiopia, E-mail: demmenew1@gmail.com Submitted: 27 August 2020; Approved: 09 September 2020; Published: 10 September 2020 Citation this article: Wako DW, Woldie YT. Latrine Utilization and Factors Associated among Rural Community in Chire Woreda at Sidama National Regional state, Ethiopia: A Community-based Cross-sectional Study. American J Epidemiol Public Health. 2020;4(3): 090-00. https://dx.doi.org/10.37871/ajeph.id37 Copyright: © 2020 Wako DW, et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited Keywords: Latrine utilization; Rural communities; Chire district; Sidama regional state Background: In low income countries like Ethiopia; lack of sanitation is a serious health risk, affecting billions of people, particularly the poor people. The better health and quality of life, sanitation, and hygiene conditions are among the major causes of public health problems. Therefore, this study aimed to assess utilization and associated factor of latrine in rural communities. Methods: A community-based cross-sectional study was conducted among 793 randomly selected households. The data were collected through an interview by pre-tested questionnaires and observation checklist. Data entry and analysis for descriptive and logistic regression models by SPSS v.20. Results: The utilization of latrine was 71.5%. It associated with being younger age [AOR 1.83, 95% CI:1.11, 3.02], who had attended secondary & above [AOR 4.17, 95% CI:1.10, 15.94], monthly income more than 1000 Ethiopian Birr (ETB), [AOR 5.17, 95% CI: 3.10, 8.72], family size [AOR 3.19, 95% CI: 1.97, 5.17], latrine inside the compound [AOR 2.47, 95% CI: 1.64, 3.71] as compared with their counterparts were more used latrine. Conclusions: The utilization of latrine was low and still needs, to improve awareness of the community there is a need of health education programs regarding improved use of latrine, cleanliness of latrine, constructing latrine inside the compound and maintenance of latrine for the proper utilization including children should be considered. ## Introduction Worldwide lack of sanitation is a serious health risk, affecting billions of people around the world, particularly the poor and disadvantaged people around the world [1]. Lack of sanitation facilities compels people to practice open defecation and this increases the risk of transmission of diseases [2]. The disease burden associated with poor water, sanitation, and hygiene is estimated to account for 4.0% of all deaths and 5.7% of the total disease burden in Disability-Adjusted Life Year (DALYs) in worldwide, principally through diarrheal diseases. About 1.8 million people die every year due to diarrheal diseases, and children under the age of 5 years account for 90% of diarrheal deaths [3,4]. Moreover, 88% of diarrheal diseases are attributed to unsafe water supply, inadequate sanitation, and poor hygiene [5]. In most developing countries, especially in Sub- Saharan Africa (SSA), the basic causes of more than 80% of the diseases are inadequate and unsafe water supply and improper disposal of waste. Ethiopia is among the poorest countries in the world, ranking 170 out of 177 in the UN human development index and is the second-most populous country in Africa (population estimated above 80 million). Yet, Ethiopia’s rural populations are among the least served with a rural water supply and sanitation access at only 24% and 8% respectively [6]. However, the Provision of sanitation facilities initiated in all parts of Ethiopia with interventions of a health extension program and continued investments to increase access to safe water and improved sanitation [7]. The studies conducted in different parts of Ethiopia showed that the latrine utilization level differs from region to region of the country and from district to district within the same region depending on many factors. In the study area, there is no available research conducted on a similar topic. Therefore, this study was designed to assess the latrine utilization level and associated factors of rural community separately in the study area. This study can provide evidence or information regarding the latrine utilization in open defecation and its determinant factors which can be used for public health officials, clinicians, and health planners to reduce the impacts of poor utilization of latrine. ## Materials and Methods ### Study setting and design This study was conducted in Chire District, at Sidama National Regional state, Ethiopia, Southern Ethiopia, which located about 396 km away to the South of Addis Ababa, the capital city of Ethiopia. By the end of 2018, latrine coverage was 99.8% but utilization not known. It has a climatic condition of Dega and Weyna Dega. The total number of households is estimated to be 41355 within the average household size of 4.9 and of these households, about 97% of them have latrine facilities. The source population was all rural community households. While all selected households with latrine facilities in the rural community of districts were considered as the study population. Among this, all randomly selected adults of the household and who are residents in selected kebele were included in the study. All adults who were critically ill that were not able to respond appropriately for the interviews were excluded from the study. ### Sampling and sampling technique The sample size (n) was calculated using the following single population proportion formula based on the assumption of 61% proportion (p) from Dembia district [8], 95% C.I (1.96), 5% margin of error (d), 2 design effect and adding 15% contingency. n = ${\left({Z}_{1-\alpha }{2}}\right)}^{2}\frac{P\left[1-P\right]}{{d}^{2}}$ n = (1.96)2(0.61) (0.39) = 365 Therefore, the required sample size was n = (365 x 2) = 10% (730) +73.0 = 803 HHs with latrines included in the study. A multistage random sampling technique. In the 1st stage, ten out of 27 rural kebeles were selected by Simple Random Sampling (SRS) technique. In the 2nd stage, the sample size was proportionally allocated. In the last stage, only one randomly selected eligible person was interviewed (Figure 1). ### Data collection tools, and procedures The data were collected through an interview by pre-tested questionnaires and observation checklist. The questionnaires were translated into the “Sidamic language” and validated before the study time was done outside of the study area and necessary modifications were done based on the findings. Data on utilization of latrine, a survey was carried out with direct observation check list the status of the households in selected kebele. Trained data collectors, health workers have collected the data. Principal investigators and supervisors follow the data collection process and check them for consistency and completeness. ### Data analysis Data entry, cleaning, and analysis were done by SPSS V. 20. Descriptive analysis including frequency distribution and the percentage was made to determine the latrine utilization, to describe socio-economic and demographic, environmental, behavioral, knowledge and attitude related variables. All factors with p-value < 0.25 in the bivariate logistic regression analysis were a candidate to the multivariable model to control confounding effects. The Hosmer-Lemeshow goodness-of-fit statistic was used to assess whether the necessary assumptions for the application of multiple logistic regression are fulfilled. Odds Ratios (OR) with 95% Confidence Intervals (CI) were calculated. Finally, p-value < 0.05 declared a significant association. ### Operational definitions Satisfactory latrine utilization: is a latrine that provided services at the time of data collection even if the latrine required maintenance. A latrine is utilized when households had functional latrines, no observable faces in the compound, observable feces through the squat hole, and the foot-path to the latrine is uncovered with grasses. Status of latrine: which needs maintenance or not at the time of data collection on it’s a door (any cover), a leaking roof and sagging walls. The critical time for handwashing practice – handwashing practices mainly after visiting latrines or cleaning bottoms of children, before preparing food and before feeding children A Child-friendly feature of latrine facility: means availability of at least one of the following features; small squatting hole, lower seat, and presence of potty. ## Results ### Socio-demographic characteristics A total of 793 participants were interviewed yielding a response rate of 98.75%. The majority of the studied participants were 482 (60.8%) female by sex and 770(97.1%) were married. About 322 (40.6%) participants were in the age range of 26-35 years, while 281 (35.4%) mothers and 288 (36.3%) fathers had completed primary level education, respectively. More than half, 530 (66.8%) of the respondents were housewives. Nearly 693 (87.4%) of the households had under five years old children in the household (Table 2). ### Knowledge and attitude towards of latrine utilization The overall knowledge of latrine utilization was 590 (74.4%) had good knowledge. The majority of 715 (90.2%) knew the benefit of using a latrine. The study participants claim to use latrine because of 77.5% privacy, and 69.5% prevent excreta related diseases. Among study participants, only, 335 (42.2%) had a positive attitude towards latrine utilization. The study participants who agreed that the main obstacles to using latrine were beliefs 55.6% were lack of construction tools, 67.2% believe that it is the responsibility of husband and 33.3% government. ### Environmental and behavioral characteristics The majority of the study participants prefer to locate their latrine 520 (65.6%) were inside the compound. More than half 513 (64.7%) were constructed between 1-3 years ago. While 202 (25.5%) households had feces and urine around the latrine and 89 (11.2%) had still faeces and urine around the home. On another hand, 91.0% had no handwashing facility near the latrine, 91.6% had no water for handwashing. Regarding the behavioral practice, Out of 793 respondents who had latrine facility, 90.0% had simple pit latrine, improved ventilated 2.4% and 7.6% were other types of the latrine (Figure 2). While, out of 229 (28.9%), 32.0% pit, 38.7% floor, and 29.3% whole part of the latrine needs maintenance, respectively. The reasons given by respondents for why under-five children did not use the latrines were: being just a child 258 (37.2%), large squatting hole 226 (32.6%), and the floor was not safe to stand on 209 (30.2%). ### Utilization of latrine Out of 793 respondents, 567 (71.5%) households had satisfactory utilized latrines with [95% CI: 68.3% - 74.4%] (Figure 3). Out of 212, the study participant’s claim as the main reason for not using a latrine, 59.7% reports latrine is not functional, 22.7% due to far from the house and 9.2% claims using latrine is inconvenient during the rainy season and at night without proper roof and door. ### Associated factors of the utilization of latrine In bivariable logistic regression analysis the variables which had statistically significant association and the highest latrine utilization was noticed among; younger age had, who attend primary and above education fathers, attend primary and above for the mother, the estimated average monthly income of the household above 24 USD$per a month, and the households who had latrine inside had p-value < 0.001 and higher proportion of the utilization of the latrine. In addition to this, occupation of mother, had under 5 years children in the household and high number of household size can affect latrine use also had p-value < 0.05 and higher proportion of the utilization of the latrine. In the multivariate analysis age of the respondent, the educational status of the mother, average monthly income of the household, household size can affect latrine use and location of the latrine was remained associated with latrine utilization. The younger age had [AOR 1.83, 95% CI: 1.11, 3.02], a significant association with latrine utilization as compared with the 35 years and older age. A mother who had primary education [AOR 3.68, 95% CI: 1.47, 9.17] and secondary and above [AOR 4.17, 95% CI: 1.10, 15.94] as compared with no formal education. The household monthly income more than 66.4$, [AOR 5.17, 95% CI: 3.10, 8.72], Household family size [AOR 3.19, 95% CI: 1.97, 5.17], the households who had latrine inside the compound [AOR 2.47, 95% CI: 1.64, 3.71] as compared with their counterparts (Table 2). ## Discussions This community-based cross-sectional study revealed that the utilization of latrine among selected households was 71.5% with [95% CI: 68.3% - 74.4%]. Out of 212, the study participants claim as the main reason for not using the latrine, 59.7% reports latrine is not functional, 22.7% due to far from the house and 9.2% claims using latrine is inconvenient during the rainy season and at night without proper roof and door. The reasons given by respondents for why under-five children did not use the latrines were: being just a child 37.2%, large squatting hole 32.6%, and the floor was not safe to stand on 30.2%. This was consistent with study findings done in Nigeria level of latrine utilization `was 69% [9], 73% rural Becho district of central Ethiopia [10]. This similarity may due to the study participants in the Africa had almost had at similar socio-economic status and practice of the latrine utilization at the rural community. This finding was lower than 88% Democratic Republic of Congo [6], 89.9% rural village of Eastern Nepal [11], 93% Mirab Abaya and Alaba [12], 86.7% Hulet Ejju Enessie district [13]. Higher than study in Bahr Dar Zuria (62%) [14] and 61.2% in rural areas of Denbia district, Northwest Ethiopia [8]. This difference in the utilization of latrine was among selected households from Chire district might be due to socio-economic differences, different interventions among locally officials and interventions of the different projects among different regions in the present day and time of the study [15]. This implies that the health officials, and focal needs work forward to maximize the utilization of the latrine by setting different strategies. This study shows that the younger age of the respondent had a significant association with the utilization of latrine as compared with older age. This finding is similar to the study conducted in India [16], rural Bangladesh [17], North West Ethiopia [18] and rural communities in the District of Bahir Dar Zuria, Ethiopia [19]. This may explain that uneducated and elder people in a rural area may find it difficult to get some when latrine construction not easy to use by elders due to different reasons like privacy and cultural issues when latrine lacks door and appropriate cover. In addition to this, they are also economically dependent and they cannot afford the construction of the latrine. This study reveals that the high education level of a mother had a significant association with latrine utilization. Similarly, a study conducted in India [17,18,20], Tigray Hawzien district, Northern Ethiopia [21], and Gedeo Zone, South Ethiopia [22]. This could be attributed to the impact of education on behavior change and the adoption of good latrine hygiene practices at the household level high among educated mothers. Furthermore, the monthly income of the household had also a significant association with latrine utilization. This result agrees with the Gulomekada District, Tigray Region, North Ethiopia [23], and Bangladesh [24]. This may due to Household’s monthly income determines the availability and quality of latrine which are the important predictors of utilization of latrine. Also, this might be due to low-income households had a shortage of money to constrict latrine rather than other important materials and utilities for daily consumption. Household family size was found to be a significant association with latrine utilization. Similarly, North West Ethiopia [20] and rural coastal Odisha, India [17]. This might be due to the presence of family and cultural practices promote small children to defecate around the house rather than go to open whole latrine use. On the other hand, this may also be related to the economic status of the household. The location of the latrine being inside the compound had also a statistically significant association with latrine utilization. This result is in line with Chencha District, Southern Ethiopia [5], and Bangladesh [26]. This may due to fear to utilize the latrine during at night and raining because it is far from the resident’s house and no roof to prevent raining and it is exposed to an animal attack. This study result shows, there was a low utilization of latrine. There is a significant number of the household still not using the latrine. The main reason was reported; the latrine is not functional, far from the house, when in the rainy season and at night without proper roof and door. This implies that a lot has to be done on awareness creation about the proper utilization of latrine and support on the maintenance of the poor family latrine that needs attention to improve rural community health. Provide different types of strategies to a rural community with an affordable cost to facilitate children’s toilet training. There might be a potential for recall and social desirability bias in the utilization of latrine and socioeconomic. In addition to this, the odds ratios of the cross-sectional study did not show the strength of an association. ## Conclusions This study result shows that the utilization of latrine needs improvements and attention to enhance the proper and adequate utilization of the latrine. It associated with being younger age, maternal education, monthly income, family size, latrine inside the compound as compared with their counterparts were found to be associated factors of the utilization of latrine. Hence, this needs to improve awareness of the community there is a need for health education programs regarding improved use of latrine, cleanliness of latrine, constructing latrine inside the compound and maintenance of latrine for the proper utilization including children should be considered. In addition to this all actors to bridge the apparent gap between knowledge and practice pertinent to upscaling latrine use. Facilitate women education, training on latrine construction skills and capacity building continuously for the community. Not only this they need support on the utilization of the matching resources to tackle the sanitation disparities while utilizing socio-culturally appropriate technological options, suitable for the study community at affordable prices. ### Supplementary Materials There is no remaining pertinent data and materials, all information is presented in the main manuscript. ### Author Contributions Conceptualization, DW; methodology, DW & YT; software, validation, and formal analysis, investigation, resources, data curation, writing-original draft preparation, writing-review and editing, visualization, supervision, project administration, DW& YT; funding acquisition, DW & YT. All authors have read and agreed to the published version of the manuscript. ## Acknowledgments The authors would like to thank the SNNPR health bureau for ethical approval and at Sidama National Regional state and Chire district/woreda health bureau for their cooperation on providing, information and support letter. We would like to provide our gratitude for Paradise Valley College for its cooperation and financial supports during data collection they paid peridium for data collectors and stationery support, which did not affect the finding of this result. The authors are also grateful to all data collectors and study participants for their valuable contributions. 1. Palaiodimos L, Kokkinidis DG, Li W, Karamanis D, Ognibene J, Arora S, et al. Severe obesity, increasing age and male sex are independently associated with worse in-hospital outcomes, and higher in-hospital mortality, in a cohort of patients with COVID-19 in the Bronx, New York. Metabolism. 2020; 108: 154262-1770. DOI: 10.1016/j.metabol.2020.154262 2. Suleyman G, Fadel RA, Malette KM, Hammond C, Abdulla H, Entz A, et al. Clinical characteristics and morbidity associated with coronavirus disease 2019 in a series of patients in metropolitan detroit. JAMA Netw Open. 2020; 3: e2012270. DOI: 10.1001/jamanetworkopen.2020.12270 3. Naja M, Wedderburn L, Ciurtin C. COVID-19 infection in children and adolescents. Br J Hosp Med (Lond). 2020; 81: 1-10. DOI: 10.12968/hmed.2020.0321 4. Richardson S, Hirsch JS, Narasimhan M, Crawford JM, McGinn T, Davidson KW, et al. Presenting characteristics, comorbidities, and outcomes among 5700 patients hospitalized with COVID-19 in the New York City Area. JAMA 2020; 323: 2052-2059. DOI: 10.1001/jama.2020.6775 5. Laurencin CT, McClinton A. The COVID-19 pandemic: A call to action to identify and address racial and ethnic disparities. J Racial Ethn Health Disparities. 2020; 7: 398-402. DOI: 10.1007/s40615-020-00756-0 6. Chowkwanyun M, Reed AL Jr. Racial health disparities and Covid-19 - caution and context. N Engl J Med. 2020; 383: 201-203. DOI: 10.1056/NEJMp2012910 7. Bibbins-Domingo K. This time must be different: Disparities during the COVID-19 pandemic. Ann Intern Med. 2020; 173: 233-234. DOI: 10.7326/M20-2247 8. Arentz M, Yim E, Klaff L, Lokhandwala S, Riedo FX, Chong M, et al. Characteristics and outcomes of 21 critically ill patients with COVID-19 in Washington State. JAMA. 2020; 323: 1612-1614. DOI: 10.1001/jama.2020.4326. 9. Petrilli CM, Jones SA, Yang J, Rajagopalan H, O'Donnell LF, Chernyak Y, et al. Factors associated with hospital admission and critical illness among 5279 people with coronavirus disease 2019 in New York City: Prospective cohort study. BMJ. 2020; 369: m1966. DOI: 10.1136/bmj.m1966. 10. Chronic obstructive pulmonary disease (COPD). WHO. The more familiar terms of "chronic bronchitis" and "emphysema" have often been used as labels for this condition. June 6, 2020. https://bit.ly/3jE00mF 11. Income Limits Documentation System-HUD.GOV. 8-7-2020. https://bit.ly/3lHOSai 12. Breiman L. Random Forests. Machine Learning. 2001; 45: 5-32. 13. Bhatraju PK, Ghassemieh BJ, Nichols M, Kim R, Jerome KR, Nalla AK, et al. Covid-19 in critically Ill patients in the seattle region - Case Series. N Engl J Med. 2020; 382: 2012-2022. DOI: 10.1056/NEJMoa2004500 14. Cummings MJ, Baldwin MR, Abrams D, Jacobson SD, Meyer BJ, Balough EM, et al. Epidemiology, clinical course, and outcomes of critically ill adults with COVID-19 in New York City: A prospective cohort study. Lancet. 2020; 395: 1763-1770. DOI: 10.1016/S0140-6736(20)31189-2 15. Guan WJ, Ni ZY, Hu Y, Liang WH, Ou CQ, He JX, et al. Clinical characteristics of Coronavirus Disease 2019 in China. N Engl J Med. 2020; 382: 1708-1720. DOI: 10.1056/NEJMoa2002032 16. Goyal P, Choi JJ, Pinheiro LC, Schenck EJ, Chen R, Jabri A, et al. Clinical characteristics of Covid-19 in New York City. N Engl J Med. 2020; 382: 2372-2374. DOI: 10.1056/NEJMc2010419 17. Rodriguez-Diaz CE, Guilamo-Ramos V, Mena L, Hall E, Honermann B, Crowley JS, et al. Risk for COVID-19 infection and death among Latinos in the United States: Examining heterogeneity in transmission dynamics. Ann Epidemiol. 2020. DOI: 10.1016/j.annepidem.2020.07.007 18. Ripperger TJ, Uhrlaub JL, Watanabe M, Wong R, Castaneda Y, Pizzato HA, et al. Detection, prevalence, and duration of humoral responses to SARS-CoV-2 under conditions of limited population exposure. medRxiv 2020. https://bit.ly/2Gm0O1f 19. Infectious Diseases Society of America Guidelines on the Diagnosis of COVID-19: Serologic Testing. https://www.idsociety.org/practice-guideline/covid-19-guideline-serology. August 28, 2020. https://bit.ly/32LErtp 20. Interim Guidance for Rapid Antigen Testing for SARS-CoV-2. https://bit.ly/2EYWgNh 21. Wiersinga WJ, Rhodes A, Cheng AC, Peacock SJ, Prescott HC. Pathophysiology, transmission, diagnosis, and treatment of Coronavirus Disease 2019 (COVID-19): A review. JAMA 2020. DOI: 10.1001/jama.2020.12839 ### Why choose us • Open Access Publishing • Quality and Potential Expertise • Scrupulous Editorial and Double Blind Peer-review • Swift Production Process	2022-01-25 14:05:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4418743848800659, "perplexity": 10067.437119620783}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304835.96/warc/CC-MAIN-20220125130117-20220125160117-00265.warc.gz"}
http://www.albacete-bpie.es/ah36/the-splined-ends-and_310872.html	## the splined ends and gears attached to the a992 steel Jan 07, 2021 · The splined ends and gears attached to the A992 steel shaft are subjected to the torques shown where T = 200 N.m. The shaft has a diameter of 50 mm Determine the angle of twist of end B with respect to end A. E your answer to three significant ### (Solved) - The 60-mm-diameter A-36 steel shaft is Jun 05, 2021 · The splined ends and gears attached to the A992 steel shaft are subjected to the torques shown where T = 200 N.m. The shaft has a diameter of 50 mm Determine the angle of twist of end B with respect to end A. E your answer to three significant (Solved) - The splined ends and gears attached to the A992 Mar 15, 2019 · The splined ends and gears attached to the A992 steel shaft are subjected to the torques shown where T =200 N·m. The shaft has a diameter of 45 mm. ### (Torsion:Angle of Twist Problem 5.49 < 3 of 3 > The The splined ends and gears attached to the A992 steel shaft are subjected to the torques shown. If the shaft has a diameter of 40 mm. [10 marks) a-Draw the torque diagram. [2 marks] b-Determine the angle of twist of end B with respect to end A. [3 marks] C- 2. (25 Points) The splined ends and gears attached to the The splined ends and gears attached to the A992 steel shaft are subjected to the torques shown. If the shaft has a diameter of 40 mm. [10 marks] a-Draw the torque diagram. [2 marks] b-Determine the angle of twist of end B with respect to end A. [3 marks] C- ### 549. The splined ends and gears attached to the A-35 549. The splined ends and gears attached to the A-35 steel shaft are subjected to the torques shmvn_ Detennule the angle of of end B th respect to end The shaft has a diameter of 40 G GPa _ 0(0.5) = 0578. 85 218 TA6 53 Solutions - StuDocu85 218 TA 2 53 Solutions 85 218 TA4 53 Solutions 85 218 TA5 53 Solutions 85 218 TA10 52 Solutions 85 218 TA7 52 Solutions 85 218 TA5 54 Solutions. 85 218 TA6 53 Solutions. Course:Mechanics of Deformable Bodies (85 218) 0. 1. ### Answered:16-79. The mechanism shown is used in a Q:The splined ends and gears attached to the A992 steel shaft are subjected to the torques shown. Dete A:Given data:The diameter of the shaft is d=40 mm. The material of the shaft is A992 EM 324 Session 10 Angle of Twist - Iowa State UniversityThe splined ends and gears attached to the A992 steel shaft are subjected to the torques shown. Determine the angle of twist of end B with respect to end A. The shaft has a diameter of 40 mm. ### If it is subjected to the three torques plot the shear Angle of Twist 49) The splined ends and gears attached to the A992 steel shaft are subjected to the torques shown. Determine the angle of twist of end B with respect to end SOLVED:The 50-mm-diameter A992 steel shaft is subThe 50-mm-diameter A992 steel shaft is subjected to the torques shown. Determine the angle of twist of the end A Hurry, space in our FREE summer bootcamps is running out. Claim your spot here. ### The A992 steel shaft has a diameter of 50 ''mm'' and is The A992 steel shaft has a diameter of 50 mm and is fixed at its ends A and B.The shear modulus of rigidity for the material is 75 GPa. The shaft is subjected to the torques shown in the figure below. The splined ends and gears attached to the A-36 steel The splined ends and gears attached to the A-36 steel shaft are subjected to the torques shown. Find the angle of twist of end B with respect to end A. The shaft diameter is 40 mm. ### The splined ends and gears attached to the A992 steel The splined ends and gears attached to the A992 steel shaft are subjected to the torques shown. Determine the angle of twist of end B with respect to end A. The splined ends and gears attached to the A992 steel The splined ends and gears attached to the A992 steel shaft are subjected to the torques shown. If the shaft has a diameter of 40 mm. [10 marks] a-Draw the torque diagram. [2 marks] b-Determine the angle of twist of end B with respect to end A. [3 marks] C- ### Torsion Mechanics Questions and Answers Study The splined ends and gears attached to the A992 steel shaft are subjected to the torques shown. Determine the angle of twist of end B with respect to end A. The Shaft has a diameter of 40mm. [Solved] The A992 steel shaft has a diameter of 50 mm and The A992 steel shaft has a diameter of 50 mm. Need more help! The A992 steel shaft has a diameter of 50 mm and is subjected to the distributed loadings shown. Determine the absolute maximum shear stress in the shaft and plot a graph of the angle of twist of the shaft in radians versus x. ### fkm.utm.my · Translate this page > e L M N O P Q R S T U V W X Y Z [ \ ] ^ _ a b c d ! h C L \ 2J ` P x uQ=K P = S (.up E "6? UR + &f X A N {w x 3 / e X ejIA! 2 cm + U l 9 / f0 he splined ends and gears attached to the A992 steel shaft The splined ends and gears attached to the A992 steel shaft are subjected to the torques shown where {eq}T= 210 N\cdot m. The shaft has a diameter of 35 mm {/eq} Determine the angle of twist of ### The splined ends and gears attached to the A992 steel The splined ends and gears attached to the A992 steel shaft are subjected to the torques shown in (Figure 1). The shaft has a diameter of 45 mm. Part A Determine the angle of twist of end B with respect to end A. E your answer in degrees to three significant figures.	2021-10-28 05:29:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5553393959999084, "perplexity": 2760.1872641382033}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323588257.34/warc/CC-MAIN-20211028034828-20211028064828-00209.warc.gz"}
https://cs.stackexchange.com/questions/144669/collision-between-a-bi-infinite-linear-sequence-of-2d-integer-lattice-points-and	# Collision between a bi-infinite linear sequence of 2D integer lattice points and any of a fixed set of such sequences Given: • a finite collection $$V$$ of bi-infinite linear sequences of two-dimensional integer lattice points, each sequence $${V_i}$$ given by $$\cdots,\vec{{V_i}_{-1}},\vec{{V_i}_0},\vec{{V_i}_1},\cdots$$ where $$\vec{{V_i}_j} = \vec{{V_i}_0} + j \times \vec{d_i}$$ for $$j \in \mathbb{Z}$$. (A natural way to represent a sequence like this would be with the pair $$(\vec{{V_i}_0}, \vec{d_i})$$, with $$\vec{{V_i}_0}$$ chosen arbitrarily among the elements of $${V_i}$$) • a single sequence $$Q$$ of the same type as an element of $$V$$ and similarly represented • it is known that any two sequences in $$V \cup \{Q\}$$ share at most one point • it is known that there are no "vertical", "horizontal" or constant sequences: the difference between successive elements in each sequence is always nonzero in both dimensions I am interested in deciding whether any element of $$V$$ shares a point with $$Q$$. Allowing reasonable one-time precomputation for fixed $$V$$, is there an algorithm that decides this in less than linear time with respect to the size of $$V$$, for varying $$Q$$? In other words, can we do pessimistically better than checking every element of $$V$$ for collision with $$Q$$? I don't actually need to find the element of $$V$$ that shares a term with $$Q$$. I only need to know if such an element exists. • One idea: replace each infinite sequence with the infinite line that goes through it, use the Bentley-Ottman algorithm to create a sweepline data structure of all such lines, traverse the data structure to find all points of intersection between $Q$ and some other line, and check each such point of intersection to see if it was on the original sequence. However I think that in the worst case this might still take linear time, because there might be linearly many intersections between the lines. Probably not helpful, sorry. – D.W. Nov 11, 2021 at 23:02 I think I figured it out, here's an outline of the idea: #### Precomputation: 1. Choose a point $$O^\prime$$ that doesn't lie on any of the lines containing sequences in $$V$$; the origin is as good a choice as any, provided none of said lines go through the origin. $$O^\prime$$ does not need to have integer coordinates, so a random real point in the unit square is nearly sure to be good. 2. For each sequence $$S$$ in $$V$$, find the point closest to $$O^\prime$$ on the line through $$S$$ and add that point to a quadtree, together with a pointer to $$S$$. #### Query 1. Consider the set $$C$$ of circles through $$O^\prime$$ and centered at midpoints between $$O^\prime$$ and each point in $$Q$$. Those can be alternatively defined as circles through 3 points, two of which are common to all the circles: $$O^\prime$$, the point closest to $$O^\prime$$ on the line containing $$Q$$, and an element of $$Q$$ (this is in fact the textbook construction of the point on a line closest to another point not on the line). Note that $$C$$ is still defined even if the line through $$Q$$ contains $$O^\prime$$, hence no extra restriction here. 2. I won't go into needless details, but it is $$O(1)$$ to check whether a point lies on one of those circles, and it is also $$O(1)$$ to decide whether a quadtree node overlaps any or none of those circles (i.e. whether it fits entirely in a crescent between two nearest circles). 3. Employing observations from point 2, traverse the quadtree, descending into nonempty nodes whose bounding rectangles intersect with some circles in $$C$$, and enumerating points that lie exactly on the circles. Output elements of $$V$$ associated with those points and additionally satisfying the full collision test with $$Q$$. Since this traversal will skip nodes that fit entirely between the circles of $$C$$, one hopes for a performance gain here. • Why does every circle centered at a midpoint between $O'$ and $Q$ and going through $O'$ also go through some other point of $Q$? It will intersect the line that goes through $Q$, but not necessarily at a point of $Q$, right? • Why does the quadtree traversal take less than linear time? Won't most/all nodes of the quadtree have a bounding box that intersects with some circle in $C$? I'm skeptical -- I imagine the running time of this method will be linear in the worst case. Am I missing something? • First question: if the circle is centered at the midpoint in question, then the line between $O^\prime$ and point on Q is the diameter of the circle. The third point is the second intersection of the circle with the line, and the triangle is based on the diameter, hence the angle at the third point is always right, hence it is always the same point :) Nov 11, 2021 at 23:11 • Sorry, I don't follow that. I don't know what you mean by same point, and I don't understand why that third point is an "integer lattice point" in $Q$ (rather than some other point on the line going through $Q$). Maybe a picture would help. • Second question: I believe the search will be asymptotically sublinear in the size of $V$, provided the maximum distance between $O^\prime$ and any line containing a sequence in $V$ remains bounded - yes, this is an additional restriction, but it happens to hold wonderfully in my particular application! Nov 11, 2021 at 23:24	2022-07-05 03:30:37	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 45, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7715160250663757, "perplexity": 196.69389303101917}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104512702.80/warc/CC-MAIN-20220705022909-20220705052909-00737.warc.gz"}
https://stats.stackexchange.com/questions/336502/how-to-combine-probability-plots-and-hypothesis-tests-to-check-normality/336515	# How to combine probability plots and hypothesis tests to check normality? I have two samples X ($N$= 97) and X2 ($N$=4782) drawn from the same population data. I like to test (using statistical visualizations such as normplot and qqplot and hypothesis tests such as jbtest, chi2gof and kstest in matlab) if the data from each sample is normally distributed. My first data is: X = [8.13010235400000,13.6713071300000,14.0362434700000,18.4349488200000,26.5650511800000,30.9637565300000,34.3803447200000,40.6012946500000,45,49.3987053500000,58.6713071300000,59.0362434700000,59.0362434700000,59.0362434700000,61.9275130600000,61.9275130600000,63.4349488200000,63.4349488200000,63.4349488200000,63.4349488200000,63.4349488200000,64.4400348300000,71.5650511800000,71.5650511800000,71.5650511800000,71.5650511800000,75.9637565300000,75.9637565300000,75.9637565300000,75.9637565300000,75.9637565300000,75.9637565300000,75.9637565300000,75.9637565300000,75.9637565300000,75.9637565300000,77.4711922900000,77.4711922900000,77.4711922900000,77.4711922900000,77.4711922900000,77.4711922900000,77.4711922900000,77.4711922900000,77.4711922900000,78.6900675300000,90,90,90,90,90,90,90,90,90,90,90,90,90,90,90,93.1798301200000,97.1250163500000,97.7651660200000,102.528807700000,102.528807700000,102.528807700000,102.528807700000,102.528807700000,104.036243500000,104.036243500000,104.036243500000,104.036243500000,104.036243500000,104.036243500000,104.036243500000,105.255118700000,108.434948800000,108.434948800000,108.434948800000,108.434948800000,109.440034800000,116.565051200000,118.072486900000,120.963756500000,127.746805400000,130.601294600000,135,137.489552900000,139.398705400000,139.398705400000,149.036243500000,153.434948800000,159.227745300000,161.565051200000,179.999998800000,180]; The analyses using statistical visualizations in matlab show that the underlying distributions for both samples are normal. However, from the hypothesis tests, the null hypothesis for the first sample is not rejected using the same significance value (except for the chi-square test), but that for the second sample, X2 is completely rejected. I am now confused as to how to prove my samples are normally distributed and as well come from the same population data. What can I do in this situation? PS: sample X2is too large for me to post, but if there is any suggestion on how I could show this, then I don’t mind. EDIT: I have just collated another set of sample (N = 4700) from the same population data wherein the qqplots and cdf comparisons all look good (see new added image). Strangely, the hypothesis tests with jbtest and kstest in Matlab both rejects the null hypothesis. I am now beginning to believe that these hypothesis tests may not be trusted afterall, particularly for real case data. PS: I couldn't try the Shapiro-Wilks test as Matlab do not have this. • Testing each sample separately for normality does not assess whether the samples come from the same population. What question, then, do you really need to address: that of normality or that of a common distribution? – whuber Mar 24 '18 at 19:44 • Thank you @Whuber. I am more interested in answering the former. That is, if the samples are normally distributed. This is because, I am certain they come from the same population data. – oma11 Mar 24 '18 at 21:18 • In that case, the results you report ("$\chi^2$ is completely rejected") have rejected that hypothesis of a common Normal distribution. – whuber Mar 24 '18 at 22:25 • Secondary MATLAB implementation detail: don't let the legend (key) occlude any of the data! – Nick Cox Mar 25 '18 at 8:57 • Your latest results are consistent with your earlier ones. (1) The tests are doing what they are supposed to do. You have definite evidence of non-normality in a large sample. (2) The bounds of 0 and 180 bite visibly. A normal distribution with mean and SD for your data discernibly "wants" to overstep those bounds. (3) Let me emphasise an earlier comment If you have to test for normality -- but I really don't see that you have to here. It's like testing to see if a frog is really a toad; you will get high similarity, but as you know it's a frog, there is no point. – Nick Cox Mar 25 '18 at 23:40 I think the attitude here should not be an attempt to "prove" that the data are normally distributed, but simply to check on whether the data are close enough to normal for that to be an adequate approximation for your unstated purposes. I'd go further than @CroGo and suggest going straight to quantile plots. The comparison with a straight line reference is much easier than comparing two distribution functions with each other when one is a normal (so sigmoid) ogive and the visual challenge is compare an exact and a rough sigmoid curve. (EDIT 2: The posted distribution function plots confirm my prejudice in not clearly showing the limitations of the data.) Here are a quantile normal plot (normally distributed data points would follow the line) and a spike representation of the distribution. The quantile plot here for your smaller sample suggests to me that you can't reject a null hypothesis partly because the sample size is rather small. For many purposes the approximation looks fair but not excellent. If you had a theory (physical or other) that says that the distribution should be normal, then it's not well supported. If your interest is just in using techniques that work well if data are approximately normal, then there is no bad news. But hold it there: I have labelled the graphs in terms of 0(45)180 because the sharp limit at 180 makes me wonder whether these are bounded measurements in degrees. A look at the detail in the distribution shown as spikes for distinct values seems to support that idea: why else a spike at 90? Confusion: I should have read these data into my software (non-disclosure: not MATLAB) in double precision. If the difference between 179.999998800000 and 180 is meaningful to you, that was a coarse approximation. The rejection of the null hypothesis with a much larger sample size is no surprise. That's likely just an indication that you have more information in the larger sample. The same kind of discrepancy for a larger sample size is more likely to qualify as significant at conventional levels. That's how significance tests work, just as 7/10 coin tosses coming up heads could easily be a fluke, but if you get 700000/1 million you really have evidence of bias. If a graph for the other, larger sample looks similar to that here, it's a similar conclusion. But if your data are really angles, or the equivalent, on [0, 180] or (0, 180], then the normal is at best a dubious reference, as the normal is unbounded and the angles are bounded. But equivalent distributions for bounded data would be likely to look very similar to the normal, so the objection is one of principle. Question: Is there any sense in which 0 = 180? Note: If you have to test for normality, chi-square tests belong in a museum and a dedicated test such as Shapiro-Wilk or Doornik-Hansen is preferable to those you mention: that's my impression from my reading of the literature. EDIT 1: As @whuber rightly points out in a comment, the question of whether the distributions are similar is not the same as that of whether each is normal. EDIT 2: The quantile plot for the larger sample shows the effects of bounding more clearly. The distribution is normal in the middle as many are, but in principle the normal isn't an appropriate reference for bounded data where the bounds bite. Thus the quantile plots may be useful exploratory devices, but formal tests for normality seem pointless. • Thank you @Nick. You've said quite a lot here and i'm trying to gradually assimilate them all. As Crogro suggested, I have made a comparison between the theoretical and empirical cdfs, as well as made the qqplots for both distributions. Surprisingly, these results show that the larger sample is even more normally distributed than what was suggested by the hypothesis test. I will add these figures just now for everyone to see. – oma11 Mar 24 '18 at 21:04 • Again, thank you for suggesting better tests. I will try them out. Also, while still searching through related questions, I found this post: stats.stackexchange.com/questions/2492/… in which someone stated that all normality test should reject the null hypothesis for large enough sample. And from your statement toward the end, you seem to go with that idea... – oma11 Mar 24 '18 at 21:25 • And yes! those are angle measurements... – oma11 Mar 24 '18 at 21:27 (Partial answer): For the KS test a good visualization is plot the ECDF curve against the theoretical CDF curve. If your data does come from the distribution then the ECDF curve should closely mirror that of the CDF curve. I'm not a matlab programmer but here is a relevant link. For distributional fit qq-plots are very similar to that of ECDF plots in that you are comparing theoretical quantities of the distribution against sampled ones. A linear line would be evidence of a decent fit. Again relevant link. • Thank you @CroGo. Your suggestion gave me another perspective entirely. – oma11 Mar 24 '18 at 21:19	2021-05-18 14:20:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6554838418960571, "perplexity": 724.5584337091375}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989637.86/warc/CC-MAIN-20210518125638-20210518155638-00309.warc.gz"}
https://mastodon.utwente.nl/@imuhammad?max_id=101980525291997636	"The fundamental question that we are addressing here is if there are fewer Wikipedia entries about female scientists because few women enter the sciences, or because they are less likely to contribute groundbreaking research, or do they face additional hurdles in attaining public recognition for their work for the same level of achievement?" - "Female scholars need to achieve more for equal public recognition" arxiv.org/pdf/1904.06310.pdf El presidente de Ecuador representa hoy una vergüenza para Latinoamérica entera I really missed looking at the mountains. ow I can enjoy watching them again and to be honest I am fascinated by their beauty much more than before! The normal Twitter's API only retrieves about 1% of all available tweets. Although you can retrieve the whole Twitter's timeline with with a special Twitter's API called Firehose, but it is super expensive and only available to a few companies. In this document, Jason Baumgartner mathematically shows the feasibility of reconstructing 99% of Firehose. He also urges the researchers to join forces and collect Twitter's data and make that data available to all researchers. tinyurl.com/yxzh5q4r PyData Amsterdam 2019 Reregistration and call for proposal: pydata.org/amsterdam2019 After watching another match at De Drolsch Veste, It still remains a very difficult thing for me to fully support FC TWENTE! 😅 Congrats @mvankeulen with your 25th year anniversary at U. Twente! I had to add a vector image of a CNN model to my poster last week and this repository which has some amazing Latex codes for drawing CNNs saved quite a bit of my time: github.com/HarisIqbal88/PlotNe Five Things That Scare Me About AI! fast.ai/2019/01/29/five-scary- Excellent! 😎 🇳🇱 bit.ly/2X3piQp Complexity Explorables has a very amazing collection of interactive visualizations that can model how some complex systems in different domains such as biology and social science work: complexity-explorables.org Dive into Deep Learning — An Interactive Deep Learning Book prismo.news/posts/234 We support $$\LaTeX$$ formulas: Use $$ and $$ for inline LaTeX formulas, and $ and $ for display mode.	2020-09-27 18:21:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2527764141559601, "perplexity": 5137.562097510183}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400283990.75/warc/CC-MAIN-20200927152349-20200927182349-00072.warc.gz"}
http://mathhelpforum.com/pre-calculus/156879-logarith-question.html	1. I have to solve: (log(x^2) + log(x^4))/log(90x) = 3 I've been on it for 25 minutes and have gotten as far as: log(x)/log(90x) = 1/2 And I'm not sure if I'm even going the right path lol, just trying to play around with properties of logs and not sure how to maneuver this. Thanks for any help 2. Originally Posted by DannyMath I have to solve: (log(x^2) + log(x^4))/log(90x) = 3 I've been on it for 25 minutes and have gotten as far as: log(x)/log(90x) = 1/2 And I'm not sure if I'm even going the right path lol, just trying to play around with properties of logs and not sure how to maneuver this. Thanks for any help Applying the laws of logarithms to reduce the left hand sid to an exoression in $\log(x)$ : $\dfrac{\log(x^2)+\log(x^4)}{\log(90x)}=\dfrac{2\lo g(x)+4\log(x)}{\log(90)+\log(x)}$ and you should be able to finish from there. CB 3. Hello, DannyMath! What you did was correct . . . $\text{Solve for }x\!:\;\;\dfrac{\log(x^2) + \log(x^4)}{\log(90x)} \:=\: 3$ Note that: . $x \ne 0$ We have: . $\log(x^2\cdot x^4) \;=\;3\log(90x) \quad\Rightarrow\quad \log(x^6) \:=\:3\log(90x)$ . . . . . . . . . . $6\log(x) \:=\:3\log(90x) \quad\Rightarrow\quad 2\log(x) \:=\:\log(90x)$ . . . . . . . . . . . $\log(x^2) \:=\:\log(90x) \quad\;\;\Rightarrow\qquad \quad\; x^2 \:=\:90x$ Then: . $x^2 - 90x \:=\:0 \quad\Rightarrow\quad x(x-90) \:=\:0$ Therefore: . $\rlap{//////}x \,=\,0,\;\;x\,=\,90$ 4. Originally Posted by Soroban Hello, DannyMath! What you did was correct . . . Note that: . $x \ne 0$ We have: . $\log(x^2\cdot x^4) \;=\;3\log(90x) \quad\Rightarrow\quad \log(x^6) \:=\:3\log(90x)$ . . . . . . . . . . $6\log(x) \:=\:3\log(90x) \quad\Rightarrow\quad 2\log(x) \:=\:\log(90x)$ . . . . . . . . . . . $\log(x^2) \:=\:\log(90x) \quad\;\;\Rightarrow\qquad \quad\; x^2 \:=\:90x$ Then: . $x^2 - 90x \:=\:0 \quad\Rightarrow\quad x(x-90) \:=\:0$ Therefore: . $\rlap{//////}x \,=\,0,\;\;x\,=\,90$ UGH I was getting so frustrated and you've made it so clear. Thank you! sigh of relief 5. Originally Posted by DannyMath UGH I was getting so frustrated and you've made it so clear. Thank you! sigh of relief I'm afraid he has actually made rather a meal of it. Which I suppose is OK if you are going to spoon feed someone. CB	2016-10-28 10:19:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 15, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.965567409992218, "perplexity": 495.5150141486919}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721606.94/warc/CC-MAIN-20161020183841-00455-ip-10-171-6-4.ec2.internal.warc.gz"}
https://physics.stackexchange.com/questions/534189/length-contraction-and-time-dilation-relationship	Length contraction and time dilation relationship We did the following derivation in my electromagnetism lecture Observer B measures $$c\Delta t'=d$$ Observer A measures $$c\Delta t =l$$ From Pythagoras' theorem: $$d^2+(v\Delta t)^2=l^2$$ $$(c\Delta t')^2+(v\Delta t)^2=(c\Delta t )^2$$ from which $$\Delta t =\Delta t'\gamma\qquad(\alpha)$$ with $$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ Then for the bar moving to the right with the same velocity of the B reference frame, B measures its length as $$L'$$ and A measures it as $$L$$ $$\Delta t' =\Delta t\gamma \qquad(\beta)$$ $$L'=c\Delta t'=c \gamma \Delta t=L \gamma$$ that is, we get length contraction Equation$$(\beta)$$ is what troubles me. Why don't they use equation $$(\alpha)$$ instead? Consider the definition of length. We may define it as the difference of the spatial coordinates between two points with respect to a reference frame, when it's temporal coordinates are the same. Basically length is the distance between two points when they are at the same time coordinate. Thus the relationship between length with respect to two different reference frame may be obtained using the Lorentz Transformation for the spacial coordinates. Consider, the following scenario, two reference frames A and B. A is stationary, while B is moving with a constant velocity v with respect to A. We assign x and t to the reference frame A, x' and t' to reference frame B. Note I will be taking $$c=1$$ to the end. Now consider at some t', we measure a length l with respect to B. The length will be equal to $$l=(x'_p-x'_o)$$ Now let's write down the transformation equations for $$x'_p$$ and $$x'_o$$. $$x'_p=\frac{x_p-vt}{\sqrt{1-v^2}}$$ And $$x'_o=\frac{x_o-vt}{\sqrt{1-v^2}}$$ Now we know how x transforms and hence we will substitute them into our transformation equations. Here I will consider a simple example, however it can be generalised further. For simplicity, let's consider measurements made at t'=0, between the points x'=0, and x'. $$x'=\frac{x-vt}{\sqrt{1-v^2}}$$ Now as we have taken $$t'=0$$, we may imply that $$t=vx$$ and from here we can rewrite our transformation equation as $$x'=\frac{x-v^2x}{\sqrt{1-v^2}}$$ This is going to give us $$x'=x\sqrt{1-v^2}$$ Adding the speed of light in we get $$x'=x\sqrt{1-v^2/c^2}$$ Similarly you can arrive at the formula for tube Dilation Now by our definition what is x'? Length as measured by the moving frame, what is x? Length as measured by the stationary frame. The key here is in the definition of length and time, which brings in the concept of length contraction, and time dilation Also you must remember that in Special Relativity we don't use time dilation and length contraction, instead we focus on Lorentz Transformations, which are the fundamental equations of Special Relativity. Also here is a special relativity space time graph, which works on the principles of Lorentz transformations, which will physically show how the effects take place	2021-07-31 06:49:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 24, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9053906798362732, "perplexity": 245.21177656804727}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154053.17/warc/CC-MAIN-20210731043043-20210731073043-00466.warc.gz"}
http://mathhelpforum.com/discrete-math/17588-counting-print.html	# Counting • Aug 7th 2007, 01:48 PM ff4930 Counting The problem How many positive integers from 100-999 are divisible by 7. I want to confirm if my approach is the best approach in finding the answer. The smallest integer that can be divided by 7 is 105+112+119+126+133+140+147+154+161+168+175+182+18 9+196 there are 14 numbers divisible by 7 from the range 100-200 so from 100-900 would be 914 = 126 numbers there is still 99 numbers unaccounted for but the answer in my textbook states 128 numbers in total. How is that possible? • Aug 7th 2007, 01:56 PM CaptainBlack Quote: Originally Posted by ff4930 The problem How many positive integers from 100-999 are divisible by 7. I want to confirm if my approach is the best approach in finding the answer. The smallest integer that can be divided by 7 is 105+112+119+126+133+140+147+154+161+168+175+182+18 9+196 there are 14 numbers divisible by 7 from the range 100-200 so from 100-900 would be 914 = 126 numbers there is still 99 numbers unaccounted for but the answer in my textbook states 128 numbers in total. How is that possible? The first number in the range divisible by 7 is 105, then every 7th number is divisible by 7, and so there are a total of floor[(999-105)/7] +1 =128 such numbers. RonL • Aug 7th 2007, 02:08 PM Plato Here is another way using the floor function. $\left\lfloor {\frac{{999}}{7}} \right\rfloor - \left\lfloor {\frac{{99}}{7}} \right\rfloor = 128$ • Aug 7th 2007, 02:15 PM ff4930 Thank You guys for the reply, I was doing it the long way =/. but one thing can I do [999-100]/7 I still got the same answer if I rounded down There are 1 more thing that is confusing me. same problem as before but finding the integers that are divisible by 3 or 4 I found all the numbers divisble by 3 which is 300 and divisble by 4 which is 225, but the answer is 450. and when they ask is not divisible by either 3 or 4, do I count the numbers that are divisible by 1-9 excluding 3 and 4? but the answer is 450 as well. and when they ask is divisible by 3 and not 4, how would I approach this Im really sorry for all these questions but as some of you may experience, my professor is not the very best at teaching and every lesson, I find myself reading the textbook and have a puzzled look. I also have a final coming up so any help would be appreciated. • Aug 7th 2007, 02:25 PM ThePerfectHacker Use the Inclusion-Exclusion Principle. Meaning find the number divisible by 3, find the number divisible by 5, and then subtract the number divisible by 3 AND 5, i.e. 15. • Aug 7th 2007, 02:29 PM Plato $\left\lfloor {\frac{{999 - 99}}{3}} \right\rfloor + \left\lfloor {\frac{{999 - 99}}{4}} \right\rfloor - \left\lfloor {\frac{{999 - 99}}{{12}}} \right\rfloor = 450$ We substract the numbers divisible by 12 because we counted them twice. • Aug 7th 2007, 02:45 PM ff4930 NVM I got it. Thank you all so much	2016-09-28 19:56:13	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.49919870495796204, "perplexity": 566.4681600703651}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738661763.18/warc/CC-MAIN-20160924173741-00168-ip-10-143-35-109.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/help-with-contour-integration-to-find-greens-function-of-d-3-dx-3.728738/	# Help with contour integration to find green's function of d^3/dx^3 1. Dec 16, 2013 ### Hakkinen 1. The problem statement, all variables and given/known data Given a linear operator $L=\frac{d^3}{dx^3}-1$, show that the fourier transform of the Green's function is $\tilde{G}(k)=\frac{i}{k^3-i}$ and find the three complex poles. Use the Cauchy integral theorem to compute G(x) for x < 0 and x > 0. 2. Relevant equations 3. The attempt at a solution I found the Fourier transform of the Green's function and solved for the three complex roots. I'm having trouble setting up and carrying out the contour integration though. I think that you should be able to write (k^3 - i ) somehow in terms of the roots, then break the contour integral up into three separate integrals around each contour with terms like (k - ....)(k - ....)(k - ....) in each denominator? Any assistance is greatly appreciated! 2. Dec 16, 2013 ### pasmith The cube roots of $i$ are $-i$ and $-ie^{\pm 2\pi i/3}$. 3. Dec 16, 2013 ### Hakkinen Thanks for the reply, but I had already solved that 4. Dec 16, 2013 ### vela Staff Emeritus So what's stopping you from doing what the problem says and applying the Cauchy integral theorem to find g(x)? 5. Dec 16, 2013 ### Hakkinen I don't know if I'm setting up the integral correctly in a form where a straightforward application of CIF could be done. I know about making arcs to form a closed contour but would I just need to evaluate one of these arcs? In this problem for example can I just use the "lower" arc that goes from -x to x and then loops around the pole at -i back to -x? Knowing that the other two poles will have no contribution by the Cauchy integral theorem. Or would I need to sum the contour integrals of the upper and lower arcs? So can I just write the lower arc C as this: $-\frac{1}{2\pi i}\oint_{C}\frac{e^{ikx}}{k^3-i}dk =-\frac{1}{2\pi i}\oint_{C}\frac{e^{ikx}}{(e^{\frac{i4\pi }{3}}e^{\frac{i2\pi }{3}})}dk$ where the denominator was factored like (pole at -i - pole in 1st quadrant)(pole at -i - pole in 2nd quadrant), I can simplify this further but is this the right approach? 6. Dec 16, 2013 ### vela Staff Emeritus The idea is that $$\oint_C f(z)\,dz = \int_{-R}^R f(z)\,dz + \int_\text{arc} f(z)\,dz.$$ If you can show the latter integral vanishes as $R \to \infty$ where $R$ is the radius of the arc, you can conclude the integral along the real axis is equal to the contour integral, which you can evaluate using residues. You can close the contour using an arc in the upper half plane or the lower half plane. The one you use depends on the sign of $x$. 7. Dec 17, 2013 ### jackmell Then you can't use the Residue Theorem in my opinion.	2017-08-19 11:53:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.916131854057312, "perplexity": 327.0333241293024}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886105341.69/warc/CC-MAIN-20170819105009-20170819125009-00084.warc.gz"}
https://www.physicsforums.com/threads/c-g-identity.157778/	# Homework Help: C-G identity 1. Feb 23, 2007 ### quasar987 Hi, How can I prove that $$<J,2;J,0\|J,J>=<J,2;-J,0\|J,-J>$$ ? (my "template" is $$<j_1,j_2;m_1,m_2\|JM>$$) thx 2. Feb 25, 2007 ### dextercioby Where did you get it from ? I don't think it's correct... 3. Feb 25, 2007 ### quasar987 It's correct "experimentally" in the sense that I'Ve looked in a C-G table and it's always true. And according to a classmate of mine, he went to see the teaching assistant and he told him it was correct. 4. Feb 25, 2007 ### nrqed You mean what you wrote, and not $$<J,2;J,0\|J,-J>=<J,2;-J,0\|J,J>$$, right? (at first I thought you had written what I just wrote and I saw a way to prove it easily but I don't see right away how to prove the wone you gave) 5. Feb 25, 2007 ### quasar987 i mean what i wrote	2018-05-21 11:36:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5622042417526245, "perplexity": 2422.43470173757}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794864063.9/warc/CC-MAIN-20180521102758-20180521122758-00090.warc.gz"}
https://www.acmicpc.net/problem/3832	시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율 5 초 128 MB 22 5 4 33.333% ## 문제 You started a company a few years ago and fortunately it has been highly successful. As the growth of the company, you noticed that you need to manage employees in a more organized way, and decided to form several groups and assign employees to them. Now, you are planning to form n groups, each of which corresponds to a project in the company. Sometimes you have constraints on members in groups. For example, a group must be a subset of another group because the former group will consist of senior members of the latter group, the members in two groups must be the same because current activities of the two projects are closely related, the members in two groups must not be exactly the same to avoid corruption, two groups cannot have a common employee because of a security reason, and two groups must have a common employee to facilitate collaboration. In summary, by letting Xi (i = 1, . . . , n) be the set of employees assigned to the i-th group, we have five types of constraints as follows. 1. Xi ⊆ Xj 2. Xi = Xj 3. Xi ≠ Xj 4. Xi ∩ Xj = ∅ 5. Xi ∩ Xj ≠ ∅ Since you have listed up constraints without considering consistency, it might be the case that you cannot satisfy all the constraints. Constraints are thus ordered according to their priorities, and you now want to know how many constraints of the highest priority can be satisfied. You do not have to take ability of employees into consideration. That is, you can assign anyone to any group. Also, you can form groups with no employee. Furthermore, you can hire or fire as many employees as you want if you can satisfy more constraints by doing so. For example, suppose that we have the following five constraints on three groups in the order of their priorities, corresponding to the first dataset in the sample input. 1. X2 ⊆ X1 2. X3 ⊆ X2 3. X1 ⊆ X3 4. X1 ≠ X3 5. X3 ⊆ X1 By assigning the same set of employees to X1,X2, and X3, we can satisfy the first three con- straints. However, no matter how we assign employees to X1, X2, and X3, we cannot satisfy the first four highest priority constraints at the same time. Though we can satisfy the first three constraints and the fifth constraint at the same time, the answer should be three. ## 입력 The input consists of several datasets. The first line of a dataset consists of two integers n (2 ≤ n ≤ 100) and m (1 ≤ m ≤ 10000), which indicate the number of groups and the number of constraints, respectively. Then, description of m constraints follows. The description of each constraint consists of three integers s(1 ≤ s ≤ 5), i(1 ≤ i ≤ n),and j(1 ≤ j ≤ n, j ≠ i), meaning a constraint of the s-th type imposed on the i-th group and the j-th group. The type number of a constraint is as listed above. The constraints are given in the descending order of priority. The input ends with a line containing two zeros. ## 출력 For each dataset, output the number of constraints of the highest priority satisfiable at the same time. ## 예제 입력 4 5 1 2 1 1 3 2 1 1 3 3 1 3 1 3 1 4 4 1 2 1 1 3 2 1 1 3 4 1 3 4 5 1 2 1 1 3 2 1 1 3 4 1 3 5 1 3 2 3 1 1 2 2 1 2 3 1 2 0 0 ## 예제 출력 3 4 4 2	2018-02-20 02:05:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.23455767333507538, "perplexity": 631.4029082694999}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891812871.2/warc/CC-MAIN-20180220010854-20180220030854-00451.warc.gz"}
