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[ "# How to execute expression trees (C#)\n\nThis topic shows you how to execute an expression tree. Executing an expression tree may return a value, or it may just perform an action such as calling a method.\n\nOnly expression trees that represent lambda expressions can be executed. Expression trees that represent lambda expressions are of type LambdaExpression or Expression<TDelegate>. To execute these expression trees, call the Compile method to create an executable delegate, and then invoke the delegate.\n\nNote\n\nIf the type of the delegate is not known, that is, the lambda expression is of type LambdaExpression and not Expression<TDelegate>, you must call the DynamicInvoke method on the delegate instead of invoking it directly.\n\nIf an expression tree does not represent a lambda expression, you can create a new lambda expression that has the original expression tree as its body, by calling the Lambda<TDelegate>(Expression, IEnumerable<ParameterExpression>) method. Then, you can execute the lambda expression as described earlier in this section.\n\n## Example\n\nThe following code example demonstrates how to execute an expression tree that represents raising a number to a power by creating a lambda expression and executing it. The result, which represents the number raised to the power, is displayed.\n\n// The expression tree to execute.\nBinaryExpression be = Expression.Power(Expression.Constant(2D), Expression.Constant(3D));\n\n// Create a lambda expression.\nExpression<Func<double>> le = Expression.Lambda<Func<double>>(be);\n\n// Compile the lambda expression.\nFunc<double> compiledExpression = le.Compile();\n\n// Execute the lambda expression.\ndouble result = compiledExpression();\n\n// Display the result.\nConsole.WriteLine(result);\n\n// This code produces the following output:\n// 8\n\n\n## Compiling the Code\n\n• Include the System.Linq.Expressions namespace." ]

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[ "Now, the best pellet grills come with more advanced features like auto-ignition using igniting rods, Wi-Fi Support, automatic turn on/off, and much more. One of the key safety features added by pellet smokers (designed by Green Mountain Grills), is a cooling fan to cool off the grill after it has been turned off. Even the shape of the body has been changed to have peaked lids so that it can accommodate turkeys and other kinds of roasts. Some come with multiple levels of racks for warming and cooling.\nAccurate Temperature Control: The Davy Crockett has better temperature control that permits us to program a great deal more correct temperatures (one-degree changes possible) using their application (the control board permits us to alter temperatures 5 degrees at any given moment). It uses a PID controller […] controller to expect and fine micromanage the temperature by controlling the wood screw speed and the fan speed.\nAdd humidity to your grilling environment while also Add humidity to your grilling environment while also adding smoker flavor. This dual smoker/humidifier can accommodate liquids and wood chips or pellets for delicious succulent foods. The even heat of the cast iron heats the liquid in its reservoir creating steam that permeates the food on the grill. The open ...  More + Product Details Close\n```We purchased our wood pellet grill (Model TFB57CLB) from Costco in August 2017. It came with one bag of the gourmet blend pellets. We used the Traeger cookbook to try a few different recipes in the fall. Most of these involve a period of smoking followed by a period of grilling. The grill seemed to work fine at first. We used it a couple of times on warmer days in the winter and it had some trouble getting up to the desired heat level. We started using it regularly again in April 2018. I was careful to clean it and check the pellets etc. The smoking feature seemed to work alright, but when I would turn the heat up for grilling, the grill would usually struggle to get much over 225. I would resort to turning it up to 375 or more just to get it to 225 and then often it would sink back down to 180 or 160. I went to the troubleshooting guide on the manufacturer's website and checked all the components, which seemed to be working fine.\n```\n\n!function(n,t){function r(e,n){return Object.prototype.hasOwnProperty.call(e,n)}function i(e){return void 0===e}if(n){var o={},s=n.TraceKit,a=[].slice,l=\"?\";o.noConflict=function(){return n.TraceKit=s,o},o.wrap=function(e){function n(){try{return e.apply(this,arguments)}catch(e){throw o.report(e),e}}return n},o.report=function(){function e(e){l(),h.push(e)}function t(e){for(var n=h.length-1;n>=0;--n)h[n]===e&&h.splice(n,1)}function i(e,n){var t=null;if(!n||o.collectWindowErrors){for(var i in h)if(r(h,i))try{h[i].apply(null,[e].concat(a.call(arguments,2)))}catch(e){t=e}if(t)throw t}}function s(e,n,t,r,s){var a=null;if(w)o.computeStackTrace.augmentStackTraceWithInitialElement(w,n,t,e),u();else if(s)a=o.computeStackTrace(s),i(a,!0);else{var l={url:n,line:t,column:r};l.func=o.computeStackTrace.guessFunctionName(l.url,l.line),l.context=o.computeStackTrace.gatherContext(l.url,l.line),a={mode:\"onerror\",message:e,stack:[l]},i(a,!0)}return!!f&&f.apply(this,arguments)}function l(){!0!==d&&(f=n.onerror,n.onerror=s,d=!0)}function u(){var e=w,n=p;p=null,w=null,m=null,i.apply(null,[e,!1].concat(n))}function c(e){if(w){if(m===e)return;u()}var t=o.computeStackTrace(e);throw w=t,m=e,p=a.call(arguments,1),n.setTimeout(function(){m===e&&u()},t.incomplete?2e3:0),e}var f,d,h=[],p=null,m=null,w=null;return c.subscribe=e,c.unsubscribe=t,c}(),o.computeStackTrace=function(){function e(e){if(!o.remoteFetching)return\"\";try{var t=function(){try{return new n.XMLHttpRequest}catch(e){return new n.ActiveXObject(\"Microsoft.XMLHTTP\")}},r=t();return r.open(\"GET\",e,!1),r.send(\"\"),r.responseText}catch(e){return\"\"}}function t(t){if(\"string\"!=typeof t)return[];if(!r(j,t)){var i=\"\",o=\"\";try{o=n.document.domain}catch(e){}var s=/(.*)\\:\\/\\/([^:\\/]+)([:\\d]*)\\/{0,1}([\\s\\S]*)/.exec(t);s&&s===o&&(i=e(t)),j[t]=i?i.split(\"\\n\"):[]}return j[t]}function s(e,n){var r,o=/function ([^(]*)\\(([^)]*)\\)/,s=/['\"]?([0-9A-Za-z\\$_]+)['\"]?\\s*[:=]\\s*(function|eval|new Function)/,a=\"\",u=10,c=t(e);if(!c.length)return l;for(var f=0;f0?s:null}function u(e){return e.replace(/[\\-\\[\\]{}()*+?.,\\\\\\^\\$|#]/g,\"\\\\\\$&\")}function c(e){return u(e).replace(\"<\",\"(?:<|<)\").replace(\">\",\"(?:>|>)\").replace(\"&\",\"(?:&|&)\").replace('\"','(?:\"|\")').replace(/\\s+/g,\"\\\\s+\")}function f(e,n){for(var r,i,o=0,s=n.length;or&&(i=s.exec(o[r]))?i.index:null}function h(e){if(!i(n&&n.document)){for(var t,r,o,s,a=[n.location.href],l=n.document.getElementsByTagName(\"script\"),d=\"\"+e,h=/^function(?:\\s+([\\w\\$]+))?\\s*\\(([\\w\\s,]*)\\)\\s*\\{\\s*(\\S[\\s\\S]*\\S)\\s*\\}\\s*\\$/,p=/^function on([\\w\\$]+)\\s*\\(event\\)\\s*\\{\\s*(\\S[\\s\\S]*\\S)\\s*\\}\\s*\\$/,m=0;m]+)>|([^\\)]+))\\((.*)\\))? 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We were fortunate enough to be able to see all of the manufacturers and models I was most interested in, at Sam’s NW BBQ. (Yoder, MAK, Fast Eddy, GMG’s, a few others, and then, the one we ended up with, Memphis.) Sam took a good amount of time, providing us fantastic information about each model. We ended up with the Memphis Pro for the following reasons: 1) The construction, mainly the double-walled construction; 2) the preciseness of temperature, and being able to hold the temperature (due to reason #1); 3) the extreme ease of changing from smoking to grilling without a lot of effort or reconfiguration; 4) the double hoppers; 5) and another BIGGIE, the fact that grilled or baked foods taste like that, and not like smoked cookies, bread, pizza, etc.\nKeep your Pellet Grill and Smoker fired up Keep your Pellet Grill and Smoker fired up with Camp Chef’s Premium Hardwood Pellets. Made of 100% natural hardwood these food grade pellets are an excellent way to add that unique smoky flavor. Choose from 4 distinct flavors. Smokehouse Hickory impart a rich and smoky bacon-like flavor to all meats ...  More + Product Details Close" ]

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[ "Thread: A Guide for Python View Single Post", null, "2020-01-09, 02:13 #15 Dylan14   \"Dylan\" Mar 2017 3×191 Posts", null, "For loops part 2: The break and continue commands So far, we just dealt with how for loops work on a high level, and we constructed a few ways to set up a loop (namely, using either an iterable object or a range object). But say we want to control a loop more succinctly, and in particular, we want to have a way to \"escape\". This can be done using the break and continue commands. We will apply these to an example that is very prevalent at mersenneforum: trial factoring of Mersenne numbers. The break command: Break is a very simple command: it breaks out of a loop (which can be a for loop or a while loop (*)). The syntax is as follows: Code: break In psuedocode, this would look like this: Code: #stuff before the loop for object in a sequence: #stuff to run in the loop if condition: break #then jump out the loop and then go to the next piece of code outside the loop #stuff following the break statement #stuff after the loop If one has a nested for loop, then a break will terminate the loop in which it appears, and move to the next iteration of the next inner loop. Example of a break statement - one loop: Code: string = \"mersenneforum\" for i in string: if i != \"s\": print(i) else: break What this does is checks if a character in the string is not s. If it is not s, we print the character. Otherwise, we break out of the loop, and since nothing follows, execution stops. Nested loops and break: Code: characters = \"abcdefghijklmnopqrstuvwxyz\" for i in range(6): print(\"Outer loop iteration:\" + str(i)) for j in characters: print(j) if j == \"c\": print(\"Inner loop interrupted...\") break This code runs the outer loop, which prints the current iteration, and then the inner loop runs, which prints the alphabet up to c, and at that point the inner loop breaks, and we go to the next iteration of the outer loop. The continue command: The continue command is similar to the break command, but instead of leaving the loop, we skip whatever is left in the body of the loop and start the next iteration. In psuedocode: Code: #stuff before the loop for object in a sequence: #stuff to run in the loop if condition: continue #then go to the beginning of the code block in the loop #stuff after the continue statement #stuff after the loop The syntax is the same as with break. Example with continue: Code: string = \"mersenneforum\" for i in string: if i != \"n\": print(i) else: continue Here, we check if the character is not n, and if it is not, we print the character. Otherwise, we go to the next letter in the string. The result is that we get all the characters in mersenneforum, one character at a time, except for the two n's in that. An application: Mersenne trial factoring: We can use for loops, the continue statement, and the break statement to create a simple program to trial factor Mersenne numbers (**). First, some facts that we can use: 1. Factors of Mersennes must have the form 2*k*p+1, where p is the exponent, which is a prime number. 2. The factor must be 1 or 7 (mod 8). This reduces the search space quite a bit. So how can we implement this? Well, the first statement tells us that we could iterate through all k's from some lower bound to some upper bound, via a for loop. The second statement states that we should check if q = 2*k*p+1 is 1 or 7 mod 8 and if it isn't, discard it. So we can code this: Code: #A simple trial factor program for Mersennes p = input(\"Enter a prime number: \") mer = 2**int(p)-1 mink = 1 maxk = 10000000 for k in range(mink, maxk+1): #test all k's from mink to maxk #candidate factor is in the form 2*k*p+1 candidate = 2*k*int(p)+1 #check if the candidate is 1 or 7 (mod 8) if candidate%8 == 1 or candidate%8 == 7: #divide the mersenne number by the candidate and see if it factors it if mer%candidate == 0: #the candidate divides the mersenne print(\"M\" + str(p) + \" has a factor: \" + str(candidate)) break else: #if not, go to the next k continue This code first requests an exponent from the user. The program then generates the Mersenne number corresponding to that exponent, and then for all k's running from kmin to kmax, it first creates the corresponding candidate factor, tests if it is 1 or 7 mod 8 and if it is, checks if it divides the mersenne. If it does, it prints the factor to the screen and breaks. If not, it continues until it finds a candidate that works, or it reaches the end of the range. Test on a known factor: Code: Enter a prime number: 7112003 M7112003 has a factor: 355600151 corresponding to k = 25 (see https://www.mersenne.ca/exponent/7112003). (*) We will cover while loops in the next section. (**) Note I say simple, not fast. If you want speed, I recommend either using prime95/mprime if you have an x86_64 processor, mfactor from the mlucas suite if you have an ARM processor, factor5 for higher Mersenne's, mfaktc/mfakto for GPU's, or mmff, dm or dmdsieve + pfgw/LLR for Double Mersennes.", null, "" ]

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[ "## Tips in constructing vlookup\n\nWe talked about the syntax of vlookup and we had written a vlookup formula successfully in the previous post.  YEAH!", null, "Now we want to ride on the formula by copying it DOWN and to the RIGHT so that we could extend the vlookup for other ranks.  However, all we get is “#N/A”… why????\n\nSimply because we forgot to give Dollar to the formula so that it did not perform in the way we wanted.", null, "Tips for writing vlookup:\n\nlookup_value is usually set to Absolute Column and Relative Row (e.g. \\$G3) ==> when you copy your formula to the right to get other information, the lookup_value stays at the LEFTMOST column\n\ntable_array should always be set to both Absolute Column and Absolute Row (e.g. \\$A\\$2:\\$E\\$12) ==> logically you always look at the same data source (for expandable datasource, you should use Dynamic NamedRange or Table in 2007 or above; both topics are more advanced and may be discussed in the future)\n\ncol_index_num is a bit tricky.  If we input number into the formula, we have to change it one by one when we copy the formula to the right as number won’t change with the position of a cell.", null, "To overcome that, we first put the corresponding column numbers on the top of the table_array (which could be done fairly easy enough and be hidden later).  Then we input the reference B\\$1 as the col_index_number instead of hardcoding it by 2.  As a result, when the formula is copied to the right, it becomes C\\$1 (i.e. 3), and D\\$1 (i.e. 4) etc.", null, "With every argument set with appropriate Absolute and Relative references, you are ready to copy the formula to the Right and Down to get the correct result.  Pls observe the changes in the formula in different cells.\n\nNote: This trick works fine for retrieving data from consecutive columns.  For non-consecutive columns, we may use MATCH which will be discussed in the next post.\n\n### A brief explanation of \\$ in formulation:\n\nDollar sign \\$ is for switching a reference from Relative to Absolute, meaning it will NOT change with the position of a cell.\n\n• If you put a \\$ in front of a Letter (absolute column reference, e.g. \\$A1), it will always refer to column A even when you move your cell to the RIGHT or LEFT\n• If you put a \\$ in front of a Number (absolute row reference, e.g. A\\$1), it will always refer to row 1 even when you move your cell UP or DOWN\n• If you put \\$ in front of both Letter (absolute column) and Number (absolute row), e.g. \\$A\\$1, it will always refer to the cell A1 regarding the movement of the cell.\n• Without \\$, it is Relative and it moves along with the position of a cell: If you move your cell one column to the right, A1 will change to B1; if you move your cell one row down, A1 will become A2.\n\nAs a good habit – always consider the Relative and Absolute reference when writing a formula", null, "" ]

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[ "", null, "[ Home ] [ Up ]", null, "Welcome to the Teachers' Choice Software web site. Our FREE on line mathematics 'How-To' library is open 24 hours a day.  The ever-growing library now has 71 helpful topics, and every topic includes a free download!\n\nThis is a sample from the Maths Helper Plus on-line help applications...\n\n# Differentiate from first principles\n\nThis topic shows how to use Maths Helper Plus to differentiate a function y = f(x) numerically from first principles.\n\nThe derivative of a function y = f(x) at the point (x,f(x)) equals the gradient of the tangent line to the graph at that point. It can be defined as:", null, "where 'h' approaches zero as a limit. This diagram illustrates this concept graphically:", null, "The derivative formula (above) gives the gradient of the secant line between the two points. As the value of 'h' gets smaller, the two points get closer and the gradient of the secant approaches that of the tangent line to the curve at (x,f(x)):", null, "", null, "## 1. Load the 'Differentiate from first principles.tpl' template file\n\nNote: It is recommended that you begin with an empty Maths Helper Plus document before proceeding further. You can create a new empty document by holding down a Ctrl key and pressing the 'N' key.", null, "From the 'File' menu, select the 'Use Template' command.", null, "Choose the 'Differentiate from first principles.tpl' template file, then click the 'Open' button.\n\n## 2. Enter the function to differentiate", null, "Press the F5 key to display the parameters box.", null, "Click in the f(x) edit box, and type your function. (Leave off the 'y=' part. Just type the right side of the equation of the function.) Click the 'Update' button to update the data sets.\n\nTip: The graph scale is already set up for the function y = x² - 2x + 1. If you just want to see the demonstration, then leave this function as it is.\n\n## 3. Calculate the derivatives\n\nSymbols from the gradient formula used in this demonstration:\n\n1. 'A' is used for: 'h'\n\n2. 'X' is used for 'x'\n\nData sets are as follows:\n\n y = f(x) This data set plots a graph of the original function defined in the parameters box. (X,f(X)) (X+A, f(X+A)) These two (x,y) points draw the secant line on the graph, and display a table of calculations which include the gradient of the secant. Ignore all but the top row of values in the table. y = (f(x+A) - f(x))/A This data set plots a graph of gradient values calculated all along the y=f(x) curve. Small 'x' is used because we are not restricting the plot to just one 'x' value.", null, "Demonstration: In the parameters box, click on the 'A' edit box, then on the slider. Now use the up arrow and down arrow keys on the keyboard to change the value or 'A'. Note the effect on the derivative approximations when 'A' is: large, small, positive, negative, and zero.\n\nReturn to step '2' above and enter other functions. Investigate the form of the derivative graph and how it relates to the original function. Examples of functions to investigate would be: y=x³, y=sinx, y=cosx.\n\nÂ" ]

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[ "# 8 & 8 8 8 9 \"}40 0 8 2\n\n###### Question:\n\n8 & 8 8 8 9 \"} 40 0 8 2", null, "", null, "#### Similar Solved Questions\n\n##### Point) Use Taylor series to find the function that satisfies the following second order differential equation and initial conditions:2y + 2y = 9x, Y(0) = 0.If the solution isy = @0 + a1* + a2x2 a3x3 +a4xt +asxs a6+6 +a7x7 +enter the following coefficients:@oa2 a3 a4 asa6\npoint) Use Taylor series to find the function that satisfies the following second order differential equation and initial conditions: 2y + 2y = 9x, Y(0) = 0. If the solution is y = @0 + a1* + a2x2 a3x3 +a4xt +asxs a6+6 +a7x7 + enter the following coefficients: @o a2 a3 a4 as a6...\n##### Chapter 2, Problem 2.032 The 10-V source absorbs 2.5 mW of power. Calculate (a) Vba and...\nChapter 2, Problem 2.032 The 10-V source absorbs 2.5 mW of power. Calculate (a) Vba and (b) the power absorbed by the dependent voltage source in the figure. We were unable to transcribe this imageWe were unable to transcribe this imageWe were unable to transcribe this imag...\n##### Consider the independent investment projects in the table below. Compute the project worth of each project...\nConsider the independent investment projects in the table below. Compute the project worth of each project at the end of six years with variable MARRs as follows: 10% for n = 0 to n= 3 and 15% for n = 4 to n=6. B Click the icon to view the information about the independent investment projects. Click...\n##### (4%) Problem 9: A light bulb is immersed in a tank of liquid with refractive index...\n(4%) Problem 9: A light bulb is immersed in a tank of liquid with refractive index 1.52. A ray of light from the bulb is incident on the water/air interface at an angle of 56 from the horizontal interface. > At what angle, in degrees from the normal, does the refracted ray exit into the air? Grad...\n##### 2. (a) The data below is the annual sales, in millions of Ghana Cedis, for 21...\n2. (a) The data below is the annual sales, in millions of Ghana Cedis, for 21 pharmaceutical companies in Ghana. 8408 1374 1872 8879 2459 11413 608 14138 6452 1850 2818 1356 10498 7478 4019 4341 739 2127 3653 5794 8305 i. Provide a five-number summary (5 marks] ii. Compute the lower and upper limits...\n##### We were unable to transcribe this imagefficient rotomold oven would operate on 15,000 therms of natural...\nWe were unable to transcribe this imagefficient rotomold oven would operate on 15,000 therms of natural gas price quotes from a few suppliers, Diane determined that it would cost approximately energy-efficient rotomold oven. She determines that the expected useful life of the new oven it would have ...\n##### The figure shows the graphs of f. f', and f\": Identify each curve. a=f,buf, c =f\"...\nThe figure shows the graphs of f. f', and f\": Identify each curve. a=f,buf, c =f\" O a = f,b=f,c=f\" O a=f\", b =fc=f a = f, b=f\",c=f...\n##### 1.  What is the total number of valence electrons in one molecule of S2Cl2? A) 13                          B) 20&nb\n1.  What is the total number of valence electrons in one molecule of S2Cl2? A) 13                          B) 20        ...\n##### RR 16-29(b). Please verify your result with AMPL. Which solver will you choose to solve this problem? You don't have to submit your AMPL code: 16-29 Use convexitylconcavily t0 establish thal each of the following solutions 1 is either an un constrained global maximum Or an unconstrained global minimum of the f indicated, and explain which.(b) f(x1.X2) 500 _ S(*+ 1)2 _ 2(82 1)2 + 4r1\"_ atr (-1,0)\nRR 16-29(b). Please verify your result with AMPL. Which solver will you choose to solve this problem? You don't have to submit your AMPL code: 16-29 Use convexitylconcavily t0 establish thal each of the following solutions 1 is either an un constrained global maximum Or an unconstrained globa...\n##### Consider the following: (Letc, 9.60 pF and Cz 3.60 pF.)009.00 VFind the equivalent capacitance of the capacitors in the figure_(b) Find the charge on each capacitor: on the right 9 60 pF capacitor 28.8 PC on the left 9.60 HF capacitor 28.8 HC on the 3.60 HF capacitor 10.8 VC 21,6 on the 6.00 HF capacitor Your response differs from the correct answer by more than 100_ Double check your calculations. HC(c) Find the potential difference across each capacitor; on the right 9.60 pF capacitor on the l\nConsider the following: (Letc, 9.60 pF and Cz 3.60 pF.) 00 9.00 V Find the equivalent capacitance of the capacitors in the figure_ (b) Find the charge on each capacitor: on the right 9 60 pF capacitor 28.8 PC on the left 9.60 HF capacitor 28.8 HC on the 3.60 HF capacitor 10.8 VC 21,6 on the 6.00 HF ...\n##### An air mixture is intended to moisturize. A temperature of 20 oC with a relative humidity...\nAn air mixture is intended to moisturize. A temperature of 20 oC with a relative humidity of 70%. If you increase the relative humidity by 90%. What molar fraction of water is needed in air at 70% and 90% humidity? (Use saturation pressure)...\n##### 2. Compute the first four non-zero Taylor coefficients of the function 1/cos(3) from the Taylor coefficients...\n2. Compute the first four non-zero Taylor coefficients of the function 1/cos(3) from the Taylor coefficients of cos(x)....\n##### Evaluate using the rules of exponents.$rac{8^{5}}{8^{7}}$\nEvaluate using the rules of exponents. $\\frac{8^{5}}{8^{7}}$...\n##### Queetlon 4 manutctnng duncu thet Merburd Jmnu Nolbaics moen aiz0 0.9 Th bpo of C1p uiod 48re preoability Ehat tha semplo allne chadt arcoudi 0 Oimm? 0.9351end atindeid da-uton 0 2 mnAOU opn0,036304inst0aJ-\nQueetlon 4 manutctnng duncu thet Merburd Jmnu Nolbaics moen aiz0 0.9 Th bpo of C1p uiod 48re preoability Ehat tha semplo allne chadt arcoudi 0 Oimm? 0.9351 end atindeid da-uton 0 2 mn AOU opn 0,0363 04inst 0aJ-...\n##### 2 and 3) 1 Required information The following information applies to the questions displayed below) Delph...\n2 and 3) 1 Required information The following information applies to the questions displayed below) Delph Company uses a job-order costing system and has two manufacturing departments --Molding and Fabrication. The company provided the following estimates at the beginning of the year: Machine-hours ...\n##### Draw the product you expect from the reaction of (R) - 2 -bromooctane with CH_3CO_2. Use...\nDraw the product you expect from the reaction of (R) - 2 -bromooctane with CH_3CO_2. Use the wedge/hash bond tools to indicate stereochemistry. Include H atoms at chiral centers only. If a group is achiral, do not use wedged or hashed bonds on it....\n##### Using the following information, caiculatc the amount of you would have schedule variable cost staff: Available labor dollars 515,000,cost benefits- 52,000, and exempt (salared staff) 54,000 59,000 B.S11,000 513,000 S15,000QUESTION 21Which of the following provides US with the relatonship between the cost Payroll cost B.Payroll cost percentage Benefits percentage Labor cost percentagelabor and the sales produced?QUESTION 22The cost of the following employee benefits (iixed and variable wages med\nUsing the following information, caiculatc the amount of you would have schedule variable cost staff: Available labor dollars 515,000,cost benefits- 52,000, and exempt (salared staff) 54,000 59,000 B.S11,000 513,000 S15,000 QUESTION 21 Which of the following provides US with the relatonship between ...\n##### Test the indicaled claim about the means of two populations Assuuin that the two samples &re independent simple random samples selected from normally distributed populations Do not assume (hat the population standard deviations are equal. Use the traditional method r value method 49 indicated: rosearcher wa Interested comparing the response times of two different cab oru Cach called 50 randomly selected times The calls companics Companies Uec Iaa independenty ot the calls t0 company The res\nTest the indicaled claim about the means of two populations Assuuin that the two samples &re independent simple random samples selected from normally distributed populations Do not assume (hat the population standard deviations are equal. Use the traditional method r value method 49 indicated: r...\n##### Rationalize each denominator and simplify. Assume that all variables represent positive numbers. $\\sqrt{\\frac{2 u^{4}}{9 v}}$\nRationalize each denominator and simplify. Assume that all variables represent positive numbers. $\\sqrt{\\frac{2 u^{4}}{9 v}}$...\n##### 7. Calculate the number of moles contalning 2.22 grams of sample of KNC;HzOz: Show calculation; with correct number = significant figures and units to get credit:Calculate the number of grams from 1.710 moles of KNC H_Oz Show calculation, with correct number of significant figures and units t0 get credit:How many molccules are there in 5.72 grams of KNC H,Oz- Show calculation, with correct number of significant figures and units t0 get credit;10.How many molecule: are there in SC00 mole of KNC;H\n7. Calculate the number of moles contalning 2.22 grams of sample of KNC;HzOz: Show calculation; with correct number = significant figures and units to get credit: Calculate the number of grams from 1.710 moles of KNC H_Oz Show calculation, with correct number of significant figures and units t0 get ...\n##### APPENDEX & AUTOCORRELATIORS OF THE MULTIPLICATIE (u;dk- Wuz MODELThe dlletence equzLoa (ur Lus mlouc[urcn byMultiplyumg Kxh saAu; Lli7e expccIalions keadsPyuz 03' 0(8 0Je' 0p(- a [( 0*) + 0(0 @J( 1 97 |0kioPy01 0\"(1 0*)[0 A(a Muluplying bath sda Equatlon (I} ohtainand LiuLapakuon8a( Oo' -02*(Solving Equations (2) and (J), = 0 (1 0)! 0 = Zte0(1 e27FuhertuugcEick_eOE(4;-**,-i) ~ Fae(& Te)8a(m A)a'El*-W)08*Sokmg Equations (S) und (6) Icad ~%\"['-= 2%'\nAPPENDEX & AUTOCORRELATIORS OF THE MULTIPLICATIE (u;dk- Wuz MODEL The dlletence equzLoa (ur Lus mlouc[ urcn by Multiplyumg Kxh sa Au; Lli7e expccIalions keads Pyuz 03' 0(8 0Je' 0p(- a [( 0*) + 0(0 @J( 1 97 | 0kio Py01 0\"(1 0*)[0 A(a Muluplying bath sda Equatlon (I} ohtain and LiuL...\n##### Question (4 marks in total) You are conducting the following _ ~[est:H:0'20 ; H;:o +01 What is the conclusion ol the lest undcr the followIng Menarins (1 5 marks) 0 O0001 . (15 rarks) '298 (}mark) [s possible lind = Iest stalislic _ <07 Why?\nQuestion (4 marks in total) You are conducting the following _ ~[est: H:0'20 ; H;:o +01 What is the conclusion ol the lest undcr the followIng Menarins (1 5 marks) 0 O0001 . (15 rarks) '298 (}mark) [s possible lind = Iest stalislic _ <07 Why?...\n##### I spcfatro 13c Specfhum Nsi spechum1 N-o+t9 3ciz\nI spcfatro 13c Specfhum Nsi spechum 1 N-o+t 9 3 ciz...\n##### Researchers have sought to examine the effect of various types of music on agitation levels in patients who are in the early and middle stages of Alzheimer $disease . Patients were selected t0 participate in the study based on their stage of Alzheimer$ disease Three forms of music were tested: Easy listening, Mozart , and piano interludes. While listening tO music . agitation levels were recorded for the patients with high score indicating higher level of agitation Scores are recorded below_ P\nResearchers have sought to examine the effect of various types of music on agitation levels in patients who are in the early and middle stages of Alzheimer $disease . Patients were selected t0 participate in the study based on their stage of Alzheimer$ disease Three forms of music were tested: Eas...\n##### 9. At what distance would the repulsive force between two electrons have a magnitude of 4.0...\n9. At what distance would the repulsive force between two electrons have a magnitude of 4.0 N? (1 point) O8.0 x 10-20 m O7.6 x 10-15 m 5.8 x 10-29 m 01.9x10.5 m 10. Two like charges of the same magnitude are 1.0 mm apart. If the force of repulsion they exert (1point) upon each other is 5.0 N, what i...\n##### Nptnuaan Chae 6ne COOH 21446 CH; Caal fEdtlen Question Tot thc = theoretical yield . COOH7 W tne CH; belox 19.1 \" anart (Oint Setect ~OLate 017~etSTablllatean uenty€ F0ald 4EL4 4a544 Aadd Lhkl- CN(EDAO cONTAT moduted 'aduk= reon_ Vuc\nnpt nuaan Chae 6ne COOH 21446 CH; Caal fEdtlen Question Tot thc = theoretical yield . COOH7 W tne CH; belox 19.1 \" anart (Oint Setect ~OLate 017 ~etSTablllatean uenty€ F0ald 4EL4 4a544 Aadd Lhkl- CN (EDAO cONTAT moduted 'aduk= reon_ Vuc...\nq2.3...\n##### S e s national is considering a olne The try to g et r id or...\ns e s national is considering a olne The try to g et r id or productos de ons based on with and without the new products out the co very excited ab company's CFO, has provided the following We New Prats $11, 0 00 37, 14000 rage Botas$5.824.00 um (a) Compute the company return on p rofit magand ...\n##### Q XwlmukgieilntuIkaHUMH lapidl 7(9 zhneilzitzan uoqenba Bulmoiio} 34104 J4aueh 3u4 Wyrhalpdenkalahrkki\nq #XwlmukgieilntuIkaHUMH lapidl 7(9 zhneilzitzan uoqenba Bulmoiio} 34104 J4aueh 3u4 Wyrhalpdenkalahrkki...\n##### Cookie Creations 17 (Part Level Submission) Natalie has prepared the balance sheet and income statement of...\nCookie Creations 17 (Part Level Submission) Natalie has prepared the balance sheet and income statement of Cookie & Coffee Creations Inc. for the first year of operations, but does not understand how to prepare the statement of cash flows. The income statement and balance sheet appear below. Rec...\n##### Figure below illustrates five strong promoters and their recognition by RNA polymerasc in the process of transcription initiation B E coh sigma (nctors Hame J5 atqutnct Funclion TCACA CHTe MAa HAna Lk lu mutu Gan MCnCACh Furean ccorcT LMaalal Ban ordanmuTuc CMCu Jrenoadnio Aelenecnaa Maehens Aeaaae Guc Prpona lo m 'durd Ena palcnnpccuaT KGGue TTTD AUI crlar Oatam Eor Kaneo1. Write down consensus sequences for -35 and 10 regions_ 35 sequence: 10 sequcncc: 2. If you have promoter upstream of\nFigure below illustrates five strong promoters and their recognition by RNA polymerasc in the process of transcription initiation B E coh sigma (nctors Hame J5 atqutnct Funclion TCACA CHTe MAa HAna Lk lu mutu Gan MCnCACh Furean ccorcT LMaalal Ban ordanmuTuc CMCu Jrenoadnio Aelenecnaa Maehens Aeaaae ...\n##### Which of the following are two linearly independent solutions to the differential y\" + 2y + 3y = 0? equationy1 (t) = e ' sin(Vzt); y2(t) = e cos( V Ft) y1 (t) = e ' sin( V2t); V2 (t) = 3e sin( V2t) 91 (t) = 'sin(VBt); y2 (t) cos( V3t) 91 (t) Fe ' sin(t); y2 (t) = e ' cos(t)\nWhich of the following are two linearly independent solutions to the differential y\" + 2y + 3y = 0? equation y1 (t) = e ' sin(Vzt); y2(t) = e cos( V Ft) y1 (t) = e ' sin( V2t); V2 (t) = 3e sin( V2t) 91 (t) = 'sin(VBt); y2 (t) cos( V3t) 91 (t) Fe ' sin(t); y2 (t) = e ' c...\n##### E) none of the above un equilibrium occurs les intersect. 26. In the Keynesian model, short-run...\nE) none of the above un equilibrium occurs les intersect. 26. In the Keynesian model, short-run egun A) where the IS and LA curves intersect. Where the IS curve. Meurve. and FE lines inters C) where the IS curve intersects the FB fine. D) where the LM curve intersects the Fence he money supply will ..." ]

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[ "# Using a Linear Programming Relaxation Example\n\nIf you’re anything like me, you’ve done a lot of programming where you must “play the game” using linear and/or non-deterministic rules. I’ve found myself stuck in situations where I had to solve a problem based on a set of predefined parameters. And oftentimes when I’m programming in my Datajack Notebook, I can spend many hours – sometimes days or weeks – depending on how complex the problem is, only to come up with little more than a fraction of what I was hoping to accomplish. In this article, I will present some linear programming relaxation example to show how one can easily overcome such difficulties. Hopefully, you will find these examples to be inspiring and useful in your programming endeavors.\n\nBefore we get started, let’s take a look at what linear programming relaxation is. According to Wikipedia,” Linear programming is a probabilistic process for solving linear problems, which are characterized by a series of inputs that generate an output value. This process yields a desired output value given the input data, without requiring any knowledge of the actual external physical conditions”. This definition may seem quite abstract, but the reality is that this idea has been well documented in scientific literature for decades.\n\nIn a nutshell, linear programming relaxation describes the use of mathematical functions that generate output values independently of the actual inputs they are evaluated against. This is most commonly used in machine learning and optimization. For instance, we can easily generate positive results from linear programs by feeding random numbers into the program; the results will then be used to create predictions about future outcomes. This concept of linear programming can also be applied to non-linear systems where the output function would somehow be affected by the input to the system, or even vice-versa. For instance, if we want to know if our stock prices are set to go up or down in the near future, we could use linear programming to infer this information, or to simply make an educated prediction based on past trends.\n\nOf course, there are many more applications for linear programming, such as in finance, optimization and algorithms. In these cases, the function being linear is not necessarily a single function, but rather a series of linear programs that are loosely-coupled to one another. This allows the developer to more easily visualize the relationship between the variables of the linear programs, and the real world financial parameters. Of course, in this case, the developer is also able to make more detailed predictions about how these factors will behave over time.\n\nSo how does the linear programming relaxation example actually work? Basically, a relaxation example starts out with some initial data and then begins to generate random numbers that satisfy some arbitrary equation. Once these numbers are generated, they are then fed into the linear programming code. This code uses some well-known optimization techniques to identify areas of low profitability and then uses the resulting information to further refine the algorithm.\n\nThe beauty of the example used above is that it’s completely synthetic – it uses completely random numbers, so the results cannot be compared to any financial indicator you’re likely to use in the real world. It’s only job is to generate numbers that are likely to give you an optimal result when coupled with certain inputs. As a result, the final output is a random number sequence that can’t fail to give you a good result.\n\nThis relaxation example is not the only application for linear programming. In fact, the random number generator is usually generated using a greedy algorithm. This simply means that the more numbers you pass through the function – the higher the probability that your final result will be a profitable investment. Of course, even if your numbers are profitable – it doesn’t necessarily mean that you’ll make money from them.\n\nObviously, this isn’t a stock trading simulation, and the financial results are definitely not what you want to look at. However, it still serves as a good way to get a feel for linear programming. It allows you to think about how the numbers generate the outcome, and what you can do with them to get the highest returns. This is the basis of many types of trading, and it’s the basis for many types of quantitative trading as well. If you find that you’ve found a method that you’re comfortable with, but can’t put your finger on exactly what it is – then it might be worth looking into linear programming as a possible solution." ]

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[ "# APTS module: Applied Stochastic Processes\n\nModule leader: N Georgiou & M Roberts\n\nPlease see the full Module Specifications for background information relating to all of the APTS modules, including how to interpret the information below.\n\nAims: This module will introduce students to two important notions in stochastic processes — reversibility and martingales — identifying the basic ideas, outlining the main results and giving a flavour of some of the important ways in which these notions are used in statistics.\n\nLearning outcomes: A student successfully completing this module will be able to:\n\n• describe and calculate with the notion of a reversible Markov chain, both in discrete and continuous time;\n• describe the basic properties of discrete-parameter martingales and check whether the martingale property holds;\n• recall and apply significant concepts from martingale theory (indicative list: optional stopping, martingale convergence);\n• explain how to use Foster-Lyapunov criteria to establish recurrence and speed of convergence to equilibrium for Markov chains.\n\nPrerequisites: Preparation for this module should include a review of the basic theory and concepts of Markov chains as examples of simple stochastic processes (transition and rate matrices, irreducibility and aperiodicity, equilibrium equations and results on convergence to equilibrium), and with the definition and basic properties of the Poisson process (as an example of a simple counting process).\n\nFurther reading: Various useful textbooks at increasing levels of mathematical sophistication:\n\n• Haggstrom (2002) “Finite Markov chains and algorithmic applications”.\n• Grimmett and Stirzaker (2001) “Probability and random processes”.\n• Breiman (1992) “Probability”.\n• Norris (1998) “Markov chains”.\n• Ross (1996) “Stochastic processes”.\n• Williams (1991) “Probability with martingales”.\n\nSome useful texts that are free on the web:\n\nTopics:\n\n• Reversibility of Markov chains in both discrete and continuous time, computation of equilibrium distributions for such chains, application to important examples.\n• Discrete time martingales, examples, application, super-martingales, sub-martingales.\n• Stopping times, statements and applications of optional stopping theorem, martingale convergence theorem.\n• Recurrence and rates of convergence for Markov chains, application to important examples.\n• Statements and applications of Foster-Lyapunov criteria, viewed using the language of martingales.\n• Statistical applications and relevance (highlighted where appropriate throughout).\n\nAssessment:\n\n• Complete appropriate exercises that are simple developments or extensions of aspects of the results in the module." ]

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[ "Hi, I want to create multiplication table in PHP - how can I do it? From 1 to 10\n\n10x\n\nHere's mine...In line #19 I've used concatanation, instead of \\$result variable creation.\n\n```<!doctype html>\n<html lang=\"en\">\n<meta charset=\"UTF-8\">\n<title>Multiplication Table in Php</title>\n<body>\n\n<?php\n\n\\$x = 1; //rows\necho \"<table border = '2'>\\n\";\n\nwhile ( \\$x <= 10 ) {\necho \"\\t<tr>\\n\";\n\n\\$y = 1;//columns\nwhile ( \\$y <= 10 ) {\necho \"\\t\\t<td>\\$x * \\$y = \" . \\$x * \\$y . \"</td>\\n\";//the result of the multiplication\n\\$y++;\n}\n\necho \"\\t</tr>\\n\";\n\\$x++;\n}\n\necho \"</table>\";\n\n?>\n\n</body>\n</html>```", null, "selected by user golearnweb\n\nHere is my solution:\n\n```<!doctype html>\n<html lang=\"en\">\n<meta charset=\"UTF-8\">\n<title>PHP Multiplication Table</title>\n<body>\n\n<?php\n\n\\$i = 1;\necho \"<table border='1'>\\n\";\nwhile ( \\$i <= 10 ) {\necho \"\\t<tr>\\n\";\n\\$n = 1;\nwhile ( \\$n <= 10 ) {\n\\$result = \\$n * \\$i;\necho \"\\t\\t<td>\\$n * \\$i = \\$result</td>\\n\";\n\\$n++;\n}\necho \"\\t</tr>\\n\";\n\\$i++;\n}\n\n?>\n\n</body>\n</html>```" ]

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[ "## Questions and Answers", null, "Time-dependent NEGF\n\nIn the time-dependent NEGF equation, given a sigma_in(t,t’) due to the dot, I am getting an I-V equation that is making it difficult for me to group terms. For instance, looking at the analogue of the first term of I = 2q/hbar. int dE TrGin I get I1 = q/ihbar. int dt1 TrSigma_in.G^dagger-G.Sigma_in\n\nNow, because of the different order of the time arguments in the G and G^dagger terms, I cannot pool these to write it as a spectral function A = i(G-G^dagger). Similarly, I can’t seem to take Sigma – Sigma^dagger to get Gamma. Is this correct?" ]

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[ "# A Bisecting Angle\n\nThis is a bit of fun geometry that doesn’t have much to do with what’s going on in class, but does reflect on mathematical thinking.\n\nAn article promoting the use of technology in the classroom began: “Draw a perfect circle. Now bisect that with a 45-degree angle, the perfect slice of geometric pizza. Now, using your drawing, find the area of the rest of the circle. ”\n\nWhat they meant was that the 45-degree angle divides the circle (or, more properly speaking, the region contained by the circle*). “Bisect” has a rigorous definition in mathematics: It means to divide an object into two objects of the same exact size. So if we bisect a circular region with anything, then the area of the two pieces will have half the area of the circular region, by definition.\n\nRather than accepting this and moving on, though, I asked myself: Can you bisect a circular region with a 45-degree angle? I shifted this to: Given a circle and an angle whose vertex is on or in the circle, what is the smallest angle that will cut out half of its region?\n\nI’m going to start with a conjecture: Our target angle is an inscribed angle bisected by a diameter. In other words, here’s a picture of circle A, which has a radius of one unit:", null, "My conjecture is that the shaded region, where angles BEA and CEA are congruent,  represents the largest portion of the circular region that can be covered by angle BEC. Let $$\\alpha = m\\angle BEC$$.\n\nSo what is the area of the shaded region? First, we have sector BAC. The area of a sector is $$\\beta r^2/2$$, where $$\\beta$$ is the measure of the central angle (in radians). Since it’s a unit circle, $$r^2 = 1$$. Since angle BEC is the inscribed angle that corresponds to the central angle BAC, it has half the measure, and the area of the sector is $$\\beta/2 = \\alpha$$.\n\nSince AE bisects angle BEC, the two triangles are congruent. Since AC, AE, and AB are all radii, the two triangles are isosceles. We can determine the area of triangle BAE by first dropping a line perpendicular to BE. This divides the triangle into two congruent right triangles, AFE and AFB.", null, "Call $$\\gamma = m\\angle AEF$$, so $$\\gamma = \\alpha/2$$. To find the area of $$\\Delta AFE$$, we need a height and a width. Since AE is a radius, the height AF is $$\\sin\\gamma$$ and the width FE is $$\\cos\\gamma$$. This gives an area of $$\\sin\\gamma\\cos\\gamma/2$$.\n\nI’ll write about the double- and half-angle trigonometric formulas in a separate post, but one of these is: $\\sin 2\\theta = 2\\sin\\theta\\cos\\theta$\n\nApplying this gives us an area of triangle AEF of $$(\\sin 2\\gamma)/4 = (\\sin\\alpha)/4$$. Since there are four such congruent triangles, the total area of the two larger shaded triangles is $$\\sin\\alpha$$, and the area of the entire shaded region is $$\\sin\\alpha + \\alpha$$.\n\nIf the shaded region is half the area bounded by the circle, and since the area bounded by a unit circle is $$\\pi$$, $$\\sin\\alpha + \\alpha = \\pi/2$$. I’m not sure how or if that can be solved analytically, but we can use a calculator to graph $$\\sin\\alpha + \\alpha – \\pi/2$$ and find its solution.", null, "The solution function of the calculator gives a zero at 0.8317112.\n\nThroughout this post, I have worked in radians. For the Algebra 2 students, we haven’t gotten there yet. Radians represent a different way to measure angles, but it’s a straightforward conversion: $$2\\pi = 360^o$$, and $$1 = 180^o/\\pi$$. So we have $$0.8317112 \\times 180^o/\\pi \\approx 47.65^o$$.\n\nSo: An inscribed angle slightly larger than 47.65 degrees bisected by a diameter will create a rounded wedge that bounds half as many points as a circle.\n\nGoing back to the original inspiration for this item, this means that the 45 degree angle with a vertex that is inside a circle cannot bisect the circle’s region.\n\nI started with a conjecture. This conjecture can be proven, but the formal proof requires more trigonometry, so I’ll leave it for the reader, or for another time.\n\n* Since this article is about rigorous definitions, I’m using “circle” to refer to the set of points equidistant from a point (i.e., $$(x-h)^2 + (y-k)^2 = r^2$$) and “circular region” to refer to all points satisfying $$(x-h)^2 + (y-k)^2 \\le r^2$$. In high school, we often conflate these two with the word “circle”.\n\n# The Derivative of the Sine and Cosine Functions\n\nIn AP Calculus, we saw the following derivatives: $\\sin’ x = \\cos x \\\\ \\cos’ x = -\\sin x$\n\nThat is, the derivative of the sine is the cosine, and the derivative of the cosine is the opposite of the sine.\n\nI showed one way to see why this is true using Desmos. The purple line is the tangent, while the black dot is the slope of that tangent. As you move the slider a from side to side, the black dot traces the cosine function.\n\nThe book has an algebraic proof of the first rule (p. 112), using the limit process. In this post, I’ll discuss a more visual, geometry-based explanation.\n\nWe’ll start with a few triangles and a portion of the unit circle.", null, "The angle $$\\theta$$ represents our original angle, while $$\\delta = \\Delta\\theta$$. Note that we are NOT looking at the change in the $$x$$ value; we’re looking at the change in the angle measurement. This is key!\n\nSo what we want to know is: How does $$\\sin\\theta$$ change as $$\\theta$$ changes?\n\n$$\\overline{AC}$$ is the radius of the unit circle, so its length is 1; $$\\sin\\theta = m\\overline{CF}/m\\overline{AC} = f$$. Likewise, $$\\cos\\theta = g$$.\n\n$$\\overline{EC}$$ is on the tangent to the circle at point $$C$$. $$\\overline{CD}$$ is collinear with $$\\overline{CF}$$; $$\\overline{ED}\\perp\\overline{CD}$$.\n\nSo let’s consider the blue and green triangles. Using geometry, we know that since the blue triangle is a right triangle, $$\\theta$$ and $$\\gamma$$ are complementary. Since $$\\overline{DF}$$ is on a line, and since $$\\angle ECA$$ is right, $$\\alpha$$ and $$\\gamma$$ are complementary… meaning that $$\\alpha\\cong\\theta$$.\n\nMore about the green triangle: $$\\cos\\alpha = \\cos\\theta = e/c \\Rightarrow e = c \\cos\\theta$$. But $$c$$ is also opposite $$\\delta$$ in $$\\triangle ECA$$, meaning that $$\\sin\\delta = c/a \\Rightarrow c = a\\sin\\delta$$ and $$e = a\\sin\\delta\\cos\\theta$$.\n\nWe’re ready to write our limit now. We want to know what happens to $$e$$ when $$\\delta$$ gets smaller and smaller. Specifically: $\\lim_{\\Delta\\theta\\rightarrow 0}\\frac{\\Delta y}{\\Delta\\theta} = \\lim_{\\delta\\rightarrow 0}\\frac{e}{\\delta} \\\\ = \\lim_{\\delta\\rightarrow 0}\\frac{a\\sin\\delta\\cos\\theta}{\\delta} \\\\ = \\lim_{\\delta\\rightarrow 0} a \\cdot \\lim_{\\delta\\rightarrow 0}\\frac{\\sin\\delta}{\\delta} \\cdot \\lim_{\\delta\\rightarrow 0} \\cos\\theta$\n\nConsider each in turn. As $$\\delta$$ gets smaller, the difference between $$a$$ and the radius of the unit circle gets smaller. That is, $\\lim_{\\delta\\rightarrow 0} a = 1$\n\nFrom our examination of the Squeeze Theorem, we know the important identity $\\lim_{x\\rightarrow 0}\\frac{\\sin x}{x} = 1$ which we can apply here (where $$x = \\delta$$).\n\nFinally, since $$\\cos\\theta$$ is not affected by $$\\delta$$ at all, $\\lim_{\\delta\\rightarrow 0} \\cos\\theta = \\cos\\theta$\n\nPutting these together, we get  $\\lim_{\\delta\\rightarrow 0} a \\cdot \\lim_{\\delta\\rightarrow 0}\\frac{\\sin\\delta}{\\delta} \\cdot \\lim_{\\delta\\rightarrow 0} \\cos\\theta = 1 \\cdot 1 \\cdot \\cos\\theta = \\cos\\theta$ QED.\n\nThe process for demonstrating the cosine is very similar (using $$d = a\\sin\\delta\\sin\\theta$$), but the question that comes up is: Why is the cosine’s derivative the opposite of the sine? Notice what happens in the green triangle: While the height goes upward, the width (corresponding to $$\\sin\\alpha = \\sin\\theta$$) goes to the left. This is why the cosine’s derivative is the opposite of the sine.\n\n# Welcome to 2018!\n\nThis website has been updated for the new school year.\n\nIf you want to know what you missed or review the notes, select the appropriate class.\n\nThis tab will contain announcements and some additional thoughts.\n\nAll posts that are older than this one are for previous years. You’re welcome to go through them if you’re curious.\n\n# Equations, expressions, and functions\n\nThree things that students of algebra often confuse are the notions of equations, expressions, and functions.\n\n### Expressions\n\nAn expression is any set of constant numeric values, variables, and operations. The idea is that, if we know the values of all the variables at a given point, we can determine the numeric value of the entire expression.\n\nSome expressions don’t have any variables at all. This usually means that they have a specific, unchanging mathematical value. The only exception to this is when some portion of the expression breaks math. Dividing by zero, for instance, always breaks math, so $$5/0$$ has no mathematical value. Taking square roots of negative numbers leads to mathematical values, but not real ones, so depending on our goal, $$\\sqrt(-5)$$ may not have a valid value.\n\nSince some values can create problems, this means that some expressions with variables have specific values or sets of values that create problems. For instance, if we’re limiting ourselves to real numbers, $$\\sqrt(x)$$ prohibits $$x$$ from being negative. And $$1/x$$ prohibits $$x$$ from being zero.\n\n### Equations\n\nAn equation is a statement of fact about two expressions. An equation is true for any values of variables that make the two expressions have the same mathematical value, and otherwise it is false.\n\nFor instance, consider $$x^2 = 4$$. This is true for any numbers which, when squared, have a value of 4. These are 2 and -2. Those are the only values that make that statement true.\n\nWe can have variables on both sides of the equation. Consider $$x^2 = x + 2$$. This is also true for two values, 2 and -1.\n\nIf an equation involves two or more variables, it can be true for an infinite number of values. For instance, $$y = x$$ is true for all pairs $$(x, y)$$ where $$x$$ and $$y$$ have the same value.\n\n### Functions\n\nA function is a relationship between sets of data. It is often described as being a machine: If you put a specific value into the machine, you can predict exactly what output you’ll get.\n\nIt can very often by described by an expression, but it is rarely described by an equation. This is confusing because we normally state a function by giving an equation. But let’s take a careful look at a function. Here’s an example: $f(x) = x^2 + 7$\n\nThis is an equation, but it’s connecting two expressions. One expression consists of the name of a function ($$f$$) and its input ($$x$$). The other expression describes what operations are going to be applied to the input. The expression on the right is the actual function; the expression on the left gives its name.\n\nConsider this: “My car is a silver Honda.” What is my car? It’s not the entire sentence “My car is a silver Honda.” That would be silly. It’s the object that both “my car” and “a silver Honda” refer to. In a similar way, $$f(x)$$ and $$x^2 + 7$$ are two ways to refer to a machine that takes any value, squares it, and adds seven to the result.\n\nWe use $$f(x)$$ when we want to make general comments, or when we want to make other statements about the function. We use the specific form of the function ($$x^2 + 7$$) when we want to see what it actually does.\n\nToday in class I presented a technique for factoring quadratics. Here are the steps. Remember that we’re starting from the right. This only works when the coefficient on the $$x^2$$ term is 1.\n\n1. List the pairs of numbers that multiply to the constant (ignoring the sign!).\n2. The sign on the constant tells us whether we’re looking for a sum or a difference.\n3. The coefficient on the $$x$$ term (ignoring the sign!) tells us what sum or difference we’re looking for.\n4. Now, using the sign on the $$x$$ term, decide the signs for the factors.\n1. The higher factor will use the same sign.\n2. The lower factor will use the same sign if it’s a sum, and the opposite sign if it’s a difference.\n\nHere are some examples.\n\n$x^2 + 7x + 12$\n\n1. $$12 = 1 \\times 12 = 2 \\times 6 = 3 \\times 4$$\n2. We’re looking for a sum. Our options are:\n1. $$1 + 12 = 13$$\n2. $$2 + 6 = 8$$\n3. $$3 + 4 = 7$$\n3. Our sum is 7. That means we want to use 3 and 4.\n4. We’re using +. So our factors are:\n1. $$x + 4$$\n2. $$x + 3$$\n\nThat means that $$x^2 + 7x + 12 = (x + 3)(x + 4)$$.\n\n$x^2 – 2x – 63$\n\n1. $$63 = 1 \\times 63 = 3 \\times 21 = 7 \\times 9$$\n2. We’re looking for a difference. Our options are:\n1. $$63 – 1 = 62$$\n2. $$21 – 3 = 18$$\n3. $$9 – 7 = 2$$\n3. Our difference is 2. That means we want to use 7 and 9.\n4. We’re using -. So our factors are:\n1. $$x – 9$$\n2. $$x + 7$$\n\nThat means that $$x^2 – 2x – 63 = (x – 9)(x + 7)$$.\n\nRemember that the solutions will have the opposite signs to the factors, because solutions are values that make the factors equal to zero. So, in our first example, our solutions are $$x = {-3, -4}$$, while in the second example, they’re $$x = {9, -7}$$.\n\nIf this method doesn’t work, it means that at least one of the solutions of the quadratic expression isn’t an integer.\n\n# Week of Apr 23, 2018\n\nPowerpoints: Apr 25 Apr 24 Apr 23\n\nMonday\nTopic: Graphing Rational Functions\nHomework: Rational Functions Graphing 2\n\nTuesday\nTopic: Simplifying and Multiplying Rational Functions\nHomework: Simplifying and Multiplying Rational Functions\n\nWednesday\nTopic: Simplifying, Multiplying, and Dividing Rational Functions\n\n# Week of Apr 23, 2018\n\nPowerpoints: Apr 25 Apr 24 Apr 23\n\nMonday\nTopic: Graphing Rational Functions\nHomework: Rational Functions Graphing 2\n\nTuesday\nTopic: Simplifying and Multiplying Rational Functions\nHomework: Simplifying and Multiplying Rational Functions\n\nWednesday\nTopic: Simplifying, Multiplying, and Dividing Rational Functions\n\n# Week of Apr 16, 2018\n\nPowerpoints: Apr 20 Apr 19 Apr 16\n\nMonday\nTopic: Direct and Inverse Variation\nClasswork: Direct and Inverse Variation\n\nTuesday\nGraphing (Mr. Hartzer absent)\nClasswork: Graphing Polynomials and Rational Functions (Kuta)\n\nWednesday\nPolynomials (Mr. Hartzer absent)\nClasswork: Polynomial Review (Kuta)\n\nThursday\nTopic: Rational Functions\n\nFriday\nTopic: Graphing Rational Functions\nClasswork: Rational Functions Graphing 1" ]

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[ "Formal Education\nInformal Education\nEducation Collaborations\nWeb Shortcuts\nSpace Math @ NASA\n\nSpaceMath@NASA introduces students to the use of mathematics in today's scientific discoveries. Through press releases and articles, Space Math explores how many kinds of mathematics skills come together in exploring the Universe.", null, "A Pulsar Shot Out from a Supernova Explosion! (problem #680)Students study the speed of a pulsar ejected from a supernova explosion, and describe what would happen if the dense star collided with a star like the sun. [Grade: 6-8 | Topics: Scientific notation; speed=distance/time; unit conversions; density] (PDF) [Press Release]", null, "Exploring the Evaporating Exoplanet HD189733b (problem #561)Students estimate how quickly this planet will lose its atmosphere and evaporate at its present loss rate of 6 million tons/second [Grade: 6-8 | Topics:mass=densityx volume; rates; volume of a sphere ] (PDF) [Press Release]", null, "Giant Gas Cloud in System NGC 6240 (problem #511)Students use scientific notation and volume of sphere to estimate the density of the gas cloud, and the number of hydrogen atoms per cubic meter. [Grade: 8-10 | Topics:Volume of a sphere; scientific notation; unit conversion ] (PDF) [Press Release]", null, "Chandra Sees a Distant Planet Evaporating (problem #439)The planet CoRot2b is losing mass at a rate of 5 million tons per second. Students estimate how long it will take for the planet to lose its atmosphere.[Grade: 6-8 | Topics: Scientific Notation; RAte = Amount/Time] (PDF) [Press Release]", null, "Estimating the Size and Mass of a Black Hole (problem #417)Students use a simple formula to estimate the size of a black hole located 3.8 billion light years from Earth, recently studied by NASA's Chandra and Swift satellites. [Grade: 8-10 | Topics: distance=speed x time] (PDF) [Press Release]", null, "The Crab Nebula - Exploring a pulsar up close! (problem #398)Students work with a photograph to determine its scale and the time taken by light and matter to reach a specified distance. [Grade: 6-8 | Topics: Scale drawings; unit conversion; distance = speed x time] (PDF) [Press Release]", null, "X-rays from hot gases near the black hole SN1979c (problem #390) Students use two functions to estimate the size of a black hole from the gas emitting x-rays which is flowing into it. [Grade: 8-10 | Topics: Functions; substitution; evaluation] (PDF) [Press Release]", null, "Estimating the diameter of the SN1979c black hole (problem #389) Students use simple equations to learn about the various definitions for the sizes of black holes in terms of their event horizons, last photon orbit, and last stable particle orbit radii, and apply this to the recently discovered 'baby' black hole in the galaxy M-100. [Grade: 6-8 | Topics: evaluating linear functions; integer math; metric units] (PDF) [Press Release]", null, "Chandra Studies an Expanding Supernova Shell (problem #314) Using a millimeter ruler and a sequence of images of a gaseous shell between 2000 and 2005, students calculate the speed of the material ejected by Supernova 1987A. [Grade: 6-9 | Topics: Measuring; Metric Units; speed=distance/time] (PDF) [Press Release]", null, "Chandra Spies the Longest Sound Wave in the Universe (problem #289)Students use an image of sound waves produced by a massive black hole to determine wavelength, and comparisons with musical scale to find how many octaves this sound wave is below the wavelength of middle-C. [Grade: 6-8 | Topics: metric measurement; scaling; Scientific Notation; exponents] (PDF) [Press Release]", null, "Chandra Sees the Most Distant Cluster in the Universe (problem #285) Students work with kinetic energy and escape velocity to determine the mass of a distant cluster of galaxies by using information about its x-ray light emissions. [Grade: 9-12 | Topics: Algebra I; Solving for X; Scientific notation] (PDF) [Press Release]", null, "Chandra Observatory Sees the Atmosphere of a Neutron Star (problem #283) Students determine the mass of the carbon atmosphere of the neutron star Cas-A. [Grade: 8-10 | Topics: Volume of spherical shell; mass = density x volume] (PDF) [Press Release]", null, "The Hand of Chandra (problem #234) Students use an image from the Chandra Observatory to measure a pulsar ejecting a cloud of gas. [Grade: 6-8 | Topics: Scientific Notation; proportions; angle measure] (PDF) [Press Release]", null, "Exploring Angular Size (problem #144) Students examine the concept of angular size and how it relates to the physical size of an object and its distance. A Chandra Satellite x-ray image of the star cluster NGC-6266 is used, along with its distance, to determine how far apart the stars are based on their angular separations. [Grade: 7 - 10 | Topics: Scientific Notation; degree measurement; physical size=distance x angular size.] (PDF) [Press Release]" ]

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[ "Select Page\n\n# PHYS 501 Mathematical Methods in Physics (Graduate)\n\nThis course develops a mathematical foundation to succeed in graduate level courses in classical mechanics, electrodynamics, thermodynamics/statistical physics, and modern and quantum physics. It encompasses algorithmic skills but aims higher to develop the ability to relate mathematics and phenomena as well as the ability to analyze solutions for limitations and prediction of behavior. Note: This course is designed for those seeking the credentials required by many regional accrediting bodies in order to be able to teach advanced placement, concurrent early college, and community college physics courses.\n\nUpon successful completion of this course, you should be able to:\n\n1. Demonstrate problem solving competency in the mathematical concepts and techniques used in theoretical physics.\n2. Translate physical problems to mathematical formulations and mathematical solutions to physical behavior.\n3. Identify recurring patterns of mathematical concepts and techniques across the areas of theoretical physics.\n4. Evaluate the suitability and limitations of a range of mathematical approaches to physical problems.\n5. Use technology (e.g. Mathematica, Maple) to solve differential equations and linear algebra problems and visualize Fourier, Legendre and Bessel that arise in applications to waves, heat flow and quantum mechanics using technology (e.g. Mathematica, Maple).\n6. Express the relevance of mathematics to the physical world in terms of creation and God’s providence (no grading associated with this topic).\n\nPrerequisite Courses: None\n\nPrerequisite Skills and Knowledge: A bachelor’s degree with a physics major or state certification (in any state) to teach physics at a secondary school level. Undergraduate coursework must include calculus (through multivariate) and ordinary differential equations.\n\nSequential Offerings:\nPhysics 501 Mathematical Methods in Physics (Spring 2020 & Fall I 2021)\nPhysics 502 Classical Mechanics (Summer I 2020)\nPhysics 503 Electromagnetism (Summer II 2020)\nPhysics 504 Intro to Quantum Mechanics (Fall 2020)\nPhysics 505 Quantum Mechanics II (Spring 2021)\nPhysics 506 Thermodynamics and Statistical Mechanics (Summer 2021)\n\nIndiana Wesleyan University" ]

[ null ]

[ "", null, "", null, "• Matrices\n• Algebra\n• Geometry\n• Funciones\n• Trigonometry\n• Coordinate geometry\n• Combinatorics\n Suma y resta Producto por escalar Producto Inversa\n Monomials Polynomials Special products Equations Quadratic equations Radical expressions Systems of equations Sequences and series Inner product Exponential equations Matrices Determinants Inverse of a matrix Logarithmic equations Systems of 3 variables equations\n 2-D Shapes Areas Pythagorean Theorem Distances\n Graphs Definition of slope Positive or negative slope Determine slope of a line Ecuación de una recta Equation of a line (from graph) Quadratic function Posición relativa de dos rectas Asymptotes Limits Distancias Continuity and discontinuities\n Sine Cosine Tangent Cosecant Secant Cotangent Trigonometric identities Law of cosines Law of sines\n Ecuación de una recta Posición relativa de dos rectas Distancias Angles in space Inner product\n\n# Law of sines\n\nThe Law of Sines is the relationship between sides and angles in any triangle.\n\nThe sides of a triangle are to one another in the same ratio as the sines of their opposite angles.\n\nLet's see...\n\nConsider this triangle:", null, "Then, the Law of Sines states that:", null, "The law can also be written as the reciprocal:", null, "Proof:\n\nThe perpendicular, oc, splits this triangle into two right-angled triangles. This lets us calculate h in two different ways\n\n• Using the triangle cao gives", null, "• Using the triangle cbo gives", null, "• Eliminate h from these two equations", null, "• Rearrange", null, "By using the other two perpendiculars the full law of sines can be proved. QED.\n\nFind c", null, "= ; c·sin50=10·sin88; c= ; c=13.05\n\nFind the value of the missing angle (Round the solution to hundredths).", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "=10", null, "=3", null, "", null, "", null, "", null, "", null, "=30 =?", null, "The value of the missing side is:" ]

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[ "# How to Calculate the KL Divergence for Machine Learning\n\nLast Updated on November 1, 2019\n\nIt is often desirable to quantify the difference between probability distributions for a given random variable.\n\nThis occurs frequently in machine learning, when we may be interested in calculating the difference between an actual and observed probability distribution.\n\nThis can be achieved using techniques from information theory, such as the Kullback-Leibler Divergence (KL divergence), or relative entropy, and the Jensen-Shannon Divergence that provides a normalized and symmetrical version of the KL divergence. These scoring methods can be used as shortcuts in the calculation of other widely used methods, such as mutual information for feature selection prior to modeling, and cross-entropy used as a loss function for many different classifier models.\n\nIn this post, you will discover how to calculate the divergence between probability distributions.\n\nAfter reading this post, you will know:\n\n• Statistical distance is the general idea of calculating the difference between statistical objects like different probability distributions for a random variable.\n• Kullback-Leibler divergence calculates a score that measures the divergence of one probability distribution from another.\n• Jensen-Shannon divergence extends KL divergence to calculate a symmetrical score and distance measure of one probability distribution from another.\n\nKick-start your project with my new book Probability for Machine Learning, including step-by-step tutorials and the Python source code files for all examples.\n\nLet’s get started.\n\n• Update Oct/2019: Added a description of the alternative form of the equation (thanks Ori).", null, "How to Calculate the Distance Between Probability Distributions\nPhoto by Paxson Woelber, some rights reserved.\n\n## Overview\n\nThis tutorial is divided into three parts; they are:\n\n1. Statistical Distance\n2. Kullback-Leibler Divergence\n3. Jensen-Shannon Divergence\n\n## Statistical Distance\n\nThere are many situations where we may want to compare two probability distributions.\n\nSpecifically, we may have a single random variable and two different probability distributions for the variable, such as a true distribution and an approximation of that distribution.\n\nIn situations like this, it can be useful to quantify the difference between the distributions. Generally, this is referred to as the problem of calculating the statistical distance between two statistical objects, e.g. probability distributions.\n\nOne approach is to calculate a distance measure between the two distributions. This can be challenging as it can be difficult to interpret the measure.\n\nInstead, it is more common to calculate a divergence between two probability distributions. A divergence is like a measure but is not symmetrical. This means that a divergence is a scoring of how one distribution differs from another, where calculating the divergence for distributions P and Q would give a different score from Q and P.\n\nDivergence scores are an important foundation for many different calculations in information theory and more generally in machine learning. For example, they provide shortcuts for calculating scores such as mutual information (information gain) and cross-entropy used as a loss function for classification models.\n\nDivergence scores are also used directly as tools for understanding complex modeling problems, such as approximating a target probability distribution when optimizing generative adversarial network (GAN) models.\n\nTwo commonly used divergence scores from information theory are Kullback-Leibler Divergence and Jensen-Shannon Divergence.\n\nWe will take a closer look at both of these scores in the following section.\n\n### Want to Learn Probability for Machine Learning\n\nTake my free 7-day email crash course now (with sample code).\n\nClick to sign-up and also get a free PDF Ebook version of the course.\n\n## Kullback-Leibler Divergence\n\nThe Kullback-Leibler Divergence score, or KL divergence score, quantifies how much one probability distribution differs from another probability distribution.\n\nThe KL divergence between two distributions Q and P is often stated using the following notation:\n\n• KL(P || Q)\n\nWhere the “||” operator indicates “divergence” or Ps divergence from Q.\n\nKL divergence can be calculated as the negative sum of probability of each event in P multiplied by the log of the probability of the event in Q over the probability of the event in P.\n\n• KL(P || Q) = – sum x in X P(x) * log(Q(x) / P(x))\n\nThe value within the sum is the divergence for a given event.\n\nThis is the same as the positive sum of probability of each event in P multiplied by the log of the probability of the event in P over the probability of the event in Q (e.g. the terms in the fraction are flipped). This is the more common implementation used in practice.\n\n• KL(P || Q) = sum x in X P(x) * log(P(x) / Q(x))\n\nThe intuition for the KL divergence score is that when the probability for an event from P is large, but the probability for the same event in Q is small, there is a large divergence. When the probability from P is small and the probability from Q is large, there is also a large divergence, but not as large as the first case.\n\nIt can be used to measure the divergence between discrete and continuous probability distributions, where in the latter case the integral of the events is calculated instead of the sum of the probabilities of the discrete events.\n\nOne way to measure the dissimilarity of two probability distributions, p and q, is known as the Kullback-Leibler divergence (KL divergence) or relative entropy.\n\n— Page 57, Machine Learning: A Probabilistic Perspective, 2012.\n\nThe log can be base-2 to give units in “bits,” or the natural logarithm base-e with units in “nats.” When the score is 0, it suggests that both distributions are identical, otherwise the score is positive.\n\nImportantly, the KL divergence score is not symmetrical, for example:\n\n• KL(P || Q) != KL(Q || P)\n\nIt is named for the two authors of the method Solomon Kullback and Richard Leibler, and is sometimes referred to as “relative entropy.”\n\nThis is known as the relative entropy or Kullback-Leibler divergence, or KL divergence, between the distributions p(x) and q(x).\n\n— Page 55, Pattern Recognition and Machine Learning, 2006.\n\nIf we are attempting to approximate an unknown probability distribution, then the target probability distribution from data is P and Q is our approximation of the distribution.\n\nIn this case, the KL divergence summarizes the number of additional bits (i.e. calculated with the base-2 logarithm) required to represent an event from the random variable. The better our approximation, the less additional information is required.\n\n… the KL divergence is the average number of extra bits needed to encode the data, due to the fact that we used distribution q to encode the data instead of the true distribution p.\n\n— Page 58, Machine Learning: A Probabilistic Perspective, 2012.\n\nWe can make the KL divergence concrete with a worked example.\n\nConsider a random variable with three events as different colors. We may have two different probability distributions for this variable; for example:\n\nWe can plot a bar chart of these probabilities to compare them directly as probability histograms.\n\nThe complete example is listed below.\n\nRunning the example creates a histogram for each probability distribution, allowing the probabilities for each event to be directly compared.\n\nWe can see that indeed the distributions are different.", null, "Histogram of Two Different Probability Distributions for the Same Random Variable\n\nNext, we can develop a function to calculate the KL divergence between the two distributions.\n\nWe will use log base-2 to ensure the result has units in bits.\n\nWe can then use this function to calculate the KL divergence of P from Q, as well as the reverse, Q from P.\n\nTying this all together, the complete example is listed below.\n\nRunning the example first calculates the divergence of P from Q as just under 2 bits, then Q from P as just over 2 bits.\n\nThis is intuitive if we consider P has large probabilities when Q is small, giving P less divergence than Q from P as Q has more small probabilities when P has large probabilities. There is more divergence in this second case.\n\nIf we change log2() to the natural logarithm log() function, the result is in nats, as follows:\n\nThe SciPy library provides the kl_div() function for calculating the KL divergence, although with a different definition as defined here. It also provides the rel_entr() function for calculating the relative entropy, which matches the definition of KL divergence here. This is odd as “relative entropy” is often used as a synonym for “KL divergence.”\n\nNevertheless, we can calculate the KL divergence using the rel_entr() SciPy function and confirm that our manual calculation is correct.\n\nThe rel_entr() function takes lists of probabilities across all events from each probability distribution as arguments and returns a list of divergences for each event. These can be summed to give the KL divergence. The calculation uses the natural logarithm instead of log base-2 so the units are in nats instead of bits.\n\nThe complete example using SciPy to calculate KL(P || Q) and KL(Q || P) for the same probability distributions used above is listed below:\n\nRunning the example, we can see that the calculated divergences match our manual calculation of about 1.3 nats and about 1.4 nats for KL(P || Q) and KL(Q || P) respectively.\n\n## Jensen-Shannon Divergence\n\nThe Jensen-Shannon divergence, or JS divergence for short, is another way to quantify the difference (or similarity) between two probability distributions.\n\nIt uses the KL divergence to calculate a normalized score that is symmetrical. This means that the divergence of P from Q is the same as Q from P, or stated formally:\n\n• JS(P || Q) == JS(Q || P)\n\nThe JS divergence can be calculated as follows:\n\n• JS(P || Q) = 1/2 * KL(P || M) + 1/2 * KL(Q || M)\n\nWhere M is calculated as:\n\n• M = 1/2 * (P + Q)\n\nAnd KL() is calculated as the KL divergence described in the previous section.\n\nIt is more useful as a measure as it provides a smoothed and normalized version of KL divergence, with scores between 0 (identical) and 1 (maximally different), when using the base-2 logarithm.\n\nThe square root of the score gives a quantity referred to as the Jensen-Shannon distance, or JS distance for short.\n\nWe can make the JS divergence concrete with a worked example.\n\nFirst, we can define a function to calculate the JS divergence that uses the kl_divergence() function prepared in the previous section.\n\nWe can then test this function using the same probability distributions used in the previous section.\n\nFirst, we will calculate the JS divergence score for the distributions, then calculate the square root of the score to give the JS distance between the distributions. For example:\n\nThis can then be repeated for the reverse case to show that the divergence is symmetrical, unlike the KL divergence.\n\nTying this together, the complete example of calculating the JS divergence and JS distance is listed below.\n\nRunning the example shows that the JS divergence between the distributions is about 0.4 bits and that the distance is about 0.6.\n\nWe can see that the calculation is symmetrical, giving the same score and distance measure for JS(P || Q) and JS(Q || P).\n\nThe SciPy library provides an implementation of the JS distance via the jensenshannon() function.\n\nIt takes arrays of probabilities across all events from each probability distribution as arguments and returns the JS distance score, not a divergence score. We can use this function to confirm our manual calculation of the JS distance.\n\nThe complete example is listed below.\n\nRunning the example, we can confirm the distance score matches our manual calculation of 0.648, and that the distance calculation is symmetrical as expected.\n\nThis section provides more resources on the topic if you are looking to go deeper.\n\n## Summary\n\nIn this post, you discovered how to calculate the divergence between probability distributions.\n\nSpecifically, you learned:\n\n• Statistical distance is the general idea of calculating the difference between statistical objects like different probability distributions for a random variable.\n• Kullback-Leibler divergence calculates a score that measures the divergence of one probability distribution from another.\n• Jensen-Shannon divergence extends KL divergence to calculate a symmetrical score and distance measure of one probability distribution from another.\n\nDo you have any questions?\nAsk your questions in the comments below and I will do my best to answer.\n\n## Get a Handle on Probability for Machine Learning!", null, "#### Develop Your Understanding of Probability\n\n...with just a few lines of python code\n\nDiscover how in my new Ebook:\nProbability for Machine Learning\n\nIt provides self-study tutorials and end-to-end projects on:\nBayes Theorem, Bayesian Optimization, Distributions, Maximum Likelihood, Cross-Entropy, Calibrating Models\nand much more...\n\n#### Finally Harness Uncertainty in Your Projects\n\nSkip the Academics. Just Results.\n\n### 75 Responses to How to Calculate the KL Divergence for Machine Learning\n\n1.", null, "Ori October 18, 2019 at 2:22 pm #\n\nHi Jason, you mentioned it’s the negative sum in the formula, but the in the code is a positive sum.\n\n•", null, "Jason Brownlee October 18, 2019 at 2:59 pm #\n\nRight!\n\nWe can also use an alternative form: KL(P || Q) = sum x in X P(x) * log(P(x) / Q(x))\n\nThis is the version that we implement in practice.\n\nI have updated the post, thanks!\n\n•", null, "Montu April 14, 2021 at 10:37 pm #\n\nHi Jason,\nThis might be answered in your blog but I am struggling to figure it out. Suppose you are given from start histograms of two samples. Given this much, how should I proceed to calculate the KL divergence of the two data sets?\n\n•", null, "Jason Brownlee April 15, 2021 at 5:25 am #\n\nYou need the data, histograms (plots) are insufficient.\n\n2.", null, "Saurabh October 18, 2019 at 6:37 pm #\n\nHello Jason,\n\nDo you have written an article on the “correlation coefficient calculation for categorical as well as numerical/continuous data”?\n\nSometimes it seems to be misleading for me and I got confused about what is the right way to calculate the correlation coefficient when data contains categorical as well as continuous features?\n\nIf yes then kindly provide me the pointer to the blog.\n\nThanking you,\nSaurabh\n\n•", null, "Jason Brownlee October 19, 2019 at 6:30 am #\n\nYes, I will schedule a post on the topic of cross-type correlation.\n\nThanks for the suggestion.\n\n3.", null, "Raj October 24, 2019 at 12:38 pm #\n\nHi Jason,\n\nCan you please provide a code for KL and JS divergence if the given distributions are continuous probability distributions?\n\n•", null, "Jason Brownlee October 24, 2019 at 2:02 pm #\n\nThanks for the suggestion.\n\n4.", null, "bendaizer November 17, 2019 at 5:01 pm #\n\nHi Jason,\n\nit would be more pythonic to use the zip method instead of range(len()) :\n\n[pi * np.log2(qi/pi) for pi, qi in zip(p,q)]\n\n•", null, "Jason Brownlee November 18, 2019 at 6:44 am #\n\nYes, thanks!\n\n5.", null, "Nicolas November 26, 2019 at 11:17 am #\n\nHi Jason, do you know if there is a threshold value for each of these two metrics that corresponds to an acceptable approximation of one distribution by another? I feel that all probability metrics are pretty subjectives in the end. Thanks!\n\n•", null, "Jason Brownlee November 26, 2019 at 1:31 pm #\n\nWhen the distribution contains certain probabilities (0 and 1), then the KL divergence will match the cross entropy, as the entropy for the distribution will be zero.\n\nDoes that help?\n\nI don’t think they are subjective, they are measuring real data.\n\n6.", null, "K.S.Wong June 6, 2020 at 4:29 pm #\n\nHi Jason, thank you for the article! I would like to ask if the following makes sense:\n\nI have performed unsupervised clustering on several images using the Gaussian Mixture Modelling method and I want to find out the similarity between 2 images, where each image is from a different cluster.\nDoes it make sense to measure the similarity between the 2 images by calculating the Jensen-Shannon Distance using the 2 images and treating each image as a different distribution (https://docs.scipy.org/doc/scipy/reference/generated/scipy.spatial.distance.jensenshannon.html)?\n\nThank you.\n\n•", null, "Jason Brownlee June 7, 2020 at 6:18 am #\n\nPerhaps try it and see if it is appropriate for your data.\n\nI believe image similarity is a large field of study, I recommend performing a review of the literature to see what methods work well generally.\n\n•", null, "Dhruvanshu Parmar April 16, 2021 at 5:56 am #\n\nSo I am working on a problem similar to stuffs what I am asking. I have a data sample, and it’s histogram is plotted following some random uncommon distribution. Since I need probabilities for calculating KL divergence, I tried the Kernel non parametric estimation(about which I read in your blog on probability density) to estimate a density function. Then using the density function I obtained the KL divergence.\nSo my concern is if I am doing something wrong or not.\n\n•", null, "Jason Brownlee April 17, 2021 at 6:01 am #\n\nYour approach sounds reasonable.\n\n7.", null, "Varsha July 9, 2020 at 7:45 pm #\n\nHi,\nIs it possible to find KL divergence for dependent variable’s probability distribution in multivariate dataset?If yes how? Please give example . Here dependent variable depends on multiple independent variables.\nThank you\n\n•", null, "Jason Brownlee July 10, 2020 at 5:53 am #\n\nSorry, I don’t understand, can you please elaborate what you mean?\n\n8.", null, "Varsha July 10, 2020 at 4:50 pm #\n\nHello,\nI mean if dataset is multivariate how do we compute KL divergence ? suppose every instance is in the form of (x,y) where x={x1,x,2,…xn} i.e.number of x attributes(independent) and y depends on these x values . How to compute KL divergence for such type of multivariate dataset\n\n•", null, "Jason Brownlee July 11, 2020 at 6:04 am #\n\nThe approach is general as far as I understand, you can calculate divergence between univariate and multivariate probability distributions.\n\n•", null, "Varsha Khandekar July 11, 2020 at 8:37 pm #\n\nok thanks\n\n9.", null, "Arunima July 13, 2020 at 4:25 am #\n\nHi,\nYour post was great! But the kl_divergence(p, q) function you wrote, this is only for discrete variable, right? What to do for continuous variable?\n\n•", null, "Jason Brownlee July 13, 2020 at 6:07 am #\n\nThanks.\n\nI described it for a discrete variable, but you can use it for continuous as well.\n\n•", null, "Arunima July 13, 2020 at 8:43 am #\n\nOkay, another thing, I know the values in p or q need to be between 0-1. But should the sum of all the values of p be 1?\nAnd also, using the rel_entr() function, I got negative value for some example I was trying, is it possible? If it is, what does it mean?\n\n•", null, "Jason Brownlee July 13, 2020 at 1:38 pm #\n\nThe sum of the probabilities for all evens must sum to one.\n\nP and Q are distributions of events. All events for P must sum to 1. All events for Q must sum to 1.\n\nI don’t undertand a negative kl divergenve, sorry.\n\n•", null, "Arunima July 13, 2020 at 3:05 pm #\n\nThank you so much for your reply. After searching many resources, at one place I found that the value of KL divergence can’t be negative. There was some facts about my data which may not be acceptable by KL divergence that’s why was generating the negative value. Thank you.\n\n•", null, "Jason Brownlee July 14, 2020 at 6:14 am #\n\nI’m happy to hear that.\n\n10.", null, "Erfan July 13, 2020 at 4:57 pm #\n\nHi Jason,\nDo you have any idea on estimating KL (P(x)’ll Q(x)) when we just have Monte Carlo samples from P, and Q is a know normal distribution, say N (0,1)?\nThanks!\n\n•", null, "Jason Brownlee July 14, 2020 at 6:15 am #\n\nAs long as you have a probability for the same events in each distribution, I think you’re good to go.\n\nPerhaps you need to interpolate some samples.\n\n•", null, "Erfan July 14, 2020 at 10:12 am #\n\nThe thing is that for P distribution, I just have the samples, so I do not know how to compute the probability. unless I discretize the distribution to use the normalized frequency of the samples in each bin as the probability values, but it causes additional estimation error due to the discretization.\n\n11.", null, "syed Rafee August 5, 2020 at 7:08 am #\n\nHi Jason\n\nDo you have any tutorial on how to approximate the KL-Divergence between two GMMs?\n\n•", null, "Jason Brownlee August 5, 2020 at 1:41 pm #\n\nSorry I do not.\n\n12.", null, "G Venkataraman August 27, 2020 at 8:38 pm #\n\nHI Jason, Great article!\n\nIn my project i have 10 discrete probability distributions. I have calculated the KL divergence for each pair of distributions and have created a KL divergence score matrix [10X10]. With this matrix, i am then applying hierarchical clustering to cluster distributions that are similar.\n\nMy question is: Is there any way the KL divergence score be used to understand how much similar the pair of distributions are? Understand 0 means identical and the KL value can go upto infinity. For example, What does a KL value of 0.5 represent? Can this be equated to some sort of 90% or 80% similarity between the distributions?\n\nThe question is basically to identify a threshold KL similarity score between two probability distributions to understand whether they can be clustered. The threshold may depend on the business problem we are trying to solve, but how do we justify any threshold KL divergence score?\n\n•", null, "Jason Brownlee August 28, 2020 at 6:42 am #\n\nIf you use log base 2, you can interpret the result in bits.\n\nE.g. how much more information or surprise there is in one distribution of events compared to another.\n\n•", null, "G Venkataraman August 28, 2020 at 2:54 pm #\n\nThank you very much Jason!\n\n•", null, "Jason Brownlee August 29, 2020 at 7:55 am #\n\nYou’re welcome.\n\n•", null, "Nandana S September 15, 2022 at 9:22 pm #\n\nHow do we interpret the results in bits, as in how do we find an ideal threshold for this method?\n\n13.", null, "Nikhil August 30, 2020 at 12:27 pm #\n\nThis is a great post Jason! thank you!\nOne question – if JS can be used to calculate the “distance” between two distributions, can you explain when do I use this distance metric vs using something like cosine distance?\n\nI have been using the cosine distance for quite some time now, and i would like to understand if there are specific situations when it is more appropriate to use JS over cosine.\n\nThanks again!\n\nNIkhil\n\n•", null, "Jason Brownlee August 31, 2020 at 6:07 am #\n\nCosine distance is between two vectors. The above method are for the distance between two distributions.\n\n14.", null, "Curiously Coding Foxah October 10, 2020 at 4:30 am #\n\nI’ve got four (non-linear, tree-based) models in production and using the average of them as the served prediction. We get ground truth data immediately.\n\nDuring training the optimized candidate models had very similar performance, so I decided to deploy all of them and take the average and served that as the prediction. With the intention of figuring which one would really be best at a later point.\n\nThat later point has come.\n\nOut of these four models two seem fit the ground truth distribution quite well, at least by examining the KDE plot of the predicted values against the ground truth distribution\n\nI was initially thinking of doing a pairwise 2-sample KS Test (each model against ground truth) to see if the predicted values from the models come from the same distribution as the ground truth.\n\nBut I talked myself out of it. Mostly because I have 50,000+ predictions and I figured that a sample size that large will result in a small p-value anyway.\n\nI then turned my attention to KL divergence. Something I have never used before.\n\nWould comparing the KL-Divergence between each model’s prediction and the ground truth be a good way to assess model fit?\n\nIf so, how would I go about doing this? scipy.stats.kl_div outputs an array which contains (I assume) the divergence between the prediction and ground truth.\n\nWould I just take the sum of the array and call that the Divergence between the model and ground truth?\n\nHopefully I don’t sound crazy or too much like I don’t know what I’m doing or talking about. Because I kinda don’t, but not too much.\n\nThank you\n\n•", null, "Jason Brownlee October 10, 2020 at 7:10 am #\n\nYes, perhaps try it and see.\n\nYou would average the divergence for each prediction.\n\nThis will give you ideas:\nhttps://machinelearningmastery.com/cross-entropy-for-machine-learning/\n\n•", null, "Curiously Coding Foxah October 14, 2020 at 7:02 am #\n\nHi Jason! Thanks for the response. I have another version of this question\nWould it make sense to use KL-Divergence to measure the difference in predictions versus ground truth for a regression problem?\n\nI’ve tuned four models and serve the average as a prediction in the production environment.\n\nI plotted the ecdf and kde for each model’s prediction versus the ground truth, and I want a way to capture the closeness of the distributions in one number that I can track over time.\n\nI use an evaluation metric for MAE for assessing performance, but I also wanted a way to capture how similar or different the shapes of the distributions for prediction and ground truth are in a single number.\n\n•", null, "Jason Brownlee October 14, 2020 at 7:38 am #\n\nHmm, great question.\n\nCautiously, I would say yes, but double check the literature to see if there is a more appropriate divergence measure.\n\n15.", null, "Islam Hassan Ahmed January 4, 2021 at 2:04 pm #\n\nHi Jason, thank you for this great article!\n\nOne question – Could you please explain in more details how LK-Divergence is not the same for both cases (p to q – q to p)? In other word, I can’t understand the intuitive (quoted below); the subtraction of probabilities for each event is the same regardless the direction (p to q – q to p)\n\n”This is intuitive if we consider P has large probabilities when Q is small, giving P less divergence than Q from P as Q has more small probabilities when P has large probabilities.”\n\n•", null, "Jason Brownlee January 5, 2021 at 6:15 am #\n\nThink of it as “relative entropy”, e.g. the first distribution relative to the second. If the order is changed, the relative entropy must change.\n\n•", null, "Islam Hassan Ahmed January 5, 2021 at 7:28 am #\n\nThanks Jason! it’s much clearer now.\n\n•", null, "Jason Brownlee January 5, 2021 at 7:31 am #\n\nYou’re welcome.\n\n16.", null, "Nafees Dipta January 9, 2021 at 5:13 pm #\n\nHi Jason,\nWhat if the p and q are multidimensions like shape (32,50)? Then how do we calculate the kl?\n\n•", null, "Jason Brownlee January 10, 2021 at 5:37 am #\n\nNot sure what you mean, sorry.\n\np and q are probability distributions for events.\n\n17.", null, "Ouss January 11, 2021 at 2:06 pm #\n\nHi Jason, the article is well written. I just have a remark about the KL divergence. when you calculate it in the example you got\nKL(P || Q): 1.927 bits\nbased on the definition the result should be negative, did you use the absolute value to infer the number of bits?\n\n•", null, "Jason Brownlee January 12, 2021 at 7:47 am #\n\nGood question, please note the we are using the alternate positive calculation that is more common in practice, discussed in the section titled “Kullback-Leibler Divergence”.\n\n18.", null, "Karina Samvelyan January 21, 2021 at 9:40 pm #\n\nHi Jason,\nThanks for the post!\nYou write the the KL is not symmetrical, “for example …”. Mathematically speaking, non-symmetric means precisely that KL(P,Q) = KL(Q, P) is not always true, so imho it’s better to replace “for example” with “that is, the equality … doesn’t always hold” or “that is, usually KL(P,Q) != KL(Q,P).\n\n•", null, "Jason Brownlee January 22, 2021 at 7:19 am #\n\nThanks for the suggestion.\n\n19.", null, "Manal February 28, 2021 at 8:30 pm #\n\nHello Jason,\nIf I want to use the KL for evaluating my GAN and calculate it between a real distribution and generated distribution, How can I do that?\n\nI used your script and make P as a real distribution and Q as a generated distribution but it returns nan values.\n\nSo, Can you help me?\n\n20.", null, "Shanel April 23, 2021 at 5:46 am #\n\nHey there. Could you possibly speak to how this might be expanded to a Markov process?\n\n•", null, "Jason Brownlee April 24, 2021 at 5:12 am #\n\nThanks for the suggestion.\n\n21.", null, "Paul August 6, 2021 at 7:21 pm #\n\nHi Jason,\nThanks for the great post. Is there a preferred way to compute KL if p(k)=0 (division by zero) or q(k) (log of 0)?\n\n•", null, "Jason Brownlee August 7, 2021 at 5:40 am #\n\nTypically we add tiny values to parameters to avoid zeros when coding the thing up for real.\n\n22.", null, "Juraj September 3, 2021 at 9:44 am #\n\n* KL(P || Q) = sum x in X P(x) * log(P(x) / Q(x))\nThe intuition for the KL divergence score is that when the probability for an event from P is large, but the probability for the same event in Q is small, there is a large divergence. When the probability from P is small and the probability from Q is large, there is also a large divergence, but not as large as the first case.\n\nIt seems to be, that if P(p) is small and P(q) for the same event is large, then the fraction in the log is < 1. That would give a negative result, and as such, shouldn't it actually cancel out some of the divergence? Given we are not taking an absolute value of the terms of the sum.\nTherefore, yes, if P is large but Q is small, there is large divergence, but in the inverse situation there is actually \"negative\" divergence.\n\nIs there something I am missing here?\n\nThank you for this post, very useful, just wanted to clear up some confusion regarding this point.\n\n•", null, "Jason Brownlee September 4, 2021 at 5:15 am #\n\nPerhaps you can test your intuition with some worked examples/real numbers.\n\n23.", null, "Juraj Jursa September 4, 2021 at 12:34 am #\n\n* KL(P || Q) = sum x in X P(x) * log(P(x) / Q(x))\nThe intuition for the KL divergence score is that when the probability for an event from P is large, but the probability for the same event in Q is small, there is a large divergence. When the probability from P is small and the probability from Q is large, there is also a large divergence, but not as large as the first case.\n\nWouldn’t the divergence in the second case be negative, meaning it would be cancelling out somewhat with the other cases where P > Q? Why do we not take absolute value of all the terms in the sum instead?\n\n•", null, "Jason Brownlee September 4, 2021 at 5:22 am #\n\nRun real numbers through and see.\n\n24.", null, "Mona November 25, 2021 at 12:05 pm #\n\nI am not sure why I get NAN for kl. y_train is 0.7 of y and y_val is 0.3 of y.\np = y_train\nq = y_val\nkl_pq = rel_entr(p, q)\nprint(‘KL(P || Q): %.3f nats’ % sum(kl_pq))\nkl_qp = rel_entr(q, p)\nprint(‘KL(Q || P): %.3f nats’ % sum(kl_qp))\n\nresult:\n\nKL(P || Q): nan nats\nKL(Q || P): nan nats\n\nI don’t have any nan in y itself.\n\n•", null, "Adrian Tam November 25, 2021 at 2:44 pm #\n\nBoth p and q must be all non-negative or otherwise your rel_entr will give you nan. Then sum will be nan too.\n\n25.", null, "Mona Jalal November 25, 2021 at 1:23 pm #\n\nHi Jason,\nI am not sure why I get one of my KLs as a negative value. I have provided for you the p and q arrays (I converted them to numpy arrays)\n\np: [6.33527306 0.17195741 0.01810078 0.01810078]\nq: [7.36958404 0.09665028 0.02416257 0.02416257]\n\nKL(p, q): -1.2543610466991473\nKL(q, p): 1.547671142537377\n\nJS(P || Q) Distance: 0.042\nJS(Q || P) Distance: 0.042\n\n•", null, "Adrian Tam November 25, 2021 at 2:51 pm #\n\nKL formula involves log, and log can return negative value\n\n26.", null, "Dipankar Porey March 19, 2022 at 4:42 pm #\n\nHow can someone find convergence in Generative adversarial networks ?\n\n27.", null, "wesam June 16, 2022 at 11:36 pm #\n\nHi,\n\nIn case we have the same type of data (PDFs values), and we need to do classification using the Random forest and the KL-divergence. Is there any way to check which one outperforms the other?\nOr just I need to try both methods and check which one outperforms the other?" ]

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[ "", null, "This version of this document is no longer maintained. For the latest documentation, see http://www.qnx.com/developers/docs.\n\n# modf(), modff()\n\nBreak a number into integral and fractional parts\n\n## Synopsis:\n\n```#include <math.h>\n\ndouble modf( double value,\ndouble* iptr );\n\nfloat modff( float value,\nfloat* iptr );```\n\n## Arguments:\n\nvalue\nThe value that you want to break into parts.\niptr\nA pointer to a location where the function can store the integral part of the number.\n\n## Library:\n\nlibm\n\nUse the -l m option to qcc to link against this library.\n\n## Description:\n\nThe modf() and modff() functions break the given value into integral and fractional parts, each of which has the same sign as the argument. They store the integral part as a double in the object pointed to by iptr.\n\n## Returns:\n\nThe signed fractional part of value.", null, "If an error occurs, these functions return 0, but this is also a valid mathematical result. If you want to check for errors, set errno to 0, call the function, and then check errno again. These functions don't change errno if no errors occurred.\n\n## Examples:\n\n```#include <stdio.h>\n#include <stdlib.h>\n#include <math.h>\n\nint main( void )\n{\ndouble integral_value, fractional_part;\n\nfractional_part = modf( 4.5, &integral_value );\nprintf( \"%f %f\\n\", fractional_part, integral_value );\n\nfractional_part = modf( -4.5, &integral_value );\nprintf( \"%f %f\\n\", fractional_part, integral_value );\n\nreturn EXIT_SUCCESS;\n}```\n\nproduces the output:\n\n```0.500000 4.000000\n-0.500000 -4.000000```\n\n## Classification:\n\nSafety:\nCancellation point No\nInterrupt handler No\nSignal handler No" ]

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[ "## Gradients Weights improve Regression and Classification\n\nSamory Kpotufe, Abdeslam Boularias, Thomas Schultz, Kyoungok Kim; 17(22):1−34, 2016.\n\n### Abstract\n\nIn regression problems over $\\mathbb{R}^d$, the unknown function $f$ often varies more in some coordinates than in others. We show that weighting each coordinate $i$ according to an estimate of the variation of $f$ along coordinate $i$ -- e.g. the $L_1$ norm of the $i$th-directional derivative of $f$ -- is an efficient way to significantly improve the performance of distance-based regressors such as kernel and $k$-NN regressors. The approach, termed Gradient Weighting (GW), consists of a first pass regression estimate $f_n$ which serves to evaluate the directional derivatives of $f$, and a second-pass regression estimate on the re-weighted data. The GW approach can be instantiated for both regression and classification, and is grounded in strong theoretical principles having to do with the way regression bias and variance are affected by a generic feature-weighting scheme. These theoretical principles provide further technical foundation for some existing feature-weighting heuristics that have proved successful in practice. We propose a simple estimator of these derivative norms and prove its consistency. The proposed estimator computes efficiently and easily extends to run online. We then derive a classification version of the GW approach which evaluates on real-worlds datasets with as much success as its regression counterpart.\n\n[abs][pdf][bib]" ]

[ null ]

[ "Solutions by everydaycalculation.com\n\n## Multiply 35/28 with 5/6\n\n1st number: 1 7/28, 2nd number: 5/6\n\nThis multiplication involving fractions can also be rephrased as \"What is 35/28 of 5/6?\"\n\n35/28 × 5/6 is 25/24.\n\n#### Steps for multiplying fractions\n\n1. Simply multiply the numerators and denominators separately:\n2. 35/28 × 5/6 = 35 × 5/28 × 6 = 175/168\n3. After reducing the fraction, the answer is 25/24\n4. In mixed form: 11/24\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

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[ "# HermiteH\n\nHermiteH[n,x]\n\ngives the Hermite polynomial", null, ".\n\n# Details", null, "• Mathematical function, suitable for both symbolic and numerical manipulation.\n• Explicit polynomials are given for nonnegative integers n.\n• The Hermite polynomials satisfy the differential equation", null, ".\n• They are orthogonal polynomials with weight function", null, "in the interval", null, ".\n• For certain special arguments, HermiteH automatically evaluates to exact values.\n• HermiteH can be evaluated to arbitrary numerical precision.\n• HermiteH automatically threads over lists.\n• HermiteH[n,x] is an entire function of x with no branch cut discontinuities.\n\n# Examples\n\nopen all close all\n\n## Basic Examples(2)\n\nCompute the 10", null, "Hermite polynomial:\n\n In:=", null, "Out=", null, "In:=", null, "Out=", null, "## Neat Examples(4)\n\nIntroduced in 1988\n(1.0)" ]

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[ "", null, "# Conformal Fractals: Ergodic Theory Methods", null, "Conformal Fractals: Ergodic Theory Methods\nby\n\nPublisher: Cambridge University Press\nISBN/ASIN: 0521438004\nISBN-13: 9780521438001\nNumber of pages: 362\n\nDescription:\nThis is a one-stop introduction to the methods of ergodic theory applied to holomorphic iteration. The authors begin with introductory chapters presenting the necessary tools from ergodic theory thermodynamical formalism, and then focus on recent developments in the field of 1-dimensional holomorphic iterations and underlying fractal sets, from the point of view of geometric measure theory and rigidity. Detailed proofs are included.\n\n(2.7MB, PDF)\n\n## Similar books", null, "Math Alive\nby - Princeton University\nDesigned for those who haven't had college mathematics but would like to understand some applications: Cryptography; Error correction and compression; Probability and Statistics; Birth, Growth, Death and Chaos; Graph Theory; Voting and Social Choice.\n(8783 views)", null, "Mathematical Methods\nby - University of Notre Dame\nMultidimensional calculus, linear analysis, linear operators, vector algebra, ordinary differential equations. Directed at first year graduate students in engineering and undergraduates who wish to become better prepared for graduate studies.\n(19928 views)", null, "Elementary Fuzzy Matrix Theory and Fuzzy Models for Social Scientists\nby - arXiv\nThis book aims to assist social scientists to analyze their problems using fuzzy models. The basic and essential fuzzy matrix theory is given. The authors have only tried to give those essential basically needed to develop the fuzzy model.\n(10705 views)", null, "The Mathematics of Investment\nby - D.C Heath and Company\nThis book provides an elementary course in the theory and the application of annuities certain and in the mathematical aspects of life insurance. The book is particularly adapted to the needs of students in colleges of business administration.\n(29919 views)" ]

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[ "# Prove that Each Angle of an Equilateral Triangle is 60°. - Mathematics\n\nProve that each angle of an equilateral triangle is 60°.\n\n#### Solution\n\nGiven to prove that each angle of an equilateral triangle is 60°\nLet us consider an equilateral triangle ABC\nSuch that AB= BC= CA\n\nNow,\n\nAB=BC ⇒ ∠A=∠C            ................(1) [Opposite angles to equal sides are equal]\n\nand BC = AC ⇒∠B = ∠A ……..(2)\n\nFrom (1) and (2), we get\n\n∠A = ∠B = ∠C          ..............(3)\n\nWe know that\nSum of angles in a triangle =180°\n\n⇒∠A+∠B+∠C=180°\n\n⇒ ∠A+∠A+∠A=180°\n\n⇒3 ∠A=180°\n\n⇒ ∠A=(180°)/3=60°\n\n∴∠S=∠B=∠C=60°\n\nHence, each angle of an equilateral triangle is 60°.", null, "Concept: Properties of a Triangle\nIs there an error in this question or solution?\n\n#### APPEARS IN\n\nRD Sharma Mathematics for Class 9\nChapter 12 Congruent Triangles\nExercise 12.3 | Q 4 | Page 47\n\nShare" ]

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[ "# Supercapacitor Charger with Adjustable Output Voltage and Adjustable Charging Current Limit\n\nIntroduction\n\nFor applications using larger value supercapacitors (tens to hundreds of farads), a charger circuit with a relatively high charging current is needed to minimize the recharge time of the system. Supercapacitors are used as energy hold up devices in applications such as solid state RAID disks, where information stored in high speed volatile memory must be transferred to non-volatile flash memory when power is lost. This transfer time may take minutes, requiring hundreds of farads to hold up the power supply until the transfer is complete. The requirement for the recharge time of these banks of supercapacitors is typically less than one hour. To accomplish this, a high charging current is required. This article describes a supercapacitor charging circuit using the LT3663 that meets these difficult requirements.\n\nThe LT3663 is a 1.2A, 1.5MHz step-down switching regulator with output current limit ideal for supercapacitor applications. The part has an input voltage range of 7.5V to 36V, has adjustable output voltage and adjustable output current limit. The output voltage is set with a resistor divider network in the feedback loop while the output current limit is set by a single resistor connected from the ILIM pin to ground. With its internal compensation network and internal boost diode, the LT3663 requires a minimal number of external components.\n\nPower Ride-Through Application\n\nA procedure for selecting the size of the supercapacitor is outlined in the September 2008 edition of Linear Technology, in an article titled “Replace Batteries in Power Ride-Through Applications with Supercaps and 3mm × 3mm Capacitor Charger.” The procedure determines the effective supercapacitor (CEFF) capacitance at 0.3Hz, based on the power level to be held up, the minimum operating voltage of the DC/DC converter supporting the load, the distributed circuit resistances including the ESR of the supercapacitors, and the required hold up time.\n\nOnce the size of the supercapacitor is known, the charging current can be determined to meet the recharge time requirements. The recharge time (TRECHARGE) is the time required to recharge the supercapacitors from the minimum operating voltage (VUV) of the DC/DC converter to the full charge voltage (VFC) of the supercapacitors. The voltage on the individual supercapacitors at the start of the recharge cycle is the minimum operating voltage divided by the number (N) of supercapacitors in series. From here on, this article describes an application with two supercapacitors in series. The recharge current (ICHARGE) is determined by the capacitor charge control law:\n\nThis assumes that the voltage across the supercapacitor doesn’t discharge below the VUV/N value. This assumption is valid if the time period while input power isn’t available is such that the supercapacitor’s leakage current hasn’t significantly reduced the voltage across the capacitor. The voltage across the supercapacitor may actually rise slightly after the DC/DC converter shuts down due to the dielectric absorption effect. The initial charge time TCHARGE for a fully discharged bank of supercapacitors is:\n\nFigure 1 shows a block diagram of the components for this supercapacitor charger application.\n\nCharging Circuit Using the LT3663\n\nTo set the charging current, a resistor RILIM is connected from the ILIM pin of the LT3663 to ground. Table 1 shows the nominal charging currents for various values of RILIM.\n\n Charging Current (A) RILIM Value (kΩ) 0.4 140 0.6 75 0.8 48.7 1.0 36.5 1.2 28.7\n\nThe full charge voltage is set by the resistor divider network in the feedback loop. Table 2 shows various full charge voltages versus the value of RFB2 (resistor from the FB pin to ground) when resistor RFB1 (resistor tied between the VOUT pin and the FB pin) is 200k. Figure 2 shows the charging circuit for each supercapacitor.\n\n Full Charge Voltage (V) RFB2 (kΩ) 2.65 86.6 2.5 93.1 2.4 100 2.2 115 2.0 133\n\nControl Circuit for Charging Supercapacitors\n\nThe control circuit in Figure 3 is used to balance the voltages of the supercapacitors while they are charging. This is accomplished by prioritizing charge current to the lower voltage supercapacitor—specifically by enabling the charging circuit for the supercapacitor with the lower voltage while disabling the circuit for the other supercapacitor." ]

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[ "#", null, "Path-Integral of Charged Particle in Chern-Simons Gauge Fields\n\n+ 0 like - 0 dislike\n179 views\n\nFrom the paper \"Fermi-Bose Transmutations Induced by Gauge Fields\" by Polyakov,\n\nhttp://inspirehep.net/record/22956/citations\n\nthe theory in 3D,\n\n$$\\mathcal{L}=\\sum_{k=1}^{2}|\\partial_{\\mu}z_{k}+iA_{\\mu}z_{k}|^{2}+\\frac{\\theta}{16\\pi^{2}}\\epsilon_{\\mu\\nu\\rho}A^{\\mu}\\partial^{\\nu}A^{\\rho}$$\n\nwith a constraint $|z_{1}|^{2}+|z_{2}|^{2}=1$ for $z_{1}(x^{0},x^{1},x^{2})$ and $z_{2}(x^{0},x^{1},x^{2})$,\n\nhas the transition amplitude given by\n\n$$G(x,y)=\\int\\mathcal{D}x \\, e^{-m\\int ds}\\left\\langle\\exp\\left\\{i\\int dx^{\\mu}A_{\\mu}\\right\\} \\right\\rangle$$\n\nwhere the average is given by\n\n$$\\left\\langle\\exp\\left\\{i\\int dx^{\\mu}A_{\\mu}\\right\\} \\right\\rangle=\\int\\mathcal{D}A \\exp\\left\\{i\\int dx^{\\mu}A_{\\mu}\\right\\}\\exp\\left\\{ iS_{CS}[A]\\right\\}$$\n\nHow to derive this two-point function?\n\n Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the \"link\" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\\varnothing$ in the following word:p$\\hbar$ysi$\\varnothing$sOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register." ]

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[ "# [FFmpeg-devel] [PATCH] Simplify IMDCT in dca.c\n\nAlexander E. Patrakov patrakov\nSat Aug 30 09:07:13 CEST 2008\n\n```Michael Niedermayer wrote:\n\n> Hi\n>\n> patch below replaces the duplicated IMDCT in dca.c by the standard one.\n> code is faster, simpler nicer, ....\n> yes the QMF of dca is a windowed IMDCT\n\nYes, here it brings the time to decode a ripped CD down from 1m11s to 51s.\nUnfortunately, I could not test the SSE patch by David Conrad and compare.\n\nYour change is not bit-exact WRT the previous version, but sounds right. The\ndifference is of the form of 1-LSB-clicks (several random clicks per\nsecond - when amplified, sounds like a Geiger counter).\n\nSome cleanup notes:\n> float praXin, *raXin = &praXin;\n<...>\n> praXin = 0.0;\n\npraXin is no longer needed, and we can just have float raXin.\n\n> + for (i = 0; i < 16; i++){\n> + float a= subband_fir_hist2[i ];\n> + float b= subband_fir_hist2[i+16];\n> + float c= 0;\n> + float d= 0;\n> for (j = 0; j < 512-hist_index; j += 64){\n> + a += prCoeff[i+j ]*(-subband_fir_hist[15-i+j]);\n> + b += prCoeff[i+j+16]*( subband_fir_hist[ i+j]);\n> + c += prCoeff[i+j+32]*( subband_fir_hist[16+i+j]);\n> + d += prCoeff[i+j+48]*( subband_fir_hist[31-i+j]);\n> }\n\nThis time-reversing business (15-i+j) still looks like an attempt to do some\nseparation of the results of two transforms done in parallel. I mean,\nyou'll find something similar if you look how people push one real signal\nin the real part of the FFT input, another real signal in the imaginary\npart, do the FFT and then use the symmetry properties of the Fourier\ntransform of the real input in order to separate the two original inputs.\nIt would be interesting to figure out what exactly is going on.\n\n--\nAlexander E. Patrakov\n\n```" ]

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[ "Search a number\nBaseRepresentation\nbin11110010111100001111…\n…11110010101011011011\n310200202021002212021021112\n433023300333302223123\n5114043413341340443\n62115202030130535\n7135246024024125\noct17136077625333\n93622232767245\n101043425340123\n11372572447741\n1214a281050a4b\n1377518ba6b7a\n14387058da215\n151c21dd0db18\n\n1043425340123 has 2 divisors, whose sum is σ = 1043425340124. Its totient is φ = 1043425340122.\n\nThe previous prime is 1043425340081. The next prime is 1043425340131. The reversal of 1043425340123 is 3210435243401.\n\nIt is a strong prime.\n\nIt is a cyclic number.\n\nIt is not a de Polignac number, because 1043425340123 - 214 = 1043425323739 is a prime.\n\nIt is a self number, because there is not a number n which added to its sum of digits gives 1043425340123.\n\nIt is not a weakly prime, because it can be changed into another prime (1043425300123) by changing a digit.\n\nIt is a polite number, since it can be written as a sum of consecutive naturals, namely, 521712670061 + 521712670062.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (521712670062).\n\nAlmost surely, 21043425340123 is an apocalyptic number.\n\n1043425340123 is a deficient number, since it is larger than the sum of its proper divisors (1).\n\n1043425340123 is an equidigital number, since it uses as much as digits as its factorization.\n\n1043425340123 is an evil number, because the sum of its binary digits is even.\n\nThe product of its (nonzero) digits is 34560, while the sum is 32.\n\nThe spelling of 1043425340123 in words is \"one trillion, forty-three billion, four hundred twenty-five million, three hundred forty thousand, one hundred twenty-three\"." ]

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[ "# Processing LiDAR w/ R – Talladega National Forest\n\nLets continue with the Talladega National Forest LiDAR dataset… I have provided a link to a KML of the LiDAR panel 625065 extent. Give the site a look in Google Earth to get a better feel for the site (Figure 1). A stream in the northern part of this site flows in an east-west direction. Terrain in both the southern and northern portions slope inward towards this stream (Figure 2). Land cover consists of mature hardwood and mature natural pine.\n\nLet me step through the code from Monday piece by piece. I am following the general work flow for “Rasterizing perfect canopy height models” ( https://github.com/Jean-Romain/lidR/wiki/Rasterizing-perfect-canopy-height-models ). I will provide links to the docs as comments in the code if you need help with syntax.\n\n```###Load required libraries\nlibrary(\"lidR\")\nlibrary(\"rgdal\")\n\n###You need to change this path to match where you save the data\nmypath<- \"V:\\\\Krista_LiDAR\\\\Tripp_LiDAR\"\n\n###Store LAS filename as a variable\nttyfn<- \"625065.las\"\n\n###Read LAS file and store as the R object 'tty'\n\nsummary(tty)\n###General statistics of LiDAR attributes, two examples...\nunique(tty\\$Z)\nmax(tty\\$Z)\n\n###The available LiDAR attributes are (see names section of the summary)\n###See ASPRS documentation, table 4.7\n### X Y Z gpstime Intensity ReturnNumber NumberOfReturns ScanDirectionFlag EdgeOfFlightline Classification Synthetic_flag Keypoint_flag Withheld_flag ScanAngleRank UserData PointSourceID ```\n\nNothing new up to this point – you have loaded the point cloud into R and stored it as the R object tty.\n\nNext, to speed processing, lets clip out a small part of the point cloud. I created this shapefile in ArcMap. I loaded the LAS into a blank ArcMap document; this matches the data frame coordinate system with the LAS file (Alabama State Plane East, NAD 1983). Next, I created a new polygon shapefile, making sure to set the spatial reference to Alabama SPE, and drew the polygon myself.\n\nA practical application… Imagine that we have sample plots located in this area and we want to relate the ground measurements to the LiDAR data. We can use this same approach to extract the LiDAR metrics (average height, max/min height, Xth percentile height, slope, aspect, etc) for our samples.\n\n```###shapefile is stored in the same folder as the other data\nshppath<- mypath\n###This is the name of the clip shapefile\nshpname<- \"ClipSection_TNF\"\nshape_spdf<- readOGR(dsn = shppath, layer = shpname)\n###reference the first polygon\nshppoly<- shape_spdf@polygons[]@Polygons[]\n\nlascliptty<- lasclip(tty, shppoly)\nplot(lascliptty)```\n\nSomebody else flew this LiDAR mission and I do not have documentation about how this layer has been processed. Part of this workflow relies on each point’s Classification (recall this is one of the LAS attributes). To ensure we are working with proper classifications, we’ll reset everything to 0.\n\n`lascliptty\\$Classification<- as.integer(0)`\n\nBefore we can normalize the point cloud, we must first identify the ‘ground’ points. In this code, we’re using the PMF algorithm ( https://rdrr.io/cran/lidR/man/lasground.html ).\n\n```###https://rdrr.io/cran/lidR/man/pmf.html\nws<- seq(3,12,3)\nth<- seq(0.1, 1.5, length.out = length(ws))\n###lasground is the lidR function to classify the ground points\nlasdpmf<- lasground(lascliptty,pmf(ws,th))\nplot(lasdpmf, color=\"Classification\")\n\nunique(lasdpmf\\$Classification)\n\n###Count the number of fround-classified points\nsum(lasdpmf\\$Classification == 2)```\n\nNotice the output of the last line. The unique Classification values are “unassigned” (1) and “ground” (2). Also, there are 138,473 ground points.\n\nIn the next segment of code, you create the digital terrain model. Recall that you classified the ground points in the section above. The grid_terrain command uses the ground-classified points to interpolate the raster surface.\n\nThe DTM is then used to normalize the original point cloud. Think of this step as subtracting the DTM values from the original Z values – moving from a measure of elevation to a measure of height above the immediate ground surface.\n\n```###create the digital terrain model\n###https://rdrr.io/cran/lidR/man/grid_terrain.html\ndtm<- grid_terrain(lasdpmf, 1, knnidw(k=6L, p=2))\nplot(dtm)\n###save the DTM out as a raster you can load into ArcMap\nwriteRaster(dtm,filename=file.path(mypath,\"dtm.tif\"), format=\"GTiff\",overwrite=TRUE)\n\n###normalize the point cloud\n###https://rdrr.io/cran/lidR/man/lasnormalize.html\nlasdpmf_norm<- lasnormalize(lasdpmf,dtm)\nplot(lasdpmf_norm)```\n\nWe expect that all values in the normalized point cloud are >= 0. A quick view of the point cloud headers, use the summary(lasdpmf_norm) reveals that the minimum Z value is -2.8. In the code below, we use the lasfilter function to create a new R object, lasdpmf_norm_gt0, that stores all non-negative points.\n\n```###remove all of the negative heights by creating a new R object that contains all points whose Z is >= 0\n###https://rdrr.io/cran/lidR/man/lasfilter.html\nlasdpmf_norm_gt0<- lasfilter(lasdpmf_norm, Z >= 0)\nsummary(lasdpmf_norm_gt0)\n\n###Canopy height model (CHM)\n###https://rdrr.io/cran/lidR/man/grid_canopy.html\nchm_pmf0205<- grid_canopy(lasdpmf_norm_gt0, 02.5, p2r(0.5))\nplot(chm_pmf0205)\n\n###export the CHM as a raster file to load into ArcMap\nwriteRaster(chm_pmf0205, filename = file.path(mypath,\"chm_pmf0205.tif\"),format=\"GTiff\",overwrite=TRUE)```\n\n```######detect individual tree tops and crowns\n###https://rdrr.io/cran/lidR/man/lmf.html\nttops <- tree_detection(lasdpmf_norm_gt0, lmf(20, 35, \"circular\"))\nplot(ttops)\n\n###https://rdrr.io/cran/lidR/man/lastrees.html\n##############\n############## This next line was the reason the plots did not work out\n############## I reference the wrong R object. Instead of lasd_n<- ... it\n############## should be lasdpmf_norm_gt0<-...\n###NOT THIS >>>>> lasd_n<- lastrees(lasdpmf_norm_gt0,mcwatershed(chm_pmf,ttops))\n###but this>>>\ncol<- pastel.colors(250)\nlasdpmf_norm_gt0<- lastrees(lasdpmf_norm_gt0,mcwatershed(chm_pmf0205,ttops))\n\n####if mcwatershed above throws an error, then try this below\nlasdpmf_norm_gt0<- lastrees(lasdpmf_norm_gt0, li2012(R=3, speed_up = 5))\n\n###############\n###############\n###https://rdrr.io/cran/lidR/man/silva2016.html\ncrowns = silva2016(chm_pmf0205,ttops)()\n###plot the point cloud and colorize by segmented crowns\nx = plot(lasdpmf_norm_gt0, color = \"treeID\", colorPalette = col)\n\n###Export the crown segments as a raster\nwriteRaster(crowns,filename=file.path(mypath,\"treecrown.tif\"), format=\"GTiff\", overwrite=TRUE)\n###Export the tree tops\nwriteOGR(obj=ttops,dsn=mypath,layer=\"treetops\",driver=\"ESRI Shapefile\", overwrite = TRUE)\n\n```\n\nOne last thing… Lets calculate statistics for each of our tree segments. Calculate the mean, maximum, and minimum values for each segment based on the CHM value (r) and the un-normalized values (zr). This is similar to using ArcMap’s Zonal Statistics As Table command.\n\n```#####Statistics for groups of trees\n###z references the CHM value\n###zr references the original elevation value\nmyMetrics = function(z, zr)\n{\nmetrics = list(\nzmean = mean(z),\nzmax = max(z),\nzmin = min(z),\nzrmean = mean(zr),\nzrmax = max(zr),\nzrmin = min(zr)\n)\n\nreturn(metrics)\n}\ntreestats<- tree_metrics(lasdpmf_norm_gt0, myMetrics(Z, Zref))\nmetricsOut<- \"TreeMetrics.csv\"\nwrite.csv(treestats, file= file.path(mypath,metricsOut))```" ]

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[ "# 财务预测与估值建模之五：绝对估值模型—股利贴现模型\n\n## 二、现金流贴现模型的形式与步骤\n\n$$V=\\sum_{t=1}^{N}\\frac{CF_t}{(1+r)^t}+\\frac{TV}{(1+r)^N}$$\n\n$$TV=\\frac{CF_N(1+g)}{r-g}$$\n\n$$V=\\sum_{t=1}^N\\frac{CF_t}{(1+r)^t}+\\frac{CF_N(1+g)}{r-g}\\cdot \\frac{1}{(1+r)^N}$$\n\n## 三、股利贴现模型\n\n$$g=b\\cdot ROE=(1-PO) \\cdot ROE$$" ]