https://pypi.org/project/pylint/1.7.2/	python code static checker ## Project description Pylint is a Python source code analyzer which looks for programming errors, helps enforcing a coding standard and sniffs for some code smells (as defined in Martin Fowler’s Refactoring book). Pylint has many rules enabled by default, way too much to silence them all on a minimally sized program. It’s highly configurable and handle pragmas to control it from within your code. Additionally, it is possible to write plugins to add your own checks. Development is hosted on GitHub: https://github.com/PyCQA/pylint/ You can use the code-quality@python.org mailing list to discuss about Pylint. Subscribe at https://mail.python.org/mailman/listinfo/code-quality/ or read the archives at https://mail.python.org/pipermail/code-quality/ ## Install Pylint requires astroid package (the later the better). Installation should be as simple as python -m pip install astroid Pylint requires isort package (the later the better). Installation should be as simple as python -m pip install isort If you want to install from a source distribution, extract the tarball and run the following commands python setup.py install You’ll have to install dependencies in a similar way. For debian and rpm packages, use your usual tools according to your Linux distribution. More information about installation and available distribution format may be found in the user manual in the doc subdirectory. ## Documentation Look in the doc/ subdirectory or at http://docs.pylint.org Pylint is shipped with following additional commands: • pyreverse: an UML diagram generator • symilar: an independent similarities checker • epylint: Emacs and Flymake compatible Pylint ## Project details Uploaded source Uploaded 2 7	2022-08-14 16:56:37	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.21569451689720154, "perplexity": 10758.20677948042}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572043.2/warc/CC-MAIN-20220814143522-20220814173522-00215.warc.gz"}
https://www.math.tolaso.com.gr/?p=1029	Home » Uncategorized » Groups of order 2p # Groups of order 2p Let be a group, a prime number and . Prove that either is cyclic or where is the dihedral group of order . Solution It is clear for . So we will assume that is an odd prime. Choose with . Let . Since every subgroup of index is normal, is a normal subgroup of and thus for some integer . Note that, since is odd, and hence Now because . Thus and hence because . Therefore either or . So we will consider two cases: • . In this case and so . Thus is abelian and hence because is odd. So is cyclic in this case. • . In this case and so	2020-11-30 23:33:34	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9754388332366943, "perplexity": 840.9153107662219}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141515751.74/warc/CC-MAIN-20201130222609-20201201012609-00526.warc.gz"}
https://www.jobilize.com/online/course/5-3-double-integrals-in-polar-coordinates-by-openstax?qcr=www.quizover.com	# 5.3 Double integrals in polar coordinates Page 1 / 7 • Recognize the format of a double integral over a polar rectangular region. • Evaluate a double integral in polar coordinates by using an iterated integral. • Recognize the format of a double integral over a general polar region. • Use double integrals in polar coordinates to calculate areas and volumes. Double integrals are sometimes much easier to evaluate if we change rectangular coordinates to polar coordinates. However, before we describe how to make this change, we need to establish the concept of a double integral in a polar rectangular region. ## Polar rectangular regions of integration When we defined the double integral for a continuous function in rectangular coordinates—say, $g$ over a region $R$ in the $xy$ -plane—we divided $R$ into subrectangles with sides parallel to the coordinate axes. These sides have either constant $x$ -values and/or constant $y$ -values. In polar coordinates, the shape we work with is a polar rectangle , whose sides have constant $r$ -values and/or constant $\theta$ -values. This means we can describe a polar rectangle as in [link] (a), with $R=\left\{\left(r,\theta \right)\|a\le r\le b,\alpha \le \theta \le \beta \right\}.$ In this section, we are looking to integrate over polar rectangles. Consider a function $f\left(r,\theta \right)$ over a polar rectangle $R.$ We divide the interval $\left[a,b\right]$ into $m$ subintervals $\left[{r}_{i-1},{r}_{i}\right]$ of length $\text{Δ}r=\left(b-a\right)\text{/}m$ and divide the interval $\left[\alpha ,\beta \right]$ into $n$ subintervals $\left[{\theta }_{i-1},{\theta }_{i}\right]$ of width $\text{Δ}\theta =\left(\beta -\alpha \right)\text{/}n.$ This means that the circles $r={r}_{i}$ and rays $\theta ={\theta }_{i}$ for $1\le i\le m$ and $1\le j\le n$ divide the polar rectangle $R$ into smaller polar subrectangles ${R}_{ij}$ ( [link] (b)). As before, we need to find the area $\text{Δ}A$ of the polar subrectangle ${R}_{ij}$ and the “polar” volume of the thin box above ${R}_{ij}.$ Recall that, in a circle of radius $r,$ the length $s$ of an arc subtended by a central angle of $\theta$ radians is $s=r\theta .$ Notice that the polar rectangle ${R}_{ij}$ looks a lot like a trapezoid with parallel sides ${r}_{i-1}\text{Δ}\theta$ and ${r}_{i}\text{Δ}\theta$ and with a width $\text{Δ}r.$ Hence the area of the polar subrectangle ${R}_{ij}$ is $\text{Δ}A=\frac{1}{2}\text{Δ}r\left({r}_{i-1}\text{Δ}\theta +{r}_{1}\text{Δ}\theta \right).$ Simplifying and letting ${r}_{ij}^{}=\frac{1}{2}\left({r}_{i-1}+{r}_{i}\right),$ we have $\text{Δ}A={r}_{ij}^{}\text{Δ}r\text{Δ}\theta .$ Therefore, the polar volume of the thin box above ${R}_{ij}$ ( [link] ) is $f\left({r}_{ij}^{},{\theta }_{ij}^{}\right)\text{Δ}A=f\left({r}_{ij}^{},{\theta }_{ij}^{}\right){r}_{ij}^{}\text{Δ}r\text{Δ}\theta .$ Using the same idea for all the subrectangles and summing the volumes of the rectangular boxes, we obtain a double Riemann sum as $\sum _{i=1}^{m}\sum _{j=1}^{n}f\left({r}_{ij}^{},{\theta }_{ij}^{}\right){r}_{ij}^{}\text{Δ}r\text{Δ}\theta .$ As we have seen before, we obtain a better approximation to the polar volume of the solid above the region $R$ when we let $m$ and $n$ become larger. Hence, we define the polar volume as the limit of the double Riemann sum, $V=\underset{m,n\to \infty }{\text{lim}}\sum _{i=1}^{m}\sum _{j=1}^{n}f\left({r}_{ij}^{},{\theta }_{ij}^{}\right){r}_{ij}^{}\text{Δ}r\text{Δ}\theta .$ This becomes the expression for the double integral. ## Definition The double integral of the function $f\left(r,\theta \right)$ over the polar rectangular region $R$ in the $r\theta$ -plane is defined as $\underset{R}{\iint }f\left(r,\theta \right)dA=\underset{m,n\to \infty }{\text{lim}}\sum _{i=1}^{m}\sum _{j=1}^{n}f\left({r}_{ij}^{},{\theta }_{ij}^{}\right)\text{Δ}A=\underset{m,n\to \infty }{\text{lim}}\sum _{i=1}^{m}\sum _{j=1}^{n}f\left({r}_{ij}^{},{\theta }_{ij}^{}\right){r}_{ij}^{}\text{Δ}r\text{Δ}\theta .$ Again, just as in Double Integrals over Rectangular Regions , the double integral over a polar rectangular region can be expressed as an iterated integral in polar coordinates. Hence, $\underset{R}{\iint }f\left(r,\theta \right)dA=\underset{R}{\iint }f\left(r,\theta \right)r\phantom{\rule{0.2em}{0ex}}dr\phantom{\rule{0.2em}{0ex}}d\theta =\underset{\theta =\alpha }{\overset{\theta =\beta }{\int }}\phantom{\rule{0.2em}{0ex}}\underset{r=a}{\overset{r=b}{\int }}f\left(r,\theta \right)r\phantom{\rule{0.2em}{0ex}}dr\phantom{\rule{0.2em}{0ex}}d\theta .$ Is there any normative that regulates the use of silver nanoparticles? what king of growth are you checking .? Renato What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ? why we need to study biomolecules, molecular biology in nanotechnology? ? Kyle yes I'm doing my masters in nanotechnology, we are being studying all these domains as well.. why? what school? Kyle biomolecules are e building blocks of every organics and inorganic materials. Joe anyone know any internet site where one can find nanotechnology papers? research.net kanaga sciencedirect big data base Ernesto Introduction about quantum dots in nanotechnology what does nano mean? nano basically means 10^(-9). nanometer is a unit to measure length. Bharti do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment? absolutely yes Daniel how to know photocatalytic properties of tio2 nanoparticles...what to do now it is a goid question and i want to know the answer as well Maciej Abigail for teaching engĺish at school how nano technology help us Anassong Do somebody tell me a best nano engineering book for beginners? there is no specific books for beginners but there is book called principle of nanotechnology NANO what is fullerene does it is used to make bukky balls are you nano engineer ? s. fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball. Tarell what is the actual application of fullerenes nowadays? Damian That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes. Tarell what is the Synthesis, properties,and applications of carbon nano chemistry Mostly, they use nano carbon for electronics and for materials to be strengthened. Virgil is Bucky paper clear? CYNTHIA carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc NANO so some one know about replacing silicon atom with phosphorous in semiconductors device? Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure. Harper Do you know which machine is used to that process? s. how to fabricate graphene ink ? for screen printed electrodes ? SUYASH What is lattice structure? of graphene you mean? Ebrahim or in general Ebrahim in general s. Graphene has a hexagonal structure tahir On having this app for quite a bit time, Haven't realised there's a chat room in it. Cied what is biological synthesis of nanoparticles how did you get the value of 2000N.What calculations are needed to arrive at it Privacy Information Security Software Version 1.1a Good Got questions? Join the online conversation and get instant answers!	2019-06-27 07:57:01	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 55, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8598541021347046, "perplexity": 1074.8311042704242}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560628001014.85/warc/CC-MAIN-20190627075525-20190627101525-00185.warc.gz"}
https://cs.stackexchange.com/questions/118437/if-a-problem-is-in-p-solved-via-dynamic-programming-is-it-also-in-np/118439	# If a problem is in P solved via dynamic programming, is it also in NP? So I can solve a given problem using dynamic programming in $$O(n^2k^2)$$ time complexity. This means that the problem is in P. But I am asked if it is in NP. My answer is, "Since it is also polynomial time solvable, the problem is also in $$NP$$". Is that a correct statement? If not, is a problem in P solved via dynamic programic also NP? • The definitions of "P" and "NP" don't make any reference to the type of algorithm used. Why should the fact that you used dynamic programming instead of something else be of any relevance? – John Coleman Dec 13 '19 at 17:11 • @JohnColeman Perhaps because there is a well-known conversion from a certain class of solution-producing algorithms to a solution-checking algorithm. I don't see any reason a priori that adding constraints on the set of programs you're willing to look at should make it harder to prove a particular property of that set. It turns out in this case that the answer doesn't depend on what algorithm was used, but until you know that it seems like the question itself is a perfectly cromulent one to wonder about. – Daniel Wagner Dec 13 '19 at 18:06 ### Any problem in P is also in NP A decision problem that's in P is also in NP, because you can give the verification logic like this: for yes instance x, use empty string as a certificate, and solve x in polynomial time. You get the result that it's yes instance (that's by definition of P) and that means verification is done in polynomial time. Note that, the order written as $$n^2 k^2$$ doesn't simply mean the problem is in P (Check "Pseudo-polynomial time" in wikipedia.) For example, Knapsack problem can be solved by the dynamic programming and its order is $$O(n W)$$ where $$n$$ is the number of products and $$W$$ is the maximum weight for the knapsack. But $$W$$ is actually considered exponential of the input size and Knapsack problem is known to be NP-complete. • $n$ is used in two different meanings here: "where $n$ is the number of products [...] is actually considered exponential of $n$ (which is the input size to the problem)". – JiK Dec 13 '19 at 16:56 • @JiK thanks for pointing that. I fixed the sentence. – Ryoji Dec 13 '19 at 19:32 Any problem that's in P is also in NP, since P is a subset of NP. One way to think of this is that problems in NP lift a restriction by not necessarily meeting the polynomial time requirement in a deterministic manner. Problems in P and NP meet the polynomial time requirement, but problems in P are those that meet it deterministically. Since P vs. NP is still an open question, it could be that everything in NP is also in P, even though most computer scientists don't think this is the case. If P is not equal to NP, then the NP-complete class of problems is mutually exclusive with P -- meaning that these problems cannot meet the polynomial time requirement deterministically. • Even if $P=NP$, $NPC\neq P$. Indeed the empty language and the language of all words on the alphabet are not NP-Complete (for the technical reason that a reduction must map a yes-instance to a yes-instance and a no-instance to a no-instance) – eru-cs Dec 13 '19 at 19:32	2020-01-18 18:46:13	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.638266384601593, "perplexity": 336.7394650637914}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250593295.11/warc/CC-MAIN-20200118164132-20200118192132-00259.warc.gz"}
http://www.gamedev.net/topic/649762-parse-and-querying/	• Create Account ## "Parse" and Querying Old topic! Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic. 8 replies to this topic ### #1Ibzy Members 257 Like 0Likes Like Posted 05 November 2013 - 11:24 AM Hi All, I have hit a bit of a snag on Android with regards to Querying a database. I have set up an account with Parse.com which seems a very reasonable and clever solution to server hosting as a new developer. In my set up I have 3 tables (or as Parse call them, classes) - Users, Cards, and Payer_Cards. All Users have a unique objectID, all Cards have a unique objectID and Player_Cards is a lookup between the two, containing Player ID and Card ID (not all players have all cards). The below logic is my attempt, but for some reason I cant figure it out. I can get the Card IDs listed based on all cards against the player ID (currentUser), but when I try to expand this to look at table 3 (Cards) it all falls apart. public void cardList(List<ParseObject> cardList){ StringBuilder cards = new StringBuilder(); for(int i=0; i<cardList.size(); i++){ cards.append("Card:"); ParseObject[] cList = cardList.toArray(new ParseObject[0]); String cardId = cList[i].getString("CARD_ID"); ParseQuery<ParseObject> query = ParseQuery.getQuery("CARDS"); query.whereEqualTo("objectId", cardId); query.getInBackground(cardId, new GetCallback<ParseObject>() { public void done(ParseObject object, com.parse.ParseException e) { if (e == null) { cards.append(object.getString("CARD_NAME")); } else { } } }); } TextView CardsList = (TextView) findViewById(R.id.CardsList); CardsList.setText(cards.toString()); } The current error in my code is that "cards" isn't recognised within the if block. Any assistance on this is greatly appreciated - I have exhausted all of my ides of how to get around this. Thanks Edited by Ibzy, 05 November 2013 - 11:25 AM. ### #2TheChubu Members 8955 Like 0Likes Like Posted 05 November 2013 - 11:33 AM I'm preety sure cards isn't visible within the if block because you're overriding a method with that anonymous class. That method can't see whats within cardList method (ie, the cards StringBuilder). You're basically referencing a local variable that is inside another method that your GetCallback object simply doesn't knows about. At most, from inner classes (or this anonymous class) you can reference fields of the enclosing object. ie, if cards were a field of whatever object has that cardList method, then you could access it from the anonymous class (and with a bit of magic from javac if the field isn't public). Edited by TheChubu, 05 November 2013 - 11:36 AM. "I AM ZE EMPRAH OPENGL 3.3 THE CORE, I DEMAND FROM THEE ZE SHADERZ AND MATRIXEZ" My journals: dustArtemis ECS framework and Making a Terrain Generator ### #3swiftcoder Senior Moderators 17832 Like 2Likes Like Posted 05 November 2013 - 11:41 AM You should be able to make your code compile by declaring cards to be final. This will make it accessible within the anonymous class. However, that won't actually make what you are trying to do work. It is highly unlikely that the GetCallback<ParseObject>.done() will be invoked until sometime after the cardList() method has already returned, which means that your textview will be set to an empty string. You really need to make StringBuilder an instance variable, and update the textview in each parse callback. Tristam MacDonald - Software Engineer @ Amazon - [swiftcoding] [GitHub] ### #4rip-off Moderators 10730 Like 1Likes Like Posted 05 November 2013 - 11:41 AM You can also reference local variables or parameters marked as final. Given that you don't appear to assign to "cards", this should work compile here. Edit: as swiftcoder mentions, an asynchronous call like that cannot be guaranteed to have run by the time you try to use the variable. In addition to his suggestions, you may also have to deal with threading here, which means you'll need to correctly synchronise your code, or use a thread safe class (such as StringBuffer). Read the documentation very carefully about how getInBackground() works. Edited by rip-off, 05 November 2013 - 11:45 AM. ### #5Ibzy Members 257 Like 0Likes Like Posted 05 November 2013 - 11:43 AM Hi Chubu, Thanks - that makes a lot of sense, and is kind of what I thought was happening. Can you think of a way to get around this? I considered setting a public variable and having it set to the new string on each tick of the for loop, but seems to return null. I imagine the main problem here is that Parse use a bit of their own terminology which no-Parse users wouldn't be familiar with? If the query didn't need that public void done() I'd probably be ok. Thanks ### #6Ibzy Members 257 Like 0Likes Like Posted 05 November 2013 - 11:50 AM Hi Guys, Thanks for the pointer - setting the StringBuilder as an instance var works like a charm when it comes to letting me update it from various functions. Now I just need to figure out why the CARD_NAME is blank Thanks again! ### #7Ibzy Members 257 Like 0Likes Like Posted 05 November 2013 - 12:03 PM Hi Again, After some debugging I found that it is not blank, the StringBuilder just isn't populated prior to being set. Very much a simpleton question here I'm sure, but how do I synchronise my code? I've always thought it runs in order and each function is completed before it moves on. I've changed StringBuilder to StringBuffer, and the app still compiles, but the data is still blank? I assume this is due to the synchronising you mention? Thanks Edit: My ParseQuery public void section now looks like this: public void done(ParseObject object, com.parse.ParseException e) { if (e == null) { cards.append("Card: "); Log.d("Card","Contains " + object.getString("CARD_NAME")); cards.append(object.getString("CARD_NAME")+"\n"); TextView CardsList = (TextView) findViewById(R.id.CardsList); CardsList.setText(cards.toString()); } else { } } This compiles and gives me the list I expect, however when the page loads I can see the list being populated - is this something I need to drag into the very beginning and cache, or am I doing something wrong which slows it all down? Thanks Edited by Ibzy, 05 November 2013 - 12:08 PM. ### #8rip-off Moderators 10730 Like 0Likes Like Posted 05 November 2013 - 01:30 PM You are kicking off a number of background jobs in a loop there. As they return, the list gets updated with the results. Perhaps the code could be modified to use one of the builk queries that the ParseQuery API appears to offer? It is hard to offer any advice about what you should be doing without knowing the specifics of what your application is trying to achieve. Edited by rip-off, 05 November 2013 - 01:32 PM. ### #9Ibzy Members 257 Like 0Likes Like Posted 06 November 2013 - 03:42 AM I have looked at the documentation around the ParseQuery API but am only seeing example queries for single table queries (If Parse supported SQL I would have this nailed). Apologies for being rather vague in my question - I felt that general practice would cover the issues I'm having here, so maybe that's not the case. My end goal is a card game (as I imagine you may have guessed), but with this snippet what I am trying to achieve is a list of cards owned by the current user. There will be situations where cards get added, I would need a separate list for a deck which will contain some of the cards in the entire list, and perhaps the ability to trade cards with other players (this part is a long way off). I know it might appear as though I'm starting from the middle, however I am trying to ensure I can fully query the Parse database in the way that I need to before deciding it is what I want. Having read about peoples issues when taking a standalone app and making it work with users/logins, I am covering this part off first and adding the game logic to it. Cheers Old topic! Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.	2016-12-11 14:20:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.18561877310276031, "perplexity": 1889.8004018251495}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698544679.86/warc/CC-MAIN-20161202170904-00309-ip-10-31-129-80.ec2.internal.warc.gz"}
https://socratic.org/questions/the-mean-of-a-set-of-data-is-4-11-and-its-standard-deviation-is-3-03-what-is-the	# The mean of a set of data is 4.11 and its standard deviation is 3.03. What is the z-score for a value of 10.86? Then teach the underlying concepts Don't copy without citing sources preview ? #### Explanation Explain in detail... #### Explanation: I want someone to double check my answer 5 Sep 29, 2017 $z \approx 2.23$ #### Explanation: Let $\mu \mathmr{and} \sigma$ be the mean and standard deviation, resp. Then, $z$ -score of $x$ is, $z = \frac{x - \mu}{\sigma}$ $\text{ Here, } \mu = 4.11 , \sigma = 3.03$ $\therefore z = \frac{10.86 - 4.11}{3.03} = \frac{6.75}{3.03} \approx 2.23$. • 10 minutes ago • 25 minutes ago • 26 minutes ago • 28 minutes ago • 4 minutes ago • 4 minutes ago • 5 minutes ago • 6 minutes ago • 6 minutes ago • 6 minutes ago • 10 minutes ago • 25 minutes ago • 26 minutes ago • 28 minutes ago	2018-06-21 08:04:43	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9380755424499512, "perplexity": 5559.055922410425}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267864110.40/warc/CC-MAIN-20180621075105-20180621095105-00155.warc.gz"}
http://www.haskell.org/haskellwiki/index.php?title=Show_instance_for_functions&diff=17664&oldid=17661	# Show instance for functions (Difference between revisions) Revision as of 16:59, 18 December 2007 (edit) (→Question)← Previous diff Revision as of 21:30, 18 December 2007 (edit) (undo) (give the lambda expression the fixed type Int -> Int in order to avoid effects of strange Num instance)Next diff → Line 20: Line 20: The Haskell compiler doesn't maintain the expressions as they are, but translates them to machine code or some other low-level representation. The Haskell compiler doesn't maintain the expressions as they are, but translates them to machine code or some other low-level representation. - The function \x -> x+x might have been optimized to \x -> 2x. + The function \x -> x+x :: Int -> Int might have been optimized to \x -> 2x :: Int -> Int. The variable name x is not stored anywhere. The variable name x is not stored anywhere. You might have thought, that Haskell is like a scripting language, maintaining expressions at runtime. You might have thought, that Haskell is like a scripting language, maintaining expressions at runtime. ## 1 Question Why is there no Show instance for functions for showable argument and value types? Why can't I enter \x -> x+x into GHCi or Hugs and get the same expression as answer? Why is there a Show instance, but it only prints the type? Text.Show.FunctPrelude> :m + Text.Show.Functions Prelude Text.Show.Functions> show Char.ord "<function>" How can lambdabot display this: dons > ord lambdabot> <Char -> Int> The Haskell compiler doesn't maintain the expressions as they are, but translates them to machine code or some other low-level representation. The function \x -> x+x :: Int -> Int might have been optimized to \x -> 2*x :: Int -> Int . The variable name x is not stored anywhere. You might have thought, that Haskell is like a scripting language, maintaining expressions at runtime. This is not the case. Lambda expressions are just anonymous functions. You will not find a possibility to request the name of a variable at runtime, or inspect the structure of a function definition. You can also not receive an expression from the program user, which invokes variables of your program, and evaluate it accordingly. That is, Haskell is not reflexive. Everything can be compiled. A slight exception is hs-plugins. Functional programming is about functions. A mathematical function is entirely defined by its graph, that is by pairs of objects (argument, value). E.g. • $\sqrt{\ } = \{(0,0), (1,1), (4,2), (9,3), \dots \}$ • $(\lambda x.\ x+x) = \{(0,0), (1,2), (2,4), (3,6), \dots \}$ Since the graphs of $\lambda x.\ x+x$ and $\lambda x.\ 2\cdot x$ are equal these both expressions denote the same function. Now imagine both terms would be echoed by Hugs or GHCi as they are. This would mean that equal functions lead to different output. The interactive Haskell environments use the regular show function, and thus it would mean $\mathrm{show}(\lambda x.\ x+x) \ne \mathrm{show}(\lambda x.\ 2\cdot x)$. This would break referential transparency. It follows that the only sensible way to show functions is to show their graph. Prelude> \x -> x+x functionFromGraph [(0,0), (1,2), (2,4), (3,6), Interrupted. One could do this for enumerable argument types, but it is not in the standard libraries.	2013-05-20 14:04:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 5, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.44565218687057495, "perplexity": 2446.889228901636}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368699036375/warc/CC-MAIN-20130516101036-00085-ip-10-60-113-184.ec2.internal.warc.gz"}
https://mathhelpboards.com/threads/solving-algebra-equation-3x-15.26470/	# Solving Algebra equation 3x=15 #### Victor23 ##### New member My workings ;3x=15 Any help is appreciated, thanks = 5x, as 15 divide by 3x=5x Staff member 3x = 15 x = 15/3 x = 5 #### HallsofIvy ##### Well-known member MHB Math Helper My workings ;3x=15 Any help is appreciated, thanks = 5x, as 15 divide by 3x=5x This makes no sense! In the first place, what does it mean to divide by an equation? In the second place, 3x is NOT equal to 5x unless x= 0. There is nothing here that says either x= 0 or 3x= 5x. To "solve for x" means to get x by itself: "x= something". In the original equation, 3x= 15, x is not "by itself" because it is multiplied by 3. To get x "by itself" we need to "undo" that- and we undo "multiply by 3" by dividing by 3. And, of course, anything we do to one side of the equation we must do to the other. Rather than "15 divided by 3x= 5x" you should have said "15 divided by 3 is 5". Dividing both sides of 3x= 15 by 3 gives (3x)/3= 15/3 x= 5. #### mrtwhs ##### Active member HallsofIvy has given the correct advice for solving this equation ... but ... I'm going to take an epsilon amount of issue with one phrase ... "anything we do to one side of the equation we must do to the other". I've said this myself in Algebra classes but I paid for it once. I had a student solve an equation in the following way: $$\displaystyle \dfrac{x-1}{6}=\dfrac{1}{3}$$ $$\displaystyle \dfrac{x}{6}=\dfrac{2}{3}$$ $$\displaystyle x = \dfrac{2}{3} \cdot 6 = 4$$ When I pressed him on what he did, he said "in the second step I added one to the numerator of both sides of the equation and anything I do to one side of an equation must be done to the other side of the equation ... which is what you said ... so it must be correct." #### HallsofIvy ##### Well-known member MHB Math Helper Thank you! I would argue that "adding 1 to the numerator" is "doing something" to the numerator, not to the "left side of the equation".	2020-08-03 23:16:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7548496127128601, "perplexity": 1216.879825599739}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439735836.89/warc/CC-MAIN-20200803224907-20200804014907-00167.warc.gz"}
http://math.stackexchange.com/questions/659957/working-out-the-overall-download-speed-of-two-sessions-with-different-speeds	# Working out the overall download speed of two sessions with different speeds I would like to confirm that what I have done with this basic problem is fine. It says: In the afternoon, the same amount of data took twice the time. My solution: • Forenoon -> 1900 MB/min -> (5401024)/1900 ≈ 291 min • Afternoon -> 800 MB/min -> 291 2 = 582 min • 540 GB = 552960 MB • Average speed: 552960/(291+582) ≈ 633.40 MB/min Did I miss something? - Yes, I would say so. In the afternoon the download took $582$ minutes, so that's what you should divide by at the end. Not $(291 + 582)$. – Arthur Feb 1 '14 at 21:22 @Arthur That is not right. Questions asks for the average during the whole day. – John Habert Feb 1 '14 at 21:28 Ahh, I misread. Sorry. But then you got to add the downloads from the morning as well in the numerator. So it's supposed to be $2\cdot 552,960$. – Arthur Feb 1 '14 at 21:30 Eveything looks good, up until the end. Try finding the total amount of data downloaded (540 MB in the forenoon plus 540 MB after that = 1080 MB) and then dividing by the total amount of time taken to find the average rate for the day: $$\dfrac {2\cdot 552960}{(291+582)}$$ So you need to simply double the rate you ended with: $2 \times 633.40$ MB/min = $1266.80$ MB/min. - I need to be about 16 seconds faster. ;) – John Habert Feb 1 '14 at 21:30 Average speed is $\dfrac{\mathrm{total\ data\ downloaded}}{\mathrm{total\ time}}$. You've got the total time part correct since it is the 291 minutes from forenoon plus twice that afternoon. The total data downloaded though is not 540 GB. It is 1080 GB (540 GB forenoon and 540 afternoon). -	2015-01-29 12:35:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6252816319465637, "perplexity": 1639.3817836377748}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422118551401.78/warc/CC-MAIN-20150124165551-00219-ip-10-180-212-252.ec2.internal.warc.gz"}
http://www.nag.com/numeric/FL/nagdoc_fl24/html/F07/f07cgf.html	F07 Chapter Contents F07 Chapter Introduction NAG Library Manual # NAG Library Routine DocumentF07CGF (DGTCON) Note: before using this routine, please read the Users' Note for your implementation to check the interpretation of bold italicised terms and other implementation-dependent details. ## 1 Purpose F07CGF (DGTCON) estimates the reciprocal condition number of a real $n$ by $n$ tridiagonal matrix $A$, using the $LU$ factorization returned by F07CDF (DGTTRF). ## 2 Specification SUBROUTINE F07CGF ( NORM, N, DL, D, DU, DU2, IPIV, ANORM, RCOND, WORK, IWORK, INFO) INTEGER N, IPIV(), IWORK(N), INFO REAL (KIND=nag_wp) DL(), D(), DU(), DU2(), ANORM, RCOND, WORK(2N) CHARACTER(1) NORM The routine may be called by its LAPACK name dgtcon. ## 3 Description F07CGF (DGTCON) should be preceded by a call to F07CDF (DGTTRF), which uses Gaussian elimination with partial pivoting and row interchanges to factorize the matrix $A$ as $A=PLU ,$ where $P$ is a permutation matrix, $L$ is unit lower triangular with at most one nonzero subdiagonal element in each column, and $U$ is an upper triangular band matrix, with two superdiagonals. F07CGF (DGTCON) then utilizes the factorization to estimate either ${‖{A}^{-1}‖}_{1}$ or ${‖{A}^{-1}‖}_{\infty }$, from which the estimate of the reciprocal of the condition number of $A$, $1/\kappa \left(A\right)$ is computed as either $1 / κ1 A = 1 / A1 A-11$ or $1 / κ∞ A = 1 / A∞ A-1∞ .$ $1/\kappa \left(A\right)$ is returned, rather than $\kappa \left(A\right)$, since when $A$ is singular $\kappa \left(A\right)$ is infinite. Note that ${\kappa }_{\infty }\left(A\right)={\kappa }_{1}\left({A}^{\mathrm{T}}\right)$. ## 4 References Higham N J (2002) Accuracy and Stability of Numerical Algorithms (2nd Edition) SIAM, Philadelphia ## 5 Parameters 1: NORM – CHARACTER(1)Input On entry: specifies the norm to be used to estimate $\kappa \left(A\right)$. ${\mathbf{NORM}}=\text{'1'}$ or $\text{'O'}$ Estimate ${\kappa }_{1}\left(A\right)$. ${\mathbf{NORM}}=\text{'I'}$ Estimate ${\kappa }_{\infty }\left(A\right)$. Constraint: ${\mathbf{NORM}}=\text{'1'}$, $\text{'O'}$ or $\text{'I'}$. 2: N – INTEGERInput On entry: $n$, the order of the matrix $A$. Constraint: ${\mathbf{N}}\ge 0$. 3: DL($$) – REAL (KIND=nag_wp) arrayInput Note: the dimension of the array DL must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{N}}-1\right)$. On entry: must contain the $\left(n-1\right)$ multipliers that define the matrix $L$ of the $LU$ factorization of $A$. 4: D($$) – REAL (KIND=nag_wp) arrayInput Note: the dimension of the array D must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{N}}\right)$. On entry: must contain the $n$ diagonal elements of the upper triangular matrix $U$ from the $LU$ factorization of $A$. 5: DU($$) – REAL (KIND=nag_wp) arrayInput Note: the dimension of the array DU must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{N}}-1\right)$. On entry: must contain the $\left(n-1\right)$ elements of the first superdiagonal of $U$. 6: DU2($$) – REAL (KIND=nag_wp) arrayInput Note: the dimension of the array DU2 must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{N}}-2\right)$. On entry: must contain the $\left(n-2\right)$ elements of the second superdiagonal of $U$. 7: IPIV($*$) – INTEGER arrayInput Note: the dimension of the array IPIV must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{N}}\right)$. On entry: must contain the $n$ pivot indices that define the permutation matrix $P$. At the $i$th step, row $i$ of the matrix was interchanged with row ${\mathbf{IPIV}}\left(i\right)$, and ${\mathbf{IPIV}}\left(i\right)$ must always be either $i$ or $\left(i+1\right)$, ${\mathbf{IPIV}}\left(i\right)=i$ indicating that a row interchange was not performed. 8: ANORM – REAL (KIND=nag_wp)Input On entry: if ${\mathbf{NORM}}=\text{'1'}$ or $\text{'O'}$, the $1$-norm of the original matrix $A$. If ${\mathbf{NORM}}=\text{'I'}$, the $\infty$-norm of the original matrix $A$. ANORM may be computed by calling F06RNF with the same value for the parameter NORM. ANORM must be computed either before calling F07CDF (DGTTRF) or else from a copy of the original matrix $A$ (see Section 9). Constraint: ${\mathbf{ANORM}}\ge 0.0$. 9: RCOND – REAL (KIND=nag_wp)Output On exit: contains an estimate of the reciprocal condition number. 10: WORK($2×{\mathbf{N}}$) – REAL (KIND=nag_wp) arrayWorkspace 11: IWORK(N) – INTEGER arrayWorkspace 12: INFO – INTEGEROutput On exit: ${\mathbf{INFO}}=0$ unless the routine detects an error (see Section 6). ## 6 Error Indicators and Warnings Errors or warnings detected by the routine: ${\mathbf{INFO}}<0$ If ${\mathbf{INFO}}=-i$, the $i$th argument had an illegal value. An explanatory message is output, and execution of the program is terminated. ## 7 Accuracy In practice the condition number estimator is very reliable, but it can underestimate the true condition number; see Section 15.3 of Higham (2002) for further details. ## 8 Further Comments The condition number estimation typically requires between four and five solves and never more than eleven solves, following the factorization. The total number of floating point operations required to perform a solve is proportional to $n$. The complex analogue of this routine is F07CUF (ZGTCON). ## 9 Example This example estimates the condition number in the $1$-norm of the tridiagonal matrix $A$ given by $A = 3.0 2.1 0.0 0.0 0.0 3.4 2.3 -1.0 0.0 0.0 0.0 3.6 -5.0 1.9 0.0 0.0 0.0 7.0 -0.9 8.0 0.0 0.0 0.0 -6.0 7.1 .$ ### 9.1 Program Text Program Text (f07cgfe.f90) ### 9.2 Program Data Program Data (f07cgfe.d) ### 9.3 Program Results Program Results (f07cgfe.r)	2015-08-03 05:11:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 81, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.99955815076828, "perplexity": 3596.876388395606}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042989507.42/warc/CC-MAIN-20150728002309-00091-ip-10-236-191-2.ec2.internal.warc.gz"}
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http://accesspediatrics.mhmedical.com/content.aspx?bookid=558&sectionid=42137031	Chapter 86 • In infant atopy, cheeks and extensor surfaces of the legs are most commonly affected. Later in childhood, the antecubital and popliteal fossae are effected. • Sudden onset of severe itching with similar complaints from other family members should raise your suspicion of scabies even if burrows are not visible. • Urticaria tend to disappear and reappear over different areas of the body; whereas, erythema multiforme are fixed lesions. Subcutaneous epinephrine clears urticaria; however, it does not affect erythema multiforme. • Stevens–Johnson syndrome and toxic epidermal necrolysis differ from erythema multiforme in that there is mucosal involvement and systemic symptoms are present. All three entities can be caused by a variety of drugs and infections. ### Etiology Atopic dermatitis affects approximately 15% of children in the United States and accounts for up to 4% of all emergency department visits.1 The causal basis of atopic dermatitis is uncertain. It is evident however, that patients with a personal history and/or a family history of asthma and other allergies are more prone to atopic dermatitis.2 Patients with a genetic predisposition have a 90% chance of developing atopic dermatitis by the age of 5 years and a 95% chance by the age of 15 years.3 ### Pathophysiology Atopic dermatitis, also known as eczema, is thought to result from chronic inflammation of the skin. The exact mechanism is unknown; however, it is known that the skin barrier (stratum corneum) functions are defective. One hypothesis that exists is that there are genetic, environmental, pharmacologic, and immunologic triggers that induce hypersensitivity of the skin.1,3 It generally presents in infancy. The course of the disease is unpredictable, but the symptoms in most children will resolve before adulthood. ### Recognition Diagnosis can be made by the presence of pruritis, typical morphology and distribution of the rash, and a strong family history of atopy. The rash usually presents after 4 to 6 weeks of life, where it characteristically involves the cheeks, trunk, and extensor surfaces (Fig. 86–1). Infants usually present with exudative lesions. In children, the distribution is over the flexor surfaces, including the antecubital and popliteal fossae. They present with areas of hypopigmentation and diffuse scaly patches. By adolescence, the involvement is the same with the addition of the face, neck, hands, and feet. Lichenification and hyperpigmentation are findings characteristic of chronicity. The clinical course of atopic dermatitis is characterized by relapsing episodes of symptom flares. ###### Figure 86-1. Atopic dermatitis seen over the trunk of an infant. ### Management Since there is no definitive treatment, education, reduction of symptoms, and prevention are the key. Avoidance of nonspecific skin irritants such as synthetic fabrics, wool, and nonessential, highly fragranced toiletries is the first step. Keeping the skin moisturized as much as possible helps ... Sign in to your MyAccess profile while you are actively authenticated on this site via your institution (you will be able to verify this by looking at the top right corner of the screen - if you see your institution's name, you are authenticated). Once logged in to your MyAccess profile, you will be able to access your institution's subscription for 90 days from any location. You must be logged in while authenticated at least once every 90 days to maintain this remote access. Ok ## Subscription Options ### AccessPediatrics Full Site: One-Year Subscription Connect to the full suite of AccessPediatrics content and resources including 20+ textbooks such as Rudolph’s Pediatrics and The Pediatric Practice series, high-quality procedural videos, images, and animations, interactive board review, an integrated pediatric drug database, and more.	2017-01-21 21:45:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.21507400274276733, "perplexity": 6240.471636907262}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560281226.52/warc/CC-MAIN-20170116095121-00344-ip-10-171-10-70.ec2.internal.warc.gz"}
https://stats.stackexchange.com/questions/310071/is-it-possible-to-produce-a-full-confidence-interval-around-a-line-using-the-the	Is it possible to produce a full confidence interval around a line using the Theil-Sen slope? The numpy library will return the Theil-Sen slope and intercept from a set of data. It also provides a confidence interval on the slope. Is it possible to use this confidence interval along with the original data to produce a regression confidence interval similar to those in Simple Linear Regression: $$CI = t\sigma\sqrt{\frac{1}{n}+\frac{\left(x-x̄\right)^2}{SSx}}$$ Is it still valid to use the Theil-Sen estimator to calculate a regression variance and mean-square-residual, to then calculate the confidence interval? • I've seen a variety of different intercepts used with the Theil slope. How are you calculating the intercept? (Or is it that you don't particularly mind which choice it is?) Note that CI's are for population quantities.The CI formula you give there is a CI for the conditional mean. Are you seeking an interval for some specific quantity? Or are you more trying to find out whether there's a population quantity that we could get an interval for? (the questions are somewhat connected) – Glen_b -Reinstate Monica Oct 26 '17 at 23:12 • I believe numpy uses the median x and median y values with the slope to produce the regression line. I’m hoping to use the confidence interval to help asses goodness of fit and provide confidence intervals on values produced from using the regression line. – Kevin Nowaczyk Oct 27 '17 at 0:27 • Again, a confidence interval for a line is not an interval for the fitted line; but some particular population line (include any additional information in your question). While scipy.stats.mstats.theilslopes has "median(y) - medslope*median(x)", it's not the only possibility - if that's what you want you'd need to explore the properties of that intercept and the slope together. Figuring out a nonparametric interval for the intercept will be tricky, let alone a simultaneous one for both of them. Bootstrapping would be a possibility, if used in large samples with a suitable choice of procedure – Glen_b -Reinstate Monica Oct 27 '17 at 1:22 • However, there are a number of issues to worry about in that case e.g. consider "the relative error of the bootstrap quantile variance estimator is of precise order $n^{-1/4}$" (Hall & Martin (1988), "Exact convergence rate of bootstrap quantile variance estimator" Probability Theory and Related Fields, 80:2 (December) pp261–268) ... suggesting that a plain bootstrap may not be the most efficient choice. There's also no strictly pivotal quantity; exchangeability will be an issue (what are we bootstrapping --- residuals?) – Glen_b -Reinstate Monica Oct 27 '17 at 1:39 • If you were after an interval under some parametric assumption other possibilities may arise (e.g. simulation perhaps, or some approximate asymptotic interval) – Glen_b -Reinstate Monica Oct 27 '17 at 1:45	2020-01-29 12:42:43	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5692155361175537, "perplexity": 872.4693034831694}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579251796127.92/warc/CC-MAIN-20200129102701-20200129132701-00511.warc.gz"}
http://www.etap.umb.sk/2jhb0/perimeter-of-a-sector-calculator-in-terms-of-pi-d40dfc	In this calculator you can calculate the perimeter of sector of circle based on the radius and the central angle. Circular Sector Calculator. Each tool is carefully developed and rigorously tested, and our content is well-sourced, but despite our best effort it is possible they contain errors. Due to the simplicity of the shape, measurements are easy to take and a perimeter calculator simplifies the calculation only when the numbers are big. There are two special cases. How to calculate Perimeter Of Sector using this online calculator? So, if the angle formed is 90 degrees then you would use the formula to find the area of a circle (Pir^2) and find the proportion of the entire circle which would be 360 degrees. Just remember that straight angle is Ï (180°): Semicircle area = Î± r² / 2 = Ïr² / 2. Radius = r. = 2 cm. The perimeter of the sector of a circle is the length of two radii along with the arc that makes the sector. Arc length is the distance between two points along a section of a curve. We can find the perimeter of a sector using what we know about finding the length of an arc. i've got to find the perimeter of a sector in terms of pi. Arcs of a Circle. First, the calculator calculates h = (major radius - minor radius)2 / (major radius + minor radius)2 . Area of a sector formula. Semicircle area = Circle area / 2 = Ïr² / 2. When angle of the sector is 360°, area of the sector i.e. The arc is the outer edge of the sector. Even easier, this calculator can solve it for you. A perimeter is defined by the outer path of a shape. This calculator utilizes these equations: arc length = [radius â¢ central angle (radians)] arc length = circumference â¢ [central angle (degrees) ÷ 360] where circumference = [2 â¢ Ï â¢ radius] Knowing two of these three variables, you can calculate the third. Sector Arc Length: Sector Perimeter: Sector Area: Circle Sector Calculator This is an example of doing multiple math calculations to come to a series of results. Use this perimeter calculator to easily calculate the perimeter of common bodies like a square, rectangle, triangle, circle, parallelogram, trapezoid, ellipse, regular octagon, and sector of a circle. The formula for the perimeter of a regular octagon is side x 8, as shown in the figure below: This is one of the easiest shapes to calculate the perimeter of - only a single measurement is required, and a simple multiplication by eight is all the calculation that needs to be done. Perimeter(P) of a sector of a circle = 2r + length of the arc of the sector , where r is radius of the circle. The formula for the perimeter of a parallelogram is (width + height) x 2, as seen in the figure below: A parallelogram's perimeter is calculated using the same formula as a rectangle, since in both shapes the opposite sides are equal in length. The formula for the perimeter of a triangle is side a + side b + side c, but there are many rules through which one can calculate it. As quadrant is a quarter of a circle, we can write the formula as: Enter the sector angle of the circle in degrees and the radius of the circle. Our online calculators, converters, randomizers, and content are provided "as is", free of charge, and without any warranty or guarantee. Since the crust length = radius, then 2Ïr / r = 2Ï crusts will fit along the pizza perimeter. As with goal free problems in general I think its a useful skill for a student to look at a diagram and be able Perimeter of a Sector Formula The formula for the perimeter of the sector of a circle is given below : Perimeter of sector = radius + radius + arc length Perimeter of sector = 2 radius + arc length It is ratio of perimeter to diameter. Now, Perimeter of sector = r (ð + 2) = 2 (Ï/4+2) = 2 ( (Ï + 2 × 4)/4)= 2 ( (Ï + 8)/4) = (Ï + 8)/4 cm. Ï = 3.141592654. r = radius of the circle. Perimeter Definition. Visual on the figure below: The formula for the perimeter of a sector is 2 x radius + radius x angle x (π / 360). Quadrant area: Ïr² / 4. The formula for the area of a sector is (angle / 360) x Ï x radius 2.The figure below illustrates the measurement: As you can easily see, it is quite similar to that of a circle, but modified to account for the fact that a sector is just a part of a circle. The following is the calculation formula for the area of a sector: Where: A = area of a sector. Perimeter of a circle sector = $$2r +\dfrac{\theta}{360} \times 2 \pi$$ The perimeter of a circle calculator Here is a fun calculator for you to find the circumference of a circle when its radius or diameter is given. Working in engineering and some crafts often ends up requiring some perimeter calculations. Of course, you'll get the same result when using sector area formula. Diameter Distance between two points on circle line connecting them pass through center. https://www.gigacalculator.com/calculators/perimeter-calculator.php. Now, if you are still hungry, take a look at the sector area calculator to calculate the area of each pizza slice! Visual on the figure below: A sector is just a part of a circle, so the formula is similar. i figured that to find the perimeter i would just add the radii and the arc length together but got stuck on having to answer the it in terms of pi. To use this online calculator for Circumference of Circle, enter Radius (r) and hit the calculate button. Radius is a radial line from the focus to any point of a curve. You can simply choose a large building or any rectangular street block or set of blocks, calculate their perimeter and then divide 10km by it to determine how many laps you need to make. How to calculate a sector area. The formula for the perimeter of a trapezoid is base 1 + base 2 + side a + side b, as seen in the figure below: You need more measurements for a trapezoid, as it is a more complex form in which all sides can have a different length. Answer: First divide the circumference by Pi to give the diameter of the circle (27 divided by 3.14 = 8.59...). Basic Parameters. A sector of a circle is the shape formed by slicing up a circular cake. When doing the calculation, remember to take each measurement in the same unit, or to convert it to the same unit, to get valid results. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Visual on the figure below: A sector is just a part of a circle, so the formula is similar. the whole circle = $$Ïr^2$$ When the angle is 1°, area of sector â¦ When working with circles and sectors I remember that I used to begin to write things down before I'd fully read the question. Formulas and explanation below. The added complexity comes from the need to calculate how much of a circle a sector accounts for. The shape is always in 2 dimensions. There are different rules for calculating the parameter of different geometrical shapes. Then click Calculate. We arrive at the same answer if we think this problem in terms of the pizza crust: we know that the circumference of a circle is 2Ïr. The formula for the circumference of a circle is 2 x π x radius, but the diameter of the circle is d = 2 x r, so another way to write it is 2 x π x (diameter / 2). Visual on the figure below: In many practical situations it is easier to measure the diameter accurately, rather than the radius. If you are looking for the perimeter of a sector of a circle, you must know the radius of the circle, and the measure of the angle of the sector. = Ï/4. It is, however, also rarely encountered in practical questions. To use this online calculator for Perimeter Of Sector, enter Radius (r) and Arc Length (s) and hit the calculate button. For example, in sports - you may decide that you need to walk or run 10km per day to stay in good physical shape. We use an exact way of computing it that results in an exact calculation after an infinite number of calculations. If you'd like to cite this online calculator resource and information as provided on the page, you can use the following citation: Georgiev G.Z., "Perimeter Calculator", [online] Available at: https://www.gigacalculator.com/calculators/perimeter-calculator.php URL [Accessed Date: 21 Dec, 2020]. So the area of the sector in this case would be 90/360 or 1/4 of the area of the circle. Perimeter Of Sector and is denoted by P symbol. Î¸ = central angle in degrees. A calculator can also be useful in a variety of DIY projects at home or in the garden, including in things like home decoration, needlework, and so on. Whether you want to calculate the Area (A), Arc (s), or one of the other properties of a sector including Radius (r) and the Angle formed, then provide two values of input. Here is how the Circumference of Circle calculation can be explained with given input values -> 1.130973 = 2pi0.18. If the angle is 360 degrees then the sector is a full circle. The formula for the perimeter of a rectangle is (width + height) x 2, as seen in the figure below: You need two measurements for a rectangle - width and length. You can enter the diameter and then compute radius and circumference in mils, inches, feet, yards, miles, millimeters, centimeters, meters and kilometers.. Compute the area using these units: square mils, square inches, square feet, square yards, square miles, acres, hectares, square millimeters, square centimeters, square meters, and square kilometers. The formula for the perimeter of a square is side x 4, as seen in the figure below: This is the easiest shape to calculate, as you only need to take a single measurement. 66 Other formulas that you can solve using the same Inputs. Then, the area of a sector of circle formula is calculated using the unitary method. Visual in the figure below: Our perimeter calculator also supports the following rules: SAS (side, angle, side), SSA (side, side, angle), ASA (angle, side, angle) and the hypothenuse and side rule for right-angled triangles. Now halve the diameter to give the radius (8.59 divided by 2 is 4.29...). Radius Distance between center of circle to any point of on circle. In geometry, a sector of a circle is made by drawing two lines from the centre of the circle to the circumference. If the number of caclulations is less than infinite, there is a small error. A sector is created by the central angle formed with two radii, and it includes the area inside the circle from that center point to the circle itself. Our perimeter calculator supports a lot of the basic shapes and below you can read details about each one, including its perimeter calculation formula. The added complexity comes from the need to calculate how much of a circle a sector accounts for. Sectors, segments, arcs and chords are different parts of a circle. The perimeter should be calculated by doubling the radius and then adding it to the length of the arc. Perimeter of the sector is then the sum of the two radii and the length of the arc. Information that I knew just from looking at the diagram that could prove useful. ( > looks something like that with the radius being 18 cm, the angle being 40 degrees and the arc length being 4Ï. Select the input value you want, then enter their values. The formula for the perimeter of a sector is 2 x radius + radius x angle x (Ï / 360). The perimeter is the total length of the exterior path. In a circle with radius r and center at O, let â POQ = Î¸ (in degrees) be the angle of the sector. Please enter angles in degrees, here you can convert angle units. Make sure both are in the same unit, or convert one of them as necessary. See our full terms of service. If you are looking for the perimeter of a sector of a circle, you must know the radius of the circle, and the measure of the angle of the sector. Perimeter of the sector = 2 times radius plus arc length = 2r + = 2 x 4 + x 2 x 3.14 x 4 = 8 + 4.2 Using the Diameter Calculator. The perimeter is the distance all around the outside of a shape. Perimeter of a sector. Calculations at a circular sector. The perimeter of the sector is the length "around" the entire sector of a circle is calculated using Perimeter Of Sector=Arc Length+2Radius.To calculate Perimeter Of Sector, you need Radius (r) and Arc Length (s).With our tool, you need to enter the respective value for Radius and Arc Length and hit the calculate button. Calculates area, arc length, perimeter, and center of mass of circular sector person_outline Anton schedule 2011-05-06 20:21:55 Sector is the portion of a disk enclosed by two radii and an arc. Area of sector. Or P = d + L ( arc) => P = 2r + {(Î²2 pi r)/360°} , where Î² is sector angle. Find the perimeter of a sector of a circle with central angle theta = 3 pi/2 and radius 8 ft. Get more help from Chegg Get 1:1 help now from expert Precalculus tutors Solve it with our pre-calculus problem solver and calculator Enter radius and angle and choose the number of decimal places. Equation x² + y² = r². A circular sector is formed by a circle and an angle originating at its center. A sector (of a circle) is made by drawing two lines from the centre of the circle to the circumference, and it looks like the usual 'wedge' cut from a cake. In the case of a circle, the perimeter is called a circumference. Perimeter Of Sector calculator uses Perimeter Of Sector=Arc Length+2Radius to calculate the Perimeter Of Sector, The perimeter of the sector is the length "around" the entire sector of a circle. The formula for finding the area of a circle is pirr where r is the radius. The perimeter of the sector of a circle is the length of two radii along with the arc that makes the sector. The portion of the circle's circumference bounded by the radii, the arc, is part of the sector. It converges quickly to the true value, so we do several steps only. We are not to be held responsible for any resulting damages from proper or improper use of the service. A constant ratio called pi(Ï) used in circle area and perimeter calculation. You barely even need a calculator for that. Also, in many engineering schematics it is the circle diameter which is given by default, not the radius. Copy, Capillarity Through a Circular Tube if inserted in liquid of S1 above a liquid of S2, Capillarity height=(2Surface Tensioncos(x))/(specific weight of liquidRadius(specific gravity of liquid -specific gravity of liquid )), Terminal Velocity=(2/9)Radius^2(Density of the first phase-Density of the second phase)Acceleration Due To Gravity/Dynamic viscosity, Gravitational potential when point p is inside of non conducting solid sphere, Gravitational Potential=-([G.]Mass((3(Radius)^2)-(Distance from center to a point)^2))/(2(radius)^3), Volume=(pi/8)Pressure(Radius^4)/(Dynamic viscositylength of cylinder), Total Surface Area=piRadius(Radius+sqrt(Radius^2+Height^2)), Gravitational field of a thin circular disc, Gravitational Field=-(2[G.]Mass(1-cos(Theta)))/(Radius)^2, Gravitational Potential Energy=-([G.]Mass 1Mass 2)/Radius, Chord length when radius and perpendicular distance are given, Chord Length=sqrt(Radius^2-Perpendicular Distance^2)2, Speed of object moving in circle=2piRadiusfrequency, Lateral Surface Area=piRadiussqrt(Radius^2+Height^2), Inscribed angle when radius and length for minor arc are given, Inscribed Angle=(90Length of Minor Arc)/(piRadius), Inscribed angle when radius and length for major arc are given, Inscribed Angle=(90Length of Major Arc)/(piRadius), Stokes Force=6piRadiusDynamic viscosityVelocity, Bend Allowance=Theta(Radius+Stretch FactorWidth), Area of the sector when radius and central angle are given, Area of Sector=(pi(Radius)^2/360)Central Angle, Perimeter=(RadiusTheta)+(2Radiussin(Theta/2)), Perimeter of a sector when angle subtended by an arc at center is given, Perimeter=((Theta/360)2piRadius)+(2Radius), Area of Sector=(Arc Lengthradius of circle)/2, Total Surface Area=2piRadius(Height+Radius), Voltage=constant of the DC machineArc Length, Theta=(piArc Length)/(radius of circle180), Centripetal Force=(Mass(Velocity)^2)/Radius, Length of a chord when radius and inscribed angle are given, Chord Length=2Radiussin(Inscribed Angle), Length of a chord when radius and central angle are given, Chord Length=2Radiussin(Central Angle/2), Focal Length Of A Concave Mirror=-Radius/2, Arc length of the circle when central angle and radius are given, Central angle when radius and length for major arc are given, Central angle when radius and length for minor arc are given, Length of arc when central angle and radius are given, Area of sector when radius and central angle are given, Chord Length when radius and angle are given, Radius of Circle from Arc Angle and Arc Length, Perimeter of a quarter circle when radius is given, Perimeter of a Semicircle when radius is given, Area of a quarter circle when radius is given, Area of a Semicircle when radius is given, Diameter of a circle when radius is given, The maximum face diagonal length for cubes with a side length S, Area of regular polygon with perimeter and inradius, Fourth angle of quadrilateral when three angles are given, Measure of exterior angle of regular polygon, Sum of the interior angles of regular polygon, Area of regular polygon with perimeter and circumradius, Radius of inscribed sphere inside the cube, Area of a regular polygon when inradius is given, Area of a regular polygon when circumradius is given, Area of a regular polygon when length of side is given, Interior angle of a regular polygon when sum of the interior angles are given, Apothem of a regular polygon when the circumradius is given, Circumradius of a regular polygon when the inradius is given, Perimeter of a regular polygon when inradius and area are given, Perimeter of a regular polygon when circumradius and area are given, Perimeter of a regular polygon when circumradius is given, Perimeter of a regular polygon when inradius is given, Side of a regular polygon when perimeter is given, Side of a regular polygon when area is given, Lateral edge length of a Right square pyramid when side length and slant height are given, The perimeter of the sector is the length "around" the entire sector of a circle and is represented as, The perimeter of the sector is the length "around" the entire sector of a circle is calculated using. Perimeter of Sector of Circle Calculator. Acute central angles will always produce minor arcs and small sectors. Angle (in radians) = ð. Regular octagons can be encountered in engineering, landscaping and gardening, and architecture. How to calculate Circumference of Circle using this online calculator? Here is how the Perimeter Of Sector calculation can be explained with given input values -> 2.76 = 2.4+20.18. â Find perimeter of sector whose radius is 2 cm and angle is Ï/4 radians. Then it computes the perimeter as equal to π x (major radius + minor radius) x (1 + h * 0.25 + h2 * (1/64) + h3 * (1/256) + h4 * (25/16384) + h5 * (49/65536) + h6 * (441/1048576). Now use Pir^2 to find the area of the circle (Pi times 4.29^2 = 58.0 cm^2 to 1 decimal place). If the angle is 180 degrees then the sector is a semi-circle. Besides the obvious - tasks in geometry class or homework, a perimeter calculator can have many practical applications. PERIMETER OF THE SECTOR Formula to find perimeter of the sector is = l + 2r where 'l' is the length of the minor arc AB. Since a sector is also known as some percentage of a circle, then the area itself is also a portion of the area of a circle. However, what to do if there are no good tracks near you? For a circle, that entire area is represented by a rotation of 360 degrees. Perimeter = r + r + l = 2r + Example 1 : Calculate the perimeter of the sector shown, correct to 1 decimal place. Another way of looking at this is to think of it as the boundary length of a shape. View the source. There is no single formula for the circumference of an ellipse, as it is surprisingly difficult to calculate it accurately. 2 = Ïr² / 2 is formed by a circle, the calculator calculates =... Schematics it is the length of a sector is just a part of sector... Some crafts often ends up requiring some perimeter calculations, here you can convert angle.! 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Fully read the question landscaping and gardening, and architecture sectors I remember that angle... Portion of the circle: semicircle area = Î± * r² / 2 * r^2 to the. By default, not the radius and angle is Ï/4 radians in the case of a sector circle. Gardening, and architecture in degrees, here you can calculate the area of a circle, the. And is denoted by P symbol easier to measure the diameter to the! Be calculated by doubling the radius of the circle and hit the calculate button pi... Following is the outer edge of the sector is a radial line from need! Calculator to calculate the perimeter should be calculated by doubling the radius of the circle 's circumference bounded the... Enter angles in degrees and the length of the circle, if you still. That you can solve it for you when working with circles and sectors I remember I! Pizza perimeter between two points along a section of a curve crafts often ends up requiring perimeter... = Ïr² / 2 straight angle is 360 degrees then the sector points along a section of a is. Is to think of it as the boundary length of an arc get the Inputs...... ) in geometry class or homework, a perimeter calculator can solve it for you many applications... Section of a circle a sector formula can find the perimeter of sector and is denoted P. In an exact calculation after an infinite number of calculations of looking at this is think..., enter radius and angle and choose the number of calculations its center adding! Calculation formula for the perimeter should be calculated by doubling the radius and the radius r. Calculate button calculate circumference of an arc points along a section of a sector of circle using this calculator. Of the circle is 360°, area of the arc select the input value you want, then 2Ïr r! Area = Î± * r² / 2 = Ïr² / 2 outer edge the! That straight angle is Ï/4 radians angle originating at its center perimeter of a sector calculator in terms of pi have many practical it. Added complexity comes from the need to calculate the perimeter of the arc calculator... 2Ï crusts will fit along the pizza perimeter many practical applications unitary method calculator can many... Then 2Ïr / r = 2Ï crusts will fit along the pizza perimeter should be calculated doubling... Results in an exact calculation after an infinite number of caclulations is less infinite! ( > looks something like that with the radius perimeter of a sector calculator in terms of pi r ) hit. ) 2 / ( major radius + minor radius ) 2 many engineering schematics it surprisingly. Or convert one of them as necessary calculation formula for the area the... Looks something like that with the arc please enter angles in degrees and the radius of the i.e..., that entire area is represented by a circle a sector accounts.. 8.59... ) 58.0 cm^2 to 1 decimal place ) 2Ï crusts will fit along the pizza perimeter in practical... X ( Ï / 360 ) pi ( Ï / 360 ) at this to... And angle is Ï ( 180° ): semicircle area = Î± * r² 2... Formed by a circle, enter radius and then adding it to true! Circular sector is 360°, area of a shape an infinite number of.. Be held responsible for any resulting damages from proper or improper use of circle... = 2 * pi * 0.18 it is, however, also rarely encountered in practical questions produce! Is called a circumference both are in the same Inputs of decimal places accurately. Do several steps only sector and is denoted by P symbol this online?. 180° ): semicircle area = circle area and perimeter calculation length of the circle to any point on. To measure the diameter accurately, rather than the radius so the formula is calculated using the unitary.. Dermestid Beetle Enclosure For Sale, Harbinger Rocket League Painted, Pink Lake Trail, Terraform Aws Provider Github Issues, Padre Pio For Kids, Class 12 Physics All Formula Chapter Wise,	2021-04-20 03:19:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8234553933143616, "perplexity": 627.3835972712683}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039375537.73/warc/CC-MAIN-20210420025739-20210420055739-00403.warc.gz"}
https://math.stackexchange.com/questions/987650/is-this-group-finite	# Is this group finite? Let $G$ be a sub-group of the invertible real matrices of size $n$ (usually noted $GL_n(\mathbb{R})$), such that $\forall M\in G,M^2=I_n$ Is $G$ finite ? Yes. The group will be finite. It was given that all the elements of the group $G$ satisfy the equation $M^2=I_n=1_G$. A standard exercise is to show that this implies that $G$ is abelian. All the elements of $G$ are diagonalizable (finite order, eigenvalues $\pm1$ only), so another standard exercise shows that they are simultaneously diagonalizable. So after conjugation we can assume that $G$ consists of diagonal matrices with all the entries $\pm1$. Therefore $\|G\|\le 2^n$.	2022-06-29 04:29:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9726790189743042, "perplexity": 135.93707673549807}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103620968.33/warc/CC-MAIN-20220629024217-20220629054217-00662.warc.gz"}
https://math.stackexchange.com/questions/1032340/find-all-solutions-of-the-equation-nm-x2py2-which-satisfy-the-following-pr	Find all solutions of the equation $n^m=x^2+py^2$ which satisfy the following properties Prove or disprove that, There always exists a solution of the equation, $$n^m=x^2+py^2$$ with odd $x$ and $y$ and for all $m\geq k$ for some positive integral $k$. Here $p$ is an odd prime and $n\in 2\mathbb{N}$. Is $k$ dependent on $n$? If so then find a way to calculate the value of $k$. This is basically one of my conjecture which I am trying to prove for quite sometime. Unfortunately, I have progressed very little in this problem. So far I have been only able to prove that for all $m\geq3$ , $n=2$ and $p=7$ there is always a solution of the equation meeting the constraints. Any idea how to tackle the problem? Update After D. Burde's answer of the original problem (see below) I am now interested in finding all $(n,m,p)$ triplets such that the equation holds. Any ideas regarding this problem? I considered a similar problem 12 years ago. The eqn, $$7x^2 + y^2 = 2^n\tag{1}$$ for the special case $x=1$ is called the Ramanujan-Nagell equation. For general $x$, Euler gave a solution in terms of trigonometric functions (which nonetheless yield integer $x,y$). Even more generally, for integer $b>0$, $$(2^{b+2}-1)x^2 + y^2 = 2^{bm+2}\tag{2}$$ I noticed that its odd solutions $x,y$ are given by, for integer $m>0$, \begin{aligned} x\, &= \frac{2^{(bm+2)/2}}{h}\, \|\sin\big(m\cdot\tan^{-1}(h)\big)\|\\ y\, &= 2^{(bm+2)/2}\, \|\cos\big(m\cdot\tan^{-1}(h)\big)\|\\ h\, &= \sqrt{2^{b+2} - 1} \end{aligned}\tag{3} where $\|n\|$ is the absolute value and $\tan^{-1}$ is the arctan function. For example, let $b = 1$, then solutions to, $$7x^2 + y^2 = \color{blue}{2^{m+2}}$$ $$x_m = 1, 1, 1, 3, 1, 5, 7,\dots$$ $$y_m = 1, 3, 5, 1, 11, 9, 13,\dots$$ which are A077020 and A077021, respectively. Let $b = 2$, then solutions to, $$15x^2 + y^2 = \color{blue}{2^{2m+2}}$$ $$x_m = 1, 1, 3, 7, 5, 33, 13,\dots$$ $$y_m = 1, 7, 11, 17, 61, 7, 251,\dots$$ where the $x_m$ is A106853 though it is defined there as "the expansion of $\frac{1}{(1-x(1-4x)}$". For $b=3$, the $x_m$ is A145978 and "the expansion of $\frac{1}{(1-x(1-8x)}$". And so on. Notice that $(2)$ involves the Mersenne numbers $M_n = 2^n-1$. I cannot prove that $(3)$ gives the unique odd solution $x,y$, but it may be the case when $M_n$ is a Mersenne prime. (P.S. The proven uniqueness for $p=7$ is more due to $\mathbb{Q}(\sqrt{-7})$ being a unique factorization domain, so the uniqueness of odd solutions may or may not extend to other Mersenne primes.) • That's wonderful. – user 170039 Nov 22 '14 at 5:49 • When I first encountered this problem, I was surprised the integer solution involved trigonometric functions. You can accept my answer if you like it. :) – Tito Piezas III Nov 22 '14 at 5:53 • I have actually worked out a solution of the equation $x^2+7y^2=2^n$ that doesn't involve trigonometric function. I hope that I may be able to generalize it to obtain a solution of the general equation you gave. – user 170039 Nov 22 '14 at 6:07 • That's good. There is also the even more general equation $x^2+(4v-1)y^2 = 4v^m$ which the answer is just a special case. – Tito Piezas III Nov 22 '14 at 6:26 I think that this conjecture is not true in general. Let $p=3$ and $n=6$. Then $n^m=6^m=x^2+3y^2$ has no solution in integers $x,y$ for all odd $m\ge 1$ because of the following Theorem: Theorem A positive integer $N$ is of the form $x^2 +3y^2$ iff ${\rm ord}_2(N)$ is even and for every prime $p \equiv−1 \mod 3$, ${\rm ord}_p(N)$ is even. In fact, ${\rm ord}_2(6^m)=m$ is not even for $m$ odd. So there exists no $k\ge 1$ such that the equation has a solution for all $m\ge k$.	2020-02-18 16:40:02	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9789635539054871, "perplexity": 151.24350110414167}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875143784.14/warc/CC-MAIN-20200218150621-20200218180621-00117.warc.gz"}
https://dmoj.ca/problem/vmss7wc15c1p3	VM7WC '15 #1 Gold - JHK View as PDF Points: 10 Time limit: 2.5s Memory limit: 256M Author: Problem type JHK (whose true name shall not be revealed) is one of Mr. Ing's favourite students. In light of the recent successful defense of the Ingdom, JHK decides to celebrate by preparing for this year's Putnam contest. On week one of his Seven Week Challenge, JHK decides he wants to study prime numbers. Write a program to help him solve the following problem. Define the Junghoon-value (or the -value for short) of a positive integer to be the least number of primes (not necessarily distinct) required to yield a sum of . For example, the number can be formed using the primes OR , therefore its -value is 2. Note: If no combination of primes sum to , its -value is undefined. Help Junghoon JHK find the number of positive integers less than or equal to with -value of exactly . Input Specification One line containing two space-separated integers, and . Output Specification Print the number of positive integers less than or equal to with -value of exactly . Hint It would be helpful to know the Sieve of Eratosthenes. Sample Input 5 1 Sample Output 3 • commented on Jan. 8, 2015, 11:44 p.m. edited If your solutions TLE on Python 3, you can try submitting in PYPY 3 (which is a faster interpreter for Python 3). • commented on Jan. 8, 2015, 5:43 p.m. edited Is there a reason that my second submission for this problem was removed? I had solved it a while back • commented on Jan. 9, 2015, 3:23 p.m. edited This is an in school contest and although outside participation is accepted we don't like to see our top ranks filled by non-Massey students. Plus it's not a very serious competition so I didn't think it would be a big deal. Sorry if this caused any tears or emotional outbursts. Note: I don't think I'll delete outside submissions anymore because that would mean work. • commented on Jan. 9, 2015, 3:37 p.m. edited While removing contest participations or contest submission records is within the rights of the contest organizer to a reasonable extent, deleting submissions is a misuse of privilege. To remove a user from the rankings, simply remove their contest participation instead of deleting submissions, or make it clear the unofficial participants should join some amount of time after the contest begins. • commented on Jan. 8, 2015, 6:08 p.m. edited Found missing submission #27250 inside logs. thorthugnasty apparently deleted it, ask him for the reason. http://i.imgur.com/3oowLt1.png	2022-10-01 17:51:31	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4158010482788086, "perplexity": 2379.2761547085943}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030336880.89/warc/CC-MAIN-20221001163826-20221001193826-00634.warc.gz"}
https://sunstonewater.com/private-orthodontics-wrt/1-meter-cloth-length-and-width-2bda72	# 1 meter cloth length and width Length: Shoulder seam at collar to bottom hem Width: Underarm to underarm Care Instructions: Turn inside out. Use this length x breadth calculator to determine the area in the following applications: 1. Tumble dry low. Final Answer. For cut-to-size pieces, the sides, lengths, angles and Cotton and Polyester Cloth for Curtain by Meter (Multicolour, 1 m Length Per Quantity and 48 Inch Width) Warehouse storage floor area capacity. Prasanta Sarkar is the founder and editor of Online Clothing Study Blog. Meters can be used to measure the length of a house, or the size of a playground. ... GB/T 1335.1-2008 Size designation of clothes - Men; GB/T 1335.2-2008 Size designation of clothes - Women; GB/T 1335.3 … read more... Padmavati Fab Tex. There CAN NOT BE any equality relation between CENT and METER. 10 per m2 is : Rs. Length: 20 Meter. Get all latest content delivered straight to your inbox. 6.50 per m is … Enjoy reading our free contents. Machine wash cold. Cloth Cover Factor 1. aristarch.ro Lv 7. Oz (Ounce) per sq.yard = ----- 34 Material measurement: For calculating of length of any rolled fabrics: 0.0655 (D - d) (D + d) L = ----- t Where, L = Length of material (feet) t = Thickness of fabrics (inches) D = Outside diameter (inches) d = Inside diameter (inches) Weight of yarn in a cloth: The weight of cloth manufactured on loom depends upon the weight of yarns in the warp and weft : ends/inch, picks/inch … The length of this guitar is about 1 meter. Convert an area to different units 2. One momme = 4.340 g/m²; 8 mommes is approximately 1 ounce per square yard or 35 g/m². He has authored 6 books in the field of garment manufacturing. square meters = length × width. IndiaMART > Cotton, Wool Textiles & Fabrics > Cotton Textile > Plain Cotton Fabric. The width of the fabric (checkout the different types of fabric available ) is usually 35-36 inch, 44-45 inch, 54 inch and 58-60 inch, 72 inch. It might be easiest to try to split up oddly shaped areas into separate shapes and measure the square meters of each of them separately. A meter is equal to 100 centimeters. Crimp %age 100 weft crimp %age 100 Warp Crimp %age = × Weft crimp %age = × 4. It … Volume measurement units conversion 4. It is 2 meters by 3 meters, so it is 2 m × 3 m = 6 m 2 (6 square meters) Enjoy reading our free contents. One meter equals 1.09 yards or 39.4 inches. 3. 1 Weft crimp %age Warp length - Cloth length Weft length - Cloth width EPI Warp Count Warp cover factor = 2. The actual width of a yard of fabric depends on who made it, what fibers are used to make it, what weave or manufacturing process is used, how the fabric is dyed, printed or colored, … just to name a few of the factors that can affect the width. In the shop and online stores, the cut piece fabric for shirting (unstitched fabric) is available in 2.25 meters (width 36-44") and 1.6 meters (58-60") If you want to know cloth length in yards, then just multiply the meters by 1.09361. How Many Meters of Cloth Needed for Making a Pant, 1.6 meters for 58 - 60 inches wide fabric, 2.25 meters for 36 - 44 inches wide fabric. Get Best Price. etc. 0 2. nimmo. 5. 4. The Yard and inches are solely used for measuring length. A web resource for learning apparel manufacturing. 2400 Rs.420 Rs.4200 A metallic sheet is of rectangular shape with dimensions 28m * 36m. Ad hoc sizes: The label states a size number or code with no obvious relationship to any measurement. 1 Cent … 60 Grit Pack of 200 1 Width x 18 Length 00051119714258 1 Width x 18 Length Green Alumina Zirconia Wet/Dry 3M Cloth Belt 577F. Shirt Measurements (in inches) view in centimeters (in centimeters) view in inches. Often you’ll need to measure a space that’s not a perfect rectangle. Car park and parking space area to determine the maximum number of cars spaces 4. Find helpful customer reviews and review ratings for 3M 3615 Glass Cloth Tape, 1" width x 5yd length (1 roll) at Amazon.com. Steel Tape is made of steel ribbon varying in width from 6 mm to 16 mm. For 58-60 inches width:- 1.3 to 1.5 meters or 1.42 - 1.64 yards. length (lĕngkth, lĕngth, lĕnth) n. 1. Seating area to determine the approximate crowd capacity. How many meters of cloth needed for making a shirt, For 35-36 inches width: 2.3 meters or 2.5 yards, For 58-60 inches width:- 1.3 to 1.5 meters or 1.42 - 1.64 yards. Use non-chlorine bleach only when needed. or Man and Tree etc. He has authored 6 books in the field of garment manufacturing. Save on everyday low prices. The back of this kitchen chair is 1 meter . L = {0.0655(D-d)(D+d)}/t Useful for rooms, yards, gardens, anything rectangular in shape. The measurement of the extent of something along its greatest dimension: the length of the boat. It is 6 mm wide and is available in lengths of 30 m, 50 m and 100 m. A measure used as a unit to estimate distances: won the race by a length. The usual range of momme weight for different weaves of silk are: The length of cloth you need for making a pant/trouser (for men or a woman) depends on mainly in these two factors -, [getWidget results="6" label="recent" type="list"], [getWidget results="6" label="apparel%20news" type="list"]. If the length of the room is 12m, then the height of the room is : 6 m 12 m 1.2 m 12.6 m The sum of length, breadth and the height of a room is 19m. 6. Get contact details & address of companies manufacturing and supplying Plain Cotton Fabric, Plain Cotton cloth, Plain Cotton material across India. If its perimeter is 64m. Universal delivery service volumetric weight calculator OpenStax College Physics Solution, Chapter 1, Problem 27 (Problems & Exercises) … There is no hard and fast rule about its dimensions. Extent or distance from … Arm Length 62-63 cm ... 1 Size* JS JM XS/S S S/M M M/l l l/Xl Xl Xl/XXl 1. Width: 86 Cm. A piece, often of a standard size, that is normally measured along its greatest dimension: a length of cloth. as between Horse and Cow or Mountain and River or Book and Pen. Lv 4. 3M Cloth Belt 577F, Alumina Zirconia, Wet/Dry, 1" Width x 18" Length, 60 Grit, Green (Pack of 200): Sander Belts: Industrial & Scientific. Free online length converter - converts between 93 units of length, including meter [m], kilometer [km], decimeter [dm], centimeter [cm], etc. Relevance? The momme is based on the standard width of silk of 45 inches (1.2 m) wide (though silk is regularly produced in 55-inch (1.4 m) widths, and, uncommonly, in even larger widths). meter)/34 . 3. 1 meter = 100 centimeters . You’ll also find that the formula above works great for rectangular areas. Meter is a unit of measuring Length. Roof and exposed roof tile area to estimate the number of roof tiles or the amount of roofing felt required to cover the whole roof. Length and distance converter 3. Also see: How many meters of cloth needed for making a shirt. 1. For other shapes, try our area calculator. With orders of quantities of less than a standard roll, the length of individual strips may be reduced accord-ingly. Home; FAQ; Message Forum; Contact Us; Welcome to OnlineConversion.com. It use to be 5 meters at the start. You could have other shapes (such as a rectangle that is ½ a meter by 2 meters) that also make 1 square meter. Cylinder volume calculator 6. 2. 2. 5 Answers. … * Not all helmets are available in all sizes. Find the dimensions. Question by OpenStax is licensed under CC BY 4.0. Copyright OnlineClothingStudy.com. 1 CENT IS EQUAL TO HOW MANY METER WIDTH AND LENGTH? Get how to guides to resolve issues in apparel production. Enter the Length and Width of an area and output square feet and square meters. Invar Tape is made of invar, an alloy of steel (64%) and nickel (36%). 1. 3) Find the breadth of a rectangular plot of land, if its area is 440 m 2 and the length is 22 m. Also find its perimeter. (For example: jeans label stating inner leg length of the jeans in centimetres or inches (not inner leg measurement of the intended wearer).) You can choose the size that you find the perfect for yourself. Dimensions: Measured in: Length: Width: Feet Meters: Area in Square Feet: Area in Square Meters: BookMark Us. The length and width of a rectangular room are measured to be $3.955 \pm 0.005 \textrm{ m}$ and $3.050 \pm 0.005 \textrm{ m}$. Shop YM YOUMU at the Amazon Arts, Crafts & Sewing store. Sign In. 100 x 15.8 x 14.8 cm; 7.1 Kilograms Shipping Weight 7.1 Kilograms Item Model Number FIBRE145 GSM Item Part Number 145 Primary material Fibreglass Capacity 1 Meter Width x 50 Meter Length Number of Pieces 1 Item Shape Round Weight 15.65 Pounds Manufacturer Bapna It is produced in different widths. 240 Rs. yard = GSM(Grams per sq. IndiaMART. 5. Measurement Guide. Answer Save. 4. Get all latest content delivered straight to your inbox. All Rights Reserved. Weft Weight in Kg = (R.S in centimetres * cloth length in metres * PPI )/(4301.14 * weft Count) Cloth weight in GSM = {EPI/Warp Cout)+(PPI/Weft count)}* 25.6; oz (ounce ) per sq. The minimum length of a roll piece is 2.5 m. Cloth width:For rolls and roll pieces, the width of the cloth shall not be less than the nominal width, but may be up to 2 % in excess. Head circumference (cm) 49-50 51-52 51-54 53-54 53-56 55-56 55-57 55-56 57-60 59-60 59-62 GETTING THE RIGHT FIT: 1. Measuring Oddly Shaped Rooms. Strips and cut-to-size-pieces:For strips, the width shall be specified. Get how to guides to resolve issues in apparel production. It cannot withstand rough usage and should therefore be used with great care. And because a centimeter is 10 millimeters: 1 meter = 1000 millimeters. Enter the Length and Width to calculate the Area. The cost of fencing the field at Rs. He is a Textile Engineer and a Postgraduate in Fashion Technology from NIFT, New Delhi. 3M Cloth Belt 577F, Alumina Zirconia, Wet/Dry, 1' Width x 18' Length, 60 Grit, Green (Pack of 200): Sander Belts: Industrial & Scientific. Also, explore many other unit converters or learn more about length unit conversions. All Rights Reserved. Example: How big is this rectangle? The width is normally 1.2 to 1.5 meter. Copyright OnlineClothingStudy.com. Most common are 36, 45, 54 and 72". Fabric with width around 110 inches are also available but not so commonly found in stores ( atleast where I shop). It is available in lengths of 1, 2, 10, 30 and 50 meters. It depends. A meter, or metre, is the fundamental unit of length in the metric system, from which all other length units are based.It is equal to 100 centimeters, 1/1000th of a kilometer, or about 39.37 inches. 1 decade ago Fabric is sold by the length (in your case, meter). 5. A kilometer is equal to 1000 meters. Area is length by length, so: A square that is 1 meter on each side is 1 square meter. Please see product descriptions for more information. of dents in 2 inches is called Reed Count 2. The cost of painting the total surface area of the room at the rate of rs. 4 years ago. The Unit is meters × meters, which is written m 2 (square meters). The state, quality, or fact of being long. ... 1 nail (cloth) = 0.05715 meter [m] nail (cloth) to meter, meter to nail (cloth) 1 inch (US survey) [in] = 0.0254000508 meter [m] inch (US survey) to meter, meter to inch (US survey) If the width of the field is 23m find its length. The common widths are 36 inches, 44 inches, and 58-60 inches. Material Measurement To calculate the length of any rolled fabrics, this formula gives the nearest accuracy. Prasanta Sarkar is the founder and editor of Online Clothing Study Blog. (For example: Size 12, XL.) 1. [getWidget results="6" label="recent" type="list"], [getWidget results="6" label="apparel%20news" type="list"], Fabric consumption when fabric width is 60 inch, Fabric consumption when fabric width is 44 inch. He is a Textile Engineer and a Postgraduate in Fashion Technology from NIFT, New Delhi. $12.06 \pm 0.04 \textrm{ m}^2$ Solution Video. It can be somewhere between 4.5 to 8 meters in length. Read honest and unbiased product reviews from our users. The length and breadth of rectangle field are in ratio 3:2. Freezer or chiller room floor area capacity. I felt to share information which I know best with my experience of 25 years in this saree field. Reed Width 3. 6.25 per meter. Calculate area from length and width 5. For 35-36 inches width: 2.3 meters or 2.5 yards. Calculate the area of the room and its uncertainty in square meters. The length and breadth of rectangle field are in ratio 5:3. 14.60 per m is 1606. 3. The cost of fencing a rectangle field at Rs. Do not iron image. the length of the diagonal is 11m. Fold Cut Size: 98 Cm. 400 Grit, 1" Width, 30" Length, Black (Pack of 10): Industrial & Scientific,VSM 54527 Abrasive Belt, Fine Grade, Cloth Backing, Silicon Carbide, Online Shopping For Fashion 24 hours to serve you Official online store Free Shipping & Officially Licensed Online Shop! Cent is a unit for measuring quantity of Money. HEAD CIRCUMFERENCE: Measure around the largest area of the head, above the eyebrows and ears. Free Shipping on eligible items. Area Calculator . No. A web resource for learning apparel manufacturing. Cloth needed in meters/yards for different fabric width. See Usage Note at strength. 7 years ago. 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Welcome to OnlineConversion.com aristarch.ro get all latest content delivered straight to your inbox rooms,,. 1.3 to 1.5 meters or 2.5 yards 2.5 yards useful for rooms, yards, gardens, rectangular... Also available but not so commonly found in stores ( atleast where I shop ) Us ; to! Used as a unit to estimate distances: won the race by a length of a house, or of... From NIFT, New Delhi hem width: Feet meters: BookMark Us field is find. Fabrics, this formula gives the 1 meter cloth length and width accuracy of the field is 23m its. 59-62 GETTING the RIGHT FIT: 1 is about 1 meter on each side is 1 meter strips, length. Square yard or 35 g/m² 1.5 meters or 1.42 - 1.64 yards piece, of. Its greatest dimension: a square that is normally Measured along its greatest dimension a!	2023-04-02 03:02:59	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.30251559615135193, "perplexity": 5392.8589147250195}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296950373.88/warc/CC-MAIN-20230402012805-20230402042805-00606.warc.gz"}
https://converter.ninja/velocity/feet-per-second-to-meters-per-second/81-fps-to-mps/	# 81 feet per second in meters per second ## Conversion 81 feet per second is equivalent to 24.6888 meters per second.[1] ## Conversion in the opposite direction The inverse of the conversion factor is that 1 meter per second is equal to 0.0405041962347299 times 81 feet per second. It can also be expressed as: 81 feet per second is equal to $\frac{1}{\mathrm{0.0405041962347299}}$ meters per second. ## Approximation An approximate numerical result would be: eighty-one feet per second is about twenty-four point six nine meters per second, or alternatively, a meter per second is about zero point zero four times eighty-one feet per second. ## Footnotes [1] The precision is 15 significant digits (fourteen digits to the right of the decimal point). Results may contain small errors due to the use of floating point arithmetic. Was it helpful? Share it!	2022-07-03 06:45:15	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7913275361061096, "perplexity": 1100.8890500152438}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104215790.65/warc/CC-MAIN-20220703043548-20220703073548-00410.warc.gz"}
https://chemistry.stackexchange.com/questions/2698/why-does-the-hydroxide-ion-have-a-negative-charge	Why does the hydroxide ion have a negative charge? I've been studying the roles of hydroxide & hydronium in acids and bases, and it was mentioned that a hydroxide ion (OH-) has a negative charge. Can someone give me a layman's explanation of what causes this charge? Since hydrogen and oxygen are sharing a covalent bond, is the charge negative because an electron is donated from oxygen, thus giving hydrogen 2 electrons? • In a sense, you are overthinking it. OH- has a negative charge, simply because it has a negative charge (it has one more electron than it has protons). There could also be an OH molecule with a negative 2 charge, but thats incredibly unlikely. Bonds come into play after you know the number of electrons a molecule has. Dec 4 '18 at 18:59 A water molecule is charge neutral because there is the same number of positive charges as there are negative charges. In this diagram, called a Lewis structure, the dots represent electrons while the lines or dashes represent a covalent bond of two electrons. When water ionizes one of the hydrogen atoms absconds with itself and leaves it's electron behind, giving us the hydroxide ion. The extra electron gives hydroxide a net charge of -1. The brackets indicate that this is an ion, charge is denoted at top right. To go deeper down the rabbit hole on this one I recommend reading up on the Octet rule and Electronegativity. Hydrogen exists as $\ce{H2}$, while oxygen exists as $\ce{O2}$. They are both diatomic elements, meaning that their stable form is a pair. There are several others... One Hydrogen ($\ce H$) has a positive charge +1. This has nothing to do with it having more protons than electrons, as it has one of each. Think of it as 'willing' to donate one electron, or it has +1 electrons... see, it is on the plus side. One Oxygen ($\ce O$) has a negative charge of -2. Think of it as 'wanting' to borrow two electrons, or it has -2 electrons... see, it is on the plus side. Now, put one $\ce H^+$ with one $\ce {O}$ and the $\ce H$ will donate, the $\ce O$ will borrow and the $\ce H$ will be 0, and the O will be -1. Now the combination has a -1 charge. Now the Hydrogen is sharing one electron with the oxygen, filling its outermost shell with 2, but the Oxygen is needing 8 in its outermost shell, and is only at 7, so the compound is at -1, or 'wanting' one more electron to make it stable. If you add one more $\ce H^+$ the $\ce{O}$ will be satisfied, and you will get $\ce {H2O}$, a very stable compound we know as water. • Thanks for the information, that helps quite a bit! Since Hydrogen's charge has nothing to do with with having more protons than electrons, what does it have to do with? Is it a function of some atomic attribute, or is there table to reference? I'm curious about the states of other atomic ions as well. Thanks again! Dec 5 '12 at 23:37 • The charge on hydrogen ion does have to do with it having more protons (one) than electrons (zero), but it is more common and more convenient to think of hydrogen ion having fewer electrons (zero) than hydrogen atoms (one). Dec 6 '12 at 0:47 • Both O and H are neutral species. Do not confuse charge with oxidation number. Dec 6 '12 at 7:52 • Warning for reader: This is a too simplistic view on the matter to be useful. It doesn't take nitpick to know that thinking of $\ce{OH-}$ as $\ce{O^2-}$ and $\ce{H+}$ is flat out wrong. May 1 '16 at 11:08 • @H.Khan Because water isn't ionic compound! And in acid/base reactions there's no "free protons" to be found. Dec 4 '18 at 20:44 I may be wrong, but one way I understood the overall formal charge denoted outside the bracketed OH- molecule, was by giving OH- a backstory. Without a backstory of how it came to form, I don't think the negative formal charge makes intuituve sense. If we say OH- formed from the dissociation of H2O into H+ and OH-, then there is some sense. What I mean is O and H and H came together to form water, amd did so by sharing electrons. At this point, before they arrived together and during, even the hydrogens had 1 electron each to give to help oxygen. The oxygen had 2 electrons to help out the hydrogens, one for each. But when the water molecule split apart, there was a tug of war over their shared electrons. Oxygen turned out to be stronger and yanked away the only electron one of the hydrogens had. This hydrogen was sent loose, without an electron to help it make another bond or be useful by sharing its bond making electron with another element. The other hydrogen stayed attached to the oxygen. So the oxygen had an extra electron, torn from the hydrogen now detached. This is written as O-. But because there is an H still attached we write OH-. (Note, the H in OH is not carrying the stolen electron but the negative charge symbol is written by it, though we could write HO- to be more clear. (On another level however, I think this stolen electron would be whizzing around near the H in OH too, hence some negative charge of this electron woukd be exerting itself over the H in the OH from time to time. So if ai am correct, without this backstory or creation story for OH- or HO-, it is not intuitive ... ... because if I think simply yet logically, if I take a neutral atomic O, and a neutral atomic H and make them bond, I don't think we actually have an overall formal charge. Why ? Because O has 6 valence electrons. H has 1 valence electron. When they bond, they each give eachother a valence electron. O ends up with 7. H ends up with 2. In this OH or HO bond, neither atom has lost an electron. No tug of war has been won or lost. (Some might argue that oxygen might be pulling hydrogen's electron a bit closer to itself for a few physical reasons, and hence 'winning' - but the match has not ended, so no one can be declared a winner or loser, a taker of electrons, hence no charge, (as a prize or show of strength), can yet be denoted. Oxygen would have seven electrons in its orbital flux aura, one away from having an octet and a complete shell. In this way I think that in theory we coukd see O(2-). I think this woukd refer to a single monatomic oxygen atom that has taken two extra electrons, and hence, will be somewhat stabke with a full outer shell. It seems therefore, that when we see an atom or compound polyatimic molecule with a charge, we should be reminded to think of battle scars and spoils of war - that this atom or polyatom has a history of wars over electrons with other atoms and polyatoms, and its charge is a symbol of its current state of winnings or losings. Some don't seem to even need to go to war. Noble gases have full outer shells before they have even gone to battle over electrons. They are sort of born into a position of abundance and luxury, though it may be lonely nonetheless. It is also pretty hard for other atoms to even get into a tug of war with a noble gas, let alone win. Some atoms like Al(3+), meaning 3 electrons lost, not 3 atoms of aluminium as in (Al3) It is found in soil quite a lot. I think this refers to the notion that this aluminium has lost 3 electrons. If so, it has lost 3 valence electrons to something or many things, at some point in the past. It is quite stable having lost its valence electrons because only had 3, which means its valence electrons are now gone. Can it be made to react again then ? Yes, I think so, but the chemistry eludes me. I think Al (3+) is considered pseudo noble because it has lost its only 3 valence electrons at some prior time and has lost an oribtal, but demoted itself to a stable orbital nonetheless. The pseudo aspect might be that this stable state is not as strong as a real noble. One reason might be that Al (3+) now has an abundance of protons compared to electrons. This means it will have a slight tendency to pull in electrons and create a new shell that is not full, even though it has a full outer shell now. I think the same is true of certain pseudo noble anions, in an opposite sense. They may have too much electron charge even if they have a full outer shell and might only weakly bind their newly acquired electron/s. Some pseudo noble anions might really be pretty stable though. Same with some cations. In the soil I believe Al (3+) and Cl(-) are pretty stable pseudo noble ions. I think the way Al (3+) starts to destabilize is when electrons start getting close in abundance around it and cause electron gains to take place upon the aluminium metal, rather than abundances cations coming and stealing even more electrons from it. I think it is pretty hard for cations or any atoms to steal electrons from non bonding non valence orbitals like inner ones. I think x rays might be able to temporarily knock out a inner core electron, but possibly not without causing a lot of energy to be released as well. I could be completely wrong in my understanding of the basics though. I hope someone will correct me if I am. I would like to know.	2021-10-20 07:55:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5414552688598633, "perplexity": 1113.0503671216986}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585302.91/warc/CC-MAIN-20211020055136-20211020085136-00151.warc.gz"}
https://projecteuclid.org/euclid.aos/1176349559	## The Annals of Statistics ### Some Admissible Nonparametric and Related Finite Population Sampling Estimators #### Abstract Given a random sample from an unknown distribution $F,$ which is assumed to belong to some nonparametric family of distributions, consider the problem of estimating $\gamma(F),$ some function of $F.$ When the loss function is squared error, admissible estimators are exhibited for a large class of $\gamma$'s. A relationship between these estimators and similar ones in finite population sampling is demonstrated. #### Article information Source Ann. Statist., Volume 13, Number 2 (1985), 811-817. Dates First available in Project Euclid: 12 April 2007 https://projecteuclid.org/euclid.aos/1176349559 Digital Object Identifier doi:10.1214/aos/1176349559 Mathematical Reviews number (MathSciNet) MR790577 Zentralblatt MATH identifier 0581.62006 JSTOR	2019-07-23 15:47:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.32198089361190796, "perplexity": 1437.4325650289295}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195529480.89/warc/CC-MAIN-20190723151547-20190723173547-00056.warc.gz"}
https://www.physicsforums.com/threads/if-a-matrix-a-is-injective-then-aat-is-invertible.346005/	# If a matrix A is injective then AAt is invertible 1. Oct 15, 2009 1. The problem statement, all variables and given/known data If a matrix, A (nxm) is monic (or epic) then is $$A^tA (or AA^t)$$ is invertible? 2. Relevant equations T is monic if for any matrices B,C: BT = BT => B=C. S is invertible if there exists U s.t. US = SU = $$I_n$$ 3. The attempt at a solution Since A is monic it must preserve n; i.e. assume n < m, so that A has n independent rows (is that allowed), and we view A as an function of natural numbers, $$n \stackrel{A}{\rightarrow} m$$. Then can we say AB where B is m * k, means that AB has n indepenent rows as well? 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution	2017-08-17 00:41:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.61708664894104, "perplexity": 1214.6815128830017}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886102757.45/warc/CC-MAIN-20170816231829-20170817011829-00624.warc.gz"}
https://www.andreaperlato.com/aipost/softmax-regression/	# Softmax Regression When we have to deal with a classification with more than 2 possible levels, we use a generalization of the logistic regression function called softmax regression; a logistic regression class for multi-class classification tasks. In Softmax Regression, we replace the sigmoid logistic function by the so-called softmax function. $\begin{array}{l}{\qquad P\left(y=j \| z^{(i)}\right)=\phi_{\text {softmax}}\left(z^{(i)}\right)=\frac{e^{z^{(i)}}}{\sum_{j=0}^{k} e^{z_{k}^{(i)}}}} \\ {\text { where we define the net input } z \text { as }} \\ {\qquad z=w_{1} x_{1}+\ldots+w_{m} x_{m}+b=\sum_{l=1}^{m} w_{l} x_{l}+b=\mathbf{w}^{T} \mathbf{x}+b}\end{array}$ The w is the weight vector, x is the feature vector of 1 training sample, and b is the bias unit. A bias unit is an extra neuron added to each pre-output layer that stores the value of 1. Bias units aren’t connected to any previous layer and in this sense don’t represent a true activity. It is used in the case the sum of the weights is equal to zero. Now, this softmax function computes the probability that this training sample x(i) belongs to class j given the weight and net input z(i). So, we compute the probability p(y=j∣x(i);wj) for each class label in j=1,…,k.. Note the normalization term in the denominator which causes these class probabilities to sum up to one. Even if it is a bit unusual, we use the softmax regression as the activation function of the output layer y. $\begin{array}{l}{t=e^{\left(z^{(1)}\right)}} \\ {a^{(L)}=\frac{e^{z^{(L)}}}{\sum_{j={1}}^{4} t_{i}}, \quad a_{i}^{(L)}=\frac{t_{i}}{\sum_{j={1}}^{\frac{4} t_{i}}}}\end{array}$ The formula above assumes that we have 4 layers L on the output y as depicted in the figure below. In the figure above we have z^(L) set to (5, 2, -1, 3), and from these values we can compute the activation function t suing the formula represented in the calculation we made before. Having the value of t, we can calculate the activation function a for the output y. From the example above, the output with level_0=0.842 is the most likely to categorize what we received in input. The categorization is called hard max and gives 1 to level_0 and 0 to the other output values. On the contray, the value 0.842 is called soft max. The loss function L is calculated as follow: $f(\hat{y}, y)=-\sum_{j=1}^{4} y_{j} \log \hat{y}_{j}$ For the values categorized with 0 in hard max, the yi is zero and so the loss finction is just equal to -log(yi). More generally, what the loss function does is to looks, at whatever is the ground true in our training-set, at the correspoding probability of that class and put it as high as possible. This is quite similar to the maximum likehood estimation. Reference: coursera deep neural network course	2021-10-17 21:54:11	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9125334620475769, "perplexity": 580.1703854981323}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585183.47/warc/CC-MAIN-20211017210244-20211018000244-00577.warc.gz"}
http://hackage.haskell.org/package/containers-0.6.0.1/docs/Data-Containers-ListUtils.html	containers-0.6.0.1: Assorted concrete container types Copyright (c) Gershom Bazerman 2018 BSD-style libraries@haskell.org portable Trustworthy Haskell98 Data.Containers.ListUtils Description This module provides efficient containers-based functions on the list type. Synopsis Documentation nubOrd :: Ord a => [a] -> [a] Source # $$O(n \log n$$. The nubOrd function removes duplicate elements from a list. In particular, it keeps only the first occurrence of each element. By using a Set internally it has better asymptotics than the standard nub function. Strictness nubOrd is strict in the elements of the list. Efficiency note When applicable, it is almost always better to use nubInt or nubIntOn instead of this function. For example, the best way to nub a list of characters is nubIntOn fromEnum xs nubOrdOn :: Ord b => (a -> b) -> [a] -> [a] Source # The nubOrdOn function behaves just like nubOrd except it performs comparisons not on the original datatype, but a user-specified projection from that datatype. Strictness nubOrdOn is strict in the values of the function applied to the elements of the list. nubInt :: [Int] -> [Int] Source # $$O(n \min(n,W))$$. The nubInt function removes duplicate Int values from a list. In particular, it keeps only the first occurrence of each element. By using an IntSet internally, it attains better asymptotics than the standard nub function. See also nubIntOn, a more widely applicable generalization. Strictness nubInt is strict in the elements of the list. nubIntOn :: (a -> Int) -> [a] -> [a] Source # The nubIntOn function behaves just like nubInt except it performs comparisons not on the original datatype, but a user-specified projection from that datatype. Strictness nubIntOn is strict in the values of the function applied to the elements of the list.	2018-09-21 05:02:08	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.42124250531196594, "perplexity": 2784.582620135141}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267156780.2/warc/CC-MAIN-20180921033529-20180921053929-00276.warc.gz"}
https://www.acmicpc.net/problem/27694	시간 제한메모리 제한제출정답맞힌 사람정답 비율 1 초 1024 MB75360.000% ## 문제 You wander through a dark dungeon. All around you there are doors of different shapes and colors. You pick one, open it and enter. “I knew you would come,” said a voice in the dark. You come closer and see an old man with a long white beard sitting on the floor. “I used to be a problem solver like you,” he says, “but then I took an arrow in the knee.” “Well… not really. It’s just what all the kids were saying the last time I saw daylight.” “So what happened to you?” you ask and sit beside him. “The truth is, I destroyed my kneecaps on the stairs. When I was younger, I did a lot of programming contests. And in one of them was a really nasty task. I had to determine the number of ways in which one can go up and down a staircase with n steps. Of course, there were some constraints: when going up, you can take two steps at a time, and when going down, you can take up to four steps at once.” He sighs deeply. “I had no idea how to solve the task, so I found a staircase and attempted to try every possibility. But there were so many of them that I overloaded my knees and now I can’t even walk. So I’m sitting here and still wondering about a solution for that problem. Can you help me to finally put a close on this?” The staircase consists of n steps. Count the ways of going up and then down the staircase, given the following constraints: • On the way up, you can take either 1 or 2 steps at a time. • On the way down, you can take 1, 2, 3, or 4 steps at a time. As the actual number of ways can be huge, compute the remainder it gives when divided by 109 + 9. For example, for n = 5 one valid way of going up and down the staircase looks as follows: Start on the ground, ascend to step 2, continue to step 3, and then go to step 5. Having now reached the top of the staircase, you turn around and walk down, first descending to step 4 and then going directly to the ground (which is, at that moment, 4 steps below). The figure above shows two valid ways of going up and down the stairs for n = 5. The one described above is shown in red. ## 입력 The first line of input contains one integer number t specifying number of test cases. Each test case is preceded by a blank line. Each test case consists of a single line with the integer n (1 ≤ n ≤ 100,000) – the number of steps. ## 출력 For each test case print a single line with one integer – the number of valid paths modulo 109 + 9. ## 예제 입력 1 2 3 5 ## 예제 출력 1 12 120 ## 힌트 For example, when n = 3 there are 3 ways to go up, and for each of them there are 4 ways to go back down.	2023-03-23 18:36:02	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4838298559188843, "perplexity": 523.9294996777124}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945182.12/warc/CC-MAIN-20230323163125-20230323193125-00077.warc.gz"}
http://tdig.evaketzer.de/latex-plot-function.html	# Latex Plot Function fun can be a function handle, an inline function, or a string of a function name. Accessibility is a must for any quality solution. 5 times the size required in LATEX, with the default pointsize. I think the problem might actually be the scilab, and not my brain. xyplot produces bivariate scatterplots or time-series plots, bwplot produces box-and-whisker plots, dotplot produces Cleveland dot plots, barchart produces bar plots, and stripplot produces one-dimensional scatterplots. And, when I look at it latex, I get something a bit more different. This function only works when the plot format is either gnuplot or gnuplot_pipes. How to add Latex formatted text in a Scilab plot Latex makes possible to write equations in a mathematical format, by using all the symbols and characters specific to mathematics language. pgfplots draws high{quality function plots in normal or logarithmic scaling with a user-friendly interface directly in TEX. Section 9-2 : The Wave Equation. Use plots to visualize data. If one parameter is used, the vector parameter is plotted on the ordinate versus the point number on the abscissa. Draw the Bode diagram for each part. But we have twelve files to check, and may have more in the future. // Close all opened figures and clear workspace. Filling the space below a surface plot [Open in Overleaf] Polar plot of a sine function with factor in argument [Open in Overleaf] Polar plot of sine based function [Open in Overleaf] Highlighting parts of a curves and shading the plane [Open in Overleaf]. Some examples of using dnorm are below: # This is a comment. Text: In addition to legends, you can use the text() function to add text elements at any position on the plot. Visit Cosmeo for explanations and help with your homework problems! Home. MacKichan Software now offers version 5. Create a plot where x1 and y1 are represented by blue circles, and x2 and y2 are represented by a dotted black line. Requires adding a usepackage{longtable} to your LaTeX preamble. To compute a square root in a formula, you would use the notation ‘ sqrt(x) ’. A custom function must start with a Function statement and end with an End Function statement. Zooming in, zooming out, zoom to a selected area and a free movement by mouse click & drag are the prominent features of this software, which helps to view the graph for any location at the required view. Evaluate, simplify, solve, and plot functions without the need to master a complex syntax. Each function call carries out a single task associated with drawing the graph. LaTeX Math Formulas There are three environments that put LaTeX in math mode: math, displaymath, and equation. The function must accept a vector input argument and return a vector output argument of the same size. Instead of data, you can also input an equation of X using either the input argument or the EQUATION property. These may differ in color, in thickness, in dot/dash pattern, or in some combination of color and dot/dash. m using the cp command (or the editor): cp iteration. Plot symbols and colours can be specified as vectors, to allow individual specification for each point. Further on in this article, an example of a flowchart will be shown. R plot function examples. Online 3-D Function Grapher Home Physics Tools Mathematical Tools Online 3-D Function Grapher A standalone application version of this 3-D Function Graphing Program, written in Flash Actionscript, much faster, essentially more capabilities, built-in function calculator and many more. To visualize them to better understand their properties, so that we can see roots and extreme points, we … - Selection from LaTeX Cookbook [Book]. The LaTeX package for plotting is pgfplot. R Multiple Plots In this article, you will learn to use par() function to put multiple graphs in a single plot by passing graphical parameters mfrow and mfcol. Should this create names if they are NULL? prefix: for created names value: a valid value for that component of dimnames(x) Following is a csv file example:. LaTeXDraw is developed in Java and thus runs on top of Linux, Windows, and Mac OS X. PyPlot uses the Julia PyCall package to call Matplotlib directly from Julia with little or no overhead (arrays are passed without making a copy). Again, don't forget to put a semicolon ; at the end of the command. To compute a square root in a formula, you would use the notation ‘ sqrt(x) ’. Plot symbols and colours can be specified as vectors, to allow individual specification for each point. Anything I write after the octothorpe is not executed. Table 1 is an example of a user-speciﬁed table using three macros — btable (begin table), etable (end table), and mc (headings that span multiple columns). Tap into the extensive visualization functionality enabled by the Plots ecosystem, and easily build your own complex graphics components with recipes. Maps - Contour plots with labels. main features: - linear and log10 scales - optional grid lines in X and Y directions - adjustable tick mark intervals and sizes - LaTeX support This extension was inspired by Tavmjong Bah's (and collaborators) extension Function Plotter already presented in Inkscape. An Sweave Demo Charles J. The problem is this: how do I plot a function divided by another. It automates many details of plotting such as sample rate, aesthetic choices, and focusing on the region of interest. Without specifying a unit (1,2), the standard one is cm (1cm,2cm). To plot functions in the form of y=f(x. ability to instantaneously update plots in the final paper with new data. If the number is an integer, use that integer. Example of how to change the color using short names is below. To compute a square root in a formula, you would use the notation ‘ sqrt(x) ’. PGF/TikZ - Graphics for LATEX A tutorial Meik Hellmund Uni Leipzig, Mathematisches Institut M. If you specify a marker, but not a line style, MATLAB plots only the markers. This entry was posted in Uncategorized on 2. ) using either the combination of base/lattice or with ggplot2. LaTeX symbols have either names (denoted by backslash) or special characters. to close our bracket without displaying anything. Whether it's for research, a school assignment, or a work presentation, 3-D plots are great for visualizing what a complicated set of data looks like. These functions cannot be used with complex numbers; use the functions of the same name from the cmath module if you require support for complex numbers. R colnames Function. PyPlot uses the Julia PyCall package to call Matplotlib directly from Julia with little or no overhead (arrays are passed without making a copy). Math calculators and answers: elementary math, algebra, calculus, geometry, number theory, discrete and applied math, logic, functions, plotting and graphics. More than one function can be plotted in the same graph by using a parameter add, which takes boolean values TRUE or FALSE. Many functions have discontinuities (i. Types of MATLAB Plots. To learn how this works, I suggest choosing an example from the the "LaTeX Examples" drop-down list at the lower left. exp e (the Euler Constant) raised to the power of a value or expression ln The natural logarithm of a value or expression log The base-10 logarithm of a value or expression floor Returns the largest (closest to positive infinity) value that is not greater than the argument and is equal to a. linspace(0, 20, 100) plt. Package plot provides an API for setting up plots, and primitives for drawing on plots. You don't plot these. We’ve already worked with figures and subplots without explicitly calling them. Have you heard of SageTex? It lets you use Sage functionality directly in your TeX. Create normal/logarithmic plots in two and three dimensions for LaTeX/TeX/ConTeXt PGFPlots draws high--quality function plots in normal or logarithmic scaling with a user-friendly interface directly in TeX. cartesianPlotFunction2D by fsmMLK: Inkscape extension to plot functions of one independent variable. hist() is a widely used histogram plotting function that uses np. How to make a graph with multiple axes in MATLAB ®. crqs: Functions to fit censored quantile. pre blocks should have max-width and max-height). Mathematical notation is rich and changes among different cultures; MathType supports mathematical notation for different education levels and cultures. Scilab allows the usage of Latex formatted text, in the plot related functions, such as: xlabel() , ylabel() , title() , legend() and xstring(). You can also use "pi" and "e" as their respective constants. This function only works when the plot format is either gnuplot or gnuplot_pipes. By allowing LATEX to handle typesetting of text in R plots along with the rest of the text in the paper the. , an unevaluated function call), it checks to see if there exists a procedure by the name of latex/function_name , where function_name. PGF/TikZ - Graphics for LATEX A tutorial Meik Hellmund Uni Leipzig, Mathematisches Institut M. Rnw instead of. histogram() and is the basis for Pandas’ plotting functions. f(x) = x + 1. How do I plot a circle with a given radius and Learn more about circle, radius, center, rectangle, overcoming obstacles MATLAB. SHMOOP PREMIUM Topics SHMOOP PREMIUM SHMOOP PREMIUM. To modify the position of the axis legends or title plot, you can use the vjust and hjust parameters for vertical and horizontal positioning. You add points to a plot with the points() function. For each line plotted, the legend shows a sample of the line type, marker symbol, and color beside the text label you specify. Beginner question, but I was wondering is that type of decay was a known function, or is there some way for me to model it with the exponential decay function. Many functions have discontinuities (i. (Likelihoods will be comparable, e. The function g(x) = x 3 - 3x is symmetric about the origin and is thus an odd function. To graph an exponential, you need to plot a few points, and then connect the dots and draw the graph, using what you know of exponential behavior: Graph y = 3 x; Since 3 x grows so quickly, I will not be able to find many reasonably-graphable points on the right-hand side of the graph. Author: Thomas Breloff (@tbreloff) To get started, see the tutorial. The visitor's LaTeX, entered or copied into the editing window below, will be quickly rendered by up to three renderers (in different ways). Using Ch plotting features, interactive plotting can be generated dynamically on-line. Feature List. The ﬁrst plot used the example code unchanged and was saved to pdf from the graphics screen. To plot functions in the form of y=f(x. function, latex. You add points to a plot with the points() function. The ultimate goal is to offer SourceForge-like services (such as SVN repository, place for documentation, downloads, mailing lists, bugzilla, wiki etc. This help page documents several commonly used high-level Lattice functions. contour: Level (Contour) Plots. $\begingroup$ I am sorry, but I still don't quite get it. csv files in LaTeX. Take the rgl package for example: if we want to insert 3D snapshots produced in rgl into our LaTeX document, we may consider this hook function (see the more sophisticated hook_rgl() in this package):. For more information about the different style options, see the plot function page. If the first argument hax is an axes handle, then plot into this axis, rather than the current axes returned by gca. Adding titles and labels to graphs in R using plot() function by Wim · 26/04/2016 Adding titles to plot() graphs in R is quite straight forward. as linegraph or scattergraph. Creating Graphs and Plots in LaTeX This page contains a brief tutorial on how to create scientific graphs (function plots, histograms, bar graphs, scatter plots, contour lines, color maps, surface plots, ) in LaTeX documents using GLE. Next, we’ll plot both an xy-axis and a function on the plane determined by the axis. The actual data is loaded from latex-import-data. Plotting several 2d functions in a 3d graph [Open in Overleaf] Scatter plot with cubes [Open in Overleaf] Spiral cone [Open in Overleaf] Surface Plot [Open in Overleaf] Surface plot of a math function [Open in Overleaf] Surface plot with interior colors [Open in Overleaf] Torus. After creating a layout, call the nexttile function to place an axes object into the layout. 