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[ "hep-th/0310008\n\nBlack Rings, Supertubes, and a Stringy Resolution of Black Hole Non-Uniqueness\n\nHenriette Elvang111E-mail: and Roberto Emparan222E-mail:\n\nDepartment of Physics\n\nUCSB, Santa Barbara, CA 93106\n\n[.5em]\n\nDepartament de Física Fonamental, and\n\nC.E.R. en Astrofísica, Física de Partícules i Cosmologia,\n\nUniversitat de Barcelona, Diagonal 647, E-08028 Barcelona, Spain\n\n[.5em]\n\nInstitució Catalana de Recerca i Estudis Avançats (ICREA)\n\n[.5em]\n\nABSTRACT\n\nIn order to address the issues raised by the recent discovery of non-uniqueness of black holes in five dimensions, we construct a solution of string theory at low energies describing a five-dimensional spinning black ring with three charges that can be interpreted as D1-brane, D5-brane, and momentum charges. The solution possesses closed timelike curves (CTCs) and other pathologies, whose origin we clarify. These pathologies can be avoided by setting any one of the charges, e.g., the momentum, to zero. We argue that the D1-D5-charged black ring, lifted to six dimensions, describes the thermal excitation of a supersymmetric D1-D5 supertube, which is in the same U-duality class as the D0-F1 supertube. We explain how the stringy microscopic description of the D1-D5 system distinguishes between a spherical black hole and a black ring with the same asymptotic charges, and therefore provides a (partial) resolution of the non-uniqueness of black holes in five dimensions.\n\n## 1 Introduction\n\nRotating black holes typically exhibit much richer dynamics than their static counterparts, especially in more than four dimensions. Static black holes present relatively small qualitative differences as we increase the number of dimensions beyond four, but the inclusion of rotation brings in a larger diversity that depends on the dimensionality of spacetime . A recent surprise has been the discovery of a rotating black ring solution in five-dimensional vacuum gravity . The horizon of the black ring has topology , and its tension and self-gravitational attraction are precisely balanced by the rotation of the ring. Perhaps more strikingly, when the mass and spin are within a certain range of values, it is possible to find a black hole and two black rings with the same spin and mass111We follow and often use the term “spherical black hole”, or even simply “black hole”, to abbreviate the phrase “topologically spherical black hole”, as opposed to the black ring. Technically speaking, the black ring is also a black hole, but we hope to cause no confusion.. Hence, the simplicity of four-dimensional black holes implied by the celebrated uniqueness theorems does not extend to five-dimensional stationary solutions. It is quite possible that the non-uniqueness of higher-dimensional black holes is indeed wider than indicated by the existence of the black ring solution [3, 4].\n\nString theory has provided a remarkably successful description of black holes, including an account of their entropy in microscopic terms , so it is natural to investigate in this context the implications of these novel features of black holes. In particular we can infer a striking consequence: to consistently account for these new solutions, there must exist string states that so far have gone unnoticed in the microscopic analysis of certain configurations with fixed conserved charges.\n\nOur current understanding of neutral black holes within string theory is based on the “correspondence principle” . According to it, a black hole is identified with a highly excited string state that is obtained by adiabatically decreasing the gravitational (string) coupling. When the black hole horizon shrinks to string-scale size, stringy corrections to gravity become too large to be neglected. At this point, the black hole can be matched to a string state with (parametrically) the same value for the entropy. So one may ask what are the string states that correspond to a black hole and to a black ring with the same spin and mass, and how they differ. Conversely, if we take a highly excited string and increase the coupling, at some point gravitational collapse will occur. For certain values of the initial mass and spin, the string will be confronted with the dilemma of what object it collapses into: a black hole or a black ring.\n\nUnfortunately, the details of the string/black hole transition are still too poorly understood to enable us to see how, or even whether, string theory can distinguish between a neutral black hole and a black ring. The problem is further compounded by the fact that there do exist black holes and black rings with the same mass, spin, and area. From this perspective, the black ring is perhaps an unexpected complication.\n\nThe map between black holes and string states becomes much more precise if we consider charged BPS states (extremal black holes), which are protected by supersymmetry as we vary the coupling, or even for systems that are slightly excited above the BPS ground state (near-extremal black holes)222More than on supersymmetry, the success of the stringy description relies on an AdS structure near the horizon. This is also present in the solutions we construct.. Therefore, in order to investigate the issues discussed above, we construct in this paper a family of charged black ring solutions that, near the BPS extremal limit, can be identified with configurations of D-branes.\n\nWhen lifted to six dimensions, black rings become black tubes. The BPS ground states of the tubes that appear in our study belong to a class that has received recent attention. They can be referred to, generically, as supertubes. Perhaps the best known example is a configuration of D0-branes and fundamental strings (F1) bound to a cylindrical D2-brane tube . Its supergravity description, in different dimensions, was obtained in . From the viewpoint of the worldvolume theory of the D2 brane, the dissolved D0 and F1 appear as crossed magnetic and electric fields. These produce a Poynting vector tangent to the circular sections of the tube, and this gives the system an angular momentum. Other supertubes are obtained from this one via U-duality transformations. For instance, T-duality along the direction of the tube gives a helical D-string, which carries linear momentum as it coils around the tube axis . Further dualizations give a six-dimensional configuration where a D1-D5 bound state rotates on a cylinder. The D1 and the D5 share the longitudinal direction of the tube, with the other four directions of the D5-brane wrapping a four-torus. In the map from the supertube with D0-F1 charges to the one with D1-D5 charges, the circular section of the original D2 tube is mapped into a ring of Kaluza-Klein monopoles, and the longitudinal direction of the tube, which is compactified, corresponds to the fiber of the KK-monopoles. So the D1 and D5 branes are bound to a KK-monopole tube. A supergravity solution for a BPS D1-D5 supertube was first obtained in , and has been generalized and extensively studied in a series of papers [11, 12, 13, 14, 15] (even if only the last among these refers to them as supertubes).\n\nThe D1-D5 bound state is capable of carrying a momentum wave along the D1-D5 intersection. We construct a three-charge black ring, which, in addition to D1 and D5 charges, carries momentum charge (along the sixth-dimensional direction of the tube). However, this extra charge results in pathologies such as closed timelike curves (CTCs) at all points. In fact we find that the boost that generates the momentum is incompatible with the KK-monopole structure in the tube. So the way in which these CTCs appear is different from other recently studied instances of CTCs in rotating systems. Setting one of the charges, e.g., momentum, to zero, eliminates the pathologies, so in order to study the microscopic string description we focus on D1-D5-charged black tubes. The solution with equal D1 and D5 charges is dual to the charged black ring obtained recently in (which can be interpreted as carrying equal F1 and momentum charges).\n\nThe D1-D5-charged black tubes can be regarded as thermally excited supertubes. The non-extremal solutions have regular, non-degenerate horizons of finite area and, as in the neutral case, there exist black holes and black rings with the same values of the mass, spin, and charges. By exploiting the detailed knowledge of the microscopics of the D1-D5 system, we find that the stringy descriptions of black holes and black rings are indeed different. This indicates how string theory can, in this case, resolve the duplicity between such objects. We must add, though, that we do not understand yet how string theory distinguishes between the two black rings that have equal asymptotic charges.\n\nThe remainder of the paper is organized as follows: In Section 2 we introduce the neutral black ring, with some improvements of the original description in . The new three-charge black rings are given in Section 3, and its physical parameters are computed in Section 4. In Section 5 we study further the structure of the three-charge black rings, and find pathologies, including CTCs, from combining a boost with a KK-monopole. Thereafter we set the momentum charge to zero. Section 6 describes the extremal limit of the D1-D5-charged ring, and the connection with supertubes. In Section 7 we discuss the microscopic description of spinning black holes and rings with D1-D5 charges, and show that they have a very different description as states of string theory. Section 8 contains the conclusions and outlook of this work. Appendix A provides the intermediate steps leading from the neutral black ring to the three-charge black ring, while appendix B shows how the BMPV black hole [17, 18] can be obtained as an extremal limit of our solutions.\n\n## 2 Neutral black ring\n\nWe generate the new family of charged black rings by performing a series of transformations on the five-dimensional neutral black ring of , so we begin with an extended review of this solution. It is a vacuum solution with metric\n\n ds2 = −F(x)F(y)(dt+R√λν(1+y)dψ)2 (2.1) +R2(x−y)2[−F(x)(G(y)dψ2+F(y)G(y)dy2)+F(y)2(dx2G(x)+G(x)F(x)dϕ2)]\n\nwith\n\n F(ξ)=1−λξ,G(ξ)=(1−ξ2)(1−νξ). (2.2)\n\nThis form of the solution is slightly different from the original one used in [2, 16]. First, we have made the minor notational change , since it is more sensible to have a radius scale for the ring rather than the “acceleration parameter” (which came from the C-metric origin of this solution). Below we will see in what sense is a radius of the ring. Second, we have adopted the form of the cubic proposed in , which yields simple explicit forms for its three roots,\n\n ξ2=−1,ξ3=+1,ξ4=1ν. (2.3)\n\nFinally, having the roots of in this form suggests also to rename the root of as . The solution then has dimensionless parameters and , and a length scale .\n\nThe variables and take values in\n\n −1≤x≤1,−∞\n\nAs shown in , in order to balance forces in the ring one must identify and with equal period\n\n Δϕ=Δψ=4π√F(−1)|G′(−1)|=2π√1+λ1+ν. (2.5)\n\nThis eliminates the conical singularities at the fixed-point sets and of the Killing vectors and , respectively. There is still the possibility of conical singularities at . These can be avoided in two manners. Fixing\n\n λ=λc≡2ν1+ν2(blackring) (2.6)\n\nmakes the circular orbits of close off smoothly also at . Then parametrize a two-sphere, parametrizes a circle, and the solution describes a black ring. Alternatively, if we set\n\n λ=1(blackhole) (2.7)\n\nthen the orbits of do not close at . Then parametrize an at constant . The solution is the same as the spherical black hole of with a single rotation parameter. Both for black holes and black rings, is an ergosurface, is the event horizon, and the inner, spacelike singularity is reached as from above.\n\nThe parameter varies in\n\n 0≤ν<1. (2.8)\n\nAs we recover a non-rotating black hole, or a very thin black ring. At the opposite limit, , both the black hole and the black ring get flattened along the plane of rotation, and at result into the same solution with a naked ring singularity.\n\nWe will often find it convenient to work with as an independent parameter, to be eventually fixed to the equilibrium value . If we allow for values of other than (2.6) or (2.7), then whenever\n\n ν<λ<1 (2.9)\n\nwe find a black ring solution regular on and outside the horizon, except for a conical singularity on the disk bounded by the inner rim of the ring (). If then the ring is rotating faster than the equilibrium value, and there is a conical deficit balancing the excess centrifugal force. If instead then the rotation is too slow and a conical excess appears. If the horizon is replaced by a naked singularity. We shall refer to the solution with as the equilibrium, or balanced black ring. All these qualitative features will remain unchanged for the non-extremal charged rings that we study below.\n\nThe mass, spin, area, temperature and angular velocity at the horizon for these solutions are given by333The temperature here is defined as the surface gravity of the horizon divided by . We prefer not to use the Euclidean approach, because there is no analytic continuation of the black ring that produces a non-singular, real Euclidean solution.\n\n M0=3πR24Gλ(λ+1)ν+1,J0=πR32G√λν(λ+1)5/2(1+ν)2, (2.10)\n A0=8π2R3λ1/2(1+λ)(λ−ν)3/2(1+ν)2(1−ν),T0=14πR1−νλ1/2(λ−ν)1/2, (2.11)\n Ω0=1R√νλ(1+λ). (2.12)\n\nWe use the subscript for quantities that refer to the neutral ring, as opposed to the charged rings below. Formulas (2.10), (2.11) and (2.12) are valid in general for and in the ranges (2.8), (2.9), as long as there is no conical defect at infinity, i.e., when (2.5) is satisfied.\n\nFocusing on the dimensionless quantity\n\n 27π32GJ20M30=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩2νν+1(black hole)(1+ν)38ν(balanced black ring) (2.13)\n\none easily sees that for black holes it grows monotonically from to , while for (equilibrium) black rings it is infinite at , decreases to a minimum value at , and then grows to at . This implies that in the range\n\n 2732≤27π32GJ20M30<1 (2.14)\n\nthere exist one black hole and two black rings with the same value of the spin for fixed mass. This regime of non-uniqueness occurs when the parameter takes values in\n\n √5−2≤ν<1 (2.15)\n\nfor black rings, and in\n\n 2737≤ν<1 (2.16)\n\nfor black holes. Bear in mind that when we speak about non-uniqueness we always refer to equilibrium black rings.\n\nThere is a limit in which many features of the black ring become particularly clear. Take and as independent parameters, and let both of them approach zero at the same rate (so remains finite). Then we obtain a ring with a large spin for a given mass. The ring is very thin (‘hula-hoop’-like), and locally it approaches the geometry of a boosted black string. To recover this limit, we focus on a region near the horizon, of size small compared to , and scale\n\n R→∞,λ,ν→0 (2.17)\n\nwhile keeping and finite. Define new parameters , , and coordinates , , that remain finite in the limit,\n\n νR=r0sinh2σ,λR=r0cosh2σ, r=−RF(y)y,cosθ=x,w=Rψ, (2.18)\n\n(so is large) to find that the black ring (2.1) goes over to\n\n ds2=−¯f(dt−r0sinh2σ2r¯fdw)2+f¯fdw2+dr2f+r2dΩ22, (2.19)\n\nwith\n\n f=1−r0r,¯f=1−r0cosh2σr. (2.20)\n\nThis a boosted black string, with boost parameter . When the boost becomes light-like. For the equilibrium ring we require instead , and the boost is . This happens to be a rather special value. The ADM stress-energy tensor (see e.g., ) of the boosted black string is\n\n Ttt = r04G(1+cosh2σ), Ttw = r04Gsinhσcoshσ, (2.21) Tww = r04G(sinh2σ−1).\n\nHere and are the energy and momentum linear densities of the black string, while is its pressure density. The limiting boost of the balanced black ring is , so its limit is a pressureless black string. This absence of pressure reflects the delicate balance of forces that the black ring represents. Furthermore, one can check that for the balanced black ring the limiting mass and spin are such that\n\n M02πR→Ttt,J02πR2→Ttw. (2.22)\n\nSince and are densities of mass and momentum per unit length, we see that must be interpreted as the circle radius of very thin rings, in the sense that, in the limit, the coordinate is identified as . So is the proper length measured at a large transverse distance from the string. This is not the same as the proper length at the horizon, which is , and is expanded by the pressure of the boost. Consistently with this interpretation, the area per unit length of a thin large ring is , i.e., it becomes equal to the area per unit length of the boosted black string. Also, in this limit the angular velocity (2.12) is , so becomes the boost velocity.\n\nEq. (2.22) for the mass assumes that the ring is balanced, i.e., pressureless. For unbalanced rings the mass gets an additional contribution from the pressure of the conical defect disk.\n\n## 3 Charging up the ring\n\nThe method to build the three-charge black ring does not differ in essence from the one that gave the three-charge rotating black hole in five dimensions [21, 22]. The final solution is given below in eqs. (3)-(3.14), for the case where the charges correspond to fundamental strings (F1), NS5-branes and momentum . Here and in appendix A we provide an outline and some of the intermediate steps leading to it.\n\nA rotating charged black ring with a single electric charge was constructed in as a solution to low energy heterotic string theory. The charged solution was generated from the neutral solution (2.1) by application of a Hassan-Sen transformation . This involves a boost in an internal direction, and the first step in generalizing the charged black ring of is to realize that such a boost can be imitated by Lorentz boosts and T-duality in an extra spatial direction followed by a Kaluza-Klein reduction to five dimensions. We elaborate on this in the following as we derive a solution for a black ring with two charges.\n\nAdd to the metric (2.1) a flat sixth dimension . We also add four flat directions , , wrapped on a , but these will play a more passive role. A Lorentz boost, and , gives the solution linear momentum in the -direction, and subsequent T-dualization of the -direction exchanges the momentum with a fundamental string charge from the 3-form flux . The choice of subscripts in the boosts will become clearer later. We can then apply another Lorentz boost in the -direction to get a solution with both charge and momentum. The resulting six-dimensional black ring (or rather black tube) solution is given in appendix A in eqs. (A)-(A.5). It is a solution to the classical equations of motion obtained from the action of the low energy NS-NS sector of the superstring compactified on ,\n\n S6=12κ26∫d6x√−^ge−2Φ(R(6)+4(∇Φ)2−112^H2). (3.1)\n\nCompactify the -direction on a circle of radius . A Kaluza-Klein reduction along the -direction, using the ansatz\n\n ^gMNdxMdxN=gμνdxμdxν+e2σ(dz+A(n)μdxμ)2, (3.2)\n\nwhere and , leads to a five-dimensional action\n\n S5 = 12κ25∫d5x√−ge−2Φ+σ(R(5)+4(∇Φ)2−4∇Φ∇σ−112H2 (3.3)\n\nwith . In (3.3), , and we have defined , so that . The 3-form flux involves a Chern-Simons term and is given by\n\n Hμνρ = (∂μ^Bνρ−A(n)μF(1)νρ)+cyclic permutations. (3.4)\n\nBy setting and one obtains a consistent truncation of the theory (3.3) and the resulting action is that of the heterotic string in five dimensions with only one subgroup included (see also ). Defining the effective dilaton , the Einstein frame metric is\n\nThe five-dimensional solution obtained by KK reducing the solution (A)-(A.5) describes a rotating black ring with two charges, one from the KK gauge field and the other from the gauge field . Including in the metric (A) the four flat directions , , we can view it as a solution with fundamental string charge (corresponding to ) and momentum in the -direction (from the boost ).\n\nFor the special case where the boost parameters are , the gauge fields become identical and , so this is a solution of the low energy heterotic string action. Up to normalizations of the gauge fields, this is exactly the solution with one charge found in .\n\nWe now proceed to find a solution with three charges. We continue from the solution (A), plus the four flat directions , viewed as a Type IIB solution with and F1 charges. A sequence of dualities maps this solution to a supergravity solution describing the D1-D5-system. Boosting the resulting system in the direction before KK reducing will then give us a black ring with three charges. We describe the procedure briefly in this section and leave the details for appendix A.\n\nS-dualizing takes the black tube solution with [, F1] charges to a solution with charges [, D1]. T-dualizing the four directions , then gives charges [, D5], and S-dualizing again takes us to a [, NS5] system. Now T-dualizing converts the momentum into F1 charge, giving a solution [F1, NS5] of Type IIA. A T-dualization in any one of the direction is trivial and takes us to Type IIB, and we can then S-dualize the solution to get a tube solution with [D1, D5] charges.\n\nFor either of the solutions [F1, NS5] or [D1, D5] we can apply a boost with parameter in the -direction and then KK reduce to five dimensions. In appendix A, we give the metric and fields for the [, F1, NS5] solution. Reducing (A) to five dimensions we find a black ring solution of (3.3) with Einstein frame metric\n\n ds25 = −1(h5h1hn)2/3F(x)F(y)(dt−√λνR(1+y)coshα5coshα1coshαndψ +√λνR(1+x)sinhα5sinhα1sinhαndϕ)2 +(h5h1hn)1/3R2(x−y)2[−F(x)(G(y)dψ2+F(y)G(y)dy2) +F(y)2(dx2G(x)+G(x)F(x)dϕ2)],\n\nwhere we have defined\n\n hi(x,y) ≡ F(y)cosh2αi−F(x)sinh2αiF(y) (3.6) = 1+λ(x−y)F(y)sinh2αi.\n\nThe dilaton and the extra scalar are given by\n\n e−2Φ=h1(x,y)h5(x,y),e2σ=hn(x,y)h1(x,y). (3.7)\n\nThe gauge fields are\n\n A(1)t = (x−y)λsinh2α12F(y)h1(x,y) (3.8) A(1)ψ = √λνR(1+y)F(x)coshα5sinhα1coshαnF(y)h1(x,y) (3.9) A(1)ϕ = −√λνR(1+x)sinhα5coshα1sinhαnh1(x,y), (3.10)\n\nand\n\n A(n)μ=A(1)μ[α1↔αn]. (3.11)\n\nThere is an antisymmetric tensor field with components\n\n Btψ = −√λνR(1+y)F(x)coshα5sinhα1sinhαnF(y)h1(x,y) (3.12) Btϕ = √λνR(1+x)sinhα5coshα1coshαnh1(x,y) (3.13) Bψϕ = −12λR2sinh2α5(G(x)x−y+k(x)+νF(x)(1+x)(1+y)sinh2α1F(y)h1(x,y)), (3.14)\n\nwhere . In five dimensions, the 3-form flux given in (3.4) is dual to a 2-form field strength . can be obtained from the gauge potential\n\n A(5)=A(1)μ[α1↔α5]. (3.15)\n\nEach boost is associated to one kind of charge,\n\n α5 ⟶NS5, α1 ⟶F1, (3.16) αn ⟶P.\n\nUnder S-duality, the NS5 and F1 transform into D5 and D1. Setting any one of the charges to zero, we recover the two-charge black ring. Originally, this was obtained by a reduction of the [, F1]-system, but the three charges [, F1, NS5] can be permuted by U-duality transformations of the ten-dimensional solution. The Einstein-frame metric (3) is invariant under such transformations, however, the radii of compactification and the dilaton do change. If we regard the [, F1, NS5]-ring as a solution to type IIA supergravity, then it can be lifted to a solution of 11D supergravity, with an additional coordinate , of the form [, M2, M5]. Reduction along the coordinate yields another type IIA solution, with D0, F1 and D4 charges, which is also T-dual to the D1--D5 solution.\n\nThe particular case of the solution (3) where all charges are equal, , yields a solution to minimal supergravity in five dimensions444This solution was obtained previously by E. Teo, and then discarded due to the pathological CTCs to be described in sec. 5 ..\n\nOne remarkable feature of the new solution (3) is that, with the inclusion of the third charge the black ring has acquired angular momentum along a second independent axis, the direction of the 2-sphere. We will see, however, that this is more a problem than a virtue.\n\n## 4 Physical parameters\n\nMany of the qualitative properties of non-extremal black rings are the same as for neutral black rings — a main distinguishing feature will be discussed in the next section. The parameters and , and the coordinates and , vary in the same ranges as in Sec. 2, and and are identified with the same periods (2.5). Again, the event horizon is located at , and defines the ergoregion.\n\nIf we make the choice of the spherical black hole class of solutions (see (2.7)), we recover a particular case of the three-charge rotating black hole in five dimensions of [21, 22]555The solutions in [21, 22] have an additional parameter corresponding to the second angular momentum of the seed neutral Myers-Perry black hole.. The change from the coordinates used in this paper (appropriate for the ring) to the more conventional Boyer-Lindquist-type coordinates, can be found in . When all charges are nonzero this solution has a smooth inner horizon at , but if either one of the charges is set to zero, the inner horizon becomes a curvature singularity.\n\nWhen , and up to an important issue to be discussed in next section, the horizon is topologically and the solution is a spinning black ring, which is balanced and free of conical singularities only when (2.6). For the ring there is no smooth inner horizon; behind the event horizon at the curvature blows up at .\n\nThe physical quantities for the charged black ring are computed in the Einstein frame. The ADM mass and angular momenta are\n\n M = 13M0(cosh2α5+cosh2α1+cosh2αn), (4.1) Jψ = J0coshα5coshα1coshαn, (4.2) Jϕ = −J0sinhα5sinhα1sinhαn, (4.3)\n\nexpressed in terms of the mass and the angular momentum of the neutral solution (2.10).\n\nThe ring couples electrically to the gauge fields and the three corresponding charges are computed as\n\n (4.4)\n\nwhere is the area of a unit three-sphere, is a field strength, , and , , and , as can be seen from the action (3.3). We find\n\n Q5=4G3πM0sinh2α5,Q1=4G3πM0sinh2α1,Qn=4G3πM0sinh2αn. (4.5)\n\nWith this definition the six-dimensional momentum is . Note that for all values of the parameters, the charges and the mass satisfy the inequality\n\n |Q1|+|Q5|+|Qn|≤4GπM, (4.6)\n\ngeneralizing the bound found in .\n\nIt is clear that, again, there is a range of parameters where there exist a black hole and two black rings with the same values of the mass, spin, and the three charges . This happens for parameters in the ranges (2.14), (2.15), (2.16).\n\nThe area of the horizon and its temperature admit also simple expressions in terms of their values (2.11) for the neutral solution,\n\n A=A0coshα5coshα1coshαn, (4.7)\n T=T0coshα5coshα1coshαn. (4.8)\n\nThe charges of the black ring are proportional to the total number of branes and momentum quanta. To obtain these numbers, first we choose units where , and take for simplicity the volume of the compact four dimensions to be , so the five-dimensional Newton’s constant is\n\n G=πg2s4Rz, (4.9)\n\nwhere is the radius of the direction along the string/fivebrane intersection, and is the string coupling constant. Since the solution is not extremal, there can be both branes and antibranes, as well as left and right movers. The integer charges measure the difference between their numbers. For the D1-D5- black ring, the integer-quantized total numbers of D1-branes, D5-branes, and momentum units are\n\n n1−¯n1=g−1sQ1,n5−¯n5=g−1sQ5,nL−nR=n=RzP=R2zg2sQn. (4.10)\n\nFor the F1-NS5- ring the integer charges of F1 and NS5 are , and , respectively.\n\nThe black ring also carries other local (dipole-type) charges. The full structure of the solution will be uncovered in the next section.\n\nIn the same manner as we did for thin neutral rings, we can blow up the region near the ring by taking the limits (2.17), (2.18). This yields a charged black string in five dimensions, or a charged black 2-brane if lifted to six dimensions. If we take the limit for the F1-NS5- six-dimensional black tube (A), then the tube is straightened out to a 2-brane spanning the coordinates ,\n\n ds2 = hnh1(dz+r0cosh2σsinh2αn2rhndt−r0sinh2σcoshα5coshα1sinhαn2rhndw +r0sinh2σsinhα5sinhα1coshαn2hn(cosθ+1)dϕ)2 −¯fh1hn(dt+r0sinh2σ2r¯fcoshα5coshα1coshαndw +r0sinh2σ2sinhα5sinhα1sinhαn(cosθ+1)dϕ)2 +h5(f¯fdw2+dr2f+r2dΩ22).\n\nHere\n\n hi=1+r0rcosh2σsinh2αi. (4.12)\n\nObserve that we have left the boost parameters fixed, so this is not an extremal, nor near-extremal, limit. The charges have been simply scaled up, like the mass, as , so as to keep a finite charge per unit length of the ring666Close to extremality, this is different than the near-horizon limit of the near-extremal black ring. The details will be discussed elsewhere..\n\nA calculation of the ADM stress-energy tensor as in (2), yields again vanishing pressure for the limit of the balanced ring. The mass, momentum and charge densities agree, as in (2.22), with those for a string of (asymptotic) length . The solution (4) can also be obtained by direct application of the transformations of sec. 3 to the boosted black string.\n\n## 5 Closed Timelike Curves, and Kaluza-Klein-monopole tube\n\nWhen all three charges are non-zero the black ring metric (3) contains the one-form\n\n C≡dt−√λνR(1+y)coshα5coshα1coshαndψ+√λνR(1+x)sinhα5sinhα1sinhαndϕ (5.13)\n\nwith the effect that the orbits of are non-trivially fibered over the surfaces parametrized by . We should be careful to avoid Dirac-Misner string singularities777A Dirac-Misner string is a gravitational analogue of the Dirac string of gauge fields, first discussed by Misner in for the Taub-NUT solution. in the geometry, which may arise at the fixed-points of the orbits of and . Since vanishes at the only possible fixed-point set of , , there is no singularity there. However, if is to describe a two-sphere then will have fixed-points at . Now, at , but not at . We can remove the Dirac-Misner singularity at in a standard manner, taking a different coordinate patch to cover this region. For some , take as the coordinate in the patch , and in . In the overlap region , relate them via\n\n t′=t+(2√λνRsinhα5sinhα1sinhαn)ϕ. (5.14)\n\nWith this construction, the Dirac-Misner singularities are absent from both poles . But we have to pay a price: the matching (5.14) requires that (and ) be identified with periodicity\n\n Δt=(2√λνRsinhα5sinhα1sinhαn)Δϕ (5.15)\n\n(or an integer fraction of this). This introduces closed timelike curves outside the horizon, even at infinity. So, either we have string singularities, or we introduce naked CTCs: The solution with three non-zero charges is pathological. This problem is instead absent from the charged black hole solution, where has a fixed-point set only at .\n\nTo further clarify the origin of the problem, let us set the momentum charge to zero, , so the pathology disappears, and consider the solution lifted to six dimensions, as in (A). The relevant structure of the solution is the same whether we consider the F1-NS5 or D1-D5 black tubes. For later convenience we give the full ten-dimensional D1-D5 solution, with (string frame) metric\n\n ds2 = 1√h1h5(dz−√λνR(1+x)sinhα5sinhα1dϕ)2 (5.16) −1√h1h5F(x)F(y)(dt−√λνR(1+y)coshα5coshα1dψ)2 +√h1h5R2(x−y)2[−F(x)(G(y)dψ2+F(y)G(y)dy2)+F(y)2(dx2G(x)+G(x)F(x)dϕ2)] +√h1h59∑i=6dxidxi,\n\ndilaton\n\n e−2Φ=h5(x,y)h1(x,y), (5.17)\n\nand the nonzero components of the RR 2-form are\n\n Ctϕ = −√λνR(1+x)sinhα5coshα1h1(x,y) (5.18) Cψϕ = 12R2λsinh2α5(G(x)x−y+k(x)+νF(x)(1+x)(1+y)sinh2α1F(y)h1(x,y)) (5.19) Ctz = −(x−y)λsinh2α12F(y)h1(x,y) (5.20) Cψz = −√λνR(1+y)F(x)coshα5sinhα1F(y)h1(x,y). (5.21)\n\nNow we see that the direction is non-trivially fibered over the two-sphere. This fibration is actually quite familiar: it is a Hopf fibration due to the presence of a KK-monopole. The KK-monopole extends along the direction (and also along the four compact ), so we have a KK-monopole tube. In fact this is not unexpected. The D1-D5 black tube, can be transformed via U-dualities into a D0-F1 black tube. As will become clearer later, the extremal limit of all these solutions are supertubes. It is known that the D0-branes and F-strings in a supertube configuration live in the worldvolume of a cylindrical D2-brane. Mapping this D2 cylinder back into the D1-D5 tube, one obtains a KK-monopole along the circle with KK-direction . This is the structure we have identified directly above. The KK monopoles in the D1-D5 supertube have also been noted in . They are also present in the non-extremal solutions, as evidenced by the fibration, even if a horizon appears before the ‘nut’ is reached. Note that the KK-monopoles are absent when we consider spherical black hole solutions.\n\nAs before, removal of Dirac-Misner strings from the fibration requires that we identify with period\n\n Δz≡2πRz=2√λνRsinhα5sinhα1nKKΔϕ, (5.22)\n\nwhere is an integer that counts the number of KK-monopoles that make up the tube (or the number of times the KK-monopole winds around the tube). Since the KK-monopole is distributed on a circle, its charge is not one of the conserved charges of the black ring. Instead, it appears in the asymptotic region as a magnetic dipole of the field . The local charge can be measured by integrating the KK magnetic flux across an that intersects the ring at a single point, analogous to the calculation of the local D2 charge for the supertube in .\n\nIt is actually the combination of the KK-monopole and the boost along the KK-direction that is responsible for the pathological Dirac-Misner strings that we have found. To see this, consider the solution for a KK-monopole in five dimensions,\n\n ds2=−dt2+H−1[dz+q(cosθ+1)dϕ]2+H(dr2+r2dΩ22) (5.23)\n\nwith . The string at can be removed with adequately chosen coordinate patches. Now boost it along the direction with boost parameter , to find\n\n ds2 = HnH(dz−qsinh2αn2rHndt+qcoshαnHn(cosθ+1)dϕ)2 (5.24) −H−1n(dt−qsinhαn(cosθ+1)dϕ)2+H(dr2+r2dΩ22)\n\nwhere . The essence of the fiber structure of and over the in this solution can be recognized in (4), and in general it is clear how the Dirac-Misner strings at the poles of the force the periodicity of the time in the same manner as we found for the three-charge black ring. As a matter of fact, the KK-monopole fibration imposes identifications in the geometry that are incompatible with the boost. So besides the CTCs, the geometry (5.24) does not even describe a consistent fibration. This same problem is present for the three-charge black tube.\n\nThe dual tube with D0, F1 and D4 charges (the D4 wraps the four internal directions ) exhibits the same inconsistencies even if, as a solution to IIA supergravity, it does not contain a KK-monopole. When lifted to eleven dimensions, this system is described by M2-branes and M5-branes on a tube, carrying momentum along a common intersection. Exactly the same solution is obtained by lifting the type IIA ring with F1, NS5 and momentum charges. This implies that the M-branes are again in the background of a KK-monopole tube, which is incompatible with the momentum wave.\n\nNote that these problems are of a different nature, and more serious, than the CTCs found for the over-rotating charged black hole in five dimensions, which appear as a result of the identifications that compactify the solution from six to five dimensions . Undoing these identifications and going to the universal covering then removes the CTCs. In our case, the presence of the KK-monopole in the D1-D5 or F1-NS5 black tube forces the direction to be compact — cannot be zero — and also forbids the possibility of boosting the solution.\n\nSo three-charge black rings appear to be unphysical. However, the two-charge solutions obtained by setting any one of the charges to zero are perfectly sensible: the D1-D5 solution does not have momentum, and the D1- and D5- do not have a KK-monopole tube. The three-charge solution allows us to obtain any of these two-charge solutions immediately, and is at least a useful way of encoding all these U-dual solutions.\n\n## 6 Extremal limit\n\nAn extremal limit of the solution is obtained by sending at least one of the boost parameters to , while keeping the metric and the corresponding charges finite. From (4.5) we see that we must send while the products remain finite. One possibility is to take , with and finite. We show in appendix B that in this way one always recovers the extremal three-charge spherical black hole.\n\nHere we are more interested in the limit where while stays finite, which preserves the ring-like structure of the solution. Since , the possibility that this limit be taken for spherical black holes is excluded. Also, it is possible to have an extremal limit for the three-charge ring if and go to zero at a different rate. But the extremal three-charge solution that results presents the same pathologies that we found for generic three-charge rings, and therefore we will not consider it further. In the following we set the momentum charge to zero. It is easy to redo the analysis when, instead, any of the other two charges is set to zero.\n\nTo obtain this extremal limit for the D1-D5 black tube, send and while keeping\n\n λe2αi = 2QiR2, √λνeα5+α1 = 8GπJR3=2g2sRzJR3, (6.1)\n\nfixed. Note in the last line we have used (4.9).\n\nThe extremal solution can be put in a convenient form with a change to new coordinates ,\n\n r2=R21−xx−y,cos2θ=1+xx−y. (6.2)\n\nThen the metric, written down as a ten-dimensional D1-D5 solution in string frame, is\n\n ds2 = −1√h1h5(dt+g2sRzJsin2θΣdψ)2+1√h1h5(dz−g2sRzJcos2θΣdϕ)2 + √h1h5[Σ(dr2r2+R2+dθ2)+(r2+R2)sin2θdψ2+r2cos2θdϕ2] + √h1h59∑i=6dxidxi.\n\nwith\n\n hi=1+QiΣ,Σ≡r2+R2cos2θ. (6.4)\n\nWhen this extremal solution is obtained as a limit of balanced black rings, is not a free parameter, instead in the limit, so from (6.1) we find that the spin of the extremal ring is fixed to be\n\n J2 = 12R2zg4sQ1Q5R2 (6.5) = 12R2zg2sn1n5R2,\n\nwhere we use (4.10) and the fact that for the extremal solution there are no anti-D1 or anti-D5 branes. As described in sec. 5, the D1 and D5 branes are rotating on top of a KK-monopole tube, which imposes the periodicity condition (5.22) on the coordinate. In the extremal limit (6.1), equation (5.22) results in\n\n nKKR2=g2sR2zJ. (6.6)\n\nTogether with (4.10), this allows us to write the equation (6.5) for the spin of the extremal ring in terms of the numbers of strings, five-branes, and KK-monopoles, as\n\n J=12n1n5nKK. (6.7)\n\nThis two-charge extremal ring, or better, extremal tube in 6D, is in fact a D1-D5 supertube, within the class of -supersymmetric solutions studied in [14, 15]. Another coordinate form, that allows to make contact with the (six-dimensional) supertube solutions in , is obtained by changing ,\n\n ρ1=√r2+R2sinθ,ρ2=rcosθ. (6.8)\n\nThen take the same form as in (6.4), with\n\n Σ=√(ρ21+ρ22+R2)2−4R2ρ21, (6.9)\n\nand the metric is\n\n ds2 = −1√h1h5(dt+4Gπ2Jρ21Σ(ρ22+R2+Σ+ρ21)dψ)2 + 1√h1h5(dz−4GπJ(ρ22+R2+Σ−ρ21)Σ(ρ22+R2+Σ+ρ21)dϕ)2 + √h1h5(dρ21+ρ21dψ2+dρ22+ρ22dϕ2) + √h1h59∑i=6dxidxi.\n\nComparing to , we see that are harmonic functions in the flat of coordinates , with sources on a circle at , . These are then D1-D5 branes distributed on a tube.\n\nIn it was found that an upper bound on the spin of supertubes follows from the requirement that does not become timelike and hence CTCs do not appear. The place where this could happen is close to the tube, where the norm of becomes approximately, up to a positive factor,888See for the expression including all terms (in this reference, ).\n\n gψψ∝" ]

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[ "# Mathematics (MATH)\n\nMATH 7240  Advanced Group Theory  3 cr\n\nRepresentation theory of finite groups, presentations of finite and infinite groups, or other topics. May not be held with MATH 4240.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: permission of department.\n\nEquiv To: MATH 4240\n\nMATH 7260  Abstract Measure Theory  3 cr\n\nLebesgue and abstract measures, measurable functions, convergence theorems, absolutely continuous functions, measure spaces, the Radon-Nikodym theorem, Fubini's and Tonnelli's theorems. May not be held with MATH 4260 and the former MATH 4750.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: Permission of department.\n\nEquiv To: MATH 4260, MATH 4750\n\nMATH 7270  Algebraic Topology  3 cr\n\nThis course will serve as an introduction to elements of homotopy or homology theory. May not be held with MATH 4270 and the former MATH 4230.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: Permission of department.\n\nEquiv To: MATH 4230, MATH 4270\n\nMATH 7280  Basic Functional Analysis  3 cr\n\nBanach spaces, Hahn-Banach, open mapping and closed graph theorems, linear operators and functionals, dual space, Hilbert spaces and compact operators. May not be held with MATH 4280 and the former MATH 4750.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: Permission of department.\n\nEquiv To: MATH 4280, MATH 4750\n\nMATH 7290  Complex Analysis 2  3 cr\n\nConformal mappings, normal families, harmonic and subharmonic functions, Perron's family, Dirichlet problem and Green's function. May not be held with MATH 4290 and the former MATH 4710.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: Permission of department.\n\nEquiv To: MATH 4290, MATH 4710\n\nMATH 7300  Combinatorial Geometry  3 cr\n\nTopics in combinatorial geometry, including arrangements of convex bodies, introduction to polytopes, problems in discrete geometry, repeated distances, and geometric graphs. May not be held with MATH 4300.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: Permission of department.\n\nEquiv To: MATH 4300\n\nMATH 7320  Dynamical Systems  3 cr\n\nTechniques for the qualitative analysis of nonlinear systems of ordinary differential equations and discrete-time systems. May not be held with MATH 4320 and the former MATH 4800.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: Permission of department.\n\nEquiv To: MATH 4320, MATH 4800\n\nMATH 7330  Fundamentals of Approximation Theory  3 cr\n\nTheoretical aspects of approximation theory: density, existence, uniqueness; direct and inverse theorems for polynomial approximation. May not be held with MATH 4330.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: Permission of department.\n\nEquiv To: MATH 4330\n\nMATH 7340  Introduction to Algebraic Geometry  3 cr\n\nThis course will introduce students to the basics of affine and projective varieties through a combination of basic theoretical tools and elementary examples. May not be held with MATH 4340.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: Permission of department.\n\nEquiv To: MATH 4340\n\nMATH 7360  Introduction to Differential Geometry  3 cr\n\nManifolds and submanifolds. One of: exterior calculus and Stokes' theorem, Riemannian or symplectic geometry, and Hamiltonian mechanics. May not be held with MATH 4360 and the former MATH 4730.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: Permission of department.\n\nEquiv To: MATH 4360, MATH 4730\n\nMATH 7370  Linear Algebra and Matrix Analysis  3 cr\n\nNorms, matrix factorizations, eigenvalues/eigenvectors, theory of non-negative matrices. Applications to differential equations, math biology, numerical analysis, graph theory, etc. May not be held with MATH 4370 and the former MATH 4310.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: Permission of department.\n\nEquiv To: MATH 4310, MATH 4370\n\nMATH 7380  Mathematical Biology  3 cr\n\nFormulation, analysis and simulation of models in math biology. Applications will be chosen from population dynamics, epidemiology, ecology, immunology and cellular dynamics. May not be held with MATH 4380 and the former MATH 3530.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: Permission of department.\n\nEquiv To: MATH 3530, MATH 4380\n\nMATH 7390  Numerical Approximation Theory  3 cr\n\nComputational aspects of approximation by interpolatory polynomials, convolutions, artificial neural networks, splines and wavelets. May not be held with MATH 4390.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: Permission of the department.\n\nEquiv To: MATH 4390\n\nMATH 7440  Numerical Analysis of Partial Differential Equations  3 cr\n\nFinite difference method, theory of Elliptic PDEs, finite element method, iterative solution of linear systems. Emphasis will be on the error analysis. May not be held with MATH 4440 and the former MATH 8150.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: Permission of department.\n\nEquiv To: MATH 4440, MATH 8150\n\nMATH 7450  Number Theory 2  3 cr\n\nAlgebraic number theory, arithmetic geometry and analytic number theory, Diophantine equations, examples such as arithmetic of elliptic curves and Dirichlet L- functions. May not be held with MATH 4450 and the former MATH 3450.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: Permission of department.\n\nEquiv To: MATH 3450, MATH 4450\n\nMATH 7460  Partial Differential Equations 2  3 cr\n\nGreen's function, Poisson, heat, Schrodinger and wave equations, Fourier and Laplace transforms, introduction to functional analytic techniques. May not be held with MATH 4460 and the former MATH 4810.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: Permission of department.\n\nEquiv To: MATH 4460, MATH 4810\n\nMATH 7470  Rings and Modules  3 cr\n\nThe general theory of (non-commutative) rings, modules and algebras. May not be held with MATH 4470.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: Permission of department.\n\nEquiv To: MATH 4470\n\nMATH 8010  Advanced Matrix Computations  3 cr\n\nMatrix computation, decomposition of matrices, iterative methods, sparse matrices, eigenvalue problems.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisites: linear algebra, computing, numerical analysis, and consent of instructor.\n\nMATH 8110  Applied Finite Element Analysis  3 cr\n\nTheory and practice of the finite element method of the solution of partial differential equations and its application to engineering and scientific problems. It includes the h, p and h-p versions, a priori and a posteriori error estimates, adaptability and the structure of finite element software.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: numerical analysis and partial differential equations or consent of the instructor.\n\nMATH 8140  Advanced Numerical Analysis of Differential & Integral Equations  3 cr\n\nContinuation of MATH 4440/7440. Topics include spectral methods, time dependent equations, multigrid, domain decomposition methods, problems on infinite domains, methods for boundary integral equations, Riemann-Hilbert problems and integrable systems.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: Permission of the department.\n\nMATH 8210  Topics in Combinatorics 1  3 cr\n\nTopics will be chosen from the areas of algebraic combinatorics, coding theory, design theory, enumerative combinatorics, graph theory,\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: approval of department.\n\nMATH 8310  Partial Differential Equations 3  3 cr\n\nContinuation of MATH 4460/7460. Topics include functional analytic techniques for linear and nonlinear partial differential equations, conservation laws, KdV equation, singular perturbation, viscosity solutions.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisites: Permission of the department.\n\nMATH 8410  Seminar in Applied and Computational Mathematics 1  3 cr\n\nDesigned to accommodate special topics in applied or computational areas of mathematics not included in other course offerings. Students are advised to consult the department as to availability.\n\nMATH 8420  Seminar in Applied and Computational Mathematics 2  6 cr\n\nDesigned to accommodate special topics in applied or computational areas of mathematics not included in other course offerings. Students are advised to consult the department as to availability.\n\nMATH 8430  Seminar in Mathematics 1  3 cr\n\nDesigned to accommodate special topics not included in topics courses.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: approval of department.\n\nMATH 8440  Seminar in Mathematics 2  6 cr\n\nDesigned to accommodate special topics not included in topics courses.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: approval of department.\n\nMATH 8510  Topics in Algebra 1  3 cr\n\nDesigned to accommodate special topics not included in topics courses.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: approval of department.\n\nMATH 8520  Topics in Algebra 2  6 cr\n\nTopics will be chosen from the areas of associative and non-associative algebras, Boolean algebra and lattice theory, category theory, group theory, ring theory and universal algebra.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: approval of department.\n\nMATH 8610  Topics in Analysis 1  3 cr\n\nTopics will be chosen from the areas of asymptotics, functional analysis, operator theory, real and complex variables, summability theory, topological vector spaces.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: approval of department.\n\nMATH 8620  Topics in Analysis 2  6 cr\n\nTopics will be chosen from the areas of asymptotics, functional analysis, operator theory, real and complex variables, summability theory, topological vector spaces.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: approval of department.\n\nMATH 8720  Topics in Foundations 2  6 cr\n\nTopics will be chosen from the areas of logic, model theory, recursive functions, set theory.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: approval by department\n\nMATH 8810  Topics in Geometry 1  3 cr\n\nTopics will be chosen from the areas of algebraic curves, combinatorial geometry, Euclidean geometry, fractal geometry, groups and geometrics, projective geometry.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: approval of department.\n\nMATH 8910  Topics in Topology 1  3 cr\n\nTopics will be chosen from the areas of compactifications and related extensions, covering properties, rings of continuous functions, set-theoretic topology, topological groups, uniformities and related structures.\n\nPR/CR: A minimum grade of C is required unless otherwise indicated.\nPrerequisite: approval of department.\n\nMATH 8996  MSc project 1  6 cr\n\nThis is a project course exclusively for students enrolled in the Course-based MSc program. Students must submit a written report, on the order of 40 to 60 pages, which can be a survey of a topic in mathematics, for instance. This course is taken under the supervision of a faculty member. Course graded pass/fail.\n\nMATH 8998  MSc project 2  6 cr\n\nThis is a project course exclusively for students enrolled in the teaching track of the Course-based MSc program. Students must submit a written report, on the order of 20-30 pages, which can be a survey of a topic in mathematics, for instance. In addition, students are required to teach one undergraduate course. This course is taken under the supervision of a faculty member. Course graded pass/fail." ]

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[ "# FIN 534 WEEK 4 Quiz 3 A U.S. Treasury bond will pay a lump sum of \\$1,000 exactly 3 years from today. The nominal interest rate is 6%, semiannual compounding. Which of the following statements is CORRECT?Answer The PV of the \\$1,000 lump sum has a higher\n\nA U.S. Treasury bond will pay a lump sum of \\$1,000 exactly 3 years from today. The nominal interest rate is 6%, semiannual compounding. Which of the following statements is CORRECT?Answer\n\n The PV of the \\$1,000 lump sum has a higher present value than the PV of a 3-year, \\$333.33 ordinary annuity. The periodic interest rate is greater than 3%. The periodic rate is less than 3%. The present value would be greater if the lump sum were discounted back for more periods. The present value of the \\$1,000 would be smaller if interest were compounded monthly rather than semiannually.\n\nWhich of the following statements regarding a 20-year (240-month) \\$225,000, fixed-rate mortgage is CORRECT? (Ignore taxes and transactions costs.)Answer\n\n The outstanding balance declines at a slower rate in the later years of the loan's life. The remaining balance after three years will be \\$225,000 less one third of the interest paid during the first three years. Because it is a fixed-rate mortgage, the monthly loan payments (which include both interest and principal payments) are constant. Interest payments on the mortgage will increase steadily over time, but the total amount of each payment will remain constant. The proportion of the monthly payment that goes towards repayment of principal will be lower 10 years from now than it will be the first year.\n\nYour bank offers a 10-year certificate of deposit (CD) that pays 6.5% interest, compounded annually. If you invest \\$2,000 in the CD, how much will you have when it matures?Answer\n\n \\$3,754.27 \\$3,941.99 \\$4,139.09 \\$4,346.04 \\$4,563.34\n\nOf the following investments, which would have the lowest present value? Assume that the effective annual rate for all investments is the same and is greater than zero.Answer\n\n Investment A pays \\$250 at the end of every year for the next 10 years (a total of 10 payments). Investment B pays \\$125 at the end of every 6-month period for the next 10 years (a total of 20 payments). Investment C pays \\$125 at the beginning of every 6-month period for the next 10 years (a total of 20 payments). Investment D pays \\$2,500 at the end of 10 years (just one payment). Investment E pays \\$250 at the beginning of every year for the next 10 years (a total of 10 payments).\n\nWhich of the following statements regarding a 30-year monthly payment amortized mortgage with a nominal interest rate of 8% is CORRECT?Answer\n\n Exactly 8% of the first monthly payment represents interest. The monthly payments will decline over time. A smaller proportion of the last monthly payment will be interest, and a larger proportion will be principal, than for the first monthly payment. The total dollar amount of principal being paid off each month gets smaller as the loan approaches maturity. The amount representing interest in the first payment would be higher if the nominal interest rate were 6% rather than 8%.\n\nA U.S. Treasury bond will pay a lump sum of \\$1,000 exactly 3 years from today. The nominal interest rate is 6%, semiannual compounding. Which of the following statements is CORRECT?Answer\n\n The PV of the \\$1,000 lump sum has a smaller present value than the PV of a 3-year, \\$333.33 ordinary annuity. The periodic interest rate is greater than 3%. The periodic rate is less than 3%. The present value would be greater if the lump sum were discounted back for more periods. The present value of the \\$1,000 would be larger if interest were compounded monthly rather than semiannually.\n\nWhich of the following statements is CORRECT?Answer\n\n The time to maturity does not affect the change in the value of a bond in response to a given change in interest rates. You hold two bonds. One is a 10-year, zero coupon, bond and the other is a 10-year bond that pays a 6% annual coupon. The same market rate, 6%, applies to both bonds. If the market rate rises from the current level, the zero coupon bond will experience the smaller percentage decline. The shorter the time to maturity, the greater the change in the value of a bond in response to a given change in interest rates. The longer the time to maturity, the smaller the change in the value of a bond in response to a given change in interest rates. You hold two bonds. One is a 10-year, zero coupon, issue and the other is a 10-year bond that pays a 6% annual coupon. The same market rate, 6%, applies to both bonds. If the market rate rises from the current level, the zero coupon bond will experience the larger percentage decline.\n\nWhich of the following statements is CORRECT?Answer\n\n An indenture is a bond that is less risky than a mortgage bond. The expected return on a corporate bond will generally exceed the bond's yield to maturity. If a bond's coupon rate exceeds its yield to maturity, then its expected return to investors exceeds the yield to maturity. Under our bankruptcy laws, any firm that is in financial distress will be forced to declare bankruptcy and then be liquidated. All else equal, senior debt generally has a lower yield to maturity than subordinated debt\n\nWhich of the following statements is CORRECT?Answer\n\n Most sinking funds require the issuer to provide funds to a trustee, who saves the money so that it will be available to pay off bondholders when the bonds mature. A sinking fund provision makes a bond more risky to investors at the time of issuance. Sinking fund provisions never require companies to retire their debt; they only establish \"targets\" for the company to reduce its debt over time. If interest rates have increased since a company issued bonds with a sinking fund, the company is less likely to retire the bonds by buying them back in the open market, as opposed to calling them in at the sinking fund call price. Sinking fund provisions sometimes turn out to adversely affect bondholders, and this is most likely to occur if interest rates decline after the bond has been issued.\n\nWhich of the following statements is CORRECT?Answer\n\n On an expected yield basis, the expected capital gains yield will always be positive because an investor would not purchase a bond with an expected capital loss. On an expected yield basis, the expected current yield will always be positive because an investor would not purchase a bond that is not expected to pay any cash coupon interest. If a coupon bond is selling at par, its current yield equals its yield to maturity. The current yield on Bond A exceeds the current yield on Bond B; therefore, Bond A must have a higher yield to maturity than Bond B. If a bond is selling at a discount, the yield to call is a better measure of return than the yield to maturity.\n\nIf its yield to maturity declined by 1%, which of the following bonds would have the largest percentage increase in value?Answer\n\n A 1-year bond with an 8% coupon. A 10-year bond with an 8% coupon. A 10-year bond with a 12% coupon. A 10-year zero coupon bond. A 1-year zero coupon bond.\n\nWhich of the following statements is CORRECT?Answer\n\n If a coupon bond is selling at a discount, its price will continue to decline until it reaches its par value at maturity. If interest rates increase, the price of a 10-year coupon bond will decline by a greater percentage than the price of a 10-year zero coupon bond. If a bond's yield to maturity exceeds its annual coupon, then the bond will trade at a premium. If a coupon bond is selling at a premium, its current yield equals its yield to maturity. If a coupon bond is selling at par, its current yield equals its yield to maturity.\n\nAssume that interest rates on 15-year noncallable Treasury and corporate bonds with different ratings are as follows: T-bond = 7.72% A = 9.64%AAA = 8.72% BBB = 10.18% The differences in rates among these issues were most probably caused primarily by:Answer\n\n Tax effects. Default risk differences. Maturity risk differences. Inflation differences. Real risk-free rate differences\n\nWhich of the following statements is CORRECT?Answer\n\n If the risk-free rate rises, then the market risk premium must also rise. If a company's beta is halved, then its required return will also be halved. If a company's beta doubles, then its required return will also double. The slope of the security market line is equal to the market risk premium, (rM - rRF). Beta is measured by the slope of the security market line.\n\nIf you randomly select stocks and add them to your portfolio, which of the following statements best describes what you should expect?Answer\n\n Adding more such stocks will increase the portfolio's expected rate of return. Adding more such stocks will reduce the portfolio's beta coefficient and thus its systematic risk. Adding more such stocks will have no effect on the portfolio's risk. Adding more such stocks will reduce the portfolio's market risk but not its unsystematic risk. Adding more such stocks will reduce the portfolio's unsystematic, or diversifiable, risk.\n\nAssume that the risk-free rate is 6% and the market risk premium is 5%. Given this information, which of the following statements is CORRECT?Answer\n\n If a stock has a negative beta, its required return must also be negative. An index fund with beta = 1.0 should have a required return less than 11%. If a stock's beta doubles, its required return must also double. An index fund with beta = 1.0 should have a required return greater than 11%. An index fund with beta = 1.0 should have a required return of 11%.\n\nAssume that the risk-free rate is 5%. Which of the following statements is CORRECT?Answer\n\n If a stock's beta doubled, its required return under the CAPM would also double. If a stock's beta doubled, its required return under the CAPM would more than double. If a stock's beta were 1.0, its required return under the CAPM would be 5%. If a stock's beta were less than 1.0, its required return under the CAPM would be less than 5%. If a stock has a negative beta, its required return under the CAPM would be less than 5%.\n\nWhich of the following statements is CORRECT?Answer\n\n Diversifiable risk can be reduced by forming a large portfolio, but normally even highly-diversified portfolios are subject to market (or systematic) risk. A large portfolio of randomly selected stocks will have a standard deviation of returns that is greater than the standard deviation of a 1-stock portfolio if that one stock has a beta less than 1.0. A large portfolio of stocks whose betas are greater than 1.0 will have less market risk than a single stock with a beta = 0.8. If you add enough randomly selected stocks to a portfolio, you can completely eliminate all of the market risk from the portfolio. A large portfolio of randomly selected stocks will always have a standard deviation of returns that is less than the standard deviation of a portfolio with fewer stocks, regardless of how the stocks in the smaller portfolio are selected.\n\nWhich of the following statements is CORRECT?Answer\n\n The higher the correlation between the stocks in a portfolio, the lower the risk inherent in the portfolio. An investor can eliminate almost all risk if he or she holds a very large and well diversified portfolio of stocks. Once a portfolio has about 40 stocks, adding additional stocks will not reduce its risk by even a small amount. An investor can eliminate almost all diversifiable risk if he or she holds a very large, well-diversified portfolio of stocks. An investor can eliminate almost all market risk if he or she holds a very large and well diversified portfolio of stocks.\n\nWhich of the following statements is CORRECT?Answer\n\n The preferred stock of a given firm is generally less risky to investors than the same firm's common stock. Corporations cannot buy the preferred stocks of other corporations. Preferred dividends are not generally cumulative. A big advantage of preferred stock is that dividends on preferred stocks are tax deductible by the issuing corporation. Preferred stockholders have a priority over bondholders in the event of bankruptcy to the income, but not to the proceeds in a liquidation.\n\nStocks X and Y have the following data. Assuming the stock market is efficient and the stocks are in equilibrium, which of the following statements is CORRECT?\n\n X Y Price \\$30 \\$30 Expected growth (constant) 6% 4% Required return 12% 10%\n\n Stock Y has a higher dividend yield than Stock X. One year from now, Stock X's price is expected to be higher than Stock Y's price. Stock X has the higher expected year-end dividend. Stock Y has a higher capital gains yield. Stock X has a higher dividend yield than Stock Y.\n\nA stock just paid a dividend of D0 = \\$1.50. The required rate of return is rs = 10.1%, and the constant growth rate is g = 4.0%. What is the current stock price?Answer\n\n \\$23.11 \\$23.70 \\$24.31 \\$24.93 \\$25.57\n\nStocks A and B have the same price and are in equilibrium, but Stock A has the higher required rate of return. Which of the following statements is CORRECT?Answer\n\n Stock B must have a higher dividend yield than Stock A. Stock A must have a higher dividend yield than Stock B. If Stock A has a higher dividend yield than Stock B, its expected capital gains yield must be lower than Stock B's. Stock A must have both a higher dividend yield and a higher capital gains yield than Stock B. If Stock A has a lower dividend yield than Stock B, its expected capital gains yield must be higher than Stock B's.\n\nA share of Lash Inc.'s common stock just paid a dividend of \\$1.00. If the expected long-run growth rate for this stock is 5.4%, and if investors' required rate of return is 11.4%, what is the stock price?Answer\n\n \\$16.28 \\$16.70 \\$17.13 \\$17.57 \\$18.01\n\nThe required returns of Stocks X and Y are rX = 10% and rY = 12%. Which of the following statements is CORRECT?Answer\n\n If Stock Y and Stock X have the same dividend yield, then Stock Y must have a lower expected capital gains yield than Stock X. If Stock X and Stock Y have the same current dividend and the same expected dividend growth rate, then Stock Y must sell for a higher price. The stocks must sell for the same price. Stock Y must have a higher dividend yield than Stock X. If the market is in equilibrium, and if Stock Y has the lower expected dividend yield, then it must have the higher expected growth rate.\nField of study:" ]

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[ "# Systems of Linear Equations Worksheets\n\n1. Worksheets >\n2. Math >\n3. Algebra >\n4. Systems of Equations\n\nWalk through our printable solving systems of equations worksheets to learn the ins and outs of solving a set of linear equations. Ensure students are thoroughly informed of the methods of elimination, substitution, matrix, cross-multiplication, Cramer's Rule, and graphing that are crucial for arriving at the solutions. Reaffirm skills in graphing the equation of a line as a prerequisite to solve the pairs of simultaneous equations by graphing them.\n\nThis array of free pdf worksheets is meticulously designed for 8th grade and high school students.\n\nCCSS: HSA-REI.6\n\nSolving Systems of Equations - Any Method | #Worksheet 1\n\nBolster skills in various methods of solving systems of linear equations with this practice worksheet for grade 8. You are free to use elimination, substitution, or any other appropriate method to find the value of the variables.\n\nSolving Systems of Equations - Any Method | #Worksheet 2\n\nEnriched with eight systems of linear equations, this printable worksheet provides essential practice in solving pairs of simultaneous equations with two variables.\n\nSolving Systems of Equations - Graphing Method | #Worksheet 1\n\nHone your skills in graphing systems of linear equations with this free eighth grade worksheet. If you graph the given equations using their slope and y-intercept, you will find two lines. The point of intersection of the two lines is the required solution.\n\nSolving Systems of Equations - Graphing Method | #Worksheet 2\n\nTrain high school students in solving the systems of linear equations using the graphing method with this exclusive worksheet. Find the solution of the systems of simultaneous equations by graphing the two lines, finding their point of intersection, and writing the x and y coordinates from each ordered pair." ]

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[ "Calculating the n-th term of the series expansion of a special function [closed]\n\nI am trying to calculate the $n^{\\text{th}}$ term of the following polynomial:\n\n$$\\, _2F_1\\left(-n,n+3;\\frac{3}{2};x\\right)$$\n\nTo do this I calculate:\n\nc[k_] = SeriesCoefficient[\nHypergeometric2F1[-n, n+3, 3/2, x], {x, 0, k},\nAssumptions -> k >= 0\n];\n\nand get:\n\nc[k] //TeXForm\n\n$\\frac{\\sqrt{\\pi } (k-n-1)! (k+n+2)!}{2 k! \\left(k+\\frac{1}{2}\\right)! (-n-1)! (n+2)!}$\n\nand the problem is when I try to calculate for $n=10$:\n\nBlock[{n = 10}, Sum[c[k] x^k, {k, 0, 10}]]\n\nInfinity::indet: Indeterminate expression (0 2 Sqrt[π] ComplexInfinity)/Sqrt[π] encountered.\n\nInfinity::indet: Indeterminate expression (0 4 Sqrt[π] ComplexInfinity)/(3 Sqrt[π]) encountered.\n\nInfinity::indet: Indeterminate expression (0 8 Sqrt[π] ComplexInfinity)/(15 Sqrt[π]) encountered.\n\nGeneral::stop: Further output of Infinity::indet will be suppressed during this calculation.\n\nIndeterminate\n\nI get error messages and an incorrect answer. The correct result is:\n\nHypergeometric2F1[-n, n+3, 3/2, x] /. n->10\n\n1/33 (33 - 2860 x + 72072 x^2 - 823680 x^3 + 5125120 x^4 - 19009536 x^5 + 43868160 x^6 - 63504384 x^7 + 56033280 x^8 - 27525120 x^9 + 5767168 x^10)\n\nOther manifestations of problems with c:\n\nBlock[{n=10}, c[k]]\nBlock[{n=10}, c]\n\n0\n\nInfinity::indet: Indeterminate expression -((128 0 64 316234143225 Sqrt[π] Sqrt[π] ComplexInfinity)/(135135 10395 4096 Sqrt[π] Sqrt[π])) encountered.\n\nIndeterminate\n\nc does not give a useful symbolic result for the $k^{\\text{th}}$ term of the series.\n\nI try to use assumptions but it does not help.\n\nclosed as unclear what you're asking by bbgodfrey, rhermans, Daniel Lichtblau, MarcoB, José Antonio Díaz NavasSep 5 '18 at 11:17\n\nPlease clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.\n\n• It will be impossible to determine why you are getting an error without code. – N.J.Evans Aug 30 '18 at 12:30\n• please how to paste the code i copy from mathematica +crtl K but do not work – Clerk Aug 30 '18 at 12:47\n• The code corresponding to the question gives me the correct answer with no error messages. I suggest saving your notebook (without any extraneous material), restarting Mathematica, and then executing the notebook. If that does not work, check your code for errors. – bbgodfrey Aug 30 '18 at 13:49\n• Do the Sum first (with n left undefined, then substitute n->10 in the result. This will avoid the indeterminate forms. – Daniel Lichtblau Aug 30 '18 at 17:04\n• If the comment by Daniel ( %/.n->10 ) doesn't solve your problem, when using a fresh kernel as suggested by bbgodfrey, then it's not clear what you are asking. Please edit your question to clarify. – rhermans Aug 31 '18 at 7:51\n\nIt seems that the output generated by SeriesCoefficient is difficult for Mathematica to simplify into a version that can evaluate properly for integer n. So, I recommend using the new symbolic order derivatives introduced in M11.1:\n\nc[k_] = Assuming[\nn>=1,\nSimplify @ D[Hypergeometric2F1[-n, n+3, 3/2, x], {x, k}]/k! /. x->0\n]\n\n(Pochhammer[-n, k] Pochhammer[3 + n, k])/(k! Pochhammer[3/2, k])\n\nNote that this version of the symbolic coefficient of the series evaluates correctly for explicit values of n and k:\n\nBlock[{n=10}, c]\n\n-6336512/11\n\nLet's check:\n\nr1 = Block[{n=10}, Sum[c[k] x^k, {k, 0, n}]];\nr1 // TeXForm\n\n$\\frac{524288 x^{10}}{3}-\\frac{9175040 x^9}{11}+\\frac{18677760 x^8}{11}-\\frac{21168128 x^7}{11}+\\frac{14622720 x^6}{11}-\\frac{6336512 x^5}{11}+\\frac{465920 x^4}{3}-24960 x^3+2184 x^2-\\frac{260 x}{3}+1$\n\nCompare to the exact answer:\n\nr2 = Block[{n=10}, Expand @ Hypergeometric2F1[-n,n+3,3/2,x]];\nr2 // TeXForm\n\n$\\frac{524288 x^{10}}{3}-\\frac{9175040 x^9}{11}+\\frac{18677760 x^8}{11}-\\frac{21168128 x^7}{11}+\\frac{14622720 x^6}{11}-\\frac{6336512 x^5}{11}+\\frac{465920 x^4}{3}-24960 x^3+2184 x^2-\\frac{260 x}{3}+1$\n\nThey are the same:\n\nr1 === r2\n\nTrue\n\nThe OP asks in a comment about a different hypergeometric function argument:\n\nc[k_]=Assuming[\nn>=1,\nSimplify @ D[Hypergeometric2F1[3/2+n, -(3/2)-n, 3/2, x], {x, k}]/k! /. x->0\n];\n\nr1 = Block[{n=10}, Sum[c[k] x^k, {k, 0, n}]];\nr1 //TeXForm\n\n$\\frac{515830463005 x^{10}}{262144}-\\frac{264205846905 x^9}{65536}+\\frac{165491574435 x^8}{32768}-\\frac{8448518815 x^7}{2048}+\\frac{2304141495 x^6}{1024}-\\frac{2304141495 x^5}{2816}+\\frac{24775715 x^4}{128}-\\frac{452295 x^3}{16}+\\frac{18515 x^2}{8}-\\frac{529 x}{6}+1$\n\nr2 = Block[{n=10}, Hypergeometric2F1[3/2+n, -(3/2)-n, 3/2, x] //Expand];\nr2 //TeXForm\n\n$-\\frac{524288}{3} \\sqrt{1-x} x^{11}+\\frac{33292288}{33} \\sqrt{1-x} x^{10}-\\frac{27852800}{11} \\sqrt{1-x} x^9+\\frac{39845888}{11} \\sqrt{1-x} x^8-\\frac{35790848}{11} \\sqrt{1-x} x^7+\\frac{20959232}{11} \\sqrt{1-x} x^6-\\frac{24134656}{33} \\sqrt{1-x} x^5+\\frac{540800}{3} \\sqrt{1-x} x^4-27144 \\sqrt{1-x} x^3+\\frac{6812}{3} \\sqrt{1-x} x^2-\\frac{263}{3} \\sqrt{1-x} x+\\sqrt{1-x}$\n\nThe difference between them is that r1 is a series approximation of r2. When r2 is not a degree 10 polynomial, than the two expressions will not be the same. Instead compare r1 with the series approximation of r2:\n\nr2 + O[x]^11 //TeXForm\n\n$1-\\frac{529 x}{6}+\\frac{18515 x^2}{8}-\\frac{452295 x^3}{16}+\\frac{24775715 x^4}{128}-\\frac{2304141495 x^5}{2816}+\\frac{2304141495 x^6}{1024}-\\frac{8448518815 x^7}{2048}+\\frac{165491574435 x^8}{32768}-\\frac{264205846905 x^9}{65536}+\\frac{515830463005 x^{10}}{262144}+O\\left(x^{11}\\right)$\n\n• Thanks Carlk for your help it is works when n is neative interger but i try with n=3/2-n and do not work for example c[k_] = Assuming[n >= 1, Simplify@D[Hypergeometric2F1[3/2 + n, -(3/2) - n, 3/2, x], {x, k}]/ k! /. x -> 0] r1 = Block[{n = 10}, Sum[c[k] x^k, {k, 0, n}]]; r1 r2 = Block[{n = 10}, Hypergeometric2F1[3/2 + n, -(3/2) - n, 3/2, x] // Expand]; r2 gives different result check it if you will ,(sorry i could paste de code cause i paste from mathematica +Crtl K but do not work for me thanks anyway – Clerk Aug 31 '18 at 9:59\n• Thanks Carlk now it is very clear – Clerk Aug 31 '18 at 13:42\n\nInstead = write := at function c[k_]\n\n$Version (* \"11.3.0 for Microsoft Windows (64-bit) (March 7, 2018)\"*) ClearAll[\"Global*\"]; Remove[\"Global`*\"];(* Clears the kernel *) c[k_] := SeriesCoefficient[Hypergeometric2F1[-n, n + 3, 3/2, x], {x, 0, k}] Block[{n = 10}, Sum[c[k] x^k, {k, 0, 10}]]$\\frac{524288 x^{10}}{3}-\\frac{9175040 x^9}{11}+\\frac{18677760 x^8}{11}-\\frac{21168128 x^7}{11}+\\frac{14622720 x^6}{11}-\\frac{6336512 x^5}{11}+\\frac{465920 x^4}{3}-24960 x^3+2184 x^2-\\frac{260 x}{3}+1\\$" ]

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[ "", null, "Sooner or later every enthusiast programs a Lego Mindstorms Ev3 robot to follow a line. Indeed the line following task is the first that I tried myself, and there is a post on this blog where I show the results obtained with the Sup3r Car.\n\nHere I show how to program the SNATCH3R to follow a line with C# and the Monobrick firmware library; the main difference between the Sup3r Car and the SNATCH3R program is that the Sup3r Car uses a PID (Proportional Integrative Derivative) controller to stay on the track, while the SNATCH3R uses a much simpler look up table.\n\n### The line following task for the SNATCH3R\n\nThe task consists in programming an Ev3 robot to follow a line on the floor usually marked with a coloured tape; to sense the line on the floor the robot uses an Ev3 Color Sensor in Reflection Mode.\n\nThe SNATCH3R carries also an Ev3 IR Sensor set in proximity mode to detect the obstacles along the line. Once an obstacle is detected, the robot uses the gripper to grab and remove it.\n\n### The state machine\n\nThe SNATCH3R state machine has the following states:", null, "$S_1$: Follow the line until an obstacle is found", null, "$S_2$: Grab the obstacle", null, "$S_3$: Turn right", null, "$S_4$: Move forward", null, "$S_5$: Release the obstacle", null, "$S_6$: Move backward", null, "$S_7$: Turn left and when done get back to state", null, "$S_1$\n\nThis state machine is very simple in the sense that it is just a sequence. Once the only action inside a state is completed, the current state changes to the following one in the sequence.\n\nThe only exception to the rule above is the", null, "$S_1$ state: the robot changes from", null, "$S_1$ to", null, "$S_2$ only when the Ev3 IR Sensor detects an obstacle along the way.\n\nThe states of the state machine are defined as follows:\n\nEach state has a dedicated method which updates the robot behaviour, so the state machine looks clean:\n\nTo stay on the track the robot must sense the line on the floor with the Ev3 Color Sensor; the sensor feedback is then compared to a predefined Set-Point value. The Set-Point value is the value that the Ev3 Color Sensor reads when the red light it emits is half on the line and half on the floor.\n\nThe algorithm is simple: if the difference between the Set-Point and the sensor feedback is positive the robot must steer toward the line, while if the sensor feedback is negative the robot must steer toward the floor. With this statement I’m assuming that the line is darker than the floor.\n\nTo avoid sharp turn, the amount of correction is not constant but depends on how far the robot is from the Set-Point: if the feedback is close to the Set-Point the robot makes a small correction, while if the feedback is far from the Set-Point the robot makes a greater correction to its course.\n\nHave a look at the table on the left for the values used in the program proposed. For the black tape I used, and the floor in my home a good value for the SetPoint is 40, however these values should be tuned for your environment condition.\n\nIn the following code, observe also how the current state changes when an obstacle is detected by the Ev3 IR Sensor.\n\nThe look up table is a simple class which performs a linear search in an array as follows:\n\n### Grab the obstacle\n\nThe SNATCH3R performs this action in open loop: it simply rotates the Ev3 Medium Motor until its tachimeter count equals or exceeds 3500 hits, then the current state is changed to turn right.\n\n### Turn right\n\nThe action is performed in open loop. The two caterpillars move in opposite direction with equal speed until the left Ev3 Large Motor tacho count equals or exceeds 400. Actually this method turns the robot once to the left and once to the right for each obstacle encountered.\n\n### Move forward\n\nThis state contains an action performed in open loop. Both Ev3 Large Motors move forward until the left tacho count reaches or exceeds 400 (same value as turn right).\n\n### Release the obstacle\n\nThe SNATCH3R performs this action in open loop: the Ev3 Medium Motor rotates until its tachimeter count returns to zero, then the current state is changed to move backward.\n\n### Move backward\n\nThe SNATCH3R performs this action in open loop: the two Ev3 Large Motors rotate until left motor tachimeter count equals or exceeds -400 hits, then the current state is changed to turn left.\n\n### Turn left\n\nThe action is performed in open loop. The two caterpillars move in opposite direction with equal speed until the robot is back on track. To give the Ev3 Color Sensor a bit of tolerance in finding the black tape again, the SNATCH3R counter rotate 20% less than the amount of initial rotation. After that the state machine is back again to the one\n\nHappy coding!\n\nLine following C# program with the SNATCH3R\n\n### One thought on “Line following C# program with the SNATCH3R”\n\n•", null, "November 9, 2018 at 10:55 am" ]