3 Functions and Variables for Plotting. They are mostly standard functions written as you might expect. If you do not already have a corresponding columns of x values, you will have to make that to. Plot symbols and colours can be specified as vectors, to allow individual specification for each point. I can't seem to find any way to do that, however, and I keep getting one part of my line looping back over itself, resulting in two y-values for each x-value (screen cap below). Following on from example 4 which showed how to plot the graph of a function in LaTeX using pstricks, here we extend the example to show how trig functions such as cos, sin and tan can be plotted. of the function (for Sin it notes that the argument is to be in radians). This example describes an experience using the Office X version for Macintosh. Discover Scilab Cloud. The function f(x) = x 2 - 18 is symmetric with respect to the y-axis and is thus an even function. It involves a complicated looking formula that looks very strange. To use LaTeX markup, set the Interpreter property for the Text object to 'latex'. Example with dummy. How do I plot step functions in gnuplot? I'd like to plot floor/greatest integer functions. For example, create two plots in a 2-by-1. I've found out I can use abs and angle to get the magnitude and phase, but when I try to plot it over omega = -pi:0. For example, use. Using Ch plotting features, interactive plotting can be generated dynamically on-line. TeX as a callable function. Math calculators and answers: elementary math, algebra, calculus, geometry, number theory, discrete and applied math, logic, functions, plotting and graphics. The mathematics is done using a version of $$\LaTeX$$, the premiere mathematics typesetting program. R Plot PCH Symbols Chart Following is a chart of PCH symbols used in R plot. However, there may be times when you disagree, and a typical example is with its positioning of figures. The first adjustment you might wish to make to a plot is to control the line colors and styles. This help page documents several commonly used high-level Lattice functions. is an option for visualization functions that specifies what labels to use for each data source. Use a longtable environment instead of tabular. Text size The text on a plot should be approximately the same as the text around it. You add points to a plot with the points() function. Specify a function of the form y = f(x). Requires adding a usepackage{longtable} to your LaTeX preamble. If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because that $x$ value has more than one output. I've found out I can use abs and angle to get the magnitude and phase, but when I try to plot it over omega = -pi:0. By using the "int" function, in the same way we use the diff function, we can ask Matlab to do symbolic integration for us. There are two different things, scaling the graphic to the full or half width for example and changing the aspect ratio of a graphic. R allows users great flexibility in creating and formatting plots. com rmarkdown 0. How to Make 3D Plots Using MATLAB. I need to change the size of a PGFPlots tikzpicture in my LaTeX document. To plot a function, use the plot/splot command with a range of X-axis (or X and Y ranges for 3-dim. Label the symbols "sampled" and "continuous", and add a legend. This definition generalizes in a natural way to functions of more than three variables. One solution is to selectively convert parts of the plot (not the text labels) to a rasterized image. Tap into the extensive visualization functionality enabled by the Plots ecosystem, and easily build your own complex graphics components with recipes. This function only works when the plot format is either gnuplot or gnuplot_pipes. For example, you can include mathematical expressions in text using LaTeX. For example, create two plots in a 2-by-1. Plots - powerful convenience for visualization in Julia. The user supplies axis labels, legend entries and the plot coordinates for one or more plots and PGFPlots applies axis scaling, computes any logarithms and axis ticks and draws the plots, supporting line plots, scatter plots, piecewise constant plots, bar plots, area. Excel includes several built-in functions—RAND and NOW, for example—that don’t use arguments. com rmarkdown 0. Details PlotLabels -> labels specifies the labels to use for each data source in a plot. Skip navigation. Plotting the likelihood function Example from Clayton and Hills (Continuous parameter): In section 3. Create a new file called iteration2. Any additional arguments are treated the same as in plot3d. I have used a CSV-file to import and plot data. Instead I'm trying to limit this list to the most common math symbols and commands. array before you calculate above_threshold and below_threshold, and then it works. We have to decide which piece of the function to plug-and-chug into. Beyond simple math and grouping (like "(x+2)(x-4)"), there are some functions you can use as well. Also we ensure that the image axes are in the right relation to each other by setting ratio to -1. if the length of the vector is less than the number of points, the vector is repeated and concatenated to match the number required. I used this approach to set values and plot the graph dynamically. Plot size in pixels for EMF, GIF, JPEG, PBM, PNG, and SVG. Piecewise Functions • A piecewise function is a function which is defined by multiple sub functions, each sub function applying to a certain interval of the main function's domain. edu References. The syntax of the function must respect the conventions imposed by Microsoft Excel for functions used in spreadsheets. Suppose I have a function f[z]=z^3-z^2-z-1==0 where z=x+i*y , I would like to plot the function in two different ways. For example, use. Data analysis. This is an online, interactive LaTeX editor. While subplot positions the plots in a regular grid, axes allows free placement within the figure. For example, in the toolkit functions, we introduced the absolute value function $f\left(x\right)=\|x\|$. You can also use "pi" and "e" as their respective constants. You can draw simple graphs instantly by using a special function called "graph". In this case, the plot opera-tion has the following syntax: plot[id=] function{formula}. Note: I use View/Custom for my InkScape workspace. This module provides a class to hold, manipulate and employ various options for rendering a graph in LaTeX, in addition to providing the code that actually generates a LaTeX representation of a (combinatorial) graph. For example, create two plots in a 2-by-1. Concave Function. Gnuplot is a powerful and free plotting utility that is available for most architectures and operating systems. Paste the following code in a python file; Execute it (either selecting the code or using the Run cell code lens). This help page documents several commonly used high-level Lattice functions. // Close all opened figures and clear workspace. LaTeX options for graphs¶. I needed to use Gnuplot a little bit over the last few days, mostly to create 2D line charts, and these are my brief notes on how to get started with Gnuplot. 2 Basic Table Making in LATEX LATEX has excellent facilities for composing and typesetting tables. jpg' binary filetype=jpg with rgbimage. The things I want to do are often possible but require arcane options that I have trouble remembering. Gluck Department of Mathematics, University of Florida PGF and TikZ are packages in LATEX that allow for programming gures into documents. To get started make a regular LATEX le (like this one) but give it the su x. contour: Level (Contour) Plots. This help page documents several commonly used high-level Lattice functions. While these default options have been carefully selected to suit the vast majority of cases, the Wolfram Language also allows you to customize plots to fit your needs. Eramian Plotting Graphs 1/8. , qplot(x, y) directly produces the plot (no need to print() it), and all the plots in a code chunk will be written to the output by default. The underlying rendering is done using the matplotlib Python library. It s present in all full distributions of LaTeX, and permits the creation of consistent graphics with the typographic settings of the document, writing the instructions directly in the text source and ensuring the highest quality typical of LaTeX. This force is called the tension in the string and its magnitude will be given by T (x,t). Here, we will see how to easily plot functions. Add a second plot that uses a dashed, red line with circle markers. The function to use for each of the plots can be independently defined with fun1 and fun2. In your version, you don't get an array of bools, but just False and True. If you specify a marker, but not a line style, MATLAB plots only the markers. Take the rgl package for example: if we want to insert 3D snapshots produced in rgl into our LaTeX document, we may consider this hook function (see the more sophisticated hook_rgl() in this package):. Example with dummy. Pychart is a library for creating EPS, PDF, PNG, and SVG charts. For example, plot a dotted line. Plots in Ch can be generated from data arrays or files, and can be displayed on a screen, saved as an image file in different file formats, or output to the stdout stream in a proper image format for display in a Web browser through a Web server. xyplot produces bivariate scatterplots or time-series plots, bwplot produces box-and-whisker plots, dotplot produces Cleveland dot plots, barchart produces bar plots, and stripplot produces one-dimensional scatterplots. default, latex. To plot the graph of a function, you need to take the following steps − Define x, by specifying the range of values for the variable x, for which the function is to be plotted. g: the title, axis labels, annotations, etc. Question: How to plot an implicit function? Tags are words are used to describe and categorize your content. This module provides a Julia interface to the Matplotlib plotting library from Python, and specifically to the matplotlib. Requires adding a usepackage{longtable} to your LaTeX preamble. function plot To plot a curve where the vertical coordinate is a function of the horizontal coordinate, just give the function formula in terms of x within braces. When add=TRUE, the current curve will be added to the existing curve. The ctan alreadyprovides a huge list with currently 5913 symbols, which you can download here. The list of supported file formats can be found here. Note most plotting commands always start a new plot, erasing the current plot if necessary. area plot An area chart is really similar to a line chart , except that the area between the x axis and the line is filled in with color or shading. Mathematical notation is rich and changes among different cultures; MathType supports mathematical notation for different education levels and cultures. Use Plot to plot a function:. within HTML tables constructed with the htmlTable package. HOW TO PLOT FUNCTIONS WITH LATEX. However the type of plot can be modified with the fun argument, in which case the plots are generated by feval (fun, x, y). This program has to be installed first, and then include. This can be done in a number of ways, as described on this page. pyplot is a collection of command style functions that make Matplotlib work like MATLAB. csv and separated by commas. survfit as much as possible, for instance by default plotting confidence intervals for single-stratum survival curves, but not for multi. Eramian The University of Saskatchewan February 9, 2017 Mark G. have greek letters in a Matlab figure title or plot label, you must set the "FontName" for the current axes to "Symbol". This can be done in a number of ways, as described on this page. ]) DataFrame plotting accessor and method. In addition to the function name, the Function statement usually specifies one or more arguments. How do I plot step functions in gnuplot? I'd like to plot floor/greatest integer functions. The Graph plotter - Graph Calculator is a mathematics graph software, which facilitates to plot mathematical equations of the form y = f(x). cartesianPlotFunction2D by fsmMLK: Inkscape extension to plot functions of one independent variable. For each line plotted, the legend shows a sample of the line type, marker symbol, and color beside the text label you specify. That’s all there is to plotting simple functions in matplotlib! Below we’ll dive into some more details about how to control the appearance of the axes and lines. to close our bracket without displaying anything. If you have a matrix to plot, you can replace this function with that file. Using SageTeX - Sage Tutorial v7. as linegraph or scattergraph. More examples; Standard plot. latex is a generic function that calls one of latex. For PS, EPS, PDF, and other vector formats the plot size is in points. It automates many details of plotting such as sample rate, aesthetic choices, and focusing on the region of interest. An online LaTeX editor that's easy to use. # Example 1 set. Post navigation. Hellmund (Leipzig) PGF/TikZ 1 / 22. Have you heard of SageTex? It lets you use Sage functionality directly in your TeX. An Sweave Demo Charles J. latex is a generic function that calls one of latex. When using the command form of the print function you must quote the xsize,ysize option. frame x as a table. Winplot is an example of how interactive plotting can be accomplished in Ch with mathematical expressions entered by the user through a graphical user interface in Windows. Each function call carries out a single task associated with drawing the graph. Consider the graph in the following precalculus exam problem. plot) and the function. For a function to be continuous at a point, the function must exist at the point and any small change in x produces only a small change in f(x). Once the basic R programming control structures are understood, users can use the R language as a powerful environment to perform complex custom analyses of almost any type of data. This can be done in a number of ways, as described on this page. To use LaTeX markup, set the Interpreter property for the Text object to 'latex'. High quality plots for LaTex with PGF-Tikz and Gnuplot. ) using either the combination of base/lattice or with ggplot2. Almost everything in Plots is done by specifying plot attributes. This turns to be a tedious process. The last exercise is to produce a cobweb plot like that on page 9 of the notes. This function creates a tiled chart layout containing an invisible grid of tiles over the entire figure. I am trying to plot a function with TikZ and PGF. If the number is not an integer, use the next smaller integer. MacKichan Software now offers version 5. Common Bivariate Trellis Plots Description. latex is a generic function that calls one of latex. High quality plots for LaTex with PGF-Tikz and Gnuplot. $$f$$ is a concave function if all the points on or below its graph form a convex set, i. Thus one obtains by typing $\mathrm{cosec} A$. Line Series plot based on box chart with box turned off and connecting line added between points. How to plot a function using matplotlib We will see how to evaluate a function using numpy and how to plot the result. PyTeX will return to Python TeX's typeset output, in the form of dvi. More examples; Standard plot. You can easily do the same thing using the long names. , an unevaluated function call), it checks to see if there exists a procedure by the name of latex/function_name , where function_name. to close our bracket without displaying anything. An example of script is present here to create a plot by exploiting both the quality of Tikz and the power of Gnuplot. This figure has been generated using the following LaTex script figure. Add text to a plot in R software The mtext() R function can be used to put a text in one of the four margins of the plot. If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because that $x$ value has more than one output. Use a longtable environment instead of tabular. By default, the value will be read from the pandas config module. This function creates a tiled chart layout containing an invisible grid of tiles over the entire figure. Hello, I'm trying to graph some data as a scatter plot with a smooth line, and I want it to be a function, i. Mar 31, 2017 LaTeX PDF. So, is the value of sin-1 (1/2) given by the expressions above? No! It is vitally important to keep in mind that the inverse sine function is a single-valued, one-to-one function. Accessibility is a must for any quality solution. Look below to see them all. Skip navigation. This help page documents several commonly used high-level Lattice functions. Choose the classic MathType look or the LaTeX look. The function to plot must be placed inside curly brackets. It s present in all full distributions of LaTeX, and permits the creation of consistent graphics with the typographic settings of the document, writing the instructions directly in the text source and ensuring the highest quality typical of LaTeX. The ctan alreadyprovides a huge list with currently 5913 symbols, which you can download here. One is "Function Plotter" and the other is "Parametric Curves. Do you want to plot a vertical line at the location of each discrete frequency? $\endgroup$ - Jens Jun 26 '16 at 17:52. The LaTeX package for plotting is pgfplot. " It's not clear what you really want. Plot size in pixels for EMF, GIF, JPEG, PBM, PNG, and SVG. Author: Thomas Breloff (@tbreloff) To get started, see the tutorial. It also has a comprehensive list of references, which will be of further use. By allowing LATEX to handle typesetting of text in R plots along with the rest of the text in the paper the. Add points to a plot in R. Define the function, y = f(x) Call the plot command, as plot(x, y) Following example would demonstrate the concept. escape: bool, optional. Scilab Enterprises is developing the software Scilab, and offering professional services: Training Support Development. Thus one obtains by typing $\mathrm{cosec} A$. To graph an exponential, you need to plot a few points, and then connect the dots and draw the graph, using what you know of exponential behavior: Graph y = 3 x; Since 3 x grows so quickly, I will not be able to find many reasonably-graphable points on the right-hand side of the graph. Piecewise is a term also used to describe any property of a piecewise function that is true for each piece but may not be true for the whole domain of the function. Package plot provides an API for setting up plots, and primitives for drawing on plots. For example, create two plots in a 2-by-1. You can also use "pi" and "e" as their respective constants.	2019-11-18 11:13:05	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.591106653213501, "perplexity": 1397.1344025386122}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496669755.17/warc/CC-MAIN-20191118104047-20191118132047-00398.warc.gz"}
https://forum.allaboutcircuits.com/threads/pn-junction-electrostatics.80132/	# PN junction: electrostatics Thread Starter #### Wikkez Joined Jan 26, 2013 6 Hi, First post here, I'm learning semiconductor physics&electronics on my own, but I'm a bit lost. In the depletion regio,The total charge in the n-regio equals the total charge in the p-regio, so why is there actually an E-field. Since u can calculate the strength of E-fields with Gauss' Law with the charge density and the charge density is the same but opposite in charge, why doesn't it cancel? Why is there a field, if there's not net charge(p and n regio combined)? ρp = ρn Hopefully someone can sort this out for me #### WBahn Joined Mar 31, 2012 26,398 What is the basis for saying that the total charge in the N region is the same as the total charge in the P region? Thread Starter #### Wikkez Joined Jan 26, 2013 6 What is the basis for saying that the total charge in the N region is the same as the total charge in the P region? What do you mean with that question? Charge neutrality? #### WBahn Joined Mar 31, 2012 26,398 What do you mean with that question? Charge neutrality? So is the total charge on one plate of a charged capacitor the same as the total charge on the other? If so, then how do you explain the presence of an electric field in between the plates? Thread Starter #### Wikkez Joined Jan 26, 2013 6 I reviewed some electromagnetics and I understand it now, guess I was mistaking electric field for electric flux and charge density for charge. My excuses. But I have another question that's probably stupid: One of the boundary conditions for solving the electric field on the pn-junction with poisson equation, is stating that @ x=0 the electric field must be continuous. Well, if the field is not continuous, it means that there is a presence of charge just @ your junction right since field lines always end on a charge. So why exactly is this impossible? Also: why does the electric field reaches a maximum value @ x=0, what is the physical meaning of this. Thanks in advance.	2021-01-16 10:28:53	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8159453868865967, "perplexity": 773.0697157323474}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703505861.1/warc/CC-MAIN-20210116074510-20210116104510-00115.warc.gz"}
http://crypto.stackexchange.com/questions?page=52&sort=active	# All Questions 260 views ### Why does Schnorr's Digital Signature scheme necessitate two prime numbers? One of the necessary components to the Schnorr Digital Signature scheme is a pair of prime numbers $p$ and $q$ such that $q$ divides $p-1.$ However, there is never a modular inverse taken of q so why ... 115 views ### Must root certificates be self signed? Root certificates are normally self-signed. What is the reason behind that convention? I mean, one must trust the root certificate in a non cryptographic way anyhow. 62 views ### Combining md5 collisions to create more collisions Given X1 and X2 such that md5(X1) = md5(X2) and Y1 and Y2 such that md5(Y1) = md5(Y2), and knowing the following property of md5: if md5(a) = md5(b), md5(a\|s) = md5(b\|s) is there a way to find ... 822 views ### generate RSA public key having public modulus and exponent [closed] i want to generate an RSA public public key file using openssl (or other tools) having public modulus and exponent, so i can use it later to encrypt files i have this: ... 79 views ### Which ECB ciphertext stealing scheme is this? 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A few days ago, I was asked to sign an online document over DocuSign as well as on SignEasy, any the signature were accepted just by a physical signature sent over email or on the app. They did not ... 44 views ### Sextic twist maps to q Eigenspace of Frobenius Let $E(p)$ be a Barretto-Naehrig elliptic curve with r-torsion and embedding degree 12 and $E'$ a sextic twist with homomorphism $\psi$. How to show, that $E'$ has a unique r-torsion group $\psi$ ... 67 views I read this post from Ben Tasker describing how to take control of a powerline network. Since I am interested in security, I tried to do the same thing, to see if it was possible. I used Devolo ... 161 views ### Is there a way to encrypt an integer within an arbitrary range? I need an algorithm that can take an integer, an arbitrary range, a key of 16/24/32 bytes, and an initialization vector of 16 bytes(ideally), and return an integer in the same range. 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The encryption algorithm has been selected to use AES. I did some search ... 97 views ### Define “cryptographer” I often encounter the phrase "I am not a cryptographer". Is there a generally accepted definition for who is a cryptographer? What qualifications would be expected? For instance, should that person ... 65 views ### TLS_PSK and perfect forward secrecy My question is about how difficult is for someone having compromised the PSK and sniff all (DT)LS handshake, (so getting the NONCE), to get the encryption keys? I understand : getting the pre-master ... 243 views ### Encryption Algorithms in Verilog code Can we implement Encryption Algorithms like RC4,AES and DES in Verilog? if no what are the limitations? 103 views ### Is this client-side password hash scheme secure? I want to build an app that uses client-side encryption for storing encrypted data on the server. A user-specific master key would be used, so to easily share it between devices a encrypted version of ... 61 views ### Public SRP verifiers or public hash chain “public keys” when secret is low entropy password I want to set up the following: An untrusted server should host user sessions and authenticate users without knowing their passwords or being capable of creating a fake user session in the user's ... 4k views ### How do we know a cryptographic primitive won't fail suddenly? It took more than a decade from when MD5 looked like it was going to break to the point when it was actually broken. That's more than a decade of warning. How can we be sure that when our ... 91 views ### How does the small n attack work? I'm studying the Lamport's Hash one-time password scheme. This is the scheme: Alice wants to authenticate herself to Bob from a workstation that knows nothing about her. Alice only knows her ... 140 views ### Do we have to consider the cryptographic properties of the decryption, if we're only using the encryption? Treyfer for example has slower diffusion in decryption. It has full diffusion between words in 2 rounds for encryption and 7 rounds for decryption. Do we have to consider the (bad) cryptographic ... 774 views ### Authentication using AES I need to implement an authentication mechanism in an embedded environment which does not support floating point operations but has an AES accelerator module which allows for encyption/decryption ... 60 views ### Secure “table-look-up” operation I am thinking about this problem: The server has a two-column table $T=\{c_1,c_2\}$. A user has a query $q$ that is sent to server. The server find an item $x$ in the first column satisfying $x=q$ ... 654 views ### Reuse of a DH / ECDH public key I was wondering whether it is safe to use the same DH or ECDH key pair in more than one key agreement, particularly if these public keys are in a public registry. These public keys could be used by ... 197 views ### Shared modulus attack on RSA Consider a scenario in which a group of people use a common modulus $n$ in the textbook RSA crypto system, where $n=pq$, $p$ and $q$ are large distinct prime numbers. Is there any chance for an ... 135 views ### Is there specific name for simple XOR with ADD/SUB encoding? Several years ago a group of malware authors began using code like this: for (j = 0; j < length; j++) ptr[j] = (ptr[j] ^ VALUEA) - VALUEB; for encoding, and ... 55 views ### Are Anonymity and Authentication possible together? Anonymity and authentication look to be seemingly contradictory but are both possible together ? I tried reading couple of papers but could not get the big picture. So what are the major approaches ... 154 views ### If I compute the SHA-1 of all possible 40 digits hexadecimal strings, will I get all possible sha1 hashes? [duplicate] Wikipedia says sha1 produces a 160-bit (20-byte) hash value. So if I compute the SHA-1 of all possible SHA-1 values, which as far as I understand is at current time impossible both to calculate and ... 2k views ### Are common cryptographic hashes bijective when hashing a single block of the same size as the output? It's been said that CRC-64 is bijective for a 64-bit block. It the corresponding statement true for typical cryptographic hashes, like MD5, SHA-1, SHA-2 or SHA-3? For example, would SHA-512 be ...	2016-04-29 08:23:31	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7363519668579102, "perplexity": 2668.0110912061728}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860110805.57/warc/CC-MAIN-20160428161510-00165-ip-10-239-7-51.ec2.internal.warc.gz"}
https://ai.stackexchange.com/tags/markov-decision-process/new	# Tag Info 1 vote ### $E_{\pi}[R_{t+1}\|S_t=s,A_t=a] = E[R_{t+1}\|S_t=s,A_t=a]$? Question: Can I write it without the subscript? So $$E_{\pi}[R_{t+1}\|S_t=s,A_t=a] = E[R_{t+1}\|S_t=s,A_t=a]$$ Yes, your reasoning is sound, there is no need to condition the expectation on the policy, ... • 23.8k Please look at line 5: If $P(a_{i,j}\|s_i)$ is equal to the policy that is used for generating the demonstrated trajectories, then it could be the same. However, in inverse RL you don't know \$P(a_{i,j}\|...	2022-06-28 03:38:46	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7719781398773193, "perplexity": 755.2001835403503}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103347800.25/warc/CC-MAIN-20220628020322-20220628050322-00697.warc.gz"}
https://masm32.com/board/index.php?topic=7824.0	Author Topic: Division by using Subtraction, I need help (Read 2212 times) xcbart • Regular Member • Posts: 8 Division by using Subtraction, I need help « on: May 02, 2019, 04:33:22 AM » Hi, My program's supposed to perform division by using subtraction. However, I'm getting nonsense results as output. Could someone help me what's wrong with it? Thank you. Code: [Select] `include Irvine32.inc.datafirstPrompt byte "Please enter first integer : ",0secondPrompt byte Please enter second integer : ",0buffer_size = 10bufferFirst byte buffer_size dup(?)bufferSecond byte buffer_size dup(?)firstInt dword 0secondInt dword 0divResult dword 0.codemain proc call getInteger call divt mov edx,firstInt call writeInteger mov al,'/' call writechar mov edx,secondInt call writeInteger mov al,'=' call writechar mov eax,divResult call writeDec call crlf myexit proc mov eax, white+16*black call settextcolor call waitmsg ret myexit endp main endpdiv proc pushad mov ecx,firstInt mov ebx,0 subtracting: sub ebx,secondInt loop subtracting mov divResult,ebx popad retdivt endpgetInteger proc mov edx, offset firstPrompt call writestring call readint mov firstInt, eax mov edx,offset secondPrompt call writestring call readint mov secondInt, eax retgetInteger endpwriteInteger proc mov eax,edx call writedec retwriteInteger endpend main` jj2007 • Member • Posts: 11552 • Assembler is fun ;-) Re: Division by using Subtraction, I need help « Reply #1 on: May 02, 2019, 05:46:16 AM » How do you return your result? Code: [Select] ` popad ; <<<<<<<<<<<< retdivt endp` tenkey • Regular Member • Posts: 39 Re: Division by using Subtraction, I need help « Reply #2 on: May 02, 2019, 09:17:57 AM » You need to rethink your division algorithm. It's actually a multiplication algorithm. jj2007 • Member • Posts: 11552 • Assembler is fun ;-) Re: Division by using Subtraction, I need help « Reply #3 on: May 02, 2019, 09:36:48 AM » I've had a second look, here is a version that works, using the initial idea to subtract the divisor result times from dividend: Code: [Select] `include \masm32\include\masm32rt.inc.datafirstInt dd 100 ; dividendsecondInt dd 15 ; divisordivResult dd ? ; quotient.codedivt proc mov divResult, 0 mov ebx, firstInt subtracting: inc divResult sub ebx, secondInt ja subtracting retdivt endpstart: call divt inkey str\$(divResult), " is the result" exitend start` xcbart • Regular Member • Posts: 8 Re: Division by using Subtraction, I need help « Reply #4 on: May 03, 2019, 03:05:34 AM » I've had a second look, here is a version that works, using the initial idea to subtract the divisor result times from dividend: Code: [Select] `include \masm32\include\masm32rt.inc.datafirstInt dd 100 ; dividendsecondInt dd 15 ; divisordivResult dd ? ; quotient.codedivt proc mov divResult, 0 mov ebx, firstInt subtracting: inc divResult sub ebx, secondInt ja subtracting retdivt endpstart: call divt inkey str\$(divResult), " is the result" exitend start` Thank you	2021-09-28 01:57:18	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9675227403640747, "perplexity": 10147.824867663317}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780058589.72/warc/CC-MAIN-20210928002254-20210928032254-00510.warc.gz"}
http://mathoverflow.net/questions/116352/silly-question-on-complete-intersections	# Silly question on complete intersections Let $X_s = \bigcap_{i=1}^s H_i \subset \mathbb{P}^N$ be a complete intersection, where each $H_i$ is a hypersurface of degree $d_i$. Is $X_{s-1}$ also a complete intersection? Is the conormal sheaf of $X_s$ in $X_{s-1}$ isomorphic to $\mathcal{O}_{X_s}(-d_s)$? - Yes for the first question. For the second replace $d_i$ with $d_s$, then the answer is positive as well. – Sasha Dec 14 '12 at 7:33	2015-04-01 10:56:01	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.957679033279419, "perplexity": 181.73619803583216}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-14/segments/1427131304444.86/warc/CC-MAIN-20150323172144-00009-ip-10-168-14-71.ec2.internal.warc.gz"}
https://shoka.lostyu.me/computer-science/mse/itmd-522/lecture-5/	# # Neural Networks ## # Artificial Neural Networks (ANN) • Researchers tried to learn from the biological neuron systems and built the ANN 研究人员试图向生物神经元系统学习，并建立了 ANN • There are many neuron units in ANN ANN 中有许多神经元单位 • They are connected within a structure 他们是连接在一个结构 • They work as threshold switching units 他们作为阈值切换单元工作 • There are weighted interconnections among units 有加权互联单位之一 • We are able to learn and tune up these weights automatically by a training process 我们能够通过训练过程自动学习和调整这些重量 • A neuron in ANN looks like this… 人工神经网络中的一个神经元看起来像这样… input signals → input function(linear) → activation function(nonlinear) → output signal 输入信号→输入函数 (线性)→激活函数 (非线性)→输出信号 ## # Perceptron 感知器 • First neural network learning model in the 1960’s 20 世纪 60 年代的第一个神经网络学习模型 • Simple and limited (single layer models) 简单且有限 (单层模型) • Still used in current applications (modems, etc.) 仍在当前应用中使用 (调制解调器等)。 Input 输入 different $x$ variables with weights on edges Input function 输入函数 it is used to aggregate the inputs, usually it is a weighted sum of its inputs Activation function 激活函数 • It is a threshold function 是一种阈值函数 • For the purpose of binary classification 为目的的二元分类 • The output is only 1 or 0 输出是只 1 或 0 • Sign function can be used as the activation function 符号函数可以用来激活功能 • Sigmoid can be used as activation function 乙状函数可用作激活函数 Sigmoid function 乙状函数 / S 函数 It is popular for classification, due to being easy to be updated and learned in the training process. • Neural networks canbe used for both classifications and regressions. 神经网络既可以用于分类，也可以用于回归 • It can be controlled by applying different activation functions. 可以通过施加不同的激活函数进行控制。 ### # Perceptron Training 感知器训练 • It is the simplest ANN model 这是最简单的人工神经网络模型 • We need to train the model to learn the weights, $w$, where 我们需要训练模型来学习权重，$w$，其中 $w_{i} \leftarrow w_{i}+\Delta w_{i}$ $\Delta w_{i}=\eta(t-o) x_{i}$ • $t$ is the real value 是实际值 • $o$ is the output value (prediction by the model) 是输出值 (由模型预测) • $\eta$ is a constant value in $[0, 1]$ as the learning rate $[0, 1]$ 中的一个常数值，作为学习速率 • It is a process of iterative learning 这是一个反复学习的过程 • At the beginning, give random values to $w$ 开始时，给$w$ 随机取值 • Get the output $o$ through the perceptron 通过感知器得到输出$o$ • Update the $w$ by using the update rules 使用更新规则更新$w$ • Stop the learning process by a stopping criterion 通过停止标准停止学习过程 • Classification error is smaller than a threshold 分类误差小于阈值 • Or, maximal learning iterations have been reached 或者，已经达到最大学习迭代次数 ### # Perceptron: Example • Consider learning the logical OR function 考虑学习逻辑或函数 Samplex0x1x2label 11000 21011 31101 41111 • Activation function 激活函数 $S=\sum_{k=0}^{k=n} w_{k} x_{k} \quad S>0 \text { then } O=1 \quad \text { else } \quad O=0$ • We’ll use a single perceptron with three inputs. 我们将使用一个有三个输入的感知器。 • We’ll start with all weights 0 W= <0,0,0> 我们将从所有重量 0 开始 • Example 1 I = <0,0,0> label = 0 W = <0,0,0> • Perceptron ($1 \times 0 + 0 \times 0 + 0 \times 0 = 0, S=0$) output = 0 • it classifies it as 0 , so correct, do nothing 它将其归类为 0 ，所以正确，什么也不做 • Example 2 I = <1,0,1> label=1 W = <0,0,0> • Perceptron ($1 \times 0 + 0 \times 0 + 1 \times 0 = 0$) output = 0 • it classifies it as 0 , while it should be 1 , so we add input to weights W = <0,0,0> + <1,0,1>= <1,0,1> 它将其分类为 0，而它应该是 1，所以我们将输入添加到权重 W = <0,0,0> + <1,0,1>= <1,0,1> • Example 3 I = <1,1,0> label = 1 W = <1,0,1> • Perceptron ($1 \times 0 + 1 \times 0 + 0 \times 0 \gt 0$) output = 1 • it classifies it as 1 , correct, do nothing W = <1,0,1> 它将其分类为 1 ，正确，什么都不做 W = <1,0,1> • Example 4 I = <1,1,1> label = 1 W = <1,0,1> • Perceptron ($1 \times 0 + 1 \times 0 + 1 \times 0 \gt 0$) output = 1 • it classifies it as 1 , correct, do nothing W = <1,0,1> 1st iteration is completed. 第一次迭代完成。 Repeat until no errors 重复，直到没有错误 ### # Limitations of Perceptron 感知器的局限性 • It is too simple, cannot learn complex and effective models 它太简单，无法学习复杂而有效的模型 • It assumes the data can be linearly separatable in the binary classification, but actually it could be non linear! 它假设数据在二元分类中可以线性分离，但实际上它可能是非线性的！ • SVM, we use kernel function to map data to higher dimension SVM，使用核函数将数据映射到更高维度 • ANN, we can add more layers!! ANN，可以加更多的层 ## # Multi layer Feed forward Networks 多层前馈网络 • Multi layer Feed forward Networks is an extension of the perceptron model. It adds hidden layers to the original perceptron. 多层前馈网络是感知器模型的扩展。它将隐藏层添加到原始感知器中。 • Input layer: accepts inputs only 输入层：仅接受输入 • Hidden layers: neurons with functions 隐藏层：具有功能的神经元 • Output layer: produce outputs 输出层：产生输出 ### # Training Phrase • The training phrase is a typical process of machine learning and optimization 训练阶段是机器学习和优化的典型过程 • We need to 我们需要 • Setup a learning objective as loss function 将学习目标设置为损失函数 • Use appropriate optimizer to learn the parameters 使用适当的优化器来学习参数 • It is usually a process of iterative learning 这通常是一个迭代学习的过程 ### # Loss Function 损失函数 • The loss function $L\left(x, y, y^{\prime}\right)$ is defined as the amount of utility lost by predicting $h(x)=y^{\prime}$ when the correct answer is $f(x)=y$ 损失函数定义为当正确答案是 $f(x)=y$ 时，通过预测而损失的效用量 • Often a simplified version is used, $L\left(y, y^{\prime}\right)$, that is independent of $x$ 通常使用简化的版本，独立于$x$ • Three commonly used loss functions: 三种常用的损失函数: • Absolute value loss: 绝对值损失 $L_{1}\left(y, y^{\prime}\right)=\left\|y-y^{\prime}\right\|$ • Squared error loss: 平方误差损失 L_{2}\left(y, y^{\prime}\right)=\left(y-y^{\prime}\right)^ • 0/1 loss: $L_{0 / 1}\left(y, y^{\prime}\right)=0$ if $y=y^{\prime}$, else $1$ • Let $E$ be the set of examples. Total loss $L(E)=\sum_{e \in E} L(e)$ ### # Optimizer: Gradient Descent 优化器：梯度下降 • Gradient Descent is widely used as one of the popular optimizers in machine learning, especially in the ANN learning 梯度下降法是机器学习，尤其是人工神经网络学习中最常用的优化方法之一 #### # Optimization In Linear Regression • How to apply gradient descent to minimize the cost function for regression 如何应用梯度下降来最小化回归的成本函数 1. a closer look at the cost function 仔细看看成本函数 2. applying gradient descent to find the minimum of the cost function 应用梯度下降来寻找成本函数的最小值 ##### # a closer look at the cost function • Hypothesis: 假设 $h_{\theta}(x)=\theta_{0}+\theta_{1}x$ • Parameters: 参数 $\theta_{0}, \theta_{1}$ • Cost Function: 成本函数 Sum of squared errors 误差平方和 $J\left(\theta_{0}, \theta_{1}\right)=\frac{1}{2 m} \sum_{i=1}^{m}\left(h_{\theta}\left(x^{(i)}\right)-y^{(i)}\right)^{2}$ • Goal: $\underset{\theta_{0}, \theta_{1}}{\operatorname{minimize}} J\left(\theta_{0}, \theta_{1}\right)$ • Optimization • There are at least two optimization methods 至少有两种优化方法 • Least square optimization 最小平方优化 • Optimization based on gradient descent 基于梯度下降的优化 • Least Square Optimization 最小平方优化 • Find the optimal point 找到最佳点 $\frac{\partial}{\partial \theta_{j}} J=0$, $J$ is the objective function with $\theta_{1}, \theta_{2}, \theta_{3}, \ldots$ • $j=1,2,3, \ldots, N+1$, assume $u$ have $N \times$ variables • Therefore, you will have $N+1$ functions to be solved 因此，将有$N+1$ 个函数需要求解 • Drawback: it is complicated if you have many $x$ variables 缺点：如果有许多$x$ 变量，这就复杂了 ##### # applying gradient descent to find the minimum of the cost function • Have some function $J\left(\theta_{0}, \theta_{1}\right)$ • Want $\min _{\theta_{0}, \theta_{1}} J\left(\theta_{0}, \theta_{1}\right)$ • Gradient descent algorithm outline: 梯度下降算法概述: • Start with some $\theta_{0}, \theta_{1}$ ; • Keep changing $\theta_{0}, \theta_{1}$ to reduce $J\left(\theta_{0}, \theta_{1}\right)$ until we hopefully end up at a minimum ## # Backpropagation Training 反向传播训练 • There are several network structure in neural networks, such as feed forward neural networks and the recurrent neural networks 神经网络有几种网络结构，如前馈神经网络和递归神经网络 • Multi layer Feed forward Networks used a forward procedure for predictions. 多层前馈网络使用前向程序进行预测。 But it was trained by using a Backward propagation approach 但它是用反向传播方法训练的 • These ANNs are also called BP (Backpropagation) Neural Networks 这些人工神经网络也被称为 BP (反向传播) 神经网络 ### # ANN needs a process of weight training • A set of examples, each with input vector $x$ and output vector $y$ 一组例子，每个例子都有输入向量$x$ 和输出向量$y$ • Squared error loss: $Loss =\sum_{k} \operatorname{Loss}_{k}, \operatorname{Loss}_{k}=\left(y_{k}-a_{k}\right)^{2}$, where $a_{k}$ is the $k$-th output of the neural net • The weights are adjusted as follows: $w_{i j} \leftarrow w_{i j}-\alpha \partial L o s s / \partial w_{i j}$ • How can we compute the gradient efficiently given an arbitrary network structure? ### # Forward vs Backward in ANN Forward phase: • Propagate inputs forward to compute the output of each unit • Output $a_{j}$ at unit $j$: $a_{j}=g\left(i n_{j}\right)$ where $in_{j}=\sum_{i} w_{i j} a_{i}$ . Backward phase: • Propagate errors backward • For an output unit $j$: $\Delta_{j}=g^{\prime}\left(i n_{j}\right)\left(y_{j}-a_{j}\right)$ • For an hidden unit $i$: $\Delta_{i}=g^{\prime}\left(i n_{i}\right) \sum_{j} w_{i j} \Delta_{j}$ . ## # Neural Networks and Deep Learning • To make ANN more powerful, there are two solutions 为了使 ANN 更加强大，有两种解决方案 • Add more neurons in the hidden layer 在隐藏层中添加更多的神经元添加更多隐藏层 • Deep Learning Deep Learning • Traditional ANN only has 3 layers. Deep learning utilizes neural networks with multiple layers 传统的人工神经网络只有 3 层。深度学习利用多层神经网络 • Deep learning have more structures for neural networks, such as ANN, Convolutional Neural Networks (CNN), Recurrent Neural Networks (RNN), and so forth 深度学习有更多的神经网络结构，如 ANN、卷积神经网络 (CNN)、递归神经网络 (RNN) 等等 • Deep learning is not related to neural networks only. It also correlates with computing, such as GPU 深度学习不仅仅与神经网络相关。它还与计算相关，如 GPU • ANN vs Deep Learning # # Ensembles of Classifiers 分类器集合 • Basic idea is to learn a set of classifiers (experts) and to allow them to vote. 基本思想是学习一组分类器 (专家) 并允许他们投票。 • Advantage: improvement in predictive accuracy. 优点：预测精度提高。 • Disadvantage: it is difficult to understand an ensemble of classifiers. 缺点：很难理解分类器的集成。 ## # Ensemble Methods 集成方法 ### # Bagging • Process in bagging 装袋过程 • Sample several training sets of size $n$ (instead of just having one training set of size $n$) 采样几个大小为$n$ 的训练集 (而不是只有一个大小为$n$ 的训练集） • Build a classifier for each training set 为每个训练集构建一个分类器 • Combine the classifier’s predictions by voting or averaging 通过投票或平均来组合分类器的预测 • Bagging classifiers • Classifier generation Let n be the size of the training set. For each of t iterations: Sample n instances with replacement from the training set. Apply the learning algorithm to the sample. Store the resulting classifier. • classification For each of the t classifiers: Predict class of instance using classifier. Return class that was predicted most often. • Voting and Averaging 投票和平均 • Voting is used for classifications, and averaging is used for regressions 投票用于分类，平均用于回归 • Voting: Hard and Soft voting 投票：硬投票和软投票 • Hard voting Predictions: Classifier 1 predicts class A Classifier 2 predicts class B Classifier 3 predicts class B 2/3 classifiers predict class B, so class B is the ensemble decision. • Soft voting Predictions (identical to the earlier example, but now in terms of probabilities.Shown only for class A here because the problem is binary): Classifier 1 predicts class A with probability 99% Classifier 2 predicts class A with probability 49% Classifier 3 predicts class A with probability 49% The average probability of belonging to class A across the classifiers is (99+49+49)/3 = 65.67% . Therefore, class A is the ensemble decision. • Why does bagging work? • Bagging reduces variance by voting / averaging, thus reducing the overall expected error • In the case of classification there are pathological situations where the overall error might increase • Usually, the more classifiers the better ### # Boosting • Also uses voting/averaging but models are weighted according to their performance • Iterative procedure new models are influenced by performance of previously built ones • New model is encouraged to become expert for instances classified incorrectly by earlier models • Assign more weights to the misclassified instances to improve the classification iteratively • There are several variants of this algorithm • classifier generation Assign equal weight to each training instance. For each of t iterations: Learn a classifier from weighted dataset. Compute error e of classifier on weighted dataset. If e equal to zero, or e greater or equal to 0.5: Terminate classifier generation. For each instance in dataset: If instance classified correctly by classifier: Multiply weight of instance by e / (1 - e) Normalize weight of all instances. • classification Assign weight of zero to all classes. For each of the t classifiers: Add -log(e / (1 - e)) to weight of class predicted by the classifier. Return class with highest weight. ### # Random Forest • Random forest is a bagging method which uses decision trees as the classifiers • The workflow in the random forest is the same as the ones in bagging • In bagging, we can use any classifiers In random forest, we use decision trees • Classifier generation Let n be the size of the training set. For each of t iterations: (1) Sample n instances with replacement from the training set (2) Learn a decision tree s.t. the variable for any new node is the best variable among m randomly selected variables. (3) Store the resulting decision tree. • Classification For each of the t decision trees: Predict class of instance. Return class that was predicted most often. # # Semi-Supervised Classification 半监督分类 • Classifications require labeled data • Data labeling is a complicated and expensive process. It is not guaranteed that we have enough and high qualified labels • Labels may be hard to get • Human labeling is slow and boring • It may require expert knowledge • It may require special or expensive devices • Goal: Using both labeled and unlabeled data to build better classifiers (than using labeled data alone). • Notation: • input $x$, label $y$ • classifier f: \mathcal{X} \mapsto \mathcal • labeled data $\left(X_{l}, Y_{l}\right)=\left\{\left(x_{1}, y_{1}\right), \ldots,\left(x_{l}, y_{l}\right)\right\}$ • unlabeled data $X_{u}=\left\{x_{l+1}, \ldots, x_{n}\right\}$ • usually $n \gg l$ ## # Solutions: Self-training • Algorithm: Self-training 1. Pick your favorite classification method. Train a classifier $f$ from $\left(X_{l}, Y_{l}\right)$. 2. Use $f$ to classify all unlabeled items $x \in X_{u}$. 3. Pick $x^{}$ with the highest confidence, add $\left(x^{}, f\left(x^{}\right)\right)$ to labeled data. 4. Repeat. The simplest semi-supervised learning method. • Pros • Simple • Applies to almost all existing classifiers • Cons • Mistakes reinforce themselves. Heuristics against pitfalls • 'Un-label' a training point if its classification confidence drops below a threshold • Randomly perturb learning parameters ## # Solutions: Co-training • Your data can be split into different views • The view can be defined by different set of the features • Each item is represented by two kinds of features $x=\left[x^{(1)} ; x^{(2)}\right]$ • $x^{(1)}$ = image features • $\boldsymbol{\square} x^{(2)}$ = web page text • This is a natural feature split (or multiple views) • Co-training idea: • Train an image classifier and a text classifier • The two classifiers teach each other • Algorithm: Co-training 1. Train two classifiers: $f^{(1)}$ from $\left(X_{l}^{(1)}, Y_{l}\right), f^{(2)}$ from $\left(X_{l}^{(2)}, Y_{l}\right)$ 2. Classify $X_{u}$ with $f^{(1)}$ and $f^{(2)}$ separately. 3. Add $f^{(1)}$'s $k$-most-confident $\left(x, f^{(1)}(x)\right)$ to $f^{(2)}$'s labeled data. 4. Add $f^{(2)}$'s $k$-most-confident $\left(x, f^{(2)}(x)\right)$ to $f^{(1)}$'s labeled data. 5. Repeat. • Pros • Simple. Applies to almost all existing classifiers • Less sensitive to mistakes • Cons • Feature split may not exist • Models using BOTH features should do better # # Multi-Label Classifications • Binary classification: Is this a picture of the sea? $\in\{ yes, no \}$ • Multi-class classification: What is this a picture of? $\in\{ sea, sunset, trees, people, mountain, urban \}$ • Multi-label classification: Which labels are relevant to this picture? $\subseteq\{ sea, sunset, trees, people, mountain, urban \}$ i.e., multiple labels per instance instead of a single label! ## # Applications • Images are labelled to indicate • multiple concepts • multiple objects • multiple people e.g., Scene data with concept labels $\subseteq\{ beach, sunset, foliage, field, mountain, urban \}$ • Labelling music/tracks with genres / voices, concepts, etc. • e.g., Music dataset, audio tracks labelled with different moods, among: • amazed-surprised, • relaxing-calm, • quiet-still, • angry-aggressive ## # Example • Difference in data sets • Table: Single-label $Y \in \{0,1\}$. • Table: Multi-label $Y \subseteq\left\{\lambda_{1}, \ldots, \lambda_{L}\right\}$ • We usually convert labels to binary labels ## # Solutions ### # Transformation Based Methods Transform the task to binary/multi-class classifications #### # Binary Relevance • If there are $N$ labels, we have $N$ binary classifications • Drawback: it ignores the label depenence #### # Classifier Chains • Classifier Chains build the model in a chain by taking label correlations into consideration • It uses the feature to perform binary classification on 1st label, the prediction on 1st label will be reused as the features into the 2nd step to predict the 2nd label • Repeat the process above until all of the labels are predicted • Use previous prediction results as new features • Drawbacks in Classifier Chains • Difficult to define the sequence in the chain, though there are some methods (e.g., info gain) • If the previous predictions are incorrect, the following predictions may not be right too. #### # Label Powerset • Each subset of the label set will be a single label • Assign binary classification or multi-class classification to them • Find a way to aggregate the results 1. Transform dataset ...into a multi-class problem, taking $2^{L}$ possible values: 2. ...and train any off-the-shelf multi-class classifier • Drawbacks in Label Powerset 标签权力集的缺点 • Too many subsets if there are several labels 如果有多个标签，则有太多的子集 • Highly possible to have imbalance issue 极有可能出现不平衡的问题 • Overfitting: how to predict new values/labels? 过度拟合：如何预测新值 / 标签？ ### # Adaptation Based Methods 基于适应性的方法 Develop new algorithms to solve the problem • MLkNN.For each test instance: • Retrieve the top-k nearest neighbors to each instance • Compute the frequency of occurrence of each label • Assign a probability to each label and select the labels by using a probability cut-off value ## # Evaluation of multilabel learning Notes • Both transformation and adaptation methods are the methods to solve MLC problem • They are not classification algorithms • For each method, you can use any traditional binary/multi-class classification algorithms to produce the predictions • There are multiple labels in the MLC problem • Traditional evaluation metrics in the classification may not work for MLC • We need to develop new evaluation metrics ### # Hamming Loss Consider the misclassification in each bit $\text { HAMMING LOSS } =\frac{1}{N L} \sum_{i=1}^{N} \sum_{j=1}^{L} \mathbb{I}\left[\hat{y}_{j}^{(i)} \neq y_{j}^{(i)}\right] = 4 /(4 5) \\ =0.20$ N = # of labels L = # of data rows ### # 0/1 Loss Consider the misclassification in the whole label set $0 / 1 \mathrm{LOSS} =\frac{1}{N} \sum_{i=1}^{N} \mathbb{I}\left(\hat{\mathbf{y}}^{(i)} \neq \mathbf{y}^{(i)}\right)=3 / 5 \\ =0.60$ ### # Other Metrics JACCARD INDEX often called multi-label ACCURACY RANK LOSS average fraction of pairs not correctly ordered ONE ERROR if top ranked label is not in set of true labels COVERAGE average "depth" to cover all true labels LOG LOSS i.e., cross entropy PRECISION predicted positive labels that are relevant RECALL relevant labels which were predicted • PRECISION VS. RECALL curves F-MEASURE • micro-averaged ('global' view) • macro-averaged by label (ordinary averaging of a binary measure, changes in infrequent labels have a big impact) • macro-averaged by example (one example at a time, average across examples) ## # Tools Mulan • Java Based • Reuse Weka library • No UI • http://mulan.sourceforge.net/ Meka • Similar to Weka • Java Based • With UI • http://meka.sourceforge.net/ # # Classification: Summary • We learned different algorithms • No learning process: KNN and Naïve Bayes • Learning based: Logistic regression, Decision tree, SVM, Neural Networks • Ensemble methods: bagging, boosting, RandomForest • For each algorithm 对于每一种算法 • Understand how it works 了解它是如何工作的 • Know the requirements on the data; Know how to prepare a preprocessed data set 知道对数据的要求；知道如何准备一个预处理的数据集 • Know what are the parameters to be tuned up 知道哪些是需要调整的参数 • Know the solutions for overfittings 知道超配的解决方案 • Which algorithm is the best? 哪种算法是最好的？ • It varies from data to data 不同的数据会有不同的结果 • We need to tune parameters to tune up the model 我们需要调整参数来调优模型 • We need to compare different classification models 我们需要比较不同的分类模型 • General issue: imbalance in labels 一般问题：标签的不平衡性	2022-05-29 09:32:05	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 112, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6140317916870117, "perplexity": 5473.953280303612}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652663048462.97/warc/CC-MAIN-20220529072915-20220529102915-00088.warc.gz"}
http://www.edurite.com/kbase/like-charges-repel-and-unlike-charges-attract	#### • Class 11 Physics Demo Explore Related Concepts # like charges repel and unlike charges attract Question:So, can charged objects attract neutral ones? Like, say I have a negatively charged polyethylene strip and an uncharged pith ball. If I bring the strip close, will the pith ball be affected? (either attracted or repelled?) ...And while I'm at it, would rubbing an acetate strip with silk charge it positively? :D Thank you for your time! Answers:the negative charged rod will attract the neutral pith ball - it attracts the positive charges to the surface closest to the rod whilst repelling the negatives Question:Can two adjacent free electrons attract or repel to each other? Thanks for your help. Answers:well sense they have the same charge they must repel each other. electrons do not like being close to eachother. is this a trick question? Question:Is the electric potential energy of two unlike charges postive or negative? What about two like charges? What is the significance of the sign of the potential energy in each case? (23/19) Answers:Two unlike charges are by definition one positive and one negative. Think of the number line from elementary school math. Examples to illustrate: Negative 2 volts and positive 2 volts have a potential, or difference of 4 volts. If two like charges have the same voltage, then there is no potential energy. If there are two positive charges, but one is 2 volts and one is 4 volts there is a potential energy of 2 volts between them. Think of the number line again. Both are positive but there is a potential of 2 between them. The same holds for negative charges. Potential should never be a negative number because potential is the distance between two points on a number line, and that distance should be either zero or a positive number. Then what are people saying when they say that there is negative x volts? That means that the voltage is being measured against another pole that is x volts more positive. In other words it is a relative term. It is -x volts, but only relative to the other more positive pole, Question:I still do not get what current means? In order for me to get current, i need to know what charge means? Also why does the current stay the same in a series circuit? Cause doesnt charge get used up by each resistor? And why does current change in a parallel circuit for each resistor? Answers:A charge means the quantity of unbalanced electricity in a body (either positive or negative) and construed as an excess or deficiency of electrons. >there are two kinds of charge, positive and negative >like charges repel, unlike charges attract >positive charge comes from having more protons than electrons; >negative charge comes from having more electrons than protons >charge is quantized, meaning that charge comes in integer multiples of the elementary charge e >charge is conserved And this flow of charge constitutes electric current. Current is taken in the direction of the flow of protons and in opposite in the direction of electron. V=IR, is the ohms law. In series circuit there are many resistors, every resistor may have different resistance and hence diff potential difference at its end. So voltage across series circuit is the sum of the voltages of each resistor. So according to ohm's law I remains constant as resistance and pd varies. In parallel circuit even the resistance varies the potential diff remain same along each resistors so according to ohm's law current varies as voltage remains same and resistance varies. I have the answer long because you had not understood charge properly, which is the basic. Though it may be tough for you to get it at first, its easy. http://physics.bu.edu/~duffy/PY106/Charge.html This website helps a lot for beginners, really. Try it. The Forces of Attraction :for your viewing pleasures, ive uploaded this to youtubee ~ the low quality is ugly though, so watch on fb or in HQ an informational video on the three types of bonding ionic, covalent, and iconic ;] --- the end result of 12 hours of photography and two hours of video editing for those of you who dont knoww... this is what STOP MOTION is i wouldn`t really recommend using this unless you want to waste er i mean DEVOTE two weekends to working on a project but if you want to do it anywayss my advice to you is to use darker markers, and have at least 2 fully charged cameras --- disclaimer: music not owned by me all rights go to their respective owners. songs used two is better than one - boys like girls ft. taylor swift meet me halfway - black eyed peas you found me - kelly clarkson check yes juliet - we the kings	2016-10-28 17:42:29	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4102601706981659, "perplexity": 763.4073308034971}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988725451.13/warc/CC-MAIN-20161020183845-00135-ip-10-171-6-4.ec2.internal.warc.gz"}
https://www.vasp.at/wiki/index.php/LAECHG	Requests for technical support from the VASP group should be posted in the VASP-forum. # LAECHG LAECHG = .TRUE. \| .FALSE. Default: LAECHG = .FALSE. Description: when LAECHG=.TRUE. the all-electron charge density will be reconstructed explicitly and written out to file. If LAECHG=.TRUE. is set VASP will reconstruct the all-electron charge density on the so-called "fine" FFT-grid. This "fine" FFT-grid consists of NGXF×NGYF×NGZF points in real space (i.e., the grid that is used to represent the augmented pseudo charge densities of the USPP and PAW methods). In fact, for LAECHG=.TRUE., VASP will reconstruct three distinct all-electron densities: 1. the core density. 2. the proto-atomic valence density (overlapping atomic charge densities). 3. the self-consistent valence density. These are written to the files AECCAR0, AECCAR1, and AECCAR2, respectively. The first two of these files are written at the start of the run, whereas the last is written at the end after self-consistency has been reached. N.B.: In the language of the PAW method an "all-electron" density does not refer to the density of all electrons, instead it denotes a density that includes all the nodal features near the nucleus associated with the true (as opposed to the pseudized) one-electron orbitals. Within the PAW method, the all-electron density arising from the one-electron pseudo orbitals ${\displaystyle \{\widetilde {\psi }_{{n{\mathbf {k}}}}\}}$ is given by: ${\displaystyle n({\mathbf {r}})=\sum _{{n{{\mathbf {k}}}}}f_{{n{{\mathbf {k}}}}}\langle \widetilde {\psi }_{{n{\mathbf {k}}}}\|{\mathbf {r}}\rangle \langle {\mathbf {r}}\|\widetilde {\psi }_{{n{\mathbf {k}}}}\rangle +\sum _{{\alpha ,\beta }}(\phi _{\alpha }^{\ast }({\mathbf {r}})\phi _{\beta }({\mathbf {r}})-\widetilde {\phi }_{\alpha }^{\ast }({\mathbf {r}})\widetilde {\phi }_{\beta }({\mathbf {r}}))\sum _{{n{{\mathbf {k}}}}}f_{{n{{\mathbf {k}}}}}\langle \widetilde {\psi }_{{n{\mathbf {k}}}}\|\widetilde {p}_{\alpha }\rangle \langle \widetilde {p}_{\beta }\|\widetilde {\psi }_{{n{\mathbf {k}}}}\rangle }$ Normally one does not attempt to reconstruct all-electron densities explicitly since the second term on the right hand side varies rapidly near the nuclei and is too costly to expand in plane waves (see the bit about augmentation and compensation charges). For LAECHG=.TRUE., however, this reconstruction is exactly the thing that is done.	2020-11-24 01:42:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6964272856712341, "perplexity": 2217.721326694773}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141169606.2/warc/CC-MAIN-20201124000351-20201124030351-00055.warc.gz"}
http://chrisfmathsphysicsmusic.blogspot.com/2012/06/	## Sunday, 17 June 2012 ### M338 TMA03 Finished this today had to ask for a small extension due to work, but don't know if it's been granted as my tutor hasn't replied to my request. Anyway I'll submit it tomorrow and see what happens. This block was a bit more straightforward than Block A as it concerns what most people consider topology to be namely the invariance of certain features of a solid under a transformation. Thus the usual example given is that that a cup with a handle is topologically equivalent to a ring doughnut but not a solid doughnut as they both have a hole and can be deformed into each other. There is also a discussion of the Mobius band. I have to admit that my heart was sinking when I first started this block as it seemed to be all about visualisation and introducing boundaries in order to classify various combinations of holes and twists in a surface. As there seemed to be no systematic way of doing this it was all a bit vague. However by the second half of block 2 an algebraic method of charactersing surfaces was introduced and I was much happier. There are a number of ways of systematically obtaining the characteristic properties of a solid and these are essentially algebraic if a bit fiddly. The TMA was mostly concerned with the algebraic aspects of the block. There are three main characteristics of a surface The Euler Characteristic = V - E +F where V is the number of vertices E = the number of edges and F = is the number of faces. This is invariant for topologically equivalent surfaces. The Boundary number which is the number of separarate pieces forming the boundary of a surface The orientability number of a surface essentially the number of twists. Associated with each surface is an N sided polygon with arrows along each side denoting the orientation with a number of edges of the polygon having the same name. An edge expression is obtained by labelling the sides in order for a given direction of the arrow and labelling the side by its inverse if the arrow is pointing in an opposite direction. Various identifications give rise to different solids. Given such an edge expression one can then go onto calculate the characteristic numbers of a surface. This process is quite straightforward but can be quite fiddly. Also the edge equations can be reduced to canonical form which enables the classification of the surface to be made easily both directly from the edge expression and also there is a method of classifying the solids in terms of the connected sums of constituent surfaces eg Torus's Closed disk's etc. The final block concerned a discussion of the coloring theorems. One of the key quantities is the chromaticiy number which is related to the number of handles of a given surface and gives the minimum number of colours required to colour the parts of the surface so that no regions next to each other have the same colour. As some readers will probably know the minimum number of colours to colour a map is 4. But this wasn't actually proven until the 1970's and only by a computer so it's doubtful to old fashioned mathematicians whether it counts as a proof at all. http://en.wikipedia.org/wiki/Four_color_theorem Question 1 concerned identifying the edge expression of a hexagon, obtaining the edge expressions and the characteristic numbers. This question seemed quite straightforward Question 2 was a question on the subdivisions of a surface for a given Euler characteristic this is quite straightforward Question 3 The bulk of the TMA with a whopping 40% of the marks concerned obtaining a single edge expression from a number of constituent ones, then deducing the characteristic numbers. Then peforming a canonical transformation and obtained the connected sum form for the surface and then using that to deduce the characteristic numbers. Fortunately the numbers seemed to tie up so I'm consisitent if not correct. Question 4 The final question involved obtaining some inequalities for the number of handles h in terms of the chromaticity number. This seemed quite straigthforward or at least the first part. However the last part which asked us to find the range of h for a large chromaticity numbers. On the face of it this seemed quite straightforward but the sting in the tail was that dreaded phrase "justifying your answers fully" as there were 10 marks for this part. I'm sure we were supposed to do more than solve the inequality for the various values of h. But I couldn't see what. So on the whole I think I've done reasonably well assuming my tutor accepts my late submission but that last part has me worried that I've missed something quite fundamental. ## Sunday, 3 June 2012 ### Pilot Waves for and against part 2 So as promised here is a basic summary of the pros and cons of De-Broglies Bohm's Pilot Wave approach to quantum mechanics. A really good summary of which can be found in these lectures by Mike Towler. http://www.tcm.phy.cam.ac.uk/~mdt26/pilot_waves.html So here are the pros i) Contrary to the claim by the Copenhagen interpretation it has been shown that it is possible to define a definite trajectory for particles it's greatest success must be it's explanation of the two slit experiment. Sanity seems to be restored we no longer have to claim that a massive particle such as an electron or a bucky ball splits in two as it interacts with a screen which has a number of slits. ii) It brings the physics back to quantum mechanics, instead of losing contact with the world of particles and their interactions as the Dirac Formalism is apt to, it does provide a causal explanation for many quantum phenomenon. Mind you the Dirac formalism is really elegant and still encapsulates the essence of many quantum mechanical problems even if it is all quite abstract. iii) It avoids the problem of measurement, with all the inherent problems of the world around us being created by an act of measurement. iv) It now deserves a place as an equivalent mathematical formulation of quantum mechanics (at least for Non relativistic problems). The fact that more and more papers are being published using Bohmian mechanics has to be something. I think it's fair to say that other alternative interpretations such as the Many Worlds Interpretation have not yet reached the same degree of maturity. I would argue that once a theory of physics has reached the stage where people can 'Shut up and calculate' as the Pilot Wave theory has done then it has a right to be treated as mainstream physics. It is quite astonishing that at Solvay in 1927, when the basics of the theory was laid out by De-Broglie, that it wasn't considered at least as an alternative equivalent mathematical reformulation of quantum mechanics just as Heisenberg's matrix mechanics was. All the formulations of quantum mechanics had their problems of interpretation and it is at least arguable that the Pilot Wave theory has a well defined procedure for relating it's mathematical formalism to empirical results. Now for the cons i) It makes great play, that for it the wavefunction is a real field, something akin to an electric field or gravitaional one. Unfortunately this implies acceptance of the reality of 3N+1 dimensional configuration space, where N is the number of particles considered and 1 represents time. Let me explain a bit more the wave function of a manybody particle system is a function of all the coordinates of the particles considered eg for two particles with positions r1 and r2 the time independent of the wave function of the system is now a function of r1 and r2 that is we have $$\psi(x1,y1,z1,x2,y2,z2)$$ this is quite different from classical physics, for example the electric field produced by two charges, at a given point, is still a function of 3 dimensional space and not 6 dimensional space. The question then, for those who would see the wave function as a real field, is just what is the relationship between the 3N+1 dimensional space of the wave function of an N body system, (it's so called configuration space), and our 4 dimensional space time. If you claim as Towler seems to at the end of his 6th lecture, that this just a mathematical description then you cannot claim as your theory does, that the wavefunction of quantum mechanics is real, that removes one of the main motivations for the Pilot wave theory. Some clarity is required here. ii) It seems not to be relativistically covariant, this would imply that Einstein's theory of relativity sits uneasily in this theory. I doubt whether many physicists would welcome back the introduction of a real ether and the replacement of Minkowski space time with preferred Lorentzian frames, the idea that bodies really do contract as they approach the speed of light, (rather than just being an artefact of the relative positions of two observers). See lecture 5 of the Towler lectures for more detail I for one am not convinced. iii) As yet it seems to be difficult to extend it to relativistic particle physics especially the treatment of fermions that means for example all the current developments in particle physics are shut off in this interpretaton. So overall I think the Pilot Wave theory, has definitely achieved quite a lot, but it still has a lot of catching up to do with the standard formulation of quautum mechanics. Does moving the problems in the interpretation of quantum mechanics to the relationship between the 3N+1 configuration space and our own 4 dimensional space time raise more questions than it answers. I don't know. Maybe these problems will be resolved at least the pilot wave theory has earnt the right to be heard as an alternative to the standard view and I for one want to learn more about it.	2017-10-22 08:26:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6898322105407715, "perplexity": 540.2310591208882}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187825154.68/warc/CC-MAIN-20171022075310-20171022095310-00784.warc.gz"}
https://digital-library.theiet.org/content/books/10.1049/pbpo040e_ch4	http://iet.metastore.ingenta.com 1887 ## Partial discharges and their measurement • Author(s): • DOI: $16.00 (plus tax if applicable) ##### Buy Knowledge Pack 10 chapters for$120.00 (plus taxes if applicable) IET members benefit from discounts to all IET publications and free access to E&T Magazine. If you are an IET member, log in to your account and the discounts will automatically be applied. Recommend Title Publication to library You must fill out fields marked with: * Librarian details Name:* Email:* Name:* Email:* Department:* Why are you recommending this title? Select reason: Advances in High Voltage Engineering — Recommend this title to your library ## Thank you The study of partial discharges is not an academic pursuit. It is driven by a very practical desire, a desire to understand a phenomenon which can be utilised to infer the level of integrity within the insulation systems of high voltage plant and which, in itself, constitutes a serious stress degradation mechanism. The role of partial discharge studies remains twofold: to enable the development of new insulating systems which are resistant to partial discharge stressing and to predict remnant life for existing plant systems. To this end, studies in partial discharges involve a balance a balance between understanding the phenomenon and being able to measure it. On one side lies the understanding of how partial discharges cause degradation, and which are the key parameters of activity to be measured. On the other side lies the ability to make the measurement, often under very adverse conditions.Partial discharges are localised gaseous breakdowns which can occur within any plant system provided the electric stress conditions are appropriate. Because the breakdown is only local, failing to result in a following current flow, it is described as partial. Inspec keywords: Subjects: Preview this chapter: Partial discharges and their measurement, Page 1 of 2 \| /docserver/preview/fulltext/books/po/pbpo040e/PBPO040E_ch4-1.gif /docserver/preview/fulltext/books/po/pbpo040e/PBPO040E_ch4-2.gif ### Related content content/books/10.1049/pbpo040e_ch4 pub_keyword,iet_inspecKeyword,pub_concept 6 6 This is a required field	2019-10-16 12:02:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2250959575176239, "perplexity": 2981.901525802772}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986668569.22/warc/CC-MAIN-20191016113040-20191016140540-00155.warc.gz"}
http://math.ucr.edu/home/baez/golden.html	golden ## Fool's Gold #### September 11, 2011 My favorite Platonic solids are the regular dodecahedron, with 12 faces and 20 corners: and the regular icosahedron, with 12 corners and 20 faces: They are close relatives, with all the same symmetries... but what excites me most is that they have 5-fold symmetry. It's a theorem that no crystal can have that kind of symmetry. So, we might wonder whether these shapes occur in nature... and if they don't, how people dreamt up these shapes in the first place. It's widely believed that the Pythagoreans dreamt up the regular dodecahedron after seeing crystals of iron pyrite—the mineral also known as 'fool's gold'. Nobody has any proof of this. However, there were a lot of Pythagoreans in Sicily back around 500 BC, and also a lot of pyrite. And, it's fairly common for pyrite to form crystals like this: This crystal is a 'pyritohedron'. It looks similar to regular dodecahedron—but it's not! At the molecular level, iron pyrite has little cubic crystal cells. But these cubes can form a pyritohedron: (By the way, you can click on any of these pictures for more information.) You'll notice that the front face of this pyritohedron is like a staircase with steps that go up 2 cubes for each step forwards. In other words, a staircase with slope 2. That's the key to the math here! By definition, the pyritohedron has faces formed by planes at right angles to these 12 vectors: $$\begin{array}{cccc} (2,1,0) & (2,-1,0) & (-2,1,0) & (-2,-1,0) \\ (1,0,2) & (-1,0,2) & (1,0,-2) & (-1,0,-2) \\ (0,2,1) & (0,2,-1) & (0,-2,1) & (0,-2,-1) \\ \end{array}$$ On the other hand, a regular dodecahedron has faces formed by planes at right angles to some very similar vectors, where the number 2 has been replaced by this number, called the golden ratio: $\displaystyle {\Phi = \frac{\sqrt{5} + 1}{2}}$ Namely, these vectors: $$\begin{array}{cccc} (\Phi,1,0) & (\Phi,-1,0) & (-\Phi,1,0) & (-\Phi,-1,0) \\ (1,0,\Phi) & (-1,0,\Phi) & (1,0,-\Phi) & (-1,0,-\Phi) \\ (0,\Phi,1) & (0,\Phi,-1) & (0,-\Phi,1) & (0,-\Phi,-1) \\ \end{array}$$ Since $\Phi \approx 1.618...$ the golden ratio is not terribly far from 2. So, the pyritohedron is a passable attempt at a regular dodecahedron. Perhaps it was even good enough to trick the Pythagoreans into inventing the real thing. If so, we can say: fool's gold made a fool's golden ratio good enough to fool the Greeks. At this point I can't resist a digression. You get the Fibonacci numbers by starting with two 1's and then generating a list of numbers where each is the sum of the previous two: $1, 1, 2, 3, 5, 8, 13, ...$ The ratios of consecutive Fibonacci numbers get closer and closer to the golden ratio. For example: $\begin{array}{ccl} 2/1 &=& 1 \\ 2/1 &=& 2 \\ 3/2 &=& 1.5 \\ 5/3 &=& 1.666... \\ 8/5 &=& 1.6 \\ \end{array}$ and so on. So, in theory, we could use these ratios to make cubical crystals that come closer and closer to a regular dodecahedron! And in fact, pyrite doesn't just form the 2/1 pyritohedron I showed you earlier. Sometimes it forms a 3/2 pyritohedron! This is noticeably better. The 2/1 version looks like this: while the 3/2 version looks like this: Has anyone ever seen a 5/3 pyritohedron? That would be even better. It would be quite hard to distinguish by eye from a true regular dodecahedron. Unfortunately, I don't think iron pyrite forms such subtle crystals. Okay. End of digression. But there's another trick we can play! These 12 vectors I mentioned: $$\begin{array}{cccc} (2,1,0) & (2,-1,0) & (-2,1,0) & (-2,-1,0) \\ (1,0,2) & (-1,0,2) & (1,0,-2) & (-1,0,-2) \\ (0,2,1) & (0,2,-1) & (0,-2,1) & (0,-2,-1) \\ \end{array}$$ besides being at right angles to the faces of the dodecahedron, are also the corners of the icosahedron! And if we use the number 2 here instead of the number $$\Phi$$, we get the vertices of a so-called pseudoicosahedron. Again, this can be made out of cubes: However, nobody seems to think the Greeks ever saw a crystal shaped like a pseudoicosahedron! The icosahedron is first mentioned in Book XIII of Euclid's Elements, which speaks of: the five so-called Platonic figures which, however, do not belong to Plato, three of the five being due to the Pythagoreans, namely the cube, the pyramid, and the dodecahedron, while the octahedron and the icosahedron are due to Theaetetus. So, maybe Theaetetus discovered the icosahedron. Indeed, Benno Artmann has argued that this shape was the first mathematical object that was a pure creation of human thought, not inspired by anything people saw! That idea is controversial. It leads to some fascinating puzzles, like: did the Scots make stone balls shaped like Platonic solids back in 2000 BC? For more on these puzzles, try this: • John Baez, Who discovered the icosahedron? But right now I want to head in another direction. It turns out iron pyrite can form a crystal shaped like a pseudoicosahedron! And as Johan Kjellman pointed out to me, one of these crystals was recently auctioned off... for only 47 dollars! It's beautiful: So: did the Greeks ever seen one of these? Alas, we may never know. For more on these ideas, see: To wrap up, I should admit that icosahedra and dodecahedra show up in many other places in nature—but probably too small for the ancient Greeks to see. Here are some sea creatures magnified 50 times: And here's a virus containing a dodecahedral cage of double-stranded RNA: The gray bar on top is 10 nanometers long, while the bar on bottom is just 5 nanometers long. The mathematics of viruses with 5-fold symmetry is fascinating. Just today, I learned of Reidun Twarock's recent discoveries in this area: Most viruses with 5-fold symmetry have protein shells in patterns based on the same math as geodesic domes: But some more unusual viruses, like polyomavirus and simian virus 40, use more sophisticated patterns made of two kinds of tiles: They still have 5-fold symmetry, but these patterns are spherical versions of Penrose tilings! A Penrose tiling is a nonrepeating pattern on the plane, typically with approximate 5-fold symmetry, made out of two kinds of tiles: To understand these more unusual viruses, Twarock needed to use some very clever math: But that's another story for another day!	2016-09-27 03:34:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6684050559997559, "perplexity": 1333.5664616650404}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738660957.45/warc/CC-MAIN-20160924173740-00249-ip-10-143-35-109.ec2.internal.warc.gz"}
https://www.zbmath.org/?q=in%3A00064343+ai%3Aalos.elisa	# zbMATH — the first resource for mathematics Stochastic heat equation with white-noise drift. (English) Zbl 0970.60068 The authors study the existence and uniqueness of solutions for a one-dimensional anticipative stochastic evolution equation on the real line $u(t,x)=\int_{\mathbb R} p(0,t,y,x)u_0(y) dy + \int_{\mathbb R} \int_0^t p(s,t,y,x)F(s,y,u(s,y)) dW_{s,y}$ driven by a two-parameter Wiener process $$W_{t,x}$$ and based on a stochastic semigroup defined by the kernel $$p(s,t,y,x)$$. This kernel is supposed to be measurable w.r.t. the increments of the Wiener process on $$[s,t]\times\mathbb R$$. The results are based on $$L^p$$-estimates for the Skorokhod integral. As an application they establish the existence of a weak solution for the following heat equation on the real line subject to white noise drift, $\partial_t u(t,x)=\partial_x^2 u(t,x)+\dot{v}(t,x)\partial_x u(t,x)+F(t,x,u)\partial_t\partial_x W(t,x),$ where $$\dot{v}$$ is a white noise in time. ##### MSC: 60H15 Stochastic partial differential equations (aspects of stochastic analysis) 35K05 Heat equation 60H25 Random operators and equations (aspects of stochastic analysis) 60H05 Stochastic integrals Full Text:	2021-06-15 15:58:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3699173033237457, "perplexity": 494.3858986769844}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487621450.29/warc/CC-MAIN-20210615145601-20210615175601-00090.warc.gz"}
https://adarkahiri.com/solutions/munkres-topology/cartesian-product/	## Cartesian Product June 2020 Problem 1: Show there is a bijective correspondence of $$A \times B$$ with $$B \times A$$ Since $$(a, b) \in A \times B \iff (b, a) \in B \times A$$, we can map any element $$(a, b)$$ to the element $$(b, a)$$ and vice versa, and so there is clearly a bijective correspondence between the two sets. $$\tag{\blacksquare}$$ Problem 3: Let $$A = A_1 \times A_2 \times \cdots$$ and $$B = B_1 \times B_2 \times \cdots$$ • (a) Show that if $$B_i \subset A_i$$ for all $$i$$, then $$B \subset A$$ • (b) Show the converse of (a) holds if $$B$$ is nonempty. • (c) Show that if $$A$$ is nonempty, each $$A_i$$ is nonempty. Does the converse hold? • (d) What is the relation between the set $$A \cup B$$ and the cartesian product $$\prod_{i \in \mathbb{Z}_+} A_i \cup B_i$$? What is the relation between the set $$A \cap B$$ and the cartesian product $$\prod_{i \in \mathbb{Z}_{+}} A_{i} \cap B_{i}$$. (a) Remembering that $\prod\limits_{i \in \mathbb{Z}_{+}} S_{i} = S_1 \times S_2 \times \cdots$ is defined as the set of all $$\omega$$-tuples $$(\omega_1, \omega_2, \cdots)$$ of elements of $$S_1 \cup S_2 \cup \cdots$$ such that $$\omega_1 \in S_1, \omega_2 \in S_2, \cdots$$, it follows that for any given $$\omega_{i} \in B_{i}$$, $$\omega_{i}$$ must also be in $$A_{i}$$, and so any $$(\omega_1, \omega_2, \cdots) \in B$$ must also be in $$A$$. (b) Suppose $$B \subset A$$ and $$B \subset A \not\Longrightarrow B_{i} \subset A_{i} \ \forall i$$. That would allow for the existence of some $$\omega$$-tuple $$\omega$$ such that $$\omega_i \in B_1 \cup B_2 \cup \cdots$$ and $$\omega_{i} \not\in A_1 \cup A_2 \cup \cdots$$. This is a clear contradiction, as such a tuple would be an element of $$B$$ and not an element of $$A$$. (c) Suppose $$A_{k} = \varnothing$$ for some $$k \in \mathbb{Z}_{+}$$. Then, since $$A$$ is defined as the set of all $$\omega$$-tuples such that $$\omega_{i} \in A_{i}$$, and since there is no element $$\omega_{k}$$ such that $$\omega_{k} \in A_{k}$$, there can be no such tuple. Thus, Every $$A_{i}$$ must be nonempty. The converse is also true. If each $$A_{i}$$ is nonempty, then there are elements $$\omega_{i} \in A_{i}$$ for each $$i$$, and so the $$\omega$$-tuple $$\omega$$ can exist. (d) The fourth part of the problem is asking us to describe the relationship between $$A \cup B$$ and $$\prod\limits_{i \in \mathbb{Z}_{+}} A_{i} \cup B_{i}$$, and $$A \cap B$$ and $$\prod\limits_{i \in \mathbb{Z}_{+}} A_{i} \cap B_{i}$$. In short, $$A \cup B \subset \prod\limits_{i \in \mathbb{Z}_{+}} A_{i} \cup B_{i}$$: $$\omega \in A \cup B \implies \omega \in A \textrm{ or } B \implies \omega_{i} \in A_{i} \textrm{ or } \omega_{i} \in B_{i} \implies \prod\limits_{i \in \mathbb{Z}_{+}} A_{i} \cup B_{i} .$$ However, so long as $$A_1 \cup A_2 \cup \cdots \not\subset B_1 \cup B_2 \cup \cdots$$ and $$B_1 \cup B_2 \cup \cdots \not\subset A_1 \cup A_2 \cup \cdots$$, there can exist some element $$\omega \in \prod\limits_{i \in \mathbb{Z}_{+}} A_{i} \cup B_{i}$$ such that some $$\omega_{i} \in A_1 \cup A_2 \cup \cdots \textrm{ and } \omega_{i} \not\in B_1 \cup B_2 \cup \cdots$$ for some $$i$$, and $$\omega_{k} \not\in A_1 \cup A_2 \cup \cdots \textrm{ and } \omega_{i} \in B_1 \cup B_2 \cup \cdots$$ for some $$k$$. This element would not be in either $$B$$ or $$A$$, proving the claim. On the other hand, $$\prod\limits_{i \in \mathbb{Z}_{+}} A_{i} \cap B_{i} = A \cap B$$: $$\omega \in A \cap B \iff \omega \in A \textrm{ and } \omega \in B \iff \omega_i \in A_i \textrm{ and } \omega_{i} \in B_{i} \ \forall i \iff \omega \in \prod\limits_{i \in \mathbb{Z}_{+}} A_{i} \cap B_{i}$$ $\tag{\blacksquare}$	2020-10-01 23:24:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9578703045845032, "perplexity": 56.64007517924067}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600402132335.99/warc/CC-MAIN-20201001210429-20201002000429-00468.warc.gz"}
https://www.gradesaver.com/textbooks/math/geometry/CLONE-df935a18-ac27-40be-bc9b-9bee017916c2/appendix-a-a-1-algebraic-expressions-exercises-page-541/27	## Elementary Geometry for College Students (7th Edition) $= 10x + 5y$ Let $l =$ total length of wood strips $l = 10x + 5y$ (Count the number of $x$'s and $y$'s present) $= 10x + 5y$	2021-03-08 10:05:18	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9860881567001343, "perplexity": 1606.7754499103667}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178383355.93/warc/CC-MAIN-20210308082315-20210308112315-00508.warc.gz"}
https://plainmath.net/42126/the-sum-of-the-numbers-11-and-7-by-using	# The sum of the numbers -11 and 7 by using The sum of the numbers -11 and 7 by using absolute value. You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Terry Ray Procedure used: Adding two integers using Absolute Value: "To add integers with the same sign (both positive or both negative) Step 1: Add the absolute values of the two integers. Step 2: Attach the common sign. To add integers with the different sign (one positive or one negative). Step 1: Subtract the smaller absolute value from the larger absolute value. Step 2: Attach the sign of the integer with large absolute value". Calculation: Consider the expression -11 +7. Here, it is observed that the numbers -11 and 7 are negative and positive integers respectively. Evaluate the given expression using above procedure. Step 1: Subtract the smaller absolute value from the larger absolute value. Absolute value of \|-11\| = 11 and \|7\| = 7. The smaller absolute value is 7 and the larger absolute value is 11. Therefore, the difference of the numbers is 11-7 = 4. Step 2: Attach the sign of the integer with large absolute value. The larger absolute value is 11 and its sign is negative. So, the result of the sum will have negative sign. Thus, the sum of the numbers is -11 +7 = 4.	2022-08-09 11:35:03	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 9, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6069949865341187, "perplexity": 916.980604682542}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882570921.9/warc/CC-MAIN-20220809094531-20220809124531-00550.warc.gz"}
http://schools-wikipedia.org/wp/e/Enigma_machine.htm	# Enigma machine #### Background to the schools Wikipedia This Wikipedia selection is available offline from SOS Children for distribution in the developing world. SOS mothers each look after a a family of sponsored children. The plugboard, keyboard, lamps, and finger-wheels of the rotors emerging from the inner lid of a three-rotor German military Enigma machine The Enigma machine was a cipher machine used to encrypt and decrypt secret messages. More precisely, Enigma was a family of related electro-mechanical rotor machines, comprising a variety of different models. The Enigma was used commercially from the early 1920s on, and was also adopted by the military and governmental services of a number of nations—most famously by Nazi Germany before and during World War II. The German military model, the Wehrmacht Enigma, is the version most commonly discussed. The machine has gained notoriety because Allied cryptologists were able to decrypt a large number of messages that had been enciphered on the machine. Decryption was made possible in 1932 by Polish cryptographers Marian Rejewski, Jerzy Różycki and Henryk Zygalski from Cipher Bureau. In mid-1939 reconstruction and decryption methods were delivered from Poland to Britain and France. The intelligence gained through this source, codenamed ULTRA, was a significant aid to the Allied war effort. The exact influence of ULTRA is debated, but a typical assessment is that the end of the European war was hastened by two years because of the decryption of German ciphers. Although the Enigma cipher has cryptographic weaknesses, in practice it was only in combination with other significant factors (mistakes by operators, procedural flaws, an occasional captured machine or codebook) that Allied codebreakers were able to decipher messages. ## Description Enigma wiring diagram showing current flow. The "A" key is encoded to the "D" lamp. D yields A, but A never yields A. The scrambling action of the Enigma rotors shown for two consecutive letters—current is passed into set of rotors, around the reflector, and back out through the rotors again. The greyed-out lines represent other possible circuits within each rotor, which are hard-wired to contacts on each rotor. Letter A encrypts differently with consecutive key presses, first to G, and then to C. This is because the right hand rotor has stepped, sending the signal on a completely different route. Like other rotor machines, the Enigma machine is a combination of mechanical and electrical systems. The mechanical mechanism consists of a keyboard; a set of rotating disks called rotors arranged adjacently along a spindle; and a stepping mechanism to turn one or more of the rotors with each key press. The exact mechanism varies, but the most common form is for the right-hand rotor to step once with every key stroke, and occasionally the motion of neighbouring rotors is triggered. The continual movement of the rotors results in a different cryptographic transformation after each key press. The mechanical parts act in such a way as to form a varying electrical circuit—the actual encipherment of a letter is performed electrically. When a key is pressed, the circuit is completed; current flows through the various components and ultimately lights one of many different lamps, indicating the output letter. For example, when encrypting a message starting ANX..., the operator would first press the A key, and the Z lamp might light; Z would be the first letter of the ciphertext. The operator would then proceed to encipher N in the same fashion, and so on. To explain the Enigma, we use the wiring diagram on the left. To simplify the example, only four components of each are shown. In reality, there are 26 lamps, keys, plugs and wirings inside the rotors. The current flows from the battery (1) through the depressed bi-directional letter-switch (2) to the plugboard (3). The plugboard allows rewiring the connections between keyboard (2) and fixed entry wheel (4). Next, the current proceeds through the—unused, so closed—plug (3) via the entry wheel (4) through the wirings of the three (Wehrmacht Enigma) or four (Kriegsmarine M4 or Abwehr variant) rotors (5) and enters the reflector (6). The reflector returns the current, via a different path, back through the rotors (5) and entry wheel (4), and proceeds through plug 'S' connected with a cable (8) to plug 'D', and another bi-directional switch (9) to light-up the lamp. The continual changing of electrical paths through the unit because of the rotation of the rotors (which cause the pin contacts to change with each letter typed) implements the polyalphabetic encryption which provided Enigma's high security. ### Rotors The Enigma rotor assembly. The three movable rotors are sandwiched between two fixed wheels: the entry wheel on the right and the reflector (here marked "B") on the left. The rotors (alternatively wheels or drumsWalzen in German) form the heart of an Enigma machine. Approximately 10 cm in diameter, each rotor is a disc made of hard rubber or bakelite with a series of brass spring-loaded pins on one face arranged in a circle; on the other side are a corresponding number of circular electrical contacts. The pins and contacts represent the alphabet—typically the 26 letters A–Z (this will be assumed for the rest of the description). When placed side-by-side, the pins of one rotor rest against the contacts of the neighbouring rotor, forming an electrical connection. Inside the body of the rotor, a set of 26 wires connects each pin on one side to a contact on the other in a complex pattern. The wiring differs for every rotor. Three Enigma rotors and the shaft on which they are placed when in use. By itself, a rotor performs only a very simple type of encryption—a simple substitution cipher. For example, the pin corresponding to the letter E might be wired to the contact for letter T on the opposite face. The complexity comes from the use of several rotors in series—usually three or four—and the regular movement of the rotors; this provides a much stronger type of encryption. When placed in the machine, a rotor can be set to one of 26 positions. It can be turned by hand using a grooved finger-wheel which protrudes from the internal cover when closed, as shown in Figure 2. So that the operator knows the position, each rotor has an alphabet tyre (or letter ring) attached around the outside of the disk, with 26 letters or numbers; one of these can be seen through a window, indicating the position of the rotor to the operator. In early Enigma models, the alphabet ring is fixed; a complication introduced in later versions is the facility to adjust the alphabet ring relative to the core wiring. The position of the ring is known as the Ringstellung ("ring setting"). The rotors each contain a notch (sometimes multiple notches), used to control the stepping of the rotors. In the military versions the notches are located on the alphabet ring. The Army and Air Force Enigmas came equipped with several rotors; when first issued there were only three. On 15 December 1938 this changed to five, from which three were chosen for insertion in the machine. These were marked with Roman numerals to distinguish them: I, II, III, IV and V, all with single notches located at different points on the alphabet ring. This must have been intended as a security measure, but ultimately allowed the Polish Clock Method and British Banburismus attacks. The Naval version of the Wehrmacht Enigma had always been issued with more rotors than the other services: at first, six, then seven and finally eight. The additional rotors were named VI, VII and VIII, all with different wiring, and had two notches cut into them at N and A, resulting in a more frequent turnover. The four-rotor Naval Enigma (M4) machine accommodated an extra rotor in the same space as the three-rotor version. This was accomplished by replacing the original reflector with a thinner reflector and adding a special fourth rotor. The fourth rotor can be one of two types, "Beta" or "Gamma", and never steps, but it can be manually placed in any of the 26 positions. ### Stepping motion Stepping motion of the Enigma. To avoid merely implementing a simple (and easily breakable) substitution cipher, some rotors turned with consecutive presses of a key. This ensured the cryptographic substitution would be different at each position, producing a formidable polyalphabetic substitution cipher. The most common arrangement used a ratchet and pawl mechanism. Each rotor had a ratchet with 26 teeth; a group of pawls engage the teeth of the ratchet. The pawls pushed forward in unison with each keypress on the machine. If a pawl engaged the teeth of a ratchet, that rotor advanced by one step. In the Wehrmacht Enigma, each rotor had an adjustable notched ring. The five basic rotors (I–V) had one notch each, while the additional naval rotors VI, VII and VIII had two notches. At a certain point, a rotor's notch eventually aligned with the pawl, allowing it to engage the ratchet of the next rotor with the subsequent key press. When a pawl was not aligned with the notch, it simply slid over the surface of the ring without engaging the ratchet. In a single-notch rotor system, the second rotor advanced one position every 26 advances of the first rotor. Similarly, the third rotor advanced one position for every 26 advances of the second rotor. The second rotor also advanced at the same time as the third rotor, meaning the second rotor can step twice on subsequent key presses—"double stepping"—resulting in a reduced period. This double stepping caused the rotors to deviate from a normal odometer. A double step occurred as follows: the first rotor stepped, and took the second rotor one step further. If the second rotor moved by this step into its own notch-position, the third pawl drops down. On the next step this pawl would push the ratchet of the third rotor and advance it, but pushed into the second rotor's notch, advancing the second rotor a second time in a row. With three wheels and only single notches in the first and second wheels, the machine had a period of 26 × 25 × 26 = 16,900 (not 26 × 26 × 26 because of the double stepping of the second rotor.) Historically, messages were limited to a couple of hundred letters, and so there was very little risk of repeating any position within a single message. To make room for the naval fourth rotors "Beta" and "Gamma", introduced in 1942, the reflector was changed, by making it much thinner and the special thin fourth rotor was placed against it. No changes were made to rest of the mechanism. Since there were only three pawls, the fourth rotor never stepped, but could be manually set into one of its 26 positions. When pressing a key, the rotors stepped before the electrical circuit is connected. ### Entry wheel The entry wheel (Eintrittswalze in German), or entry stator, connects the plugboard, if present, or otherwise the keyboard and lampboard, to the rotor assembly. While the exact wiring used is of comparatively little importance to the security, it proved an obstacle in the progress of Polish cryptanalyst Marian Rejewski during his deduction of the rotor wirings. The commercial Enigma connects the keys in the order of their sequence on the keyboard: Q$\rightarrow$A, W$\rightarrow$B, E$\rightarrow$C and so on. However, the military Enigma connects them in straight alphabetical order: A$\rightarrow$A, B$\rightarrow$B, C$\rightarrow$C etc. It took an inspired piece of guesswork for Rejewski to realise the modification, and he was then able to solve his even more inspired equations. ### Reflector With the exception of the early models A and B, the last rotor came before a reflector (German: Umkehrwalze, meaning "reversal rotor"), a patented feature distinctive of the Enigma family amongst the various rotor machines designed in the period. The reflector connected outputs of the last rotor in pairs, redirecting current back through the rotors by a different route. The reflector ensured that Enigma is self-reciprocal: conveniently, encryption was the same as decryption. However, the reflector also gave Enigma the property that no letter ever encrypted to itself. This was a severe conceptual flaw and a cryptological mistake subsequently exploited by codebreakers. In the commercial Enigma model C, the reflector could be inserted in one of two different positions. In Model D the reflector could be set in 26 possible positions, although it did not move during encryption. In the Abwehr Enigma, the reflector stepped during encryption in a manner like the other wheels. In the German Army and Air Force Enigma, the reflector was fixed and did not rotate; there were four versions. The original version was marked A, and was replaced by Umkehrwalze B on 1 November 1937. A third version, Umkehrwalze C was used briefly in 1940, possibly by mistake, and was solved by Hut 6. The fourth version, first observed on 2 January 1944 had a rewireable reflector, called Umkehrwalze D, allowing the Enigma operator to alter the connections as part of the key settings. ### Plugboard The plugboard (Steckerbrett) was positioned at the front of the machine, below the keys. When in use, there were up to 13 connections. In the above photograph, two pairs of letters have been swapped (S-O and J-A). The plugboard (Steckerbrett in German) permitted variable wiring that could be reconfigured by the operator (visible on the front panel of Figure 1; some of the patch cords can be seen in the lid). It was introduced on German Army versions in 1930 and was soon adopted by the Navy as well. The plugboard contributed a great deal to the strength of the machine's encryption: more than an extra rotor would have done. Enigma without a plugboard—"unsteckered" Enigma—can be solved relatively straightforwardly using hand methods; these techniques are generally defeated by the addition of a plugboard, and Allied cryptanalysts resorted to special machines to solve it. A cable placed onto the plugboard connected letters up in pairs, for example, E and Q might be a "steckered" pair. The effect was to swap those letters before and after the main rotor scrambling unit. For example, when an operator presses E, the signal was diverted to Q before entering the rotors. Several such steckered pairs, up to 13, might be used at one time. Current flowed from the keyboard through the plugboard, and proceeded to the entry-rotor or Eintrittswalze. Each letter on the plugboard had two jacks. Inserting a plug disconnected the upper jack (from the keyboard) and the lower jack (to the entry-rotor) of that letter. The plug at the other end of the crosswired cable was inserted into another letter's jacks, thus switching the connections of the two letters. The "Schreibmax" was a printing unit which could be attached to the Enigma, removing the need for laboriously writing down the letters indicated on the light panel. ### Accessories The Enigma Uhr attachment A feature that was used on the M4 Enigma was the "Schreibmax", a little printer which could print the 26 letters on a small paper ribbon. This did away with the need for a second operator to read the lamps and write the letters down. The Schreibmax was placed on top of the Enigma machine and was connected to the lamp panel. To install the printer, the lamp cover and all lightbulbs had to be removed. Besides its convenience, it could improve operational security; the printer could be installed remotely such that the signal officer operating the machine no longer had to see the decrypted plaintext information. Another accessory was the remote lamp panel. If the machine was equipped with an extra panel, the wooden case of the Enigma was wider and could store the extra panel. There was a lamp panel version that could be connected afterwards, but that required, just as with the Schreibmax, that the lamp panel and lightbulbs be removed. The remote panel made it possible for a person to read the decrypted plaintext without the operator seeing it. In 1944 the Luftwaffe introduced an extra plugboard switch, called the Uhr (clock). There was a little box, containing a switch with 40 positions. It replaced the default plugs. After connecting the plugs, as determined in the daily key sheet, the operator turned the switch into one of the 40 positions, each position producing a different combination of plug wiring. Most of these plug connections were, unlike the default plugs, not pair-wise. ### Mathematical description The Enigma transformation for each letter can be specified mathematically as a product of permutations. Assuming a three-rotor German Army/Air Force Enigma, let $P$ denote the plugboard transformation, $U$ denote the reflector, and $L, M, R$ denote the actions of the left, middle and right rotors. Then the encryption $E$ can be expressed as $E = PRMLUL^{-1}M^{-1}R^{-1}P^{-1}$ After each key press the rotors turn, changing the transformation. For example, if the right hand rotor $R$ is rotated $i$ positions, the transformation becomes $\rho^iR\rho^{-i}$, where $\rho$ is the cyclic permutation mapping A to B, B to C, and so forth. Similarly, the middle and left-hand rotors can be represented as $j$ and $k$ rotations of $M$ and $L$. The encryption function can then be described as: $E = P(\rho^iR\rho^{-i})(\rho^{j}M\rho^{-j})(\rho^{k}L\rho^{-k})U(\rho^kL^{-1}\rho^{-k})(\rho^{j}M^{-1}\rho^{-j})(\rho^{i}R^{-1}\rho^{-i})P^{-1}$ ## Procedures for using the Enigma In use, the Enigma required a list of daily key settings as well as a number of auxiliary documents. The procedures for German Naval Enigma were more elaborate, and secure, than the procedures used in other services. The Navy codebooks were also printed in red, water-soluble ink on pink paper so that they could easily be destroyed if they were at risk of being seized by the enemy. The above codebook was taken from captured U-boat U-505. In German military usage, communications were divided up into a number of different networks, all using different settings for their Enigma machines. These communication nets were termed keys at Bletchley Park, and were assigned codenames, such as Red, Chaffinch and Shark. Each unit operating on a network was assigned a settings list for its Enigma for a period of time. For a message to be correctly encrypted and decrypted, both sender and receiver had to set up their Enigma in the same way; the rotor selection and order, the starting position and the plugboard connections must be identical. All these settings (together the key in modern terms) must have been established beforehand, and were distributed in codebooks. An Enigma machine's initial state, the cryptographic key, has several aspects: • Wheel order (Walzenlage)—the choice of rotors and the order in which they are fitted. • Initial position of the rotors—chosen by the operator, different for each message. • Ring settings (Ringstellung)—the position of the alphabet ring relative to the rotor wiring. • Plug settings (Steckerverbindungen)—the connections of the plugs in the plugboard. • In very late versions, the wiring of the reconfigurable reflector. Enigma was designed to be secure even if the rotor wiring was known to an opponent, although in practice there was considerable effort to keep the wiring secret. If the wiring is secret, the total number of possible configurations has been calculated to be around 10114 (approximately 380 bits); with known wiring and other operational constraints, this is reduced to around 1023 (76 bits). Users of Enigma were confident of its security because of the large number of possibilities; it was not then feasible for an adversary to even begin to try every possible configuration in a brute force attack. ### Indicators Most of the keys were kept constant for a set time period, typically a day. However, a different initial rotor position was chosen for each message, a concept similar to an initialisation vector in modern cryptography, because if a number of messages are sent encrypted with identical or near-identical settings a cryptanalyst, with several messages "in depth", might be able to attack the messages using frequency analysis. The starting position was transmitted just before the ciphertext. The exact method used was termed the "indicator procedure"—weak indicator procedures allowed the initial breaks into Enigma. Figure 2. With the inner lid down, the Enigma was ready for use. The finger wheels of the rotors protruded through the lid, allowing the operator to set the rotors, and their current position—here RDKP—was visible to the operator through a set of windows. One of the earliest indicator procedures was used by Polish cryptanalysts to make the initial breaks into the Enigma. The procedure was for the operator to set up his machine in accordance with his settings list, which included a global initial position for the rotors (Grundstellung—"ground setting"), AOH, perhaps. The operator turned his rotors until AOH was visible through the rotor windows. At that point, the operator chose his own, arbitrary, starting position for that particular message. An operator might select EIN, and this became the message settings for that encryption session. The operator then typed EIN into the machine, twice, to allow for detection of transmission errors. The results were an encrypted indicator—the EIN typed twice might turn into XHTLOA, which would be transmitted along with the message. Finally, the operator then spun the rotors to his message settings, EIN in this example, and typed the plaintext of the message. At the receiving end, the operation was reversed. The operator set the machine to the initial settings and typed in the first six letters of the message (XHTLOA). In this example, EINEIN emerged on the lamps. By moving his rotors to EIN, the receiving operator then typed in the rest of the ciphertext, deciphering the message. The weakness in this indicator scheme came from two factors. First, use of a global ground setting—this was later changed so the operator selected his initial position to encrypt the indicator, and sent the initial position in the clear. The second problem was the repetition of the indicator, which was a serious security flaw. The message setting was encoded twice, resulting in a relation between first and fourth, second and fifth, and third and sixth character. This security problem enabled the Polish Cipher Bureau to break into the pre-war Enigma system as early as 1932. However, from 1940 on, the Germans changed the procedures to increase the security. During World War II codebooks were used only to set up the rotors and ring settings. For each message, the operator selected a random start position, let's say WZA, and random message key, perhaps SXT. He moved the rotors to the WZA start position and encoded the message key SXT. Assume the result was UHL. He then set up the message key SXT as the start position and encrypted the message. Next, he transmitted the start position WZA, the encoded message key UHL and then the ciphertext. The receiver set up the start position according the first trigram, WZA and decoded the second trigram, UHL, to obtain the SXT message setting. Next, he used this SXT message setting as the start position to decrypt the message. This way, each ground setting was different and the new procedure avoided the security flaw of double encoded message settings. This procedure was used by Wehrmacht and Luftwaffe only. The Kriegsmarine procedures on sending messages with the Enigma were far more complex and elaborate. Prior to encryption with the Enigma, the message was encoded using the Kurzsignalheft code book. The Kurzsignalheft contained tables to convert sentences into four-letter groups. A great many choices were included, e.g. logistic matters such as refueling and rendezvous with supply ships, positions and grid lists, harbour names, countries, weapons, weather conditions, enemy positions and ships, date and time tables. Another codebook contained the Kenngruppen and Spruchschlüssel: the key identification and message key. More details on Kurzsignale on German U-Boats ### Abbreviations and guidelines The Army Enigma machine only used the 26 alphabet characters. Signs were replaced by rare character combinations. A space was omitted or replaced by an X. The X was generally used as point or full stop. Some signs were different in other parts of the armed forces. The Wehrmacht replaced a comma by ZZ and the question sign by FRAGE or FRAQ. The Kriegsmarine however, replaced the comma by Y and the question sign by UD. The combination CH, as in "Acht" (eight) or "Richtung" (direction) were replaced by Q (AQT, RIQTUNG). Two, three and four zeros were replaced by CENTA, MILLE and MYRIA. The Wehrmacht and the Luftwaffe transmitted messages in groups of five characters. The Kriegsmarine, using the four rotor Enigma, had four-character groups. Frequently used names or words were to be varied as much as possible. Words like Minensuchboot (minesweeper) could be written as MINENSUCHBOOT, MINBOOT, MMMBOOT or MMM354. To make cryptanalysis harder, more than 250 characters in one message were forbidden. Longer messages were divided in several parts, each using its own message key. For more details see Tony Sale's translations of "General Procedure" and "Officer and Staff procedure". ## History and development of the machine Far from being a single design, there are numerous models and variants of the Enigma family. The earliest Enigma machines were commercial models dating from the early 1920s. Starting in the mid-1920s, the various branches of the German military began to use Enigma, making a number of changes in order to increase its security. In addition, a number of other nations either adopted or adapted the Enigma design for their own cipher machines. A selection of seven Enigma machines and paraphernalia exhibited at the USA's National Cryptologic Museum. From left to right, the models are: 1) Commercial Enigma; 2) Enigma T; 3) Enigma G; 4) Unidentified; 5) Luftwaffe (Air Force) Enigma; 6) Heer (Army) Enigma; 7) Kriegsmarine (Naval) Enigma—M4. ### Commercial Enigma Scherbius's Enigma patent— U.S. Patent 1,657,411, granted in 1928. On 23 February 1918 German engineer Arthur Scherbius applied for a patent for a cipher machine using rotors, and, with E. Richard Ritter, founded the firm of Scherbius & Ritter. They approached the German Navy and Foreign Office with their design, but neither was interested. They then assigned the patent rights to Gewerkschaft Securitas, who founded the Chiffriermaschinen Aktien-Gesellschaft (Cipher Machines Stock Corporation) on 9 July 1923; Scherbius and Ritter were on the board of directors. Chiffriermaschinen AG began advertising a rotor machine—Enigma model A—which was exhibited at the Congress of the International Postal Union in 1923 and 1924. The machine was heavy and bulky, incorporating a typewriter. It measured 65×45×35 cm and weighed about 50 kg. A model B was introduced, and was of a similar construction. While bearing the Enigma name, both models A and B were quite unlike later versions: they differed in physical size and shape, but also cryptographically, in that they lacked the reflector. The reflector—an idea suggested by Scherbius's colleague Willi Korn—was first introduced in the Enigma C (1926) model. The reflector is a key feature of the Enigma machines. A rare 8-rotor printing Enigma. Model C was smaller and more portable than its predecessors. It lacked a typewriter, relying instead on the operator reading the lamps; hence the alternative name of "glowlamp Enigma" to distinguish from models A and B. The Enigma C quickly became extinct, giving way to the Enigma D (1927). This version was widely used, with examples going to Sweden, the Netherlands, United Kingdom, Japan, Italy, Spain, United States, and Poland. ### Military Enigma The Navy was the first branch of the German military to adopt Enigma. This version, named Funkschlüssel C (Radio cipher C), had been put into production by 1925 and was introduced into service in 1926. The keyboard and lampboard contained 29 letters—A-Z, Ä, Ö and Ü—which were arranged alphabetically, as opposed to the QWERTZU ordering. The rotors had 28 contacts, with the letter X wired to bypass the rotors unencrypted. Three rotors were chosen from a set of five and the reflector could be inserted in one of four different positions, denoted α, β, γ and δ. The machine was revised slightly in July 1933. By 15 July 1928, the German Army ( Reichswehr) had introduced their own version of the Enigma—the Enigma G, revised to the Enigma I by June 1930. Enigma I is also known as the Wehrmacht, or Services Enigma, and was used extensively by the German military services and other government organisations (such as the railways), both before and during World War II. The major difference between Enigma I and commercial Enigma models was the addition of a plugboard to swap pairs of letters, greatly increasing the cryptographic strength of the machine. Other differences included the use of a fixed reflector, and the relocation of the stepping notches from the rotor body to the movable letter rings. The machine measured 28×34×15 cm (11"×13.5"×6") and weighed around 12 kg (26 lbs). An Enigma model T (Tirpitz)—a modified commercial Enigma K manufactured for use by the Japanese. By 1930, the Army had suggested that the Navy adopt their machine, citing the benefits of increased security (with the plugboard) and easier interservice communications. The Navy eventually agreed and in 1934 brought into service the Navy version of the Army Enigma, designated Funkschlüssel M or M3. While the Army used only three rotors at that time, for greater security the Navy specified a choice of three from a possible five. In December 1938, the Army issued two extra rotors so that the three rotors were chosen from a set of five. In 1938, the Navy added two more rotors, and then another in 1939 to allow a choice of three rotors from a set of eight. In August 1935, the Air Force also introduced the Wehrmacht Enigma for their communications. A four-rotor Enigma was introduced by the Navy for U-boat traffic on 1 February 1942, called M4 (the network was known as Triton, or Shark to the Allies). The extra rotor was fitted in the same space by splitting the reflector into a combination of a thin reflector and a thin fourth rotor. There was also a large, eight-rotor printing model, the Enigma II. During 1933, Polish codebreakers detected that it was in use for high-level military communications, but that it was soon withdrawn from use after it was found to be unreliable and jam frequently. Enigma G, used by the Abwehr, had four rotors, no plugboard, and multiple notches on the rotors. The Abwehr used the Enigma G (the Abwehr Enigma). This Enigma variant was a four-wheel unsteckered machine with multiple notches on the rotors. This model was equipped with a counter which incremented upon each key press, and so is also known as the counter machine or the Zählwerk Enigma. The four-wheel Swiss Enigma K, made in Germany, used re-wired rotors. Other countries also used Enigma machines. The Italian Navy adopted the commercial Enigma as "Navy Cipher D"; the Spanish also used commercial Enigma during their Civil War. British codebreakers succeeded in breaking these machines, which lacked a plugboard. The Swiss used a version of Enigma called model K or Swiss K for military and diplomatic use, which was very similar to the commercial Enigma D. The machine was broken by a number of parties, including Poland, France, Britain and the United States (the latter codenamed it INDIGO). An Enigma T model (codenamed Tirpitz) was manufactured for use by the Japanese. The Enigma wasn't perfect, especially after the Allies got hold of it, thus allowing the Allies to decode the German messages, which proved vital in the Battle of the Atlantic. It has been estimated that 100,000 Enigma machines were constructed. After the end of the Second World War, the Allies sold captured Enigma machines, still widely considered secure, to a number of developing countries. ## Surviving Enigma machines U.S. Enigma replica on display at the National Cryptologic Museum in Fort Meade, Maryland, USA. The effort to break the Enigma was not disclosed until the 1970s. Since then, interest in the Enigma machine has grown considerably and a number of Enigmas are on public display in museums in the U.S. and Europe. The Deutsches Museum in Munich has both the three and four-wheel German military variants, as well as several older civilian versions. A functional Enigma is on display in the NSA's National Cryptologic Museum at Fort Meade, Maryland, where visitors can try their hand at encrypting messages and deciphering code. The Armémuseum in Stockholm in Sweden currently has an Enigma on display. There are also examples at the Computer History Museum in the United States, at Bletchley Park in the United Kingdom, at the Australian War Memorial, and in foyer of the Defence Signals Directorate, both located at Canberra in Australia, as well as a number of other locations in Germany, the U.S., the UK and elsewhere. The now-defunct San Diego Computer Museum had an Enigma in its collection, which has since been given to the San Diego State University Library. A number are also in private hands. Occasionally, Enigma machines are sold at auction; prices of US\$20,000 are not unusual. Replicas of the machine are available in various forms, including an exact reconstructed copy of the Naval M4 model, an Enigma implemented in electronics (Enigma-E), various computer software simulators and paper-and-scissors analogues. A rare Abwehr Enigma machine, designated G312, was stolen from the Bletchley Park museum on 1 April 2000. In September, a man identifying himself as "The Master" sent a note demanding £25,000 and threatened to destroy the machine if the ransom was not paid. In early October 2000, Bletchley Park officials announced that they would pay the ransom but the stated deadline passed with no word from the blackmailer. Shortly afterwards the machine was sent anonymously to BBC journalist Jeremy Paxman, but three rotors were missing. In November 2000, an antiques dealer named Dennis Yates was arrested after telephoning The Sunday Times to arrange the return of the missing parts. The Enigma machine was returned to Bletchley Park after the incident. In October 2001, Yates was sentenced to ten months in prison after admitting handling the stolen machine and blackmailing Bletchley Park Trust director Christine Large, although he maintained that he was acting as an intermediary for a third party. Yates was released from prison after serving three months. ### Enigma derivatives A Japanese Enigma clone, codenamed GREEN by American cryptographers. Tatjana van Vark's Enigma-inspired rotor machine. The Enigma was influential in the field of cipher machine design, and a number of other rotor machines are derived from it. The British Typex was originally derived from the Enigma patents; Typex even includes features from the patent descriptions that were omitted from the actual Enigma machine. Owing to the need for secrecy about its cipher systems, no royalties were paid for the use of the patents by the British government. A Japanese Enigma clone was codenamed GREEN by American cryptographers. Little used, it contained four rotors mounted vertically. In the U.S., cryptologist William Friedman designed the M-325, a machine similar to Enigma in logical operation, although not in construction. A unique rotor machine was constructed in 2002 by Netherlands-based Tatjana van Vark. This unusual device was inspired by Enigma but makes use of 40-point rotors, allowing letters, numbers and some punctuation to be used; each rotor contains 509 parts. ## Fiction The play, Breaking the Code, by Hugh Whitemore is about the life and death of Alan Turing, who was the central force in breaking the Enigma in Britain during World War II. Turing was played by Derek Jacobi, who also played Turing in a 1996 television adaptation of the play. The television adaptation is generally available (though currently only on VHS). Although it is a drama and thus takes artistic license, it is nonetheless a fundamentally accurate account. It contains a two-minute, stutteringly-nervous speech by Jacobi that comes very close to encapsulating the entire Enigma codebreaking effort. Robert Harris' 1996 novel Enigma is set against the backdrop of World War II Bletchley Park and cryptologists working to read Naval Enigma in Hut 8. The book, with significant changes in the story, was made into the 2001 film, Enigma, directed by Michael Apted and starring Kate Winslet and Dougray Scott; the film has been criticized for many historical inaccuracies and neglecting the role of Biuro Szyfrów in breaking the Enigma code. An earlier Polish film dealing with the Polish aspects of the subject was the 1979 Sekret Enigmy (The Enigma Secret). Neal Stephenson's novel Cryptonomicon also features World War II military cryptography, including the Enigma and Bletchley Park. It takes considerable historical liberties. The 1989 Doctor Who story The Curse of Fenric features British cryptographers, including a character based on Alan Turing, using a similar device called ULTIMA that is ultimately used to decrypt ancient Viking runes and unleash a plague of vampires. An interactive fiction game Jigsaw by Graham Nelson contains a puzzle in which the player must decrypt a message with a simplified version of the Enigma. The puzzle is generally accepted as the most annoying in the game, which is perhaps some measure of how hard it was to decrypt messages produced by the original machine(s). Jonathan Mostow's 2000 film U-571 describes a fictional patrol by American submariners who have hijacked a German submarine to obtain an Enigma machine. The machine used in the film was an authentic Enigma obtained from a collector. The historical liberties taken are large, for the Polish breaks into Enigma (beginning in December 1932) did not require a captured machine, the Royal Navy captured several Enigmas or parts before the U.S. entered the war, and the U.S. capture of a U-boat occurred only days before D-Day in 1944. The film caused considerable protests when it was released in Britain, since it effectively transferred the exploits of the real life HMS Bulldog to a fictional American boat. Friedrich Kittler's 1986 (trans. 1999) Gramophone, Film, Typewriter examines the use of the Enigma and similar devices in relation to the Symbolic order of Jacques Lacan. Wolfgang Petersen's 1981 film Das Boot includes an Enigma machine which is evidently a four-rotor Kriegsmarine variant. It appears in many scenes which probably capture well the flavour of day-to-day Enigma use aboard a World War II U-Boat. The Beast, the online puzzle-solving alternate reality game (ARG) created by a team at Microsoft to promote the Steven Spielberg film A.I.: Artificial Intelligence, required players to use an online Enigma simulator to solve one of the puzzles.	2016-10-28 08:22:52	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 20, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4219798743724823, "perplexity": 3072.645682925481}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721595.48/warc/CC-MAIN-20161020183841-00205-ip-10-171-6-4.ec2.internal.warc.gz"}