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[ "/\n\n# Kane’s Method in Physics/Mechanics¶\n\nmechanics provides functionality for deriving equations of motion using Kane’s method [Kane1985]. This document will describe Kane’s method as used in this module, but not how the equations are actually derived.\n\n## Structure of Equations¶\n\nIn mechanics we are assuming there are 5 basic sets of equations needed to describe a system. They are: holonomic constraints, non-holonomic constraints, kinematic differential equations, dynamic equations, and differentiated non-holonomic equations.\n\n$\\begin{split}\\mathbf{f_h}(q, t) &= 0\\\\ \\mathbf{k_{nh}}(q, t) u + \\mathbf{f_{nh}}(q, t) &= 0\\\\ \\mathbf{k_{k\\dot{q}}}(q, t) \\dot{q} + \\mathbf{k_{ku}}(q, t) u + \\mathbf{f_k}(q, t) &= 0\\\\ \\mathbf{k_d}(q, t) \\dot{u} + \\mathbf{f_d}(q, \\dot{q}, u, t) &= 0\\\\ \\mathbf{k_{dnh}}(q, t) \\dot{u} + \\mathbf{f_{dnh}}(q, \\dot{q}, u, t) &= 0\\\\\\end{split}$\n\nIn mechanics holonomic constraints are only used for the linearization process; it is assumed that they will be too complicated to solve for the dependent coordinate(s). If you are able to easily solve a holonomic constraint, you should consider redefining your problem in terms of a smaller set of coordinates. Alternatively, the time-differentiated holonomic constraints can be supplied.\n\nKane’s method forms two expressions, $$F_r$$ and $$F_r^*$$, whose sum is zero. In this module, these expressions are rearranged into the following form:\n\n$$\\mathbf{M}(q, t) \\dot{u} = \\mathbf{f}(q, \\dot{q}, u, t)$$\n\nFor a non-holonomic system with $$o$$ total speeds and $$m$$ motion constraints, we will get o - m equations. The mass-matrix/forcing equations are then augmented in the following fashion:\n\n$\\begin{split}\\mathbf{M}(q, t) &= \\begin{bmatrix} \\mathbf{k_d}(q, t) \\\\ \\mathbf{k_{dnh}}(q, t) \\end{bmatrix}\\\\ \\mathbf{_{(forcing)}}(q, \\dot{q}, u, t) &= \\begin{bmatrix} - \\mathbf{f_d}(q, \\dot{q}, u, t) \\\\ - \\mathbf{f_{dnh}}(q, \\dot{q}, u, t) \\end{bmatrix}\\\\\\end{split}$\n\n## Kane’s Method in Physics/Mechanics¶\n\nThe formulation of the equations of motion in mechanics starts with creation of a KanesMethod object. Upon initialization of the KanesMethod object, an inertial reference frame needs to be supplied. along with some basic system information, suchs as coordinates and speeds\n\n>>> from sympy.physics.mechanics import *\n>>> N = ReferenceFrame('N')\n>>> q1, q2, u1, u2 = dynamicsymbols('q1 q2 u1 u2')\n>>> q1d, q2d, u1d, u2d = dynamicsymbols('q1 q2 u1 u2', 1)\n>>> KM = KanesMethod(N, [q1, q2], [u1, u2])\n\n\nIt is also important to supply the order of coordinates and speeds properly if there are dependent coordinates and speeds. They must be supplied after independent coordinates and speeds or as a keyword argument; this is shown later.\n\n>>> q1, q2, q3, q4 = dynamicsymbols('q1 q2 q3 q4')\n>>> u1, u2, u3, u4 = dynamicsymbols('u1 u2 u3 u4')\n>>> # Here we will assume q2 is dependent, and u2 and u3 are dependent\n>>> # We need the constraint equations to enter them though\n>>> KM = KanesMethod(N, [q1, q3, q4], [u1, u4])\n\n\nAdditionally, if there are auxiliary speeds, they need to be identified here. See the examples for more information on this. In this example u4 is the auxiliary speed.\n\n>>> KM = KanesMethod(N, [q1, q3, q4], [u1, u2, u3], u_auxiliary=[u4])\n\n\nKinematic differential equations must also be supplied; there are to be provided as a list of expressions which are each equal to zero. A trivial example follows:\n\n>>> kd = [q1d - u1, q2d - u2]\n\n\nTurning on mechanics_printing() makes the expressions significantly shorter and is recommended. Alternatively, the mprint and mpprint commands can be used.\n\nIf there are non-holonomic constraints, dependent speeds need to be specified (and so do dependent coordinates, but they only come into play when linearizing the system). The constraints need to be supplied in a list of expressions which are equal to zero, trivial motion and configuration constraints are shown below:\n\n>>> N = ReferenceFrame('N')\n>>> q1, q2, q3, q4 = dynamicsymbols('q1 q2 q3 q4')\n>>> q1d, q2d, q3d, q4d = dynamicsymbols('q1 q2 q3 q4', 1)\n>>> u1, u2, u3, u4 = dynamicsymbols('u1 u2 u3 u4')\n>>> #Here we will assume q2 is dependent, and u2 and u3 are dependent\n>>> speed_cons = [u2 - u1, u3 - u1 - u4]\n>>> coord_cons = [q2 - q1]\n>>> q_ind = [q1, q3, q4]\n>>> q_dep = [q2]\n>>> u_ind = [u1, u4]\n>>> u_dep = [u2, u3]\n>>> kd = [q1d - u1, q2d - u2, q3d - u3, q4d - u4]\n>>> KM = KanesMethod(N, q_ind, u_ind, kd,\n... q_dependent=q_dep,\n... configuration_constraints=coord_cons,\n... u_dependent=u_dep,\n... velocity_constraints=speed_cons)\n\n\nA dictionary returning the solved $$\\dot{q}$$‘s can also be solved for:\n\n>>> mechanics_printing(pretty_print=False)\n>>> KM.kindiffdict()\n{q1': u1, q2': u2, q3': u3, q4': u4}\n\n\nThe final step in forming the equations of motion is supplying a list of bodies and particles, and a list of 2-tuples of the form (Point, Vector) or (ReferenceFrame, Vector) to represent applied forces and torques.\n\n>>> N = ReferenceFrame('N')\n>>> q, u = dynamicsymbols('q u')\n>>> qd, ud = dynamicsymbols('q u', 1)\n>>> P = Point('P')\n>>> P.set_vel(N, u * N.x)\n>>> Pa = Particle('Pa', P, 5)\n>>> BL = [Pa]\n>>> FL = [(P, 7 * N.x)]\n>>> KM = KanesMethod(N, [q], [u], [qd - u])\n>>> (fr, frstar) = KM.kanes_equations(FL, BL)\n>>> KM.mass_matrix\nMatrix([])\n>>> KM.forcing\nMatrix([])\n\n\nWhen there are motion constraints, the mass matrix is augmented by the $$k_{dnh}(q, t)$$ matrix, and the forcing vector by the $$f_{dnh}(q, \\dot{q}, u, t)$$ vector.\n\nThere are also the “full” mass matrix and “full” forcing vector terms, these include the kinematic differential equations; the mass matrix is of size (n + o) x (n + o), or square and the size of all coordinates and speeds.\n\n>>> KM.mass_matrix_full\nMatrix([\n[1, 0],\n[0, 5]])\n>>> KM.forcing_full\nMatrix([\n[u],\n])\n\n\nExploration of the provided examples is encouraged in order to gain more understanding of the KanesMethod object." ]

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[ "Support us and cryptocurrency!\nTry a browser that's faster, safer, ad-free, and earns you cryptocurrency for using it!", null, "More Than 475 Free MCAT Questions with Detailed Answers!\n\n## MCAT Question A Day - 2/17/14 - Answer!\n\nThe speed of sound in air at a given temperature is 360 m/s. What is the shortest closed-end pipe that will produce an air column length that can resonate at a frequency of 450 Hz?\n\nA.  20 cm\nB.  80 cm\nC.  90 cm\nD.  125 cm\n\nThe correct answer is (A). The shortest pipe length will be associated with the wavelength for the fundamental tone. For a closed-end pipe, this is: wavelength = 4L. The wavelength is the ratio of the speed to the frequency: wavelength = velocity/frequency. Therefore, the shortest length is:\n\nL = velocity / (4*frequency) = (360 m/s) / (4*450/s) = 0.20 m = 20 cm\n\nWant more questions? - Click here to check out our MCAT Question A Day archive" ]

[ null, "https://2.bp.blogspot.com/-O7PgH9cW0Yo/XMX88Pz-EwI/AAAAAAAACyo/kwyOrH3G4ukxC1wVS30kTbGoc3ZEE8FnACLcBGAs/s1600/BraveBat.png", null ]

[ "# Rotating Tower Rendering\n\nWed, Oct 30, 2013", null, "In the last article I walked through the foundation for my rotating tower platformer game. This time around I want to talk about how its rendered…\n\nThe important thing to realise is that the rendering is completely independent of the game logic. The actual game `update()` loop, along with all the collision detection logic is a simple, traditional, 2 dimensional platform game that doesn’t care or know about the fact that its a rotating tower.\n\nWe can easily switch between the flat or the rotating tower by making a few simple changes in our rendering code.\n\nYou can play both versions here\n\nIf you are curious, once you’ve read through this article to learn how the rotating tower is rendered, you can look at the source code on github to see just how few changes are required (about 14 lines) to render a flat version instead of a rotating tower:\n\n## The Tower\n\nEven though the game logic needs no knowledge of how it is rendered, that is obviously not true the other way around. So before we can talk about rendering we need to take a quick look at the data we use to represent the tower we are about to render.\n\nStarting off with some constants to represent the viewport width and height, along with column and row sizes for our 2 dimensional grid:\n\n``````var WIDTH = 720, // multiple of 360 keeps things lined up nicely...\nHEIGHT = 540, // ... and 4:3 w:h ratio matches canvas css\nHORIZON = HEIGHT/5, // how much ground to show below the tower\nMETER = HEIGHT/20, // how many pixels represent 1 meter\nCOL_WIDTH = METER * 3, // 2D column width\nROW_HEIGHT = METER, // 2D row height\n``````\n\nOur `Tower` is initialized with:\n\n• the number of rows and columns defined in the level map\n• an inner radius - how wide should the tower be (walls)\n• an outer radius - how wide should the tower be (including platforms)\n• a total width\n• a total height\n• a map (2-d array of cell types e.g. platform, ladder, coin, air, etc)\n``````var Tower = Class.create({\n\ninitialize: function(level) {\nthis.rows = level.map.length;\nthis.cols = level.map.length;\nthis.ir = WIDTH/4; // inner radius (walls)\nthis.or = this.ir * 1.2; // outer radius (walls plus platforms)\nthis.w = this.cols * COL_WIDTH;\nthis.h = this.rows * ROW_HEIGHT;\nthis.map = this.createMap(level.map);\n},\n\n...\n``````\n\n## Wrapping Around\n\nThe only concession the game logic has to make to allow rendering as a rotating tower is to ensure that movement along the horizontal axis wraps when the left/right boundaries of the map are crossed.\n\nThis can be achieved by normalizing the result whenever we increment or decrement a horizontal axis related value. This generic method is provided within the common.js module:\n\n``````normalize: function(n, min, max) {\nwhile (n < min)\nn += (max-min);\nwhile (n >= max)\nn -= (max-min);\nreturn n;\n},\n``````\n\nGiven this generic method, we can normalize our game x-coordinates, or map column indices with two additional helper methods:\n\n``````function normalizex(x) { return normalize(x, 0, tower.w); } // wrap x-coord around to stay within tower boundary\nfunction normalizeColumn(col) { return normalize(col, 0, tower.cols); } // wrap column around to stay within tower boundary\n``````\n\n## Utility Methods\n\nIn addition to normalizing, we provide more utility methods for frequently needed conversions:\n\n• x-coordinate to column index (and vice versa)\n• y-coordinate to row index (and vice versa)\n• x-coordinate to 0-360 degree angle around tower\n``````function x2col(x) { return Math.floor(normalizex(x)/COL_WIDTH); } // convert x-coord to tower column index\nfunction y2row(y) { return Math.floor(y/ROW_HEIGHT); } // convert y-coord to tower row index\nfunction col2x(col) { return col * COL_WIDTH; } // convert tower column index to x-coord\nfunction row2y(row) { return row * ROW_HEIGHT; } // convert tower row index to y-coord\nfunction x2a(x) { return 360 * (normalizex(x)/tower.w); } // convert x-coord to an angle around the tower\n``````\n\n## Transforming Coordinates\n\nThe following two utility methods, tx and ty are the core of how to render a rotating tower. They take the traditional 2-dimensional, x and y coordinates and transform them to viewport coordinates.\n\n``````function tx(x, r) { return r * Math.sin((normalizex(x-camera.x)/tower.w) *2*Math.PI); }\nfunction ty(y) { return HEIGHT - HORIZON - (y - camera.y); }\n``````\n\nNOTE: in this demo our HTML5 canvas size is set to our viewport size so that no further transformation is required.\n\n## Transforming Y\n\nWe would like our logical y coordinate to represent the distance up from the ground, but then we must transform that into a canvas coordinate that is down from the top of the canvas (s.y):", null, "• HEIGHT is the viewport/canvas height\n• HORIZON is a fixed distance from the bottom of the viewport for the camera\n• y is the y coordinate of an entity, e.g. a platform\n• c.y is the y coordinate of the camera\n• s.y is the physical screen y coordinate we are trying to calculate\n\nTherefore…\n\n• y - c.y is the distance from the camera to the entity\n• HEIGHT - HORIZON is the distance from the camera to the viewport top\n\nGiving us our final implementation of ty:\n\n``````function ty(y) {\nreturn HEIGHT - HORIZON - (y - camera.y);\n}\n``````\n\n## Transforming X\n\nNow, let’s tackle tx(). It’s harder to visualize so let’s just walk through the math.\n\nWe want to transform an x-coordinate (that is between 0 and tower.width) into a position around a tower with radius r.\n\nIf we assume the screen x-origin is in the center of the viewport (achieved by translating the canvas by WIDTH/2) then we will need to transform our logical x-coordinate into a value between -r and +r.\n\n• normalizex(x - camera.x) - horizontal distance between entity and camera\n• …/tower.w to get a % (0 to 1)\n• Math.sin(…) on the % gives a position around the tower (between -1 to 1)\n• multiply by r to get the final screen x coordinate.\n\nGiving us our final implementation of tx:\n\n``````function tx(x, r) {\nreturn r * Math.sin((normalizex(x-camera.x)/tower.w) * 2*Math.PI);\n}\n``````", null, "It is important to note that this equation will work on entities both in front of, and behind, the tower. Both will end up with a transformed x coordinate between -r and +r. We will have to ensure we render them in the correct order so that entities behind the tower are hidden by the rendered tower wall, which leads us to…\n\n## The Painters Algorithm\n\nOk, so we can transform 2 dimensional coordinates into rotating tower viewport coordinates, great, now what ?\n\nSince this is a simple canvas demo game, and we’re not using any fancy webgl, opengl, directx, three.js fancy shmantzy technologies, how do we render the tower and all its entities ?\n\nWe fall back on a traditional Painters Algorithm where we simply render the furthest items first, then the nearer items render over the top of them.\n\n• render the background stars\n• render any visible entities behind the tower (at the edges)\n• render the tower wall itself\n• render entities in front of the tower\n\nWe render the tower with a gradient to simulate a curved surface, and position our entities at the correct transformed coordinates and we can trick the eye into thinking its looking at a rotating tower!\n\n## Rendering the Tower\n\nThe trick to knowing which items are further than others is to recognize that each column in the 2 dimensional map is also a column around the tower.\n\nIf we imagine what the tower might look like from above, we can see that the column directly behind the tower (labelled 0) is the furthest, and the columns to either side (labelled 1) are a little nearer, and the columns to either side of those are a little nearer still, and so on:", null, "", null, "So, for example, when we render the back of the tower, we can walk from the center-back column to the edges, and when we render the front of the tower, we can do the opposite and walk from the outer edges towards the center-front column.\n\nWe end up rendering 4 different quadrants, 2 for the back, and 2 for the front, each quadrant is rendered 1 vertical column/slice at a time walking from the furthest away to the nearest.\n\n``````renderFront: function() {\n\nvar left = x2col(camera.x - tower.w/4),\ncenter = x2col(camera.x),\nright = x2col(camera.x + tower.w/4);\n\n},\n``````\n\nAs we walk through the columns and rows for each quadrant, we can identify the contents of each cell and render it appropriately:\n\n``````renderQuadrant: function(min, max, dir) {\nvar y, cell, row, col = min;\nwhile (col != max) {\nfor(row = 0 ; row <= tower.rows ; row++) {\ny = ty(row * ROW_HEIGHT);\ncell = tower.getCell(row, col);\nif (cell.platform)\nthis.renderPlatform(col, y);\nelse if (cell.coin)\nthis.renderCoin(col, y);\nif (cell.monster)\nthis.renderMonster(col, y, cell.monster);\n}\ncol = normalizeColumn(col + dir);\n}\n},\n``````\n\nFor example, to render a platform in a particular column at a (transformed) y position:\n\n``````renderPlatform: function(col, y) {\n\nvar x = col2x(col + 0.5), // center of the column\nx0 = tx(x, tower.or), // transformed center\nx1 = x0 - this.platformWidth/2, // left edge\nx2 = x0 + this.platformWidth/2; // right edge\n\nctx.fillRect( x1, y - ROW_HEIGHT, x2 - x1, ROW_HEIGHT);\nctx.strokeRect(x1, y - ROW_HEIGHT, x2 - x1, ROW_HEIGHT);\n\n},\n``````\n\nOther entities, monsters, ladders, coins, etc, are rendered in a similar fashion. The code in this article has been simplified a little for clarity, so as always, refer to the source code for the final implementation.\n\n## Next Time…\n\nI have shown that how the game is rendered can (and should) be completely independent of how the core game logic is implemented. The only concession needed to render as a rotating tower is to ensure that the game’s horizontal coordinates wrap around when the boundaries are reached.\n\nGiven a traditional 2-dimensional grid of cells, rendering the platform game onto a rotating tower becomes largely a question of being able to transfrom x and y game coordinates to screen coordinates using a little Trigonometry.\n\nIn the next (and probably final) article I’ll take a closer look at the collision detection process.\n\nAgain, it is independent of how the game is rendered, so the fact that its a rotating tower is irrelevant, but its using a slightly more complex collision detection process than my previous tiny platformer, so it’s worth a closer look.\n\nIn the mean time, you can…\n\nEnjoy!" ]

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[ "# Reliability-Based Design Optimization in SPICE Circuit Analysis\n\nSeptember 25, 2019", null, "Without reliability-based design optimization, a fire will be the least of your worries\n\nThe last thing anyone needs in a mission-critical system is for an important system to operate at different levels than designed. As part of a larger system, variations in the behavior of one subsystem may need to be suppressed and compensated in a downstream portion of a system. This is where reliability-based design optimization becomes important for systems design. Lives could be on the line when a portion of a system fails, and systems designer need to consider all the possible ways they can prevent tragedy.\n\n## What is Reliability-Based Design Optimization?\n\nReliability-based design optimization refers to a class of analysis techniques that are useful in a number of engineering fields. In this set of processes, the mean value of some aspect of the system is taken as the design variables, and the objective function is taken as some output from the system. These design variables can be component values with predefined tolerances, and the objective could be the voltage or current in a specific portion of a circuit. The objective function is then maximized or minimized subject to some predefined probabilistic constraints (such as failure probability or reliability index).\n\nIf this sounds abstract, think of reliability-based design optimization like this: your goal is to determine the component values that will produce the desired voltage/current in some portion of a complex circuit in the presence of natural variations in the component values and input signal. Once you have calculated the output voltage/current in the circuit and its variation, you can determine the probability that a component is overstressed. The output voltage/current will have some variation due to variations in component values and the inputs, and this will determine the probability that some portion of the circuit will operate above some rated voltage/current, ultimately leading to failure.\n\nFor linear time-invariant DC circuits, you can easily determine analytic expressions that define the output variations as a linear combination of the input variations. Even with circuits driven with periodic or arbitrary signals, you could determine expressions that define this relationship between input and output variances, although it may not be a simple linear combination. Instead, this is more likely to be a complicated function of time or frequency. With more complicated circuits that include feedback and/or nonlinear elements, a numerical technique can help you quickly determine this relationship between input and output variances.\n\nAs a first step, it is a good idea to create your circuit, optimize its behavior, and determine appropriate operating limits using smoke analysis. You can then determine the relationship between variances using Monte Carlo analysis.\n\n## Monte Carlo as Part of Reliability-Based Design Optimization\n\nThere are a number of methods that can be used to determine how the output from a circuit responds to random variations in the input signal or in the component values. Circuit simulators are purely deterministic, meaning they do not work with pre-defined probability distributions on component values and output a probability function describing variations in the output or the behavior of a specific net in a circuit.\n\nInstead, random variations need to be sampled from a user-defined probability distribution. It is most common to define component variations as being normally distributed or uniformly distributed. As most circuit simulators are not symbolic simulators, meaning they do not use algorithms to manipulate mathematical expressions analytically, random variations in component tolerances values can be simulated using random sampling, i.e., as a Monte Carlo simulation.", null, "Take yourself to Monte Carlo in reliability-based design optimization\n\nA Monte Carlo technique in a circuit simulation involves drawing random numbers from the distribution you define for your components using a random number generation technique. Random number generators in computers are semi-deterministic and their samples are often treated as uniformly distributed, which requires converting the uniformly distributed random number to your desired distribution for your components (i.e., the Box-Muller transform for converting uniform random variables to normally random variables).\n\nDefining the distribution for a component’s value requires simply defining the tolerance around some mean value. Different simulators may have different definitions of how tolerances are used to create a probability distribution for your components, so you should check your simulator’s documentation before starting your simulation. Note that, in circuit analyses, variations in component values are treated as independent (i.e., tolerances are uncorrelated), so you won’t need to use a technique like a Cholesky decomposition to generate a batch of correlated random variables.\n\nUsing the random draws for your component values, your circuit simulator simply calculates the voltage and current throughout the circuit and shows you a set of curves for voltage/current in the net you specify. This set of curves can then be plotted as a histogram, giving you the data you need to generate a probability distribution for the voltage/current in the net you’ve examined.\n\nAlthough we’ve been talking about this in terms of variations in your component values, the same process can include variations in the voltage/current input to a circuit, or the power given to active components. If you want to examine how these results change for different input voltages, you can incorporate a DC sweep into your simulation and examine how the output curve changes for different inputs.\n\n### Monte Carlo in Linear vs. Nonlinear Circuits\n\nThis is an important point to consider as variations in one or more components or operating parameters may not be linearly related to the voltage/current you are examining in your circuit. This is particularly true in nonlinear circuits, as well as time-variant linear or nonlinear circuits. With linear time-invariant circuits, the distribution in the output will have the same shape as the distribution you define for your components, assuming you define component and input signal variations to have the same general distribution, albeit with different specific values for statistical moments.\n\nThis is not the case with nonlinear circuits; the distributions will not match. The same can be said for time-variant circuits, although the distribution in the output may be linearly related to the distribution governing variations in the input at some earlier time. Properly analyzing the relationship between component value/input signal variations and the output from the nonlinear circuit you are examining requires some knowledge of probability theory, which goes beyond the scope of this article.", null, "These Zener diodes are common nonlinear circuit elements\n\n### Failure and Reliability\n\nOnce you determine the relationship between variations in the output voltage/current in some portion of the circuit, you can interpolate the results from your histogram to determine a probability distribution to determine the probability that your circuit will exceed some operating limits you’ve defined. This can be done at the component level (i.e., by looking at the current and voltage drop across a component) and compared with your results from smoke analysis. If your voltage/current exceeds your desired derating, then you need to change some aspect of the circuit so that the voltage and current do not exceed the operating limits you specify.\n\nThe design choices you make can include changing some component values, changing the input voltage (this is where combining Monte Carlo with a DC sweep is an excellent choice), or adding or removing some components from the circuit. The design decision you make is up to you and should be informed by your own experience. After you make your design changes, you’ll need to rerun the Monte Carlo analysis to see how you changes affect the probability of failure. As you make changes to your circuit, you can use parameter optimization to check that your desired voltage/current meet your design goals. Remember, your objective is to decrease your probability of stressing your components and failure.\n\n## Using the Results in Other Analyses\n\nNote that this analysis can be brought into the time-domain as part of transient analysis, within a Bode plot or transfer function plot as part of a frequency sweep, or other analyses. You can also examine how the output from the circuit changes when working with an arbitrary source in the time domain. As an example, when working with transistors, you can examine how variations in component parameters will influence when a transistor switches. This allows you to examine how variations in a large number of components influence timing jitter. Finally, you can use a Monte Carlo temperature sweep to examine the temperature of a circuit.\n\nNote that reliability-based design optimization should be distinguished from sensitivity analysis, although the two techniques are similar. In Sensitivity analysis, you understand the worst-case and best-case scenarios of your circuit. Also, you determine which component/s is having the maximum or minimum impact on your output. This helps you decide on the tolerances of each component. Thereby, reducing cost. The Monte Carlo portion of reliability-based design optimization examines many (often thousands) of possible variations in more than one component, and the results are a histogram of possible voltage/current in a specific net.", null, "## Final Thoughts\n\nFrom reading the above, you have probably noticed that reliability-based design optimization is really a combination of smoke analysis, sensitivity analysis, parameter optimization, and finally a calculation of failure probability. This series of analyses ultimately helps you optimize your circuit for reliability, rather than simply designing to meet a specific design goal.\n\nAs part of circuit design, reliability-based design optimization requires a circuit simulation package that helps you determine how the behavior of your circuit changes due to component tolerances and variations on the circuit’s input.\n\nPSpice Designer for OrCAD includes a number of analysis tools that help a designer balance between yield and cost and increase reliability at the same time. Of course, all this is in addition to the PSpice Optimizer which improves productivity when you perform reliability-based design optimization as standard functions and can incorporate thousands of component models. This package gives you a statistical view of the behavior of your circuit as your designs become more complex.\n\n###### Previous Article", null, "PCB Layout for the Multilayer PCB Fabrication Process\n\nKnowing what goes into the multilayer PCB fabrication process will help you to design a better PCB layout. ...\n\n###### Next Article", null, "Best PCB Routing Methods for BGA Escape Routing\n\nThe best way to set up your high-density boards for effective routing is to plan your BGA escape routing pa..." ]

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[ "Poor control loop stability results in degraded power supply rejection ratio, transient response, output noise, and, for mixed signal circuits, degraded jitter. Datasheet application examples and reference designs do not often include stability information, and measurements often conclude the stability is poor.\n\nOne method of designing stable control loops is to create an accurate broadband simulation model and to use pole zero compensation to optimize phase margin, gain margin and stability margin.\n\nThis is a time-consuming process and much of the data needed to create an accurate model is unpublished. Another issue is that many circuits are completely integrated without access to the control loops for measurement. Information on stability performance in data sheets and reference design literature is often limited or non-existent.\n\nThis article explains a shortcut to stabilizing a control loop in under 5 minutes without requiring any knowledge of the part’s internal circuitry.\n\n### Stability Assessment\n\nThere are well-established methods of assessing stability of black box systems. One method is impedance-based stability, which is popular for assessing the stability of switching converters combined with input filters.1\n\nAnother method of assessing stability is the non-invasive stability margin, which is an impedance-based mathematical conversion algorithm that I developed and is available from Picotest for many modern vector network analyzers (VNA).\n\nOpamp manufacturers generally provide an isolation resistance vs. load capacitance curve. This is an indirect way of assuring stability based on their knowledge of the opamp output impedance.2\n\nAll these assessments work on the same principle and compute stability from the interaction of impedances between two elements. Once we define stability, we can then use the principles of the above referenced assessment methods, in reverse, to design a stable solution. RF engineers use this method for designing oscillators by defining stability as zero and solving for the negative resistance required to exactly counteract the real resistance of a crystal. Basing our solution on these well-proven and well-published assessment methods puts us on solid ground.\n\n### Equivalent Circuit\n\nVoltage output devices, including opamps, voltage references, and both switching and linear regulators, appear inductive when looking at their outputs. The equivalent circuit is a voltage source in series with a resistor and inductor. The destabilizing load appears as a capacitor with equivalent series resistance (ESR). Current regulators also appear as a L-C resonance circuit. In this case, the current regulator is capacitive. The current regulator’s destabilizing load appears as an inductor with a series resistance. An equivalent circuit supporting both current regulators and voltage regulators is shown in Figure 1.", null, "Figure 1. Equivalent circuit showing the inductance and capacitance relating to the voltage regulator inductance, current regulator capacitance, and the reactive loading.\n\nIn this simulation, L is set to 1 uH and C to 1 uF. This choice normalizes the characteristic impedance of the resonance at 1Ω.", null, "The circuit in Figure 1 is simulated while sweeping the resistor R1 and then separately sweeping the resistor R2. The results, shown in Figure 2, show a minima when R1 or R2 are equal to 1.22. And since this is normalized, R1 or R2 would be set to 1.21*Zo.", null, "Figure 2. The minimum impedance occurs when the total series resistance is 1.22* the characteristic impedance.\n\nTwo other cases worth noting are shown in Figure 3. One case where each resistor is set to half of 1.22 or 0.61Ω each, resulting in the red trace (1.1Ω pk). A second case, where each resistor is set to match the 1Ω characteristic impedance, results in flat impedance (1Ω pk). This is the basic principal of connecting 50Ω RF sources to 50Ω RF loads using 50Ω characteristic impedance cable.", null, "Figure 3. There is a certain dependency on where the resistor is located. Note there is a special case where the series resistor and the ESR both match the characteristic impedance (sqrt(L/C)) resulting in flat impedance. With the resistors equally distributed between ESR and series resistance, the peak impedance is 10 percent above the characteristic impedance.\n\nWe can apply this knowledge to stabilize any control loop in just a few minutes. We only need to determine the source (opamp, reference, voltage regulator) output inductance value. The fastest way to determine this is by measuring the impedance, either with or without the destabilizing load. You might have several options depending on the VNA you use for the measurement.\n\nAn example measurement, performed on a DC-DC converter, is shown in Figure 4. The OMICRON Lab Bode 100 can display the inductance directly; it can be estimated from the impedance 3 dB point, shown in cursor 1, or from the 157 kHz resonant frequency with the 15 uF installed capacitor.", null, "Figure 4. The inductance of the voltage regulator can be obtained in a multitude of ways, depending on the VNA used. The Bode 100 used for this example can display it directly, which is preferred. It can also be approximated by the resonant frequency with a capacitor, or from the impedance 3 dB point as noted with the cursors.\n\nOnce the inductance is determined, set the total resistance to 1.4* the characteristic impedance. This value varies from 1 to 1.4 depending on the precise circuit but using 1.4 will always be a good choice.\n\nThe total resistance can be solved as:", null, "In the case of voltage regulators and voltage references, the capacitance and the ESR can both be selected, so there are many choices. Larger capacitors will result in lower maximum impedance. The voltage regulator output impedance from the circuit in Figure 4 is 57 nH and 26 mΩ. The capacitor is 15 uF with an ESR of 10 mΩ. This can be determined from the datasheet, assuming it specifies ESR, but it is more accurately seen from the impedance minima at 700 kHz, in Figure 3, in the impedance measurement.", null, "The series resistance is 26 mΩ, which can be subtracted from the total resistance required:", null, "The equivalent circuit from Figure 4 is simulated using the 10 mΩ measured capacitor ESR with the calculated 61 mΩ ESR. The results, shown in Figure 5, confirm that setting the ESR to 61 mΩ reduced the maximum at resonance to less than 10 percent above the calculated 62 mΩ characteristic impedance.", null, "Figure 5. A simulation using the inductance and series resistance from the measurement of Figure 4, along with the load capacitor, is simulated here in the RED trace. Setting the capacitor ESR as directed here results in 10 percent peaking above the characteristic impedance, which is a good balance between stability and performance.\n\n### Summary\n\nThis method of stabilizing the control loop requires only one quick measurement and is simple to apply to all types of circuits.\n\n1. Determine the inductance of the voltage regulator circuit or capacitance of the current regulator by measuring it directly\n\n2. If the capacitor is defined, solve for the resistor\n\n3. If the capacitor is not defined, select it based on the desired characteristic impedance\n\n4. Set the capacitor ESR to 1.4*Zo—series resistance\n\nThe stability can then be confirmed using a traditional Bode plot, NISM, or step load response.\n\nThe example I used here is available for free from the Keysight “How To” video channel.4 This full simulation model (see Figure 6) allows direct simulation of the Bode plot for the two cases shown in Figure 5 (see Figures 7 and 8).", null, "Figure 6. Schematic of the LM20143 example DC-DC converter from the Keysight EEs of Power Integrity video channel.", null, "Figure 7. The simulated Bode plot using the 15 uF 10 mΩ capacitor shows a phase margin of 22 degrees, which would be considered very poor. This is the cause of the impedance peak seen in Figure 4.", null, "Figure 8. The simulated Bode plot using the 15 uF capacitor, but with the 61 mΩ ESR, determined from the proposed method, improved the phase margin from 20 to 60 degrees. 60 degrees is generally seen as the optimum balance between stability and performance.\n\n### Tips\n\nBe sure to measure the active circuit inductance or capacitance at several operating conditions. Many circuits will vary with operating voltage or operating current.\n\nIt is a good idea to measure many devices before selecting one for your circuit. The active circuit inductance or capacitance can vary drastically by manufacturer and/or by part. The lower the capacitance of the current regulator and the lower the inductance of the voltage regulator, the better. These reduce the characteristic impedance, requiring less series resistance for stability.\n\nThere are two relationships you might find helpful:\n\n1. The output inductance of a switching regulator is not the same as the output filter inductance, but it is proportional to the filter inductance. Reducing the filter inductance value will result in a proportional reduction in the output impedance used for stability assessment.\n\n2. The inductance of a linear regulator, voltage reference, or opamp is inversely proportional to the operating current. Adding a load resistor can greatly reduce the inductance presented, resulting in smaller capacitors. This is particularly true for circuits that can operate at very low output current.\n\n### References\n\n1. C. M. Wildrick, F. C. Lee, B. H. Cho, and B. Choi, “A Method of Defining the Load Impedance Specification for a Stable Distributed Power System,” IEEE Transactions on Power Electronics, Vol. 10, No. 3 , May 1995, DOI: 10.1109/63.387992.\n\n2. P. Semig and T. Claycomb, “Capacitive Load Drive Solution Using an Isolation Resistor,” Texas Instruments, December 2014, www.ti.com/lit/ug/tidu032c/tidu032c.pdf.\n\n3. S. Sandler, “The Inductive Nature of Voltage-Control Loops,” EDN Network, Feburary 5, 2015, www.edn.com/electronics-blogs/impedance-measurement-rescues/4438578/The-inductive-nature-of-voltage-control-loops.\n\n4. “How To Design for Power Integrity: DC-DC Converter Simulation and Modeling,” Keysight EEsof EDA, April 7, 2017, www.youtube.com/watch?v=CNyi4XU9xpY&list=PLtq84kH8xZ9FNXAsf-odoGNe6h5A6D3in&index=6&t=34s." ]

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[ "# mars.tensor.signbit¶\n\nmars.tensor.signbit(x, out=None, where=None, **kwargs)[source]\n\nReturns element-wise True where signbit is set (less than zero).\n\nx : array_like\nThe input value(s).\nout : Tensor, None, or tuple of Tensor and None, optional\nA location into which the result is stored. If provided, it must have a shape that the inputs broadcast to. If not provided or None, a freshly-allocated tensor is returned. A tuple (possible only as a keyword argument) must have length equal to the number of outputs.\nwhere : array_like, optional\nValues of True indicate to calculate the ufunc at that position, values of False indicate to leave the value in the output alone.\n\n**kwargs\n\nresult : Tensor of bool\nOutput tensor, or reference to out if that was supplied.\n>>> import mars.tensor as mt\n\n>>> mt.signbit(-1.2).execute()\nTrue\n>>> mt.signbit(mt.array([1, -2.3, 2.1])).execute()\narray([False, True, False])" ]

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[ "What Are the Steps to Calculating a Percent?\n\nThe steps to calculate a percent of a given number are first converting the percent to a decimal value by dividing it by 100, then multiplying the given number by that decimal value. The steps to calculate what percent one number is of another are to divide the first number by the second number, then multiply by 100 to convert the decimal result to a percentage value.\n\nFor example, to calculate what 25 percent of 80 is, first convert 25 percent to a decimal value by dividing by 100. To do this, move the decimal place two places to the left, for a decimal value of 0.25. Next multiply the original number, 80, by the decimal value, 0.25. The result is 20, so 25 percent of 80 is 20.\n\nIf starting with the whole numbers and needing to find the percent value, reverse the operation. To calculate what percentage 20 is of 80, first divide 20 by 80. The result is a decimal value of 0.25, which must be multiplied by 100 to convert it to a percentage value. To do this, move the decimal point two spaces to the right. The result is a percentage value of 25, so 20 is 25 percent of 80.\n\nSimilar Articles" ]

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[ "# Difference between revisions of \"1990 AIME Problems/Problem 14\"\n\n## Problem\n\nThe rectangle", null, "$ABCD^{}_{}$ below has dimensions", null, "$AB^{}_{} = 12 \\sqrt{3}$ and", null, "$BC^{}_{} = 13 \\sqrt{3}$. Diagonals", null, "$\\overline{AC}$ and", null, "$\\overline{BD}$ intersect at", null, "$P^{}_{}$. If triangle", null, "$ABP^{}_{}$ is cut out and removed, edges", null, "$\\overline{AP}$ and", null, "$\\overline{BP}$ are joined, and the figure is then creased along segments", null, "$\\overline{CP}$ and", null, "$\\overline{DP}$, we obtain a triangular pyramid, all four of whose faces are isosceles triangles. Find the volume of this pyramid.\n\n## Solution\n\n### Solution 1\n\nOur triangular pyramid has base", null, "$12\\sqrt{3} - 13\\sqrt{3} - 13\\sqrt{3} \\triangle$. The area of this isosceles triangle is easy to find by", null, "$\\frac{1}{2}bh$, where we can find", null, "$h$ to be", null, "$\\sqrt{399}$ by the Pythagorean Theorem. Thus", null, "$A = \\frac 12(12\\sqrt{3})\\sqrt{399} = 18\\sqrt{133}$.\n\nTo find the volume, we want to use the equation", null, "$\\frac 13Bh = 6\\sqrt{133}h$, so we need to find the height of the tetrahedron. By the Pythagorean Theorem,", null, "$AP = CP = DP = \\frac{\\sqrt{939}}{2}$. If we let", null, "$P$ be the center of a sphere with radius", null, "$\\frac{\\sqrt{939}}{2}$, then", null, "$A,C,D$ lie on the sphere. The cross section of the sphere is a circle, and the center of that circle is the foot of the perpendicular from the center of the sphere. Hence the foot of the height we want to find occurs at the circumcenter of", null, "$\\triangle ACD$.\n\nFrom here we just need to perform some brutish calculations. Using the formula", null, "$A = 18\\sqrt{133} = \\frac{abc}{4R}$ (", null, "$R$ being the circumradius), we find", null, "$R = \\frac{12\\sqrt{3} \\cdot (13\\sqrt{3})^2}{4\\cdot 18\\sqrt{133}} = \\frac{13^2\\sqrt{3}}{2\\sqrt{133}}$. By the Pythagorean Theorem,", null, "$\\begin{eqnarray*}h^2 &=& PA^2 - R^2 \\\\ &=& \\left(\\frac{\\sqrt{939}}{2}\\right)^2 - \\left(\\frac{13^2\\sqrt{3}}{2\\sqrt{133}}\\right)^2\\\\ &=& \\frac{939 \\cdot 133 - 13^4 \\cdot 3}{4 \\cdot 133} \\\\ h^2 &=& \\frac{13068 \\cdot 3}{4 \\cdot 133} = \\frac{99^2}{133}\\\\ h &=& \\frac{99}{\\sqrt{133}} \\end{eqnarray*}$\n\nFinally, we substitute", null, "$h$ into the volume equation to find", null, "$V = 6\\sqrt{133}\\left(\\frac{99}{\\sqrt{133}}\\right) = \\boxed{594}$.\n\n### Solution 2\n\nLet", null, "$\\triangle{ABC}$ (or the triangle with sides", null, "$12\\sqrt {3}$,", null, "$13\\sqrt {3}$,", null, "$13\\sqrt {3}$) be the base of our tetrahedron. We set points", null, "$B$ and", null, "$C$ as", null, "$(6\\sqrt {3}, 0, 0)$ and", null, "$( - 6\\sqrt {3}, 0, 0)$, respectively. Using Pythagoras, we find", null, "$A$ as", null, "$(0, \\sqrt {399}, 0)$. We know that the vertex of the tetrahedron (", null, "$D$) has to be of the form", null, "$(x, y, z)$, where", null, "$z$ is the altitude of the tetrahedron. Since the distance from", null, "$D$ to points", null, "$A$,", null, "$B$, and", null, "$C$ is", null, "$\\frac {\\sqrt {939}}{2}$, we can write three equations using the distance formula:", null, "$\\begin{eqnarray*} x^{2} + (y - \\sqrt {399})^{2} + z^{2} &=& \\frac {939}{4}\\\\ (x - 6\\sqrt {3})^{2} + y^{2} + z^{2} &=& \\frac {939}{4}\\\\ (x + 6\\sqrt {3})^{2} + y^{2} + z^{2} &=& \\frac {939}{4} \\end{eqnarray*}$\n\nSubtracting the last two equations, we get", null, "$x = 0$. Solving for", null, "$y,z$ with a bit of effort, we eventually get", null, "$x = 0$,", null, "$y = \\frac {291}{2\\sqrt {399}}$,", null, "$z = \\frac {99}{\\sqrt {133}}$. Since the area of a triangle is", null, "$\\frac {1}{2}\\cdot bh$, we have the base area as", null, "$18\\sqrt {133}$. Thus, the volume is", null, "$V = \\frac {1}{3}\\cdot18\\sqrt {133}\\cdot\\frac {99}{\\sqrt {133}} = 6\\cdot99 = 594$." ]

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[ "# Learning Objectives\n\n### Learning Objectives\n\nBy the end of this section, you will be able to do the following:\n\n• Define quantum states and their relationship to modern physics\n• Calculate the quantum energy of lights\n• Explain how photon energies vary across divisions of the electromagnetic spectrum\n blackbody quantized quantum ultraviolet catastrophe\n\n# Blackbodies\n\n### Blackbodies\n\nOur first story of curious significance begins with a T-shirt. You are likely aware that wearing a tight black T-shirt outside on a hot day provides a significantly less comfortable experience than wearing a white shirt. Black shirts, as well as all other black objects, will absorb and re-emit a significantly greater amount of radiation from the sun. This shirt is a good approximation of what is called a blackbody.\n\nA perfect blackbody is one that absorbs and re-emits all radiated energy that is incident upon it. Imagine wearing a tight shirt that did this! This phenomenon is often modeled with quite a different scenario. Imagine carving a small hole in an oven that can be heated to very high temperatures. As the temperature of this container gets hotter and hotter, the radiation out of this dark hole would increase as well, re-emitting all energy provided it by the increased temperature. The hole may even begin to glow in different colors as the temperature is increased. Like a burner on your stove, the hole would glow red, then orange, then blue, as the temperature is increased. In time, the hole would continue to glow but the light would be invisible to our eyes. This container is a good model of a perfect blackbody.\n\nIt is the analysis of blackbodies that led to one of the most consequential discoveries of the twentieth century. Take a moment to carefully examine Figure 21.2. What relationships exist? What trends can you see? The more time you spend interpreting this figure, the closer you will be to understanding quantum physics!\n\nFigure 21.2 Graphs of blackbody radiation (from an ideal radiator) at three different radiator temperatures. The intensity or rate of radiation emission increases dramatically with temperature, and the peak of the spectrum shifts toward the visible and ultraviolet parts of the spectrum. The shape of the spectrum cannot be described with classical physics.\n\n### Tips For Success\n\nWhen encountering a new graph, it is best to try to interpret the graph before you read about it. Doing this will make the following text more meaningful and will help to remind yourself of some of the key concepts within the section.\n\n# Understanding Blackbody Graphs\n\n### Understanding Blackbody Graphs\n\nFigure 21.2 is a plot of radiation intensity against radiated wavelength. In other words, it shows how the intensity of radiated light changes when a blackbody is heated to a particular temperature.\n\nIt may help to just follow the bottom-most red line labeled 3,000 K, red hot. The graph shows that when a blackbody acquires a temperature of 3,000 K, it radiates energy across the electromagnetic spectrum. However, the energy is most intensely emitted at a wavelength of approximately 1000 nm. This is in the infrared portion of the electromagnetic spectrum. While a body at this temperature would appear red-hot to our eyes, it would truly appear ‘infrared-hot’ if we were able to see the entire spectrum.\n\nA few other important notes regarding Figure 21.2:\n\n• As temperature increases, the total amount of energy radiated increases. This is shown by examining the area underneath each line.\n• Regardless of temperature, all red lines on the graph undergo a consistent pattern. While electromagnetic radiation is emitted throughout the spectrum, the intensity of this radiation peaks at one particular wavelength.\n• As the temperature changes, the wavelength of greatest radiation intensity changes. At 4,000 K, the radiation is most intense in the yellow-green portion of the spectrum. At 6,000 K, the blackbody would radiate white hot, due to intense radiation throughout the visible portion of the electromagnetic spectrum. Remember that white light is the emission of all visible colors simultaneously.\n• As the temperature increases, the frequency of light providing the greatest intensity increases as well. Recall the equation $v=fλ.v=fλ.$ Because the speed of light is constant, frequency and wavelength are inversely related. This is verified by the leftward movement of the three red lines as temperature is increased.\n\nWhile in science it is important to categorize observations, theorizing as to why the observations exist is crucial to scientific advancement. Why doesn’t a blackbody emit radiation evenly across all wavelengths? Why does the temperature of the body change the peak wavelength that is radiated? Why does an increase in temperature cause the peak wavelength emitted to decrease? It is questions like these that drove significant research at the turn of the twentieth century. And within the context of these questions, Max Planck discovered something of tremendous importance.\n\n# Planck’s Revolution\n\n### Planck’s Revolution\n\nThe prevailing theory at the time of Max Planck’s discovery was that intensity and frequency were related by the equation $I=2kTλ2.I=2kTλ2.$ This equation, derived from classical physics and using wave phenomena, infers that as wavelength increases, the intensity of energy provided will decrease with an inverse-squared relationship. This relationship is graphed in Figure 21.3 and shows a troubling trend. For starters, it should be apparent that the graph from this equation does not match the blackbody graphs found experimentally. Additionally, it shows that for an object of any temperature, there should be an infinite amount of energy quickly emitted in the shortest wavelengths. When theory and experimental results clash, it is important to re-evaluate both models. The disconnect between theory and reality was termed the ultraviolet catastrophe.\n\nFigure 21.3 The graph above shows the true spectral measurements by a blackbody against those predicted by the classical theory at the time. The discord between the predicted classical theory line and the actual results is known as the ultraviolet catastrophe.\n\nDue to concerns over the ultraviolet catastrophe, Max Planck began to question whether another factor impacted the relationship between intensity and wavelength. This factor, he posited, should affect the probability that short wavelength light would be emitted. Should this factor reduce the probability of short wavelength light, it would cause the radiance curve to not progress infinitely as in the classical theory, but would instead cause the curve to precipitate back downward as is shown in the 5,000 K, 4,000 K, and 3,000 K temperature lines of the graph in Figure 21.3. Planck noted that this factor, whatever it may be, must also be dependent on temperature, as the intensity decreases at lower and lower wavelengths as the temperature increases.\n\nThe determination of this probability factor was a groundbreaking discovery in physics, yielding insight not just into light but also into energy and matter itself. It would be the basis for Planck’s 1918 Nobel Prize in Physics and would result in the transition of physics from classical to modern understanding. In an attempt to determine the cause of the probability factor, Max Planck constructed a new theory. This theory, which created the branch of physics called quantum mechanics, speculated that the energy radiated by the blackbody could exist only in specific numerical, or quantum, states. This theory is described by the equation $E=nhf,E=nhf,$ where n is any nonnegative integer (0, 1, 2, 3, …) and h is Planck’s constant, given by $h=6.626×10−34J⋅s,h=6.626×10−34J⋅s,$ and f is frequency.\n\nThrough this equation, Planck’s probability factor can be more clearly understood. Each frequency of light provides a specific quantized amount of energy. Low frequency light, associated with longer wavelengths would provide a smaller amount of energy, while high frequency light, associated with shorter wavelengths, would provide a larger amount of energy. For specified temperatures with specific total energies, it makes sense that more low frequency light would be radiated than high frequency light. To a degree, the relationship is like pouring coins through a funnel. More of the smaller pennies would be able to pass through the funnel than the larger quarters. In other words, because the value of the coin is somewhat related to the size of the coin, the probability of a quarter passing through the funnel is reduced!\n\nFurthermore, an increase in temperature would signify the presence of higher energy. As a result, the greater amount of total blackbody energy would allow for more of the high frequency, short wavelength, energies to be radiated. This permits the peak of the blackbody curve to drift leftward as the temperature increases, as it does from the 3,000 K to 4,000 K to 5,000 K values. Furthering our coin analogy, consider a wider funnel. This funnel would permit more quarters to pass through and allow for a reduction in concern about the probability factor.\n\nIn summary, it is the interplay between the predicted classical model and the quantum probability that creates the curve depicted in Figure 21.3. Just as quarters have a higher currency denomination than pennies, higher frequencies come with larger amounts of energy. However, just as the probability of a quarter passing through a fixed diameter funnel is reduced, so is the probability of a high frequency light existing in a fixed temperature object. As is often the case in physics, it is the balancing of multiple incredible ideas that finally allows for better understanding.\n\n# Quantization\n\n### Quantization\n\nIt may be helpful at this point to further consider the idea of quantum states. Atoms, molecules, and fundamental electron and proton charges are all examples of physical entities that are quantized—that is, they appear only in certain discrete values and do not have every conceivable value. On the macroscopic scale, this is not a revolutionary concept. A standing wave on a string allows only particular harmonics described by integers. Going up and down a hill using discrete stair steps causes your potential energy to take on discrete values as you move from step to step. Furthermore, we cannot have a fraction of an atom, or part of an electron’s charge, or 14.33 cents. Rather, everything is built of integral multiples of these substructures.\n\nThat said, to discover quantum states within a phenomenon that science had always considered continuous would certainly be surprising. When Max Planck was able to use quantization to correctly describe the experimentally known shape of the blackbody spectrum, it was the first indication that energy was quantized on a small scale as well. This discovery earned Planck the Nobel Prize in Physics in 1918 and was such a revolutionary departure from classical physics that Planck himself was reluctant to accept his own idea. The general acceptance of Planck’s energy quantization was greatly enhanced by Einstein’s explanation of the photoelectric effect (discussed in the next section), which took energy quantization a step further.\n\nFigure 21.4 The German physicist Max Planck had a major influence on the early development of quantum mechanics, being the first to recognize that energy is sometimes quantized. Planck also made important contributions to special relativity and classical physics. (credit: Library of Congress, Prints and Photographs Division, Wikimedia Commons)\n\n### Worked Example\n\n#### How Many Photons per Second Does a Typical Light Bulb Produce?\n\nAssuming that 10 percent of a 100-W light bulb’s energy output is in the visible range (typical for incandescent bulbs) with an average wavelength of 580 nm, calculate the number of visible photons emitted per second.\n\n### Strategy\n\nThe number of visible photons per second is directly related to the amount of energy emitted each second, also known as the bulb’s power. By determining the bulb’s power, the energy emitted each second can be found. Since the power is given in watts, which is joules per second, the energy will be in joules. By comparing this to the amount of energy associated with each photon, the number of photons emitted each second can be determined.\n\nSolution\n\nThe power in visible light production is 10.0 percent of 100 W, or 10.0 J/s. The energy of the average visible photon is found by substituting the given average wavelength into the formula\n\n$E=nhf=nhcλ.E=nhf=nhcλ.$\n\nBy rearranging the above formula to determine energy per photon, this produces\n\n21.1$E/n=(6.63×10−34J⋅s)(3.00×108m/s)580×10−9m=3.43×10−19J/photon.E/n=(6.63×10−34J⋅s)(3.00×108m/s)580×10−9m=3.43×10−19J/photon.$\n\nThe number of visible photons per second is thus\n\n$photonssec=10.0J/s3.43×10−19J/photon=2.92×1019photons/s.photonssec=10.0J/s3.43×10−19J/photon=2.92×1019photons/s.$\n\nDiscussion\n\nThis incredible number of photons per second is verification that individual photons are insignificant in ordinary human experience. However, it is also a verification of our everyday experience—on the macroscopic scale, photons are so small that quantization becomes essentially continuous.\n\n### Worked Example\n\n#### How does Photon Energy Change with Various Portions of the EM Spectrum?\n\nRefer to the Graphs of Blackbody Radiation shown in the first figure in this section. Compare the energy necessary to radiate one photon of infrared light and one photon of visible light.\n\n### Strategy\n\nTo determine the energy radiated, it is necessary to use the equation $E=nhf.E=nhf.$ It is also necessary to find a representative frequency for infrared light and visible light.\n\nSolution\n\nAccording to the first figure in this section, one representative wavelength for infrared light is 2000 nm (2.000 × 10-6 m). The associated frequency of an infrared light is\n\n21.2$f=cλ=3.00×108m/s2.000×10−6m=1.50×1014Hz.f=cλ=3.00×108m/s2.000×10−6m=1.50×1014Hz.$\n\nUsing the equation$E=nhfE=nhf$, the energy associated with one photon of representative infrared light is\n\n21.3$En=h⋅f=(6.63×10−34J⋅s)(1.50×1014Hz)=9.95×10−20Jphoton.En=h⋅f=(6.63×10−34J⋅s)(1.50×1014Hz)=9.95×10−20Jphoton.$\n\nThe same process above can be used to determine the energy associated with one photon of representative visible light. According to the first figure in this section, one representative wavelength for visible light is 500 nm.\n\n21.4$f=cλ=3.00×108m/s5.00×10−7m=6.00×1014Hz.f=cλ=3.00×108m/s5.00×10−7m=6.00×1014Hz.$\n21.5$En=h⋅f=(6.63×10−34J⋅s)(6.00×1014Hz)=3.98×10−19Jphoton.En=h⋅f=(6.63×10−34J⋅s)(6.00×1014Hz)=3.98×10−19Jphoton.$\nDiscussion\n\nThis example verifies that as the wavelength of light decreases, the quantum energy increases. This explains why a fire burning with a blue flame is considered more dangerous than a fire with a red flame. Each photon of short-wavelength blue light emitted carries a greater amount of energy than a long-wavelength red light. This example also helps explain the differences in the 3,000 K, 4,000 K, and 6,000 K lines shown in the first figure in this section. As the temperature is increased, more energy is available for a greater number of short-wavelength photons to be emitted.\n\n# Practice Problems\n\n### Practice Problems\n\nAn AM radio station broadcasts at a frequency of 1,530 kHz . What is the energy in Joules of a photon emitted from this station?\n\n1. 10.1 × 10-26 J\n2. 1.01 × 10-28 J\n3. 1.01 × 10-29 J\n4. 1.01 × 10-27 J\n\nA photon travels with energy of 1.0 eV. What type of EM radiation is this photon?\n\nExercise 1\n\nDo reflective or absorptive surfaces more closely model a perfect blackbody?\n\n1. reflective surfaces\n2. absorptive surfaces\nExercise 2\nA black T-shirt is a good model of a blackbody. However, it is not perfect. What prevents a black T-shirt from being considered a perfect blackbody?\n1. The T-shirt reflects some light.\n2. The T-shirt absorbs all incident light.\n3. The T-shirt re-emits all the incident light.\n4. The T-shirt does not reflect light.\nExercise 3\nWhat is the mathematical relationship linking the energy of a photon to its frequency?\n1. $E=hfn$\n2. $E=nhf$\n3. $E=nfh$\n4. $E=nhf$\nExercise 4\n\nWhy do we not notice quantization of photons in everyday experience?\n\n1. because the size of each photon is very large\n2. because the mass of each photon is so small\n3. because the energy provided by photons is very large\n4. because the energy provided by photons is very small\nExercise 5\nTwo flames are observed on a stove. One is red while the other is blue. Which flame is hotter?\n1. The red flame is hotter because red light has lower frequency.\n2. The red flame is hotter because red light has higher frequency.\n3. The blue flame is hotter because blue light has lower frequency.\n4. The blue flame is hotter because blue light has higher frequency.\nExercise 6\nYour pupils dilate when visible light intensity is reduced. Does wearing sunglasses that lack UV blockers increase or decrease the UV hazard to your eyes? Explain.\n1. Yes, because more high-energy UV photons can enter the eye.\n2. Yes, because less high-energy UV photons can enter the eye.\n3. No, because more high-energy UV photons can enter the eye.\n4. No, because less high-energy UV photons can enter the eye.\nExercise 7\nThe temperature of a blackbody radiator is increased. What will happen to the most intense wavelength of light emitted as this increase occurs?\n1. The wavelength of the most intense radiation will vary randomly.\n2. The wavelength of the most intense radiation will increase.\n3. The wavelength of the most intense radiation will remain unchanged.\n4. The wavelength of the most intense radiation will decrease." ]

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[ "## T-shirt measurements.\n\nA=shoulder to shoulder, B=Across the chest, C=length (inch)\n\nKids Sizes\nSize 22: A=10.5, B=11, C=15.5\nSize 24: A=11, B=11.5, C=16\nSize 26: A=11.5, B=12, C=17.5\nSize 28: A=12, B=13, C=18.5\nSize 30: A=12.6, B=14.5, C=19.5\n\nUnisex Sizes\nSize 32: A=13, B=15, C=21\nSize 34: A=15, B=16, C=22\nSize 36: A=15.5, B=17.5, C=23.5\nSize 38: A=16.5, B=19, C=24\nSize 40: A=18, B=20, C=26.5\nSize 42: A=19.5, B=21, C=28" ]

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[ "Exercise 110.65.3 (Cubic hypersurfaces). Let $F \\in \\mathbf{C}[T_0, \\ldots , T_ n]$ be homogeneous of degree $3$. Given $3$ vectors $x, y, z \\in \\mathbf{C}^{n + 1}$ consider the condition\n\n$(*)\\quad F(\\lambda x + \\mu y + \\nu z) = 0 \\text{ in } \\mathbf{C}[\\lambda , \\mu , \\nu ]$\n\n1. What is the dimension of the space of all choices of $x, y, z$?\n\n2. How many equations on the coordinates of $x$, $y$, and $z$ is condition (*)?\n\n3. What is the expected dimension of the space of all triples $x, y, z$ such that (*) is true?\n\n4. What is the dimension of the space of all triples such that $x, y, z$ are linearly dependent?\n\n5. Conclude that on a hypersurface of degree $3$ in $\\mathbf{P}^ n$ we expect to find a linear subspace of dimension $2$ provided $n \\geq a$ where it is up to you to find $a$.\n\nIn your comment you can use Markdown and LaTeX style mathematics (enclose it like $\\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar)." ]

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[ "# qiskit.pulse.library.GaussianDeriv\n\nqiskit.pulse.library.GaussianDeriv(duration, amp, sigma, angle=0.0, name=None, limit_amplitude=None)\n\nAn unnormalized Gaussian derivative pulse.\n\nThe Gaussian function is centered around the halfway point of the pulse, and the envelope of the pulse is given by:\n\n$f(x) = -\\text{A}\\frac{x-\\mu}{\\text{sigma}^{2}}\\exp \\left[-\\left(\\frac{x-\\mu}{2\\text{sigma}}\\right)^{2}\\right] , 0 <= x < duration$\n\nwhere $\\text{A} = \\text{amp} \\times\\exp\\left(i\\times\\text{angle}\\right)$, and $\\mu=\\text{duration}/2$.\n\nParameters\n\nReturns\n\nScalableSymbolicPulse instance.\n\nReturn type\n\nScalableSymbolicPulse" ]

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[ "# Using sinc as a filter\n\nI'm trying to use a cardinal sine as a lowpass filter for a cosine signal with a fundamental of 1k.\n\nIn frequency domain, what really happens is that I'm multiplying two impulses (centered at -1k and 1k) with a rectangular pulse of width equals 2 (convolution is multiplication in frecuency domain). The result would be a constant zero function. However using the above code in Matlab I'm getting again the sinc function as output. Any help would be appreciated (attached the full plot).\n\nD = 5;\nf0 = 1000;\nfm = 2*f0; % nyquist reestriction\n\nt = (-D:1/fm:D);\nx = cos(1000*2*pi*t);\nh = sinc(t);\n\nx_p = filter(h,1,x);\nplot(x_p);", null, "• Using conv(h,x) I'm getting two cardinal sines, also tested with filter(x,1,h)\n– JMFS\nApr 30 '14 at 23:36\n\nYou messed with f0/fm relationship. Try this code:\n\nD = 5;\nf0 = 1e3;\nfm = 40e3;\nt = 0:1:2*fm*D-1;\nx = cos(2*pi*t*f0/fm);\nh = sinc(-D*pi:1/10:D*pi);\nx_p = filter(h,1,x);\nplot(x_p); plot(x); plot(h);\nfigure; psd(x,2^11,fm);\nfigure; psd(h,2^11,fm);\nfigure; psd(x_p,2^11,fm);\n\n• Thanks Dagoff, I run your code but x_p is still returning a cosine, like not filtering operation was made. F0 is the fundamental frequency of the cosine and fm is the sample rate. Wasn't <br> a typo?\n– JMFS\nMay 1 '14 at 15:03\n• Yes, because I made f0 in passband of sinc filter. If you want to reject cosine, make for example f0=4e3 in my code and cosine magnitude will go down. You can observe exact frequency response of sinc filter using psd and cosine outside its passband will be rejected. Of course, residual cosine still will be there, because filter rejection is not infinite. May 1 '14 at 16:16\n• It took me time to analyze your code, thanks for your patience :D I realize that when using a sampling rate of 10Hz instead of 2000Hz (or higher) in the original code (as you did in yours) I can successfully filter the signal. Could you explain to me why did you choose 10 Hz? Is the filter's length also an important parameter?\n– JMFS\nMay 2 '14 at 19:27\n• Probably, you meant 10Hz for f0, not the sampling rate fm. Signal spectrum (f0 in our case) should be less than fm/2. That's what you are missing. May 3 '14 at 0:59\n• 1/10 is a normalized frequency. We already chose sampling frequency to be 40KHz. Read this en.wikipedia.org/wiki/… May 7 '14 at 4:21\n\nProbably the problem was happening because I was using a filter as long as my signal, so I was getting only the transient instead of the whole filtered result; as Honglei Chen graciously explained in this thread. He also suggested to me to use this version of the code:\n\nD = 5;\nf0 = 1000;\nfm = 2*f0; % nyquist reestriction\nt = (-D:1/fm:D-1/fm);\nx = cos(1000*2*pi*t);\nh = sinc(t(3*numel(t)/8:5*numel(t)/8-1));\nx_p = filter(h,1,x);\nplot(x_p);" ]

[ null, "https://i.stack.imgur.com/GCO6k.png", null ]

[ "# Friedrich : Gaussian Process Regression\n\nThis libarie implements Gaussian Process Regression in Rust. Our goal is to provide a building block for other algorithms (such as Bayesian Optimization).\n\nGaussian process have both the ability to extract a lot of information from their training data and to return a prediction and an uncertainty on their prediction. Furthermore, they can handle non-linear phenomenons, take uncertainty on the inputs into account and encode a prior on the output.\n\nAll of those properties make them an algorithm of choice to perform regression when data is scarce or when having uncertainty bars on the ouput is a desirable property.\n\nHowever, the `o(n^3)` complexity of the algorithm makes the classical implementation unsuitable for large training datasets.\n\n## Functionalities\n\nThis implementation lets you :\n\n• define a gaussian process with default parameters or using the builder pattern\n• train it on multidimensional data\n• fit the parameters (kernel, prior and noise) on the training data\n• predict the mean and variance and covariance matrix for given inputs\n• sample the distribution at a given position\n\n## Inputs\n\nMost methods of this library can currently work with the following `input -> ouput` pairs :\n\n• `Vec<Vec<f64>> -> Vec<f64>` each inner vector is a multidimentional training sample\n• `Vec<f64> -> f64` a single multidimensional sample\n• `DMatrix<f64> -> DVector<f64>` using a nalgebra matrix with one row per sample\n\nSee the `Input` trait if you want to add you own input type.\n\n## Modules\n\n gaussian_process Gaussian process kernel Kernels prior Prior\n\n## Traits\n\n Input Implemented by `Input -> Output` type pairs" ]