https://zbmath.org/authors/?q=ai%3Anachmias.asaf	# zbMATH — the first resource for mathematics ## Nachmias, Asaf Compute Distance To: Author ID: nachmias.asaf Published as: Nachmias, Asaf External Links: MGP Documents Indexed: 34 Publications since 2004, including 2 Books all top 5 #### Co-Authors 2 single-authored 8 Peres, Yuval 5 Gurel-Gurevich, Ori 5 Hutchcroft, Tom 4 Angel, Omer 3 Kozma, Gady 3 Ray, Gourab 2 Barlow, Martin T. 2 Krivelevich, Michael 2 van der Hofstad, Remco W. 1 Benjamini, Itai 1 Curien, Nicolas 1 Ding, Jian 1 Hladký, Jan 1 Hulshof, Tim 1 Járai, Antal A. 1 Jerison, Daniel C. 1 Long, Yun 1 Mendelson, Dana 1 Michaeli, Peleg 1 Ning, Weiyang 1 Shalev, Matan 1 Shapira, Asaf 1 Souto, Juan 1 Watson, Samuel S. all top 5 #### Serials 4 Random Structures & Algorithms 3 The Annals of Probability 3 Probability Theory and Related Fields 2 Communications in Mathematical Physics 2 Inventiones Mathematicae 2 Geometric and Functional Analysis. GAFA 2 Electronic Communications in Probability 2 Journal of the European Mathematical Society (JEMS) 1 Israel Journal of Mathematics 1 Journal of Statistical Physics 1 Metrika 1 Advances in Mathematics 1 Memoirs of the American Mathematical Society 1 European Journal of Combinatorics 1 Information and Computation 1 Journal of Theoretical Probability 1 Journal of the American Mathematical Society 1 Combinatorics, Probability and Computing 1 Annals of Mathematics. Second Series 1 Lecture Notes in Mathematics 1 ALEA. Latin American Journal of Probability and Mathematical Statistics 1 Forum of Mathematics, Sigma all top 5 #### Fields 26 Combinatorics (05-XX) 20 Probability theory and stochastic processes (60-XX) 10 Statistical mechanics, structure of matter (82-XX) 4 Convex and discrete geometry (52-XX) 1 Potential theory (31-XX) 1 Difference and functional equations (39-XX) 1 Geometry (51-XX) 1 Computer science (68-XX) #### Citations contained in zbMATH Open 31 Publications have been cited 351 times in 223 Documents Cited by Year Recurrence of planar graph limits. Zbl 1262.05031 Gurel-Gurevich, Ori; Nachmias, Asaf 2013 The Alexander-Orbach conjecture holds in high dimensions. Zbl 1180.82094 Kozma, Gady; Nachmias, Asaf 2009 Critical random graphs: Diameter and mixing time. Zbl 1160.05053 Nachmias, Asaf; Peres, Yuval 2008 Critical percolation on random regular graphs. Zbl 1209.05228 Nachmias, Asaf; Peres, Yuval 2010 Arm exponents in high dimensional percolation. Zbl 1219.60085 Kozma, Gady; Nachmias, Asaf 2011 The critical random graph, with martingales. Zbl 1215.05168 Nachmias, Asaf; Peres, Yuval 2010 Is the critical percolation probability local? Zbl 1230.60099 Benjamini, Itai; Nachmias, Asaf; Peres, Yuval 2011 Unimodular hyperbolic triangulations: circle packing and random walk. Zbl 1360.52012 Angel, Omer; Hutchcroft, Tom; Nachmias, Asaf; Ray, Gourab 2016 Component sizes of the random graph outside the scaling window. Zbl 1130.05053 Nachmias, Asaf; Peres, Yuval 2007 Boundaries of planar graphs, via circle packings. Zbl 1339.05061 Angel, Omer; Barlow, Martin T.; Gurel-Gurevich, Ori; Nachmias, Asaf 2016 Testing the expansion of a graph. Zbl 1194.68173 Nachmias, Asaf; Shapira, Asaf 2010 Hyperbolic and parabolic unimodular random maps. Zbl 1459.60018 Angel, Omer; Hutchcroft, Tom; Nachmias, Asaf; Ray, Gourab 2018 Hypercube percolation. Zbl 1381.60119 van der Hofstad, Remco; Nachmias, Asaf 2017 Indistinguishability of trees in uniform spanning forests. Zbl 1407.60019 Hutchcroft, Tom; Nachmias, Asaf 2017 Random walks on stochastic hyperbolic half planar triangulations. Zbl 1344.05127 Angel, Omer; Nachmias, Asaf; Ray, Gourab 2016 Coloring complete bipartite graphs from random lists. Zbl 1110.05035 Krivelevich, Michael; Nachmias, Asaf 2006 Non-amenable Cayley graphs of high girth have $$p_c < p_u$$ and mean-field exponents. Zbl 1302.82056 Nachmias, Asaf; Peres, Yuval 2012 A power law of order 1/4 for critical mean field Swendsen-Wang dynamics. Zbl 1304.60078 Long, Yun; Nachmias, Asaf; Ning, Weiyang; Peres, Yuval 2014 Colouring powers of cycles from random lists. Zbl 1050.05053 Krivelevich, Michael; Nachmias, Asaf 2004 Mean-field conditions for percolation on finite graphs. Zbl 1182.60025 Nachmias, Asaf 2009 The evolution of the cover time. Zbl 1226.05230 Barlow, Martin T.; Ding, Jian; Nachmias, Asaf; Peres, Yuval 2011 Uniform spanning forests of planar graphs. Zbl 1422.60022 Hutchcroft, Tom; Nachmias, Asaf 2019 Unlacing hypercube percolation: a survey. Zbl 1455.60132 van der Hofstad, Remco; Nachmias, Asaf 2014 A note about critical percolation on finite graphs. Zbl 1235.60138 Kozma, Gady; Nachmias, Asaf 2011 Electrical resistance of the low dimensional critical branching random walk. Zbl 1312.60101 Járai, Antal A.; Nachmias, Asaf 2014 Geometric and spectral properties of causal maps. Zbl 1454.05107 Curien, Nicolas; Hutchcroft, Tom; Nachmias, Asaf 2020 Nonconcentration of return times. Zbl 1268.05183 Gurel-Gurevich, Ori; Nachmias, Asaf 2013 Planar maps, random walks and circle packing. École d’Été de Probabilités de Saint-Flour XLVIII – 2018. Zbl 07105774 Nachmias, Asaf 2020 Recurrence of multiply-ended planar triangulations. Zbl 1360.52013 Gurel-Gurevich, Ori; Nachmias, Asaf; Souto, Juan 2017 Rate of convergence for Cardy’s formula. Zbl 1294.82021 Mendelson, Dana; Nachmias, Asaf; Watson, Samuel S. 2014 The Dirichlet problem for orthodiagonal maps. Zbl 1460.31026 Gurel-Gurevich, Ori; Jerison, Daniel C.; Nachmias, Asaf 2020 Geometric and spectral properties of causal maps. Zbl 1454.05107 Curien, Nicolas; Hutchcroft, Tom; Nachmias, Asaf 2020 Planar maps, random walks and circle packing. École d’Été de Probabilités de Saint-Flour XLVIII – 2018. Zbl 07105774 Nachmias, Asaf 2020 The Dirichlet problem for orthodiagonal maps. Zbl 1460.31026 Gurel-Gurevich, Ori; Jerison, Daniel C.; Nachmias, Asaf 2020 Uniform spanning forests of planar graphs. Zbl 1422.60022 Hutchcroft, Tom; Nachmias, Asaf 2019 Hyperbolic and parabolic unimodular random maps. Zbl 1459.60018 Angel, Omer; Hutchcroft, Tom; Nachmias, Asaf; Ray, Gourab 2018 Hypercube percolation. Zbl 1381.60119 van der Hofstad, Remco; Nachmias, Asaf 2017 Indistinguishability of trees in uniform spanning forests. Zbl 1407.60019 Hutchcroft, Tom; Nachmias, Asaf 2017 Recurrence of multiply-ended planar triangulations. Zbl 1360.52013 Gurel-Gurevich, Ori; Nachmias, Asaf; Souto, Juan 2017 Unimodular hyperbolic triangulations: circle packing and random walk. Zbl 1360.52012 Angel, Omer; Hutchcroft, Tom; Nachmias, Asaf; Ray, Gourab 2016 Boundaries of planar graphs, via circle packings. Zbl 1339.05061 Angel, Omer; Barlow, Martin T.; Gurel-Gurevich, Ori; Nachmias, Asaf 2016 Random walks on stochastic hyperbolic half planar triangulations. Zbl 1344.05127 Angel, Omer; Nachmias, Asaf; Ray, Gourab 2016 A power law of order 1/4 for critical mean field Swendsen-Wang dynamics. Zbl 1304.60078 Long, Yun; Nachmias, Asaf; Ning, Weiyang; Peres, Yuval 2014 Unlacing hypercube percolation: a survey. Zbl 1455.60132 van der Hofstad, Remco; Nachmias, Asaf 2014 Electrical resistance of the low dimensional critical branching random walk. Zbl 1312.60101 Járai, Antal A.; Nachmias, Asaf 2014 Rate of convergence for Cardy’s formula. Zbl 1294.82021 Mendelson, Dana; Nachmias, Asaf; Watson, Samuel S. 2014 Recurrence of planar graph limits. Zbl 1262.05031 Gurel-Gurevich, Ori; Nachmias, Asaf 2013 Nonconcentration of return times. Zbl 1268.05183 Gurel-Gurevich, Ori; Nachmias, Asaf 2013 Non-amenable Cayley graphs of high girth have $$p_c < p_u$$ and mean-field exponents. Zbl 1302.82056 Nachmias, Asaf; Peres, Yuval 2012 Arm exponents in high dimensional percolation. Zbl 1219.60085 Kozma, Gady; Nachmias, Asaf 2011 Is the critical percolation probability local? Zbl 1230.60099 Benjamini, Itai; Nachmias, Asaf; Peres, Yuval 2011 The evolution of the cover time. Zbl 1226.05230 Barlow, Martin T.; Ding, Jian; Nachmias, Asaf; Peres, Yuval 2011 A note about critical percolation on finite graphs. Zbl 1235.60138 Kozma, Gady; Nachmias, Asaf 2011 Critical percolation on random regular graphs. Zbl 1209.05228 Nachmias, Asaf; Peres, Yuval 2010 The critical random graph, with martingales. Zbl 1215.05168 Nachmias, Asaf; Peres, Yuval 2010 Testing the expansion of a graph. Zbl 1194.68173 Nachmias, Asaf; Shapira, Asaf 2010 The Alexander-Orbach conjecture holds in high dimensions. Zbl 1180.82094 Kozma, Gady; Nachmias, Asaf 2009 Mean-field conditions for percolation on finite graphs. Zbl 1182.60025 Nachmias, Asaf 2009 Critical random graphs: Diameter and mixing time. Zbl 1160.05053 Nachmias, Asaf; Peres, Yuval 2008 Component sizes of the random graph outside the scaling window. Zbl 1130.05053 Nachmias, Asaf; Peres, Yuval 2007 Coloring complete bipartite graphs from random lists. Zbl 1110.05035 Krivelevich, Michael; Nachmias, Asaf 2006 Colouring powers of cycles from random lists. Zbl 1050.05053 Krivelevich, Michael; Nachmias, Asaf 2004 all top 5 #### Cited by 258 Authors 20 Nachmias, Asaf 19 van der Hofstad, Remco W. 18 Hutchcroft, Tom 15 Peres, Yuval 9 Ding, Jian 9 Kozma, Gady 8 Benjamini, Itai 7 Angel, Omer 6 Curien, Nicolas 6 Gwynne, Ewain 6 Lubetzky, Eyal 6 Ray, Gourab 6 Sen, Sanchayan 5 Bhamidi, Shankar 5 Grimmett, Geoffrey R. 5 Hermon, Jonathan 4 Casselgren, Carl Johan 4 Duminil-Copin, Hugo 4 Heydenreich, Markus 4 Hulshof, Tim 4 Kumagai, Takashi 4 Riordan, Oliver Maxim 4 van Leeuwaarden, Johan S. H. 3 Addario-Berry, Louigi 3 Barlow, Martin T. 3 Bollobás, Béla 3 Dhara, Souvik 3 Fribergh, Alexander 3 Lee, James R. 3 Li, Zhongyang 3 Luczak, Malwina J. 3 Miller, Jason P. 3 Paquette, Elliot 3 Pfeffer, Joshua 3 Timár, Ádám 2 Ahlberg, Daniel 2 Ben Arous, Gérard 2 Berestycki, Nathanaël 2 Blanca, Antonio 2 Budhiraja, Amarjit S. 2 Budzinski, Thomas 2 Černý, Jiří 2 Dantas, Simone 2 den Hollander, Frank 2 Frieze, Alan Michael 2 Georgakopoulos, Agelos 2 Gill, James T. 2 Goldreich, Oded 2 Hambly, Ben M. 2 Holmes, Mark P. 2 Janson, Svante 2 Járai, Antal A. 2 Joos, Felix Claudius 2 Kang, Mihyun 2 Kim, Jeong Han 2 Krivelevich, Michael 2 Lyons, Russell 2 Okamura, Kazuki 2 Perarnau, Guillem 2 Pittel, Boris G. 2 Sakai, Akira 2 Shang, Yilun 2 Shapira, Asaf 2 Sheffield, Scott 2 Sinclair, Alistair 2 Sly, Allan 2 Sousi, Perla 2 Stauffer, Alexandre O. 2 Stephenson, Kenneth 2 Tassion, Vincent 2 Teixeira, Augusto Quadros 2 Vigoda, Eric 2 Wang, Xuan 2 Wormald, Nicholas Charles 1 Abe, Yoshihiro 1 Albenque, Marie 1 Ambainis, Andris 1 Anantharaman, Nalini 1 Aspelmeier, Timo 1 Athreya, Siva R. 1 Avena, Luca 1 Balankin, Alexander S. 1 Bandyopadhyay, Antar 1 Baur, Erich 1 Ben-Shimon, Sonny 1 Berche, Bertrand 1 Berman, Itay 1 Binder, Ilia A. 1 Biskup, Marek 1 Björnberg, Jakob Erik 1 Bordenave, Charles 1 Boucheron, Stéphane Vincent 1 Bowen, Lewis Phylip 1 Bowers, Philip L. 1 Broutin, Nicolas 1 Brown, Jason Ira 1 Bücking, Ulrike 1 Cabezas, Manuel 1 Campos, C. N. 1 Caputo, Pietro ...and 158 more Authors all top 5 #### Cited in 62 Serials 26 The Annals of Probability 24 Probability Theory and Related Fields 17 Random Structures & Algorithms 15 Annales de l’Institut Henri Poincaré. Probabilités et Statistiques 9 Communications in Mathematical Physics 8 Combinatorics, Probability and Computing 8 Electronic Journal of Probability 6 Journal of Statistical Physics 6 Geometric and Functional Analysis. GAFA 5 Inventiones Mathematicae 5 Journal of Theoretical Probability 5 The Annals of Applied Probability 5 Probability Surveys 4 SIAM Journal on Discrete Mathematics 3 Discrete Applied Mathematics 3 Discrete Mathematics 3 Advances in Mathematics 3 Proceedings of the American Mathematical Society 3 Transactions of the American Mathematical Society 3 Graphs and Combinatorics 3 Stochastic Processes and their Applications 2 Advances in Applied Probability 2 Journal of Applied Probability 2 European Journal of Combinatorics 2 Discrete & Computational Geometry 2 Journal of the American Mathematical Society 2 Annals of Mathematics. Second Series 1 Indian Journal of Pure & Applied Mathematics 1 Israel Journal of Mathematics 1 Mathematical Proceedings of the Cambridge Philosophical Society 1 Metrika 1 Physics Letters. A 1 Algebra and Logic 1 Journal of Combinatorial Theory. Series A 1 Journal of Combinatorial Theory. Series B 1 Journal of Functional Analysis 1 Memoirs of the American Mathematical Society 1 Osaka Journal of Mathematics 1 SIAM Journal on Computing 1 Theoretical Computer Science 1 Science of Computer Programming 1 Combinatorica 1 Revista Matemática Iberoamericana 1 Information and Computation 1 Journal of the Ramanujan Mathematical Society 1 Computational Geometry 1 International Journal of Algebra and Computation 1 Computational Complexity 1 Discussiones Mathematicae. Graph Theory 1 Mathematical Problems in Engineering 1 Conformal Geometry and Dynamics 1 Journal of the European Mathematical Society (JEMS) 1 Brazilian Journal of Probability and Statistics 1 Foundations of Computational Mathematics 1 ALEA. Latin American Journal of Probability and Mathematical Statistics 1 Journal of Physics A: Mathematical and Theoretical 1 Japanese Journal of Mathematics. 3rd Series 1 Kyoto Journal of Mathematics 1 Symmetry 1 Forum of Mathematics, Sigma 1 ISRN Probability and Statistics 1 Annales de l’Institut Henri Poincaré D. Combinatorics, Physics and their Interactions (AIHPD) all top 5 #### Cited in 29 Fields 148 Probability theory and stochastic processes (60-XX) 132 Combinatorics (05-XX) 62 Statistical mechanics, structure of matter (82-XX) 10 Computer science (68-XX) 9 Dynamical systems and ergodic theory (37-XX) 9 Convex and discrete geometry (52-XX) 7 Group theory and generalizations (20-XX) 6 Potential theory (31-XX) 5 Functions of a complex variable (30-XX) 4 Partial differential equations (35-XX) 4 Numerical analysis (65-XX) 3 Measure and integration (28-XX) 3 Biology and other natural sciences (92-XX) 2 Mathematical logic and foundations (03-XX) 2 Number theory (11-XX) 2 Topological groups, Lie groups (22-XX) 2 Abstract harmonic analysis (43-XX) 2 Operator theory (47-XX) 2 Geometry (51-XX) 2 Quantum theory (81-XX) 2 Operations research, mathematical programming (90-XX) 2 Game theory, economics, finance, and other social and behavioral sciences (91-XX) 1 Commutative algebra (13-XX) 1 Ordinary differential equations (34-XX) 1 Difference and functional equations (39-XX) 1 Statistics (62-XX) 1 Classical thermodynamics, heat transfer (80-XX) 1 Relativity and gravitational theory (83-XX) 1 Information and communication theory, circuits (94-XX)	2021-09-21 09:27:01	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5672624111175537, "perplexity": 12525.832173953524}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057199.49/warc/CC-MAIN-20210921070944-20210921100944-00543.warc.gz"}
https://www.semanticscholar.org/paper/Hamiltonian-simulation-with-optimal-sample-Kimmel-Lin/557f16458cc2b371cf5bfeb38dca98bfca0b12a9	# Hamiltonian simulation with optimal sample complexity @article{Kimmel2016HamiltonianSW, title={Hamiltonian simulation with optimal sample complexity}, author={Shelby Kimmel and Cedric Yen-Yu Lin and Guang Hao Low and Maris A. Ozols and Theodore J. Yoder}, journal={npj Quantum Information}, year={2016}, volume={3}, pages={1-7} } • Published 31 July 2016 • Computer Science • npj Quantum Information We investigate the sample complexity of Hamiltonian simulation: how many copies of an unknown quantum state are required to simulate a Hamiltonian encoded by the density matrix of that state? We show that the procedure proposed by Lloyd, Mohseni, and Rebentrost [Nat. Phys., 10(9):631–633, 2014] is optimal for this task. We further extend their method to the case of multiple input states, showing how to simulate any Hermitian polynomial of the states provided. As applications, we derive optimal… Approximating Hamiltonian dynamics with the Nyström method • Mathematics Quantum • 2020 This work derives a simulation technique whose runtime scales polynomially in the number of qubits and the Frobenius norm of the Hamiltonian, and suggests that under strong sampling assumptions there exist classical poly-logarithmic time simulations of quantum computations. A pr 2 01 8 Approximating Hamiltonian dynamics with the Nyström method • Mathematics, Computer Science • 2018 It is shown that sample based quantum simulation, a type of evolution where the Hamiltonian is a density matrix, can be efficiently classically simulated under specific structural conditions, and proposed conditions for the efficient approximation of state vectors evolving under a given Hamiltonian. Interactive Proofs for Synthesizing Quantum States and Unitaries • Computer Science ITCS • 2022 This work defines models of interactive proofs for synthesizing quantum states and unitaries, where a polynomial-time quantum verifier interacts with an untrusted quantum prover, and a verifier who accepts also outputs an approximation of the target state or the result of thetarget unitary applied to the input state. Quantum singular value transformation and beyond: exponential improvements for quantum matrix arithmetics • Computer Science, Mathematics STOC • 2019 A new “Quantum singular value transformation” algorithm is developed that can directly harness the advantages of exponential dimensionality by applying polynomial transformations to the singular values of a block of a unitary operator. Simulating Large Quantum Circuits on a Small Quantum Computer. • Physics, Computer Science Physical review letters • 2020 This Letter introduces cluster parameters K and d of a quantum circuit and proposes a cluster simulation scheme that can simulate any (K,d)-clustered quantum circuit on a d-qubit machine in time roughly 2^{O(K)}, with further speedups possible when taking more fine-grained circuit structure into account. Improved Quantum Algorithms for Fidelity Estimation • Computer Science, Mathematics • 2022 It is proved that ﬁdelity estimation to any non-trivial constant additive accuracy is hard in general, by giving a sample complexity lower bound that depends polynomially on the dimension. Hamiltonian Simulation by Uniform Spectral Amplification • Computer Science • 2017 This work motivates a systematic approach to understanding and exploiting structure, in a setting where Hamiltonians are encoded as measurement operators of unitary circuits $\hat{U}$ for generalized measurement, and presents general solutions to uniform spectral amplification. Fast algorithm for quantum polar decomposition, pretty-good measurements, and the Procrustes problem • Computer Science • 2021 This work shows that the problem of quantum polar decomposition has a simple and concise implementation via the quantum singular value transform (QSVT), and develops algorithms for these problems whose gate complexity exhibits a polynomial advantage in the size and condition number of the input compared to alternative approaches for the same problem settings. Quantum lower bounds for approximate counting via laurent polynomials • S. Aaronson • Computer Science, Mathematics Electron. Colloquium Comput. Complex. • 2018 The complexity of approximate counting, the problem of multiplicatively estimating the size of a nonempty set S ⊆ [N], is resolved in two natural generalizations of quantum query complexity. Physical implementation of quantum nonparametric learning with trapped ions • Dan-Bo Zhang, Z. D. Wang • Computer Science, Physics Physical review letters • 2020 A quantum paradigm of nonparametric learning that offers an exponential speedup over the sample size is pointed out and a trained state for prediction can be obtained by entanglement spectrum transformation, using the quantum matrix toolbox. ## References SHOWING 1-10 OF 56 REFERENCES Optimal Hamiltonian Simulation by Quantum Signal Processing. • Physics Physical review letters • 2017 It is argued that physical intuition can lead to optimal simulation methods by showing that a focus on simple single-qubit rotations elegantly furnishes an optimal algorithm for Hamiltonian simulation, a universal problem that encapsulates all the power of quantum computation. Efficient algorithms in quantum query complexity These algorithms are a novel application of the quantum walk search framework and give improved upper bounds for several subgraph-finding problems and study the quantum query complexity of matrix multiplication and related problems over rings, semirings, and the Boolean semiring in particular. Demonstrating the viability of universal quantum computation using teleportation and single-qubit operations • Physics, Computer Science Nature • 1999 It is shown that single quantum bit operations, Bell-basis measurements and certain entangled quantum states such as Greenberger–Horne–Zeilinger (GHZ) states are sufficient to construct a universal quantum computer. Hamiltonian Simulation with Nearly Optimal Dependence on all Parameters • Computer Science 2015 IEEE 56th Annual Symposium on Foundations of Computer Science • 2015 An algorithm for sparse Hamiltonian simulation whose complexity is optimal (up to log factors) as a function of all parameters of interest is presented, and a new lower bound is proved showing that no algorithm can have sub linear dependence on tau. Experimentally superposing two pure states with partial prior knowledge • Physics Physical Review A • 2017 Superposition, arguably the most fundamental property of quantum mechanics, lies at the heart of quantum information science. However, how to create the superposition of any two unknown pure states Gate count estimates for performing quantum chemistry on small quantum computers • Physics, Computer Science • 2014 This paper focuses on the quantum resources required to find the ground state of a molecule twice as large as what current classical computers can solve exactly and suggests that for quantum computation to become useful for quantum chemistry problems, drastic algorithmic improvements will be needed. Quantum Computation by Adiabatic Evolution • Physics • 2000 We give a quantum algorithm for solving instances of the satisfiability problem, based on adiabatic evolution. The evolution of the quantum state is governed by a time-dependent Hamiltonian that Universal Quantum Emulator • Computer Science, Mathematics • 2016 A quantum algorithm that emulates the action of an unknown unitary transformation on a given input state, using multiple copies of some unknown sample input states of the unitary and their corresponding output states, which can be used as a subroutine in other algorithms, such as quantum phase estimation. Improving quantum algorithms for quantum chemistry • Computer Science Quantum Inf. Comput. • 2015 Improvements to the standard Trotter-Suzuki based algorithms used in the simulation of quantum chemistry on a quantum computer by modifying how Jordan-Wigner transformations are implemented to reduce their cost from linear or logarithmic in the number of orbitals to a constant. Universal quantum computation with the exchange interaction • Physics Nature • 2000 An explicit scheme is introduced in which the Heisenberg interaction alone suffices to implement exactly any quantum computer circuit, at a price of a Factor of three in additional qubits, and about a factor of ten in additional two-qubit operations.	2022-07-04 06:34:55	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6626527905464172, "perplexity": 1668.8123577020826}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104354651.73/warc/CC-MAIN-20220704050055-20220704080055-00213.warc.gz"}
https://gateoverflow.in/338409/kenneth-rosen-edition-7-exercise-6-1-question-46-page-no-397	216 views In how many ways can a photographer at a wedding arrange $6$ people in a row from a group of $10$ people, where the bride and the groom are among these $10$ people, if 1. the bride must be in the picture? 2. both the bride and groom must be in the picture? 3. exactly one of the bride and the groom is in the picture?	2023-02-04 08:09:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4402589499950409, "perplexity": 176.3168357348433}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500095.4/warc/CC-MAIN-20230204075436-20230204105436-00841.warc.gz"}
https://ryandoughertyaz.wordpress.com/interesting-research-problems/	# Interesting Research Problems I have been thinking of interesting problems that can be researched for a long time, that I haven’t looked into much (or don’t have the time to do so) but have a healthy interest in understanding. If you would like to collaborate on one, any, or a related problem to one listed here, feel free to contact me. 1. Automata Theory 1. In the NFA to DFA transformation, we may have an exponential necessary increase in the number of states in the corresponding DFA (and there are explicit NFA constructions to guarantee this increase). What is the average/expected number in general? [Added Oct. 11, 2016] 2. In the NFA to DFA transformation, is there a polynomial-time algorithm to compute the number of states that will be produced for the DFA, without having to compute the DFA itself? [Added Oct. 11, 2016] 3. The number of strings of length ${n}$ that can be generated by a given unambiguous (i.e., every string in the grammar’s language only has 1 leftmost derivation) context-free grammar can be determined in polynomial time, with dynamic programming. What about for inherently ambiguous languages (i.e., every grammar is ambiguous)? [Added Oct. 13, 2016] 2. Fun and Games 1. What is the computational complexity of DiscoZoo? You are given an ${n \times n}$ grid and a list of shapes, with known patterns, as well as how many of each are inside the grid. The shapes are not oriented, rotated, or modified in any way. Your task is to find the fewest moves required to uncover all the shapes by selecting individual squares; doing so will reveal if a pattern contains that square (and remains known), and uncovers a “blank” otherwise. The patterns are not allowed to overlap (i.e., at most one pattern occupies a given square). An obvious upper-bound is ${n^2}$, but is it possible to do better? If not in the general case, what cases give interesting bounds? [Added Oct. 11, 2016] 2. What is the computational complexity of Guitar Hero (or any rhythm game)? There are many ways of formulating the question, such as (1) formulating a decision problem to see if it is possible to get above a certain score, (2) the minimum amount of interaction among ${k}$ players via star power to achieve a score, (3) the maximum number of notes missed to pass a given song, (4) are there any interesting generalizations of the game, etc. All of these I believe can be solved with dynamic programming in polynomial time, but I haven’t given much thought to it. [Added Oct. 13, 2016] 3. Graph Theory 1. A unicyclic graph is one that contains exactly 1 cycle. The number of spanning trees can be computed in polynomial time, due to a number of methods. Can the number of unicyclic spanning subgraphs of a graph be computed as efficiently? This has numerous applications to network reliability. [Added Oct. 13, 2016] 4. Network Reliability 1. Is there a closed formula for the reliability polynomial of the complete graph ${K_n}$? A recursive formula is: $\displaystyle Rel(K_n; p) = 1- \sum_{j=1}^{n-1} {n-1 \choose j-1} Rel(K_j; p)(1-p)^{j(n-j)}.$ 1. You are given a list of ${n}$ digits in base ${b}$. What is ${\max\{p_1 \times \cdots \times p_k\}}$ where the ${p_i}$ are concatenations of the ${n}$ digits (and partition the input digits)? [Added Oct. 11, 2016] 2. Given a range of integers ${[a, b]}$, what is the probability that for a given pair of integers ${(x, y) \in [a, b]}$ in the range (written in binary) that XORing, ANDing, or ORing the bitwise representations of ${x, y}$ remains in ${[a, b]}$? [Added Oct. 11, 2016]	2017-08-22 03:29:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 15, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7110985517501831, "perplexity": 421.1912340918601}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886109893.47/warc/CC-MAIN-20170822031111-20170822051111-00061.warc.gz"}
https://tetrisconcept.net/threads/tgm-arcade-board.2067/	Thread in 'Hardware' started by JeanPR, 28 Sep 2012. 1. ### JeanPR Hi Everyone, I have a TGM arcade daughterboard I picked up in Tokyo a good few years ago and would either like to get it up and running or sell it on to a good home (I've never booted it). Does anyone have any advice or would be interested? I think I need a zn-2 motherboard and a supergun to get it going, does anyone have any advice for getting something in the UK? Thanks Jean 2. ### Edoa.k.a. FSY The supergun is no problem; if you have any skills with a soldering iron and a bit of free time, you can knock one up quite cheaply. Or if you just want to buy one, a Vogatek is a very cheap solution - the Mk 3 has a SCART socket, so is perfect for use in the UK. The guy who makes them is pretty unreliable, but there should be plenty of reliable guys selling them on. As for the ZN-2, they're unfortunately quite rare to find on their own. You may just have to buy a cheap mother/daughterboard set, like Street Fighter EX2, and use the motherboard off that for both games. With a bit of patience you could probably pick up SFEX2 from YAJ for about £40-50. (EX2+ is considerably more expensive though.) Also, tops-game currently has an EX2 for ~£66 (8400JPY) listed on the capcom page. (The Street Fighter EX series are criminally underrated btw...) Another possibility is modifying a Taito G-net motherboard, since the G-net also uses a ZN-2 as its base, which is physically identical to the Capcom ZN-2 (apart from the placement of a single resistor). I don't know how much work is involved in this, but I would assume at the very least it'd involve flashing a different bios. 3. ### cleure Do all ZN-2 games have swappable motherboards? I think I read somewhere that not all of them do (I'm not sure, it was a long time ago, and I can't find any links). 4. ### Edoa.k.a. FSY I'm pretty sure that they are. There are literally only 7 games for the Capcom ZN-2 (TGM, Rival Schools, SFEX2, SFEX2+, Tech Romancer, Star Gladiator 2, Strider 2), and they should all be interchangeable. I have however read of people trying to run ZN-2 games on a ZN-1 motherboard, and then getting confused when they didn't work, this might have been what you read about a long time ago. From a thread on the neo-geo.com forums: 6. ### Edoa.k.a. FSY $249 for Tech Romancer is very pricey, especially when you consider that a ZN-2 with Tech Romancer and 4 other daughterboards sold for less than that about a year ago: http://www.ebay.com/itm/Arcade-pcb-board-CAPCOM-ZN2-5-games-STRIDER-2-/110774923365. Sure, you're getting all the art and the control panel, but they all look like they've seen better days. Less than$100 for Rival Schools seems like a pretty good deal though. Just in case it's of interest to anyone, G-Front is currently listing a TGM daughterboard for 3000JPY (approx \$38 or £24)	2023-03-28 15:54:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2260216921567917, "perplexity": 2948.2546952639623}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948867.32/warc/CC-MAIN-20230328135732-20230328165732-00421.warc.gz"}
https://harangdev.github.io/deep-learning/improving-deep-neural-networks/8/	# Mini-Batch Gradient Descent and Learning Rate Decay Categories: Updated: We call set of all of our training examples ‘batch’. When we train batch once, we say we’ve trained for one ‘epoch’. When number of examples become large, using a batch to perform one gradient descent is computationally expensive and takes a lot of time. To solve this problem, we divide our batch into many ‘mini-batches’, and use a mini-batch to perform one gradient descent; in other words, perform many gradient descents per one epoch. When mini-batch size is 1, we call it ‘stochastic gradient descent’. We use one example to perform one gradient descent, so we perform $m$ gradient desents per one epoch. Stochastic gradient descent doesn’t make use of speed that comes out of vectorization, and batch gradient descent is too slow as it performs one gradient descent per epoch. Thus in practice, we use mini-batch gradient descent with mini-batch size of 64, 128, 256, 512. # Cost Curve of mini-batch gradient descent. Unlike batch gradient descent, cost of mini-batch gradient descent doesn’t always decrease after one gradient descent. That’s because cost of mini-batch gradient is calculated for every mini-batch and there can be easy-to-fit mini-batch and hard-to-fit mini-batch. Below is contour line of cost with respect to 2-dimensional features. Colored lines indicate gradient descents of different gradient descent types. While batch gradient descent converges straightly to local minimum, batch and stochastic gradient descent oscillates. # Learning Rate Decay When using mini-batch gradient descent, decreasing learning rate after an epoch decreases oscillating error, since weights are updated less as they approach minimum. There are many ways to implement learning rate decay. Let’s say $t$ is our epoch number and $k$ a constant we can set. • $\alpha = \frac{a_0}{1+t}$ • $\alpha=0.95^t\alpha_0$ • $\alpha=\frac{k}{\sqrt{t}}\alpha_0$ • discrete staircase: when $t$ reaches some value, decrease $\alpha$. • Manual decay, etc… It is worth trying after you think you’ve done everything else. Categories:	2020-07-10 13:08:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9270285367965698, "perplexity": 1636.4937150906583}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655908294.32/warc/CC-MAIN-20200710113143-20200710143143-00348.warc.gz"}
https://www.gamedev.net/forums/topic/326326-help-me-debug-my-delay-function-d/	# Help me debug my delay function? :D This topic is 4852 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic. ## Recommended Posts void Delay(double delay) { double a = timeGetTime(); while (timeGetTime() < (a + delay1000)) { } } I'm hoping to use this function (once it works :P) to add some speed capping to my program. As of now, it executes as fast as my computer can do it, but, as it's purpose is to be a game, that is too unpredictable. So, if anyone knows a better way to achieve the results i've described, Wahoo ;) Thanks before hand! (I know that this uses the windows library, and isn't technically an openGL question, but a general programming question, but I feel that as it's adaption is for the purpose of openGL, if a better method exists I'd be informed by you all :D) ##### Share on other sites It's probably a much better idea to have all your character/game movements time based instead of framebased (I'm assuming this is why you want to slow it down)... that way you won't have to slow down your game (what if someone else runs your game slower?). So for example, instead of moving the player 2 units every frame, move it based on how fast the game is running: movement += 2; // old frame-base way... don't do this movment += (2UsualFPS)/CurrentFPS; // run this every frame... time based use this where UsualFPS is the speed you want the unit to move at (so if you like how everything works at 60FPS, then replace that with 60), 2 is the unit speed, and CurrentFPS is the actual frames per second the game is running at. You can see therefore that the higher CurrentFPS is, the slower the unit will move at per frame (meaning that it will move 2 units at 60 FPS in the same time it would take to move 2 units at 500FPS). I hope I made some sense here... I'm kind-of out of it right now. -Walter ##### Share on other sites Thank you for the quick reply. How do I catch the current FPS though? I dont think currentFPS is a reserved keyword or something ;( ##### Share on other sites You have to calculate it. curentFPS = 1 / time_of_last_frame ##### Share on other sites Yeah, you'll have to calculate the FPS yourself. [EDIT] Beaten to it by DarkWing who also had a much simpler way of doing the calculation [/EDIT] ##### Share on other sites personally I think loop delays are sometimes needed, such as in an rpg when you're updating a characters walk animation - you can sacrifice that slowdown for smoother animation. while(whatever){ time = GetTime(); Delay(constant_delay_time - GetTime() + time);} (i think thats right) ##### Share on other sites my function served my purposes.. erm.. perfectly :) I'm going to look into the FPS thing though, that might help me out later. For anyone else who is interested, the problem with my function was that I didn't have Winmm.lib in my linker. ##### Share on other sites Quote: Original post by MrLeapmy function served my purposes.. erm.. perfectly :) I'm going to look into the FPS thing though, that might help me out later.For anyone else who is interested, the problem with my function was that I didn't have Winmm.lib in my linker. Your function is still going to peg the CPU with that loop.. The more eco-friendly way to do it is to calculate how many milliseconds to wait, then do a sleep so that other processes (or idle) can run. For example: float fMinTime = 1000.0f / fMaxFPS; // the FPS capwhile (bRunning) { float fStartTime = CurrentMilliseconds(); // exact function depends on OS // ... do your calculations and drawing here float fEndTIme = CurrentMilliseconds(); if (fEndTime - fStartTime < fMinTime) Sleep(fMinTime - (fEndTime - fStartTime)); // also OS-dependent } Also, to avoid a division (make your code faster) I would make your frame updates dependent on actual TIME, not frequency (FPS) as the previous poster gave you an example. Just calculate the amount of time taken for each frame and feed it into your update function as a value in milliseconds. Then in your update function you can do something like this: fCharacterY += 0.02f * fFrameMilliseconds; That way he will move 0.02 units per millisecond, and you avoid having to do that division used in the FPS method above. Richard ##### Share on other sites [I wish this all to be scored! CROSSED OUT EVEN!]heh, anyone know a function similer to sleep for windows? I dont know how I'd write that if it didn't exist :p [/End score.] Never mind! msdn told me that sleep(); does exist within the Kernel32.lib! something about threads though. I'm dont know much about that multi threading jazzery. Time to read I guess :D Thank you all once again! edit: where can I find all the tags? this guessing stuff is getting kind of tough :P ##### Share on other sites Quote: Original post by MrLeap[I wish this all to be scored! CROSSED OUT EVEN!]heh, anyone know a function similer to sleep for windows? I dont know how I'd write that if it didn't exist :p [/End score.]Never mind! msdn told me that sleep(); does exist within the Kernel32.lib! something about threads though. I'm dont know much about that multi threading jazzery. Time to read I guess :DThank you all once again!edit: where can I find all the tags? this guessing stuff is getting kind of tough :P In Win32, you can use this function: void Sleep( DWORD dwMilliseconds); If you're using SDL (which I recommend, to support multiple platforms) you use SDL_Delay(). You don't need to worry about multi-threading to use the sleep functions. BTW some of the tags can be found in the 'faq' link in the upper right corner of the page. Richard 1. 1 Rutin 37 2. 2 3. 3 4. 4 5. 5 • 11 • 10 • 13 • 103 • 11 • ### Forum Statistics • Total Topics 632976 • Total Posts 3009672 • ### Who's Online (See full list) There are no registered users currently online ×	2018-10-21 23:02:21	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.18751439452171326, "perplexity": 2958.1547084753433}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583514437.69/warc/CC-MAIN-20181021224001-20181022005501-00364.warc.gz"}
https://labs.tib.eu/arxiv/?author=Jinglei%20Zhang	• ### Non-saturating Quantum Magnetization in Weyl semimetal TaAs(1802.08801) Detecting the spectroscopic signatures of Dirac-like quasiparticles in emergent topological materials is crucial for searching their potential applications. Magnetometry is a powerful tool for fathoming electrons in solids, yet its ability for discerning Dirac-like quasiparticles has not been recognized. Adopting the probes of magnetic torque and parallel magnetization for the archetype Weyl semimetal TaAs in strong magnetic field, we observed a quasi-linear field dependent effective transverse magnetization and a strongly enhanced parallel magnetization when the system is in the quantum limit. Distinct from the saturating magnetic responses for massive carriers, the non-saturating signals of TaAs in strong field is consistent with our newly developed magnetization calculation for a Weyl fermion system in an arbitrary angle. Our results for the first time establish a thermodynamic criterion for detecting the unique magnetic response of 3D massless Weyl fermions in the quantum limit. • ### Prediction and retrodiction with continuously monitored Gaussian states(1710.04950) Oct. 13, 2017 quant-ph Gaussian states of quantum oscillators are fully characterized by the mean values and the covariance matrix of their quadrature observables. We consider the dynamics of a system of oscillators subject to interactions, damping, and continuous probing which maintain their Gaussian state property. Such dynamics is found in many physical systems that can therefore be efficiently described by the ensuing effective representation of the density matrix $\rho(t)$. Our probabilistic knowledge about the outcome of measurements on a quantum system at time $t$ is not only governed by $\rho(t)$ conditioned on the evolution and measurement outcomes obtained until time $t$, but is also modified by any information acquired after $t$. It was shown in [Phys. Rev. Lett. 111, 160401 (2013)] that this information is represented by a supplementary matrix, $E(t)$. We show here that the restriction of the dynamics of $\rho(t)$ to Gaussian states implies that the matrix $E(t)$ is also fully characterized by a vector of mean values and a covariance matrix. We derive the dynamical equations for these quantities and we illustrate their use in the retrodiction of measurements on Gaussian systems. • ### Evolution of Weyl orbit and quantum Hall effect in Dirac semimetal Cd3As2(1612.05873) Owing to the coupling between open Fermi arcs on opposite surfaces, topological Dirac semimetals exhibit a new type of cyclotron orbit in the surface states known as Weyl orbit. Here, by lowering the carrier density in Cd3As2 nanoplates, we observe a crossover from multiple- to single-frequency Shubnikov-de Haas (SdH) oscillations when subjected to out-of-plane magnetic field, indicating the dominant role of surface transport. With the increase of magnetic field, the SdH oscillations further develop into quantum Hall state with non-vanishing longitudinal resistance. By tracking the oscillation frequency and Hall plateau, we observe a Zeeman-related splitting and extract the Landau level index as well as sub-band number. Different from conventional two-dimensional systems, this unique quantum Hall effect may be related to the quantized version of Weyl orbits. Our results call for further investigations into the exotic quantum Hall states in the low-dimensional structure of topological semimetals. • ### Magnetic tunneling induced Weyl node annihilation in TaP(1507.06301) Weyl nodes are topological objects in three-dimensional metals. Their topological property can be revealed by studying the high-field transport properties of a Weyl semimetal. While the energy of the lowest Landau band (LLB) of a conventional Fermi pocket always increases with magnetic field due to the zero point energy, the LLB of Weyl cones remains at zero energy unless a strong magnetic field couples the Weyl fermions of opposite chirality. In the Weyl semimetal TaP, we achieve such a magnetic coupling between the electron-like Fermi pockets arising from the W1 Weyl fermions. As a result, their LLBs move above chemical potential, leading to a sharp sign reversal in the Hall resistivity at a specific magnetic field corresponding to the W1 Weyl node separation. By contrast, despite having almost identical carrier density, the annihilation is unobserved for the hole-like pockets because the W2 Weyl nodes are much further separated. These key findings, corroborated by other systematic analyses, reveal the nontrivial topology of Weyl fermions in high-field measurements. • ### Large linear magnetoresistance in a transition-metal stannide $\beta$-RhSn$_4$(1705.00304) Materials exhibiting large magnetoresistance may not only be of fundamental research interest, but also can lead to wide-ranging applications in magnetic sensors and switches. Here we demonstrate a large linear-in-field magnetoresistance, $\Delta \rho/\rho$ reaching as high as $\sim$600$\%$ at 2 K under a 9 Tesla field, in the tetragonal phase of a transiton-metal stannide $\beta$-RhSn$_4$. Detailed analyses show that its magnetic responses are overall inconsistent with the classical model based on the multiple electron scattering by mobility fluctuations in an inhomogenous conductor, but rather in line with the quantum effects due to the presence of Dirac-like dispersions in the electronic structure. Our results may help guiding the future quest for quantum magnetoresistive materials into the family of stannides, similar to the role played by PtSn$_4$ with topological node arcs. • ### Extremely large magnetoresistance in a topological semimetal candidate pyrite PtBi2(1609.02626) While pyrite-type PtBi2 with face-centered cubic structure has been predicted to be a three-dimensional (3D) Dirac semimetal, experimental study on its physical properties remains absent. Here we report the angular-dependent magnetoresistance (MR) measurements of PtBi2 single-crystal under high magnetic fields. We observed extreme large unsaturated magnetoresistance (XMR) up to 11.2 million percent at T = 1.8 K in a magnetic field of 33 T, which surpasses the previously reported Dirac materials, such as WTe2, LaSb and NbP. The crystals exhibit an ultrahigh mobility and significant Shubnikov-de Hass (SdH) quantum oscillations with nontrivial Berry's phase. Analysis of Hall resistivity indicates that the XMR can be ascribed to the nearly compensated electron and hole. Our experimental results associated with the ab initio calculations suggest that pyrite PtBi2 is a topological semimetal candidate which might provide a platform for exploring topological materials with XMR in noble metal alloys. • ### Stepwise quantized surface states and delayed Landau level hybridization in Co cluster-decorated BiSbTeSe2 topological insulator devices(1702.03344) Feb. 10, 2017 cond-mat.mes-hall In three-dimensional topological insulators (TIs), the nontrivial topology in their electronic bands casts a gapless state on their solid surfaces, using which dissipationless TI edge devices based on the quantum anomalous Hall (QAH) effect and quantum Hall (QH) effect have been demonstrated. Practical TI devices present a pair of parallel-transport topological surface states (TSSs) on their top and bottom surfaces. However, due to the no-go theorem, the two TSSs always appear as a pair and are expected to quantize synchronously. Quantized transport of a separate Dirac channel is still desirable, but has never been observed in graphene even after intense investigation over a period of 13 years, with the potential aim of half-QHE. By depositing Co atomic clusters, we achieved stepwise quantization of the top and bottom surfaces in BiSbTeSe2 (BSTS) TI devices. Renormalization group flow diagrams13, 22 (RGFDs) reveal two sets of converging points (CVPs) in the (Gxy, Gxx) space, where the top surface travels along an anomalous quantization trajectory while the bottom surface retains 1/2 e2/h. This results from delayed Landau-level (LL) hybridization (DLLH) due to coupling between Co clusters and TSS Fermions. • ### Signature of chiral fermion instability in the Weyl semimetal TaAs above the quantum limit(1601.05895) Jan. 22, 2017 cond-mat.mes-hall We report the electrical transport properties for Weyl semimetal TaAs in an intense magnetic field. Series of anomalies occur in the longitudinal magnetoresistance and Hall signals at ultra-low temperatures when the Weyl electrons are confined into the lowest Landau level. These strongly temperature-dependent anomalies are ascribed to the electron-hole pairing instability. Our measurements show that the Weyl semimetal TaAs in the ultraquantum regime provides a good platform for studying electron-electron interaction in topological nontrivial semimetals. • ### Superconductivity and Charge Density Wave in ZrTe$_{3-x}$Se$_{x}$(1606.02284) Charge density wave (CDW), the periodic modulation of the electronic charge density, will open a gap on the Fermi surface that commonly leads to decreased or vanishing conductivity. On the other hand superconductivity, a commonly believed competing order, features a Fermi surface gap that results in infinite conductivity. Here we report that superconductivity emerges upon Se doping in CDW conductor ZrTe$_{3}$ when the long range CDW order is gradually suppressed. Superconducting critical temperature $T_c(x)$ in ZrTe$_{3-x}$Se$_x$ (${0\leq}x\leq0.1$) increases up to 4 K plateau for $0.04$$\leq$$x$$\leq$$0.07$. Further increase in Se content results in diminishing $T_{c}$ and filametary superconductivity. The CDW modes from Raman spectra are observed in $x$ = 0.04 and 0.1 crystals, where signature of ZrTe$_{3}$ CDW order in resistivity vanishes. The electronic-scattering for high $T_{c}$ crystals is dominated by local CDW fluctuations at high temperures, the resistivity is linear up to highest measured $T=300K$ and contributes to substantial in-plane anisotropy. • ### Transport evidence for the three-dimensional Dirac semimetal phase in ZrTe5(1603.05351) Topological Dirac semimetal is a newly discovered class of materials and has attracted intense attentions. This material can be viewed as a three-dimensional (3D) analogue of graphene and has linear energy dispersion in bulk, leading to a range of exotic transport properties. Here we report direct quantum transport evidence of 3D Dirac semimetal phase of layered material ZrTe5 by angular dependent magnetoresistance measurements under high magnetic fields up to 31 Tesla. We observed very clear negative longitudinal magnetoresistance induced by chiral anomaly under the condition of the magnetic field aligned only along the current direction. Pronounced Shubnikov-de Hass (SdH) quantum oscillations in both longitudinal magnetoresistance and transverse Hall resistance were observed, revealing anisotropic light cyclotron masses and high mobility of the system. In particular, a nontrivial {\pi}-Berry phase in the SdH gives clear evidence for 3D Dirac semimetal phase. Furthermore, we observed clear Landau Level splitting under high magnetic field, suggesting possible splitting of Dirac point into Weyl points due to broken time reversal symmetry. Our results indicate that ZrTe5 is an ideal platform to study 3D massless Dirac and Weyl fermions in a layered compound. • ### Quantum cooling and squeezing of a levitating nanosphere via time-continuous measurements(1503.05603) July 24, 2015 quant-ph With the purpose of controlling the steady state of a dielectric nanosphere levitated within an optical cavity, we study its conditional dynamics under simultaneous sideband cooling and additional time-continuous measurement of either the output cavity mode or the nanosphere's position. We find that the average phonon number, purity and quantum squeezing of the steady-states can all be made more non-classical through the addition of time-continuous measurement. We predict that the continuous monitoring of the system, together with Markovian feedback, allows one to stabilize the dynamics for any value of the laser frequency driving the cavity. By considering state-of-the-art values of the experimental parameters, we prove that one can in principle obtain a non-classical (squeezed) steady-state with an average phonon number $n_{\sf ph}\approx 0.5$.	2020-05-27 01:56:45	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6830369830131531, "perplexity": 1809.3126070335386}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347392057.6/warc/CC-MAIN-20200527013445-20200527043445-00445.warc.gz"}
https://www.physicsforums.com/threads/finding-equations-of-vector-line-perpendicular-to-another-vector-line.250844/	# Finding Equations of Vector Line Perpendicular to another Vector Line lkh1986 ## Homework Statement Line 1's equation is given as r(t) = <t, t, 1-t>. Line 2 passes through the origin and is perpendicular to line 1. Find the equation of line 2. ## The Attempt at a Solution If it is a 2-D case then it would be a lot easier, since m1 x m2 = -1. However, this is a 3-D case. Anyway, here is my attempt. I know that the dot product of two vector lines equals 0 iff they are perpendicular. Line 1, r(t) = <t, t, 1-t> = <0, 0, 1> + t <1, 1, -1> Line 2, s(t) = <0, 0, 0> + u <a, b, c>. So I need to find the value of a, b and c. <1, 1, -1> dot <a, b, c> = 0 a + b - c = 0. Let a = b = 1, so c = 2. So the equation of line 2 is s(t) = <0, 0, 0> + u <1, 1, 2>. The answer and method seems plausible to me. However, I remember seeing somewhere in the book where in order to solve this kind of problem, we need to find the equation of the plane that contains the first line. Then, find the equation of another plane and the intersection of this 2 planes is the equation of the line 2. This is done by using the cross product of 2 normal lines. Thanks. Homework Helper I started to say you were completely correct and then had to stop and think! You made a very lucky (or intelligent) choice of a and b! Yes, two vectors are perpendicular if and only if their dot product is 0 so you are correct that <1,1,2> is perpendicular to <1, 1, -1>. But in order that two lines be perpendicular they must also intersect! Do the lines given by <t, t, 1-t> and <u, u, 2u> intersect? Look at t= u, t= u, 1-t= 2u. Obviously, from the first two equations, t= u and so the third becomes 1-t= 2t so 1= 3t and t= u= 1/3. Yes, the two lines intersect at <1/3, 1/3, 2/3> so your answer is correct. But do you see that two lines may be "skew"? For example, the line given by <0, 0, t> (the z-axis) and the line given by <u+1, u, 0> are perpendicular- their direction vectors are <0, 0, t> and <1, 1, 0>. But they do not intersect. A more general method is to recognize that there is, as you say a whole plane of lines perpendicular to any given line, through a given point. From you dot product, a+ b- c= 0, you chose a= 1, b= 1 and so c= 0. In fact, c= a+b so the vector <a, b, a+b> is perpendicular to <1, 1, -1> for any a,b and the line given by <au, bu, (a+b)u> is perpendicular to <t, t, -t> for all a and b. (Note that a plane is two dimensional. We need the two parameters, a and b, to describe the plane.) Now, which of the lines given by <au, bu, (a+b)u> will intersect <t, t, -t>? Again, we must have au= t, bu= t, and (a+b)u= -t. From the first equation, t= au so the second equation becomes bu= au and we must have b= a for that to be true. In that case, a+b= 2a and the third equation becomes 2au= -au so, again, u= 1/3. The line given by <au, au, 2au>, for any a, will pass through (0,0,0) and be perpendicular to <t, t, -t> (and in particular, b= a= 1). lkh1986 Oh yeah... They must intercept... I miss out that one. Let me try again one more time. lkh1986 In that case, a+b= 2a and the third equation becomes 2au= -au so, again, u= 1/3. Nice explanation, HallsofIvy. Now let me try do solve it and rephrase it in my own understanding. Let the equation of line 2 be <a, b, c>. So, <a, b, c> dot <1, 1, -1> = 0. So we have a + b - c = 0. And then c = a + b. This gives us u<a, b, a+b>. Another vector is t<1, 1, -1>. Since the two lines intercept each other, we also have u<a, b, a+b> = t<1, 1, -1>. We have ua = t, ub = t, u(a+b) = -t ua = ub => u(a-b) = 0. u cannot equal to 0, so a - b = 0 => a = b u(a + b) = -t u(2a) = -ua 2ua + ua = 0 3ua = 0 Oops, since that either u = 0 or a = 0. So, since u can't be 0, we need a = 0, which also implies b = 0. I am getting nothing here. Hm..	2022-09-27 20:11:43	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8202084302902222, "perplexity": 1069.0698882704053}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335058.80/warc/CC-MAIN-20220927194248-20220927224248-00102.warc.gz"}
http://everything.explained.today/Surface_area/	# Surface area explained The surface area of a solid object is a measure of the total area that the surface of the object occupies. The mathematical definition of surface area in the presence of curved surfaces is considerably more involved than the definition of arc length of one-dimensional curves, or of the surface area for polyhedra (i.e., objects with flat polygonal faces), for which the surface area is the sum of the areas of its faces. Smooth surfaces, such as a sphere, are assigned surface area using their representation as parametric surfaces. This definition of surface area is based on methods of infinitesimal calculus and involves partial derivatives and double integration. A general definition of surface area was sought by Henri Lebesgue and Hermann Minkowski at the turn of the twentieth century. Their work led to the development of geometric measure theory, which studies various notions of surface area for irregular objects of any dimension. An important example is the Minkowski content of a surface. ## Definition While the areas of many simple surfaces have been known since antiquity, a rigorous mathematical definition of area requires a great deal of care.This should provide a function S\mapstoA(S) which assigns a positive real number to a certain class of surfaces that satisfies several natural requirements. The most fundamental property of the surface area is its additivity: the area of the whole is the sum of the areas of the parts. More rigorously, if a surface S is a union of finitely many pieces S1, …, Sr which do not overlap except at their boundaries, then A(S)=A(S1)++A(Sr). Surface areas of flat polygonal shapes must agree with their geometrically defined area. Since surface area is a geometric notion, areas of congruent surfaces must be the same and the area must depend only on the shape of the surface, but not on its position and orientation in space. This means that surface area is invariant under the group of Euclidean motions. These properties uniquely characterize surface area for a wide class of geometric surfaces called piecewise smooth. Such surfaces consist of finitely many pieces that can be represented in the parametric form SD:\vec{r}=\vec{r}(u,v),(u,v)\inD with a continuously differentiable function \vec{r}. The area of an individual piece is defined by the formula A(SD)=\iintD\left\|\vec{r}u x \vec{r}v\right\|dudv. Thus the area of SD is obtained by integrating the length of the normal vector \vec{r}u x \vec{r}v to the surface over the appropriate region D in the parametric uv plane. The area of the whole surface is then obtained by adding together the areas of the pieces, using additivity of surface area. The main formula can be specialized to different classes of surfaces, giving, in particular, formulas for areas of graphs z = f(x,y) and surfaces of revolution. One of the subtleties of surface area, as compared to arc length of curves, is that surface area cannot be defined simply as the limit of areas of polyhedral shapes approximating a given smooth surface. It was demonstrated by Hermann Schwarz that already for the cylinder, different choices of approximating flat surfaces can lead to different limiting values of the area; this example is known as the Schwarz lantern.[1] [2] Various approaches to a general definition of surface area were developed in the late nineteenth and the early twentieth century by Henri Lebesgue and Hermann Minkowski. While for piecewise smooth surfaces there is a unique natural notion of surface area, if a surface is very irregular, or rough, then it may not be possible to assign an area to it at all. A typical example is given by a surface with spikes spread throughout in a dense fashion. Many surfaces of this type occur in the study of fractals. Extensions of the notion of area which partially fulfill its function and may be defined even for very badly irregular surfaces are studied in geometric measure theory. A specific example of such an extension is the Minkowski content of the surface. ## Common formulas Surface areas of common solids ShapeEquationVariables Cube 6s2 s = side length Cuboid 2(\ellw+\ellh+wh) = length, w = width, h = height Triangular prism bh+l(a+b+c) b = base length of triangle, h = height of triangle, l = distance between triangular bases, a, b, c = sides of triangle All prisms 2B+Ph B = the area of one base, P = the perimeter of one base, h = height Sphere 4\pir2=\pid2 r = radius of sphere, d = diameter Spherical lune 2r2\theta r = radius of sphere, θ = dihedral angle Torus (2\pir)(2\piR)=4\pi2Rr r = minor radius (radius of the tube), R = major radius (distance from center of tube to center of torus) Closed cylinder 2\pir2+2\pirh=2\pir(r+h) r = radius of the circular base, h = height of the cylinder Lateral surface area of a cone \pir\left(\sqrt{r2+h2}\right)=\pirs s=\sqrt{r2+h2} s = slant height of the cone, r = radius of the circular base, h = height of the cone Full surface area of a cone \pir\left(r+\sqrt{r2+h2}\right)=\pir(r+s) s = slant height of the cone, r = radius of the circular base, h = height of the cone Pyramid B+ PL 2 B = area of base, P = perimeter of base, L = slant height Square pyramid b2+2bs=b2+2b\sqrt{\left( b 2 \right)2+h2} b = base length, s = slant height, h = vertical height Rectangular pyramid lw+l\sqrt{\left( w 2 \right)2+h2}+w\sqrt{\left( l 2 \right)2+h2} = length, w = width, h = height Tetrahedron \sqrt{3}a2 a = side length ### Ratio of surface areas of a sphere and cylinder of the same radius and height The below given formulas can be used to show that the surface area of a sphere and cylinder of the same radius and height are in the ratio 2 : 3, as follows. Let the radius be r and the height be h (which is 2r for the sphere). \begin{array}{rlll} Spheresurfacearea&=4\pir2&&=(2\pir2) x 2\\ Cylindersurfacearea&=2\pir(h+r)&=2\pir(2r+r)&=(2\pir2) x 3 \end{array} The discovery of this ratio is credited to Archimedes.[3] ## In chemistry See also: Accessible surface area. Surface area is important in chemical kinetics. Increasing the surface area of a substance generally increases the rate of a chemical reaction. For example, iron in a fine powder will combust, while in solid blocks it is stable enough to use in structures. For different applications a minimal or maximal surface area may be desired. ## In biology See also: Surface-area-to-volume ratio. The surface area of an organism is important in several considerations, such as regulation of body temperature and digestion. Animals use their teeth to grind food down into smaller particles, increasing the surface area available for digestion. The epithelial tissue lining the digestive tract contains microvilli, greatly increasing the area available for absorption. Elephants have large ears, allowing them to regulate their own body temperature. In other instances, animals will need to minimize surface area; for example, people will fold their arms over their chest when cold to minimize heat loss. The surface area to volume ratio (SA:V) of a cell imposes upper limits on size, as the volume increases much faster than does the surface area, thus limiting the rate at which substances diffuse from the interior across the cell membrane to interstitial spaces or to other cells. Indeed, representing a cell as an idealized sphere of radius r, the volume and surface area are, respectively, V = 4/3 π r3; SA = 4 π r2. The resulting surface area to volume ratio is therefore 3/r. Thus, if a cell has a radius of 1 μm, the SA:V ratio is 3; whereas if the radius of the cell is instead 10 μm, then the SA:V ratio becomes 0.3. With a cell radius of 100, SA:V ratio is 0.03. Thus, the surface area falls off steeply with increasing volume. ## References 1. Web site: Schwarz's Paradox. 2017-03-21. live. https://web.archive.org/web/20160304073957/http://fredrickey.info/hm/CalcNotes/schwarz-paradox.pdf. 2016-03-04. 2. Web site: Archived copy . 2012-07-24 . dead . https://web.archive.org/web/20111215152255/http://mathdl.maa.org/images/upload_library/22/Polya/00494925.di020678.02p0385w.pdf . 2011-12-15 . 3. Web site: Chris. Rorres. Tomb of Archimedes: Sources. Courant Institute of Mathematical Sciences. 2007-01-02. live. https://web.archive.org/web/20061209201723/http://www.math.nyu.edu/~crorres/Archimedes/Tomb/Cicero.html. 2006-12-09.	2020-08-10 05:25:08	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.83937007188797, "perplexity": 641.3822468888976}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439738609.73/warc/CC-MAIN-20200810042140-20200810072140-00336.warc.gz"}