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[ "LANDSAT 8 OLIP\nA time series of Landsat 8 OLIP images\nat Lizard Island, GBR, Australia\n\nwork done in 2020 at 15 m GSD\n\n Using the Panchromatic band for water column correction to derive water depth and spectral bottom signature: Landsat 8 OLIP bandset used for this work Purple=1, Blue=2, Green=3, PAN=4, Red=5, NIR=6 and SWIR1=7", null, "peer-reviewed and published in 2017", null, "home\n\n500*2000 HICO image of Bahrain, october 21rst 2013\nscene iss.2013294.1021.093734.L1B.Bahrain.v04.14946.20131021200925.100m.hico courtesy of OSU\nHICO: Hyperspectral Imager for the Coastal Ocean: GSD 100 m\nwork done march 2016\n\nhome\n\n# The data\n\nHICO bandset in nanometers for 4SM\nWLMin[ 1]=401.2 WL[ 1]=406.9 WLMax[ 1]=412.7\nWLMin[ 2]=412.7 WL[ 2]=418.4 WLMax[ 2]=424.1\nWLMin[ 3]=424.1 WL[ 3]=429.9 WLMax[ 3]=435.6\nWLMin[ 4]=435.6 WL[ 4]=441.3 WLMax[ 4]=447.0\nWLMin[ 5]=447.1 WL[ 5]=452.8 WLMax[ 5]=458.5\nWLMin[ 6]=458.5 WL[ 6]=464.2 WLMax[ 6]=469.9\nWLMin[ 7]=470.0 WL[ 7]=475.7 WLMax[ 7]=481.4    BLUE\nWLMin[ 8]=481.4 WL[ 8]=487.1 WLMax[ 8]=492.9\nWLMin[ 9]=492.9 WL[ 9]=498.6 WLMax[ 9]=504.3\nWLMin=504.3 WL=510.0 WLMax=515.8\nWLMin=515.8 WL=521.5 WLMax=527.2\nWLMin=527.2 WL=533.0 WLMax=538.7\nWLMin=538.7 WL=544.4 WLMax=550.1   GREEN\nWLMin=550.2 WL=555.9 WLMax=561.6\nWLMin=561.6 WL=567.3 WLMax=573.0\nWLMin=573.1 WL=578.8 WLMax=584.5\nWLMin=584.5 WL=590.2 WLMax=596.0\nWLMin=596.0 WL=601.7 WLMax=607.4\nWLMin=607.4 WL=613.2 WLMax=618.9\nWLMin=618.9 WL=624.6 WLMax=630.3\nWLMin=630.4 WL=636.1 WLMax=641.8\nWLMin=641.8 WL=647.5 WLMax=653.2   RED\nWLMin=653.3 WL=659.0 WLMax=664.7\nWLMin=664.7 WL=670.4 WLMax=676.1\nWLMin=676.2 WL=681.9 WLMax=687.6\nWLMin=687.6 WL=693.3 WLMax=699.1\nWLMin=699.1 WL=704.8 WLMax=710.5   PAN\nWLMin=836.6 WL=842.3 WLMax=848.0   NIR\nData preparation\n\n• Binning: HICO's original bands 1_to_52 are binned pairwise into 26 new_channels\n• visible bands are placed in channels 1 to 26 of the working database\n• and NIR band is placed into channel_28 of the working database\n• this is the maximum number of channels in the 4SM command line, even using a full HD screen!\n•\n• new_channels 10_to_26 are binned into a panchromatic new_channel\n• that's all of the Green-Red region\n• this new_channel is placed in channel_27 of the working database\n• and assigned a wavelength of 705 nm\n\n•\n\n# Deglinting\n\n• adjacency effect does not dominate the NIR signal\n•\n\n• by contrast, sea-surface clutter is a significant source of variation of the NIR signal" ]

[ null, "http://rsp-4sm.wifeo.com/images/4/4sm/4SM_graphical_abstract_100.png", null, "http://rsp-4sm.wifeo.com/images/4/4sm/4SM_graphical_abstract_100.png", null ]

[ "Cody\n\n# Problem 1227. Generate a random matrix A of (1,-1)\n\nSolution 1846173\n\nSubmitted on 11 Jun 2019\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Fail\nx = 5; assert(isequal(unique(rand_plusminus1(x)),[-1;1]))\n\ny = -1 1\n\nAssertion failed.\n\n2   Pass\nx = 1; assert(isequal(abs(rand_plusminus1(x)),1))\n\ny = 1\n\n3   Pass\nx = 0; assert(isequal(rand_plusminus1(x),[]))\n\ny = []\n\n4   Fail\nx = 999; assert(isequal(unique(rand_plusminus1(x)),[-1;1]))\n\ny = -1 1\n\nAssertion failed." ]

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[ "# Is there something behind non-commuting observables?\n\nConsider a quantum system described by the Hilbert space $\\mathcal{H}$ and consider $A,B\\in \\mathcal{L}(\\mathcal{H},\\mathcal{H})$ to be observables. If those observables do not commute there's no simultaneous basis of eigenvectors of each of them. In that case, in general if $|\\varphi\\rangle$ is eigenvector of $A$ it will not be of $B$.\n\nThis leads to the problem of not having a definite value of some quantity in some states.\n\nNow, this is just a mathematical model. It works because it agrees with observations. But it makes me wonder about something. Concerning the Physical quantities associated to $A$ and $B$ (if an example helps consider $A$ to be the position and $B$ the momentum) what is really behind non-commutativity?\n\nDo we have any idea whatsoever about why two observables do not commute? Is there any idea about any underlying reason for that?\n\nAgain I know one might say \"we don't care about that because the theory agrees with the observation\", but I can't really believe there's no underlying reason for some physical quantities be compatible while others are not.\n\nI believe this comes down to the fact that a measurement of a quantity affects the system in some way that interferes with the other quantity, but I don't know how to elaborate on this.\n\nEDIT: I think it's useful to emphasize that I'm not saying that \"I can't accept that there exist observables which don't commute\". This would enter the rather lengthy discussion about whether nature is deterministic or not, which is not what I'm trying to get here.\n\nMy point is: suppose $A_1,A_2,B_1,B_2$ are observables and suppose that $A_1$ and $B_1$ commute while $A_2$ and $B_2$ do not commute. My whole question is: do we know today why the physical quantities $A_1$ and $B_1$ are compatible (can be simultaneously known) and why the quantities $A_2$ and $B_2$ are not?\n\nIn other words: accepting that there are incompatible observables, and given a pair of incompatible observables do we know currently, or at least have a guess about why those physical quantities are incompatible?\n\n• Every physical theory is just \"a mathematical model\". If you are asking about the next higher level theory behind quantum field theory... we don't have that, yet. It may or it may not exist. If you want to find it, then you will have to make a measurement that quantum field theory can not describe. – CuriousOne Feb 29 '16 at 4:16\n• You say that if observables do not commute, they can have no eigenvectors in common. This is not true. – WillO Feb 29 '16 at 4:19\n• If you like this question you may also enjoy reading this Phys.SE post. – Qmechanic Feb 29 '16 at 18:38\n• If I tell you that X is the reason for observables not commuting, will you then ask the same question with \"observables not commuting\" replaced by X? (My point is that either the chain of reasons for things terminates at something that has no reason, or it goes on forever. So either at some point the answer is \"No, there is no reason\", or these kinds of questions go on forever, so you might want to rethink \"I can't really believe there's no underlying reason\") – ACuriousMind Mar 1 '16 at 18:41\n• @Shing: $\\pmatrix{1&1\\cr 0&2\\cr}$, $\\pmatrix{1&1\\cr 0&3\\cr}$ – WillO Mar 2 '16 at 16:04\n\nObservables don't commute if they can't be simultaneously diagonalized, i.e. if they don't share an eigenvector basis. If you look at this condition the right way, the resulting uncertainty principle becomes very intuitive.\n\nAs an example, consider the two-dimensional Hilbert space describing the polarization of a photon moving along the $z$ axis. Its polarization is a vector in the $xy$ plane.\n\nLet $A$ be the operator that determines whether a photon is polarized along the $x$ axis or the $y$ axis, assigning a value of 0 to the former option and 1 to the latter. You can measure $A$ using a simple polarizing filter, and its matrix elements are $$A = \\begin{pmatrix} 0 & 0 \\\\ 0 & 1 \\end{pmatrix}.$$\n\nNow let $B$ be the operator that determines whether a photon is $+$ polarized (i.e. polarized southwest/northeast) or $-$ polarized (polarized southeast/northwest), assigning them values 0 and 1, respectively. Then $$B = \\begin{pmatrix} 1/2 & -1/2 \\\\ -1/2 & 1/2 \\end{pmatrix}.$$\n\nThe operators $A$ and $B$ don't commute, so they can't be simultaneously diagonalized and thus obey an uncertainty principle. And you can immediately see why from geometry: $A$ and $B$ are picking out different sets of directions. If you had a definite value of $A$, you have to be either $x$ or $y$ polarized. If you had a definite value of $B$, you'd have to be $+$ or $-$ polarized. It's impossible to be both at once.\n\nOr, if you rephrase things in terms of compass directions, the questions \"are you going north or east\" and \"are you going northeast or southeast\" do not have simultaneously well-defined answers. This doesn't mean compasses are incorrect, or incomplete, or that observing a compass 'interferes with orientation'. They're just different directions.\n\nPosition and momentum are exactly the same way. A position eigenstate is sharply localized, while a momentum eigenstate has infinite spatial extent. Thinking of the Hilbert space as a vector space, they're simply picking out different directions; no vector is an eigenvector of both at once.\n\n• this is a truly excellent answer. – JPattarini Feb 29 '16 at 7:58\n• As @JamesPattarini says, this answer is great. Could you take it one step further and finish the connection to the HUP derivation, maybe the special case of $\\Delta x \\Delta p$ for simplicity? – user1717828 Feb 29 '16 at 15:23\n• As a lay enthusiast, I think this is one of the best answers I've seen on PhysicsSE. – Will Feb 29 '16 at 21:38\n• Thanks! This is my favorite way to explain uncertainty, because it just has so many simplifying properties: the space is finite-dimensional (unlike position and momentum), the photon is spin 1 (so we're dealing with ordinary vectors, not spinors) the photon is massless (so the space only has dimension 2), and it has a nice classical limit (replace the photon polarization with an electric field). I haven't seen this specific example in any textbooks, though. – knzhou Feb 29 '16 at 22:47\n• @knzhou Actually Dirac relied on the photon example in his Principles Of Quantum Mechanics, see secs.2-4 of chap. 1 in fulviofrisone.com/attachments/article/447/…. – udrv Mar 1 '16 at 4:33\n\nNon commuting observables means that a so called measurement is capable of changing the state of the system.\n\nFor instance, when there are two observables A and B that fail to commute, there is an eigenvector of A that it is not an eigenvector of B.\n\nWhen you interact with A then A then B the two results of the A interaction always agree with each other. That means the A interaction always leaves it in a special state, one that gives a specific definite results for an A interaction (the same specific result as the first one gave).\n\nBut when that state fails to be an eigenvector of B (and some eigenvector of one fails to be eigen to the other if they fail to commute) then interacting with A then B then A can give two different results for the A interactions.\n\nThis proves, definitively, that the interaction with B isn't a passive reveal of preexisting information, but is an interaction that can change the state in question.\n\nSpecifically it can change the state from one that gives a particular fixed result for an interaction with A into one capable of giving a different result for an interaction with A.\n\n• \"This proves, definitively, that the interaction with B isn't a passive reveal of preexisting information, but is an interaction that can change the state in question.\" /quote I think this is incorrect. One can postulate hidden variables, and it's only through experiments with entangled states (i.e. Bell violations) that we can rule them out. – DanielSank Feb 29 '16 at 5:32\n• @DanielSank Nope. Even hidden variables have it so that a \"measurement\" of B on an eigenvector of A that isn't eigen to B must change the state into a different state. Hidden variable theories still have states (if nothing else you can say that some things give particular values under A) they just also have hidden variables. The states without the hidden variables are just partial information. But they still are changed. And experiments don't rule out super-determinism and such. So you shouldn't worry about it if an experiment can't rule it out. Focus on what experiments can test. – Timaeus Feb 29 '16 at 5:43\n• Yes, I think I see your point about super-determinism. Actually the experimental distinction between non-commutation and entangled measurements seems murky in my head now. That usually means I'm about to learn something, so thanks! – DanielSank Feb 29 '16 at 6:07\n\nYou can make a connection to the fact that the exact physical state of an arbitrary physical system that occupies a finite volume, can be specified with only a finite amount of information. If you consider some observable then the eigenstates may be degenerate, you then need another observable commuting with the first one to lift that degeneracy, if you continue this way you'll eventually end up with a complete set of commuting observables. Since only a finite amount of information is need to specify the state of the system, this means that this set will be finite. It is then guaranteed that you can find observables that don't belong to this set.\n\nIn the classical limit, all observables commute. In this limit, the number of distinguishable physical states per unit phase phase tends to infinity.\n\nClassical mechanics can be expressed in analytical form in terms of position and \"conjugate momentum\", a term from Lagrangian mechanics. This pair gives us the P, Q variables of Hamiltonian mechanics; when any such pair is quantized you will find that they do not commute. The quantum commutator \"inherits\" this behavior from the classical Poisson brackets.\n\nSo this provides some physical background to your question; the Lagrangian ultimately derives from Newton's Laws of Motion via the Principle of Least Action, a variational principle.\n\n• -1: QM is the more fundamental theory. Classical Poisson brackets cannot be an explanation for non-vanishing commutators in QM. It's backwards to even look for such an explanation. What you need to explain is how observables can seem to commute in classical mechanics when fundamentally, in QM, they don't. \"Quantization\" does not mean taking a limit of a classical mechanical system to get a QM system, it's just a way of making educated guesses. – Robin Ekman Feb 29 '16 at 23:57\n• @Robin Ekman: you are welcome to explain it in full, including how classical mechanics appears in the limit; however, the usual introduction to quantum mechanics is preceded by a proper exposure to analytical mechanics, and the original ansatz used was indeed that of the conjugate variables and the Poisson brackets. It does, IMHO, provide physical intuition into what is going on. OTOH, I learned QM long ago, and may be out of fashion. C'est le vie! Others have mentioned the theorems of linear algebra that apply. – Peter Diehr Mar 1 '16 at 0:07\n• Yes, that's how you do it pedagogically and the history of it, but it's entirely backwards compared to how reality actually is. Quantization by Poisson bracket -> commutator is a method of generating hypotheses but it can't tell you anything about the peculiarities of QM because there is nothing in classical mechanics that corresponds to non-commuting observables. It's like trying to explain the forces between atoms in a crystal in terms of springs or rubber bands, when you should explain the spring or rubber band in terms of the atoms. – Robin Ekman Mar 1 '16 at 0:17\n• @Robin Ekman: well, I'm primarily an experimentalist, and I've found, over many years, that any source of intuition is useful. I'm perfectly happy using ray optics for one thing, polarization for another, non-linear optics for a third, and quantum entanglement for the next. In fact, that describes the current project that I am working on. And I'm also expected to make it work! – Peter Diehr Mar 1 '16 at 0:22\n\nYes, there is a fundamental reason why some observables do not commute. Some are the non-commuting generators of a group. For example, the angular momenta $J_x,J_y,J_z$ are observables. They are also the generators of rotations in the rotation group. By rotating a pencil with your fingers, you can verify that $Rot_xRot_y$ and $Rot_yRot_x$ do not yield the same pencil orientation. Rotations do not commute and by considering small rotations you deduce the commutation relations $[J_k,J_l]=i\\epsilon_{klm}J_m$.\n\nYou say that\n\nThis leads to the problem of not having a definite value of some quantity in some states.\n\nbut what sort of problem is this? It's not a problem of disagreement between theory and experiment. In fact, as I will get to, it's the other way around! If observables commute, we cannot construct theories that agree with experiment. So the problem here is that this is a psychologically or philosophically difficult thing to accept, but that's the way it is and if you don't like it you have to find another universe where the rules are simpler...\n\nBell's inequalities and GHZ experiments can be argued to show that there really isn't anything behind non-commuting observables. A theory built on only commuting observables simply can't give the correct prediction for the GHZ experiment, but QM does (very easily, too). So non-commuting observables seem to be a very fundamental part of the way our universe works.\n\nWhatever you would suggest is \"really\" behind quantum mechanics has to make the same prediction as quantum mechanics about the results of a Bell or GHZ experiment, because we have done those experiments and found the result predicted by QM. This rules out that if there is something behind quantum mechanics, all its observables commute.\n\nThe universe looks quantum mechanical because it is quantum mechanical.\n\n(There's a loophole in the above: we could allow for faster-than-light signals and then there could be hidden variables. But that's genuinely disturbing to physicists in a sense that non-commuting observables isn't, because if we allow faster-than-light communication we must, according to Einstein, allow conditions in the future to affect what happens in the present. That means I can't trust my experiments anymore because someone from the future might be interfering with them. Non-commuting observables means we have to accept only predicting probabilities in experiments but that's much better than throwing all experiments out because your rival from the future could be sabotaging them.)\n\nI'd like to expand a bit on a subtly incorrect statement at the beginning of the original post that was pointed out in the comments:\n\nOP: If those observables do not commute there's no simultaneous basis of eigenvectors of each of them. In that case, in general if $|\\phi \\rangle$ is eigenvector of $A$ it will not be of $B$. (italics $\\rightarrow$ incorrect statement)\n\nComment (WillO): You say that if observables do not commute, they can have no eigenvectors in common. This is not true.\n\nTo give the simplest concrete example, suppose our Hilbert space is finite-dimensional and $A$ and $B$ are non-commuting observables (matrices). Consider these two \"expanded\" operators on a similarly \"expanded\" Hilbert space:\n\n$$A'=\\begin{pmatrix} A & 0 \\\\ 0 & 1\\end{pmatrix},\\,\\,B'=\\begin{pmatrix} B & 0 \\\\ 0 & 1\\end{pmatrix}$$\n\nClearly $A'$ and $B'$ also do not commute, but they do share an eigenvector $v=(\\,0 \\,\\,1 \\,)^{\\text{T}}$ because of the block $1$ in each of their corners.\n\nThe moral of the story is that non-commuting observables only imply the OP's first sentence, not the second that I've italicized (see above). Non-commuting observables imply that they can't share an entire common eigenbasis.\n\n# Edit\n\nI was wrong about the statement being subtly incorrect (see @Kostya's comments below). The originally intended meaning by the OP (which is how I also first understood it, and which I know was actually the intended meaning by the OP because of their subsequent comments) was incorrect, but the way it was worded actually negated the issue resulting in a technically correct statement. In general an eigenvector of one matrix $A$ will not be an eigenvector of another matrix $B$ when $[A,B]\\neq 0$. There may be some eigenvectors of $A$ in common with $B$ (as I originally pointed out), but not all will be in common.\n\n• Would you agree that in general for two numbers $a$ and $b$, $a+b$ will not be 5? – Kostya Aug 2 '18 at 15:54\n• @Kostya Um,.. obviously yes. I take the phrase \"in general\" to be synonymous with \"for all/every ...\". Sorry I don't follow the implication. – Arturo don Juan Aug 2 '18 at 16:01\n• @Kostya Oh my gosh wow okay that slapped me in the face. So that statement is logically correct then. I'll update my post. – Arturo don Juan Aug 2 '18 at 16:04\n• Yes, its about logical meaning of English \"in general\". It is not true that for all $a,b: a+b \\ne 5$. – Kostya Aug 2 '18 at 16:04\n• @Kostya Still though, at the very least both the OP and myself were confused a bit about that sentence (see the comments to the original post), so I'll leave this up in case anybody else gets confused by it. – Arturo don Juan Aug 2 '18 at 16:07" ]

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[ "# Number of possible sentences Essay\n\nNoam Chomsky has infamously stated, “There is no such thing as the probability of a sentence. ” For that, he is roundly mocked by computational linguists, which brings us to the final argument I want to discuss in this essay is as follows: “probability of a sentence is zero, so philosophy is dead! ” Let’s not take into account the fact that word dead in context of philosophy is quite ambiguous. As the former President of the USA, Bill Clinton (August 17, 1998) said: “That depends on what the meaning of ‘is’ is”.\n\nTherefore, in order to disambiguate the argument, I am going to explicitly say what I mean by word dead, is that there does not exist any philosophy at all. Important assumption is that sentences are uniformly distributed. As in the case of a fair coin or a dice, probabilities are 1/2 and 1/6 respectively, where 2 and 6 are the number of possible outcomes. As a result, probability of a sentence is one divided by the number of outcomes (number of possible sentences). We can easily show that number of sentences in any natural language is infinite.\n\nNatural numbers are a subset of the set of all possible words,\n\nNeed essay sample on \"Number of possible sentences\"? We will write a custom essay sample specifically for you for only \\$ 13.90/page\n\nwhile the later is a subset of the set of all possible sentences. Archimedes in 250 BC proved that there is no upper bound for natural numbers; in other words, number of natural numbers is infinite. Therefore, number of possible sentences is infinite. Then probability of a sentence is mathematical limit of one divided by the number of sentences in a natural language. Since number of sentences in a natural language tends to infinity, hence the limit is zero. So we proved that probability of a sentence is zero.\n\nIf probability of a sentence is zero then we cannot make any sentence. If we cannot make any sentence then we cannot speak. If we cannot speak then philosophy is dead. By Hypothetical syllogism we claim that: “If probability of a sentence is zero then philosophy is dead”. Applying Modus ponens we get that philosophy is dead, since as it is shown above that probability of a sentence is zero. To summarise, we almost formally proved that philosophy is dead! What went wrong in the argument, where we concluded that philosophy is dead?\n\nEven though all the steps were rigorously explained and formal methods such as Modus ponens and Hypothetical syllogism were used, the crucial thing is assumption we made! We made a fairly obvious assumption that all the sentences are uniformly distributed; in other words we said that probability of a sentence is one divided by the number of possible sentences. Since number of possible sentences in any natural language is infinite, as proved above, hence we obtained that the probability of a sentence is zero.\n\nFor instance, let’s consider the following anecdote: A man and a blond were asked: “What is the probability that you will see now a real dinosaur walking towards you? “. The man said, that the probability is going to be one in a billion, whereas blond said that it is fifty-to-fifty, either I am going to see it or not. So this example showed that assuming something is quite dangerous thing to do, since we obtained a valid argument, but it is not sound. So this situation is also known as false or misleading presuppositions.\n\nIn conclusion, I argued both for and against the main argument of this article, mentioned in the introduction passage. Using common sense in reasoning about conditional probabilities is an example of false dilemma, whereas making almost obvious assumptions is an example of unjustified premises. These two arguments were carefully analysed and showed that proper way of using formal methods and rigorous proof will validate some issues, where common sense might mislead. However, the main argument is unsound, because it is not valid, which was shown using the fact that averages hide a lot of information.\n\n#### Can’t wait to take that assignment burden offyour shoulders?\n\nLet us know what it is and we will show you how it can be done!\n×\nSorry, but copying text is forbidden on this website. If you need this or any other sample, please register\n\nNo, thanks. I prefer suffering on my own\nSorry, but copying text is forbidden on this website. If you need this or any other sample register now and get a free access to all papers, carefully proofread and edited by our experts.\nNo, thanks. I prefer suffering on my own\nNot quite the topic you need?\nWe would be happy to write it\nService Open At All Times\n|" ]

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[ "", null, "Uh Oh! It seems you’re using an Ad blocker!\n\nWe always struggled to serve you with the best online calculations, thus, there's a humble request to either disable the AD blocker or go with premium plans to use the AD-Free version for calculators.\n\nOr", null, "# Hypergeometric Calculator\n\nInput the values and the calculator will calculate individual and cumulative probability distributions, with detailed calculations shown.\n\nCalculation For:\n\nSelect Function:\n\nPopulation Size (N)\n\nSuccesses in population (K)\n\nSample Size (n)\n\nSuccesses in Sample (k)\n\nIncrement\n\nRepetition\n\nTable of Content\n 1 What is hypergeometric distribution? 2 Hypergeometric Distribution Formula 3 About hypergeometric calculator 4 How to Use This hypergeometric distribution calculator? 5 FAQ's 6 Takeaway 7 References\nGet The Widget!\n\nAdd Hypergeometric Calculator to your website through which the user of the website will get the ease of utilizing calculator directly. And, this gadget is 100% free and simple to use; additionally, you can add it on multiple online platforms.\n\nAvailable on App\n\nThe hypergeometric calculator is a smart tool that allows you to calculate individual and cumulative hypergeometric probabilities.\n\nApart from it, this hypergeometric calculator helps to calculate a table of the probability mass function, upper or lower cumulative distribution function of the hypergeometric distribution, draws the chart, and also finds the mean, variance, and standard deviation of the hypergeometric distribution.\n\n## What is hypergeometric distribution?\n\nSpecifically, a hypergeometric distribution is said to be a probability distribution that simply represents the probabilities that are associated with the number of successes in a hypergeometric experiment. You can try this hypergeometric calculator to figure out hypergeometric distribution probabilities instantly.\n\nSuppose that you randomly selected 5 cards from an ordinary deck of playing cards, here you might ask: what’s the probability distribution form the number of red cards in our selection.\n\nIn this example, selecting a red card would be referred to as a success. Well, the probabilities associated with each possible outcome are an example of a hypergeometric distribution, as shown in the given chart:\n\n Outcome Hypergeo Prob Cumu Prob 0 red cards 0.025 0.025 1 red card 0.150 0.175 2 red cards 0.325 0.500 3 red cards 0.325 0.825 4 red cards 0.150 0.975 5 red cards 0.025 1.00\n\nBy given this probability distribution, you can depict at a glance that the cumulative and individual probabilities are being associated with any outcome. For instance, the cumulative probability of selecting 1 or fewer red cards would be 0.175, and when it comes to the individual probability, selecting exactly 1 red card would be 0.15.\n\n### Hypergeometric Distribution Formula:\n\nThe hypergeometric distribution probabilities or statistics can be derived from the given formula:\n\nFormula:\n\nh(k; N, n, K) = [ KCk ] [ N-KCn-k ] / [ NCn ]\n\nWhere;\n\nN is said to be the Population Size\n\nK is said to be the number of Successes in population\n\nn is said to be the Sample Size\n\nk is said to be the number of Successes in Sample\n\nC is said to be combinations\n\nh is said to be hypergeometric\n\nThe hypergeometric distribution calculator is an online discrete statistics tool that helps to determine the individual and cumulative hypergeometric probabilities. The hypergeometric calculator will assists you to calculate the following parameters and draw the chart for a hypergeometric distribution:\n\n• probability mass function\n• Lower Cumulative Distribution P\n• Upper Cumulative Distribution Q\n• Mean of hypergeometric distribution\n• Variance hypergeometric distribution\n• Standard Deviation hypergeometric distribution\n\n### How to Use This hypergeometric distribution calculator:\n\nThis hypergeometric calculator is loaded with user-friendly interface; you just have to follow the given steps to get instant results:\n\n#### Calculation for Hypergeometric Probability distribution:\n\nInputs:\n\n• First of all, you have to select the option of Hypergeometric Probability distribution from the distribution from the drop-down menu\n• Now, you have to enter the population size (N) into the designated field\n• Very next, you have to enter the number of successes in population (K) into the given field\n• Now, you have to enter the sample size (n) into the designated field\n• Finally, you have to enter number of successes in sample (k) into the designated field of this hypergeometric probability calculator\n\nOutputs:\n\nOnce done, you have to hit the calculate button, this distribute calculator will shows the following:\n\n• Hypergeometric Probability: P(X = x)\n• Cumulative Probability: P(X < x)\n• Cumulative Probability: P(X ≤ x)\n• Cumulative Probability: P(X > x)\n• Cumulative Probability: P(X ≥ x)\n• Mean\n• Variance\n• Standard Deviation\n• Hypergeometric Distribution Probability Chart\n\n#### Calculation for Hypergeometric Probability distribution (chart):\n\nInputs:\n\n• First of all, you have to choose the option of Hypergeometric Probability distribution (chart) from the drop-down menu\n• Very next, you have to select the function for which you want to calculate a table of the probability, it can either be in (probability mass f, lower cumulative distribution P, upper cumulative distribution Q)\n• Now, you have to enter the population size (N) into the designated filed of this hypergeometric distribution calculator\n• Then, you have to add the number of successes in population (K) into the given box\n• Right after, you have to add the sample size (n) into the designated filed of the above calculator\n• Then, enter the value of successes in sample (k) initial into the designated field\n• Enter the value into the increment field, tell how much you want increment in every repetition for a successes in sample (k) initial\n• Now, enter the value to tell how much steps you want to repeat\n\nOutputs:\n\nOnce done, you have to hit the calculate button, this Hypergeometric distribution (chart) Calculator will shows:\n\n• Table of probability according to the selected function\n• Mean\n• Variance\n• Standard Deviation\n• Draws the chart for a hypergeometric distribution\n\n## FAQ’s\n\n### How do you know when to use hypergeometric distribution?\n\nYou can use the hypergeometric distribution with populations that are so small, which the outcome of a trial has a large effect on the probability that the next outcome is a non-event or event. For instance, within a population of 10 people, only 7 people have A+ blood. So, try the above distribute calculator to find the hypergeometric distribution.\n\n### What is a hypergeometric experiment?\n\nThe hypergeometric experiment has two particularities that are mentioned-below:\n\n• The random selections from the finite population take place without any replacement\n• Each item in the population can either be considered as a success or failure\n\nHowever, a hypergeometric distribution indicates the probability that associated with the occurrence of a specific number of successes in a hypergeometric experiment.\n\n### What is the number of successes?\n\nWhen it comes to hypergeometric experiment, each item in the population can be represented as a success or a failure. The number of successess is said to be a count of the successes in a particular grouping. Therefore, the number of successes in the sample indicates a count of successes in the sample; and the number of successes in the population indicates a count of successes in the population.\n\n### What is a hypergeometric probability?\n\nA hypergeometric probability is said to be a probability that is associated with a hypergeometric experiment." ]

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[ "# xkcd样式页面编号\n\n65\n\nRandall Munroe的书“ xkcd，第0卷”对页码使用了一个相当奇数的系统。前几个页码是\n\n``````1, 2, 10, 11, 12, 20, 100, 101, 102, 110, 111, 112, 120, 200, 1000, 1001, ...\n``````\n\n## 测试用例\n\n``````1 => 1\n2 => 2\n3 => 10\n6 => 20\n7 => 100\n50 => 11011\n100 => 110020\n200 => 1100110\n1000 => 111110120\n10000 => 1001110001012\n100000 => 1100001101010020\n1000000 => 1111010000100100100\n1048576 => 10000000000000000001\n\n1000000000000000000 => 11011110000010110110101100111010011101100100000000000001102\n``````\n\n## 排行榜\n\n``````# Language Name, N bytes\n``````\n\n`N`您提交的文件大小在哪里。如果您提高了分数，则可以将旧分数保留在标题中，方法是将它们打掉。例如：\n\n``````# Ruby, <s>104</s> <s>101</s> 96 bytes\n``````\n\n``<script>site = 'meta.codegolf'; postID = 5314; isAnswer = true; QUESTION_ID = 51517</script><script src='https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js'></script><script>jQuery(function(){var u='https://api.stackexchange.com/2.2/';if(isAnswer)u+='answers/'+postID+'?order=asc&sort=creation&site='+site+'&filter=!GeEyUcJFJeRCD';else u+='questions/'+postID+'?order=asc&sort=creation&site='+site+'&filter=!GeEyUcJFJO6t)';jQuery.get(u,function(b){function d(s){return jQuery('<textarea>').html(s).text()};function r(l){return new RegExp('<pre class=\"snippet-code-'+l+'\\\\b[^>]*><code>([\\\\s\\\\S]*?)</code></pre>')};b=b.items.body;var j=r('js').exec(b),c=r('css').exec(b),h=r('html').exec(b);if(c!==null)jQuery('head').append(jQuery('<style>').text(d(c)));if (h!==null)jQuery('body').append(d(h));if(j!==null)jQuery('body').append(jQuery('<script>').text(d(j)))})})</script>``\n\n32\n\nAlex\n\n2\n@AlexA。我将其放在Kindle上，在这里您无法进行任何有趣的页面编号。\nLegionMammal978\n\n21\n@ LegionMammal978：悲惨的。\nAlex A.\n\n1\n@ SuperJedi224 共识似乎不是，所以很抱歉（如果您的语言可以处理的唯一输入是这种情况，我会从共识中例外，但事实并非如此）。\n\n4\n\nCyoce\n\n12\n\n# Pyth，17个字节\n\n``````Ljb3ye.f!-P-yZ01Q\n``````\n\n32\n\n# CJam，24 23 22 21 19字节\n\n``````ri_)2b,,W%{2\\#(md}/\n``````\n\n### 这个怎么运作\n\n``````ri e# Read an integer N from STDIN.\n2b, e# Push the number of binary digits of N + 1, i.e, K + 2 = log(N + 1) + 1.\n,W% e# Push the range [K+1 ... 0].\n{ e# For each J in the range:\n2\\# e# J -> 2**J\n( e# -> 2**J - 1\nmd e# Perform modular division of the integer on the stack (initially N)\ne# by 2**J - 1: N, 2**J - 1 -> N/(2**J - 1), N%(2**J - 1)\n}/ e#\n``````\n\n28\n\n# Python 2，67个字节\n\n``def f(n):x=len(bin(n+1))-3;y=2**x-1;return n and n/y*10**~-x+f(n%y)``\n\n# Python 3，65个字节\n\n``def g(n,x=1):*a,b=n>x*2and g(n,x-~x)or[n];return a+[b//x,b%x][:x]``\n\n`2and`在3 中编译？我在2中并`2and2`抛出语法错误\nTankorSmash 2015年\n\n3\n@TankorSmash `2and2`是行不通的，因为它会得到解析为`2 and2`-尝试`2and 2`，它应该工作，如果你的Python版本够新（在Python 2.7.10测试）\nSP3000\n\nTankorSmash 2015年\n\n12\n\n# CJam，22 21 20字节\n\n``````ri_me,3fb{0-W<1-!},=\n``````\n\nCJam解释器中在线尝试。\n\n### 这个怎么运作\n\n``````ri e# Read an integer N from STDIN.\n_me, e# Push [0 ... floor(exp(N))-1].\n3fb e# Replace each integer in the range by the array of its digits in base 3.\n{ e# Filter; for each array A:\n0- e# Remove all 0's.\nW< e# Remove the last element.\n1- e# Remove all 1's.\n! e# Logical NOT. Pushes 1 iff the array is empty.\n}, e# If ! pushed 1, keep the array.\n= e# Select the (N+1)th element of the filtered array.\n``````\n\n9\n\n# Pyth，20个字节\n\n``````Jt^2hslhQ#p/=%QJ=/J2\n``````\n\n``````Jt^2hslhQ#p/=%QJ=/J2\nImplicit: Q is the input.\nlhQ log(Q+1,2)\nslhQ floor(log(Q+1,2))\nhslhQ floor(log(Q+1,2))+1\n^2hslhQ 2^(floor(log(Q+1,2))+1)\nt^2hslhQ 2^(floor(log(Q+1,2))+1)-1\nJt^2hslhQ J=2^(floor(log(Q+1,2))+1)-1\n# until an error is thrown:\n=%QJ Q=Q%J\n=/J2 J=J/2\n/ The value Q/J, with the new values of Q and J.\np print that charcter, with no trailing newline.\n``````\n\nJ初始化为大于输入的最小歪斜二进制数字位置的值。然后，每次循环时，我们都会执行以下操作：\n\n• 除去的值的每个数位`J``Q``=%QJ`。例如，如果`Q=10``J=7``Q`成为`3`，其对应于二进制偏斜改变从`110``10`。这在第一次迭代中没有影响。\n• 用更改`J`为下一个较小的偏斜二进制基值`=/J2`。这是由2难倒分工，改变`J=7``J=3`，例如。因为这是在输出第一个数字之前发生的，所以将其`J`设置为比所需位置高一个数字的位置。\n• 使用`/QJ`（有效）找到实际数字值。\n• 使用`p`而不是Pyth的默认打印来打印该值，以避免尾随换行符。\n\n8\n\n# ES6，105个字节\n\n``````f=n=>{for(o=0;n--;c?o+=Math.pow(3,s.length-c):o++)s=t(o),c=s.search(2)+1;return t(o);},t=a=>a.toString(3)\n``````\n\n``````f=n=>{ //define function f with input of n (iteration)\nfor(o=0; //define o (output value in decimal)\nn--; //decrement n (goes towards falsy 0) each loop until 0\nc?o+=Math.pow(3,s.length-c):o++) //if search finds a 2, increment output by 3^place (basically moves the 2 to the left and sets the place to 0), else ++\ns=t(o), //convert output to base 3\nc=s.search(2)+1; //find the location of 2, +1, so a not-found becomes falsy 0.\nreturn t(o); //return the output in base 3\n},\n\nt=a=>a.toString(3); //convert input a to base 3\n``````\n\n5\n\nMartin Ender 2015年\n\n2\n-16个字节：`f=n=>{for(o=0;~n--;o+=c?Math.pow(3,s.length+c):1)s=o.toString(3),c=~s.search(2);return s}`\nnderscore 2015年\n\n@nderscore非常酷：D\n\n1\n\n3\n@GustavoRodrigues我什至还没有学习完ES6！@ _ @\n\n7\n\n# 视网膜，55字节\n\n``````^\n0a\n(+`12(.*a)1\n20\\$1\n0?2(.*a)1\n10\\$1\n0a1\n1a\n)`1a1\n2a\na\n<empty line>\n``````\n\n``````> echo 11111|retina -s skew\n12\n``````\n\n• 如果包含`2``^2 -> ^12`; `02 -> 12`;`12 -> 20`\n• 如果不包含`2``0\\$ -> 1\\$`;`1\\$ -> 2\\$`\n\n`2`字符串中最多可以有一个；`^``\\$`在规则中标记字符串的开头和结尾。）\n\n7\n\n# Java中，154 148\n\n``````n->{String s=\"0\";for(;n-->0;)s=s.contains(\"2\")?s.replaceAll(\"(^|0)2\",\"10\").replace(\"12\",\"20\"):s.replaceAll(\"1\\$\",\"2\").replaceAll(\"0\\$\",\"1\");return s;}\n``````\n\n``````import java.util.function.Function;\npublic class Skew {\npublic static void main(String[] args){\nFunction<Integer,String> skew = n->{String s=\"0\";for(;n-->0;)s=s.contains(\"2\")?s.replaceAll(\"(^|0)2\",\"10\").replace(\"12\",\"20\"):s.replaceAll(\"1\\$\",\"2\").replaceAll(\"0\\$\",\"1\");return s;};\n\nfor(String s:args){\nSystem.out.println(skew.apply(Integer.parseInt(s)));\n}\n}\n}\n``````\n\n5\n\n# Bash + coreutils，52岁\n\n``````dc -e3o0[r1+prdx]dx|grep -v 2.\\*|sed -n \\$1{p\\;q}\n``````\n\n### 输出：\n\n``````\\$ ./xkcdnum.sh 1000\n111110120\n\\$\n``````\n\n5\n\n# Java中，337个335 253 246 244字节\n\n``````String f(long i){List<Long>l=new ArrayList<>();l.add(0L);for(;i-->0;){int j=l.indexOf(2);if(j!=-1){l.set(j,0L);if(j==0){l.add(0,1L);}else{l.set(j-1,l.get(j-1)+1);}}else{j=l.size()-1;l.set(j,l.get(j)+1);}}String s=\"\";for(long q:l)s+=q;return s;}\n``````\n\n4\n\nGeobits，2015年\n\n2\n\ndurron597 2015年\n\n4\n\n# 哈斯克尔，73 72\n\n``````i(2:b)=1:0:b\ni[b]=[b+1]\ni(b:2:c)=b+1:0:c\ni(b:c)=b:i c\ns=(iterate i!!)\n``````\n\n4\n\n# CJam，24个字节\n\n``````Q{0\\+2a/())+a\\+0a*}ri*si\n``````\n\nCJam解释器中在线尝试。\n\n### 背景\n\n1. 最终将2替换为0\n\n2. 如果已替换2，则将数字向左递增。\n\n否则，增加最后一位数字。\n\n### 这个怎么运作\n\n``````Q e# Push an empty array.\n{ e# Define an anonymous function:\n0\\+ e# Prepend a 0 to the array.\n2a/ e# Split the array at 2's.\n( e# Shift out the first chunk of this array.\n) e# Pop the last digit.\n)+ e# Increment it and append it to the array.\na\\+ e# Prepend the chunk to the array of chunks.\n0a* e# Join the chunks, using as separator.\ne# If there was a 2, it will get replaced with a 0. Otherewise, there's\ne# only one chunk and joining the array dumps the chunk on the stack.\n} e#\nri* e# Call the function int(input()) times.\nsi e# Cast to string, then to integer. This eliminates leading 0's.\n``````\n\n4\n\n# VBA，209个147 142字节\n\n``````Sub p(k)\nFor i=1To k\na=StrReverse(q)\nIf Left(Replace(a,\"0\",\"\"),1)=2Then:q=q-2*10^(InStr(a,2)-1)+10^InStr(a,2):Else:q=q+1\nNext\nmsgbox q\nEnd Sub\n``````\n\nEdit1：感谢Manu帮助削减了62个字节\n\nEdit2：交换`debug.print``msgbox`输出。保存了5个字节\n\n1\n\nCommonGuy 2015年\n\nCommonGuy 2015年\n\n1\n\nJimmyJazzx 2015年\n\nCommonGuy 2015年\n\n4\n\n# 的Javascript ES6，99 86 78 76 72个字符\n\n``````f=n=>{for(s=\"1\";--n;s=s.replace(/.?2|.\\$/,m=>[1,2,10][+m]||20));return s}\n\n// Old version, 76 chars:\nf=n=>{for(s=\"1\";--n;s=s.replace(/02|12|2|.\\$/,m=>[1,2,10][+m]||20));return s}\n\n// Old version, 86 chars:\nf=n=>{for(s=\"1\";--n;s=s.replace(/(02|12|2|.\\$)/,m=>[1,2,10,,,,,,,,,,20][+m]));return s}\n\n// Old version, 99 chars:\nf=n=>{for(s=\"1\";--n;s=s.replace(/(^2|02|12|20|.\\$)/,m=>({0:1,1:2,2:10,12:20,20:100}[+m])));return s}\n``````\n\n``````;[1,2,3,6,7,50,100,200,1000,10000,100000,1000000,1048576].map(f) == \"1,2,10,20,100,11011,110020,1100110,111110120,1001110001012,1100001101010020,1111010000100100100,10000000000000000001\"\n``````\n\nQwertiy\n\n3\n\n# 八度，107101字节\n\n``````function r=s(n)r=\"\";for(a=2.^(uint64(fix(log2(n+1))):-1:1)-1)x=idivide(n,a);r=[r x+48];n-=x*a;end;end\n``````\n\n``````function r=s(n)\nr=\"\";\nfor(a=2.^(uint64(fix(log2(n+1))):-1:1)-1)\nx=idivide(n,a);\nr=[r x+48];\nn-=x*a;\nend\nend\n``````\n\n``````octave:83> s(uint64(1e18))\nans = 11011110000010110110101100111010011101100100000000000001102\n\noctave:84> s(uint64(1e18)-1)\nans = 11011110000010110110101100111010011101100100000000000001101\n\noctave:85> tic();s(uint64(1e18)-1);toc()\nElapsed time is 0.0270021 seconds.\n``````\n\n3\n\n## T-SQL，221个189 177字节\n\n``````DECLARE @ BIGINT=,@T VARCHAR(MAX)='';WITH M AS(SELECT CAST(2AS BIGINT)I UNION ALL SELECT I*2FROM M WHERE I<@)SELECT @T += STR(@/(I-1),1),@%=(I-1)FROM M ORDER BY I DESC SELECT @T\n``````\n\n``````DECLARE\n@ BIGINT=,\n@T VARCHAR(MAX)='';\n\nWITH M AS\n(\nSELECT\nCAST(2 AS BIGINT) I\n\nUNION ALL\n\nSELECT I * 2\nFROM M\nWHERE I < @\n)\n\nSELECT\n@T+=STR(@/(I-1),1),\n@%=(I-1)\nFROM M\nORDER BY I DESC\n\nSELECT @T\n``````\n\n``````DECLARE @ INT=,@T VARCHAR(MAX)='';WITH M AS(SELECT 2I UNION ALL SELECT I*2FROM M WHERE I<@)SELECT @T+=STR(@/(I-1),1),@%=(I-1)FROM M ORDER BY I DESC SELECT @T\n``````\n\n``````DECLARE\n@ INT=,\n@T VARCHAR(MAX)='';\n\nWITH M AS\n(\nSELECT\n2I\n\nUNION ALL\n\nSELECT\nI * 2\nFROM M\nWHERE I < @\n)\n\nSELECT\n@T+=STR(@/(I-1),1),\n@%=(I-1)\nFROM M\nORDER BY I DESC\n\nSELECT @T\n``````\n\nMichael B\n\n@MichaelB哦，我以为我曾经用过`@`，愚蠢的我。该`CHAR(8000)`建议非常好，我会尝试的。我似乎总是忘记了`STR()`要感谢的存在。\nPenutReaper 2015年\n\nMichael B\n\n@MichaelB啊。`CONCAT`，我在2008年。因此，如果不立即使用SQL提琴，就无法对其进行测试。好电话。\nPenutReaper 2015年\n\n3\n\n# 图灵机代码，333个 293字节\n\n``````0 _ _ l 1\n0 * * r 0\n1 9 8 l 2\n1 8 7 l 2\n1 7 6 l 2\n1 6 5 l 2\n1 5 4 l 2\n1 4 3 l 2\n1 3 2 l 2\n1 2 1 l 2\n1 1 0 l 2\n1 0 9 l 1\n1 _ _ r 8\n2 _ _ l 3\n2 * * l 2\n3 _ 1 r 4\n3 * * l 5\n4 _ _ r 0\n4 * * r 4\n5 * * l 5\n5 _ _ r 6\n6 _ _ l 7\n6 2 0 l 7\n6 * * r 6\n7 _ 1 r 4\n7 0 1 r 4\n7 1 2 r 4\n8 _ _ * halt\n8 * _ r 8\n``````\n\n2\n\n# Perl，66个字节\n\n``````\\$_=1;\\$c=<>;s/(.*)(.?)2(.*)/\\$1.\\$2+1 .\\$3.0/e||\\$_++while(--\\$c);print;\n``````\n\nCJ Dennis\n\n@CJDennis感谢您保存该字节。无论如何，。？除非数字中没有数字（例如20），否则将使数字紧靠两位。在例如120或10020的情况下，正则表达式组如下：（）（1）2（0）和（10）（0）2（0）。然后，简单地忽略第一组，增加第二组（如果可能，总是为一位，否则为空白），然后忽略第三组（始终由零组成）并添加零。我只是将OEIS条目用作此正则表达式的指南。\nfrederick 2015年\n\nCJ Dennis\n\nThraidh\n\n2\n\n# Pyth，19个字节\n\n``````m/=%Qtydtd^L2_SslhQ\n``````\n\n2\n\n# Perl，84 70 67字节\n\n``````\\$n=<>;\\$d*=2while(\\$d++<\\$n);\\$_.=int(\\$n/\\$d)while(\\$n%=\\$d--,\\$d/=2);print\n``````\n\n``````\\$d*=2while\\$d++<\\$_;\\$\\.=\\$_/\\$d|0while\\$_%=\\$d--,\\$d/=2}{\n``````\n\n@frederick我们在座位边上等待\n\n@RobertGrant他从两件事变成一件事！\nCJ Dennis\n\nfrederick 2015年\n\nfrederick 2015年\n\n@frederick不要那么谦虚。我们知道发生了什么！\n\n1\n\n## Mathematica，65岁\n\n``````f = (n = #;\nl = 0;\nWhile[n > 0,\nm = Floor[Log2[1 + n]];\nl += 10^(m - 1);\nn -= 2^m - 1\n]; l)&\n``````\n\n``````f\n``````\n\n``````11011110000010110110101100111010011101100100000000000001102\n``````\n\n``````Timing[Block[{\\$MaxExtraPrecision = Infinity}, f[10^8000]];]\n``````\n\n``````{1.060807, Null}\n``````\n\n1\n\n# C，95个字节\n\n``void f(unsigned long i,int*b){for(unsigned long a=~0,m=0;a;a/=2,b+=!!m)m|=*b=i/a,i-=a**b;*b=3;}``\n\n### 扩展代码\n\n``````void f(unsigned long i,int*b)\n{\nfor (unsigned long a=~0, m=0; a; a/=2, b+=(m!=0)) {\n*b = i/a; /* rounds down */\ni -= *b * a;\nm = m | *b; /* m != 0 after leading zeros */\n}\n}``````\n\n### 测试程序\n\n``````#include <stdio.h>\n#include <stdlib.h>\nint main(int argc, char**argv)\n{\nwhile (*++argv) {\nunsigned long i = strtoul(*argv, NULL, 10);\nint result;\nf(i,result);\n\n/* convert to string */\nchar s;\n{char*d=s;int*p=result;while(*p!=3)*d++=*p+++'0';*d=0;}\nprintf(\"%lu = %s\\n\", i, s);\n}\n\nreturn EXIT_SUCCESS;\n}``````\n\n### 检测结果\n\n``````\\$ ./51517 \\$(seq 20)\n1 = 1\n2 = 2\n3 = 10\n4 = 11\n5 = 12\n6 = 20\n7 = 100\n8 = 101\n9 = 102\n10 = 110\n11 = 111\n12 = 112\n13 = 120\n14 = 200\n15 = 1000\n16 = 1001\n17 = 1002\n18 = 1010\n19 = 1011\n20 = 1012``````\n\nToby Speight 2015年\n\n1\n\n# JavaScript（Node.js），40字节\n\n``v=>(g=n=>n>v?'':g(n-~n)+(v-(v%=n))/n)(1)``\n\n0\n\n## CoffeeScript，92 69字节\n\n``````f=(n)->s='1';(s=s.replace /(.?2|.\\$)/,(m)->[1,2,10][+m]||20)while--n;s\n\n# older version, 92 bytes\nf=(n)->s='1';(s=s.replace /(^2|02|12|20|.\\$)/,(m)->{0:1,1:2,2:10,12:20,20:100}[+m])while--n;s``````\n\n2\n\nQwertiy 2015年\n\n@Qwertiy我已为您的答案提供了归属，如果没有正则表达式中的括号，我将无法给出相同的结果\nrink.attendant.6 2015年\n\nQwertiy 2015年\n\n0\n\n# Japt，31个字节\n\n``````_r/?2|.\\$/g_÷C ç20 ª°Zs3}}gU['0]\n``````\n\n### 开箱及其工作方式\n\n``````X{Xr/?2|.\\$/gZ{Z÷C ç20 ||++Zs3}}gU['0]\n\nX{ Declare a function...\nXr Accept a string, replace the regex...\n/?2|.\\$/g /.?2|.\\$/ (g is needed to match *only once*, opposite of JS)\nZ{ ...with the function... (matched string will be 0,1,2,02 or 12)\nZ÷C Implicitly cast the matched string into number, divide by 12\nç20 Repeat \"20\" that many times (discard fractions)\n|| If the above results in \"\", use the next one instead\n++Z Increment the value\ns3 Convert to base-3 string (so 0,1,2 becomes 1,2,10)\n}}\ngU['0] Repeatedly apply the function on \"0\", U (input) times\n``````\n\n0\n\n# Stax，16 个字节\n\n``````üxëàè£öΦGΩ│Je5█ò\n``````\n\n``````z push []\n{ start block for while loop\n|X:2N -log2(++x)\n{^}& increment array at index (pad with 0s if necessary)\nxc:G-X unset high bit of x; write back to x register\nw while; loop until x is falsy (0)\n\\$ convert to string\n``````" ]

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[ "# Figure Count\n\nNumber of <fig> elements that appear in the document.\n\n## Related Elements\n\nInside the <counts> container element are the counts of various components of the document: the generic count element <count> (for which the @count-type names what is being counted) and the specific named counting elements: <fig-count> is the number of figures, the <table-count> is the number of tables, the <equation-count> is the number of equations, the <ref-count> is either the number of references or (more properly) the number of citations in the bibliographic reference list, the <page-count> is the page count, and the <word-count> is the number of words in the document.\n\n## Content Model\n\n`<!ELEMENT fig-count EMPTY >`\n\n## Description\n\nThis is an EMPTY element\n\n### Example\n\n``` ...\n<article-meta>...\n<abstract>...</abstract>\n<conference>\n<conf-date iso-8601-date=\"1999\">1999</conf-date>\n<conf-name>The 27th annual ACM SI/GUCCS\nconference</conf-name>\n<conf-acronym>SIGUCCS</conf-acronym>\n<conf-num>27</conf-num>\n<conf-loc>Denver, Colorado, United States</conf-loc>\n<conf-sponsor>ACM, Assoc. for Computing\nMachinery</conf-sponsor>\n<conf-theme>User services conference for\nuniversity and college computing service\norganizations</conf-theme>\n</conference>\n<counts>\n<count count-type=\"contributors\" count=\"3\"/>\n<fig-count count=\"5\"/>\n<table-count count=\"3\"/>\n<equation-count count=\"10\"/>\n<ref-count count=\"26\"/>\n<page-count count=\"6\"/>\n<word-count count=\"2847\"/>\n</counts>\n</article-meta>\n... ```" ]

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[ "Theory and Modern Applications\n\n# Finite-time synchronization problem of a class of discontinuous Cohen–Grossberg neural networks with mixed delays via new switching design\n\n## Abstract\n\nThis paper investigates a class of generalized Cohen–Grossberg neural networks (CGNNs) with discontinuous activations and mixed delays. Based on the nonsmooth analysis theory, the drive-response concept, differential inclusions theory, we give several basic assumptions to gain the finite-time synchronization issue of CGNNs. Sufficient conditions are provided without the boundedness or monotonicity of discontinuous activation functions. Moreover, one can estimate the settling time’s upper bounds of the system. At last, two numerical examples and their simulations are given to further show the benefits of the obtained control approach.\n\n## 1 Introduction\n\nRecently, the research of neural networks with discontinuous activation functions has gradually attracted the attention of many researchers, including systems oscillating under earthquake, power circuits, chaos phenomenon, and dry friction (see ). Because the non-Lipschitz phenomenon has many special advantages, the emergence of nonsmooth has greatly improved the research of neural networks. In addition, the analysis of neural networks with discontinuous activations is accompanied by many interesting practical phenomena to explore important dynamic behavior characteristics. This arouses researchers’ interest in the generalized neural networks by using discontinuous activation functions .\n\nIn 1983, Cohen and Grossberg firstly introduced the CGNNs system, which was a useful recurrent neural networks system, including evolutionary theory, population biology, neurobiology . After that, a large number of results have emerged such as the existence, dissipation, and exponential stability of the CGNNs model. However, there are few works on discontinuous CGNNs system with mixed delays. In , the authors investigated the exponential stability and exponential synchronization of a class of CGNNs. Abdurahman and his team in studied the exponential lag synchronization for both discrete time-delays and distributed delays CGNNs. It worthy to know that, in 2003, Forti introduced the global stability of a discontinuous right-hand side neural network system via the framework of the theory of Filippov differential inclusions [22, 23]. In they pointed out that the sliding mode method was used to solve constrained optimization problems, because high-gain neuron activations were often encountered in the neural networks system. In , the authors analyzed the fixed-time synchronization of a class of discontinuous fuzzy inertial neural networks with time-varying delays based on the new improved fixed-time stability lemmas.\n\nAs we know, the synchronization phenomenon has been widely used in software engineering, ecological structure, security storage, information processing system, and many other fields, which gets a lot of social attention. There are some micro motions from the view of the mathematical model, which we call qualitative or stability problems. Meanwhile, the researchers pay attention to the macroscopic topology based on the synchronization problem. The literature was the first paper to consider finite-time control of discontinuous chaotic systems, and studied the finite-time synchronization of time-delayed neural networks. There were many classifications of synchronization, such as anti-synchronization [28, 29], exponential synchronization , robust synchronization , chaos synchronization , and so on. The synchronization technologies mentioned above have many defects in a real practical environment. For instance, the existence results of the above synchronization usually are guaranteed over the infinite horizon. In addition, when the finite initial value is required and the control accuracy has great influence on the system, it is always difficult to estimate. Moreover, even if stringent convergence time is given, the neural networks model may not be available in a real experimental environment. For the sake of convergence time, one needs to propose a concept named finite-time synchronization, which means that the settling time function of any finite initial value is bounded. Song and his team provided a novel and effective techniques method in , then in they investigated the finite-time synchronization problem of a class of discontinuous neural networks with nonlinear coupling and mixed delays. Peng and his team in investigated the finite-time synchronization control methodology for the CGNNs system. Yang in verified that the considered neural networks can gain the synchronization in a finite time. In , the authors ensured that the target model realized the finite-time synchronization goal of the coupled neural networks. In , the author solved the challenging issues in the field of finite-time synchronization of the cellular neural networks.\n\nMotivated by the aforementioned works on finite-time synchronization of CGNNs system, this paper aims to realize the finite-time synchronization issue for the considered system CGNNs. Our main contributions of this paper include the following three aspects.\n\n• The CGNNs discussed in this brief are state-dependent discontinuous systems; based on the properties of differential inclusion and set-analysis theory, the drive-response CGNNs can be transformed into a synchronization error system. Theoretical analysis can be extended to other fields.\n\n• When both mixed delays and discontinuities exist in the dynamical CGNNs, how to deal with the discrepancy within the scope of the Filippov solutions of the drive system and the response system?\n\n• Because the system has special discontinuous characteristics, in order to shorten the settling time of the drive-response CGNNs system, the more ingenious switching controller should be devised.\n\n## 2 Model description and some basic definitions\n\nThis section considers a general class of discontinuous CGNNs with mixed delays. Based on the previous works , one can describe the model by the following equation:\n\n\\begin{aligned} \\frac{\\mathrm{d}\\pi _{i}(t)}{\\mathrm{d}t}&=-\\varpi _{i}\\bigl(\\pi _{i}(t)\\bigr) \\Biggl[a_{i}\\bigl(\\pi _{i}(t) \\bigr)-\\sum^{n}_{j=1}b_{ij}(t)f_{j} \\bigl(\\pi _{j}(t)\\bigr) -\\sum^{n}_{j=1}c_{ij}(t)f_{j} \\bigl(\\pi _{j}\\bigl(t-\\tau _{ij}(t)\\bigr)\\bigr) \\\\ &\\quad {}-\\sum^{n}_{j=1} \\int ^{+\\infty }_{0}K_{ij}(t,s)f_{j} \\bigl(\\pi _{j}(s)\\bigr) \\,\\mathrm{d}s-I_{i}(t) \\Biggr],\\quad i=1,2,\\ldots ,n, \\end{aligned}\n(1)\n\nthe state vector is $$\\pi (t)=(\\pi _{1}(t),\\pi _{2}(t),\\ldots ,\\pi _{n}(t))^{T}\\in {\\mathbb {R}}$$; $$\\varpi _{i}(\\cdot )$$ is the amplification function of the system; $$a_{i}(\\cdot )$$ is the function with proper behavior; $$\\tau _{ij}(t)$$ and $$K_{ij}: [0,\\infty )\\rightarrow [0,\\infty )$$ denote the discrete delay and the distributed delay, respectively; $$b_{ij}(t)$$ and $$c_{ij}(t)$$ are the connection strength and the delayed feedbacks of two different neurons; $$f_{j}(\\cdot )$$ is the neuron input–output activation of the ith neuron; $$I_{i}(\\cdot )$$ is an input signal function of the external factors;\n\nThe neuron activation functions $$f_{j}(\\cdot )$$ in the above model satisfy the following conditions:\n\n1. (H1)\n\n$$f_{i}:{\\mathbb {R}}\\rightarrow {\\mathbb {R}}$$ is discontinuous on a countable set of isolate point $$\\{\\rho ^{i}_{k}\\}$$ for each $$i=1,2,\\ldots ,n$$.\n\n2. (H2)\n\nThere are two nonnegative constants $$L_{i}$$ and $$h_{i}$$ that satisfy the following inequality:\n\n$$\\bigl\\Vert \\mathbb{F}\\bigl[f_{i}(x)-f_{i}(y)\\bigr] \\bigr\\Vert =\\sup_{\\zeta _{i}\\in \\mathbb{F}[f_{i}(x)-f_{i}(y)]} \\Vert \\zeta _{i} \\Vert \\leq L_{i} \\Vert x-y \\Vert +h_{i},\\quad i=1,2, \\ldots ,n,$$\n\nwhere $$\\mathbb{F}(f_{i}(x))=K[f_{i}(x)]= [\\min \\{f_{i}(x_{i}^{-}),f_{i}(x_{i}^{+}) \\},\\max \\{f_{i}(x_{i}^{-}),f_{i}(x_{i}^{+})\\} ]$$.\n\n3. (H3)\n\nThere exist nonnegative constants $$K_{ij}$$ satisfying\n\n$$\\int ^{+\\infty }_{0}K_{ij}(\\cdot ) \\,\\mathrm{d}s\\leq K_{ij},\\quad i,j=1,2 \\ldots ,n.$$\n\nFor every $$i,j=1,2,\\ldots ,n$$, we assume that $$a_{i}(t)$$, $$b_{ij}(t)$$, $$c_{ij}(t)$$, $$\\tau _{ij}(t)$$ are continuous ω-periodic functions and $$I_{i}(t)$$ are almost periodic functions; $$0\\leq \\tau _{ij}(t)\\leq \\tau _{ij}$$, $$\\dot{\\tau }_{ij}(t)\\leq \\sigma _{ij}<1$$; $$\\varpi _{i}(\\cdot )$$ is continuous and $$0<\\underline{\\omega }\\leq \\varpi _{i}(\\cdot )\\leq \\overline{\\omega }$$; $$\\dot{a}_{i}(\\cdot )\\geq a_{i}$$, where $$\\tau _{ij}$$, $$\\sigma _{ij}$$, $$\\underline{\\omega }$$, ω̅, and $$a_{i}$$ are nonnegative constants. Moreover, we denote $$a^{\\max }=\\max_{1\\leq i\\leq n}\\sup_{t\\in {\\mathbb {R}}}|a_{i}(t)|$$, $$b^{\\max }=\\max_{1\\leq i\\leq n, 1\\leq j\\leq n}\\sup_{t\\in {\\mathbb {R}}}|b_{ij}(t)|$$, $$c^{\\max }=\\max_{1\\leq i\\leq n, 1\\leq j\\leq n}\\sup_{t\\in {\\mathbb {R}}}|c_{ij}(t)|$$.\n\nChoose a transformation function $$\\Phi ^{-1}_{i}(\\cdot )$$, which satisfies\n\n$$\\frac{\\mathrm{d}}{\\mathrm{d}u}\\bigl(\\Phi ^{-1}_{i}(u)\\bigr)= \\frac{1}{\\varpi _{i}(u)}.$$\n\nFrom the above discussion we know that $$\\frac{1}{\\varpi _{i}(u)}>0$$, which yields that $$\\Phi ^{-1}_{i}(\\cdot )$$ is strictly monotone increasing, then the inverse function of $$\\Phi ^{-1}_{i}(\\cdot )$$ exists, we denote $$(\\Phi ^{-1}_{i}(\\cdot ))^{-1}=\\Phi _{i}(\\cdot )$$. Let $$x_{i}(t)=\\Phi ^{-1}_{i}(\\pi _{i}(t))$$, one can have $$\\pi _{i}(t)=\\Phi _{i}(x(t))$$ and $$\\frac{\\mathrm{d}{x}_{i}(t)}{\\mathrm{d}t}= \\frac{\\mathrm{d}\\Phi ^{-1}_{i}(\\pi _{i}(t))}{\\mathrm{d}\\pi _{i}(t)} \\dot{\\pi }_{i}(t)=\\frac{1}{\\varpi _{i}(\\pi _{i}(t))}\\dot{\\pi }_{i}(t)$$, then we can obtain that\n\n\\begin{aligned} \\frac{\\mathrm{d}{x}_{i}(t)}{\\mathrm{d}t}&=-a_{i}\\bigl(\\Phi _{i}\\bigl(x_{i}(t)\\bigr)\\bigr)+ \\sum ^{n}_{j=1}b_{ij}(t)f_{j} \\bigl(\\Phi _{i}\\bigl(x_{i}(t)\\bigr)\\bigr) +\\sum ^{n}_{j=1}c_{ij}(t)f_{j} \\bigl(\\Phi _{i}\\bigl(x_{i}\\bigl(t-\\tau _{ij}(t)\\bigr)\\bigr)\\bigr) \\\\ &\\quad {}+\\sum^{n}_{j=1} \\int ^{+\\infty }_{0}K_{ij}(t,s)f_{j} \\bigl(\\Phi _{i}\\bigl(x_{i}(s)\\bigr)\\bigr) \\,\\mathrm{d}s+I_{i}(t),\\quad i=1,2,\\ldots ,n. \\end{aligned}\n(2)\n\nObviously, in the framework of differential inclusions, system (2) can be rewritten as follows:\n\n\\begin{aligned} \\frac{\\mathrm{d}x_{i}(t)}{\\mathrm{d}t}&\\in -a_{i}\\bigl(\\Phi _{i} \\bigl(x_{i}(t)\\bigr)\\bigr)+ \\sum^{n}_{j=1}b_{ij}(t)K \\bigl[f_{j}\\bigl(\\Phi _{i}\\bigl(x_{i}(t) \\bigr)\\bigr)\\bigr] +\\sum^{n}_{j=1}c_{ij}(t)K \\bigl[f_{j}\\bigl(\\Phi _{i}\\bigl(x_{i}\\bigl(t- \\tau _{ij}(t)\\bigr)\\bigr)\\bigr)\\bigr] \\\\ &\\quad {}+\\sum^{n}_{j=1} \\int ^{+\\infty }_{0}K_{ij}(t,s)K \\bigl[f_{j}\\bigl(\\Phi _{i}\\bigl(x_{i}(s) \\bigr)\\bigr)\\bigr] \\,\\mathrm{d}s+I_{i}(t). \\end{aligned}\n\nFor any compact interval of $$[0,\\tau )$$, the vector function $$x=(x_{1},x_{2},\\ldots ,x_{n})^{T}$$ is continuous and absolutely continuous. According to the Filippov framework, one can find a measurable function $$\\gamma =(\\gamma _{1},\\gamma _{2},\\ldots ,\\gamma _{n})^{T}:(-\\infty , \\tau )\\rightarrow {\\mathbb {R}}^{n}$$ such that $$\\gamma _{i}(t)\\in K[f_{i}(\\Phi _{i}(x_{i}(t)))]$$, then one can obtain that x is a state solution of CGNNs and\n\n\\begin{aligned} \\frac{\\mathrm{d}x_{i}(t)}{\\mathrm{d}t} =& -a_{i}\\bigl(\\Phi _{i}\\bigl(x_{i}(t)\\bigr)\\bigr)+ \\sum ^{n}_{j=1}b_{ij}(t)\\gamma _{j}(t)+\\sum^{n}_{j=1}c_{ij}(t) \\gamma _{j}(t-\\tau ) \\\\ &{}+\\sum^{n}_{j=1} \\int ^{+\\infty }_{0}K_{ij}(t,s) \\gamma _{j}(s)\\,\\mathrm{d}s+I_{i}(t). \\end{aligned}\n(3)\n\nThrough the above discussion, consider CGNNs system (1) as the drive system. By giving the initial value of CGNNs $$\\phi (s)=(\\phi _{1}(s),\\phi _{2}(s),\\ldots ,\\phi _{n}(s))^{T}$$, we can obtain the corresponding response system as follows:\n\n\\begin{aligned} \\frac{\\mathrm{d}\\xi _{i}(t)}{\\mathrm{d}t}&=-\\varpi _{i}\\bigl(\\xi _{i}(t)\\bigr) \\Biggl[a_{i}\\bigl(\\xi _{i}(t) \\bigr)-\\sum^{n}_{j=1}b_{ij}(t)f_{j} \\bigl(\\xi _{j}(t)\\bigr) -\\sum^{n}_{j=1}c_{ij}(t)f_{j} \\bigl(\\xi _{j}\\bigl(t-\\tau _{ij}(t)\\bigr)\\bigr) \\\\ &\\quad {}-\\sum^{n}_{j=1} \\int ^{+\\infty }_{0}K_{ij}(t,s)f_{j} \\bigl(\\xi _{j}(s)\\bigr) \\,\\mathrm{d}s-I_{i}(t) \\Biggr]+u_{i}(t),\\quad i=1,2,\\ldots ,n, \\end{aligned}\n(4)\n\nwhere $$u_{i}(t)$$ is the appropriate controller.\n\nSimilarly, let $$y_{i}(t)=\\Phi ^{-1}_{i}(\\xi _{i}(t))$$, $$i=1,2,\\ldots ,n$$, we can derive that\n\n\\begin{aligned} \\frac{\\mathrm{d}y_{i}(t)}{\\mathrm{d}t}&= -a_{i}\\bigl(\\Phi _{i}\\bigl(y_{i}(t)\\bigr)\\bigr)+ \\sum ^{n}_{j=1}b_{ij}(t)\\widetilde{\\gamma }_{j}(t) +\\sum^{n}_{j=1}c_{ij}(t) \\widetilde{\\gamma }_{j}(t-\\tau ) \\\\ &\\quad {}+\\sum^{n}_{j=1} \\int ^{+\\infty }_{0}K_{ij}(t,s) \\widetilde{ \\gamma }_{j}(s)\\,\\mathrm{d}s+I_{i}(t)+ \\frac{u_{i}(t)}{\\varpi _{i}(\\Phi _{i}(y_{i}(t)))}, \\end{aligned}\n(5)\n\nwhere $$\\widetilde{\\gamma }_{i}(t)\\in K[f_{i}(\\Phi _{i}(y_{i}(t)))]$$ and $$\\widetilde{\\gamma }(t)=(\\gamma _{1}(t),\\gamma _{2}(t),\\ldots ,\\gamma _{n}(t))^{T}$$.\n\n### Lemma 2.1\n\n(See [23, 39])\n\nIf $$V(y(t)):{\\mathbb {R}}^{n}\\times {\\mathbb {R}}$$ is a C-regular function, for any compact interval of $$[0,+\\infty )$$, $$y(t):[0,+\\infty )\\rightarrow {\\mathbb {R}}^{n}$$ is an absolutely continuous function. For a continuous function $$\\Upsilon :(0,\\infty )\\rightarrow {\\mathbb {R}}$$ with $$\\Upsilon (\\varrho )>0$$ for $$\\varrho \\in (0,+\\infty )$$, if it satisfies that\n\n$$\\frac{dV(t)}{dt}\\leq -\\Upsilon \\bigl(V(t)\\bigr) \\quad \\textit{for a.e. } t\\geq 0$$\n\nand\n\n$$\\int ^{V(0)}_{0}\\frac{1}{\\Upsilon (\\varrho )}=t^{*}< + \\infty ,$$\n\nthen we have $$V(t)=0$$ for $$t\\geq t^{*}$$; especially, we have:\n\n1. (1)\n\nIf $$\\Upsilon =K_{1}\\varrho +K_{2}\\varrho ^{\\mu }$$ for all $$\\varrho >0$$, where $$\\mu \\in (0,1)$$ and $$K_{1},K_{2}>0$$, then one can estimate the settling time as\n\n$$t^{*}=\\frac{1}{K_{1}(1-\\mu )}\\ln \\frac{K_{1}V^{1-\\mu }(0)+K_{2}}{K_{2}}.$$\n2. (2)\n\nIf $$\\Upsilon (\\varrho )=K\\varrho ^{\\mu }$$ and $$K>0$$, then one can estimate the settling time as\n\n$$t^{*}=\\frac{V^{1-\\mu }(0)}{K(1-\\mu )}.$$\n\n### Remark 2.2\n\nSince $$\\Phi (\\cdot )$$ is strictly monotone increasing and $$a_{i}(\\Phi _{i}(t))$$ is an abstract function which contains linear functions as special cases. In other words, we can express $$a_{i}(\\Phi _{i}(t))$$ as a common function $$-D_{i}(t)\\Phi _{i}(t)$$. Compared with the existing papers, our model which considers CGNNs is more general and common of previous results.\n\n## 3 Main results\n\nFirstly, we focus on ensuring the finite-time synchronization issue between the above response model (4) and the drive model (1). Let $$e_{i}(t)=y_{i}(t)-x_{i}(t)$$, $$i=1,2,\\ldots ,n$$, one can obtain\n\n\\begin{aligned} \\frac{\\mathrm{d}e_{i}(t)}{\\mathrm{d}t}&= -\\bigl(a_{i}\\bigl(\\Phi _{i}\\bigl(y_{i}(t)\\bigr)\\bigr)-a_{i}\\bigl( \\Phi _{i}\\bigl(x_{i}(t)\\bigr)\\bigr)\\bigr)+\\sum ^{n}_{j=1}b_{ij}(t)\\gamma _{j}^{*}(t) +\\sum^{n}_{j=1}c_{ij}(t) \\gamma _{j}^{*}(t-\\tau ) \\\\ &\\quad {}+\\sum^{n}_{j=1} \\int ^{+\\infty }_{0}K_{ij}(t,s)\\gamma _{j}^{*}(s) \\,\\mathrm{d}s+\\frac{u_{i}(t)}{\\varpi _{i}(\\Phi _{i}(y_{i}(t)))}, \\end{aligned}\n(6)\n\nwhere $$\\gamma ^{*}_{j}(t)=\\widetilde{\\gamma }_{j}(t)-\\gamma _{j}(t)$$.\n\nThen we consider the following two kinds of important controllers to achieve the finite-time synchronization issue.\n\nCase (1). The state-feedback controller $$u_{i}(t)$$:\n\n$$u_{i}(t)=-k_{1}\\bigl(\\pi _{i}(t)-\\xi _{i}(t)\\bigr)-k_{2}\\operatorname{sign} \\bigl(\\pi _{i}(t)- \\xi _{i}(t)\\bigr),$$\n(7)\n\nwhere $$i=1,2,\\ldots ,N$$, $$k_{1}, k_{2}>0$$.\n\nCase (2). The corresponding adaptive controller $$u_{i}(t)$$s of Case (1):\n\n$$u_{i}(t)=-p_{i}\\bigl(\\pi _{i}(t)-\\xi _{i}(t)\\bigr)-q_{i}\\operatorname{sign} \\bigl(\\pi _{i}(t)- \\xi _{i}(t)\\bigr),$$\n(8)\n\nwhere $$P=\\operatorname{diag}(p_{1},p_{2},\\ldots ,p_{n})$$, $$Q=\\operatorname{diag}(q_{1},q_{2},\\ldots ,q_{n})$$, and the controller rules of $$p_{i}$$ and $$q_{i}$$ are as follows:\n\n\\begin{aligned}& \\dot{p}_{i}=e^{T}_{i}(t) \\frac{\\rho _{i}}{\\varpi _{i}(y_{i}(t))}\\bigl(\\pi _{i}(t)- \\xi _{i}(t)\\bigr) \\quad \\mbox{and} \\\\& \\dot{q}_{i}=e^{T}_{i}(t) \\frac{\\varrho _{i}}{\\varpi _{i}(y_{i}(t))}\\operatorname{sign}\\bigl(\\pi _{i}(t)- \\xi _{i}(t)\\bigr), \\quad i=1,2,\\ldots ,n, \\end{aligned}\n\n$$\\rho _{i}$$ and $$\\varrho _{i}$$ are arbitrary positive constants.\n\n### Theorem 3.1\n\nIf conditions (H1)(H3) are supported, the response system (4) with state-feedback controller (7) can synchronize to the corresponding drive system (1) in a finite-time if the following assumption holds:\n\n1. (H4)\n\n$$\\underline{\\omega }+ \\frac{k_{1}\\cdot \\underline{\\omega }}{\\overline{\\omega }}> (b^{\\max }+c^{ \\max }+K^{\\max } )\\cdot nL^{\\max }$$ and $$\\frac{k_{2}}{\\overline{\\omega }}> (b^{\\max }+c^{\\max }+K^{\\max } ) \\cdot nh^{\\max }$$.\n\n### Proof\n\nLet\n\n$$V(t) =\\frac{1}{2}e^{T}(t)e(t)+\\frac{n}{2}\\cdot L^{\\max }c^{\\max } \\int ^{t}_{t- \\tau }e^{T}(s)\\cdot e(s) \\,\\mathrm{d}s,$$\n\nwhere $$e(t)=(e_{1}(t),e_{2}(t),\\ldots ,e_{n}(t))^{T}$$.\n\nObviously $$V(t)$$ is C-regular. The derivative of Lyapunov function $$V(t)$$ can be obtained along the error system (6) as follows:\n\n\\begin{aligned} \\frac{\\mathrm{d}V(t)}{\\mathrm{d}t} =&e^{T}(t)\\dot{e}(t)=\\sum ^{n}_{i=1}e^{T}_{i}(t) \\dot{e}_{i}(t) \\\\ =&\\sum^{n}_{i=1}e^{T}_{i}(t) \\Biggl(-\\bigl(a_{i}\\bigl(\\Phi _{i}\\bigl(y_{i}(t) \\bigr)\\bigr)-a_{i}\\bigl( \\Phi _{i}\\bigl(x_{i}(t) \\bigr)\\bigr)\\bigr)+\\sum^{n}_{j=1}b_{ij}(t) \\gamma _{j}^{*}(t) \\\\ &{}+\\sum^{n}_{j=1}c_{ij}(t) \\gamma _{j}^{*}(t-\\tau )+\\sum^{n}_{j=1} \\int ^{+\\infty }_{0}K_{ij}(t,s)\\gamma _{j}^{*}(s) \\,\\mathrm{d}s \\\\ &{}-\\frac{k_{1}}{\\varpi _{i}(y_{i}(t))}\\bigl(\\pi _{i}(t)-\\xi _{i}(t)\\bigr)- \\frac{k_{2}}{\\varpi _{i}(y_{i}(t))} \\operatorname{sign}\\bigl(\\pi _{i}(t)-\\xi _{i}(t)\\bigr) \\Biggr) \\\\ &{}+\\frac{n}{2}\\cdot L^{\\max }c^{\\max }e^{T}(t)e(t)- \\frac{n}{2}\\cdot L^{ \\max }c^{\\max }e^{T}(t- \\tau )e(t-\\tau ) \\\\ =&-\\sum^{n}_{i=1}e^{T}_{i}(t) \\bigl(a_{i}\\bigl(\\Phi _{i}\\bigl(y_{i}(t) \\bigr)\\bigr)-a_{i}\\bigl( \\Phi _{i}\\bigl(x_{i}(t) \\bigr)\\bigr) \\bigr)+\\sum^{n}_{i=1}\\sum ^{n}_{j=1}e^{T}_{i}(t)b_{ij}(t) \\gamma ^{*}_{j}(t) \\\\ &{}+\\sum^{n}_{i=1} \\sum^{n}_{j=1}e^{T}_{i}(t)c_{ij}(t) \\gamma ^{*}_{j}(t-\\tau )+\\sum^{n}_{i=1}\\sum ^{n}_{j=1}e^{T}_{i}(t) \\int ^{+ \\infty }_{0}K_{ij}(t,s)\\gamma _{j}^{*}(s)\\,\\mathrm{d}s \\\\ &{}-\\sum ^{n}_{i=1}e^{T}_{i}(t)k_{1} \\frac{1}{\\varpi _{i}(y_{i}(t))} \\bigl(\\Phi _{i}\\bigl(y_{i}(t) \\bigr)-\\Phi _{i}\\bigl(x_{i}(t)\\bigr) \\bigr) \\\\ &{}-\\sum^{n}_{i=1}e^{T}_{i}(t)k_{2} \\frac{1}{\\varpi _{i}(y_{i}(t))}\\operatorname{sign}\\bigl[\\Phi _{i} \\bigl(y_{i}(t)\\bigr))- \\Phi _{i}\\bigl(x_{i}(t) \\bigr)\\bigr] \\\\ &{}+\\frac{n}{2}\\cdot L^{\\max }c^{\\max }e^{T}(t)e(t)- \\frac{n}{2}\\cdot L^{ \\max }c^{\\max }e^{T}(t- \\tau )e(t-\\tau ). \\end{aligned}\n(9)\n\nBased on the definition of function $$\\Phi (\\cdot )$$ and generalized mean value theorem, one can have\n\n$$-\\sum^{n}_{i=1}e^{T}_{i}(t) \\bigl(a_{i}\\bigl(\\Phi _{i}\\bigl(y_{i}(t) \\bigr)\\bigr)-a_{i}\\bigl( \\Phi _{i}\\bigl(x_{i}(t) \\bigr)\\bigr)\\bigr)\\leq -\\sum^{n}_{i=1}e^{T}_{i}(t) \\underline{\\omega }e_{i}(t).$$\n(10)\n\nFrom assumptions (H1)–(H2), we can obtain that\n\n\\begin{aligned} \\sum^{n}_{i=1}\\sum ^{n}_{j=1}e^{T}_{i}(t)b_{ij}(t) \\gamma ^{*}_{j}(t)&\\leq \\sum ^{n}_{i=1}\\sum^{n}_{j=1} \\bigl\\vert e^{T}_{i}(t) \\bigr\\vert \\cdot \\bigl\\vert b_{ij}(t) \\bigr\\vert \\cdot \\bigl\\vert \\gamma ^{*}_{j}(t) \\bigr\\vert \\\\ &\\leq \\sum^{n}_{i=1}\\sum ^{n}_{j=1} \\bigl\\vert e^{T}_{i}(t) \\bigr\\vert \\cdot \\bigl\\vert b_{ij}(t) \\bigr\\vert \\cdot \\bigl(L_{j} \\bigl\\vert e_{j}(t) \\bigr\\vert +h_{j}\\bigr) \\\\ &\\leq n b^{\\max }L^{\\max }\\sum^{n}_{i=1}e^{T}_{i}(t)e_{i}(t)+nb^{ \\max }h^{\\max } \\sum^{n}_{i=1}\\sum ^{n}_{k=1} \\bigl\\vert e_{ik}(t) \\bigr\\vert , \\end{aligned}\n(11)\n\nand\n\n\\begin{aligned} \\sum^{n}_{i=1}\\sum ^{n}_{j=1}e^{T}_{i}(t)c_{ij}(t) \\gamma ^{*}_{j}(t-\\tau )&\\leq \\sum ^{n}_{i=1}\\sum^{n}_{j=1} \\bigl\\vert e^{T}_{i}(t) \\bigr\\vert \\cdot \\bigl\\vert c_{ij}(t) \\bigr\\vert \\cdot \\bigl\\vert \\gamma ^{*}_{j}(t- \\tau ) \\bigr\\vert \\\\ &\\leq \\sum^{n}_{i=1}\\sum ^{n}_{j=1} \\bigl\\vert e^{T}_{i}(t) \\bigr\\vert \\cdot \\bigl\\vert c_{ij}(t) \\bigr\\vert \\cdot \\bigl(L_{j} \\bigl\\vert e_{j}(t-\\tau ) \\bigr\\vert +h_{j}\\bigr) \\\\ &\\leq c^{\\max }L^{\\max }\\sum^{n}_{i=1} \\sum^{n}_{j=1}e^{T}_{i}(t)e_{j}(t- \\tau )+nc^{\\max }h^{\\max }\\sum^{n}_{i=1} \\sum^{n}_{i=1} \\bigl\\vert e_{ik}(t) \\bigr\\vert \\\\ &\\leq c^{\\max }L^{\\max } \\Biggl(\\frac{n}{2}\\sum ^{n}_{i=1}e^{T}_{i}(t)e_{i}(t)+ \\frac{n}{2}\\sum^{n}_{j=1}e^{T}_{j}(t- \\tau )e_{j}(t-\\tau ) \\Biggr) \\\\ &\\quad {}+nc^{\\max }h^{\\max }\\sum ^{n}_{i=1}\\sum ^{n}_{k=1} \\bigl\\vert e_{ik}(t) \\bigr\\vert . \\end{aligned}\n(12)\n\nFrom condition (H3), similarly we have\n\n\\begin{aligned} \\sum^{n}_{i=1}\\sum ^{n}_{j=1}e^{T}_{i}(t) \\int ^{+ \\infty }_{0}K_{ij}(t,s)\\gamma _{j}^{*}(s)\\,\\mathrm{d}s&\\leq \\sum ^{n}_{i=1}\\sum^{n}_{j=1}K_{ij}e^{T}_{i}(t) \\gamma _{j}^{*}(s) \\\\ &\\leq n K^{\\max }L^{\\max }\\sum^{n}_{i=1}e^{T}_{i}(t)e_{i}(t) \\\\ &\\quad {}+nK^{ \\max }h^{\\max } \\sum^{n}_{i=1}\\sum ^{n}_{k=1} \\bigl\\vert e_{ik}(t) \\bigr\\vert . \\end{aligned}\n(13)\n\nMoreover, because $$\\Phi _{i}(\\cdot )$$ is strictly monotone increasing, which implies that $$\\operatorname{sign}(\\Phi _{i}(y_{i}(t))-\\Phi _{i}(x_{i}(t)))=\\operatorname{sign}(y_{i}(t)-x_{i}(t))= \\operatorname{sign}e_{i}(t)$$, then we obtain\n\n\\begin{aligned}& -\\sum^{n}_{i=1}e^{T}_{i}(t)k_{1} \\frac{1}{\\varpi _{i}(y_{i}(t))} \\bigl(\\Phi _{i}\\bigl(y_{i}(t) \\bigr)-\\Phi _{i}\\bigl(x_{i}(t)\\bigr) \\bigr)\\leq -\\sum ^{n}_{i=1} \\frac{k_{1}\\cdot \\underline{\\omega }}{\\overline{\\omega }}e^{T}_{i}(t)e_{i}(t), \\end{aligned}\n(14)\n\\begin{aligned}& -\\sum^{n}_{i=1}e^{T}_{i}(t)k_{2} \\frac{1}{\\varpi _{i}(y_{i}(t))}\\operatorname{sign}\\bigl[\\Phi _{i} \\bigl(y_{i}(t)\\bigr)- \\Phi _{i}\\bigl(x_{i}(t) \\bigr)\\bigr]\\leq -\\sum^{n}_{i=1}\\sum ^{n}_{k=1} \\frac{k_{2}}{\\overline{\\omega }} \\bigl\\vert e_{ik}(t) \\bigr\\vert . \\end{aligned}\n(15)\n\nRecalling controller (9) and combined with equations (10)–(15), based on the basic inequalities of Jensen’s inequality, one can gain\n\n\\begin{aligned} \\frac{\\mathrm{d}V(t)}{\\mathrm{d}t}&\\leq - \\biggl[\\underline{\\omega }+ \\frac{k_{1}\\cdot \\underline{\\omega }}{\\overline{\\omega }}- \\bigl(b^{\\max }+c^{ \\max }+K^{\\max } \\bigr)\\cdot nL^{\\max } \\biggr]\\cdot \\sum^{n}_{i=1}e_{i}^{T}(t)e_{i}(t) \\\\ &\\quad {}- \\biggl[\\frac{k_{2}}{\\overline{\\omega }}- \\bigl(b^{\\max }+c^{\\max }+K^{\\max } \\bigr)\\cdot nh^{\\max } \\biggr]\\cdot \\sum^{n}_{i=1} \\sum^{n}_{k=1} \\bigl\\vert e_{ik}(t) \\bigr\\vert \\\\ &\\leq -\\kappa _{1}V(t)-\\kappa _{2}V^{\\frac{1}{2}}(t). \\end{aligned}\n(16)\n\nThen, according to assumption (H4) in this theorem, one can see $$\\kappa _{1}>0$$ and $$\\kappa _{2}>0$$, then we know that the origin of error system (6) is finite-time stable with feedback controller (7), and the settling time is obtained by\n\n$$t^{*}_{1}\\leq \\frac{2}{\\kappa _{1}}\\ln \\frac{\\kappa _{1}V^{\\frac{1}{2}}(0)+\\kappa _{2}}{\\kappa _{2}} = \\frac{2}{\\kappa _{1}}\\ln \\frac{\\kappa _{1} \\Vert e(0) \\Vert _{2}+\\kappa _{2}}{\\kappa _{2}}.$$\n\n□\n\n### Theorem 3.2\n\nIf conditions (H1)(H3) are supported, then the response system (4) with adaptive controller (8) can synchronize to the corresponding drive system (1).\n\n### Proof\n\nIn this proof, we let the new Lyapunov functional be as follows:\n\n\\begin{aligned} V(t) =&\\frac{1}{2}\\sum^{n}_{i=1}e^{T}_{i}(t)e_{i}(t)+ \\frac{n}{2}\\cdot L^{\\max }c^{\\max } \\int ^{t}_{t-\\tau }e^{T}(s)\\cdot e(s) \\,\\mathrm{d}s \\\\ &{}+\\sum^{n}_{i=1} \\frac{1}{2\\rho _{ik}} (p_{i}- \\theta )^{2} +\\sum ^{n}_{i=1}\\frac{1}{2\\varrho _{i}} (q_{i}- \\vartheta )^{2}, \\end{aligned}\n\nwhere θ and ϑ are positive constants to be determined.\n\nThe derivative of Lyapunov function $$V(t)$$ can be obtained along the error system (6) as follows:\n\n\\begin{aligned} \\dot{V}(t) =&\\sum^{n}_{i=1}e^{T}_{i}(t) \\Biggl[-\\bigl(a_{i}\\bigl(\\Phi _{i}\\bigl(y_{i}(t) \\bigr)\\bigr)-a_{i}\\bigl( \\Phi _{i}\\bigl(x_{i}(t) \\bigr)\\bigr)\\bigr)+\\sum^{n}_{j=1}b_{ij}(t) \\gamma _{j}^{*}(t) +\\sum^{n}_{j=1}c_{ij}(t) \\gamma _{j}^{*}(t-\\tau ) \\\\ &{}+\\sum^{n}_{j=1} \\int ^{+\\infty }_{0}K_{ij}(t,s)\\gamma _{j}^{*}(s) \\,\\mathrm{d}s-\\frac{p_{i}}{\\varpi _{i}(y_{i}(t))}\\bigl(\\pi _{i}(t)-\\xi _{i}(t)\\bigr) \\\\ &{}- \\frac{q_{i}}{\\varpi _{i}(y_{i}(t))} \\operatorname{sign}\\bigl(\\pi _{i}(t)-\\xi _{i}(t)\\bigr) \\Biggr] \\\\ &{}+\\sum^{n}_{i=1}(p_{i}- \\theta )e^{T}_{i}(t) \\frac{1}{\\varpi _{i}(y_{i}(t))}\\bigl(\\pi _{i}(t)-\\xi _{i}(t)\\bigr) \\\\ &{}+\\sum ^{n}_{i=1}(q_{i}-\\vartheta )e^{T}_{i}(t) \\frac{1}{\\varpi _{i}(y_{i}(t))}\\operatorname{sign}\\bigl(\\pi _{i}(t)-\\xi _{i}(t)\\bigr) \\\\ &{}+\\frac{n}{2}\\cdot L^{\\max }c^{\\max }e^{T}(t)e(t)- \\frac{n}{2}\\cdot L^{ \\max }c^{\\max }e^{T}(t- \\tau )e(t-\\tau ). \\end{aligned}\n(17)\n\nSince\n\n\\begin{aligned}& -\\sum^{n}_{i=1}e^{T}_{i}(t) \\frac{\\theta }{\\varpi _{i}(y_{i}(t))} \\bigl(\\Phi _{i}\\bigl(y_{i}(t) \\bigr)-\\Phi _{i}\\bigl(x_{i}(t)\\bigr) \\bigr)\\leq -\\sum ^{n}_{i=1} \\frac{\\theta \\cdot \\underline{\\omega }}{\\overline{\\omega }}e^{T}_{i}(t)e_{i}(t), \\end{aligned}\n(18)\n\\begin{aligned}& -\\sum^{n}_{i=1}e^{T}_{i}(t) \\frac{\\vartheta }{\\varpi _{i}(y_{i}(t))}\\operatorname{sign}\\bigl[\\Phi _{i} \\bigl(y_{i}(t)\\bigr)- \\Phi _{i}\\bigl(x_{i}(t) \\bigr)\\bigr]\\leq -\\sum^{n}_{i=1}\\sum ^{n}_{k=1} \\frac{\\vartheta }{\\overline{\\omega }} \\bigl\\vert e_{ik}(t) \\bigr\\vert . \\end{aligned}\n(19)\n\nIt follows from (18)–(19), recalling (10)–(13), one can deduce that\n\n\\begin{aligned} \\frac{\\mathrm{d}V(t)}{\\mathrm{d}t}&\\leq - \\biggl[\\underline{\\omega }+ \\frac{\\theta \\cdot \\underline{\\omega }}{\\overline{\\omega }}- \\bigl(b^{ \\max }+c^{\\max }+K^{\\max } \\bigr)\\cdot nL^{\\max } \\biggr]\\cdot \\sum^{n}_{i=1}e_{i}^{T}(t)e_{i}(t) \\\\ &\\quad {}- \\biggl[\\frac{\\vartheta }{\\overline{\\omega }}- \\bigl(b^{\\max }+c^{\\max }+K^{ \\max } \\bigr)\\cdot nh^{\\max } \\biggr]\\cdot \\sum^{n}_{i=1} \\sum^{n}_{k=1} \\bigl\\vert e_{ik}(t) \\bigr\\vert \\\\ &\\leq -\\kappa _{3}V(t)-\\kappa _{4}V^{\\frac{1}{2}}(t). \\end{aligned}\n(20)\n\nBased on the definition of θ and ϑ, we can choose suitable values of θ and ϑ to make $$\\kappa _{3}=\\underline{\\omega }+ \\frac{\\theta \\cdot \\underline{\\omega }}{\\overline{\\omega }}- (b^{ \\max }+c^{\\max }+K^{\\max } )\\cdot nL^{\\max }>0$$ and $$\\kappa _{4}=\\frac{\\vartheta }{\\overline{\\omega }}- (b^{\\max }+c^{ \\max }+K^{\\max } )\\cdot nh^{\\max }>0$$. Then we prove that the origin of error system (6) is finite-time stable with adaptive controller (8), and the settling time is obtained by\n\n$$t^{*}_{1}\\leq \\frac{2}{\\kappa _{3}}\\ln \\frac{\\kappa _{3}V^{\\frac{1}{2}}(0)+\\kappa _{4}}{\\kappa _{4}} = \\frac{2}{\\kappa _{3}}\\ln \\frac{\\kappa _{3} \\Vert e(0) \\Vert _{2}+\\kappa _{4}}{\\kappa _{4}}.$$\n\n□\n\n### Remark 3.3\n\nThis paper considers that the distributed delays are unbounded, which is more difficult to verify than the bounded case. In the pervious results, the delay kernels satisfy\n\n$$K_{ij}= \\textstyle\\begin{cases} 1,& 0\\leq t\\leq \\tau _{ij}, \\\\ 0,& t>\\tau _{ij}, \\end{cases}$$\n\nwhere $$\\tau _{ij}>0$$ are constants, then the CGNNs can be rewritten as a special case in this paper.\n\n### Remark 3.4\n\nIn fact, the finite time synchronization problem is very complex and difficult to calculate when there exist the discontinuity phenomenon, mixed delays, and switching controllers in the traditional neural network model. This paper overcomes these difficulties and has some innovation. By using Theorem 3.1 and Theorem 3.2, the finite-time synchronization problem can be generalized, that is, by choosing the appropriate controller, the stability time of the synchronization error system can be estimated more easily. On the other hand, the controller selected in this paper is more widely used as the estimation of stability time. It provides a theoretical basis for solving complex problems in engineering application.\n\n## 4 Examples\n\n### Example 4.1\n\nConsider the following two-dimensional discontinuous CGNNs with and mixed delays:\n\n\\begin{aligned} \\frac{\\mathrm{d}\\pi _{i}(t)}{\\mathrm{d}t}&=-\\varpi _{i}\\bigl(\\pi _{i}(t)\\bigr) \\Biggl[a_{i}\\bigl(\\pi _{i}(t) \\bigr)-\\sum^{n}_{j=1}b_{ij}(t)f_{j} \\bigl(\\pi _{j}(t)\\bigr) -\\sum^{n}_{j=1}c_{ij}(t)f_{j} \\bigl(\\pi _{j}\\bigl(t-\\tau _{ij}(t)\\bigr)\\bigr) \\\\ &\\quad {}-\\sum^{n}_{j=1} \\int ^{+\\infty }_{0}K_{ij}(t,s)f_{j} \\bigl(\\pi _{j}(s)\\bigr) \\,\\mathrm{d}s-I_{i}(t) \\Biggr],\\quad i=1,2. \\end{aligned}\n(21)\n\nLet $$\\varpi _{1}(\\pi _{1}(t))=0.5+0.1\\cos (\\pi _{1}(t))$$, $$\\varpi _{2}(\\pi _{2}(t))=0.5-0.1\\sin (\\pi _{2}(t))$$, $$a_{1}(\\pi _{1}(t))=-0.4\\pi _{1}(t)$$, $$a_{2}(\\pi _{2}(t))=-0.4\\pi _{2}(t)$$, $$b_{11}(t)=b_{22}(t)=0.1$$, $$b_{12}(t)=b_{21}(t)=0$$, $$c_{11}(t)=c_{22}(t)=0.2$$, $$c_{12}(t)=c_{21}(t)=0$$, $$K_{11}=K_{12}=K_{21}=K_{22}=1$$, and $$I_{1}(t)=0.2\\sin \\sqrt{2}t+0.1\\sin \\sqrt{5}t$$, $$I_{2}(t)=0.3 \\cos \\sqrt{3}t-0.2\\sin 2t$$.\n\nThe two neuron activation functions which satisfy the two conditions (H1)–(H2) are designed as follows:\n\n$$f_{1}(\\cdot )=f_{2}(\\cdot )= \\textstyle\\begin{cases} x-0.1,& x< 0, \\\\ x+0.1,& x\\geq 0. \\end{cases}$$\n\nLet $$L^{\\max }=h^{\\max }=1$$, then we consider the control rule $$u_{i}(t)=-k_{1}(\\pi _{i}(t)-\\xi (t))-k_{2}\\operatorname{sign}(\\pi _{i}(t)- \\xi (t))$$ with $$k_{1}=k_{2}=3.5$$. It is easy to check that\n\n\\begin{aligned}& 2.7=\\underline{\\omega }+ \\frac{k_{1}\\cdot \\underline{\\omega }}{\\overline{\\omega }}> \\bigl(b^{\\max }+c^{ \\max }+K^{\\max } \\bigr)\\cdot nL^{\\max }=2.6, \\\\& 5.8=\\frac{k_{2}}{\\overline{\\omega }}> \\bigl(b^{\\max }+c^{\\max }+K^{\\max } \\bigr)\\cdot nh^{\\max }=2.6, \\end{aligned}\n\nwhich show that the assumption in Theorem 3.1 is satisfied. Figure 1 and Fig. 2(c) indicate the simulation results.\n\n### Example 4.2\n\nConsider the following discontinuous CGNNs with mixed delays:\n\n\\begin{aligned} \\frac{\\mathrm{d}\\pi _{i}(t)}{\\mathrm{d}t}&=-\\varpi _{i}\\bigl(\\pi _{i}(t)\\bigr) \\Biggl[a_{i}\\bigl(\\pi _{i}(t) \\bigr)-\\sum^{n}_{j=1}b_{ij}(t)f_{j} \\bigl(\\pi _{j}(t)\\bigr) -\\sum^{n}_{j=1}c_{ij}(t)f_{j} \\bigl(\\pi _{j}\\bigl(t-\\tau _{ij}(t)\\bigr)\\bigr) \\\\ &\\quad {}-\\sum^{n}_{j=1} \\int ^{+\\infty }_{0}K_{ij}(t,s)f_{j} \\bigl(\\pi _{j}(s)\\bigr) \\,\\mathrm{d}s-I_{i}(t) \\Biggr], \\quad i=1,2, \\end{aligned}\n(22)\n\nwhere the parameters have the same meanings as in equation (21).\n\nWe consider the novel adaptive controller $$u_{i}(t)=-p_{i}(\\pi _{i}(t)-\\xi (t))-q_{i}\\operatorname{sign}(\\pi _{i}(t)- \\xi (t))$$ with $$\\dot{p}_{i}=e^{T}_{i}(t)\\frac{\\varrho _{i}}{\\varpi _{i}(y_{i}(t))}( \\pi _{i}(t)-\\xi _{i}(t))$$, $$\\dot{q}_{i}=e^{T}_{i}(t)\\frac{\\varrho _{i}}{\\varpi _{i}(y_{i}(t))} \\operatorname{sign}(\\pi _{i}(t)-\\xi _{i}(t))$$. For CGNNs (22), we can choose suitable parameters $$\\theta =3$$ and $$\\vartheta =3$$ to make the condition in Theorem 3.2 be satisfied. With given random initial state, Fig. 2(d) shows that the two state trajectories approach the zero solution, then the simulation result is presented to illustrate the obtained theoretical findings.\n\n## 5 Conclusions\n\nThe finite-time synchronization problem of discontinuous CGNNs with mixed delay is studied in this brief. By using the Lyapunov functional framework, new mathematical analysis techniques, Filippov theory, and inequality techniques, a new state feedback controller and an adaptive controller are constructed to realize finite-time synchronization of complex neural networks. Compared with previous results, we overcome the problem of non-Lipschitz continuity system, so how to deal with the right discontinuous system is a challenge. 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Clarke, F.: Optimization and Nonsmooth Analysis. SIAM, Philadelphia (1990)\n\n43. Polyakov, A.: Nonlinear feedback design for fixed-time stabilization of linear control systems. IEEE Trans. Autom. Control 57(8), 2106–2110 (2012)\n\n44. Strogatz, S., Stewart, I.: Coupled oscillators and biological synchronization. Sci. Am. 269(6), 102–109 (1993)\n\n45. Milanovi, V., Zaghloul, M.: Synchronization of chaotic neural networks and applications to communications. Int. J. Bifurc. Chaos 6(12b), 2571–2585 (1996)\n\nNot applicable.\n\n## Funding\n\nThis work is supported by the National Natural Science Foundation of China (11801042) and Training Program for Excellent Young Innovators of Changsha (kq2009023).\n\n## Author information\n\nAuthors\n\n### Contributions\n\nZX and RL initiated and discussed the research problem; ZX verified the correctness of the experimental results; RL performed the experiments and took figures, analyzed the data; ZX drafted the paper. All authors have read and approved the paper.\n\n### Corresponding author\n\nCorrespondence to Zhaohong Xiang.\n\n## Ethics declarations\n\n### Competing interests\n\nThe authors declare that they have no competing interests.\n\n## Rights and permissions", null, "" ]

[ null, "https://advancesincontinuousanddiscretemodels.springeropen.com/track/article/10.1186/s13662-021-03542-2", null ]

[ "# BZOJ 1030 【JSOI2007】 文本生成器\n\n## Description\n\nJSOI交给队员ZYX一个任务，编制一个称之为“文本生成器”的电脑软件：该软件的使用者是一些低幼人群，\n\n## Input\n\n输入文件的第一行包含两个正整数，分别是使用者了解的单词总数N (<= 60)，GW文本生成器 v6生成的文本固\n\n## Output\n\n一个整数，表示可能的文章总数。只需要知道结果模10007的值。\n\n像我这样成天刷水题吃枣药丸\n这道题就是问有多少个 包含至少一个给定串，并且长度为$m$的串。然后，显然这种多串处理的题是要AC自动机的。构出AC自动机后在上面dp就可以了。\n其实感觉直接dp也可以做，但其实把问题转化一下更好做。求出所有不合法的串然后用总数量减一下就可以了。于是$f_{i,j}$表示在AC自动机上走了$i$步后到达节点$j$的方案数，转移的时候不经过单词的结束节点即可。\nAC自动机都写错的我吃枣药丸\n下面贴代码：\n#include<iostream>\n#include<cstdio>\n#include<cstring>\n#include<algorithm>\n#include<cmath>\n#define File(s) freopen(s\".in\",\"r\",stdin),freopen(s\".out\",\"w\",stdout)\n#define maxn 6010\n#define mod 10007\n\nusing namespace std;\ntypedef long long llg;\n\nint n,m,fl[maxn],d[maxn],ans;\nint ch[maxn],sz,f[maxn];\nbool val[maxn],w[maxn];\n\nint getint(){\nint w=0;bool q=0;\nchar c=getchar();\nwhile((c>'9'||c<'0')&&c!='-') c=getchar();\nif(c=='-') c=getchar(),q=1;\nwhile(c>='0'&&c<='9') w=w*10+c-'0',c=getchar();\nreturn q?-w:w;\n}\n\nvoid gi(int &x){if(x>=mod) x%=mod;}\nvoid insert(){\nchar c=getchar();\nwhile(c>'Z'||c<'A') c=getchar();\nint u=0;\nwhile(c>='A' && c<='Z'){\nif(!ch[u][c-'A']) ch[u][c-'A']=++sz;\nu=ch[u][c-'A']; c=getchar();\n}\nval[u]=1;\n}\n\nvoid getf(){\nint l=0,r=0,u;d[r++]=0;\nwhile(l!=r){\nu=d[l++];\nfor(int i=0,j;i<26;i++)\nif(ch[u][i]){\nj=fl[u];\nwhile(!ch[j][i] && j) j=fl[j];\nif(u!=j){\nfl[ch[u][i]]=ch[j][i];\nval[ch[u][i]]|=val[ch[j][i]];\n}\nd[r++]=ch[u][i];\n}\nelse ch[u][i]=ch[fl[u]][i];\n}\n}\n\nint main(){\nFile(\"a\");\nn=getint(); m=getint();\nwhile(n--) insert();\ngetf(); f=ans=1;\nfor(int i=0;i<m;i++)\nfor(int u=0;u<=sz;u++)\nif(!val[u])\nfor(int j=0;j<26;j++)\nf[i+1][ch[u][j]]+=f[i][u],gi(f[i+1][ch[u][j]]);\nfor(int i=1;i<=m;i++) ans*=26,gi(ans);\nfor(int u=0;u<=sz;u++) if(!val[u]) ans-=f[m][u],ans+=mod,gi(ans);\nprintf(\"%d\",ans);\nreturn 0;\n}\nposted @ 2016-10-28 21:51  lcf2000  阅读(161)  评论(0编辑  收藏" ]

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[ "Home | | Electronic Circuits II | The RC Integrator\n\n# The RC Integrator\n\nThe Integrator is basically a low pass filter circuit operating in the time domain that converts a square wave \"step\" response input signal into a triangular shaped waveform output as the capacitor charges and discharges.\n\nThe RC Integrator\n\nThe Integrator is basically a low pass filter circuit operating in the time domain that converts a square wave \"step\" response input signal into a triangular shaped waveform output as the capacitor charges and discharges.\n\nA Triangular waveform consists of alternate but equal, positive and negative ramps. As seen below, if the RC time constant is long compared to the time period of the input waveform the resultant output waveform will be triangular in shape and the higher the input frequency the lower will be the output amplitude compared to that of the input.", null, "This then makes this type of circuit ideal for converting one type of electronic signal to another for use in wave-generating or wave-shaping circuits.\n\nThe Low Pass Filter\n\nA simple passive Low Pass Filter or LPF, can be easily made by connecting together in series a single Resistor with a single Capacitor as shown below. In this type of filter arrangement the input signal (Vin) is applied to the series combination (both the Resistor and Capacitor together) but the output signal (Vout) is taken across the capacitor only.\n\nThis type of filter is known generally as a \"first-order filter\" or \"one-pole filter\", why first-order or single-pole, because it has only \"one\" reactive component in the circuit, the capacitor.\n\nLow Pass Filter Circuit", null, "The reactance of a capacitor varies inversely with frequency, while the value of the resistor remains constant as the frequency changes. At low frequencies the capacitive reactance, (Xc) of the capacitor will be very large compared to the resistive value of the resistor, R and as a result the voltage across the capacitor, Vc will also be large while the voltage drop across the resistor, Vr will be much lower. At high frequencies the reverse is true with Vc being small and Vr being large.\n\nHigh Pass Filters\n\nA High Pass Filter or HPF, is the exact opposite to that of the Low Pass filter circuit, as now the two components have been interchanged with the output signal (Vout) being taken from across the resistor as shown.\n\nWhere the low pass filter only allowed signals to pass below its cut-off frequency point, fc. The passive high pass filter circuit as its name implies, only passes signals above the selected cut-off point, fc eliminating any low frequency signals from the waveform. Consider the circuit below.\n\nThe High Pass Filter Circuit", null, "In this circuit arrangement, the reactance of the capacitor is very high at low frequencies so the capacitor acts like an open circuit and blocks any input signals at Vin until the cut-off frequency point (fc) is reached.\n\nAbove this cut-off frequency point the reactance of the capacitor has reduced sufficiently as to now act more like a short circuit allowing all of the input signal to pass directly to the output as shown below in the High Pass Frequency Response Curve.\n\nRC Differentiator\n\nUp until now the input waveform to the filter has been assumed to be sinusoidal or that of a sine wave consisting of a fundamental signal and some harmonics operating in the frequency domain giving us a frequency domain response for the filter.\n\nHowever, if we feed the High Pass Filter with a Square Wave signal operating in the time domain giving an impulse or step response input, the output waveform will consist of short duration pulse or spikes as shown.", null, "Each cycle of the square wave input waveform produces two spikes at the output, one positive and one negative and whose amplitude is equal to that of the input. The rate of decay of the spikes depends upon the time constant, (RC) value of both components, (t = R x C) and the value of the input frequency. The output pulses resemble more and more the shape of the input signal as the frequency increases\n\nStudy Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail\n\nRelated Topics" ]

[ null, "http://www.brainkart.com/media/extra/DCA2AI0.jpg", null, "http://www.brainkart.com/media/extra/ZSpNejX.jpg", null, "http://www.brainkart.com/media/extra/dfHdlPU.jpg", null, "http://www.brainkart.com/media/extra/7KRGpRk.jpg", null ]

[ "# #00dabf Color Information\n\nIn a RGB color space, hex #00dabf is composed of 0% red, 85.5% green and 74.9% blue. Whereas in a CMYK color space, it is composed of 100% cyan, 0% magenta, 12.4% yellow and 14.5% black. It has a hue angle of 172.6 degrees, a saturation of 100% and a lightness of 42.7%. #00dabf color hex could be obtained by blending #00ffff with #00b57f. Closest websafe color is: #00cccc.\n\n• R 0\n• G 85\n• B 75\nRGB color chart\n• C 100\n• M 0\n• Y 12\n• K 15\nCMYK color chart\n\n#00dabf color description : Pure (or mostly pure) cyan.\n\n# #00dabf Color Conversion\n\nThe hexadecimal color #00dabf has RGB values of R:0, G:218, B:191 and CMYK values of C:1, M:0, Y:0.12, K:0.15. Its decimal value is 55999.\n\nHex triplet RGB Decimal 00dabf `#00dabf` 0, 218, 191 `rgb(0,218,191)` 0, 85.5, 74.9 `rgb(0%,85.5%,74.9%)` 100, 0, 12, 15 172.6°, 100, 42.7 `hsl(172.6,100%,42.7%)` 172.6°, 100, 85.5 00cccc `#00cccc`\nCIE-LAB 78.404, -50.342, 0.756 34.472, 53.901, 57.874 0.236, 0.369, 53.901 78.404, 50.348, 179.139 78.404, -63.401, 9.016 73.417, -44.668, 4.654 00000000, 11011010, 10111111\n\n# Color Schemes with #00dabf\n\n• #00dabf\n``#00dabf` `rgb(0,218,191)``\n• #da001b\n``#da001b` `rgb(218,0,27)``\nComplementary Color\n• #00da52\n``#00da52` `rgb(0,218,82)``\n• #00dabf\n``#00dabf` `rgb(0,218,191)``\n• #0088da\n``#0088da` `rgb(0,136,218)``\nAnalogous Color\n• #da5200\n``#da5200` `rgb(218,82,0)``\n• #00dabf\n``#00dabf` `rgb(0,218,191)``\n• #da0088\n``#da0088` `rgb(218,0,136)``\nSplit Complementary Color\n• #dabf00\n``#dabf00` `rgb(218,191,0)``\n• #00dabf\n``#00dabf` `rgb(0,218,191)``\n• #bf00da\n``#bf00da` `rgb(191,0,218)``\n• #1bda00\n``#1bda00` `rgb(27,218,0)``\n• #00dabf\n``#00dabf` `rgb(0,218,191)``\n• #bf00da\n``#bf00da` `rgb(191,0,218)``\n• #da001b\n``#da001b` `rgb(218,0,27)``\n• #008e7c\n``#008e7c` `rgb(0,142,124)``\n• #00a792\n``#00a792` `rgb(0,167,146)``\n• #00c1a9\n``#00c1a9` `rgb(0,193,169)``\n• #00dabf\n``#00dabf` `rgb(0,218,191)``\n• #00f4d5\n``#00f4d5` `rgb(0,244,213)``\n• #0effe1\n``#0effe1` `rgb(14,255,225)``\n• #28ffe4\n``#28ffe4` `rgb(40,255,228)``\nMonochromatic Color\n\n# Alternatives to #00dabf\n\nBelow, you can see some colors close to #00dabf. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #00da89\n``#00da89` `rgb(0,218,137)``\n• #00da9b\n``#00da9b` `rgb(0,218,155)``\n``#00daad` `rgb(0,218,173)``\n• #00dabf\n``#00dabf` `rgb(0,218,191)``\n``#00dad1` `rgb(0,218,209)``\n• #00d1da\n``#00d1da` `rgb(0,209,218)``\n• #00bfda\n``#00bfda` `rgb(0,191,218)``\nSimilar Colors\n\n# #00dabf Preview\n\nThis text has a font color of #00dabf.\n\n``<span style=\"color:#00dabf;\">Text here</span>``\n#00dabf background color\n\nThis paragraph has a background color of #00dabf.\n\n``<p style=\"background-color:#00dabf;\">Content here</p>``\n#00dabf border color\n\nThis element has a border color of #00dabf.\n\n``<div style=\"border:1px solid #00dabf;\">Content here</div>``\nCSS codes\n``.text {color:#00dabf;}``\n``.background {background-color:#00dabf;}``\n``.border {border:1px solid #00dabf;}``\n\n# Shades and Tints of #00dabf\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000202 is the darkest color, while #eefffd is the lightest one.\n\n• #000202\n``#000202` `rgb(0,2,2)``\n• #001613\n``#001613` `rgb(0,22,19)``\n• #002924\n``#002924` `rgb(0,41,36)``\n• #003d36\n``#003d36` `rgb(0,61,54)``\n• #005147\n``#005147` `rgb(0,81,71)``\n• #006458\n``#006458` `rgb(0,100,88)``\n• #007869\n``#007869` `rgb(0,120,105)``\n• #008c7a\n``#008c7a` `rgb(0,140,122)``\n• #009f8b\n``#009f8b` `rgb(0,159,139)``\n• #00b39d\n``#00b39d` `rgb(0,179,157)``\n• #00c6ae\n``#00c6ae` `rgb(0,198,174)``\n• #00dabf\n``#00dabf` `rgb(0,218,191)``\n• #00eed0\n``#00eed0` `rgb(0,238,208)``\n• #02ffe0\n``#02ffe0` `rgb(2,255,224)``\n• #16ffe2\n``#16ffe2` `rgb(22,255,226)``\n• #29ffe5\n``#29ffe5` `rgb(41,255,229)``\n• #3dffe7\n``#3dffe7` `rgb(61,255,231)``\n• #51ffe9\n``#51ffe9` `rgb(81,255,233)``\n• #64ffec\n``#64ffec` `rgb(100,255,236)``\n• #78ffee\n``#78ffee` `rgb(120,255,238)``\n• #8cfff1\n``#8cfff1` `rgb(140,255,241)``\n• #9ffff3\n``#9ffff3` `rgb(159,255,243)``\n• #b3fff6\n``#b3fff6` `rgb(179,255,246)``\n• #c6fff8\n``#c6fff8` `rgb(198,255,248)``\n• #dafffa\n``#dafffa` `rgb(218,255,250)``\n• #eefffd\n``#eefffd` `rgb(238,255,253)``\nTint Color Variation\n\n# Tones of #00dabf\n\nA tone is produced by adding gray to any pure hue. In this case, #657573 is the less saturated color, while #00dabf is the most saturated one.\n\n• #657573\n``#657573` `rgb(101,117,115)``\n• #5c7e7a\n``#5c7e7a` `rgb(92,126,122)``\n• #548680\n``#548680` `rgb(84,134,128)``\n• #4b8f86\n``#4b8f86` `rgb(75,143,134)``\n• #43978d\n``#43978d` `rgb(67,151,141)``\n• #3b9f93\n``#3b9f93` `rgb(59,159,147)``\n• #32a899\n``#32a899` `rgb(50,168,153)``\n• #2ab09f\n``#2ab09f` `rgb(42,176,159)``\n• #22b8a6\n``#22b8a6` `rgb(34,184,166)``\n• #19c1ac\n``#19c1ac` `rgb(25,193,172)``\n• #11c9b2\n``#11c9b2` `rgb(17,201,178)``\n• #08d2b9\n``#08d2b9` `rgb(8,210,185)``\n• #00dabf\n``#00dabf` `rgb(0,218,191)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #00dabf is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "Recalling the perceptron\n\nIn the previous post we have seen how an artificial neuron works to classify simple functions and we have also seen its limitations when the space function to be classified is not linear, e.g. it fails to compute correctly the XOR function.\n\nWe have also seen that the way to overcome this limitation is to combine more neurons together in a network but one thing we did not yet seen was how to compute the “weights” of the network (they were hard-coded in the previous example).\n\nAs a reminder, here is the perceptron able to compute an OR function:", null, "The input values are multiplied with specific weights and combined together into an activation function which outputs the OR function.\nThe “artificial neural network” (ANN) is really just this matrix of weights. Here the matrix has only one column (the OR function is very simple) but we may have “layers” and layers of neurons and their weights.\nThe values calculated are just transient values based on the input dataset. We don’t save them, they are forgotten.\nAll of the learning is stored in the weight matrix.\n\nLike similar supervised models, the ANNs can be trained to learn the weights using some labeled training data (the “supervised” data) and once the weights matrix is learnt, we can use it to compute the function that we trained for.\n\nSigmoid neurons\n\nBefore seeing how the weights can be learned, we will slightly tune the activation function (the red circle in the figure above)  in order to make the learning phase easier.\nIn the perceptron example, this activation function was a simple threshold one, outputting 0 if the input was below a threshold and 1 otherwise. Instead, we will use a sigmoid function which has the advantage that small changes in the weights and bias cause only a small change in their output.  The sigmoid function – often called the logistic function – is defined by:", null, "$\\frac{1}{(1+e^{-z})}$\n\nThe sigmoid activation function is always positive,  bounded between 0 and 1 and strictly increasing.\n\nAnd this is its Python implementation:\n\nimport numpy as np\n\ndef sigmoid(z):\n\"\"\"\nCompute the sigmoid function\n\nArgument:\nz: float number\n\nReturns:\nfloat\n\"\"\"\n\nreturn 1 / (1 + np.exp(-z))\n\nThe sigmoid function can be thought as a “squashing function”, because it takes the input and squashes it to be between zero and one: very negative values are squashed towards zero and positive values get squashed towards one.\nFor example we have:\n\nsigmoid(-5) = 0.007;  sigmoid(5) = 0.993\n\nThe output of a sigmoid neuron\n\nThe output of a sigmoid neuron is obviously not 0 or 1 but is a real number between 0 and 1. This can be useful, for example, if we want to use the output value to represent the intensity of the pixels in an image input to a neural network. But sometimes it is not, in fact to model an OR gate we need to have as output 0 or 1, as it was in the perceptron.\nIn practice we can use a convention to deal with this, for example, by deciding to interpret any output of at least 0.5 as indicating a “1”, and any output less than 0.5 as indicating a “0”.\n\nWe can quickly verify that the original perceptron simulating the OR function is still working with this new activation function:\n\ndef sigmoidActivation(z):\n\"\"\"\nCompute the activation function for boolean operators using sigmoid\nSince a sigmoid function is always between 0 and 1,\nwe just cut off at 0.5\n\nArguments:\nz: input data points including bias, array\n\nReturns:\ninteger (0 or 1)\n\"\"\"\n\nif sigmoid(z) > 0.5:\nreturn 1\nelse:\nreturn 0\n\ndef predictOneLayer(X, weights):\n\"\"\"\nCompute the binary output of an Artificial Neural Network with single layer,\ngiven its neuron weights and an input dataset\n\nArguments:\nX: input data points without bias, list\nweights: the single-layer weights, array\n\nReturns:\ninteger (0 or 1)\n\"\"\"\nX = + X # add bias to the input list\n\n# forward propagation\ny = np.dot(X, weights).sum()\n\nreturn sigmoidActivation(y)\n\ndef forwardOR(operands):\n\"\"\"\nSimulate an OR function using a simple forward neural network\n(single layer)\n\nArguments:\noperands: input data points without bias, list\n\nReturns:\ninteger (0 or 1)\n\"\"\"\n\nm = len(operands) # number of training examples\nORweights = [-0.5] + *m # hard-coded weights\nreturn predictOneLayer(operands, ORweights)\n\nprint(\"*** Simulation of OR using one forward neuron ***\")\nprint(\"X1 | X2 | Y = X1 OR X2\")\nprint(\"-----------------------\")\nfor x1 in (0,1):\nfor x2 in (0,1):\nprint(x1,\" | \",x2, \" | \", forwardOR([x1,x2]))\n\nAnd here is its output:\n\n*** Simulation of OR using one forward neuron ***\nX1 | X2 | Y = X1 OR X2\n-----------------------\n0 | 0 | 0\n0 | 1 | 1\n1 | 0 | 1\n1 | 1 | 1\n\nThe OR function still works correctly using sigmoid neurons.\n\nAs you notice, the weights in this OR example above are hard-coded because we can get them manually very quick since the example is pretty straightforward. But this is obviously not scalable as the complexity of a function is growing.\nTherefore we will see now how we can find the appropriate weights to model a simple function like the OR.\n\nTraining the network\n\nWhat we need is an algorithm that will find the appropriate weights (remember: they ARE the artificial neural network) programmatically for us so that the output from the network approximates the expected y(x) for all training inputs x.\n\nCost function\n\nTo quantify how well we’re achieving this goal we will define a cost function, sometimes referred to as a loss or objective function.\nWe have already seen examples of cost functions in the regression models and it can be a complex function, depending on the domain – a common choice is the categorical cross-entropy loss  – but for this simple example we just stick to the difference between the network output and the training output:\n\nerror = actual output(label)  – predicted output\n\nLearning the weights\n\nThe output of the ANN is:\n\noutput = h(W * X + w0)\n\nwhere X is the input, w0 is the bias, W are the weights and h is the non-linear activation function.", null, "Learning the parameters for our network means finding weights that minimise the error on our training data.  By finding parameters that minimise the error we maximise the likelihood of our training data.\nDesigning and training a neural network is not much different from training any other machine learning model with gradient descent.\n\nSigmoid derivative\n\nThe gradient descent works by using the derivative function. The derivatives of the sigmoid with respect to the inputs and the weights are very simple:", null, "$\\sigma = \\frac{1}{(1+e^{-z})} \\rightarrow \\frac{\\partial \\sigma(z)}{\\partial z} = \\sigma(z) * (1-\\sigma(z))$\n\ntherefore its Python implementation is clear-cut:\n\ndef sigmoidDeriv(s):\n\"\"\"\nCompute the derivative of a sigmoid function\n\nArgument:\ns: float number\n\nReturns:\nfloat\n\"\"\"\n\nreturn s * (1-s)\n\nNow we have everything to explore the learning algorithm for artificial neural network.\n\nThe learning algorithm\n\nThe general idea is to forward-feed the input dataset to an initial set of weights to compute the output and the error, then back-propagate the error using the gradient (this is the “core” part!) and update the weights until the error is acceptable:\n\n0. start with X and y (input and output training data)\n1. initialise weights Wij randomly\n2. loop until error < tolerance or max iterations:\n3. for each training example Xi:\n4. forward: out = theta(Wij * Xi) # theta is the activation function\n5. compute the error\n6. backward: compute gradient\n7. update all weights Wij using the gradient\n8. return the final weights\n\nWe will see it in details line by line line but here is the entire working Python implementation in the case of one single layer of neurons:\n\ndef learnWeightsOneLayer(X, y, tolerance = 0.1, max_iterations = 1000):\n\"\"\"\nLearn the weights for an artificial neural network with single layer,\ngiven a training set of input and output values.\nUses a backpropagation algorithm.\n\nArguments:\nX: input data points without bias, matrix\ny: output data points, array\ntolerance: minimum error to stop the training, float. Optional (default=0.1)\nmax_iterations: maximum number of iterations performed by the algorithm.\nOptional. Default=1000\n\nReturns:\nweights, array\n\"\"\"\n\n# input and output training sizes\nn_inputs = X.shape\nn_outputs = y.shape\nn_examples = X.shape\n\nbias = np.ones(n_examples)\nX = np.c_[bias, X] # add bias column to X\n\n# initialization\nconverged = False\ni = 0 # iteration counter\n\n# initialize weights randomly with mean 0\nw = 2 * np.random.random((n_inputs+1, n_outputs)) - 1\n\n# main loop until converged or max iterations reached\nwhile not converged and i <= max_iterations:\n\n# note: all training examples are used together\n\n# forward propagation\npred_y = sigmoid(np.dot(X, w)) # predicted output\n\n# how much did we miss? Compute the error\nerror = y - pred_y\n\n# backpropagation: multiply how much we missed by the\n# slope of the sigmoid at the values in pred_y\ndelta = error * sigmoidDeriv(pred_y)\n\n# update weights\nw += np.dot(X.T, delta)\n\n# did we converge?\nmaxError = max(error) # biggest error among training examples\nif abs(maxError) < tolerance:\nconverged = True # yes !\nprint (\"** converged after iterations: \",i)\n\ni += 1 # no convergence yet: next iteration\n\n# end of loop\nif not converged:\nprint(\"** Not converged!\")\n\nreturn w\n\nAnd this is how it works with a simple two-values input and an OR output:\n\n# seed random numbers to make calculation\n# deterministic (just a good practice when testing / debugging)\nnp.random.seed(1)\n\nprint(\"*** Simulation of OR gate using neural network ***\")\n# input dataset: 4 training examples\nX_train = np.array([ [0,0], [0,1], [1,0], [1,1] ])\n# output dataset: the expected results for each training input\ny_train = np.array([[0,1,1,1]]).T\n\n# learn the weights through training!\nORweights = learnWeightsOneLayer(X_train, y_train)\n\n# print the truth table\nprint(\"X1 | X2 | Y = X1 OR X2\")\nprint(\"-----------------------\")\nfor x1 in (0,1):\nfor x2 in (0,1):\nprint(x1,\" | \",x2, \" | \", predictOneLayer([x1,x2], ORweights))\n\nAnd this is the output:\n\n*** Simulation of OR gate using neural network ***\n** converged after iterations: 149\nX1 | X2 | Y = X1 OR X2\n-----------------------\n0 | 0 | 0\n0 | 1 | 1\n1 | 0 | 1\n1 | 1 | 1\n\nWell, it works!\n\nLet’s explore the code piece by piece.\n\nTraining data and bias\n\n# input and output training sizes\nn_inputs = X.shape\nn_examples = X.shape\n\nThe input dataset is a numpy matrix. Each row is a single “training example”. Each column corresponds to one of our input nodes.\nIn our OR example above, we have 2 input nodes (X1 and X2) and 4 training examples.\nSimilarly the number of nodes in the output layer (the “labels” of our dataset) is determined by the number of classes, in this case we have only one output node for two classes: 0 and 1. That is the number of column for the y matrix while the rows are determined again by the number of training examples.\n\nbias = np.ones(n_examples)\nX = np.c_[bias, X] # add bias column to X\n\nThis will add a “bias” neuron, i.e. an additional column of weights. It’s always a good idea to add a bias to a neural network.\nThe bias node in a neural network is a node that is always ‘on’. That is, its value is set to 1. It is analogous to the intercept in a regression model, and serves the same function: allows you to shift the activation function to the left or right, which may be critical for successful learning.\n\nFor example, assume that you want a neuron to fire y≈1 when all the inputs are x≈0. If there is no bias no matter what weights W you have, given the equation y = σ(WX) the neuron will always fire y≈0.5.\n\nInitial weights\n\n# initialise weights randomly with mean 0\nw = 2 * np.random.random((n_inputs+1, n_outputs)) - 1\n\nThis creates our weight matrix for the neural network.\nAnd it initialises the weights to a random number, using the numpy function random() which returns a random float number or array.\nThe size of the output is controlled by the parameter. In our example it’s a tuple (n_inputs+1, n_outputs) which means it will return a matrix.\n\nWe can choose the dimensionality of the hidden layer.\nThe more nodes we put into the hidden layer the more complex functions the neural network will be able to model but higher dimensionality means more computation is required to learn the network weights.\n\nSince we only have one layer (the input values) we only need a matrix of weights with one column, to connect them. We keep its dimension here minimal: (3,1) because we have two inputs plus the bias and one output.\n\nResults from the random() function are from the “continuous uniform” distribution over the half-open interval [0.0, 1.0). It’s a best practice to sample the weights from a uniform distribution with mean zero: to sample Unif[-1,1) we just have to multiply the output of random() by 1-(-1)=2 and add -1.\n\nIterate\n\n# initialisation\nconverged = False\ni = 0 # iteration counter\n\n# main loop until converged or max iterations reached\nwhile not converged and i <= max_iterations:\n\nThis is the loop to train the network weights.\nIt  iterates over the training code until we have reached the wished error (converged) or until we have reached the maximum number of iterations.\nBoth the max_iterations and the error tolerance are optional inputs of this function.\n\nFeed-forward\n\n# forward propagation\npred_y = sigmoid(np.dot(X, w)) # predicted output\n\nThis is the forward propagation prediction step: the network predicts the output given the input and the current weights by matrix multiplication and applying the activation function.\n\ny = σ(Xw)\n\nNote that X contains all training examples (input) and we process all of them at the same time in this implementation. This is known as “full batch” training.\n\nThe X matrix is multiplied with the current weights matrix w, using the numpy function dot() and its result is then going through the sigmoid squashing function, our activation function.\nThe result will be also a matrix, in this case of 4 rows (the four training examples) and one column (the predicted output):\n\n(4 x 3) dot (3 x 1) = (4 x 1)  # 4 examples, 3 input and 1 layer of weights\n\nEach output corresponds with the network’s prediction for a given training input. This would work independently on the number of training examples, i.e. if I train the network with more examples, the matrix will be bigger but no changes in code are required.\n\nLoss function (the error)\n\n# Compute the error\nerror = y - pred_y\n\nWe can now compare how well it did by subtracting the true answer (y from the training dataset) from the prediction (pred_y).\nThe error is an array: the difference between actual and prediction for each training example.\nAgain: this is a super-simple loss function just for illustration. For more complex examples you will use e.g. a quadratic function like we have seen in regression algorithms.\n\nBack-propagation\n\nOne popular approach is an algorithm called back-propagation that has similarities to the gradient descent algorithm we looked at earlier. The back-propagation algorithm (Rumelhart et al., 1986), often simply called backprop, allows the error information to flow backwards through the network, in order to compute the gradient. Once we have the gradient of this error as a function of the neuron’s weights, we can propagate these output errors backward to infer errors for the other layers and adjust its weights in the direction that most decreases the error.", null, "The term back-propagation is often misunderstood as meaning the whole learning algorithm for neural networks but actually, it refers only to the method for computing the gradient, while another algorithm, such as stochastic or batch gradient descent, is used to perform learning using this gradient.\n\nWe use the gradient to find the minimum; this means we first need to compute the gradient. The gradient is just made up of the derivatives of all the inputs concatenated in a vector (remember that a derivative of a function measures the sensitivity to change):", null, "$\\nabla e(w) = \\frac{\\partial e(w)}{\\partial w^{(l)}_{ij}}$ for all i,j,l\n\nwhere w are the weights and e(w) is the error on the example X,y.\n\nBack-propagation is the key algorithm that makes training deep models computationally tractable, basically is just a clever “trick” to efficiently calculate the gradients starting from the output. I am not going into details here but it’s an old trick and you can find the generic implementation under the name “reverse-mode differentiation”:\n\nClassical methods are slow at computing the partial derivatives of a function with respect to many inputs, as is needed for gradient-based optimization algorithms. Automatic differentiation solves all of these problems […]\nIn reverse accumulation, one first fixes the dependent variable to be differentiated and computes the derivative with respect to each sub-expression recursively. In a pen-and-paper calculation, one can perform the equivalent by repeatedly substituting the derivative of the outer functions in the chain rule:", null, "$\\frac{\\partial y}{\\partial x} = \\frac{\\partial y}{\\partial w_{1}} \\cdot \\frac{\\partial w_{1}}{\\partial x}$\n\nThe chain rule of calculus is used to compute the derivatives of functions formed by composing other functions whose derivatives are known. The chain rule very simply states that the right thing to do is to simply multiply the gradients together to chain them.\nReverse-mode differentiation is an algorithm that computes the chain rule, with a specific order of operations that is highly efficient.\n\nI won’t go into more details here but there are many excellent explanations in the web, such as the Andrew Ng or Yaser Abu-Mostafa courses.\n\nBottom line: back-propagation tracks how every neuron affects one output.\n\n# backpropagation: multiply how much we missed by the\n# slope of the sigmoid at the values in pred_y\ndelta = error * sigmoidDeriv(pred_y)\n\nThe first thing we need is a function that computes the gradient of the sigmoid function we created earlier. This is sigmoidDeriv(). The derivative will give back the slope of the predictions.\n\nThe variable error is the total error across the whole data, calculated by running the data plus current parameters through the “network” (the forward-propagate function) and comparing the output to the true labels. This is the part we discussed earlier as the cost function.\n\nWhen we multiply the “slopes” by the error, we are reducing the error of high confidence predictions.\nThe multiplication is essentially answering the question “how can I adjust my parameters to reduce the error the next time I run through the network”?\nIt does this by computing the contributions at each layer to the total error and adjusting appropriately by coming up with a gradient matrix (or, how much to change each parameter and in what direction).\n\nUpdate the weights\n\n# update weights\nw += np.dot(X.T, delta)\n\nWe are now ready to update our network! This line computes the weight updates for each weight for each training example, sums them, and updates the weights, all in a simple line. A small error and a small slope means a VERY small update. However, because we’re using a “full batch” configuration, we’re doing the above step on all four training examples.\n\nCheck if converged\n\n# did we converge?\nmaxError = max(error) # biggest error among training examples\nif abs(maxError) < tolerance:\nconverged = True # yes !\nprint (\"** converged after iterations: \",i)\n\ni += 1 # next iteration (to check if max is reached)\n\nThe last step is to check if we have reached the end of our optimisation, either because the error is smaller than our tolerance or because we reached the maximum number of iterations.\n\nIn our example above, the convergence (i.e. the error was less than 0.1) was reached after 149 iterations.\n\nWe can also print out the weights and we can see that they are different than the hard-coded ones (but follow the same pattern: the bias is negative, the inputs are similar and positive).\n\nWeights for ANN modelling OR function:\n[[-1.63295267]\n[ 3.84814604]\n[ 3.83626687]]\n\nThere are several combinations of weights that can bring to the same result and there are other equivalent solutions that gradient descent could also find.\nThe convergence point of gradient descent depends on the initial values of the parameters: the initial set of weights (that is random chosen), the gradient descent meta-parameters (not used in this simple example) and in minor part also on the error tolerance defined.\n\nIt is also possible to demonstrate that multiplying the resulting weights for a constant C will keep the ANN to output the same result but this is out of the scope for this post. You can try it yourself if you are curious.\n\nAn OR gate with 3 inputs\n\nLet´s see if the ANN modelling an OR gate can work with 3 inputs:\n\nprint(\"*** Simulation of OR gate with 3 inputs using ANN ***\")\n\n# input dataset\nX_train = np.array([ [0,0,0], [0,0,1], [0,1,0], [1,0,0] ])\n# output dataset\ny_train = np.array([[0,1,1,1]]).T\n\nORweights = learnWeightsOneLayer(X_train, y_train)\n\nprint(\"X1 | X2 | X3 | Y = X1 OR X2 OR X3\")\nprint(\"----------------------------------\")\ninputs = ((x1,x2,x3) for x1 in (0,1) for x2 in (0,1) for x3 in (0,1))\nfor x1,x2,x3 in inputs:\nprint (x1,\" | \",x2, \" | \", x3, \" | \",\npredictOneLayer([x1,x2,x3], ORweights))\n\nNote that the training set X_train is now a matrix with 3 columns for the 3 inputs and that you do not need to list ALL combinations as training examples.\nFor an OR function with 3 operators there would be a total of 2^3 = 8 possible combinations but here we train the network with only 4 examples.\nChoosing the right training dataset (large and representative enough) is a science for itself but I will not dig more here. You can refer to the previous posts about overfitting and cross-validations.\n\nThis is the output:\n\n*** Simulation of OR gate with 3 inputs using ANN ***\n** converged after iterations: 169\nX1 | X2 | X3 | Y = X1 OR X2 OR X3\n----------------------------------\n0 | 0 | 0 | 0\n0 | 0 | 1 | 1\n0 | 1 | 0 | 1\n0 | 1 | 1 | 1\n1 | 0 | 0 | 1\n1 | 0 | 1 | 1\n1 | 1 | 0 | 1\n1 | 1 | 1 | 1\n\nIt works perfectly for all 8 combinations and convergence has been reached after 169 iterations, pretty fast.\n\nNon-linear example: XOR\n\nThe real proof of truth is to model a non-linear function, such as the XOR function, a sort of Hello World for Neural Networks.\nWe have seen in the previous post that a simple perceptron (one neuron with only one single layer of input) cannot model it. But it’s possible by combining together perceptrons that model each one different functions, such as OR, AND and NOT:\n\nx1 XOR x2 = (x1 AND NOT x2) OR (NOT x1 AND x2)\n\nThe resulting multi-layer network of perceptrons is basically an artificial neural network with one additional layer – often called the hidden layer:", null, "To model such a network we need to add one layer of weights, i.e. one column to the weights matrix and to compute the hidden layer weights in the training algorithm.\n\nXOR with two inputs\n\nNote: To keep the example more clear I will keep the weights of the two layers separately.\n\ndef learnWeightsTwoLayers(X, y, tolerance = 0.1, max_iterations = 1000):\n\"\"\"\nLearn the weights for an artificial neural network with two layers,\ngiven a training set of input and output values.\nUses a backpropagation algorithm.\n\nArguments:\nX: input data points without bias, matrix\ny: output data points, array\ntolerance: minimum error to stop the training, float. Optional (default=0.1)\nmax_iterations: maximum number of iterations performed by the algorithm.\nOptional. Default=1000\n\nReturns:\nweights for input layer, array\nweights for hidden layer, array\n\"\"\"\n\n# input and output training sizes\nn_inputs = X.shape\nn_outputs = y.shape\nn_examples = X.shape\n\nbias = np.ones(n_examples)\nX = np.c_[bias, X] # add bias column to X\n\n# initialize weights randomly with mean 0\nw1 = 2 * np.random.random((n_inputs+1, n_inputs)) - 1 # input layer\nw2 = 2 * np.random.random((n_inputs+1, n_outputs)) - 1 # hidden layer\n\n# initialization\nconverged = False\ni = 0 # iteration counter\n\n# main loop until converged or max iterations reached\nwhile not converged and i <= max_iterations:\n\n# note: all training examples together (batch)\n\n# forward propagation\nhiddenLayer = sigmoid(np.dot(X, w1))\nhiddenLayer = np.c_[bias, hiddenLayer] # add bias\npred_y = sigmoid(np.dot(hiddenLayer, w2))  # predicted output\n\n# Compute the error\nerror_y = y - pred_y\n\n# backpropagation: multiply how much we missed by the\n# slope of the sigmoid at the values in y_pred\ndelta_y = error_y * sigmoidDeriv(pred_y)\n\n# repeat for hidden layer: error and delta\n# how much did each value of hidden layer contribute\n# to the final error (according to the weights)?\nerror_l1 = delta_y.dot(w2.T)\ndelta_l1 = error_l1 * sigmoidDeriv(hiddenLayer)\n# the first column refers to the bias: remove it\ndelta_l1 = np.delete(delta_l1, 0, axis=1)\n\n# update weights\nw2 += np.dot(hiddenLayer.T, delta_y)\nw1 += np.dot(X.T, delta_l1)\n\n# did we converge?\nmaxError = max(abs(error_y))\nif abs(maxError) < tolerance:\nconverged = True # yes !\nprint (\"** converged after iterations: \",i)\n\ni += 1 # next iteration (to check if max is reached)\n\nif not converged:\nprint(\"** Not converged!\")\n\nreturn w1, w2\n\nThe mainly difference is that we have now two arrays of weights w1 and w2, one storing the weights for the input layer and one for the hidden layer.\nThen accordingly the code is changed to take in considerations both sets of weights.\n\nLet´s see it in action:\n\ndef predictTwoLayers(X, w1, w2):\n\"\"\"\nCompute the binary output of an Artificial Neural Network with two layers,\ngiven its neuron weights and an input dataset\n\nArguments:\nX: input data points without bias, list\nw1: the input-layer weights, array\nw2: the hidden-layer weights, array\n\nReturns:\ninteger (0 or 1)\n\"\"\"\nbias = np.ones(1)\nX = np.append(bias, X, 0) # add bias column to X\n\n# forward propagation\nhiddenLayer = sigmoid(np.dot(X, w1))\nhiddenLayer = np.append(bias, hiddenLayer, 0) # add bias\ny = np.dot(hiddenLayer, w2)\n\nreturn sigmoidActivation(y)\n\nprint(\"*** Simulation of XOR gate using ANN ***\")\nprint(\"***\")\n\n# input dataset\nX_train = np.array([ [0,0], [0,1], [1,0], [1,1] ])\n# output dataset\ny_train = np.array([[0,1,1,0]]).T\n# note that we use tolerance = 0.4\nXORweights1, XORweights2 = learnWeightsTwoLayers(X_train, y_train, 0.4)\n\nprint(\"X1 | X2 | Y = X1 XOR X2\")\nprint(\"------------------------\")\nfor x1 in (0,1):\nfor x2 in (0,1):\nprint(x1,\" | \",x2, \" | \", predictTwoLayers(np.array([x1,x2]),\nXORweights1, XORweights2))\n\nI used a higher tolerance, since the output of a XOR gate is binary so we don’t need that much approximation. This will make the training algorithm faster.\n\nAnd here is the output:\n\n*** Simulation of XOR gate using ANN ***\n***\n** converged after iterations: 967\nX1 | X2 | Y = X1 XOR X2\n------------------------\n0 | 0 | 0\n0 | 1 | 1\n1 | 0 | 1\n1 | 1 | 0\n\nIt works but note that it takes now 967 iterations to converge.\nAdding layers to a network will normally make it slower and harder to train. Normally deeper neural network will be more optimised, e.g. will use better gradient descent algorithms, regularisations and so on.\n\nXOR with 3 inputs\n\nAnd it works with 3 inputs too:\n\nprint(\"*** Simulation of XOR gate using ANN, now with 3 inputs ***\")\nprint(\"***\")\n\n# input dataset\nX_train = np.array([ [0,0,0], [0,1,0], [1,0,1], [0,1,1], [1,0,0], [1,1,1]])\n# output dataset\ny_train = np.array([[0,1,1,1,1,0]]).T\n\nXORweights1, XORweights2 = learnWeightsTwoLayers(X_train, y_train, 0.4)\n\nprint(\"X1 | X2 | X3 | Y = X1 XOR X2 XOR X3\")\nprint(\"---------------------------------------\")\ninputs = ((x1,x2,x3) for x1 in (0,1) for x2 in (0,1) for x3 in (0,1))\nfor x1,x2,x3 in inputs:\nprint (x1,\" | \",x2, \" | \", x3, \" | \", predictTwoLayers([x1,x2,x3],\nXORweights1, XORweights2))\n\nand the output is:\n\n*** Simulation of XOR gate using ANN, now with 3 inputs ***\n***\n** converged after iterations: 360\nX1 | X2 | X3 | Y = X1 XOR X2 XOR X3\n---------------------------------------\n0 | 0 | 0 | 0\n0 | 0 | 1 | 1\n0 | 1 | 0 | 1\n0 | 1 | 1 | 1\n1 | 0 | 0 | 1\n1 | 0 | 1 | 1\n1 | 1 | 0 | 1\n1 | 1 | 1 | 0\n\nConclusion\n\nThe perceptron can model linear functions but cannot model non-linear functions, unless you combine perceptrons together. Nevertheless the perceptron convergence procedure works by ensuring that every time the weights change, they get closer to every “generously feasible” set of weights. This type of guarantee cannot be extended to more complex networks in which the average of two good solutions may be a bad solution (non-convex problems). Therefore the perceptron learning procedure cannot be generalised to hidden layers.\n\nInstead of showing the weights get closer to a good set of weights, the Artificial Neural Networks learn the weights so that the actual output values get closer the target values. The aim of learning is to minimise the error summed over all training cases.\nSigmoid neurons give a real-valued output that is a smooth and bounded function of their total input; they have nice derivatives which make learning easy.\n\nSuch an Artificial Neural Network can represent non-linear functions.\nThe Universal Approximation theorem states that a simple feed-forward neural network with a single hidden layer is enough to represent a wide variety of continuous functions.\n\nThis is a bare example. Deep neural network have dozens of layers and use many additional optimisations to improve the learning:  different activation functions, regularisations to prevent overfitting ((L1 and L2 regularization, dropout), better choice of cost function (squared difference, cross-entropy), tuned weight initialisation and choice of hyper-parameters (learning rate, number of layers and neurons, …), sophisticated gradient descent algorithms. But the principle is the same.\n\nANNs can represent efficiently almost any kind of function and studies are constantly progressing." ]