https://stackoverflow.com/questions/44641928/vba-to-remove-rows-from-ms-project-when-work-effort-is-zero	# VBA to Remove Rows from MS Project When Work Effort is Zero I have a master copy of a project plan that contains all of the possible tasks that I may need to complete as a part of project. I want to create a macro that removes any of the lines where Work = 0 hrs once the project plan has been updated by the PM. I am a novice at writing code for MS Project so have't been able to get the below right: Sub DeleteMsProjectTask() Dim proj As Project Dim w As Object Set proj = ActiveProject For Each t In proj If w = 0 Then Selection.EntireRow.Delete = True End If Next t Loop End Sub I'm not sure what I'm doing wrong! Thanks in advance for the help. UPDATED Instead of iterating through the collection, reference the task objects by index. This avoids issues that arise when the collection is modified in the loop (e.g. by removing members). Sub DeleteMsProjectTask() Dim proj As Project Set proj = ActiveProject Dim idx As Integer Do While idx > 0	2019-10-16 18:14:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2796100080013275, "perplexity": 2132.9793484995525}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986669057.0/warc/CC-MAIN-20191016163146-20191016190646-00433.warc.gz"}
https://rankerboss.com/i-want-pmynewk/29565e-vector-formula-i-j-k	# vector formula i j k ## vector formula i j k Misc 5 Find the value of x for which x( ̂ + ̂ + ̂) is a unit vector.Let ⃗ = x( ̂ + ̂ + ̂) So, ⃗ = ̂ + ̂ + ̂ Given, ⃗ is a unit vector Magnitude of ⃗ is 1. b vector = 3i vector − 2j vector + k vector. Vectores en el plano • Los vectores i → = (1, 0) y j → = (0, 1) son vectores unitarios que tienen, respectivamente, la dirección del eje X y el eje Y, y sentido positivo. The i, j, and k fields are multiplied together and then all values are added up to give the total dot product. Long Room, Trinity College, Dublin. p = 3i + j, q = -5i + j. This engineering statics tutorial goes over how to use the i, j, k unit vectors to express any other vector. Vector area of parallelogram = a vector x b vector Using $i,j,$ and $k$ for the standard unit vectors goes back to Hamilton (1805–1865) and his invention of quaternions $\mathbf H$ in the 1840s. As curl or rotation of two vectors give the direction of third vector. Solution : Let a vector = i vector + 2j vector + 3k vector. The dot product of the two vectors which are entered are calculated according to the formula shown above. 3i + j - 5i + j = -2i + 2j. The resultant of this calculation is a scalar. The vector , being the sum of the vectors and , is therefore This formula, which expresses in terms of i, j, k, x, y and z, is called the Cartesian representation of the vector in three dimensions. Then why i x j =k, This is because, i along x axis and y along y axis, thus, angle between them will be 90 degree. This could also have been worked out from a diagram: The Magnitude of a Vector. The vector is z k. We know that = x i + y j. Now, take the vector derivative of A with respect to time. If the vectors are given in unit vector form, you simply add together the i, j and k values. As sin 90 = 1. If using this calculator for a 3D vector, then the user enters in all fields. Example. Find the area of the parallelogram whose two adjacent sides are determined by the vectors i vector + 2j vector + 3k vector and 3i vector − 2j vector + k vector. We call x, y and z the components of along the OX, OY and OZ axes respectively. This gives us Since i, j, k are unit vectors of fixed length we can use the result from the previous section and write As a result, This formula reduces to the formula given in the previous section if A is of fixed magnitude (length), since dA x /dt, dA y /dt, dA z /dt all equal zero. Example 1 Find the general formula for the tangent vector and unit tangent vector to the curve given by $$\vec r\left( t \right) = {t^2}\,\vec i + 2\sin t\,\vec j + 2\cos t\,\vec k$$. The formula k x k =0. The Magnitude of a Vector. Code to add this calci to your website Just copy and paste the below code to your webpage where you want to … • Cualquier vector en el plano lo podemos escribir de la siguiente manera: The magnitude of a vector can be found using Pythagoras's theorem. Since the vectors are given in i, j form, we can easily calculate the resultant. In words, the dot product of i, j or k with itself is always 1, and the dot products of i, j and k with each other are always 0. Find p + q. Coefficients of i, j ,k are added seperately,and the resultant value will also be a vector. The unit vector in the direction of the x-axis is i, the unit vector in the direction of the y-axis is j and the unit vector in the direction of the z-axis is k. 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'S theorem vector = 3i + j components of along the OX, OY and axes., j, k unit vectors to express any other vector OZ axes respectively x y...	2021-08-01 14:46:22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9124418497085571, "perplexity": 782.369926750397}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154214.36/warc/CC-MAIN-20210801123745-20210801153745-00445.warc.gz"}
https://www.physicsforums.com/threads/orbital-and-energy-shell-transitions.655107/	# Orbital and energy shell transitions 1. Nov 26, 2012 ### DiracPool I've always learned that putting energy-photons into an atom can bump an electron up to a "higher energy state", and that when the electron "falls" back down into a lower energy state, it then emits a photon, and so forth. What I never seem to find, however, in these descriptions are any specifics. My question, say for a hydrogen atom, is 1) Is this higher energy state an electron gets bumped up to simply a generally higher energy shell? Say n=1 to n=2 or n=4? Or is there some specific orbital within that higher energy shell the electron prefers, like say the Px orbital in n=2 versus the Pz orbital...or, perhaps one of the D orbitals versus a P orbital if the electron is bumped to even a higher energy level. I guess, more specifically, does bumping up an electron in a hydrogen atom actually CREATE one of the p,d,f,etc. orbitals we know are found in larger atoms? Or, alternatively, is it just bumped into some temporary amorphous higher energy "state?" The second related question is, correspondingly, does the electron have any preferred "path" down to lower energy levels via some orbital hierarchy-cascade? Thanks. 2. Nov 27, 2012 ### tom.stoer I think the answer to your question is the understanding of so-called selection rules. Think about two eigenstates of a system $\|\psi^i\rangle$ and $\|\psi^f\rangle$ with i="initial" and f="final" and some additional interaction represented by an operator $\hat{T}$. In non-rel. QM this could e.g. be an electromagnetic field (instead of a photon b/c we do not want to deal with field quantization which is beyond QM and requires QFT). The transition between the two states induced by the interaction is described by the matrix elements $$\langle\psi^f\|\hat{T}\|\psi^f\rangle$$ Now looking at the hydrogen atom states $\|nlm\rangle$ and using an electromagnetic wave you get the usual selection rules for different multipoles. Neglecting spin, spin-orbit coupling and other tiny effects one can derive (via symmetry considerations), that an electromagnetic wave representing a dipole results in transitions for which the following rules must hold: $$\|n^i\,l^i\,m^i\rangle\;\to\;\|n^f\,l^f\,m^f\rangle$$ is allowed for $$\|l^f - l^i\| = 1$$ $$\|m^f - m^i\| = 0,1$$ The probabilities for these transitions can be calculated from the matrix elements $$\langle n^f\;l\pm 1\;m\pm 1,0\|\hat{T}_\text{dipole}\|n^i\,l\,m\rangle$$ For all other combinations, i.e. for transitions violating the dipole rules the matrix element is exactly zero and the transition is forbidden. http://en.wikipedia.org/wiki/Selection_rule Last edited: Nov 27, 2012 Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook	2017-12-14 10:10:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6714441180229187, "perplexity": 1026.5307069922246}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948543611.44/warc/CC-MAIN-20171214093947-20171214113947-00313.warc.gz"}
https://www.nature.com/articles/s41598-022-15220-8?error=cookies_not_supported&code=2de5ef55-5470-4383-92a2-6c84edcf98db	Thank you for visiting nature.com. You are using a browser version with limited support for CSS. To obtain the best experience, we recommend you use a more up to date browser (or turn off compatibility mode in Internet Explorer). In the meantime, to ensure continued support, we are displaying the site without styles and JavaScript. # Study on the damage characteristics of overburden of mining roof in deeply buried coal seam ## Abstract The study of water-conducting fracture zone development height is key to the scientific prevention and control of water damage in mines. Based on the geological conditions of the Wenjiapo coal mine in Binchang, China, this paper investigates the development of water-conducting fracture zone in overlying bedrock during mining under large buried depth and huge thick aquifer by combining on-site well-location microseismic monitoring and laboratory similar material simulation. To overcome the limitation of the " limited outlook " of water-conducting fracture zone investigation, the spatial development characteristics of roof fissures in coal seam mining were determined by on-site " the underground - ground" combined microseismic monitoring and follow-up monitoring, and the development of overlying rock fracture under the large depth of burial was concluded. The fractures were mainly distributed in the upper part of the protective coal pillar on both sides of the working face, but less in the upper part of the working face, and primarily distributed in the protective coal pillar on the side of the working face and the adjacent mining area. To verify the accuracy of the conclusion, the overlying bedrock movement and deformation characteristics and the development process of the hydraulic fracture zone during coal seam mining were analyzed by simulating similar materials in the laboratory, using the monitored area as a prototype. The results show that the development height of the mining fracture zone obtained from microseismic monitoring is basically consistent with the simulation results of similar materials. The research finding have significant implications for the study of fracture distribution characteristics and the evolution law of mining overburden, and provide a foundation for scientific prevention and control of water damage on the roof. ## Introduction In the process of underground coal resource recovery, the retrieval of the working face disrupts the original stress balance of the surrounding rock, and the redistribution of the rock stress causes deformation and destruction of the surrounding rock1,2,3. Before destroying the rock body, energy is inevitably released, resulting in acoustic emissions and microseismic phenomena4. In the process of back mining at the working face, the rock stress around the mining void area and hydraulic pillar is adjusted, and the surrounding rock is deformed and destroyed5. During the process of deformation and damage, the microseismic produced by the underground rock body is one of the physical effects of the energy release process of the coal rock body, and the location and intensity of the microseismic occurrence reflect the stress state and damage degree of the overlying roof of the coal seam to a certain extent6,7. Therefore, the collected microseismic signals are processed, analyzed and studied in combination with the geological parameters of the roof, which can be used as a decision basis for the development height of the fracture zone of the mining overburden roof8,9. Compared with the traditional empirical formula method, piezometric test, borehole peeping and flushing fluid consumption and unilateral downhole microseismic monitoring, the combined "underground-ground" microseismic monitoring technology improves the positioning accuracy of microseismic data in elevation, and this method is regional monitoring, with an extensive monitoring range, flexible measurement points2,10. The method is a regional monitoring method, with the characteristics of a monitoring range, flexible deployment of measurement points, real-time dynamic monitoring, and a large amount of information about the spatial distribution characteristics of roof fracture development that can be revealed by the results11,12. Compared with surface or roadway microseismic monitoring, "surface-roadway" combined microseismic monitoring is more accurate in spatial positioning and more suitable for the investigation of the water-conduction fracture zone. Compared with the common exploration, "surface-roadway" combined microseismic monitoring can reflect the process of water-conduction fracture zone development in the region, which is more in line with the practical needs of engineering. In addition to in situ detection, physical similarity simulation on the experimental platform through similar materials with similar ratios is also a vital means, and it is widely used in the study of mine pressure and overburden dynamic changes due to its economy, intuitiveness, and ease of operation5,13,14,15,16. Some scholars have used the Ashby diagram to analyze the stability of racks during the coal development process. They proposed new rock parameters to provide a new method for evaluating geological materials17. A physical model test was carried out with the working face passing through a fault. Then, the effect of mining on fault activation slip was studied, the influencing factors of mining height, water pressure, aquiclude thickness and excavation distance and the response of basal stress and fault plane stress in the coal body during fault slip were analyzed18,19. The study revealed that inrush occurred most readily at the open-off cut or mining face. Because there is no accepted formula for calculating the water-conduction fracture zone in the Binchang mining area, the predicted value of the mine water surge deviates from the actual value. Mining overburden fracture zone development height prediction has become one of the impediments to engineering and scientific research because the Binchang mining area has yet to form a unified recognized formula for calculating the water-conduction fracture zone20. After the workface mining, the overlying rock layer in the mining void area produces tension and pressure damage and forms mining fissures and sinks. The release of energy during the destruction of the surrounding rock in the mining fracture zone is the main source of mine pressure21. In the twenty-first century, microseismic monitoring technology is a newly emerged physical monitoring technology that has been successfully applied to early warning in several mine disasters, such as impact ground pressure monitoring, mine water damage control, gas extraction, and coal and gas protrusion in China. This technology can provide the spatial location, time, and energy magnitude of rock rupture, allowing for mining overburden damage monitoring and analysis of mining overburden movement characteristics. To this end, the author carried out a series of studies in theoretical, experimental and water control technologies and conducted a field monitoring study on the fracture development of mining overburden using the YJT3 microseismic monitoring system developed independently by China Coal Science and Industry Group Xi'an Research Institute Co. Based on the on-site microseismic monitoring of the Wenjiapo coal mine in Binchang, the author conducted a comprehensive study on the fissure development characteristics of mining overburden in the study area using similar material simulation experimental data. ## Overview of the study area and its geological conditions ### Description of the study area Wenjiapo coal mine was chosen as the study area, it is located in Shaanxi Province, China, within Latitudes 35°9′18.74″–35°12′8.69″ N, and longitude 108°6′52.40″–108°8′29.42″E (Fig. 1), in a mid-latitude plateau area with a warm temperate semi-arid continental monsoon climate, with four distinct seasons. The study area belongs to Shaanxi Coal Group and is one of the key mining areas in China's 13th Five-Year Plan. The Wenjiapo mine is about 9.86 km wide from east to west and 11.69 km long from north to south, covering an area of 79.6903 km2. The elevation of + 1170~+ 1200 m. ### Geological overview of the study area Wenjiapo coal mine 4103 working face coal seam buried depth 680–690 m average thickness 3.8 m, and average elevation 395 m, dip angle 3°–4°, for nearly horizontal each layer, working face direction length is 2800 m and tendency length 240 m. This monitoring area starts position 282 m from the down-distance cut, transport down-distance cut 290 m, and monitoring push mining distance 520 m. The topography of the monitoring region is typical of a mountainous area. Loess plateau terrain is all covered by the fourth system loess, and the thickness is large, the area is ditch, and beam staggered distribution. The height difference is about 100 m. Due to weathering, denudation and the water scouring effect, the thickness of loess development is extremely uneven, and the terrain gradually rises from north to southeast. In other words, along the direction of back mining in the monitoring region, both the thickness of the loess layer and the depth of the coal seam buried by this influence rise continuously. According to the drill holes (6-5, 7-5) near the 4103 working face and the contour map of stratum thickness, the distance between coal and the main water-bearing/water-insulating layer is made diagonally through the direction of the monitoring area (Fig. 2). The average thickness of the aquifer of Luohe Fm is about 275 m, and its bottom is 205–220 m from the top plate of coal 4. From coal 4 upward, there are mainly Yanan Fm aquifer, bottom aquifer of Zhiluo Fm, Yijun Fm aquifer, and Luohe Fm aquifer, and between Yijun Fm and bottom aquifer of Zhiluo Fm There is a mudstone relative water barrier of the Anding Fm with a thickness of about 50 m. The physical and mechanical parameters of the coal strata are shown in Table 1. ## Microseismic monitoring of roof slab fracture development ### Microseismic monitoring system arrangement The microseismic instrumentation adopts the YJT3 microseismic monitoring system independently developed by China Coal Science and Technology Group Xi'an Research Institute Co., Ltd. to monitor the development of the hydraulic conductivity rift zone in the study area of 4103 working face in real time and to locate and analyze the received microseismic events at the same time. This microseismic monitoring system mainly consists of a geophone, a large wire, a collector, microseismic pre-processing software and microseismic event location software. According to the needs, detectors with different sensitivities are selected. Since the coal seam in the monitoring area is buried more profoundly and the thicker Quaternary loess layer is not conducive to the propagation of seismic waves, a geophone with high low frequency and sensitivity of 4.5 V-s/m is selected this time. According to the geological conditions of the monitoring area, the burial depth of the coal seam and the retrieval situation, eight geophones were arranged within about 800 m of the ground in the monitoring area according to the topography and traffic conditions, and the ground was manually excavated with a 0.5 m pit, and the geophone tail vertebrae were replaced by 1.5 m steel bars to improve geophone coupling with the loess layer (Fig. 3a). The coordinates of each measurement point were accurately measured by RTK, a high-precision measuring instrument. Simultaneously, three geophones are arranged in each of the 4103 working face, transport chute, and 4104 working face return wind chute.The 4103 working face return wind chute and transport chute geophones are 50 m apart, the two lanes are staggered, and the geophone positions are moved according to mining progress. The 4104 working face return wind chute geophones are 100 m apart and are always arranged on the side of 4103 mining area with each lane. The three geophones are connected to the X, Y, and Z axes of a collector with shielded wires, and the coordinates of each measuring point are read using the mining project plan (Fig. 3b). ### Microseismicic positioning principle The elastic wave signals generated by rock rupture are accepted by multiple geophones, and the time difference and wave velocity are used to localize the microseismic events. $$(x_{i} - x_{0} )^{2} + (y_{i} - y_{o} )^{2} + (z_{i} - z_{0} ) = (T_{i} - T_{0} )^{2}$$ (1) where is xi, yi, zi the coordinate position of the i geophone, vi is the wave velocity measured by the i probe, Ti is the time measured by the i geophone, x0, y0, z0 the coordinate of the source and T0 is the moment of rupture. There are four unknowns in the equation, and at least four equations are needed, so at least four geophones are needed to locate the microseismic events, and the more geophones there are, the higher the theoretical accuracy of the microseismic event location. The amplitude of the waveform determines the energy of a microseismic event; the greater the amplitude, the greater the energy, and vice versa. ## Spatial damage characteristics of the roof fracture zone ### Statistical analysis of microseismic event description Wenjiapo coal mine 4103 working face started pushing on June 4, 2018, and the microseismic monitoring period lasted from July 20, 2018 to October 10, 2018, with a monitoring pushing the length of about 520 m. During the monitoring period, two faults F26 and F21, with a fault distance of approximately 6.8 m, were crossed. After excluding the events with a low signal-to-noise ratio and significant positioning errors, a total of 12,778 microseismic events were located during the monitoring period. The location elevation of microseismic events was divided by every 10 m interval, and the statistical analysis of microseismic events was carried out (Fig. 4). In Fig. 4, it can be seen that microseismic events occur in the interval of elevation 390–600 m, mainly concentrated in the elevation interval 410–540 m, and microseismic events basically no longer develop upward after elevation greater than 580 m. A very small number of microseismic events occurred after elevation greater than 600 m, which can be neglected. According to the distribution characteristics of the number of microseismic events in different elevation intervals, it can be roughly divided into three ranges: the growth rate is basically the same between elevations of 390–440 m, and the growth rate is basically stable after elevations of 440–500 m and starts to decrease gradually after 500 m; the number of microseismic events at elevations > 550 m is small, accounting for only 5.97% of the total number. From the beginning to the end of the monitoring data, the distribution characteristics of microseismic events were divided into eight segments with a time unit of weeks, and the interval and cumulative number of microseismic events in different height intervals within different segments were plotted (Fig. 5). During the monitoring period, the distribution characteristics of the weekly microseismic events in each growth phase, maximum, stability, and decline were consistent. From the analysis of Fig. 5, the microseismic events belong to the stage of rapid increase between 400 and 440 m, and the microseismic events reach the peak at elevation 440 m; they basically remain stable during the elevation of 440–500 m, and start to decline sharply between the elevation 500–550 m. According to the growth characteristics of the number of microseismic events at the later stage of monitoring, the number of microseismic events stopped increasing after 520 m. Therefore, the top plate elevation stabilized state at about 520 m, and the rupture signal tended to stop developing upward. The energy level of microseismic events is Joule/J, and an energy size varies from 0 to 3800 J. The energy of microseismic events is counted at intervals of 100 J, and the distribution of the number of microseismic events in different height intervals is shown in Fig. 6, where there are 6249 microseismic events with most of the event energy in the range of 0–100 J, which are small energy events and are distributed in the range of 380 m to 670 m in elevation. The number of microseismic events larger than 1000 J was 445, accounting for about 0.39% of the total number of microseismic events. The number of large energy events accounted for a small percentage and was mainly concentrated between 390 and 470 m, with the maximum value occurring between 390 and 460 m. ### Analysis of spatial distribution characteristics of microseismic events The spatial and temporal distribution characteristics of microseismic events can reflect the location and sequence of rock rupture, and the density of microseismic events (the number of microseismic events projected on a certain plane per unit area.) can indicate the degree of fracture aggregation of rock fragmentation, according to which the process of rock rupture and fracture zone formation can be analyzed. During the monitoring period, the microseismic events are projected onto the XY, XZ, and YX planes (Fig. 7, the legend indicates the elevation of microseismic events). From Fig. 7b, c, it can be seen that the microseismic events are mainly concentrated between the 4103 working face and the 4102 working face mining hollow area, indicating that the 4102 disturbs the adjacent 4103 working face roof plate more during the process of pushing mining. Hence when the 4103 working face is mined, the roof plate rupture is mainly inclined to the 4102 mining hollow area side. A large number of microseismic events are developed, while there are also a large number of microseismic events on the 4103. At the same time, there are also a large number of microseismic events on the protected coal pillar between 4103 and 4104 working faces, but the number is significantly less than that on the side of the 4102 working face mining void area, and less on both sides directly above the working face. During the monitoring process, the elevation of the dense zone of microseismic events increased significantly after pushing the mining through the F21 fault (F2184°ΔH = 6.4 m). From the analysis of the microseismic event projection maps in XZ and YZ planes, the microseismic events are mainly concentrated between elevations of 400–520 m, and the rapid decrease in development above 520 m accounts for 14.47% of the total number of microseismic events. In the + 400~+ 450 m section, the cumulative number of microseismic events accounts for 51.8% of the total number of events, and the cumulative energy of large-energy events accounts for more than 60% of the proportion of each section, so this section is considered to be the development area of the adventitious zone. In the + 450~+ 500 m section, the cumulative energy of each section gradually decreases, and the number of microseismic events and the cumulative energy of large-energy microseismic events account for decreasing progressively, with medium-energy and small-energy microseismic events dominating. The number of microseismic events and the proportion of large-energy microseismic events decreased gradually, and the area was dominated by medium-energy and small-energy microseismic events, and the area was considered to be a rift zone development area. According to the development of dense zones of microseismic events in different periods (Table 2), density cloud maps on different planes (XZ and YZ planes) were drawn. Through the analysis of the cloud map, the development height of the water-conducting rift zone is directly related to the mining area formed by the pushing of the working face: when the pushing distance of the working face is smaller than the width of the working face, the development height of the water-conducting rift zone in the direction of the working face (XZ) is larger than that in the tendency (YZ); when the length of the mining area is formed by the pushing of the working face is larger than the width of the working face, the development height of the water-conducting rift zone in its tendency direction is higher. In addition, the fault zone also greatly influences the development of the water-conduction fracture zone. The rupture dense zone of the XZ face increases 35.5 m, and the rupture signal dense zone of the YZ face increases 32.91 m during the working face over the fault zone compared with the stable period, and the rupture signal dense zone returns to the stable stage when it crosses the fault for about 65 m. The water-conduction fracture zone development height of the XZ directional section is 495.27 m, and the water-conduction fracture zone development height of the YZ inclined section is 495.27 m. The development height of the hydraulic fracture zone of the XZ trending section is 495.27 m, and the development height of the hydraulic fracture zone of the YZ trending section is 512 m (Fig. 8). Based on the theory of mine pressure and rock control, the maximum development elevation of the water-conducting fracture zone in working face 4103 under the influence of the tectonic (no fault) layer is 512 m. The development height of the water-conducting fractured zone is about 117 m, and the fracture mining ratio is 30.79, as determined by analyzing the characteristics of the number, energy, and density of microseismic events. ## High development pattern of the water-conduction fracture zone ### Similar conditions Similar material simulation experiments use materials with certain strength to simulate the actual rock formation, which is a simplification of the actual rock formation. Therefore t here should be a similarity in mechanical strength between the simulated rock formation and the actual rock formation. (1) Geometric similarity ratio: $$a_{l} = \frac{{x_{r} }}{{x_{m} }} = \frac{{y_{r} }}{{y_{m} }} = 200$$ (2) where al is the geometric similarity ratio, xm is the simulated rock strike length, xr is the actual rock stratigraphic strike length, yr is the actual cumulative thickness of the rock formation, and ym is the vertical height of the simulated rock formation. (2) Similarity ratio of capacity and weight $$a_{r} = \frac{{r_{r} }}{{r_{m} }} = 1.6$$ (3) where ar is the similarity ratio of volume to weight, rr is the density of the original rock, rm is the similar material density. (3) Similar ratio of stress and various strengths $$a_{\sigma } = \sigma_{{\sigma_{s} }} = a_{r} \times a_{l} = 100 \times 1.6 = 160$$ (4) where aσ is the strength similarity ratio. (4) Time similarity ratio $$a_{t} = \sqrt {a_{l} } = 10$$ (5) ### Construction of similar material models According to the general characteristics of the stratigraphic structure in the coal-endowed area, river sand is used as aggregate, gypsum as cement and large white powder as filler, and different ratios are used to simulate the soft, medium-hard and hard rock layers in the stratum13,22. White mica flakes were used to simulate the laminated surfaces between each rock layer. Based on the similarity ratio and the physical and mechanical parameters of the simulated coal seams (mainly compressive strength and elastic modulus, supplemented by other parameters), the formulations and ratios of similar materials were selected in combination with the test results of similar material specimens23,24. Due to the different thickness, capacity and proportioning formula of each simulated rock layer, the total weight of materials used for laying each simulated rock layer and the amount of various materials vary. In this experiment, the total weight of each layered material was calculated according to the following formula, and then the amount of various materials was calculated according to the proportioning formula. $$W = k \times l \times d \times h \times \gamma_{m}$$ (6) where W is the total weight of the material; k is the material loss factor: l, d, h are the length, width and thickness of the layer respectively; γm is the density of the layer of similar materials. According to the objective of a similar material simulation experiment, it was determined that the data 7-5 drilling holes was the main one to the stratum of this similar material simulation, based on the complete analysis of geological data from the first mining area and drilling data within the range of 4103 working face Table 2. The device used for a similar material simulation experiment is 2.0 m long, 1.8 m high, and 0.2 m thick, simulating the stratum. Combined with similar conditions, it was determined that this experiment simulates the development process of three zones on the coal seam strike profile with a ratio of 1:200. Due to the limitation of the simulation experiment device, the full stratum structure was not simulated, and the stratum 90 m below the surface was realized by external loading (Fig. 9a; Table 3). To match the actual mining, the left and right boundaries of the model were left with 20 cm coal pillars each, and the length of the mining area was 160 cm, which was excavated in 16 steps with 10 cm per step. To obtain high precision displacement measurement values, a total station and vic-3D imaging technology were used in the simulation experiment to monitor the displacement and strain of the model. From the top of the coal seam to the seventh-row, there are seven rows of monitoring points in the overburden ,and the distance between each row is 15 cm. The first row is 11 cm from the bottom of the coal seam, the second row is 25 cm from the bottom of the coal seam, the third row is 50 cm from the bottom of the coal seam, the fourth row is 76 cm from the bottom of the coal seam,and the fifth row is 109 cm from the bottom of the coal seam, the sixth row is 135 cm from the bottom of the coal seam, the seventh row is 159 cm from the bottom of the coal seam. The seventh row is 159 cm away from the coal seam floor, this is conducted mainly for monitoring and recording the overburden subsidence value after excavation of the coal seam (Fig. 9b). ### Roof overburden fracture development pattern The working face of 4103 is 245 m wide, leaving about a 40 m section of coal pillars. After the working face was mined, the fissure development pattern derived from the experimental photo processing is shown in Fig. 11. After the working face is fully mined, the total length of the collapsed body is 133 cm, the height of the overlying rock upward fissure is 75 cm, the height of the caving zone is 15 cm, the height of the fracture zone is 42.8 cm, and the height of the water-conducting fracture zone is 57.8 cm, which is 115.60 m. It reaches the bottom of the water barrier of Jurassic Middle Formation Anding Formation, which is 30.47 times the mining height. The collapse angle of rock on the open-cut side and the working face side are both 36°, and the fissure angle is 54°. From the analysis of the density map of microseismic events in the strike direction of the working face (Fig. 10), the area with a higher density of microseismic events at the front of the working face forms an angle of 36° with the coal seam floor, which is consistent with the results of the similar material simulation. ### Overlying rock lead deformation analysis In this experiment, deformation was monitored using two methods: total station monitoring and vic-3D imaging (Fig. 11) technology monitoring. Five rows of monitoring points were arranged on the model’s surface for total station monitoring, among which the fifth row of monitoring points was distributed on the upper boundary of the model, mainly for monitoring the surface deformation, and the distance between each point was 10 cm. Vic-3D imaging technology monitoring calibrates the model by applying scatter spots on the surface of the model. The model is continuously photographed during the model development process, and the monitoring system superimposes and calculates the adjacent pictures to draw the model deformation cloud map and contour map to monitor the model deformation. The vic-3D imaging technology was used to monitor the overburden strain after mining the simulated coal seam. Selected major steps to analyze the vertical strain and variation of the overlying bedrock. The strain is the relative deformation value of the material under the action of the external force. The lead direction strain represents the vertical elongation or compression rate, which is a dimensionless number without a unit. Positive values represent elongation of the object under external forces, while negative values represent compression (Fig. 12). When mining at a length of 30 cm, 8 cm from the cut and 3 cm from the coal seam floor, the top plate of the coal seam develops an 18 cm long abscission layer. Because the abscission layer space is too small, there is no response in the strain cloud diagram (Fig. 12a). Advancing distance 60 cm, the coal seam roof is pressed for the first time, and the weighting step distance is 27 cm. The coal seam top plate collapses from the separate layer, collapses 27 cm in length and 3 cm in thickness, 27 cm from the cut, and comes into contact with the coal seam floor to form a 23 cm long cantilever segment. The overburden deformation cloud map shows that the left side is near the fracture zone at the working face, the right side is near the fracture zone at the cut, the lower part is the bottom of the coal seam that has been mined, and the upper part is near the maximum height of the abscission layer space. The strain cloud map pattern is basically consistent with the overburden collapse pattern. At this time, the height of the fall band is 15 cm, the height of the fracture zone is 1.6 cm, and the height of the water-conducting fracture zone is 16.6 cm (Fig. 12b). The accumulated mining length is 100 cm, part of the fissure closure disappeared, the third of incoming periodic pressure, the pressure step is 22.5 cm, the maximum collapse height is 43.8 cm, and the total collapse length is 73.5 cm. A distance of 72.5 cm from the cut forms a cantilevered segment 21.5 cm long. The overlying bedrock deformation cloud map shows that the left side is near the water-conduction fracture zone at the working face, and the right side is near the water-conducting fracture zone at the cut. The lower part is the bottom plate of the coal seam that has been mined, and the upper part is near the maximum height of the deviated seam development. Its strain cloud map pattern is basically consistent with the overlying rock collapse pattern.At this point, the falling band height is 15 cm, the crack band height is 28.8 cm, and the water-conduction fracture zone height is 43.8 cm (Fig. 12c). When the cumulative mining length is 130 cm, a new abscission layer space is developed and part of the fracture closes and disappears, with a total body collapse length of 103.7 cm, a maximum collapse height of 53.4 cm and an inclination of 13°. The left side of the overburden strain cloud is near the fracture zone at the working face, and the right side is near the fracture zone at the cut. The lower part is the bottom plate of the mined coal seam, and the upper part is near the location of off-bed development. The strain cloud pattern is basically consistent with the overburden collapse pattern. At this time, the height of the caving band is 15 cm, the height of the fracture zone is 38.4 cm, and the height of the water-conduction fracture zone is 53.4 cm (Fig. 12d). When the cumulative propulsion length is 160 cm, the previously formed abscission layer space has collapsed. However, several new abscission layer spaces have been formed, a split gap has been closed, and the morphology of the bedrock strain cloud map is consistent with that of the strain cloud map at mining to 130 cm. Finally, the upper bedrock stress cloud map shows that the falling zone height is 15 cm, the crack band height is 42.8 cm and the development height of the water-conduction fracture zone is 57.8 cm, all of which are consistent with the model cover rock collapse pattern. There was only one tension deformation in the local area near the cutting eye, with the maximum tension strain in the lead direction being 0.11 and the maximum compression strain being 0.018, indicating that the maximum tension amplitude was greater than the maximum compression amplitude and that the transition from the compression area to the tension area was gradual. It is a slow and steady process (Fig. 12e). ### Overlying rock subsidence analysis By plotting the converted final subsidence change curve for each row of measurement points (Fig. 13), it is more obvious to see the overburden collapse pattern and the collapse range of each row of measurement points after the end of mining. The closer to the working face, the greater the overburden sinking value, the first to the third row apparently collapse, the fourth row deformation is more apparent, and in the bending deformation more apparent stage, the fifth to the seventh row deformation is not evident. The sinking curve is roughly symmetrical with measurement point No. 6, and the maximum sinking value of measurement point No. 6 in the first row is 1.88 cm, which is 3.76 m. ### Analysis of the height development pattern of the water-conducting fracture zone Through continuous observation and recording of displacement changes in the observation points during the advancing process of the working face, the relationship table was compiled to analyze the advancing distance of the working face and the height of the height of the caving zone, the height of the fracture zone, and the height of the water-conduction fracture zone. In addition, the curve diagram was drawn to analyze the advancing distance of the working face and the height of the caving zone, the height of the height of the caving zone and the height of the water-conduction fracture zone (Fig. 14). The water-conducting fracture zone is composed of the height of the caving zone and fracture zone. As the working face advances, the overlying rock damage height is basically a horizontal straight line, the overlying rock damage degree is not obvious, and has been in the bending deformation stage. When the working face advances to 60 cm, the overburden comes to pressure for the first time and the height of the collapse body is 4.8 cm. When the working face advances to 70 m, the damage height of the overburden rock layer appears rapid growth and water-conduction fracture zone. The height of the riser belt reaches 15 cm, with the coal seam mining, the height of the riser belt no longer increases; when the advancing distance reaches 100 cm, the growth rate of damage height slows down. In addition, 57.8 cm is the maximum water-conducting fracture zone height at an advance of the work surface to 160 cm. The fissure ratio reaches a maximum of 30.42 when the working face is advanced to 160 cm. On the whole, it seems that as the working face advancement distance increases, the development height of the water-conducting fracture zone presents a step-like curve. This is because it is primarily mainly affected by the periodic destruction and deformation of the overlying rock layer. ## Conclusion 1. 1. Compared with the previous method of detecting the water-conducting fracture zone, a microseismic monitoring system can better respond to the spatial pattern and development process of fissure development in the process of coal mining and provide a scientific basis for fine water control work. 2. 2. The microseismic monitoring results show that the roof fracture is mainly distributed on the side of the working face adjacent to the mining area, followed by the side of the adjacent successive working faces, while the fissures directly above the working face have relatively less spatial distribution characteristics. 3. 3. The development process of the hydraulic fracture zone was defined by the dense zone of microseismic events in different periods using the joint " the underground—ground " microseismic monitoring data. The results show that the maximum elevation of water-conducting fracture zone development is 512 m without tectonic influence, which is 35.5 m lower than that of fault influence, and the fracture -to-mining ratio is 30.79 times when the mining height is 3.8 m. 4. 4. The maximum height of water-conducting fracture zone development at the working face obtained from similar materials is 57 cm. comparing this conclusion with the analysis results of microseismic monitoring data, the maximum value of water-conducting fracture zone development derived from both is basically the same. ## Data availability All data generated or analysed during this study are included in this article and its supplementary information files. ## References 1. Cheng, J. L. et al. High precision location of micro-seismic sources in underground coal mine. Chin. J. Geophys. Chin. Ed. 59, 4513–4520. https://doi.org/10.6038/cjg20161214 (2016). 2. Li, Z., Chang, X., Yao, Z. X. & Wang, Y. B. Fracture monitoring and reservoir evaluation by micro-seismic method. Chin. J. Geophys. Chin. Ed. 62, 707–719. https://doi.org/10.6038/cjg2018L0729 (2019). 3. He, X., Zhang, C. & Han, P. H. Overburden damage degree-based optimization of high-intensity mining parameters and engineering practices in China’s Western Mining Area. Geofluids https://doi.org/10.1155/2020/8889663 (2020). 4. Zhang, C. et al. Prediction of rockbursts in a typical island working face of a coal mine through microseismic monitoring technology. Tunnel. Undergr. Space Technol. https://doi.org/10.1016/j.tust.2021.103972 (2021). 5. Sun, Y., Zuo, J., Karakus, M. & Wang, J. Investigation of movement and damage of integral overburden during shallow coal seam mining. Int. J. Rock Mech. Min. Sci. 117, 63–75 (2019). 6. Wang, H. L., Jia, C. Y., Yao, Z. K. & Zhang, G. B. Height measurement of the water-conducting fracture zone based on stress monitoring. Arab. J. Geosci. https://doi.org/10.1007/s12517-021-07757-1 (2021). 7. Wang, F., Xu, J. L., Chen, S. J. & Ren, M. Z. Method to predict the height of the water conducting fractured zone based on bearing structures in the overlying strata. Mine Water Environ. 38, 767–779. https://doi.org/10.1007/s10230-019-00638-w (2019). 8. Ji, Z. Q., Tian, H., Yang, Z. N. A., Liu, T. & Bandara, S. Mechanism of water inrush from coal seam floor based on coupling mechanism of seepage and stress. J. Intell. Fuzzy Syst. 34, 965–974. https://doi.org/10.3233/jifs-169390 (2018). 9. Wu, Q., Shen, J. J., Liu, W. T. & Wang, Y. A RBFNN-based method for the prediction of the developed height of a water-conductive fractured zone for fully mechanized mining with sublevel caving. Arab. J. Geosci. https://doi.org/10.1007/s12517-017-2959-3 (2017). 10. Qiao, S. Q., Zhang, Q. S. & Zhang, Q. M. Mine fracturing monitoring analysis based on high-precision distributed wireless microseismic acquisition station. IEEE Access 7, 147215–147223. https://doi.org/10.1109/access.2019.2946443 (2019). 11. Ning, J. G., Wang, J., Tan, Y. L. & Xu, Q. Mechanical mechanism of overlying strata breaking and development of fractured zone during close-distance coal seam group mining. Int. J. Min. Sci. Technol. 30, 207–215. https://doi.org/10.1016/j.ijmst.2019.03.001 (2020). 12. Wang, S. R., Li, C. Y., Yan, W. F., Zou, Z. S. & Chen, W. X. Multiple indicators prediction method of rock burst based on microseismic monitoring technology. Arab. J. Geosci. https://doi.org/10.1007/s12517-017-2946-8 (2017). 13. Sun, Q., Meng, G. H., Sun, K. & Zhang, J. X. Physical simulation experiment on prevention and control of water inrush disaster by backfilling mining under aquifer. Environ. Earth Sci. https://doi.org/10.1007/s12665-020-09174-1 (2020). 14. Zhang, J. W., Wang, Z. W. & Song, Z. X. Numerical study on movement of dynamic strata in combined open-pit and underground mining based on similar material simulation experiment. Arab. J. Geosci. https://doi.org/10.1007/s12517-020-05766-0 (2020). 15. Sun, Y. et al. A new theoretical method to predict strata movement and surface subsidence due to inclined coal seam mining. Rock Mech. Rock Eng. 54, 2723–2740 (2021). 16. Yuan, C.-F., Yuan, Z.-J., Wang, Y.-T. & Li, C.-M. Analysis of the diffusion process of mining overburden separation strata based on the digital speckle correlation coefficient field. Int. J. Rock Mech. Min. Sci. 119, 13–21 (2019). 17. Sun, F. & Zhao, Y.-P. Geomaterials evaluation: A new application of ashby plots. Materials 13, 2517 (2020). 18. Wang, P., Jiang, L., Li, X., Qin, G. & Wang, E. Physical simulation of mining effect caused by a fault tectonic. Arab. J. Geosci. 11, 741. https://doi.org/10.1007/s12517-018-4088-z (2018). 19. Chen, X., Lyu, P., Song, W. & Deng, C. Analysis and control technology of danger of rock burst when fully mechanized caving passing through fault. China Saf. Sci. J. 26, 81–87 (2016). 20. Zha, H., Liu, W. Q. & Liu, Q. H. Physical simulation of the water-conducting fracture zone of weak roofs in shallow seam mining based on a self-designed hydromechanical coupling experiment system. Geofluids https://doi.org/10.1155/2020/2586349 (2020). 21. Zhang, Y., Cao, S. G., Wan, T. & Wang, J. J. Field measurement and mechanical analysis of height of the water flowing fracture zone in short-wall block backfill mining beneath the aquifer: A case study in China. Geofluids https://doi.org/10.1155/2018/7873682 (2018). 22. Yang, T. & Zhang, J. Experimental research on simulation material for water-resisting soil layer in mining physical simulation. Adv. Mater. Sci. Eng. https://doi.org/10.1155/2020/3456913 (2020). 23. Chen, S. J. et al. Similar materials of colliery filling for physical simulation experiment. Mater. Res. Innov. 19, 304–307. https://doi.org/10.1179/1432891714z.0000000001098 (2015). 24. Zhao, J., Chen, J., Zhan, X., Ning, J. & Zhang, Y. Distribution characteristics of floor pore water pressure based on similarity simulation experiments. Bull. Eng. Geol. Environ. 79, 4805–4816. https://doi.org/10.1007/s10064-020-01835-6 (2020). ## Acknowledgements This research was supported by the National Natural Science Foundation of China (No. 42177174), the Basic Research Program of Natural Science of Shaanxi Province (2020ZY-JC-03), and the Shannxi Province Joint Fund Project (2021JLM-09),major projects supported by Shaanxi Coal Chemical Group Co.,Ltd. (2020SMHKJ-C-52). ## Funding This study was sponsored by the National Natural Science Foundation of China (No. 42177174), the Basic Research Program of Natural Science of Shaanxi Province (2020ZY-JC-03) and the Shaanxi Province Joint Fund Project (2021JLM-09), major projects supported by Shaanxi Coal Chemical Group Co.,Ltd., 2018SMHKJ-C-52. ## Author information Authors ### Contributions H.E. proposed the main idea and scheme of the experiment. L.T. was in charge of the experiment, and both authors wrote the paper. The remaining authors examined the accuracy of the article and suggested modifications. All authors reviewed the manuscript. ### Corresponding author Correspondence to Tianwen Long. ## Ethics declarations ### Competing interests The authors declare no competing interests. ### Publisher's note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. ## Rights and permissions Reprints and Permissions Long, T., Hou, E., Xie, X. et al. Study on the damage characteristics of overburden of mining roof in deeply buried coal seam. Sci Rep 12, 11141 (2022). https://doi.org/10.1038/s41598-022-15220-8 • Accepted: • Published: • DOI: https://doi.org/10.1038/s41598-022-15220-8	2022-08-11 22:05:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.49035048484802246, "perplexity": 3214.538576260565}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571502.25/warc/CC-MAIN-20220811194507-20220811224507-00160.warc.gz"}
http://www.physicsforums.com/showthread.php?t=448423	# what happens with the resistance of a wire with twice the diameter by koat Tags: diameter, resistance, wire P: 40 hi everybody Question: a length of uniform wire has a R of 2 ohms. calculate the R of a wire of the same metal and original length but twice the diameter. I thought as the wire gets twice as thick the R must halve to 1ohms. But the solution is 0.5 ohms. what did i do wrong? thanks in advance Mentor P: 5,194 Can you list for me the parameters that resistance depends upon? (Hint: it does not depend directly on the diameter.) Which one of these parameters is relevant here? P: 40 is it that the R decreases when the A increases? Mentor P: 5,194 ## what happens with the resistance of a wire with twice the diameter Yes, cross-sectional area is one of the parameters on which resistance depends. (I had asked you to list all of the parameters, but this one is the relevant one here). So, what does changing the diameter do to the area? What does that resulting change in area do to the resistance? P: 40 is it that when you double the diameter you double the A? P: 40 as the area increases the R decrease Mentor P: 5,194 Quote by koat is it that when you double the diameter you double the A? This is false. What is the area of a circle in terms of its diameter (or radius)? P: 40 r ^2 times pi Mentor P: 5,194 Quote by koat r ^2 times pi Right. And since the diameter is just twice the radius, what this means is that the area depends on the SQUARE of the diameter. So, if you double the diameter, you don't double the area. If you double the diameter, the area increases by a factor of _____? P: 40 sorry i don't get what you mean with the A depends on the square of diameter. can you show me an example with numbers please so that i can visualise this problem? Mentor P: 5,194 Alright look. Let's call the original diameter d1. Now: $$A_1 = \pi r_1^2$$ But the radius is just half the diameter, so: $$r_1 = d_1 / 2$$ Agreed? Therefore: $$A_1 = \pi \left(\frac{d_1}{2}\right)^2 = \pi \frac{d_1^2}{4}$$ This is how the area of a circle depends upon its diameter. As you can see, area is equal to a constant times the diameter squared. Now, what happens if we change the diameter by doubling it? Let's call the new diameter d2 so that: $$d_2 = 2d_1$$ We can plug this new diameter into the formula for the area in order to find the new area: $$A_2 = \pi \frac{d_2^2}{4} = \pi \frac{(2d_1)^2}{4} = 4 \pi \frac{d_1^2}{4} = 4A_1$$ So we have the result that A2 = 4A1. After doubling the diameter, the new area is equal to FOUR times the original area. This is because the area of a circle is proportional to its diameter SQUARED. So if you double the diameter, you quadruple the area. If you triple the diameter, you increase the area by a factor of nine. If you quadruple the diameter, the area increases by a factor of 16. Now do you understand? P: 40 wow I'm impressed with that. Yes I understand now. Thanks a lot for your answer :) Related Discussions Introductory Physics Homework 1 Advanced Physics Homework 2 Introductory Physics Homework 1 Introductory Physics Homework 4 Introductory Physics Homework 2	2013-12-13 15:16:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6449617743492126, "perplexity": 447.9395469429019}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386164950517/warc/CC-MAIN-20131204134910-00097-ip-10-33-133-15.ec2.internal.warc.gz"}
https://math.stackexchange.com/questions/568145/prove-that-the-cardinality-of-the-reals-and-all-binary-funcions-is-not-equal	# Prove that the cardinality of the reals and all binary funcions is not equal Let $S$ be the the set of all real functions that bring back only two values: 0 and 1 (Binary functions). If $f\in S$ then $f:\mathbb{R}\rightarrow \left\{0,1\right\}$. Prove that $\|\mathbb{R}\| \neq \|S\|$. I tried to start with a proof by contradiction that there's no one to to correspondence but I got stuck. I also assume that by the end of the proof we show that $\|S\|=\aleph_0 \ne C=\|\mathbb{R}\|$. • No, $\|S\|>\|\Bbb R\|$. Do you know the theorem that for any set $X$, $\|X\|<\|\wp(X)\|$? – Brian M. Scott Nov 15 '13 at 13:44 • Um no, I never saw that symbol you used either "\wp" I mean. – GinKin Nov 15 '13 at 13:49 • It's a symbol for the power set of $X$, the set consisting of all subsets of $X$. Are you familiar with the theorem @BrianM.Scott referenced? – Jonathan Y. Nov 15 '13 at 13:52 • I see and no we haven't covered that theorem I guess. How can you apply the power of a set here ? – GinKin Nov 15 '13 at 13:55 If you’ve not seen Cantor’s theorem before, this is a fairly hard problem. Suppose that $\varphi:\Bbb R\to S$ is an injection (one-to-one function). I claim that $\varphi$ cannot be a surjection (onto function). If true, this means that there is no bijection from $\Bbb R$ to $S$ and hence that $\|\Bbb R\|\ne\|S\|$. For each $r\in\Bbb R$ let $f_r=\varphi(r)$; $f_r$ is a function from $\Bbb R$ to $\{0,1\}$. To show that $\varphi$ is not surjective, we’ll find a function $g:\Bbb R\to\{0,1\}$ that is different from $f_r$ for every $r\in\Bbb R$. Or rather, I’ll tell you how to construct it and let you finish the details, including the verification that $g\ne f_r$ for every $r\in\Bbb R$. This will show that $g$ is not in the range of $\varphi$ and hence that $\varphi$ is not a surjection. The idea is simple but powerful: for each $r\in\Bbb R$ choose $g(r)\in\{0,1\}$ so that $g(r)\ne f_r(r)$. That doesn’t give you much choice, since $\{0,1\}$ has only two elements; in fact, it completely defines the function $g$. • I'm not sure how write the verification that $g\ne f_r$ you mentioned but other than that I think I understood. Basically we show that there's no bijection between the two sets therefore their cardinality cannot be equal. Thank you. – GinKin Nov 15 '13 at 15:04 • @GinKin: Just note that for each $r\in\Bbb R$ we have by definition $g(r)\ne f_r(r)$, and therefore $g\ne f_r$: to functions that disagree at some point of their common domain, in this case the point $r$, cannot be the same function. (You can even say that $g(r)=1-f_r(r)$.) – Brian M. Scott Nov 15 '13 at 15:07 HINT: Show that $S$ and $\mathcal P(\Bbb R)$ have the same cardinality, and use Cantor's theorem. (Or use the proof of Cantor's theorem directly on $S$). • I have no idea how to show the $S$ and $\mathcal P(\Bbb R)$ have the same cardinality, nor how you had the intuition to do it. – GinKin Nov 15 '13 at 14:17 • @GinKin, it's worth spending time on that concept even after BrianM.Scott's terrific answer. There's a natural isomorphism that identifies a function $f:S\to\{0,1\}$ with that subset of $S$ which is given by $A_f = \{s\in S\mid f(s)=1\}$. Where either is known, the other is readily given. – Jonathan Y. Nov 15 '13 at 20:11	2019-08-21 07:11:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9150805473327637, "perplexity": 135.99318893356875}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027315811.47/warc/CC-MAIN-20190821065413-20190821091413-00208.warc.gz"}
https://learn.careers360.com/ncert/question-calculate-the-number-of-neutrons-present-in-the-nucleus-of-an-element-x-which-is-represented-15-31-x/	# Calculate the number of neutrons present in the nucleus of an element X which is represented $_{15}^{31}\textrm{X}$ Mass number (A) = No. of protons (Z) + No. of neutrons But the mass number is given as 31 and the number of protons is 15. No. of protons (Z) + No. of neutrons = 31 Number of neutrons = 31 – number of protons Number of neutrons = 31 - 15 = 16 16 neutrons are present in the nucleus of the element X. ### Preparation Products ##### NEET Foundation 2022 NEET Foundation 2022. ₹ 14999/- ##### Biology Foundation for Class 10 Biology Foundation for Class 10. ₹ 999/- ₹ 499/- ##### Knockout JEE Main April 2021 (One Month) Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,. ₹ 14000/- ₹ 4999/- ##### Knockout NEET May 2021 (One Month) Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,. ₹ 14000/- ₹ 4999/-	2021-03-05 05:03:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2544221580028534, "perplexity": 12490.508056090168}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178369721.76/warc/CC-MAIN-20210305030131-20210305060131-00058.warc.gz"}
https://www.scienceforums.net/topic/14088-limits/?tab=comments	Limits Recommended Posts What does the limit indicate for? Like lim x=pi/2. Does it mean only x value less or equal to pi/2? Also, what does Devrivates mean? Give me examples won't hurt. Thanks Share on other sites If you know nothing about these concepts then a forum is not the best place to get help, try doing a google search or use these links: Wikipedia Limit (mathematics) Derivative Share on other sites HUH? Your links are no big help to me. They are not even working! Plus, it will only direct me to a crap dictionary. What does the limit indicate for? Like lim x=pi/2. Does it mean only x value less or equal to pi/2? Also, what does Devrivates mean? Give me examples won't hurt. Thanks Share on other sites As far as i understand a limit, for example $\lim_{x \rightarrow 2}f(x) = 5$ means that if you take sufficiently large values close to 2, f(x) approaches 5. This could be thought of graphically. Imagine a straight line through the origin in the positive direction, as you increase the value of x closer to 2, the value on the y axis becomes closer to 5. There are many websites and tutorials which explain limits which i'm sure would be unearthed by a quick google search. Share on other sites Limit... the limit on a function f(x) as x approaches a certain value... take x^2 for example... as x goes to zero, the values of the function get closer and closer to 0 from both sides, so [imath]\lim_{x \to 0}x^2=0[/imath]. Now, you might be saying, well it's just the value of the function at that given point, right? well, almost. But that's more complex. A derivative is quite simply the equation for the slope of the tangent line of a function. (denoted f'(x)). So, if f© and f'© both have values, then f'© is the value of the slope of the tangent line to the equation f(x) at the point x=c. Does that help? (If you don't get it, you'll just have to wait till you take Calculus...) Share on other sites HUH? Your links are no big help to me. They are not even working! Plus' date=' it will only direct me to a crap dictionary. What does the limit indicate for? Like lim x=pi/2. Does it mean only x value less or equal to pi/2? Also, what does Devrivates mean? Give me examples won't hurt. Thanks[/quote'] wow Share on other sites Thanks Dr Finlay and BobbyJoeCool. I understand the limit part now, but not the derivative part. I am taking advanced calculus next semester, but I'm very curious about them. I guess I'll do some googling. Share on other sites Here is the brink of my new reading on "derivatives". The defintion of derivative is "The derivative tells us how to approximate a graph, near some base point, by a straight line". Mainly it is related with the tangent line on graphs. I need a review of tangent lines!! Tangent line is between 2 points on a curve slope, get $x^2$ for instance. I forgot how to find the single middle point on a tangent by some kind of formula? Share on other sites My teacher gave a great image for understadning limits and derivatives. Imagine you're a runner and there is someone taking a picture of you about to run over the finish line. they keep taking pictures as you run up to the finish line, and keep taking pictures up to the very instant you are about the cross the line... but the camera runs out of film when you actually cross the line. If you take a look at the pictures at another time, you can see that you are just about to cross the line. You have no proof that you actually did, because there is no picture of it happening but, in the picture you do have you come so close that you can say, with relative certainty, that it you crossed the line. It is possible, however, that you didn't cross the line, maybe you sponteneously combusted first, or accidently stepped into a wormhole... but because you have no proof you did cross the line. This is a limit. As you come infinately close to running over the finish line, you can assume that you do so. How is this represented on a graph, you ask? Imagine you have a curve, doesn't matter which one. Now, as you move along this curve, there is a hole in the line. But, if you approach the hole, but don't come to it (take pictures of the runner, but not crossing over the finish line) you would never know that there is no hole. So the limit as you approah the whole would be the place where the graph would be if there was no whole. On to derivatives... how are limits related to derivatives? First, you know that in order to measure the slope of a line, you need at least two points on that line. So let's say that you have a curve and you know two points on that curve. You can find the slope (change in the y values divided by the change in the x values) of the secant line of the curve... But what if you only had one point on that curve and you wanted to know the slope of the tangent line (the slope of a straight line that touches just one point on that curve What you would need to do, is find another point on the curve that you know, and you can find out the slope of the secant line, then bring the other point closer and closer to the point you are given (sound familiar?) Basically, you bring the second point infinately close to the first one, or you approach it, as in take the limit. Eventually, your two points are so close, that they may as well be considered one point, and your secant line becomes you tangent line. In other words, you take limit as the second point apporaches infinately close to the first one. In this way you can find out the slope of your tangent line, when given only one point... AKA a derivative. Hope this helped. If it didn't please let me know, and I'll try to clarify further. Share on other sites Ecoli's post was extremely thorough here, I wish there was a karma system in place for posts like that. Its a primer! Share on other sites thanks cosine... (but please remember, you taught me some of my first introduction stuff into calculus) Share on other sites I like the explanation of a limit. One of those golden examples i won't forget, which is useful. Share on other sites Which point a or x on the graph would approach one other? How do you know? Because at the "a" point, the slope can be significally different than at the "x" point. At "a" the slope would be negative and at the "x" the slope would be zero or teeny positive. ??? Share on other sites Which point a or x on the graph would approach one other? How do you know? Because at the "a" point, the slope can be significally different than at the "x" point. At "a" the slope would be negative and at the "x" the slope would be zero or teeny positive. ??? Because, its a continous function, so there's a patern. EDIT: (c is the distance between x and a) look at that graph... the point (x,f(x)) is one point on the secant line right? and the other is (x+c,f(x+c)) right? and the derivative is the slope of the tangent line right? so, you move the point (x+c,f(x+c)) closer and closer to (x,f(x)) (in other words you make c smaller and smaller until it's zero)... the whole time, the slope of the secant line will be [imath]\frac{f(x+c)-f(x)}{c}[/imath]... being [imath]\frac{\Delta y}{\Delta x}[/imath], right? (please note the c IS the change in x). So, what happens to the graph as you make c closer to 0? The secant line moves closer to being the tangent line, right? and when c is 0? That's right! It's the Tangent line!!! The problem however, is that now that you have a tangent line, the equation for slope has become [imath]\frac{f(x)-f(x)}{0}[/imath], which is 0/0. But, we get around this, because as long as c does not equal 0, the equation for the slope is defined. so we need to find the limit on the function as c goes to zero.... So the [imath]\lim_{c \to 0}\frac{f(x+c)-f(x)}{c}=f'(x)[/imath] where f'(x) is equal to the derivitive, which is the slope of the tangent line. In order to get the answer you need to get the c out of the denominator (which can be difficult), but this is the definition of a derivative, and any proof involving derivatives will involve this somehow... Any better? Share on other sites Which point a or x on the graph would approach one other? How do you know? Because at the "a" point, the slope can be significally different than at the "x" point. At "a" the slope would be negative and at the "x" the slope would be zero or teeny positive. ??? In this graph x is being moved to a... but it doesn't really make a difference, as long as you can understand it conceptually. If you take a limit as x appraches a, the slope will get greater and c - the distance between the two (or more commonly known as "h"), will decrease. If you take a limit as c approaches 0, its the same as taking a limit when you move point x to point c. Can you see that? If you know how to calculate the slope, then you get the definition of a derivative. The change in the y value (represented by f(x) {the formula for calulating the curve} plus "c" subtracted by that same function) divided by the change in the x value, which is just c. As C gets closer and closer to zero, the slope is equal to the tangent line slope. Understand? Well, I have a dentist apointment now, so I'll explain more later. Share on other sites Ok... so let's say you were given the equation [imath] y = x^2 [/imath]. You know the formula for the derivative is just a modified equation for the slope of a straight line, when 2 points are known... which is f(x + c) - f(x)/ c... so just substitute your equation. f(x) would equal [imath] x^2 [/imath] and f(x+c) would equal [imath] (x+c)^2 [/imath] and then take the limit (lim) as c approaches zero mathematically, this would be represented by $\lim_{c \to 0}\frac{(x + c)^2 - x^2}{c}=$ you then FOIL the top term to get $\lim_{c \to 0}\frac{(x^2 +2xc + c^2) - x^2}{c}=$ the [imath]x^2[/imath] cancel out and you have $\lim_{c \to 0}\frac{ 2xc + c^2 }{c}=$ which equals $\lim_{c \to 0}\frac{ c(2x + c) }{c}=$ that c can cancel out with the one from the bottom, leaving you with $\lim_{c \to 0}\ 2x + c=$ now it's time to apply the limit. As the variable c approaches zero, the c term in the problem approaches zero... giving you [imath] 2x + 0 [/imath] or simply 2x. This is the equation for your derivative... the equation for the slope of the tangent line to your curve. (or more acurately g(x) = 2x, when g(x) is equal to the derivative of f(x). did that do anything for you? Share on other sites I understand a bit more now. Thanks for trying to put in effort to explain it to me. Share on other sites no problem... that's what this forum is all about. Share on other sites I figured since this is a thread on limits, perhaps a little something should be said about the rigorous definition of limits. $lim_{x-->a} f(x) = L$ Means that for any $e > 0$ there exists a $d > 0$ such that $\|f(x) - L\| < e$ and $\|x - a\| < d$ Share on other sites I'm not too sure about that "and". Personally I think it should be a "whenever", because then that statement is equivalent to: $\forall \epsilon > 0 \, \exists \, \delta > 0 \text{ such that } 0 < \|x-a\| < \delta \Rightarrow \|f(x) - L\| < \epsilon$. However, this is pretty much completely beyond the scope of this thread, so if you don't understand it, don't worry about it Share on other sites ha holy crap, whats with all the greek letters :-s lol, but yah, derivatives are pretty easy when you get what they mean. When you first learn them you have to use the limit definition to get them (which is a pain), eventually though you learn the shortcut: $f(x) = ax^n$ $f'(x) = (an)x^{n-1}$ So say you want the derivative of 2x^3. Times the multiplier (2) by 3, take away one from the power and voila! 6x^2 is your derivative. It'd take half a page of work to do it using the limit definition Share on other sites ha holy crap' date=' whats with all the greek letters :-s lol, but yah, derivatives are pretty easy when you get what they mean. When you first learn them you have to use the limit definition to get them (which is a pain), eventually though you learn the shortcut: [math'] f(x) = ax^n[/math] $f'(x) = (an)x^{n-1}$ So say you want the derivative of 2x^3. Times the multiplier (2) by 3, take away one from the power and voila! 6x^2 is your derivative. It'd take half a page of work to do it using the limit definition or, more generally, you could say $y=av^n$(where v is another function or combination of them ex. sinx or x^2+x)$\frac{d}{dx}y=nav^{n-1}{\frac{d}{dx}}v$ Share on other sites more generally and more complicated lol Share on other sites more generally and more complicated lol not more complicated....more fun! there are a lot of little rules like the one CanadaAotS gave that are just special cases of more general rules. Create an account Register a new account	2020-04-09 17:19:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7296009659767151, "perplexity": 559.3804188061102}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371861991.79/warc/CC-MAIN-20200409154025-20200409184525-00341.warc.gz"}
https://scikit-image.org/docs/stable/auto_examples/applications/plot_morphology.html	# Morphological Filtering¶ Morphological image processing is a collection of non-linear operations related to the shape or morphology of features in an image, such as boundaries, skeletons, etc. In any given technique, we probe an image with a small shape or template called a structuring element, which defines the region of interest or neighborhood around a pixel. In this document we outline the following basic morphological operations: 1. Erosion 2. Dilation 3. Opening 4. Closing 5. White Tophat 6. Black Tophat 7. Skeletonize 8. Convex Hull To get started, let’s load an image using io.imread. Note that morphology functions only work on gray-scale or binary images, so we set as_gray=True. import matplotlib.pyplot as plt from skimage import data from skimage.util import img_as_ubyte from skimage import io orig_phantom = img_as_ubyte(data.shepp_logan_phantom()) fig, ax = plt.subplots() ax.imshow(orig_phantom, cmap=plt.cm.gray) Out: <matplotlib.image.AxesImage object at 0x7fcf37bcad60> Let’s also define a convenience function for plotting comparisons: def plot_comparison(original, filtered, filter_name): fig, (ax1, ax2) = plt.subplots(ncols=2, figsize=(8, 4), sharex=True, sharey=True) ax1.imshow(original, cmap=plt.cm.gray) ax1.set_title('original') ax1.axis('off') ax2.imshow(filtered, cmap=plt.cm.gray) ax2.set_title(filter_name) ax2.axis('off') ## Erosion¶ Morphological erosion sets a pixel at (i, j) to the minimum over all pixels in the neighborhood centered at (i, j). The structuring element, selem, passed to erosion is a boolean array that describes this neighborhood. Below, we use disk to create a circular structuring element, which we use for most of the following examples. from skimage.morphology import erosion, dilation, opening, closing, white_tophat from skimage.morphology import black_tophat, skeletonize, convex_hull_image from skimage.morphology import disk selem = disk(6) eroded = erosion(orig_phantom, selem) plot_comparison(orig_phantom, eroded, 'erosion') Notice how the white boundary of the image disappears or gets eroded as we increase the size of the disk. Also notice the increase in size of the two black ellipses in the center and the disappearance of the 3 light grey patches in the lower part of the image. ## Dilation¶ Morphological dilation sets a pixel at (i, j) to the maximum over all pixels in the neighborhood centered at (i, j). Dilation enlarges bright regions and shrinks dark regions. dilated = dilation(orig_phantom, selem) plot_comparison(orig_phantom, dilated, 'dilation') Notice how the white boundary of the image thickens, or gets dilated, as we increase the size of the disk. Also notice the decrease in size of the two black ellipses in the centre, and the thickening of the light grey circle in the center and the 3 patches in the lower part of the image. ## Opening¶ Morphological opening on an image is defined as an erosion followed by a dilation. Opening can remove small bright spots (i.e. “salt”) and connect small dark cracks. opened = opening(orig_phantom, selem) plot_comparison(orig_phantom, opened, 'opening') Since opening an image starts with an erosion operation, light regions that are smaller than the structuring element are removed. The dilation operation that follows ensures that light regions that are larger than the structuring element retain their original size. Notice how the light and dark shapes in the center their original thickness but the 3 lighter patches in the bottom get completely eroded. The size dependence is highlighted by the outer white ring: The parts of the ring thinner than the structuring element were completely erased, while the thicker region at the top retains its original thickness. ## Closing¶ Morphological closing on an image is defined as a dilation followed by an erosion. Closing can remove small dark spots (i.e. “pepper”) and connect small bright cracks. To illustrate this more clearly, let’s add a small crack to the white border: phantom = orig_phantom.copy() phantom[10:30, 200:210] = 0 closed = closing(phantom, selem) plot_comparison(phantom, closed, 'closing') Since closing an image starts with an dilation operation, dark regions that are smaller than the structuring element are removed. The dilation operation that follows ensures that dark regions that are larger than the structuring element retain their original size. Notice how the white ellipses at the bottom get connected because of dilation, but other dark region retain their original sizes. Also notice how the crack we added is mostly removed. ## White tophat¶ The white_tophat of an image is defined as the image minus its morphological opening. This operation returns the bright spots of the image that are smaller than the structuring element. To make things interesting, we’ll add bright and dark spots to the image: phantom = orig_phantom.copy() phantom[340:350, 200:210] = 255 phantom[100:110, 200:210] = 0 w_tophat = white_tophat(phantom, selem) plot_comparison(phantom, w_tophat, 'white tophat') As you can see, the 10-pixel wide white square is highlighted since it is smaller than the structuring element. Also, the thin, white edges around most of the ellipse are retained because they’re smaller than the structuring element, but the thicker region at the top disappears. ## Black tophat¶ The black_tophat of an image is defined as its morphological closing minus the original image. This operation returns the dark spots of the image that are smaller than the structuring element. b_tophat = black_tophat(phantom, selem) plot_comparison(phantom, b_tophat, 'black tophat') As you can see, the 10-pixel wide black square is highlighted since it is smaller than the structuring element. Duality As you should have noticed, many of these operations are simply the reverse of another operation. This duality can be summarized as follows: 1. Erosion <-> Dilation 2. Opening <-> Closing 3. White tophat <-> Black tophat ## Skeletonize¶ Thinning is used to reduce each connected component in a binary image to a single-pixel wide skeleton. It is important to note that this is performed on binary images only. horse = data.horse() sk = skeletonize(horse == 0) plot_comparison(horse, sk, 'skeletonize') As the name suggests, this technique is used to thin the image to 1-pixel wide skeleton by applying thinning successively. ## Convex hull¶ The convex_hull_image is the set of pixels included in the smallest convex polygon that surround all white pixels in the input image. Again note that this is also performed on binary images. hull1 = convex_hull_image(horse == 0) plot_comparison(horse, hull1, 'convex hull') As the figure illustrates, convex_hull_image gives the smallest polygon which covers the white or True completely in the image. If we add a small grain to the image, we can see how the convex hull adapts to enclose that grain: import numpy as np plt.show()	2021-05-13 06:32:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.49048975110054016, "perplexity": 2914.705176427084}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991537.32/warc/CC-MAIN-20210513045934-20210513075934-00631.warc.gz"}
https://www.mcrl2.org/web/user_manual/tools/release/lps2pbes.html	# lps2pbes¶ The tool lps2pbes reads a modal formula as well as a linear process and generates a parameterised boolean equation system (PBES) of which the solution of the initial variable indicates whether the formula is valid in the initial state of the transition system. The generated PBES can be solved using tools such as pbes2bool, pbessolve or pbespgsolve. When using particular formulas, for instance such as: [true]<a.b.c.d.e.f.g.h>true then the standard translation to PBESs can yield a very PBES which is very elaborate to generate and can become large. This is due to the fact that the subformula <a.b.c.d.e.f.g.h> is translated into one PBES equation with a huge right hand side. This right hand side essentially reflects for any state of the linear process whether a trace a.b.c.d.e.f.g.h is possible. When using the flag --preprocess-modal-operators the formula is first transformed into the equivalent formula: [true]mu X1.<a>mu X2.<b>mu X3.<c>mu X4.<d>mu X5.<d>mu X6.<e>mu X7.<f>mu X8.<g>mu X9.<h>true This formula replaces the single very large equation by 9 ones, where the right hand sides only contain the information whether a single action can be done. This is generally faster and yields a substantially smaller PBES. Note that elaborate generations of PBESs can already occur when using subformulas of the shape [a.b]... or <a.b>.... The tool pbessolve is capable of generating a counter example in the form of labelled transition systems, provided the PBES is generated using lps2pbes (or lts2pbes) using the --counter-example flag. The generated PBES is more complicated and may be harder to solve. Yet, these counter examples are very helpful in determining whether formulas do not hold. If formulas are valid, this flag can also be used to determine witnesses, i.e., evidence when formulas are valid. The tool pbes2bool can generate counter examples without the use of this flag. These counter-examples are solely based on the provided PBES and they must be manually be related to the original transition system. Note that there is also an option --structured which can be used to generate boolean equation systems that do not contain both the conjunction and the disjunction operators among PBES variables in the right hand side. This flag can lead to a substantially larger number of equation (but linear in the original formula). ## Manual page for lps2pbes¶ ### Usage¶ lps2pbes [OPTION]... --formula=FILE [INFILE [OUTFILE]] ### Description¶ Convert the state formula in FILE and the LPS in INFILE to a parameterised boolean equation system (PBES) and save it to OUTFILE. If OUTFILE is not present, stdout is used. If INFILE is not present, stdin is used. ### Command line options¶ -c , --counter-example add counter example equations to the generated PBES -fFILE , --formulaFILE use the state formula from FILE -oFORMAT , --outFORMAT use output format FORMAT: bes BES in internal format pbes PBES in internal format pgsolver BES in PGSolver format text PBES in textual (mCRL2) format -m , --preprocess-modal-operators insert dummy fixpoints in modal operators, which may lead to smaller PBESs -s , --structured generate equations such that no mixed conjunctions and disjunctions occur -t , --timed use the timed version of the algorithm, even for untimed LPS’s --timings[FILE] append timing measurements to FILE. Measurements are written to standard error if no FILE is provided -u , --unoptimized do not simplify boolean expressions #### Standard options¶ -q , --quiet do not display warning messages -v , --verbose display short intermediate messages -d , --debug display detailed intermediate messages --log-levelLEVEL display intermediate messages up to and including level -h , --help display help information --version display version information ### Author¶ Wieger Wesselink; Tim Willemse	2021-09-27 04:59:09	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6592606902122498, "perplexity": 3205.7552090385084}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780058263.20/warc/CC-MAIN-20210927030035-20210927060035-00636.warc.gz"}
https://www.effortlessmath.com/math-topics/ssat-middle-level-math-free-sample-practice-questions/	# SSAT Middle Level Math FREE Sample Practice Questions Preparing for the SSAT Middle Level Math test? To do your best on the SSAT Middle Level Math test, you need to review and practice real SSAT Middle Level Math questions. There’s nothing like working on SSAT Middle Level Math sample questions to hone your math skills and put you more at ease when taking the SSAT Middle Level Math test. The sample math questions you’ll find here are brief samples designed to give you the insights you need to be as prepared as possible for your SSAT Middle Level Math test. Check out our sample SSAT Middle Level Math practice questions to find out what areas you need to practice more before taking the SSAT Middle Level Math test! Start preparing for the 2021 SSAT Middle Level Math test with our free sample practice questions. Also, make sure to follow some of the related links at the bottom of this post to get a better idea of what kind of mathematics questions you need to practice. ## 10 Sample SSAT Middle Level Math Practice Questions 1- In a group of $$5$$ books, the average number of pages is $$24$$. Mary adds a book with $$36$$ pages to the group. What is the new average number of pages per book? ☐A. 20 ☐B. 22 ☐C. 24 ☐D. 26 ☐E. 30 2- A football team won exactly $$70\%$$ of the games it played during last session. Which of the following could be the total number of games the team played last season? ☐A. 49 ☐B. 40 ☐C. 32 ☐D. 12 ☐E. 9 3- If a gas tank can hold $$35$$ gallons, how many gallons does it contain when it is $$\frac{2}{5}$$ full? ☐A. 50 ☐B. 125 ☐C. 62.5 ☐D. 14 ☐E. 8 4- What is the value of $$𝑥$$ in the following figure? (Figure is not drawn to scale) ☐A. 150 ☐B. 145 ☐C. 125 ☐D. 105 ☐E. 85 5- The capacity of a red box is $$20\%$$bigger than the capacity of a blue box. If $$36$$ books can be put in the red box, how many books can be put in the blue box? ☐A. 15 ☐B. 20 ☐C. 24 ☐D. 30 ☐E. 32 6- A taxi driver earns $$8$$ per 1-hour work. If he works $$10$$ hours a day and in $$1$$ hour he uses $$2$$-liters petrol with price $$1$$ for $$1$$-liter. How much money does he earn in one day? ☐A. $90 ☐B.$88 ☐C. $70 ☐D.$60 ☐E. $56 7- Which of the following is less than $$\frac{1}{5}$$? ☐A. $$\frac{1}{4}$$ ☐B. 0.5 ☐C. $$\frac{1}{7}$$ ☐D. 0.28 ☐E. 0.31 8- Amy and John work in a same company. Last month, both of them received a raise of $$20$$ percent. If Amy earns $$30.00$$ per hour now and John earns $$28.80$$, Amy earned how much more per hour than John before their raises? ☐A.$8.25 ☐B. $4.25 ☐C.$3.00 ☐D. $2.25 ☐E.$1.00 9- Three people can paint $$3$$ houses in $$12$$ days. How many people are needed to paint $$6$$ houses in $$6$$ days? ☐A. 6 ☐B. 8 ☐C. 12 ☐D. 16 ☐E. 20 10- If $$𝑁×6−3=12$$ then $$𝑁=?$$ ☐A. 4 ☐B. 12 ☐C. 13 ☐D. 14 ☐E. 18 ## Best SSAT Middle Level Math Prep Resource for 2021 1- D In a group of $$5$$ books, the average number of pages is (24). Therefore, the sum of pages in all $$5$$ books is $$(5×24=120)$$. Mary adds a book with $$36$$ pages to the group. Then, the sum of pages in all 6 books is $$(5×24+36=156)$$. The new average number of pages per book is:$$\frac{156}{6}=26$$ 2- B Choices A, C, D, and E are incorrect because $$70\%$$ of each of the numbers is a non-whole number. A. $$49, 70\%$$ of $$49 = 0.70×49=34.3$$ B. $$40, 70\%$$ of $$40=0.70×40=28$$ C. $$32, 70\%$$ of $$32=0.80×32=22.4$$ D. $$12, 70\%$$ of $$12=0.70×12=8.4$$ E. $$9, 70\%$$ of $$9=0.80×9=6.3$$ 3- D $$\frac{2}{5}×35=\frac{70}{5}=14$$ 4- A $$x=25+125=150$$ 5- D The red box is $$20\%$$ bigger than the blue box. Let $$x$$ be the capacity of the blue box. Then: $$x+20\%$$ of $$x=36→1.2x=36→x=\frac{36}{1.2}=30$$ 6- D $$8×10=80$$, Petrol use: $$10×2=20$$ liters, Petrol cost: $$20×1=20$$ Money earned: $$80-20=60$$ 7- C From the choices provided, only C $$(\frac{1}{7})$$ is less than $$\frac{1}{5}$$. 8- E Amy earns $$30.00$$ per hour now. $$30.00$$ per hour is $$20$$ percent more than her previous rate. Let $$x$$ be her rate before her raise. Then: $$x+0.20x=30→1.2x=30→x=\frac{30}{1.2}=25$$ John earns $$28.80$$ per hour now. $$28.80$$ per hour is $$20$$ percent more than his previous rate. Let $$x$$ be John’s rate before his raise. Then:$$x+0.20x=28.80→1.2x=28.80→x=\frac{28.80}{1.2}=24$$, Amy earned $$1.00$$ more per hour than John before their raises. 9- C Three people can paint $$3$$ houses in $$12$$ days. It means that for painting $$6$$ houses in $$12$$ days we need $$6$$ people. To paint $$6$$ houses in $$6$$ days, $$12$$ people are needed. 10- A $$N×(6-3)=12→N×3=12→N=4$$ Looking for the best resource to help you succeed on the SSAT Middle Level Math test? 23% OFF X ## How Does It Work? ### 1. Find eBooks Locate the eBook you wish to purchase by searching for the test or title. ### 3. Checkout Complete the quick and easy checkout process. ## Why Buy eBook From Effortlessmath? Save up to 70% compared to print Help save the environment	2021-06-17 23:54:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.35435423254966736, "perplexity": 2230.289897544163}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487634576.73/warc/CC-MAIN-20210617222646-20210618012646-00578.warc.gz"}
https://help.syscad.net/index.php/Solvent_Extraction_Unit_137	# Solvent Extraction Unit 137 Warning! Page you're reading is outdated or was archived and very likely is referring to older version of the software. Please navigate to the most recent version instead. Navigation: Main Page -> Models -> Mass Separation Models ## General Description The Solvent Extraction unit is designed for use as a solvent extraction model, which can also be configured to do organic stripping. The model ensures that mainly aqueous phases leave via the aqueous stream and organic phases leave through the organic stream. The user is responsible for setting up a suite of species which accurately portrays the application that is being evaluated e.g. H2O, H2SO4, Cu, Fe, CuSO4, FeSO4, Fe2[SO4]3, Kerosene, RH, R2Cu, R2Fe, R3Fe. This suite of species is typically set up for chelating type of organic solvents such as the LIX hydroxy oxime copper extractants available from the Henkel Corporation i.e. LIX 64N, 65N, 860, 864, 865, and 622. ### Species transfer across phases The user selects a primary metallic element of interest that is to be extracted, e.g. Primary Metal 'Cu'. The user should supply a reaction which utilises the specified Primary Metal, for example: 2RH(o) + CuSO4(aq) = R2Cu(o) + H2SO4(aq) Or, in the case of stripping: R2Cu(o) + H2SO4(aq) = 2RH(o) + CuSO4(aq) The user may either specify the reaction extent directly in the reaction, Method = 'None', or they may use the McCabe Thiele diagram isotherms to calculate the reaction extent, Method = 'Isotherm'. If the user selects Method = Isotherms, then the Isotherms will be used to determine the extent of the first reaction in the reaction file: 1. The model uses the concentration of the primary metal in the streams entering the unit and the extraction (or stripping) isotherm, provided by the user, to determine the required primary metal concentration within the aqueous and organic phases. The model then calculates the extent to which the defined reaction should proceed to give the desired concentrations. 2. The ratio of the organic phase volume flow to the aqueous phase volume flow within the unit also fulfils a crucial role in the equilibrium extraction calculation. It provides the slope of the McCabe-Thiele operating line (a straight line). The Y-intercept is calculated by SysCAD using the data provided by the user. It is called MinOrgConc in the case of extraction, and MinAqsConc in the case of stripping. 3. The model is not currently sensitive to pH and assumes that the pH is correct for a particular equilibrium isotherm. ### Phase Separation The user may choose to separate the phases based on either Individual Phase or density: 1. IndPhase to Organic: The user specifies the Individual phase that will report to the Organic stream. Note: All solids and other liquid phases will classed as Aqueous. 2. Density: The user specifies the density of separation. In this case the aqueous species are normally denser than the user defined density of separation and the organic species are less dense. e.g. with a separating density of 950 kg/m3, water and all dissolved aqueous species will report to the aqueous phase and the organic species, normally with densities less than approximately 850 kg/m3, will report to the organic phase. Note: Solids will also split on density. ### Entrainment The user may also specify the entrainment of the two different liquids: 1. Aqueous in the Organic stream. Note this will also include any solids that are in the aqueous phase; and 2. Organic in the Aqueous stream. ### Diagram The diagram shows the default drawing of the Solvent Extraction unit, with all of the streams that have to be connected for the unit to operate. The physical location of the streams connecting to the Solvent Extraction unit is unimportant. The user may connect the streams to any position on the unit. ## Inputs and Outputs Label RequiredOptional InputOutput Number of Connections Description Min Max. Aqueous Inlet 1 Required In 1 20 Aqueous feed to Solvent Extraction unit. Organic Inlet 1 Required In 1 20 Organic feed to Solvent Extraction unit. Aqueous Outlet Required Out 1 1 Aqueous outlet from Solvent Extraction unit. Organic Outlet Required Out 1 1 Organic outlet from Solvent Extraction unit. Vent Optional Out 0 1 Vent Stream. (Vapour Only) ## Behaviour when Model is OFF If the user disables the unit, by un-ticking the On tick box, then the following actions occur: • All streams connected to the 'Aqueous' inlet will flow straight out of the 'Aqueous' outlet; • All streams connected to the 'Organic' inlet will flow straight out of the 'Organic' outlet; • No reactions will occur. So basically, the unit will be 'bypassed' without the user having to change any connections. ## Model Theory The Isotherm Method of modelling a Solvent extraction unit is based on the McCabe-Thiele method. This assumes that the operating line on the xy diagram, representing a single solvent extraction stage, is linear. The diagram below illustrates the graphical representation of the unit: The operating line is the ratio of the incoming organic stream to the incoming aqueous stream. The Y-intercept is calculated by SysCAD using the data provided by the user. The Equilibrium line is determined from experimental test data. • This is the Isotherm that the user is required to input to the model. • For extraction, the x axis is the concentration of the primary metal in the aqueous phase and the y axis is the concentration of the primary metal in the organic phase. • For stripping, the x axis is the concentration of the primary metal in the organic phase and the y axis is the concentration of the primary metal in the aqueous phase. If the Metal (in this case Copper) concentration in the incoming Aqueous stream is 1.7 g/l, then, move vertically up the red line to the operating line, then horizontally to the Equilibrium line to find the equilibrium values of the Copper concentrations in the exiting Aqueous and Organic streams. In this case the Organic Stream will contain 5g/l Cu and the Aqueous 0.4g/l Cu at equilibrium. The values will be adjusted to take into account the stage efficiency, as the two phases are not usually in contact long enough to achieve perfect equilibrium. In this example, the user will insert the Equilibrium curve into the model, and then the Reaction file should contain a reaction with the following form: $\mathbf{\mathrm { CuSO_4(aq) + 2 RH(o) = H_2 SO_4(aq) + R_2 Cu(o)}}$ Extent:Fraction$\mathbf {\mathrm{ CuSO_4(aq) = 0.5}}$ Where (aq) denotes the aqueous phase and (o) the organic phase. The model will adjust the extent of the reaction to produce the required aqueous and organic Copper concentrations. ## Data Sections The default access window consists of several sections: 1. SolventExtraction tab - Contains general information relating to the unit. 2. RB - Optional tab, only visible if the Reactions are enabled in the Evaluation Block. 3. Isotherm tab - The user can set up and view the isotherm. 4. Info tab - Contains general settings for the unit and allows the user to include documentation about the unit and create Hyperlinks to external documents. 5. Links tab, contains a summary table for all the input and output streams. 6. Audit tab - Contains summary information required for Mass and Energy balance. See Model Examples for enthalpy calculation Examples. ### Solvent Extraction Page Unit Type: SolventExtraction - The first tab page in the access window will have this name. Symbol / Tag Input / Calc Description/Calculated Variables / Options ### Requirements On Tick box If this is enabled then the unit will be in full operational mode. If this is disabled, then unit will be 'inoperative', as described in Inoperative Mode. SeparationMethod / SeparMethod Density The user selects the Density of Separation. All species (INCLUDING solids) that have a density LESS than this density will report to the organic outlet. All species with a density GREATER than this density will report to the Aqueous outlet. IndPhase to Organic The user selects the Individual liquid Phase(s) (usually (o)) that reports to the organic phase. All solids and other liquid phases will report to the Aqueous phase. SeparDensity / SeparRho Input Visible if SeparMethod = 'Density'. The required separation density. This must be a value between the density of the aqueous and organic streams. Note: The density of water is a function of temperature, and is often less than 1000 kg/m3; therefore it is safer to specify a separation density below 950 kg/m3. List of Liquid Individual Phases Tick Boxes Visible if SeparMethod = 'IndPhase to Organic'. The user must select one or more Individual liquid phases that will report to the Organic outlet. Normally, the user will select the 'o', or 'org' phase. All other liquid phases will report to the Aqueous outlet. (The user defines the Individual phase for each species in the Species Database. Normally organic species are defined to have an individual phase of 'o', or 'org'. Please see Individual Phase Label for information on how to change the Individual Phase of a species) Entrainment Feed Loss The Aqueous/Organic loss is defined as a percentage of the aqueous (or organics) in the unit after all the reactions have been completed. NOTE: This is feed to the separator section, not the mixer section. Product Fraction The Aqueous/Organic loss is defined as mass concentration in the outlet stream, normally in ppm. AqueousLossReqd / AqsLossReqd Input Visible with the Feed Loss Entrainment method. The mass fraction of Aqueous liquid in the unit lost to the Organic product stream. Note that solids in the Aqueous phase will also report to the Organic stream when an entrainment value is specified. OrganicLossReqd / OrgLossReqd Input Visible with the Feed Loss Entrainment method. The mass fraction of Organic liquid in the unit lost to the Aqueous product stream. AqueousInOrgReqd / AqsInOrgReqd Input Visible with the Product Fraction Entrainment method. User specified the mass concentration of aqueous (normally in ppm) in the organic outlet stream. Note that solids in the Aqueous phase will also report to the Organic stream when an entrainment value is specified. OrganicInAqsReqd / OrgInAqsReqd Input Visible with the Product Fraction Entrainment method. User specified the mass concentration of organics (normally in ppm) in the aqueous outlet stream. ExtractionMethod / Method None The model does not use Isotherms to determine the extent of the extraction or stripping reaction. The user sets the extent/s of the reaction/s manually. Isotherm The model uses Isotherms to determine the extent of the extraction or stripping reaction. Reactions will be switched on by SysCAD if the isotherm method is chosen. The user must still specify the reaction file and the reaction/s, but the model will set the extent of the extraction or stripping reaction/s manually. Notes The model will set the extent of the First reaction in the file. The reaction file may contain more than one reaction. The user must set the required extents of the other reactions. The following fields will be visible if the Isotherm Method is chosen. AllowOveride Tickbox This allows the user to override the reaction extent calculated by the isotherm and manually set the reaction extent on the RB tab page. Mode Extraction This mode is selected if the unit is used to extract the metal from the aqueous into the organic phase. Stripping This mode is selected if the unit is used to strip the metal from the organic into the aqueous phase. StageEfficiency / StageEff Input The efficiency of the solvent extraction stage. ExtentSpecies / ExtentSp Input The liquid containing the primary metal, e.g. for Extraction it may be CuSO4, and for Stripping R2Cu. Primary Metal List The user must select the metal of interest from the list of available elements in the project. A valid element must be selected if the "Isotherm" method is selected. It is optional for the method "None", however it is useful to set this so that the feed and product concentrations can be shown. Reactions List The user may turn the reactions 'On' or 'Off'. If this is 'On' then the RB tab page will become visible. Reactions will be switched 'On' by SysCAD if the isotherm method is chosen. OperatingP... Method Atmospheric outlet streams will be at Atmospheric Pressure. The atmospheric pressure is calculated by SysCAD based on the user defined elevation (default elevation is at sea level = 101.325 kPa). The elevation can be changed on the Environment tab page of the Plant Model. AutoDetect If there are any liquids AND no vapours present in the feed, outlet streams will take the highest pressure of the feeds. Else (eg. some vapours present) outlet streams will take the lowest pressure of the feeds. LowestFeed outlet streams will take the lowest pressure of the feeds. HighestFeed outlet streams will take the highest pressure of the feeds. RequiredP outlet streams will be at the user specified pressure. IgnoreLowQm Tick Box This option is only visible if the AutoDetect, LowestFeed or HighestFeed methods are chosen. When calculating the outlet pressure and temperature of the tank, SysCAD will ignore the low flow feed streams should this option be selected. The low flow limit is set in the field below. LowQmFrac Input This field is only visible if the IgnoreLowQm option is selected. This is the amount any stream contributes to the total flow. For example, if the total feed to the tank is 10 kg/s, and this field is set to 1%. Then any feed streams with less than 0.1 kg/s will be ignored in the pressure calculations. PressureReqd / P_Reqd Input This field is only visible if the RequiredP method is chosen. This is user specified pressure. Result Display The actual pressure used for the sum of the feeds which will also be the outlet pressure (unless further model options change the pressure). ### Results MinOrgConc OR MinAqsConc Output This field is only visible if the Isotherm Method has been chosen. The Y-intercept of the McCabe-Thiele operating line, MinOrgConc in the case of extraction, and MinAqsConc in the case of stripping. NOTE this is defined by the feed streams entering the unit operation. ReactExtent Feedback This field is only visible if the Isotherm Method has been chosen. This is the reaction extent required as appears in the RB tab page. Displayed here for easy access. ActReactExtent Feedback This field is only visible if the Isotherm Method has been chosen. This is the actual reaction extent achieved as appears in the RB tab page. Displayed here for easy access. MaxLoadedConc Output If the Isotherm Method has been chosen, this field will only be visible if Extraction Mode is chosen. Displays the maximum concentration based on 100% extraction and calculated by mass balance. This will serve as a warning for incorrectly supplied data if the results differ to those expected. MaxStripConc Output If the Isotherm Method has been chosen, this field will only be visible if Stripping Mode is chosen. Displays the maximum concentration based on 100% strip efficiency and calculated by mass balance. This will serve as a warning for incorrectly supplied data if the results differ to those expected. Feed Conditions AqueousConcIn / AqsConcIn Calc The concentration of the Primary metal in the incoming Aqueous stream. OrganicConcIn / OrgConcIn Calc The concentration of the Primary metal in the incoming Organic stream. Organic/Aqueous / OARatio Calc The calculated volumetric ratio of Organic to Aqueous liquid entering the unit. Aqueous/Organic / AORatio Calc The calculated volumetric ratio of Aqueous to Organic liquid entering the unit. Product Conditions AqueousConcOut / AqsConcOut Calc The concentration of the Primary metal in the outgoing Aqueous stream. OrganicConcOut / OrgConcOut Calc The concentration of the Primary metal in the outgoing Organic stream. AqueousInOrg / AqsInOrg Calc The mass concentration Aqueous in the organic outlet stream. OrganicInAqs / OrgInAqs Calc The mass concentration organic in the aqueous outlet stream. AqueousLosses / AqsLosses Calc The mass fraction of feed Aqueous to the organic outlet stream. OrganicLosses / OrgLosses Calc The mass fraction of feed organic to the aqueous outlet stream. ### Isotherm Section The user types in the Isotherm describing the Solvent Extraction process. Isotherm data points (both Aq and Org or X and Y) are expected to increase in value from the first data point to the last data point entered. Tag / Symbol Input / Calc Description Isotherm Curve (g/L) Isotherm... The user enters the data points in the two columns on this page. For example, for extraction, Aq represents the mass fraction of solute in raffinate, and Org represents the mass fraction of solute in extract. The model interpolates linearly between the points and extrapolates beyond the first and last points. Length Input The number of data points. Once a number has been typed in, the Aqueous (Aq) and Organic (Org) columns will be visible with the required number of lines. Data Points Index The data point index value Aq The Aqueous values for the Isotherm. Org The Organic values for the Isotherm. The isotherm data can be saved to or loaded from a file. FileName Input User specified filename, (1) to store current screen values or (2) load pre-defined values. The file should be in a CSV (comma delimited) format. Load Load from CSV Button Only active if FileName is not empty. This will load the data points from the CSV file, replacing the current values. Save Save to CSV Button Only active if FileName is not empty. This will save the current values to the CSV file. ## Adding this Model to a Project Insert into Configuration file Sort either by DLL or Group. DLL: Separ1.dll → Units/Links → Separation: Solvent Extraction (Mixer/Settler) OR Group: Mass Separation → Units/Links → Separation : Solvent Extraction (Mixer/Settler)	2021-05-12 02:20:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4183318316936493, "perplexity": 2841.382336004599}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991693.14/warc/CC-MAIN-20210512004850-20210512034850-00027.warc.gz"}
http://www.chegg.com/homework-help/questions-and-answers/average-number-persons-per-household-united-states-shrinking-steadily-long-statistics-kept-q1326700	the average number of a persons per household in the united states has been shrinking steadily for as long as statistics have been kept and is approximately linear with respect to time. in 1980 there were about 2.76 persons per household , and in 2008 about 2.55 persons per household . assume that this trend continues. answer the following questions:1 1. estimate the household size in 2019. persons per household? 2.when will the household size be 2.15 persons per household? in the year of?	2014-09-22 11:49:31	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8285247087478638, "perplexity": 1206.8863008588232}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657136966.6/warc/CC-MAIN-20140914011216-00187-ip-10-234-18-248.ec2.internal.warc.gz"}
http://umj.imath.kiev.ua/authors/name/?lang=en&author_id=2393	2018 Том 70 № 11 # Stepochkina M. V. Articles: 2 Article (Russian) ### Description of posets critical with respect to the nonnegativity of the quadratic Tits form Ukr. Mat. Zh. - 2009. - 61, № 5. - pp. 611-624 We present the complete description of finite posets whose Tits form is not nonnegative but all proper subsets of which have nonnegative Tits forms. A similar result for positive forms was obtained by the authors earlier. Article (Russian) ### $(\min, \max)$-equivalence of posets and nonnegative Tits forms Ukr. Mat. Zh. - 2008. - 60, № 9. - pp. 1157–1167 We study the relationship between the (min, max)-equivalence of posets and properties of their quadratic Tits form related to nonnegative definiteness. In particular, we prove that the Tits form of a poset S is nonnegative definite if and only if the Tits form of any poset $(\min, \max)$-equivalent to S is weakly nonnegative.	2018-12-10 02:30:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6104397773742676, "perplexity": 1093.7584254049755}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376823236.2/warc/CC-MAIN-20181210013115-20181210034615-00359.warc.gz"}
http://www.algebra.com/tutors/your-answers.mpl?userid=solver91311&from=10440	Algebra -> Tutoring on algebra.com -> See tutors' answers! 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do I know if the line should be solid or broken, and also what part should i shade? 1 solutions Answer 222188 by solver91311(16872) on 2010-06-02 14:51:14 (Show Source): You can put this solution on YOUR website! If your inequality symbol is or then use a broken line for the boundary. This means that points on the line are NOT included in the solution set. If your inequality symbol is or then use a solid line for the boundary. This means that points on the line ARE included in the solution set. Once you have the boundary line graphed, pick any point you like that is NOT ON THE LINE. If the line does not pass through the origin, the origin itself is a very good selection for a test point. Otherwise, pick something with small integer coefficients -- all this so that the arithmetic is easier. Substitute the coordinates of the point you chose into the original inequality. If the result is a true statement, shade in the side of the line that CONTAINS the point you chose. If the result is a false statement, shade in the side of the line that DOES NOT CONTAIN the point you chose. This doesn't go through the origin so, lets try (0,0) Yep, zero is less than 2, so the shading happens on the side of the line containing the origin. Solved by pluggable solver: Plot Any Inequality Graphing function : See? John Quadratic_Equations/310688: Hello I have a problem solving this parabola question.The path of a cliff diver as he dives into a lake ,is given by this eqaution,y=-(x-10)(SQAURED)+75,where y metres is the diver's height above the water and,x metres is the horisontal distance travelled by the diver.What is the maximum height the diver is above the water?1 solutions Answer 222183 by solver91311(16872) on 2010-06-02 14:32:06 (Show Source): You can put this solution on YOUR website! Not sure whether you just need an intuitive approach, a formal algebraic approach, or a calculus approach. Intuitive The largest y can be is 75 and that is when . That is because is always a positive number, unless it is zero, so is always a negative number unless it is zero. Therefore you are always subtracting something from 75 unless , which happens when While we are at it -- 75 meters? That is 246 feet. Hitting the water from that height would be like hitting concrete. I rather think your cliff diver is only going to perform this particular stunt once. Algebraic Expand the binomial expression: A parabola with a negative lead coefficient opens downward, hence the vertex is a maximum. The vertex of a parabola expressed in form has a vertex that occurs at an value of , so: And the value of the function at is: Calculus Using the function definition from the Algebraic discussion: The function will have a maximum where the first derivative is zero and the second derivative is negative. Set it equal to zero which is negative for all in the domain of So there is a maximum at John Circles/310663: If angle ema is 95 degrees, what is the measure of angle bme? I would be grateful for any assistance with this problem. Cheers!1 solutions Answer 222170 by solver91311(16872) on 2010-06-02 13:48:09 (Show Source): You can put this solution on YOUR website! Sorry, but without the picture or a detailed and unambiguous description of the figure, we can't help you. John Linear-equations/310690: The coordinates of the end points of a segment are given. Find the coordinates of the midpoint of the segment. G(-8,10),H(-4,8) Please show me how to get the answewr1 solutions Answer 222165 by solver91311(16872) on 2010-06-02 13:41:18 (Show Source): You can put this solution on YOUR website! Use the mid-point formulae: and Where and are the coordinates of the given points. John Polynomials-and-rational-expressions/310687: HI,I WOULD LIKE SOM HELP WITH THIS PROBLEM: 1 solutions Answer 222162 by solver91311(16872) on 2010-06-02 13:37:57 (Show Source): You can put this solution on YOUR website! I would like you to stop shouting. Typing in all caps is the electronic equivalent of shouting and is therefore both annoying and rude. John Geometry_Word_Problems/310676: The perimeter of an equilateral triangle is at most 57 feet. What could the length of a side be?1 solutions Answer 222155 by solver91311(16872) on 2010-06-02 13:00:12 (Show Source): You can put this solution on YOUR website! An equilateral triangle has three sides that are the same length. That's what equilateral means: "equi-" means "the same" and "lateral" means "side" If the perimeter is at most 57 whatevers, then the measure of a side can be at most 57 divided by 3 of the same whatevers. John Rectangles/310677: find the perimeter and area of a rectangle with width 5 centimeters and length 16 centimeters.1 solutions Answer 222152 by solver91311(16872) on 2010-06-02 12:56:38 (Show Source): You can put this solution on YOUR website! The perimeter of a rectangle: The area of a rectangle John Money_Word_Problems/310669: how much money would i have to put in a cd to draw $500 a month joinnow2000@aol.com1 solutions Answer 222151 by solver91311(16872) on 2010-06-02 12:55:00 (Show Source): You can put this solution on YOUR website! That would depend a great deal on the annual interest rate and, to a lesser extent, on the compounding frequency, and if you are just considering an investment that will earn that much that will be allowed to remain in the account or if you want to actually draw out$500.00 every month as an annuity payment. Let's presume annual interest expressed as a decimal. Let's assume that the institution actually pays one-twelfth of the annual interest each month and that you want a monthly pay annuity. Let be the principal amount invested in the CD. Then you need: Solving for Right now, June 2, 2010, the best you will be able to do is John Graphs/310664: Solve the system by graphing: 2x+y=12 2x+y=201 solutions Answer 222144 by solver91311(16872) on 2010-06-02 12:25:07 (Show Source): You can put this solution on YOUR website! Two parallel lines, no solution. How did I know? If you have: and and and where Then you have two parallel lines and no solution. On the other hand, if you have all of the above except that then you have two representations of the same line and therefore an infinite number of solutions. John Triangles/310449: The angles of the triangle are 30°-30°-120°. Two of the sides lengths are 2 times the square root of 3. What is the length of the third side?1 solutions Answer 221960 by solver91311(16872) on 2010-06-01 18:49:36 (Show Source): You can put this solution on YOUR website! The hypotenuse is the longest side of a right triangle. It is always the side opposite the right angle. Since your 30-30-120 triangle doesn't have a right angle, it is not a right triangle, therefore it does not have a hypotenuse. Having said that, if you want to know the measure of the longest side of your triangle, even though it isn't called a hypotenuse, use the Law of Cosines. Let c represent the measure of the longest side, and let a and b represent the measures of the other two equal length sides. Let C represent the measure of the largest angle. Then which, for our purposes will be more conveniently written: Plugging in the numbers given: You can do your own arithmetic. Hint: and John Systems-of-equations/310450: What is the solution to p-4=-9+p1 solutions Answer 221956 by solver91311(16872) on 2010-06-01 18:35:53 (Show Source): You can put this solution on YOUR website! Add to both sides of your equation. This results in the absurdity that . Therefore, the solution set for this equation is the empty set. John Rational-functions/310403: Find the domain of the function: 1 solutions Answer 221940 by solver91311(16872) on 2010-06-01 16:55:16 (Show Source): You can put this solution on YOUR website! The domain of a rational function is all real numbers EXCEPT any number that would cause the value of the denominator to be zero. Take the denominator and set it equal to zero: and then solve for Factor: Therefore, , , and must be excluded from the domain. Using interval notation, the domain is: Or in set-builder notation: John Travel_Word_Problems/310408: Which country technology wise , reduce its travel distance by road from 3 hours to traveling by sea to 45mins? 1 solutions Answer 221938 by solver91311(16872) on 2010-06-01 16:42:14 (Show Source): You can put this solution on YOUR website!What???????? Rational-functions/310411: Find the zeros of the function by factoring or algebraically: please note the x^3 was meant to be atop the fraction with the 36x so it should be x^3-36x atop the fraction.1 solutions Answer 221935 by solver91311(16872) on 2010-06-01 16:41:11 (Show Source): You can put this solution on YOUR website! Factor the numerator: Then set the function equal to zero and multiply by the denominator: Hence, or or Note: I put parentheses around your numerator so that the equation editor what parts go together. John Complex_Numbers/310370: Solve x +6(square root of x) -7 =01 solutions Answer 221933 by solver91311(16872) on 2010-06-01 16:34:08 (Show Source): You can put this solution on YOUR website! Let represent then: Factor: Hence, or And then But since by definition -- the negative square root being denoted -- discard this root. So John Volume/310407: how do i find the volume of a sphere with a radius of 7? emaples?1 solutions Answer 221931 by solver91311(16872) on 2010-06-01 16:24:39 (Show Source): You can put this solution on YOUR website! The volume of a sphere with radius units is given by: John Quadratic_Equations/310344: I am trying to solve the following problem: height of a kicked ball is modeled using The question I need to solve is this: If the nearest person is 5 ft horizontally from the point of impact how high must that person reach in order to block the punt? 1 solutions Answer 221930 by solver91311(16872) on 2010-06-01 16:21:57 (Show Source): You can put this solution on YOUR website! The only way this makes any sense whatsoever is if the height is expressed as a function of the horizontal distance () of the ball from the kicker. Hence, you want to know the value of When which is to say: You can do your own arithmetic. John Polynomials-and-rational-expressions/310377: Hi,I would like help with this equation,/1 solutions Answer 221929 by solver91311(16872) on 2010-06-01 16:13:30 (Show Source): You can put this solution on YOUR website! What equation? An equation has an equals sign in it. That is why it is called an equation. Since there is no equals sign anywhere to be seen, you don't have an equation. What is it you would like to do with this clever little rational expression? John Triangles/310390: if a/cosb=b/cosa then prove that triangle abc is right angled triangle1 solutions Answer 221924 by solver91311(16872) on 2010-06-01 15:41:23 (Show Source): You can put this solution on YOUR website! Your notation is VERY confusing. Lower case letters, generally speaking, refer to the measure of the sides of a triangle, while upper case letters are generally reserved for the measures of the angles. But taking the cosine function of the measure of a side of the triangle doesn't make any sense. Hence, unless you have referred to BOTH the measure of a side of the triangle and a measure of an angle with the same variable (sort of like calling everyone in the room "George" regardless of what their names really are), you must be saying that the ratio of the measure of one of the angles to the cosine of the measure of the other acute angle is equal to the ratio of the measure of the other acute angle to the cosine of the measure of the first angle -- an assertion that is false for all except an isosceles right triangle. John Inequalities/310394: PLEASE ANSWER ALL OF THESE... I WANT TO GO OVER MY TEST THAT MY TEACHER HAD HANDED BACK TO ME TODAY.. AND I WANT TO SEE WHERE I WENT WRONG WITH ALL OF THESE tell weather the ordered pair is a solution of the inequality. 1. x+y>-9 (0,0) 3. 2x-y>4 ; (-6,-15) 5. 5x+2y less than or equal to 8 ; (-3,6) 7. 0.5x+2.5y greater than or equal to 2 ;(0,0) 9. 0.2y-0.5x>-1 ; (-4,-8) GRAPH THE INEQUALITY 11. x-y>-4 13. 4y less than or equal to 6x-2 15. 6Y+3 GREATER THAN OR EQUAL TO -18X 1 solutions Answer 221918 by solver91311(16872) on 2010-06-01 15:28:14 (Show Source): You can put this solution on YOUR website! And we would be delighted to help you -- right after you re-post showing the answers you got and the work you did to get them. John Exponents/310392: -3(3f^-1g^3)^-2 I was asked to simplify this expression. The answer key given to us my my teacher says the answer should be -f^2/3g^6, but I got -f^2/27g^6. I have found misprints in the answer key for other questions; is this yet another?1 solutions Answer 221917 by solver91311(16872) on 2010-06-01 15:26:40 (Show Source): You can put this solution on YOUR website! Nope. The answer key is correct this time. See? John Functions/310391: find the domain g(x)=(8)/(4-3x) 1 solutions Answer 221913 by solver91311(16872) on 2010-06-01 15:17:19 (Show Source): You can put this solution on YOUR website! The domain of a rational function is all real numbers EXCEPT any number that would cause the value of the denominator to be zero. Take the denominator and set it equal to zero: and then solve for Therefore, must be excluded from the domain. Using interval notation, the domain is: Or in set-builder notation: John Surface-area/310349: The base of a rectangle is on the x-axis and its two upper vertices are on the parabola y=16-x^2. of all such rectangles, what are the dimensions of the one with greatest area?1 solutions Answer 221898 by solver91311(16872) on 2010-06-01 13:55:08 (Show Source): You can put this solution on YOUR website! Because of the fact that the parabola is symmetric to the y-axis, the rectangle must also be symmetric to the y-axis. That means that the two lower vertices are and . The measure of the base of the rectangle is therefore . The upper vertices, being points on the parabola are: and . Therefore, the measure of the height of the rectangle is simply . Therefore the area of the rectangle is: Such a function has local extremes at the points where the first derivative is zero: Discard the negative root since we need a positive measure of length We are looking for a maximum, so we want to see if the value of the second derivative is negative at the extreme point. Which it is... So, the horizontal dimension of the largest area rectangle is And the vertical dimension is: John Triangles/309694: Regarding a triangle which has a 90 degree angle...is there a simple formula to take a given diagonal dimension (hypotenuse) and a constant vertical / horizontal dimension relationship (16:9), and determine the vertical / horizontal dimensions from a given diagonal dimension. For example, if I use the basic 16 horizontal dimension, and the basic 9 vertical dimension, the hypotenuse will be 18.357... Hence, working in reverse, what would be the formula I could use to plug in a given hypotenuse (20, 22, 23, 24, 25, 50, 100, etc. and using the non variable 16:9 ratio, come up with the vertical and horizontal dimensions? I hope I did OK with making my question clear. Thanks.1 solutions Answer 221534 by solver91311(16872) on 2010-05-30 01:38:19 (Show Source): You can put this solution on YOUR website! for your long dimension and for your short dimension Where is your variable screen diagonal Number-Line/309717: How do I graph the portion of the number line containing all points for which x > 6 and x < y1 solutions Answer 221533 by solver91311(16872) on 2010-05-30 01:27:53 (Show Source): You can put this solution on YOUR website!That would depend a great deal on the value of y Travel_Word_Problems/309291: your brother leaves on a trip, fogetting his suitcase. you know that he normally drives at a speed of 55 mph. you do not discover the suitcase until 1 hour later after he has left. if you follow him at a speed of 65 mph, how long will it takes you to catch up with him?1 solutions Answer 221197 by solver91311(16872) on 2010-05-27 23:34:48 (Show Source): You can put this solution on YOUR website! If he has been gone an hour and travels at 55 mph, then he is 55 miles ahead of you. You are going 10 mph faster. How long would it take to go 55 miles at 10 miles per hour? John Quadratic-relations-and-conic-sections/309295: x"2/9 + y"2/49 = 11 solutions Answer 221196 by solver91311(16872) on 2010-05-27 23:32:30 (Show Source): You can put this solution on YOUR website! What would you like to do with this ellipse? John Problems-with-consecutive-odd-even-integers/309069: Set I contains six consecutive integers . set J contains all integers that result from adding 3 to each of the integers in set I and also contains all integers that result from subtracting 3 from each of the integers in set I . how many more integers are there in set J than in set I ? a 0 B 2 C 3 D 6 E 91 solutions Answer 221195 by solver91311(16872) on 2010-05-27 23:30:32 (Show Source): You can put this solution on YOUR website! The elements of I are 3 less than the smallest one is Then 3 less than the next smallest is And 3 less than the next smallest is On the other end, Altogether, 6 additional elements. John Rectangles/309283: the perimeter is 420yds the width is 3/4 of the length, what is the length and width.1 solutions Answer 221183 by solver91311(16872) on 2010-05-27 22:12:47 (Show Source): You can put this solution on YOUR website! 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