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[ "", null, "What's the battle there? How do physicists define impedance?", null, "Ohm's law is an empirical law that only holds in certain circumstances. A classic exercise is measuring the current and voltage across a lightbulb, plotting it, and measuring the slope of the line. The slope is the impedance. Then you turn up the voltage and watch the line turn into a curve, which is where the law breaks down and doesn't apply anymore. The engineers treat it like a definition and assume linearity over all voltage.", null, "Applications are open for YC Winter 2020\n\nSearch:" ]

[ null, "https://news.ycombinator.com/y18.gif", null, "https://news.ycombinator.com/s.gif", null, "https://news.ycombinator.com/s.gif", null ]

[ "# Riesz representation theorem\n\nThere are several well-known theorems in functional analysis known as the Riesz representation theorem. They are named in honor of Frigyes Riesz.\n\nThis article will describe his theorem concerning the dual of a Hilbert space, which is sometimes called the Fréchet–Riesz theorem. For the theorems relating linear functionals to measures, see Riesz–Markov–Kakutani representation theorem.\n\n## The Hilbert space representation theorem\n\nThis theorem establishes an important connection between a Hilbert space and its continuous dual space. If the underlying field is the real numbers, the two are isometrically isomorphic; if the underlying field is the complex numbers, the two are isometrically anti-isomorphic. The (anti-) isomorphism is a particular natural one as will be described next; a natural isomorphism.\n\nLet H be a Hilbert space, and let H* denote its dual space, consisting of all continuous linear functionals from H into the field $\\mathbb {R}$", null, "or $\\mathbb {C}$", null, ". If $x$", null, "is an element of H, then the function $\\varphi _{x},$", null, "for all $y$", null, "in H defined by:\n\n$\\varphi _{x}(y)=\\left\\langle y,x\\right\\rangle ,$", null, "where $\\langle \\cdot ,\\cdot \\rangle$", null, "denotes the inner product of the Hilbert space, is an element of H*. The Riesz representation theorem states that every element of H* can be written uniquely in this form.\n\nRiesz–Fréchet representation theorem. Let $H$", null, "be a Hilbert space and $\\varphi \\in H^{*}$", null, ". Then there exists $f\\in H$", null, "such that for any $x\\in H$", null, ", $\\varphi (x)=\\langle f,x\\rangle$", null, ". Moreover $\\|f\\|_{H}=\\|\\varphi \\|_{H*}$", null, "Proof. Let $M=\\{u\\in H\\ |\\ \\varphi (u)=0\\}$", null, ". Clearly $M$", null, "is closed subspace of $H$", null, ". If $M=H$", null, ", then we can trivially choose $f=0$", null, ". Now assume $M\\neq H$", null, ". Then $M^{\\perp }$", null, "is one-dimensional. Indeed, let $v_{1},v_{2}$", null, "be nonzero vectors in $M^{\\perp }$", null, ". Then there is nonzero real number $\\lambda$", null, ", such that $\\lambda \\varphi (v_{1})=\\varphi (v_{2})$", null, ". Observe that $\\lambda v_{1}-v_{2}\\in M^{\\perp }$", null, "and $\\varphi (\\lambda v_{1}-v_{2})=0$", null, ", so $\\lambda v_{1}-v_{2}\\in M$", null, ". This means that $\\lambda v_{1}-v_{2}=0$", null, ". Now let $g$", null, "be unit vector in $M^{\\perp }$", null, ". For arbitrary $x\\in H$", null, ", let $v$", null, "be the orthogonal projection of $x$", null, "onto $M^{\\perp }$", null, ". Then $v=\\langle g,x\\rangle g$", null, "and $\\langle g,x-v\\rangle =0$", null, "(from the properties of orthogonal projections), so that $x-v\\in M$", null, "and $\\langle g,x\\rangle =\\langle g,v\\rangle$", null, ". Thus $\\varphi (x)=\\varphi (v+x-v)=\\varphi (\\langle g,x\\rangle g)+\\varphi (x-v)=\\langle g,x\\rangle \\varphi (g)+0=\\langle g,x\\rangle \\varphi (g)$", null, ". Hence $f=\\varphi (g)g$", null, ". We also see $\\|f\\|_{H}=\\varphi (g)$", null, ". From the Cauchy-Bunyakovsky-Schwartz inequality $\\varphi (x)\\leq \\|g\\|\\|x\\|\\varphi (g)$", null, ", thus for $x$", null, "with unit norm $\\varphi (x)\\leq \\varphi (g)$", null, ". This implies that $\\|\\varphi \\|_{H*}=\\varphi (g)$", null, ".\n\nGiven any continuous linear functional g in H*, the corresponding element $x_{g}\\in H$", null, "can be constructed uniquely by $x_{g}=g(e_{1})e_{1}+g(e_{2})e_{2}+...$", null, ", where $\\{e_{i}\\}$", null, "is an orthonormal basis of H, and the value of $x_{g}$", null, "does not vary by choice of basis. Thus, if $y\\in H,y=a_{1}e_{1}+a_{2}e_{2}+...$", null, ", then $g(y)=a_{1}g(e_{1})+a_{2}g(e_{2})+...=\\langle x_{g},y\\rangle .$", null, "Theorem. The mapping $\\Phi$", null, ": HH* defined by $\\Phi (x)$", null, "= $\\varphi _{x}$", null, "is an isometric (anti-) isomorphism, meaning that:\n\n• $\\Phi$", null, "is bijective.\n• The norms of $x$", null, "and $\\varphi _{x}$", null, "agree: $\\Vert x\\Vert =\\Vert \\Phi (x)\\Vert$", null, ".\n• $\\Phi$", null, "is additive: $\\Phi (x_{1}+x_{2})=\\Phi (x_{1})+\\Phi (x_{2})$", null, ".\n• If the base field is $\\mathbb {R}$", null, ", then $\\Phi (\\lambda x)=\\lambda \\Phi (x)$", null, "for all real numbers λ.\n• If the base field is $\\mathbb {C}$", null, ", then $\\Phi (\\lambda x)={\\bar {\\lambda }}\\Phi (x)$", null, "for all complex numbers λ, where ${\\bar {\\lambda }}$", null, "denotes the complex conjugation of $\\lambda$", null, ".\n\nThe inverse map of $\\Phi$", null, "can be described as follows. Given a non-zero element $\\varphi$", null, "of H*, the orthogonal complement of the kernel of $\\varphi$", null, "is a one-dimensional subspace of H. Take a non-zero element z in that subspace, and set $x={\\overline {\\varphi (z)}}\\cdot z/{\\left\\Vert z\\right\\Vert }^{2}$", null, ". Then $\\Phi (x)$", null, "= $\\varphi$", null, ".\n\nHistorically, the theorem is often attributed simultaneously to Riesz and Fréchet in 1907 (see references).\n\nIn the mathematical treatment of quantum mechanics, the theorem can be seen as a justification for the popular bra–ket notation. The theorem says that, every bra $\\langle \\psi |$", null, "has a corresponding ket $|\\psi \\rangle$", null, ", and the latter is unique." ]

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[ "Home | Physics | What is Orbital Velocity?-Definition, And Formula\n\n# What is Orbital Velocity?-Definition, And Formula\n\nAugust 22, 2022", null, "written by Adeel Abbas\n\nOrbital Velocity is the velocity at which an object travels around a celestial body. In space, objects travel at different speeds depending on their distance from the center of mass of the planet. If we were to draw a line between two points on Earth’s surface, then the point closer to the center would have a higher orbital velocity than the point further away.\n\n## What is orbital velocity?\n\nOrbital Velocity is the speed of an object around the center of a planet. The object moves at a constant speed and it has to pass through the same path or path length.\n\nThe speed at which an astronomical object like the moon or artificial satellite moves around the barycenter is the speed at which it travels around the planet or moon.\n\nIf the muzzle velocity of the cannon is increased, it will throw a projectile further up the mountain. The surface of the Earth may be thought of as curving away from the satellite as it falls towards it.\n\nThe mass of the body is directly proportional to the orbital velocity, and the radius of the body is also directly proportional to it. If the air friction is not taken into account, the Earth’s orbital velocity is around eight kilometers per second. The closer a satellite is to the center of attraction, the weaker the force of gravity and the less speed it needs to remain in space.\n\n## Orbital Velocity Formula\n\nThe Earth and other planets revolve around the sun. The artificial satellites around the Earth are nearly in circular paths. This type of motion is called orbital motion: Consider a satellite of mass m, revolving around the Earth of mass ‘M’ in an orbit of radius ‘r’ with orbital speed v. The centripetal force required to hold the satellite in orbit is given by\n\nFc=msv2/r\n\nThe gravitational force of attraction between the Earth and satellite provides this centripetal force.\n\nFg=GMm/r2\n\nBut Fc=Fg, so\n\nBy comparing\n\nVorb=√GM/r\n\nThe orbital speed of a satellite is independent of the mass of the satellite. It only depends upon the distance r of the satellite from the center of the earth.\n\nFile Under:" ]

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[ "[Solutions] Junior Balkan Mathematical Olympiad 2018\n\n1. Find all integers $m$ and $n$ such that the fifth power of $m$ minus the fifth power of $n$ is equal to $16mn$.\n2. Find max number $n$ of numbers of three digits such that\n• Each has digit sum $9$,\n• No one contains digit $0$,\n• Each $2$ have different unit digits,\n• Each $2$ have different decimal digits,\n• Each $2$ have different hundreds digits.\n3. Let $k>1$ be a positive integer and $n>2018$ an odd positive integer. The non-zero rational numbers $x_1,x_2,\\ldots,x_n$ are not all equal and $$x_1+\\frac{k}{x_2}=x_2+\\frac{k}{x_3}=x_3+\\frac{k}{x_4}=\\ldots=x_{n-1}+\\frac{k}{x_n}=x_n+\\frac{k}{x_1}$$Find the minimum value of $k$, such that the above relations hold.\n4. Let $\\triangle ABC$ and $A'$, $B'$, $C'$ the symmetrics of vertex over opposite sides. The intersection of the circumcircles of $\\triangle ABB'$ and $\\triangle ACC'$ is $A_1$. $B_1$ and $C_1$ are defined similarly. Prove that lines $AA_1$, $BB_1$ and $CC_1$ are concurent." ]

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[ "index 445722b..0d9e33d 100644 (file)\n@@ -3,24 +3,26 @@\nManipulating trees with monads\n------------------------------\n\nManipulating trees with monads\n------------------------------\n\n-This topic develops an idea based on a detailed suggestion of Ken\n-Shan's.  We'll build a series of functions that operate on trees,\n-doing various things, including replacing leaves, counting nodes, and\n-converting a tree to a list of leaves.  The end result will be an\n-application for continuations.\n-\n-From an engineering standpoint, we'll build a tree transformer that\n+This topic develops an idea based on a suggestion of Ken Shan's.\n+We'll build a series of functions that operate on trees, doing various\n+things, including updating leaves with a Reader monad, counting nodes\n+with a State monad, copying the tree with a List monad, and converting\n+a tree into a list of leaves with a Continuation monad.  It will turn\n+out that the continuation monad can simulate the behavior of each of\n+the other monads.\n+\n+From an engineering standpoint, we'll build a tree machine that\ndeals in monads.  We can modify the behavior of the system by swapping\none monad for another.  We've already seen how adding a monad can add\na layer of funtionality without disturbing the underlying system, for\ninstance, in the way that the Reader monad allowed us to add a layer\ndeals in monads.  We can modify the behavior of the system by swapping\none monad for another.  We've already seen how adding a monad can add\na layer of funtionality without disturbing the underlying system, for\ninstance, in the way that the Reader monad allowed us to add a layer\n-of intensionality to an extensional grammar, but we have not yet seen\n+of intensionality to an extensional grammar. But we have not yet seen\nthe utility of replacing one monad with other.\n\nFirst, we'll be needing a lot of trees for the remainder of the\ncourse.  Here again is a type constructor for leaf-labeled, binary trees:\n\nthe utility of replacing one monad with other.\n\nFirst, we'll be needing a lot of trees for the remainder of the\ncourse.  Here again is a type constructor for leaf-labeled, binary trees:\n\n-    type 'a tree = Leaf of 'a | Node of ('a tree * 'a tree)\n+    type 'a tree = Leaf of 'a | Node of ('a tree * 'a tree);;\n\n[How would you adjust the type constructor to allow for labels on the\ninternal nodes?]\n\n[How would you adjust the type constructor to allow for labels on the\ninternal nodes?]\n@@ -30,7 +32,7 @@ We'll be using trees where the nodes are integers, e.g.,\n\nlet t1 = Node (Node (Leaf 2, Leaf 3),\nNode (Leaf 5, Node (Leaf 7,\n\nlet t1 = Node (Node (Leaf 2, Leaf 3),\nNode (Leaf 5, Node (Leaf 7,\n-                                             Leaf 11)))\n+                                           Leaf 11)))\n.\n___|___\n|     |\n.\n___|___\n|     |\n@@ -50,9 +52,9 @@ Our first task will be to replace each leaf with its double:\n| Node (l, r) -> Node (tree_map leaf_modifier l,\ntree_map leaf_modifier r);;\n\n| Node (l, r) -> Node (tree_map leaf_modifier l,\ntree_map leaf_modifier r);;\n\n-`tree_map` takes a function that transforms old leaves into new leaves,\n-and maps that function over all the leaves in the tree, leaving the\n-structure of the tree unchanged.  For instance:\n+`tree_map` takes a tree and a function that transforms old leaves into\n+new leaves, and maps that function over all the leaves in the tree,\n+leaving the structure of the tree unchanged.  For instance:\n\nlet double i = i + i;;\ntree_map double t1;;\n\nlet double i = i + i;;\ntree_map double t1;;\n@@ -71,14 +73,15 @@ structure of the tree unchanged.  For instance:\n14   22\n\nWe could have built the doubling operation right into the `tree_map`\n14   22\n\nWe could have built the doubling operation right into the `tree_map`\n-code.  However, because we've left what to do to each leaf as a parameter, we can\n-decide to do something else to the leaves without needing to rewrite\n-`tree_map`.  For instance, we can easily square each leaf instead by\n-supplying the appropriate `int -> int` operation in place of `double`:\n+code.  However, because we've made what to do to each leaf a\n+parameter, we can decide to do something else to the leaves without\n+needing to rewrite `tree_map`.  For instance, we can easily square\n+each leaf instead, by supplying the appropriate `int -> int` operation\n+in place of `double`:\n\nlet square i = i * i;;\ntree_map square t1;;\n\nlet square i = i * i;;\ntree_map square t1;;\n-       - : int tree =ppp\n+       - : int tree =\nNode (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))\n\nNote that what `tree_map` does is take some unchanging contextual\nNode (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))\n\nNote that what `tree_map` does is take some unchanging contextual\n@@ -92,40 +95,54 @@ a Reader monad---is to have the `tree_map` function return a (monadized)\ntree that is ready to accept any `int -> int` function and produce the\nupdated tree.\n\ntree that is ready to accept any `int -> int` function and produce the\nupdated tree.\n\n-\\tree (. (. (f 2) (f 3)) (. (f 5) (. (f 7) (f 11))))\n-\n-       \\f      .\n-          _____|____\n-          |        |\n-          .        .\n-        __|___   __|___\n-        |    |   |    |\n-       f 2  f 3  f 5  .\n-                    __|___\n-                    |    |\n-                   f 7  f 11\n+       fun e ->    .\n+              _____|____\n+              |        |\n+              .        .\n+            __|___   __|___\n+            |    |   |    |\n+           e 2  e 3  e 5  .\n+                        __|___\n+                        |    |\n+                       e 7  e 11\n\nThat is, we want to transform the ordinary tree `t1` (of type `int\n\nThat is, we want to transform the ordinary tree `t1` (of type `int\n-tree`) into a reader object of type `(int -> int) -> int tree`: something\n-that, when you apply it to an `int -> int` function `f` returns an `int\n-tree` in which each leaf `i` has been replaced with `f i`.\n-\n-With previous readers, we always knew which kind of environment to\n-expect: either an assignment function (the original calculator\n-simulation), a world (the intensionality monad), an integer (the\n-Jacobson-inspired link monad), etc.  In the present case, we expect that our \"environment\" will be some function of type `int -> int`. \"Looking up\" some `int` in the environment will return us the `int` that comes out the other side of that function.\n-\n-       type 'a reader = (int -> int) -> 'a;;  (* mnemonic: e for environment *)\n+tree`) into a reader monadic object of type `(int -> int) -> int\n+tree`: something that, when you apply it to an `int -> int` function\n+`e` returns an `int tree` in which each leaf `i` has been replaced\n+with `e i`.\n+\n+[Application note: this kind of reader object could provide a model\n+for Kaplan's characters.  It turns an ordinary tree into one that\n+expects contextual information (here, the `e`) that can be\n+used to compute the content of indexicals embedded arbitrarily deeply\n+in the tree.]\n+\n+With our previous applications of the Reader monad, we always knew\n+which kind of environment to expect: either an assignment function, as\n+in the original calculator simulation; a world, as in the\n+intensionality monad; an individual, as in the Jacobson-inspired link\n+monad; etc.  In the present case, we expect that our \"environment\"\n+will be some function of type `int -> int`. \"Looking up\" some `int` in\n+the environment will return us the `int` that comes out the other side\n+of that function.\n+\n+       type 'a reader = (int -> int) -> 'a;;\nlet reader_unit (a : 'a) : 'a reader = fun _ -> a;;\nlet reader_unit (a : 'a) : 'a reader = fun _ -> a;;\n-       let reader_bind (u: 'a reader) (f : 'a -> 'b reader) : 'b reader = fun e -> f (u e) e;;\n+       let reader_bind (u: 'a reader) (f : 'a -> 'b reader) : 'b reader =\n+         fun e -> f (u e) e;;\n\nIt would be a simple matter to turn an *integer* into an `int reader`:\n\nIt would be a simple matter to turn an *integer* into an `int reader`:\n\n-       let int_readerize : int -> int reader = fun (a : int) -> fun (modifier : int -> int) -> modifier a;;\n-       int_readerize 2 (fun i -> i + i);;\n+       let asker : int -> int reader =\n+         fun (a : int) ->\n+           fun (modifier : int -> int) -> modifier a;;\n+       asker 2 (fun i -> i + i);;\n- : int = 4\n\n- : int = 4\n\n-But how do we do the analagous transformation when our `int`s are scattered over the leaves of a tree? How do we turn an `int tree` into a reader?\n+`asker a` is a monadic box that waits for an an environment (here, the argument `modifier`) and returns what that environment maps `a` to.\n+\n+How do we do the analagous transformation when our `int`s are scattered over the leaves of a tree? How do we turn an `int tree` into a reader?\nA tree is not the kind of thing that we can apply a\nfunction of type `int -> int` to.\n\nA tree is not the kind of thing that we can apply a\nfunction of type `int -> int` to.\n\n@@ -139,7 +156,7 @@ But we can do this:\nreader_unit (Node (l', r'))));;\n\nThis function says: give me a function `f` that knows how to turn\nreader_unit (Node (l', r'))));;\n\nThis function says: give me a function `f` that knows how to turn\n-something of type `'a` into an `'b reader`---this is a function of the same type that you could bind an `'a reader` to---and I'll show you how to\n+something of type `'a` into an `'b reader`---this is a function of the same type that you could bind an `'a reader` to, such as `asker` or `reader_unit`---and I'll show you how to\nturn an `'a tree` into an `'b tree reader`.  That is, if you show me how to do this:\n\n------------\nturn an `'a tree` into an `'b tree reader`.  That is, if you show me how to do this:\n\n------------\n@@ -173,17 +190,17 @@ Then we can expect that supplying it to our `int tree reader` will double all th\nIn more fanciful terms, the `tree_monadize` function builds plumbing that connects all of the leaves of a tree into one connected monadic network; it threads the\n`'b reader` monad through the original tree's leaves.\n\nIn more fanciful terms, the `tree_monadize` function builds plumbing that connects all of the leaves of a tree into one connected monadic network; it threads the\n`'b reader` monad through the original tree's leaves.\n\n-       # tree_monadize int_readerize t1 double;;\n+       # tree_monadize asker t1 double;;\n- : int tree =\nNode (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))\n\nHere, our environment is the doubling function (`fun i -> i + i`).  If\nwe apply the very same `int tree reader` (namely, `tree_monadize\n- : int tree =\nNode (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))\n\nHere, our environment is the doubling function (`fun i -> i + i`).  If\nwe apply the very same `int tree reader` (namely, `tree_monadize\n-int_readerize t1`) to a different `int -> int` function---say, the\n+asker t1`) to a different `int -> int` function---say, the\nsquaring function, `fun i -> i * i`---we get an entirely different\nresult:\n\nsquaring function, `fun i -> i * i`---we get an entirely different\nresult:\n\n-       # tree_monadize int_readerize t1 square;;\n+       # tree_monadize asker t1 square;;\n- : int tree =\nNode (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))\n\n- : int tree =\nNode (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))\n\n@@ -210,50 +227,95 @@ modification whatsoever, except for replacing the (parametric) type\n\nThen we can count the number of leaves in the tree:\n\nThen we can count the number of leaves in the tree:\n\n-       # tree_monadize (fun a -> fun s -> (a, s+1)) t1 0;;\n+       # let incrementer = fun a ->\n+           fun s -> (a, s+1);;\n+\n+       # tree_monadize incrementer t1 0;;\n- : int tree * int =\n(Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 5)\n\n- : int tree * int =\n(Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 5)\n\n-           .\n-        ___|___\n-        |     |\n-        .     .\n-       _|__  _|__\n-       |  |  |  |\n-       2  3  5  .\n-               _|__\n-               |  |\n-               7  11\n+               .\n+            ___|___\n+            |     |\n+            .     .\n+        (  _|__  _|__     ,   5 )\n+           |  |  |  |\n+           2  3  5  .\n+                   _|__\n+                   |  |\n+                   7  11\n+\n+Note that the value returned is a pair consisting of a tree and an\n+integer, 5, which represents the count of the leaves in the tree.\n+\n+Why does this work? Because the operation `incrementer`\n+takes an argument `a` and wraps it in an State monadic box that\n+increments the store and leaves behind a wrapped `a`. When we give that same operations to our\n+`tree_monadize` function, it then wraps an `int tree` in a box, one\n+that does the same store-incrementing for each of its leaves.\n+\n+We can use the state monad to annotate leaves with a number\n+corresponding to that leave's ordinal position.  When we do so, we\n+reveal the order in which the monadic tree forces evaluation:\n+\n+       # tree_monadize (fun a -> fun s -> ((a,s+1), s+1)) t1 0;;\n+       - : int tree * int =\n+         (Node\n+           (Node (Leaf (2, 1), Leaf (3, 2)),\n+            Node\n+             (Leaf (5, 3),\n+              Node (Leaf (7, 4), Leaf (11, 5)))),\n+         5)\n\n-Why does this work? Because the operation `fun a -> fun s -> (a, s+1)` takes an `int` and wraps it in an `int state` monadic box that increments the state. When we give that same operations to our `tree_monadize` function, it then wraps an `int tree` in a box, one that does the same state-incrementing for each of its leaves.\n+The key thing to notice is that instead of just wrapping `a` in the\n+monadic box, we wrap a pair of `a` and the current store.\n\n-One more revealing example before getting down to business: replacing\n-`state` everywhere in `tree_monadize` with `list` gives us\n+Reversing the annotation order requires reversing the order of the `state_bind`\n+operations.  It's not obvious that this will type correctly, so think\n+it through:\n+\n+       let rec tree_monadize_rev (f : 'a -> 'b state) (t : 'a tree) : 'b tree state =\n+         match t with\n+           | Leaf a -> state_bind (f a) (fun b -> state_unit (Leaf b))\n+           | Node (l, r) -> state_bind (tree_monadize f r) (fun r' ->     (* R first *)\n+                              state_bind (tree_monadize f l) (fun l'->    (* Then L  *)\n+                                state_unit (Node (l', r'))));;\n+\n+       # tree_monadize_rev (fun a -> fun s -> ((a,s+1), s+1)) t1 0;;\n+       - : int tree * int =\n+         (Node\n+           (Node (Leaf (2, 5), Leaf (3, 4)),\n+            Node\n+             (Leaf (5, 3),\n+              Node (Leaf (7, 2), Leaf (11, 1)))),\n+         5)\n\n-       # tree_monadize (fun i -> [ [i; square i] ]) t1;;\n-       - : int list tree list =\n-       [Node\n-         (Node (Leaf [2; 4], Leaf [3; 9]),\n-          Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))]\n+Later, we will talk more about controlling the order in which nodes are visited.\n\n-Unlike the previous cases, instead of turning a tree into a function\n-from some input to a result, this transformer replaces each `int` with\n-a list of `int`'s. We might also have done this with a Reader monad, though then our environments would need to be of type `int -> int list`. Experiment with what happens if you supply the `tree_monadize` based on the List monad an operation like `fun -> [ i; [2*i; 3*i] ]`. Use small trees for your experiment.\n+One more revealing example before getting down to business: replacing\n+`state` everywhere in `tree_monadize` with `list` lets us do:\n\n+       # let decider i = if i = 2 then [20; 21] else [i];;\n+       # tree_monadize decider t1;;\n+       - : int tree List_monad.m =\n+       [\n+         Node (Node (Leaf 20, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)));\n+         Node (Node (Leaf 21, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))\n+       ]\n\n-<!--\n-FIXME: We don't make it clear why the fun has to be int -> int list list, instead of int -> int list\n--->\n\n+Unlike the previous cases, instead of turning a tree into a function\n+from some input to a result, this monadized tree gives us back a list of trees,\n+one for each choice of `int`s for its leaves.\n\nNow for the main point.  What if we wanted to convert a tree to a list\nof leaves?\n\nNow for the main point.  What if we wanted to convert a tree to a list\nof leaves?\n\n-       type ('a, 'r) continuation = ('a -> 'r) -> 'r;;\n+       type ('r,'a) continuation = ('a -> 'r) -> 'r;;\nlet continuation_unit a = fun k -> k a;;\nlet continuation_bind u f = fun k -> u (fun a -> f a k);;\n\nlet continuation_unit a = fun k -> k a;;\nlet continuation_bind u f = fun k -> u (fun a -> f a k);;\n\n-       let rec tree_monadize (f : 'a -> ('b, 'r) continuation) (t : 'a tree) : ('b tree, 'r) continuation =\n+       let rec tree_monadize (f : 'a -> ('r,'b) continuation) (t : 'a tree) : ('r,'b tree) continuation =\nmatch t with\n| Leaf a -> continuation_bind (f a) (fun b -> continuation_unit (Leaf b))\n| Node (l, r) -> continuation_bind (tree_monadize f l) (fun l' ->\nmatch t with\n| Leaf a -> continuation_bind (f a) (fun b -> continuation_unit (Leaf b))\n| Node (l, r) -> continuation_bind (tree_monadize f l) (fun l' ->\n@@ -265,10 +327,21 @@ We use the Continuation monad described above, and insert the\n\nSo for example, we compute:\n\nSo for example, we compute:\n\n-       # tree_monadize (fun a -> fun k -> a :: k a) t1 (fun t -> []);;\n+       # tree_monadize (fun a k -> a :: k ()) t1 (fun _ -> []);;\n- : int list = [2; 3; 5; 7; 11]\n\n- : int list = [2; 3; 5; 7; 11]\n\n-We have found a way of collapsing a tree into a list of its leaves. Can you trace how this is working? Think first about what the operation `fun a -> fun k -> a :: k a` does when you apply it to a plain `int`, and the continuation `fun _ -> []`. Then given what we've said about `tree_monadize`, what should we expect `tree_monadize (fun a -> fun k -> a :: k a` to do?\n+We have found a way of collapsing a tree into a list of its\n+leaves. Can you trace how this is working? Think first about what the\n+operation `fun a k -> a :: k a` does when you apply it to a\n+plain `int`, and the continuation `fun _ -> []`. Then given what we've\n+said about `tree_monadize`, what should we expect `tree_monadize (fun\n+a -> fun k -> a :: k a)` to do?\n+\n+Soon we'll return to the same-fringe problem.  Since the\n+simple but inefficient way to solve it is to map each tree to a list\n+of its leaves, this transformation is on the path to a more efficient\n+solution.  We'll just have to figure out how to postpone computing the\n+tail of the list until it's needed...\n\nThe Continuation monad is amazingly flexible; we can use it to\nsimulate some of the computations performed above.  To see how, first\n\nThe Continuation monad is amazingly flexible; we can use it to\nsimulate some of the computations performed above.  To see how, first\n@@ -288,30 +361,107 @@ interesting functions for the first argument of `tree_monadize`:\n- : int tree =\nNode (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))\n\n- : int tree =\nNode (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))\n\n-       (* Simulating the int list tree list *)\n-       # tree_monadize (fun a -> fun k -> k [a; square a]) t1 (fun t -> t);;\n-       - : int list tree =\n-       Node\n-        (Node (Leaf [2; 4], Leaf [3; 9]),\n-         Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))\n-\n(* Counting leaves *)\n# tree_monadize (fun a -> fun k -> 1 + k a) t1 (fun t -> 0);;\n- : int = 5\n\n(* Counting leaves *)\n# tree_monadize (fun a -> fun k -> 1 + k a) t1 (fun t -> 0);;\n- : int = 5\n\n-We could simulate the tree state example too, but it would require\n-generalizing the type of the Continuation monad to\n+It's not immediately obvious to us how to simulate the List monadization of the tree using this technique.\n+\n+We could simulate the tree annotating example by setting the relevant\n+type to `(store -> 'result, 'a) continuation`.\n\n-       type ('a, 'b, 'c) continuation = ('a -> 'b) -> 'c;;\n+Andre Filinsky has proposed that the continuation monad is\n+able to simulate any other monad (Google for \"mother of all monads\").\n\nIf you want to see how to parameterize the definition of the `tree_monadize` function, so that you don't have to keep rewriting it for each new monad, see [this code](/code/tree_monadize.ml).\n\nIf you want to see how to parameterize the definition of the `tree_monadize` function, so that you don't have to keep rewriting it for each new monad, see [this code](/code/tree_monadize.ml).\n\n+The idea of using continuations to characterize natural language meaning\n+------------------------------------------------------------------------\n+\n+We might a philosopher or a linguist be interested in continuations,\n+especially if efficiency of computation is usually not an issue?\n+Well, the application of continuations to the same-fringe problem\n+shows that continuations can manage order of evaluation in a\n+well-controlled manner.  In a series of papers, one of us (Barker) and\n+Ken Shan have argued that a number of phenomena in natural langauge\n+semantics are sensitive to the order of evaluation.  We can't\n+reproduce all of the intricate arguments here, but we can give a sense\n+of how the analyses use continuations to achieve an analysis of\n+natural language meaning.\n+\n+**Quantification and default quantifier scope construal**.\n+\n+We saw in the copy-string example (\"abSd\") and in the same-fringe example that\n+local properties of a structure (whether a character is `'S'` or not, which\n+integer occurs at some leaf position) can control global properties of\n+the computation (whether the preceeding string is copied or not,\n+whether the computation halts or proceeds).  Local control of\n+surrounding context is a reasonable description of in-situ\n+quantification.\n+\n+    (1) John saw everyone yesterday.\n+\n+This sentence means (roughly)\n+\n+    forall x . yesterday(saw x) john\n+\n+That is, the quantifier *everyone* contributes a variable in the\n+direct object position, and a universal quantifier that takes scope\n+over the whole sentence.  If we have a lexical meaning function like\n+the following:\n+\n+       let lex (s:string) k = match s with\n+         | \"everyone\" -> Node (Leaf \"forall x\", k \"x\")\n+         | \"someone\" -> Node (Leaf \"exists y\", k \"y\")\n+         | _ -> k s;;\n+\n+Then we can crudely approximate quantification as follows:\n+\n+       # let sentence1 = Node (Leaf \"John\",\n+                                                 Node (Node (Leaf \"saw\",\n+                                                                         Leaf \"everyone\"),\n+                                                               Leaf \"yesterday\"));;\n+\n+       # tree_monadize lex sentence1 (fun x -> x);;\n+       - : string tree =\n+       Node\n+        (Leaf \"forall x\",\n+         Node (Leaf \"John\", Node (Node (Leaf \"saw\", Leaf \"x\"), Leaf \"yesterday\")))\n+\n+In order to see the effects of evaluation order,\n+observe what happens when we combine two quantifiers in the same\n+sentence:\n+\n+       # let sentence2 = Node (Leaf \"everyone\", Node (Leaf \"saw\", Leaf \"someone\"));;\n+       # tree_monadize lex sentence2 (fun x -> x);;\n+       - : string tree =\n+       Node\n+        (Leaf \"forall x\",\n+         Node (Leaf \"exists y\", Node (Leaf \"x\", Node (Leaf \"saw\", Leaf \"y\"))))\n+\n+The universal takes scope over the existential.  If, however, we\n+replace the usual `tree_monadizer` with `tree_monadizer_rev`, we get\n+inverse scope:\n+\n+       # tree_monadize_rev lex sentence2 (fun x -> x);;\n+       - : string tree =\n+       Node\n+        (Leaf \"exists y\",\n+         Node (Leaf \"forall x\", Node (Leaf \"x\", Node (Leaf \"saw\", Leaf \"y\"))))\n+\n+There are many crucially important details about quantification that\n+are being simplified here, and the continuation treatment used here is not\n+scalable for a number of reasons.  Nevertheless, it will serve to give\n+an idea of how continuations can provide insight into the behavior of\n+quantifiers.\n\n-The Binary Tree monad\n----------------------\n\n-Of course, by now you may have realized that we have discovered a new\n-monad, the Binary Tree monad:\n+The Tree monad\n+==============\n+\n+Of course, by now you may have realized that we are working with a new\n+monad, the binary, leaf-labeled Tree monad.  Just as mere lists are in fact a monad,\n+so are trees.  Here is the type constructor, unit, and bind:\n\ntype 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);;\nlet tree_unit (a: 'a) : 'a tree = Leaf a;;\n\ntype 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);;\nlet tree_unit (a: 'a) : 'a tree = Leaf a;;\n@@ -328,22 +478,20 @@ To check the other two laws, we need to make the following\nobservation: it is easy to prove based on `tree_bind` by a simple\ninduction on the structure of the first argument that the tree\nresulting from `bind u f` is a tree with the same strucure as `u`,\nobservation: it is easy to prove based on `tree_bind` by a simple\ninduction on the structure of the first argument that the tree\nresulting from `bind u f` is a tree with the same strucure as `u`,\n-except that each leaf `a` has been replaced with `f a`:\n-\n-\\tree (. (f a1) (. (. (. (f a2) (f a3)) (f a4)) (f a5)))\n+except that each leaf `a` has been replaced with the tree returned by `f a`:\n\n.                         .\n__|__                     __|__\n\n.                         .\n__|__                     __|__\n-                     |   |                     |   |\n+                     |   |                    /\\   |\na1  .                   f a1  .\n_|__                     __|__\na1  .                   f a1  .\n_|__                     __|__\n-                        |  |                     |   |\n+                        |  |                     |   /\\\n.  a5                    .  f a5\nbind         _|__       f   =        __|__\n.  a5                    .  f a5\nbind         _|__       f   =        __|__\n-                       |  |                    |   |\n+                       |  |                    |   /\\\n.  a4                   .  f a4\n__|__                   __|___\n.  a4                   .  f a4\n__|__                   __|___\n-                     |   |                   |    |\n+                     |   |                  /\\    /\\\na2  a3                f a2  f a3\n\nGiven this equivalence, the right identity law\na2  a3                f a2  f a3\n\nGiven this equivalence, the right identity law\n@@ -361,9 +509,6 @@ As for the associative law,\nwe'll give an example that will show how an inductive proof would\nproceed.  Let `f a = Node (Leaf a, Leaf a)`.  Then\n\nwe'll give an example that will show how an inductive proof would\nproceed.  Let `f a = Node (Leaf a, Leaf a)`.  Then\n\n-\\tree (. (. (. (. (a1) (a2)))))\n-\\tree (. (. (. (. (a1) (a1)) (. (a1) (a1)))))\n-\n.\n____|____\n.               .            |       |\n.\n____|____\n.               .            |       |\n@@ -384,7 +529,7 @@ Now when we bind this tree to `g`, we get\n\nAt this point, it should be easy to convince yourself that\nusing the recipe on the right hand side of the associative law will\n\nAt this point, it should be easy to convince yourself that\nusing the recipe on the right hand side of the associative law will\n-built the exact same final tree.\n+build the exact same final tree.\n\nSo binary trees are a monad.\n\nSo binary trees are a monad.\n\n@@ -394,135 +539,8 @@ called a\nthat is intended to represent non-deterministic computations as a tree.\n\nthat is intended to represent non-deterministic computations as a tree.\n\n-What's this have to do with tree\\_mondadize?\n+What's this have to do with tree\\_monadize?\n--------------------------------------------\n\n--------------------------------------------\n\n-So we've defined a Tree monad:\n-\n-       type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);;\n-       let tree_unit (a: 'a) : 'a tree = Leaf a;;\n-       let rec tree_bind (u : 'a tree) (f : 'a -> 'b tree) : 'b tree =\n-           match u with\n-           | Leaf a -> f a\n-           | Node (l, r) -> Node (tree_bind l f, tree_bind r f);;\n-\n-What's this have to do with the `tree_monadize` functions we defined earlier?\n-\n-       let rec tree_monadize (f : 'a -> 'b reader) (t : 'a tree) : 'b tree reader =\n-           match t with\n-           | Leaf a -> reader_bind (f a) (fun b -> reader_unit (Leaf b))\n-           | Node (l, r) -> reader_bind (tree_monadize f l) (fun l' ->\n-                              reader_bind (tree_monadize f r) (fun r' ->\n-                                reader_unit (Node (l', r'))));;\n-\n-... and so on for different monads?\n-\n-The answer is that each of those `tree_monadize` functions is adding a Tree monad *layer* to a pre-existing Reader (and so on) monad. So far, we've defined monads as single-layered things. Though in the Groenendijk, Stokhoff, and Veltmann homework, we had to figure out how to combine Reader, State, and Set monads in an ad-hoc way. In practice, one often wants to combine the abilities of several monads. Corresponding to each monad like Reader, there's a corresponding ReaderT **monad transformer**. That takes an existing monad M and adds a Reader monad layer to it. The way these are defined parallels the way the single-layer versions are defined. For example, here's the Reader monad:\n-\n-       (* monadic operations for the Reader monad *)\n-\n-       type 'a reader =\n-               env -> 'a;;\n-       let unit (a : 'a) : 'a reader =\n-               fun e -> a;;\n-       let bind (u: 'a reader) (f : 'a -> 'b reader) : 'b reader =\n-               fun e -> (fun v -> f v e) (u e);;\n-\n-We've just beta-expanded the familiar `f (u e) e` into `(fun v -> f v e) (u e)`, in order to factor out the parts where any Reader monad is being supplied as an argument to another function. Then if we want instead to add a Reader layer to some arbitrary other monad M, with its own M.unit and M.bind, here's how we do it:\n-\n-       (* monadic operations for the ReaderT monadic transformer *)\n-\n-       (* We're not giving valid OCaml code, but rather something\n-        * that's conceptually easier to digest.\n-        * How you really need to write this in OCaml is more circuitous...\n-        * see http://lambda.jimpryor.net/code/tree_monadize.ml for some details. *)\n-\n-       type ('a, M) readerT =\n-               env -> 'a M;;\n-       (* this is just an 'a M reader; but don't rely on that pattern to generalize *)\n-\n-       let unit (a : 'a) : ('a, M) readerT =\n-               fun e -> M.unit a;;\n-\n-       let bind (u : ('a, M) readerT) (f : 'a -> ('b, M) readerT) : ('b, M) readerT =\n-               fun e -> M.bind (u e) (fun v -> f v e);;\n-\n-Notice the key differences: where before we just returned `a`, now we instead return `M.unit a`. Where before we just supplied value `u e` of type `'a reader` as an argument to a function, now we instead `M.bind` the `'a reader` to that function. Notice also the differences in the types.\n-\n-What is the relation between Reader and ReaderT? Well, suppose you started with the Identity monad:\n-\n-       type 'a identity = 'a;;\n-       let unit (a : 'a) : 'a = a;;\n-       let bind (u : 'a) (f : 'a -> 'b) : 'b = f u;;\n-\n-and you used the ReaderT transformer to add a Reader monad layer to the Identity monad. What do you suppose you would get?\n-\n-The relations between the State monad and the StateT monadic transformer are parallel:\n-\n-       (* monadic operations for the State monad *)\n-\n-       type 'a state =\n-               store -> ('a * store);;\n-\n-       let unit (a : 'a) : 'a state =\n-               fun s -> (a, s);;\n-\n-       let bind (u : 'a state) (f : 'a -> 'b state) : 'b state =\n-               fun s -> (fun (a, s') -> f a s') (u s);;\n-\n-We've used `(fun (a, s') -> f a s') (u s)` instead of the more familiar `let (a, s') = u s in f a s'` in order to factor out the part where a value of type `'a state` is supplied as an argument to a function. Now StateT will be:\n-\n-       (* monadic operations for the StateT monadic transformer *)\n-\n-       type ('a, M) stateT =\n-               store -> ('a * store) M;;\n-       (* notice this is not an 'a M state *)\n-\n-       let unit (a : 'a) : ('a, M) stateT =\n-               fun s -> M.unit (a, s);;\n-\n-       let bind (u : ('a, M) stateT) (f : 'a -> ('b, M) stateT) : ('b, M) stateT =\n-               fun s -> M.bind (u s) (fun (a, s') -> f a s');;\n-\n-Do you see the pattern? Where ordinarily we'd return an `'a` value, now we instead return an `'a M` value. Where ordinarily we'd supply a `'a state` value as an argument to a function, now we instead `M.bind` it to that function.\n-\n-Okay, now let's do the same thing for our Tree monad.\n-\n-       (* monadic operations for the Tree monad *)\n-\n-       type 'a tree =\n-               Leaf of 'a | Node of ('a tree) * ('a tree);;\n-\n-       let unit (a: 'a) : 'a tree =\n-               Leaf a;;\n-\n-       let rec bind (u : 'a tree) (f : 'a -> 'b tree) : 'b tree =\n-           match u with\n-           | Leaf a -> (fun b -> Leaf b) (f a)\n-           | Node (l, r) -> (fun l' r' -> Node (l', r')) (bind l f) (bind r f);;\n-\n-       (* monadic operations for the TreeT monadic transformer *)\n-\n-       type ('a, M) treeT =\n-               'a tree M;;\n-\n-       let unit (a: 'a) : ('a, M) tree =\n-               M.unit (Leaf a);;\n-\n-       let rec bind (u : ('a, M) tree) (f : 'a -> ('b, M) tree) : ('b, M) tree =\n-           match u with\n-           | Leaf a -> M.bind (f a) (fun b -> M.unit (Leaf b))\n-           | Node (l, r) -> M.bind (bind l f) (fun l' ->\n-                                                       M.bind (bind r f) (fun r' ->\n-                                                               M.unit (Node (l', r'));;\n-\n-Compare this definition of `bind` for the TreeT monadic transformer to our earlier definition of `tree_monadize`, specialized for the Reader monad:\n-\n-       let rec tree_monadize (f : 'a -> 'b reader) (t : 'a tree) : 'b tree reader =\n-           match t with\n-           | Leaf a -> reader_bind (f a) (fun b -> reader_unit (Leaf b))\n-           | Node (l, r) -> reader_bind (tree_monadize f l) (fun l' ->\n-                              reader_bind (tree_monadize f r) (fun r' ->\n-                                reader_unit (Node (l', r'))));;\n-\n+Our different implementations of `tree_monadize` above were different *layerings* of the Tree monad with other monads (Reader, State, List, and Continuation). We'll explore that further here: [[Monad Transformers]]." ]

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[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "RE: How to display one plot as an inset in another plot ?\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg35368] RE: How to display one plot as an inset in another plot ?\n• From: \"Wolf, Hartmut\" <Hartmut.Wolf at t-systems.com>\n• Date: Tue, 9 Jul 2002 06:50:56 -0400 (EDT)\n• Sender: owner-wri-mathgroup at wolfram.com\n\n```\n> -----Original Message-----\n> From: AES [mailto:siegman at stanford.edu]\nTo: mathgroup at smc.vnet.net\n> Sent: Monday, July 08, 2002 9:20 AM\n> Subject: [mg35368] How to display one plot as an inset in\n> another plot?\n>\n>\n> How to display one (complete) plot as an inset (maybe scaled,\n> certainly\n> offset) within (or on top of) another plot?\n>\n> Kind of a\n>\n> Show[ plot1, {Scaled[0.5], Offset[{2,2}], plot2} ]\n>\n> capability.\n>\n\nYou nearly got it, just missed Rectangle, e.g.\n\nShow[g, Graphics[\nRectangle[Scaled[{.5, .3}], Offset[{162, 100}, Scaled[{.5, .3}]], g2]],\nPlotRange -> {All, {0, 11}}, Axes -> {True, False}]\n\ng, g2 are some Graphics.\n\n--\nHartmut\n\n```\n\n• Prev by Date: Re: Quantile function\n• Next by Date: Sovling integrals: non-algebraic???\n• Previous by thread: How do I generate drawings with a fixed scaling on paper?\n• Next by thread: Sovling integrals: non-algebraic???" ]

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[ "# Redefining a function\n\n+1 vote\n\nHai, I have a problem defining a new function.\n\nI am using Cadabra for perturbation analysis in which I want to calculate higher-order perturbation terms. I have to define the result of each order term as independent tensors i.e, I have\n\n$$T_{\\mu \\nu}^{\\text{eff}}= T_{\\mu \\nu}^{(0)}+\\epsilon T_{\\mu \\nu}^{(1)}+\\epsilon^{2} T_{\\mu \\nu}^{(2)}+\\epsilon^{3}T_{\\mu \\nu}^{(3)}.$$\n\nthen I redefined the equation as the form\n\n$$T_{\\mu \\nu}^{(eff)}=a^{0} T_{\\mu \\nu}^{(0)}+a^{1} T_{\\mu \\nu}^{(1)}+a^{2} T_{\\mu \\nu}^{(2)}+a^{3}T_{\\mu \\nu}^{(3)}.$$\n\nI would like to obtain the $T_{\\mu \\nu}^{0}$, $T_{\\mu \\nu}^{1}$, $T_{\\mu \\nu}^{2}$ and $T_{\\mu \\nu}^{3}$ by substituting $a^{0}=a^{2}=a^{3}=0$ like conditions, for which have used the command as\n\nEx:=T_{\\mu \\nu}^(eff)=a^{0} T_{\\mu \\nu}^(0)\n+a^{1} T_{\\mu \\nu}^(1)\n+a^{2} T_{\\mu \\nu}^(2)\n+a^{3}T_{\\mu \\nu}^(3);\nT_{\\mu \\nu}^{0}:=substitute(Ex,$a^{1}->0,a^{2}->0,a^{3}->0$);\nT_{\\mu \\nu}^{1}:=substitute(Ex,$a^{0}->0,a^{2}->0,a^{3}->0$);\nT_{\\mu \\nu}^{2}:=substitute(Ex,$a^{0}->0,a^{1}->0,a^{3}->0$);\nT_{\\mu \\nu}^{3}:=substitute(Ex,$a^{0}->0,a^{1}->0,a^{2}->0$);\n\nBut the results are showing the same. How can I rectify it?\n\nedited\n\n+1 vote\n\nAlmost all Cadabra functions modify the expression in-place. So you first need to make a copy of the expression, and then apply the substitution.\n\nThere are a few other things not quite right in your example: Ex is a class name (so you cannot use it as the name of an expression; use ex with lowercase instead), and expressions must have valid Python names (so you cannot call them T_{\\mu\\nu}^{1}).\n\nHere's something that works:\n\nex:=Teff_{\\mu \\nu} = a^{0} T0_{\\mu \\nu}\n+a^{1} T1_{\\mu \\nu}\n+a^{2} T2_{\\mu \\nu}\n+a^{3} T3_{\\mu \\nu};\n\nT0 = rhs(ex)\nsubstitute(T0, $a^{0}=1, a^{1}=0, a^{2}=0, a^{3}=0$);\n\nThis sets the Python variable T0 equal to the mathematical expression $T0_{\\mu\\nu}$.\n\nby (65.1k points)" ]

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[ "# Discretizing a Continuous Time Stochastic Volatility Model\n\nHow does the discrete time stochastic volatility model arise from the continuous time one?\n\nAlso, forgive me for cross-posting.\n\nI have the following continuous time SDE for a stochastic volatility model. $$S_t$$ is the price, and $$v_t$$ is a variance process. $$dS_t = \\mu S_tdt + \\sqrt{v_t}S_t dB_{1t} \\\\ dv_t = (\\theta - \\alpha \\log v_t)v_tdt + \\sigma v_t dB_{2t} .$$ I'm more familiar with the discrete time version: $$y_t = \\exp(h_t/2)\\epsilon_t \\\\ h_{t+1} = \\mu + \\phi(h_t - \\mu) + \\sigma_t \\eta_t \\\\ h_1 \\sim N\\left(\\mu, \\frac{\\sigma^2}{1-\\phi^2}\\right).$$ $$\\{y_t\\}$$ are the log returns, and $$\\{h_t\\}$$ are the \"log-volatilites.\" Keep in mind there might be some confusion about parameters; for example the $$\\mu$$s in each of these models are different.\n\nHow do I verify that the first discretizes into the second?\n\nHere's my work so far. First I define $$Y_t = \\log S_t$$ and $$h_t = \\log v_t$$. Then I use Ito's lemma to get \\begin{align*} dY_t &= \\left(\\mu - \\frac{\\exp h_t}{2}\\right)dt + \\exp[h_t/2] dB_{1t}\\\\ dh_t &= \\left(\\theta - \\alpha\\log v_t - \\sigma^2/2\\right)dt + \\sigma dB_{2,t}\\\\ &= \\alpha\\left(\\tilde{\\mu} - h_t \\right)dt + \\sigma dB_{2t}. \\end{align*}\n\nI got the state/log-vol process piece. I use the Euler method to discretize, setting $$\\Delta t = 1$$, to get \\begin{align*} h_{t+1} &= \\alpha \\tilde{\\mu} + h_t(1-\\alpha) + \\sigma \\eta_t \\\\ &= \\tilde{\\mu}(1 - \\phi) + \\phi h_t + \\sigma \\eta_t \\\\ &= \\tilde{\\mu} + \\phi(h_t - \\tilde{\\mu}) + \\sigma \\eta_t. \\end{align*}\n\nThe observation equation is a little bit more difficult, however:\n\n\\begin{align*} y_{t+1} = Y_{t+1} - Y_t &= (\\mu - \\frac{v_t}{2}) + \\sqrt{v_t}\\epsilon_{t+1} \\\\ &= \\left(\\mu - \\frac{\\exp h_t}{2} \\right) + \\exp[ \\log \\sqrt{v_t}] \\epsilon_{t+1} \\\\ &= \\left(\\mu - \\frac{\\exp h_t}{2}\\right) + \\exp\\left[ \\frac{h_t}{2}\\right] \\epsilon_{t+1}. \\end{align*}\n\nWhy is the mean return not $$0$$ or $$\\mu$$? How should I have defined the transformations? I suspect it might have something to do with the meaning of parameters and random variables. In the discrete time model above, $$y_t$$ is the mean-adjusted log return. In the SDE above that, $$\\mu$$ probably means the interest rate plus half the variance (Questions on continuously compounded return vs long term expected return).\n\n• Hi Taylor. First notice that the limiting behaviour of discrete time processes and their continuous time counterpart is quite tricky (see the famous 1990 paper by Neslon). Second, the first SDE prevails under a certain probability measure which you have yet to define. This measure will give you the meaning of the parmeter $\\mu$. Under $\\Bbb{P}$ (real world measure) it's the expected return of the stock, under $\\Bbb{Q}$ it's its funding rate (e.g. risk-free rate minus dividends minus repo rate). – Quantuple Jul 11 '18 at 7:45\n• Hi @Quantuple. Thanks for the note. Care to elaborate further in an answer? Also, which Nelson paper? There were two in 1990. – Taylor Jul 11 '18 at 16:55\n\nI guess you can discretize the raw price process too instead of the log price process. You get $$S_{t+1} = S_t + \\mu S_t + \\sqrt{v_t} S_t Z_t$$ (where $$Z_t$$ is a standard normal variate), or $$\\frac{S_{t+1}}{S_t} - 1 = \\mu + \\sqrt{v_t} Z_t.$$ Got the idea from: https://arxiv.org/pdf/1707.00899.pdf" ]

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[ "# big-theta notation solved examples\n\nList Of Big-Theta Notation Solved Examples 2023. This always indicates the minimum time required for any algorithm for all input values, therefore the best case of any algorithm. Upper bound of a geometric series show that t(n) = xn i=0 3i is o(3n).", null, "", null, "Solved Problem 5 Big O, Theta, Omega Notation YouTube from www.youtube.com\n\nIt is define as upper bound and upper bound on an algorithm is the most amount of time required (. For example, consider the following expression. The key is finding n₀ and c 10.", null, "www.lynda.com\n\nThe constants c that are used for the big o and big ω bounds will typically be different from each other. It represents the upper bound of the algorithm.\n\nwww.chegg.com\n\nWe write f(n) = θ(g(n)), if there are positive constants n0 and c 1 and c 2 such that, to the right of n 0 the f(n) always lies between c 1 *g(n) and c 2 *g(n) inclusive. This always indicates the minimum time required for any algorithm for all input values, therefore the best case of any algorithm.", null, "www.youtube.com\n\nIt is define as upper bound and upper bound on an algorithm is the most amount of time required (. Youtube mp3, stafaband, gudang lagu, metrolagu", null, "www.chegg.com\n\nO (g (n)) = { f (n): This always indicates the minimum time required for any algorithm for all input values, therefore the best case of any algorithm.", null, "www.chegg.com\n\nF (n) = θ (g (n)) iff there are three positive constants c1, c2 and n0 such that. Big omega notation with solved examples.", null, "www.youtube.com\n\nHowever, here we take only the tight lower bound which means the greatest lower bound, the highest value which will lower bound the function. Take this course for free.", null, "www.chegg.com\n\nWe use big theta when a program has only one case in term of runtime. The function f is said to be θ (g.", null, "www.youtube.com\n\nCheck out the course here: Make sure your password is at least 8 characters and.", null, "www.youtube.com\n\nI mean the equation on the right side confuses me. The key is finding n₀ and c 10.", null, "www.youtube.com\n\nWhen studying the runtime of an algorithm or a program, you really do have an f ( n) hiding in there somewhere ( n being the size of the input). Take this course for free.", null, "www.youtube.com\n\nFunction, f (n) = o (g (n)), if and only if positive constant c is present and thus: Take this course for free.", null, "www.youtube.com\n\nAsymptotic notations describe the function’s limiting behavior. Let us take an example and understand big theta:\n\n### When Studying The Runtime Of An Algorithm Or A Program, You Really Do Have An F ( N) Hiding In There Somewhere ( N Being The Size Of The Input).\n\nI mean the equation on the right side confuses me. C1 g (n) <= f (n) <= c2 g (n). When doing complexity analysis, the following assumptions are assumed.\n\n### Upper Bound Of A Geometric Series Show That T(N) = Xn I=0 3I Is O(3N).\n\nThe constants c that are used for the big o and big ω bounds will typically be different from each other. Both upper and lower bound. We use big theta when a program has only one case in term of runtime.\n\n### Let Us Take An Example And Understand Big Theta:\n\nYoutube mp3, stafaband, gudang lagu, metrolagu Take this course for free. In the analysis of algorithms, asymptotic notations are used to evaluate the performance of an algorithm, in its best cases and worst cases.\n\n### F (N) = Θ (G (N)) Iff There Are Three Positive Constants C1, C2 And N0 Such That.\n\nSearch for jobs related to big theta notation examples or hire on the world’s largest freelancing marketplace with 21m+ jobs. Big omega notation with solved examples. This video is part of an online course, intro to algorithms.\n\n### In Other Words, If You Want To Prove Big Theta, Then Find Out The Big O And Big Omega Separately And You Will Be Able To Prove Big Theta.\n\nThis is the currently selected item. If you try and check these values you will see satisfy the condition. Big omega notation is used to define the lower bound of any algorithm or we can say the best case of any algorithm" ]

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[ "", null, "# As an Astable Multivibrator 1. 2  An integrated chip that is used in a wide variety of circuits to generate square wave and triangular shaped single.\n\n## Presentation on theme: \"As an Astable Multivibrator 1. 2  An integrated chip that is used in a wide variety of circuits to generate square wave and triangular shaped single.\"— Presentation transcript:\n\nAs an Astable Multivibrator 1\n\n2\n\n An integrated chip that is used in a wide variety of circuits to generate square wave and triangular shaped single and periodic pulses. ◦ Examples in your home are  high efficiency LED and fluorescence light dimmers and  temperature control systems for electric stoves  tone generators for appliance “beeps” ◦ The Application Notes section of the datasheets for the TLC555 and LM555 timers have a number of other circuits that are in use today in various communications and control circuits. 3\n\n Astable – a circuit that can not remain in one state.  Monostable – a circuit that has one stable state. When perturbed, the circuit will return to the stable state.  One Shot – Monostable circuit that produces one pulse when triggered.  Flip Flop – a digital circuit that flips or toggles between two stable states (bistable). The Flip Flop inputs decide which of the two states its output will be.  Multivibrator – a circuit used to implement a simple two-state system, which may be astable, monostable, or bistable.  CMOS – complimentary MOSFET logic. CMOS logic dominates the digital industry because the power requirements and component density are significantly better than other technologies. 4\n\n Monostable ◦ A single pulse is outputted when an input voltage attached to the trigger pin of the 555 timer equals the voltage on the threshold pin.  Astable ◦ A periodic square wave is generated by the 555 timer.  The voltage for the trigger and threshold pins is the voltage across a capacitor that is charged and discharged through two different RC networks. 5 I know – who comes up with these names?\n\n We will operate the 555 Timer as an Astable Multivibrator in the circuit for the metronome. 6 http://www.williamson-labs.com/480_555.htm\n\nThe components that make up a 555 timer are shown within the gray box. Internal resistors form a voltage divider that provides ⅓V CC and ⅔ V CC reference voltages. Two internal voltage comparators determine the state of a D flip-flop. The flip-flop output controls a transistor switch. 7\n\n As a reminder, an Op Amp without a feedback component is a voltage comparator. ◦ Output voltage changes to force the negative input voltage to equal the positive input voltage.  A maximum output voltage (V o ) is against the positive supply rail (V+) if the positive input voltage (v 2 ) is greater than negative input voltage (v 1 ).  A minimum output voltage (V o ) is is against the negative supply rail (V-) if the negative input voltage (v 1 ) is greater than the positive input voltage (v 2 ). 8\n\nhttp://www.williamson-labs.com/480_555.htm The voltage comparators use the internal voltage divider to keep the capacitor voltage (V C ) between ⅓V CC and ⅔V CC. The output of the lower voltage comparator will be high (Vcc) when V C ⅓V CC (⅓V CC = the voltage across the lower resistor in the internal voltage divider). The output of the upper voltage comparator will be low (0 V) when V C ⅔V CC (⅔V CC = the voltage across the two lower resistors in the internal voltage divider). 9\n\nhttp://www.williamson-labs.com/480_555.htm The bipolar transistor (BJT) acts as a switch. NOTE: Your kit TLC555 uses a MOSFET instead of a BJT. 10\n\nAs you will learn in ECE 2204, a BJT or MOSFET transistor can be connected to act like a switch. – When a positive voltage is applied to the base or gate, the transistor acts like there is a very small resistor is between the collector and the emitter, or the drain and the source. – When ground is applied to the base or gate, the transistor acts like there is a an open circuit between the collector and the emitter, or the drain and the source. 11\n\nhttp://www.williamson-labs.com/480_555.htm The transistor inside the 555 switches the discharge pin (7) to ground (or very close to 0 V), when Qbar (the Q with a line over it) of the D flip-flop is high (V Qbar ≈ V CC ). The transistor grounds the node between external timing resistors R a and R b. The capacitor discharges through R b to ground through the transistor. Current through R a also goes to ground through the transistor. When the transistor is switched off, it acts like an open circuit. V CC now charges the capacitor through R a and R b. 12\n\n The capacitor charges through R A and R B. ◦ Because V C started 0 V, the first timing period will be longer than the periods that follow. 13\n\n The capacitor charges through R a and R b until V C = ⅔V CC. 14  When V C reaches ⅔V CC, the output of the upper voltage comparator changes and resets the D flip-flop, Qbar switches to high (≈ V CC ), and the transistor switches on.  The capacitor then begins discharging through R b & the transistor to ground.\n\nDischarging: The capacitor discharges through R b and the transistor to ground. Current through R a is also grounded by the transistor.  When V C reaches ⅓V CC, the output of the lower voltage comparator changes and sets the D flip-flop, Qbar switches to low (≈ 0 V), and the transistor switches off.  The capacitor then begins charging through R a and R b. http://www.williamson-labs.com/480_555.htm Thus, the voltage of the capacitor can be no more than ⅔V CC and no less than ⅓V CC if all of the components internal and external to the 555 are ideal. 15\n\n The output of the 555 timer, pin 3, is Q on the D flip-flop. ◦ When Qbar is 5 V and the capacitor is charging, Q is 0 V. ◦ When Qbar is 0 V and the capacitor is discharging, Q is 5 V.  Thus, the output of a 555 timer is a continuous square wave function (0 V to 5 V) where: ◦ the period is dependent the sum of the time it takes to charge the capacitor to ⅔V CC and the time that it takes to discharge the capacitor to ⅓V CC. ◦ In this circuit, the only time that the duty cycle (the time that the output is at 0 V divided by the period) will be 0.5 (or 50%) is when Ra = 0 W, which should not be allowed to occur as that would connect Vcc directly to ground when the transistor switches on. 16 http://www.williamson-labs.com/480_555.htm\n\n T H is the time it takes C to charge from ⅓V CC to ⅔V CC ◦ T H = (R a + R b )*C*[-ln(½)] (from solving for the charge time between voltages)  T L is the time it takes C to discharge from ⅔V CC to ⅓V CC ◦ T Low =R b *C*[-ln(½)] (from solving for the charge time between voltages)  The duty cycle (% of the time the output is high) depends on the resistor values.  Williamson Labs 555 astable circuit waveform animation555 astable circuit waveform animation 17\n\n The duty cycle of the standard 555 timer circuit in Astable mode must be greater than 50%. ◦ T high = 0.693(R a + R b )C [C charges through R a and R a from V CC ] ◦ T low = 0.693R b C [C discharges through R b into pin 7 ] ◦ R 1 must have a resistance value greater than zero to prevent the discharge pin from directly shorting V DD to ground. ◦ Duty cycle = T high / (T high + T low ) = (R a + R b ) / (R a + 2R b ) > 50% if R a ≠ 0  Adding a diode across R b allows the capacitor to charge directly through R a. This sets T high ≈ 0.693R a C T low = 0.693R b C (unchanged) 18\n\n TI Data Sheets and design info TI Data Sheets and design info ◦ Data Sheet (pdf) Data Sheet ◦ Design Calculator (zip) Design Calculator  Williamson Labs http://www.williamson-labs.com/480_555.htm http://www.williamson-labs.com/480_555.htm ◦ Timer tutorials with a 555 astable circuit waveform animation. ◦ Philips App Note AN170 (pdf)AN170  Wikipedia - 555 timer IC Wikipedia - 555 timer IC  NE555 Tutorials http://www.unitechelectronics.com/NE-555.htm http://www.unitechelectronics.com/NE-555.htm  Doctronics 555 timer tips http://www.doctronics.co.uk/555.htm http://www.doctronics.co.uk/555.htm  The Electronics Club http://www.kpsec.freeuk.com/555timer.htm http://www.kpsec.freeuk.com/555timer.htm  555 Timer Circuits http://www.555-timer-circuits.com http://www.555-timer-circuits.com  555 Timer Tutorial http://www.sentex.net/~mec1995/gadgets/555/555.html http://www.sentex.net/~mec1995/gadgets/555/555.html  Philips App Note AN170 http://www.doctronics.co.uk/pdf_files/555an.pdf http://www.doctronics.co.uk/pdf_files/555an.pdf 19\n\nDownload ppt \"As an Astable Multivibrator 1. 2  An integrated chip that is used in a wide variety of circuits to generate square wave and triangular shaped single.\"\n\nSimilar presentations" ]

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[ "### Implicit Conversion\n\nConversion is perfomed automatically.\nAlso known as coercion.\nThis includes arithmetic operation. Is invoked automatically when a value of one data type is assigned to a value of another.\nNo special syntax is required because the conversion is type safe and no data will be lost.\nExamples include conversions from smaller to larger integral types\n\n`short myshort = 5; int myint = myshort; `\n\nThis is an example of a widening conversion.\n\nFor classes there is always an implicit conversion from a Derived class to a Base class.\nA derived class always contains all the members of its base class.\n\n`public class Base {} public class Derived : Base {} Base B = new Derived(); `\n\nC# will automatically convert an integer variable to a string when it is concatenated using the + operator.\n\nExamples include converting from smaller to larger integral types and converting from a derived class to a base class.\nFor built-in numeric types an implicit conversion can be made when the value can fit into the variable without being truncated or rounded off\nThere are no implicit conversions to the char type\nThere are no implicit conversions between floating-point and decimal types\n\n`int value1 = 1234; long value2 = value1; `\n\n From To Sbyte short , int, long, float, double, or decimal Byte - short , ushort, int, uint, long, ulong, float, double, or decimal Short - int , long, float, double, or decimal Ushort - int , uint, long, ulong, float, double, or decimal Int - long , float, double, or decimal Uint - long , ulong, float, double, or decimal Long - float , double, or decimal Char - ushort , int, uint, long, ulong, float, double, or decimal Float - double Ulong - float , double, or decimal\n\n#### Important\n\nThere is no implicit datatype conversion in VB.NET it must be explicit using the ToString method.\n\n© 2021 Better Solutions Limited" ]

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[ "IIT JAM Follow\nJuly 7, 2021 12:23 pm", null, "30 pts\nin option C, it is given that maximum value of f(x) is 1 ,if it is false then what is the maximum value of f(x)?????????A function f(x) is continuous in the interval [O,21. It is known that f(0) = f{2) = -1 and f(1) =1. Which one of the following statements must be true? There exists a y in the interval (0,1) such that f(y) = f(y +1) (A) (B) For every y in the interval (0,1), f(y) = {(2 - y) The maximum value of the function in the interval (0,2) is 1 (C) (D) There exists a y in the interval (0, 1) such that f(y) = -f(2 y)\n• 0 Likes\n• Shares\n• Alka gupta\nhere C options must be ture because 1 is point of maxima and maximum value is f(1)=1", null, "" ]

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[ "One of my favorite mathematical proofs (not original) is to show that the\n\nInfinite Sum of 1/(kk)  =  1 + 1/4 + 1/9 + 1/16 + 1/25 + …   =  (pi.pi)/6\n\nThis problem, sometimes called the Basel problem, is named from the Swiss city where the famous mathematician Jakob Bernoulli first posed it after proving that the harmonic series,  1 + 1/2  + 1/3 + 1/4  + 1/5 + ….  diverges or does not converge to a sum.\n\nThe easiest proof of the Basel problem may be to write sin x as an infinite product of linear factors and also to find the Maclaurin series for sin x.  Comparing the coefficients of the x cubed terms from each expansion gives the result!   Unfortunately, this proof is not very inspiring.\n\nThe proof below requires a number of mathematical techniques that math majors learn as undergraduates and which are generally fun to use and thus delight the person:\n\nWe start with sin t expressed as an infinite product of its zeros\n\n(1) sin t = t.(1 – t/pi).(1 + t/pi).(1 – t/2pi).(1 + t/2pi).      or making a change in the variable, let t=pi.y\n\n(2) sin pi.y  = pi.y . (1 – y).(1 + y) . (2-y)/2 . (2+y)/2 .       or\n\n(3) sin pi.y = pi.y.(1 – yy) . (4 – yy)/4 . (9 – yy)/9 . (16 – yy)/16 . …\n\nTaking the ln of both sides of the equation to turn a product into a sum gives\n\n(4) ln (sin pi.y)  =  ln (pi) + ln(y) + ln (1 – yy) + ln (4 – yy) – ln4 + ln (9 – yy) – ln9 + ln (16 – yy) – ln16 + …\n\nTaking the derivative of both sides of the equation gives\n\n(5) pi.cos pi.y /sin pi.y   =  0 + 1/y – 2y/(1-yy) – 2y/(4-yy) – 0 – 2y/(9-yy) – 0 – 2y/(16-yy) – 0 - …\n\nDividing the equation by 2y and rearranging gives\n\n(6) 1/(1-yy) + 1/(4-yy) + 1/(9-yy) + 1/(16-yy) + …   =  1/2yy – (pi/2y)cot pi.y\n\nMaking another change in the variable, let x= -y/i  (or y = -ix) where I is the square root of -1, we then have xx = - yy and thus the equation becomes\n\n(7) 1/(1+xx) + 1/(4+xx) + 1/(9+xx) + 1/(16+xx) + …  = -1/2xx + (pi/2ix)cot (-i.pi.x)\n\nLet us now work with the last term of equation (7) and express it in one of its known equivalent forms; that is, cot z = cos z/sin z = I (e^2iz + 1)/ (e^2iz – 1) where ^ denotes an exponent\n\nThus, (pi/2ix)cot (-i.pi.x) = (pi/2ix)I(e^2pi.x + 1)/( e^2pi.x – 1) = (pi/2x)(e^2pi.x – 1 +2)/ (e^2pi.x – 1) = pi/2x  +  pi/ x(e^2pi.x – 1) and the right hand side of (7) becomes\n\n-1/2xx + (pi/2ix)cot(-i.pi.x) =  -1/2xx + pi/2x + pi/ x(e^2pi.x – 1)\n\n=  (pi.x-1)/2xx + pi/ x(e^2pi.x – 1)\n\n=  (pi.x-1)(e^2pi.x – 1)/2xx(e^2pi.x – 1) + 2x.pi/2xx(e^2pi.x -1)\n\n= (pi.xe^2pi.x – e^2pi.x – pi.x + 1 + 2x.pi) / (2xxe^2pi.x – 2xx)\n\nWe now rewrite (7) with this equivalent expression on the right hand side of the equation\n\n(8) 1/(1+xx) + 1/(4+xx) + 1/(9+xx) + 1/(16+xx) + …  = (pi.xe^2pi.x – e^2pi.x – pi.x + 1 + 2x.pi) / (2xxe^2pi.x – 2xx)\n\nAs x gets closer to 0 without limit, the LHS = 1 + 1/4 + 1/9 + 1/16 + …   while the RHS approaches 0/0\n\nUsing L’Hopital’s Rule and simplifying gives the following expression:  (2pi.xe^2pi.x – pi.e^2pi.x + pi)/ (4pi.xxe^2pi.x + 4xe^2pi.x – 4x) and this expression also approaches 0/0 as x gets close to 0\n\nUsing L’Hopital’s Rule a second time and simplifying gives:  4pi.pi.pi.xe^2pi.x / (8pi.pi.xxe^2pi.x + 16pi.xe^2pi.x + 4e^2pi.x – 4)  and this expression also goes to 0/0\n\nUsing L’Hopital’s Rule a third time   =  (4pi.pi.pi.e^2pi.x + 8pi.pi.pi.pi.xe^2pi.x)/ (16pi.pi.pi.xxe^2pi.x + 16pi.pi.xe^2pi.x + 16pi.e^2pi.x + 32pi.pi.xe^2pi.x + 8pi.e^2pi.x) and now as x goes to 0\n\n=  4pi.pi.pi / (16pi + 8pi)\n\n= pi.pi / 6\n\nIt’s amazing that all the above collapses to a simple expression involving the number pi, 3.14159…, the ratio of the circumference of a circle to its diameter.  Where did that come from given the original infinite series?  Well, we started the proof by looking at the sine curve which has a 2pi cycle.\n\nWhat is the link between sine curves and an infinite series?  We know that the distribution of prime numbers is linked to the Riemann zeta function and more recently to eigenvalues of a random Hermitian matrix and the world of modern physics and that is even more amazing.  Anyway we have shown that\n\n(9)  1 + 1/4 + 1/9 + 1/16 + …   = pi.pi / 6   first proved by the mathematician Leonhard Euler in 1735.\n\nThis result leads naturally to the Riemann zeta function and the Riemann Hypothesis, which has kept serious mathematicians busy for decades.  The Basel problem remains a good introduction or gateway to more advanced mathematics." ]

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[ "# LESSON 8-1 SIMILARITY IN RIGHT TRIANGLES PROBLEM SOLVING\n\nThey both share that angle there. Crash Course Physics 4. You can write the same ratio in three ways. Well it’s going to be vertex B. So this is BDC. Assess Way Whether you are creating leveled homework worksheets or preparing a chapter test, Glencoe Math Accelerated provides the flexibility to use the prepared templates or -6 -3 -2 -1 1 5 6 Fold lengthwise.", null, "There’s actually three different triangles that I can see here. So this is my triangle, ABC. Triangles on SAT Math: Rose Tree Media Created Date: Today in class students worked on a review packet. Meat lb 8 4 0 6 2 10 12 12 45 63 Number of People Meat Consumption Geometry Guide and Practice Problems.\n\nExperimental rihgt Theoretical Probability: What is the solution of the system? So in triangle BDC, you have one right angle. Does the graph show a proportional relationship? So we want to make sure we’re getting the similarity right. Exponents and Scientific Notation Chapter 1: But we haven’t thought about just that little angle right over there. Create your website today. And then if we look at BC on the larger triangle, BC is going to correspond to what on the smaller triangle?\n\nAnd then dimilarity go to vertex C, which is in orange. Similar Triangles in right triangles to solve problems. You can write the same ratio in three ways.\n\nHOMEWORK DAFT PUNK DISCOGS\n\nWe have a bunch of triangles here, and some lengths of sides, and a couple of right angles.\n\nA and C is going to correspond to BC. Crash Course Physics 4.\n\n# Solutions to Geometry () :: Free Homework Help and Answers :: Slader\n\nAnd so what is it going to correspond to? For example, decide whether the words in a chapter of a seventh-grade science book are generally longer than the words in a chapter of a fourth-grade science book.\n\nSo BDC looks like this. Will you write them all as fractions, decimals, or Edgewood Math 7: What fun it will be to see your students understand the concepts and get excited about learning!\n\n## Similarity In Right Triangles Practice And Problem Solving A/B\n\nThe sum of the everyday world. Projected Schedule for Chapter 5 If you are absent, please be sure to check the agenda tab for specifics!", null, "Real Numbers and the Pythagorean Theorem Chapter 8: And we know that the length of this side, which we figured out through this problem is 4. This course expands the study of numbers to include complex numbers and includes the study of exponents and radicals, rational expressions, as well as quadratic, polynomial, exponential, and logarithmic functions.\n\nAnd just to make it clear, let me actually draw these two triangles separately.\n\n# | CK Foundation\n\nTo find equal ratios, multiply or divide the numerator and denominator by the same nonzero number. And now that we know that they are similar, we can attempt to take ratios between the sides.\n\nExcept as permitted under the United States Copyright Act, no part of this publication in the everyday world. BC on our smaller triangle corresponds to AC on trianglfs larger triangle.\n\n## Solving similar triangles: same side plays different roles\n\nMath High school geometry Similarity Solving similar triangles. They both share that angle there. Decide on a strategy for ordering the numbers. The Neshaminy community builds futures by empowering each child to become a productive citizen and a probem learner.\n\nWelcome to Algebra 1. Data Analysis and Displays Chapter Friday If you need to reference any of the lessons or want some additional practice, please select the PDF below of the textbook pages for Chapter 4.", null, "", null, "" ]

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[ "# Joanne Lee – Project 07\n\n``````// Joanne Lee\n// Section C\n// [email protected]\n// Project-07\n\nvar nPoints = 100;\nvar colorR = 167;\nvar colorG = 168;\nvar colorB = 212;\nvar div = 0;\nvar w = 0;\nvar h = 0;\n\nfunction setup() {\ncreateCanvas(480,480);\n}\n\nfunction draw() {\nbackground (240);\npush();\ntranslate(0,0);\nfor (var c = 0; c < 6; c ++) {\nh = c * 96; // shifts height of flower\nfor (var d = 0; d < 7; d ++) {\nw = d * 80; // shift width of flower\ndrawEpitrochoidCurve(colorR, colorG, colorB, div, w, h);\n}\n}\nif (mouseY > 0){\ndiv = (mouseY / 15.0) + 5; // flower size scales to mouseY position\n}\npop();\n}\n\nfunction drawEpitrochoidCurve(colorR, colorG, colorB, div, w, h) {\n// Epicycloid:\n// http://mathworld.wolfram.com/Epicycloid.html\nvar x;\nvar y;\n\nvar a = 80.0;\nvar b = a / 2.0;\n\nfill(colorR, colorG, colorB);\nbeginShape();\nfor (var i = 0; i < 100; i++) {\nvar t = map(i, 0, nPoints, 0, TWO_PI);\nx = a * (6 * cos(t) - cos(6 * t)) / div + w;\ny = a * (6 * sin(t) - sin(6 * t)) / div + h;\nvertex(x, y);\n}\nendShape(CLOSE);\n\n}\n\nfunction mouseClicked() { // petal color changes upon mouseClick\nvar colNum = int(random(1,4));\nif (colNum == 1) {\ncolorR = 167;\ncolorG = 168;\ncolorB = 212;\n}\nelse if (colNum == 2) {\ncolorR = 207;\ncolorG = 61;\ncolorB = 160;\n}\nelse if (colNum == 3) {\ncolorR = 90;\ncolorG = 76;\ncolorB = 181;\n}\n}``````", null, "For this week’s project, I first thought of flower petals when I was looking at the possible curves. I looked up flower petals and liked the colors and shapes of these, so I used ranuncloid curves to create them. Ranuncloid curves are essentially epicycloids with 5 cusps. For interactivity, I thought it would be interesting to make the flower “grow” using the mouseY position as well as randomly alternate between the flower colors depicted above with mouse clicks. I purposely had the flowers overlap with each other at its largest size because I thought that it formed an interesting pattern.\n\nPosted on Categories Uncategorized" ]

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[ " Convert erg/d to erg/h | erg per day to ergs per hour\n\n# power units conversion\n\n## Amount: 1 erg per day (erg/d) of power Equals: 0.042 ergs per hour (erg/h) in power\n\nConverting erg per day to ergs per hour value in the power units scale.\n\nTOGGLE :   from ergs per hour into ergs per day in the other way around.\n\n## power from erg per day to erg per hour conversion results\n\n### Enter a new erg per day number to convert\n\n* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)\n* Precision is how many digits after decimal point (1 - 9)\n\nEnter Amount :\nDecimal Precision :\n\nCONVERT :   between other power measuring units - complete list.\n\nHow many ergs per hour are in 1 erg per day? The answer is: 1 erg/d equals 0.042 erg/h\n\n## 0.042 erg/h is converted to 1 of what?\n\nThe ergs per hour unit number 0.042 erg/h converts to 1 erg/d, one erg per day. It is the EQUAL power value of 1 erg per day but in the ergs per hour power unit alternative.\n\n erg/d/erg/h power conversion result From Symbol Equals Result Symbol 1 erg/d = 0.042 erg/h\n\n## Conversion chart - ergs per day to ergs per hour\n\n1 erg per day to ergs per hour = 0.042 erg/h\n\n2 ergs per day to ergs per hour = 0.083 erg/h\n\n3 ergs per day to ergs per hour = 0.13 erg/h\n\n4 ergs per day to ergs per hour = 0.17 erg/h\n\n5 ergs per day to ergs per hour = 0.21 erg/h\n\n6 ergs per day to ergs per hour = 0.25 erg/h\n\n7 ergs per day to ergs per hour = 0.29 erg/h\n\n8 ergs per day to ergs per hour = 0.33 erg/h\n\n9 ergs per day to ergs per hour = 0.38 erg/h\n\n10 ergs per day to ergs per hour = 0.42 erg/h\n\n11 ergs per day to ergs per hour = 0.46 erg/h\n\n12 ergs per day to ergs per hour = 0.50 erg/h\n\n13 ergs per day to ergs per hour = 0.54 erg/h\n\n14 ergs per day to ergs per hour = 0.58 erg/h\n\n15 ergs per day to ergs per hour = 0.63 erg/h\n\nCategory: main menupower menuErgs per day\n\nConvert power of erg per day (erg/d) and ergs per hour (erg/h) units in reverse from ergs per hour into ergs per day.\n\n## Power units\n\nPower units represent power physics, which is the rate at which energy is used-up, either transformed or transferred from its source to elsewhere, by various ways within the nature of physics. Conversion tool with multiple power units.\n\n# Converter type: power units\n\nFirst unit: erg per day (erg/d) is used for measuring power.\nSecond: erg per hour (erg/h) is unit of power.\n\nQUESTION:\n15 erg/d = ? erg/h\n\nANSWER:\n15 erg/d = 0.63 erg/h\n\nAbbreviation, or prefix, for erg per day is:\nerg/d\nAbbreviation for erg per hour is:\nerg/h\n\n## Other applications for this power calculator ...\n\nWith the above mentioned two-units calculating service it provides, this power converter proved to be useful also as a teaching tool:\n1. in practicing ergs per day and ergs per hour ( erg/d vs. erg/h ) measures exchange.\n2. for conversion factors between unit pairs.\n3. work with power's values and properties.\n\nTo link to this power erg per day to ergs per hour online converter simply cut and paste the following.\nThe link to this tool will appear as: power from erg per day (erg/d) to ergs per hour (erg/h) conversion.\n\nI've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting." ]

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[ "# Rotating reference frame\n\n\nRotating reference frame\n\nA rotating frame of reference is a special case of a non-inertial reference frame that is rotating relative to an inertial reference frame. An everyday example of a rotating reference frame is the surface of the Earth. (This article considers only frames rotating about a fixed axis. For more general rotations, see Euler angles.)\n\nFictitious forces\n\nAll non-inertial reference frames exhibit fictitious forces. Rotating reference frames are characterized by three fictitious forcescite book |title=Mathematical Methods of Classical Mechanics |page=p. 130 |author=Vladimir Igorević Arnolʹd |edition=2nd Edition |isbn=978-0-387-96890-2 |year=1989 |url=http://books.google.com/books?id=Pd8-s6rOt_cC&pg=PT149&dq=%22additional+terms+called+inertial+forces.+This+allows+us+to+detect+experimentally%22&lr=&as_brr=0&sig=ACfU3U1qRbkvn6x7FcBsHO8Bp4Ty95XbZw#PPT150,M1 |publisher=Springer]\n\n* the centrifugal force\n* the Coriolis forceand, for non-uniformly rotating reference frames,\n* the Euler force.\n\nScientists living in a rotating box can measure the speed and direction of their rotation by measuring these fictitious forces. For example, Léon Foucault was able to show the Coriolis force that results from the Earth's rotation using the Foucault pendulum. If the Earth were to rotate a thousand-fold faster (making each day only ~86 seconds long), these fictitious forces could be felt easily by humans, as they are on a spinning carousel.\n\nRelating rotating frames to stationary frames\n\nThe following is a derivation of the formulas for accelerations as well as fictitious forces in a rotating frame. It begins with the relation between coordinates of the position of a particle in a rotating frame and the coordinates in an inertial (stationary) frame. Then, by taking time derivatives, formulas are derived that relate the velocity of the particle as seen in the two frames, and the acceleration relative to each frame. Using these accelerations a comparison of Newton's second law as formulated in the frames identifies the fictitious forces.\n\nRelation between positions in the two frames\n\nTo derive these fictitious forces, it's helpful to be able to convert between the coordinates $left\\left( x^\\left\\{prime\\right\\},y^\\left\\{prime\\right\\},z^\\left\\{prime\\right\\} ight\\right)$ of the rotating reference frame and the coordinates $left\\left( x, y, z ight\\right)$ of an inertial reference frame with the same origin. If the rotation is about the $z$ axis with an angular velocity $Omega$ and the two reference frames coincide at time $t=0$, the transformation from rotating coordinates to inertial coordinates can be written\n\n:$x = x^\\left\\{prime\\right\\} cosOmega t - y^\\left\\{prime\\right\\} sinOmega t$:$y = x^\\left\\{prime\\right\\} sinOmega t + y^\\left\\{prime\\right\\} cosOmega t$\n\nwhereas the reverse transformation is\n\n:$x^\\left\\{prime\\right\\} = x cosleft\\left(-Omega t ight\\right) - y sinleft\\left( -Omega t ight\\right)$:$y^\\left\\{prime\\right\\} = x sinleft\\left( -Omega t ight\\right) + y cosleft\\left( -Omega t ight\\right)$\n\nThis result can be obtained from a rotation matrix.\n\nIntroduce the unit vectors representing standard unit basis vectors in the rotating frame. The time-derivatives of these unit vectors are found next. Suppose the frames are aligned at \"t = \"0 and the \"z\"-axis is the axis of rotation. Then for a counterclockwise rotation through angle \"&Omega;t\"::where the (\"x\", \"y\") components are expressed in the stationary frame. Likewise,:Thus the time derivative of these vectors, which rotate without changing magnitude, is::This result is the same as found using a vector cross product with the rotation vector pointed along the z-axis of rotation , namely,:where is either or .\n\nTime derivatives in the two frames\n\nIntroduce the unit vectors representing standard unit basis vectors in the rotating frame. As they rotate they will remain normalized. If we let them rotate at the speed of $Omega$ about an axis then each unit vector of the rotating coordinate system abides by the following equation::Then if we have a vector function , : and we want to examine its first dervative we have (using the chain rule of differentiation):cite book |url=http://books.google.com/books?as_q=&num=10&btnG=Google+Search&as_epq=The+author+likes+to+call+it+the+%22Euler+force%2C%22+in+view&as_oq=&as_eq=&as_brr=0&lr=&as_vt=&as_auth=&as_pub=&as_sub=&as_drrb=c&as_miny=&as_maxy=&as_isbn= |title=The Variational Principles of Mechanics |author=Cornelius Lanczos |year=1986 |isbn=0-486-65067-7 |publisher=Dover Publications |edition=Reprint of Fourth Edition of 1970 |page=Chapter 4, §5] cite book |title=Classical Mechanics |author=John R Taylor |page= p. 342 |publisher=University Science Books |isbn=1-891389-22-X |year=2005 |url=http://books.google.com/books?id=P1kCtNr-pJsC&pg=PP1&dq=isbn:189138922X&sig=ACfU3U0kWmspY7W8eh9g1e6AqiMP83uSGw#PPA342,M1] :::::where is the rate of change of as observed in the rotating coordinate system. As a shorthand the differentiation is expressed as:::\n\nRelation between velocities in the two frames\n\nA velocity of an object is the time-derivative of the object's position, or\n\n:$mathbf\\left\\{v\\right\\} stackrel\\left\\{mathrm\\left\\{def\\left\\{=\\right\\} frac\\left\\{dmathbf\\left\\{r\\left\\{dt\\right\\}$\n\nThe time derivative of a position in a rotating reference frame has two components, one from the explicit time dependence due to motion of the particle itself, and another from the frame's own rotation. Applying the result of the previous subsection to the displacement , the velocities in the two reference frames are related by the equation\n\n:where subscript \"i\" means the inertial frame of reference, and \"r\" means the rotating frame of reference.\n\nRelation between accelerations in the two frames\n\nAcceleration is the second time derivative of position, or the first time derivative of velocity\n\n:where subscript \"i\" means the inertial frame of reference.Carrying out the differentiations and re-arranging some terms yields the acceleration in the \"rotating\" reference frame\n\n:\n\nwhere $mathbf\\left\\{a\\right\\}_\\left\\{mathrm\\left\\{r stackrel\\left\\{mathrm\\left\\{def\\left\\{=\\right\\} left\\left( frac\\left\\{d^\\left\\{2\\right\\}mathbf\\left\\{r\\left\\{dt^\\left\\{2 ight\\right)_\\left\\{mathrm\\left\\{r$ is the apparent acceleration in the rotating reference frame.\n\nNewton's second law in the two frames\n\nWhen the expression for acceleration is multiplied by the mass of the particle, the three extra terms on the right-hand side result in fictitious forces in the rotating reference frame, that is, apparent forces that result from being in a non-inertial reference frame, rather than from any physical interaction between bodies.\n\nUsing Newton's second law of motion \"F\"$=$\"m\" \"a\", we obtain:cite book |title=Mechanics |author=LD Landau and LM Lifshitz |page= p. 128 |url=http://books.google.com/books?id=e-xASAehg1sC&pg=PA40&dq=isbn:9780750628969&sig=ACfU3U2LCcLQRZqYxDOXTg_9Ks_zp_qorg#PPA128,M1 |edition=Third Edition |year=1976 |isbn=978-0-7506-2896-9]\n\n* the Coriolis force\n\n:\n\n* the centrifugal force\n\n:\n\n* and the Euler force\n\n:\n\nwhere $m$ is the mass of the object being acted upon by these fictitious forces. Notice that all three forces vanish when the frame is not rotating, that is, when\n\nFor completeness, the inertial acceleration $mathbf\\left\\{a\\right\\}_\\left\\{mathrm\\left\\{i$ due to impressed external forces $mathbf\\left\\{F\\right\\}_\\left\\{mathrm\\left\\{imp$ can be determined from the total physical force in the inertial (non-rotating) frame (for example, force from physical interactions such as electromagnetic forces) using Newton's second law in the inertial frame:\n\n:$mathbf\\left\\{F\\right\\}_\\left\\{mathrm\\left\\{imp = m mathbf\\left\\{a\\right\\}_\\left\\{mathrm\\left\\{i$Newton's law in the the rotating frame then becomes::$mathbf\\left\\{F_r\\right\\} = mathbf\\left\\{F\\right\\}_\\left\\{mathrm\\left\\{imp +mathbf\\left\\{F\\right\\}_\\left\\{mathrm\\left\\{centrifugal +mathbf\\left\\{F\\right\\}_\\left\\{mathrm\\left\\{Coriolis+mathbf\\left\\{F\\right\\}_\\left\\{mathrm\\left\\{Euler = mmathbf\\left\\{a_r\\right\\} .$In other words, to handle the laws of motion in a rotating reference frame:cite book |title=Analytical Mechanics |author =Louis N. Hand, Janet D. Finch |page=p. 267 |url=http://books.google.com/books?id=1J2hzvX2Xh8C&pg=PA267&vq=fictitious+forces&dq=Hand+inauthor:Finch&lr=&as_brr=0&source=gbs_search_s&sig=ACfU3U33emV_6eJZihu3M6IZKurSt85_eg\nisbn=0521575729 |publisher=Cambridge University Press |year=1998\n] cite book |title=Mechanics |author=HS Hans & SP Pui |page=P. 341 |url=http://books.google.com/books?id=mgVW00YV3zAC&pg=PA341&dq=inertial+force+%22rotating+frame%22&lr=&as_brr=0&sig=ACfU3U1--cWJ02SuFZwp4Y6Uyoe4hbGFmQ |isbn=0070473609 |publisher=Tata McGraw-Hill |year=2003 ] cite book |title=Classical Mechanics |author=John R Taylor |page= p. 328 |publisher=University Science Books |isbn=1-891389-22-X |year=2005 |url=http://books.google.com/books?id=P1kCtNr-pJsC&pg=PP1&dq=isbn:189138922X&sig=ACfU3U0kWmspY7W8eh9g1e6AqiMP83uSGw#PPA328,M1]\n\nReferences and notes\n\nee also\n\n* Centrifugal force (rotating reference frame) Centrifugal force as seen from systems rotating about a fixed axis\n* Centrifugal force (planar motion) Centrifugal force exhibited by a particle in planar motion as seen by the particle itself and by observers in a co-rotating frame of reference\n* Coriolis force The effect of the Coriolis force on the Earth and other rotating systems\n* Inertial frame of reference\n* Non-inertial frame\n* Fictitious force A more general treatment of the subject of this article\n\n* [http://www.youtube.com/watch?v=49JwbrXcPjc Animation clip] showing scenes as viewed from both an inertial frame and a rotating frame of reference, visualizing the Coriolis and centrifugal forces.\n\nWikimedia Foundation. 2010.\n\n### Look at other dictionaries:\n\n• Accelerated reference frame — In theoretical physics, an accelerated reference frame is usually a coordinate system or frame of reference, that undergoes a constant and continual change in velocity over time as judged from an inertial frame. An object in an accelerated frame… …   Wikipedia\n\n• Terrestrial reference frame — A terrestrial reference frame is the reference frame as one views from earth, or from the ground of another earth like body. A terrestrial reference frame effects the way we perceive almost everything from day to day because as we live on the… …   Wikipedia\n\n• Non-inertial reference frame — A non inertial reference frame is a frame of reference that is under acceleration. The laws of physics in such a frame do not take on their most simple form, as required by the theory of special relativity. To explain the motion of… …   Wikipedia\n\n• Frame-dragging — Albert Einstein s theory of general relativity predicts that rotating bodies drag spacetime around themselves in a phenomenon referred to as frame dragging. The rotational frame dragging effect was first derived from the theory of general… …   Wikipedia\n\n• Frame of reference — Frame of reference, n. 1. an arbitrary set of spatial coordinates used to describe the position or motion of objects. The coordinates may be fixed or moving; as, a rotating frame of reference. [PJC] 2. a set of assumptions or conditions that are… …   The Collaborative International Dictionary of English\n\n• Frame of reference — A frame of reference in physics, may refer to a coordinate system or set of axes within which to measure the position, orientation, and other properties of objects in it, or it may refer to an observational reference frame tied to the state of… …   Wikipedia\n\n• Inertial frame of reference — In physics, an inertial frame of reference is a frame of reference which belongs to a set of frames in which physical laws hold in the same and simplest form. According to the first postulate of special relativity, all physical laws take their… …   Wikipedia\n\n• Mechanics of planar particle motion — Classical mechanics Newton s Second Law History of classical mechanics  …   Wikipedia\n\n• Coriolis effect — For the psychophysical perception effect, see Coriolis effect (perception). Classical mechanics Newton s Second Law …   Wikipedia" ]

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[ "(function (a, d, o, r, i, c, u, p, w, m) { m = d.getElementsByTagName(o), a[c] = a[c] || {}, a[c].trigger = a[c].trigger || function () { (a[c].trigger.arg = a[c].trigger.arg || []).push(arguments)}, a[c].on = a[c].on || function () {(a[c].on.arg = a[c].on.arg || []).push(arguments)}, a[c].off = a[c].off || function () {(a[c].off.arg = a[c].off.arg || []).push(arguments) }, w = d.createElement(o), w.id = i, w.src = r, w.async = 1, w.setAttribute(p, u), m.parentNode.insertBefore(w, m), w = null} )(window, document, \"script\", \"https://95662602.adoric-om.com/adoric.js\", \"Adoric_Script\", \"adoric\",\"9cc40a7455aa779b8031bd738f77ccf1\", \"data-key\");\nvar domain=window.location.hostname; var params_totm = \"\"; (new URLSearchParams(window.location.search)).forEach(function(value, key) {if (key.startsWith('totm')) { params_totm = params_totm +\"&\"+key.replace('totm','')+\"=\"+value}}); var rand=Math.floor(10*Math.random()); var script=document.createElement(\"script\"); script.src=`https://stag-core.tfla.xyz/pre_onetag?pub_id=34&domain=\\${domain}&rand=\\${rand}&min_ugl=0\\${params_totm}`; document.head.append(script);" ]

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[ "Missing points of repair\n\nFigure 1a : the line from F to VP, divides AV and BW so that a/b = c/d.\n\nFigure 1b and 1c : If VP cannot be created as shown in figure 1, G is best estimated by dividing BW in the relation a/b using e.g. a calculator.", null, "Figure 2a : Missing repair line ‘ ? ‘\n\nDraw a small image (figure 2b), create VP1 as crossing-point of a and b, draw HL, create crossing-point with c to obtain VP2. Connect VP2 with Q: The angle in-between PQ and P-VP2 is equal to the angle of inclination of ’?’.\n\nOr (figure 2c) : Create mini-triangle (blue) with aa parallel to a to define x/y, divide PQ in the relation x/y to create Z. Draw vertical through PQ through Z to create VP2, proceed as above." ]

[ null, "http://rst-art.com/per-s05.bmp", null ]

[ "# Salim's question at Yahoo! Answers regarding trigonometry and circular sectors\n\n#### MarkFL\n\nStaff member\nHere is the question:\n\nHow to solve this maths problem?\n\nSo we have a triangle ABC with given sides. On the BC side, there's a part of a circle (we do not know how much of a circle). We are asked to find the angle A to ensure that the part of a circle has the same area as the triangle. The problem boils down to finding how to calculate the area of the part of circle. So how would you go about doing it?\nThanks\nI have posted a link there to this thread so the OP can view my work.\n\n#### MarkFL\n\nStaff member\nRe: Salim's question at Yahoo! Answers regarding trignometry and circular sectors\n\nHello Salim,\n\nPlease refer to the following diagram:", null, "We may determine $\\theta$ using the Law of Cosines:\n\n$$\\displaystyle a^2=2r^2\\left(1-\\cos(\\theta) \\right)$$\n\n$$\\displaystyle \\theta=\\cos^{-1}\\left(\\frac{2r^2-a^2}{2r^2} \\right)$$\n\nFrom this, we may determine the area $A_S$ of circular sector $OBC$:\n\n$$\\displaystyle A_S=\\frac{1}{2}r^2\\theta=\\frac{1}{2}r^2\\cos^{-1}\\left(\\frac{2r^2-a^2}{2r^2} \\right)$$\n\nAnd we may now also determine the area $A_T$ of triangle $OBC$:\n\n$$\\displaystyle A_T=\\frac{1}{2}r^2\\sin(\\theta)=\\frac{1}{2}r^2\\frac{a\\sqrt{4r^2-a^2}}{2r^2}=\\frac{a\\sqrt{4r^2-a^2}}{4}$$\n\nThus the portion of the circle's area $A_O$ outside the triangle is:\n\n$$\\displaystyle A_O=\\pi r^2-\\frac{1}{2}r^2\\cos^{-1}\\left(\\frac{2r^2-a^2}{2r^2} \\right)+\\frac{a\\sqrt{4r^2-a^2}}{4}$$\n\nEquating this to the area of triangle $ABC$, we obtain:\n\n$$\\displaystyle \\frac{1}{2}bc\\sin(A)=\\pi r^2-\\frac{1}{2}r^2\\cos^{-1}\\left(\\frac{2r^2-a^2}{2r^2} \\right)+\\frac{a\\sqrt{4r^2-a^2}}{4}$$\n\nHence, solving for $A$, we obtain:\n\n$$\\displaystyle A=\\sin^{-1}\\left(\\frac{2r^2\\left(2\\pi-\\cos^{-1}\\left(\\dfrac{2r^2-a^2}{2r^2} \\right) \\right)+a\\sqrt{4r^2-a^2}}{2bc} \\right)$$" ]

[ null, "https://mathhelpboards.com/data/attachments/3/3704-174e0f9dc430010cfaa87c812c93c208.jpg", null ]

[ "Module java.desktop\nPackage java.awt.geom\n\n## Class Area\n\n• All Implemented Interfaces:\n`Shape`, `Cloneable`\n\n```public class Area\nextends Object\nimplements Shape, Cloneable```\nAn `Area` object stores and manipulates a resolution-independent description of an enclosed area of 2-dimensional space. `Area` objects can be transformed and can perform various Constructive Area Geometry (CAG) operations when combined with other `Area` objects. The CAG operations include area `addition`, `subtraction`, `intersection`, and `exclusive or`. See the linked method documentation for examples of the various operations.\n\nThe `Area` class implements the `Shape` interface and provides full support for all of its hit-testing and path iteration facilities, but an `Area` is more specific than a generalized path in a number of ways:\n\n• Only closed paths and sub-paths are stored. `Area` objects constructed from unclosed paths are implicitly closed during construction as if those paths had been filled by the `Graphics2D.fill` method.\n• The interiors of the individual stored sub-paths are all non-empty and non-overlapping. Paths are decomposed during construction into separate component non-overlapping parts, empty pieces of the path are discarded, and then these non-empty and non-overlapping properties are maintained through all subsequent CAG operations. Outlines of different component sub-paths may touch each other, as long as they do not cross so that their enclosed areas overlap.\n• The geometry of the path describing the outline of the `Area` resembles the path from which it was constructed only in that it describes the same enclosed 2-dimensional area, but may use entirely different types and ordering of the path segments to do so.\nInteresting issues which are not always obvious when using the `Area` include:\n• Creating an `Area` from an unclosed (open) `Shape` results in a closed outline in the `Area` object.\n• Creating an `Area` from a `Shape` which encloses no area (even when \"closed\") produces an empty `Area`. A common example of this issue is that producing an `Area` from a line will be empty since the line encloses no area. An empty `Area` will iterate no geometry in its `PathIterator` objects.\n• A self-intersecting `Shape` may be split into two (or more) sub-paths each enclosing one of the non-intersecting portions of the original path.\n• An `Area` may take more path segments to describe the same geometry even when the original outline is simple and obvious. The analysis that the `Area` class must perform on the path may not reflect the same concepts of \"simple and obvious\" as a human being perceives.\nSince:\n1.2\n• ### Constructor Summary\n\nConstructors\nConstructor Description\n`Area()`\nDefault constructor which creates an empty area.\n`Area​(Shape s)`\nThe `Area` class creates an area geometry from the specified `Shape` object.\n• ### Method Summary\n\nAll Methods\nModifier and Type Method Description\n`void` `add​(Area rhs)`\nAdds the shape of the specified `Area` to the shape of this `Area`.\n`Object` `clone()`\nReturns an exact copy of this `Area` object.\n`boolean` ```contains​(double x, double y)```\nTests if the specified coordinates are inside the boundary of the `Shape`, as described by the definition of insideness.\n`boolean` ```contains​(double x, double y, double w, double h)```\nTests if the interior of the `Shape` entirely contains the specified rectangular area.\n`boolean` `contains​(Point2D p)`\nTests if a specified `Point2D` is inside the boundary of the `Shape`, as described by the definition of insideness.\n`boolean` `contains​(Rectangle2D r)`\nTests if the interior of the `Shape` entirely contains the specified `Rectangle2D`.\n`Area` `createTransformedArea​(AffineTransform t)`\nCreates a new `Area` object that contains the same geometry as this `Area` transformed by the specified `AffineTransform`.\n`boolean` `equals​(Area other)`\nTests whether the geometries of the two `Area` objects are equal.\n`void` `exclusiveOr​(Area rhs)`\nSets the shape of this `Area` to be the combined area of its current shape and the shape of the specified `Area`, minus their intersection.\n`Rectangle` `getBounds()`\nReturns a bounding `Rectangle` that completely encloses this `Area`.\n`Rectangle2D` `getBounds2D()`\nReturns a high precision bounding `Rectangle2D` that completely encloses this `Area`.\n`PathIterator` `getPathIterator​(AffineTransform at)`\nCreates a `PathIterator` for the outline of this `Area` object.\n`PathIterator` ```getPathIterator​(AffineTransform at, double flatness)```\nCreates a `PathIterator` for the flattened outline of this `Area` object.\n`void` `intersect​(Area rhs)`\nSets the shape of this `Area` to the intersection of its current shape and the shape of the specified `Area`.\n`boolean` ```intersects​(double x, double y, double w, double h)```\nTests if the interior of the `Shape` intersects the interior of a specified rectangular area.\n`boolean` `intersects​(Rectangle2D r)`\nTests if the interior of the `Shape` intersects the interior of a specified `Rectangle2D`.\n`boolean` `isEmpty()`\nTests whether this `Area` object encloses any area.\n`boolean` `isPolygonal()`\nTests whether this `Area` consists entirely of straight edged polygonal geometry.\n`boolean` `isRectangular()`\nTests whether this `Area` is rectangular in shape.\n`boolean` `isSingular()`\nTests whether this `Area` is comprised of a single closed subpath.\n`void` `reset()`\nRemoves all of the geometry from this `Area` and restores it to an empty area.\n`void` `subtract​(Area rhs)`\nSubtracts the shape of the specified `Area` from the shape of this `Area`.\n`void` `transform​(AffineTransform t)`\nTransforms the geometry of this `Area` using the specified `AffineTransform`.\n• ### Methods declared in class java.lang.Object\n\n`equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait`\n• ### Constructor Detail\n\n• #### Area\n\n`public Area()`\nDefault constructor which creates an empty area.\nSince:\n1.2\n• #### Area\n\n`public Area​(Shape s)`\nThe `Area` class creates an area geometry from the specified `Shape` object. The geometry is explicitly closed, if the `Shape` is not already closed. The fill rule (even-odd or winding) specified by the geometry of the `Shape` is used to determine the resulting enclosed area.\nParameters:\n`s` - the `Shape` from which the area is constructed\nThrows:\n`NullPointerException` - if `s` is null\nSince:\n1.2\n• ### Method Detail\n\n`public void add​(Area rhs)`\nAdds the shape of the specified `Area` to the shape of this `Area`. The resulting shape of this `Area` will include the union of both shapes, or all areas that were contained in either this or the specified `Area`.\n``` // Example:\nArea a1 = new Area([triangle 0,0 => 8,0 => 0,8]);\nArea a2 = new Area([triangle 0,0 => 8,0 => 8,8]);\n\na1(before) + a2 = a1(after)\n\n################ ################ ################\n############## ############## ################\n############ ############ ################\n########## ########## ################\n######## ######## ################\n###### ###### ###### ######\n#### #### #### ####\n## ## ## ##\n```\nParameters:\n`rhs` - the `Area` to be added to the current shape\nThrows:\n`NullPointerException` - if `rhs` is null\nSince:\n1.2\n• #### subtract\n\n`public void subtract​(Area rhs)`\nSubtracts the shape of the specified `Area` from the shape of this `Area`. The resulting shape of this `Area` will include areas that were contained only in this `Area` and not in the specified `Area`.\n``` // Example:\nArea a1 = new Area([triangle 0,0 => 8,0 => 0,8]);\nArea a2 = new Area([triangle 0,0 => 8,0 => 8,8]);\na1.subtract(a2);\n\na1(before) - a2 = a1(after)\n\n################ ################\n############## ############## ##\n############ ############ ####\n########## ########## ######\n######## ######## ########\n###### ###### ######\n#### #### ####\n## ## ##\n```\nParameters:\n`rhs` - the `Area` to be subtracted from the current shape\nThrows:\n`NullPointerException` - if `rhs` is null\nSince:\n1.2\n• #### intersect\n\n`public void intersect​(Area rhs)`\nSets the shape of this `Area` to the intersection of its current shape and the shape of the specified `Area`. The resulting shape of this `Area` will include only areas that were contained in both this `Area` and also in the specified `Area`.\n``` // Example:\nArea a1 = new Area([triangle 0,0 => 8,0 => 0,8]);\nArea a2 = new Area([triangle 0,0 => 8,0 => 8,8]);\na1.intersect(a2);\n\na1(before) intersect a2 = a1(after)\n\n################ ################ ################\n############## ############## ############\n############ ############ ########\n########## ########## ####\n######## ########\n###### ######\n#### ####\n## ##\n```\nParameters:\n`rhs` - the `Area` to be intersected with this `Area`\nThrows:\n`NullPointerException` - if `rhs` is null\nSince:\n1.2\n• #### exclusiveOr\n\n`public void exclusiveOr​(Area rhs)`\nSets the shape of this `Area` to be the combined area of its current shape and the shape of the specified `Area`, minus their intersection. The resulting shape of this `Area` will include only areas that were contained in either this `Area` or in the specified `Area`, but not in both.\n``` // Example:\nArea a1 = new Area([triangle 0,0 => 8,0 => 0,8]);\nArea a2 = new Area([triangle 0,0 => 8,0 => 8,8]);\na1.exclusiveOr(a2);\n\na1(before) xor a2 = a1(after)\n\n################ ################\n############## ############## ## ##\n############ ############ #### ####\n########## ########## ###### ######\n######## ######## ################\n###### ###### ###### ######\n#### #### #### ####\n## ## ## ##\n```\nParameters:\n`rhs` - the `Area` to be exclusive ORed with this `Area`.\nThrows:\n`NullPointerException` - if `rhs` is null\nSince:\n1.2\n• #### reset\n\n`public void reset()`\nRemoves all of the geometry from this `Area` and restores it to an empty area.\nSince:\n1.2\n• #### isEmpty\n\n`public boolean isEmpty()`\nTests whether this `Area` object encloses any area.\nReturns:\n`true` if this `Area` object represents an empty area; `false` otherwise.\nSince:\n1.2\n• #### isPolygonal\n\n`public boolean isPolygonal()`\nTests whether this `Area` consists entirely of straight edged polygonal geometry.\nReturns:\n`true` if the geometry of this `Area` consists entirely of line segments; `false` otherwise.\nSince:\n1.2\n• #### isRectangular\n\n`public boolean isRectangular()`\nTests whether this `Area` is rectangular in shape.\nReturns:\n`true` if the geometry of this `Area` is rectangular in shape; `false` otherwise.\nSince:\n1.2\n• #### isSingular\n\n`public boolean isSingular()`\nTests whether this `Area` is comprised of a single closed subpath. This method returns `true` if the path contains 0 or 1 subpaths, or `false` if the path contains more than 1 subpath. The subpaths are counted by the number of `SEG_MOVETO` segments that appear in the path.\nReturns:\n`true` if the `Area` is comprised of a single basic geometry; `false` otherwise.\nSince:\n1.2\n• #### getBounds2D\n\n`public Rectangle2D getBounds2D()`\nReturns a high precision bounding `Rectangle2D` that completely encloses this `Area`.\n\nThe Area class will attempt to return the tightest bounding box possible for the Shape. The bounding box will not be padded to include the control points of curves in the outline of the Shape, but should tightly fit the actual geometry of the outline itself.\n\nSpecified by:\n`getBounds2D` in interface `Shape`\nReturns:\nthe bounding `Rectangle2D` for the `Area`.\nSince:\n1.2\n`Shape.getBounds()`\n• #### getBounds\n\n`public Rectangle getBounds()`\nReturns a bounding `Rectangle` that completely encloses this `Area`.\n\nThe Area class will attempt to return the tightest bounding box possible for the Shape. The bounding box will not be padded to include the control points of curves in the outline of the Shape, but should tightly fit the actual geometry of the outline itself. Since the returned object represents the bounding box with integers, the bounding box can only be as tight as the nearest integer coordinates that encompass the geometry of the Shape.\n\nSpecified by:\n`getBounds` in interface `Shape`\nReturns:\nthe bounding `Rectangle` for the `Area`.\nSince:\n1.2\n`Shape.getBounds2D()`\n• #### clone\n\n`public Object clone()`\nReturns an exact copy of this `Area` object.\nOverrides:\n`clone` in class `Object`\nReturns:\nCreated clone object\nSince:\n1.2\n`Cloneable`\n• #### equals\n\n`public boolean equals​(Area other)`\nTests whether the geometries of the two `Area` objects are equal. This method will return false if the argument is null.\nParameters:\n`other` - the `Area` to be compared to this `Area`\nReturns:\n`true` if the two geometries are equal; `false` otherwise.\nSince:\n1.2\n• #### transform\n\n`public void transform​(AffineTransform t)`\nTransforms the geometry of this `Area` using the specified `AffineTransform`. The geometry is transformed in place, which permanently changes the enclosed area defined by this object.\nParameters:\n`t` - the transformation used to transform the area\nThrows:\n`NullPointerException` - if `t` is null\nSince:\n1.2\n• #### createTransformedArea\n\n`public Area createTransformedArea​(AffineTransform t)`\nCreates a new `Area` object that contains the same geometry as this `Area` transformed by the specified `AffineTransform`. This `Area` object is unchanged.\nParameters:\n`t` - the specified `AffineTransform` used to transform the new `Area`\nReturns:\na new `Area` object representing the transformed geometry.\nThrows:\n`NullPointerException` - if `t` is null\nSince:\n1.2\n• #### contains\n\n```public boolean contains​(double x,\ndouble y)```\nTests if the specified coordinates are inside the boundary of the `Shape`, as described by the definition of insideness.\nSpecified by:\n`contains` in interface `Shape`\nParameters:\n`x` - the specified X coordinate to be tested\n`y` - the specified Y coordinate to be tested\nReturns:\n`true` if the specified coordinates are inside the `Shape` boundary; `false` otherwise.\nSince:\n1.2\n• #### contains\n\n`public boolean contains​(Point2D p)`\nTests if a specified `Point2D` is inside the boundary of the `Shape`, as described by the definition of insideness.\nSpecified by:\n`contains` in interface `Shape`\nParameters:\n`p` - the specified `Point2D` to be tested\nReturns:\n`true` if the specified `Point2D` is inside the boundary of the `Shape`; `false` otherwise.\nSince:\n1.2\n• #### contains\n\n```public boolean contains​(double x,\ndouble y,\ndouble w,\ndouble h)```\nTests if the interior of the `Shape` entirely contains the specified rectangular area. All coordinates that lie inside the rectangular area must lie within the `Shape` for the entire rectangular area to be considered contained within the `Shape`.\n\nThe `Shape.contains()` method allows a `Shape` implementation to conservatively return `false` when:\n\n• the `intersect` method returns `true` and\n• the calculations to determine whether or not the `Shape` entirely contains the rectangular area are prohibitively expensive.\nThis means that for some `Shapes` this method might return `false` even though the `Shape` contains the rectangular area. The `Area` class performs more accurate geometric computations than most `Shape` objects and therefore can be used if a more precise answer is required.\nSpecified by:\n`contains` in interface `Shape`\nParameters:\n`x` - the X coordinate of the upper-left corner of the specified rectangular area\n`y` - the Y coordinate of the upper-left corner of the specified rectangular area\n`w` - the width of the specified rectangular area\n`h` - the height of the specified rectangular area\nReturns:\n`true` if the interior of the `Shape` entirely contains the specified rectangular area; `false` otherwise or, if the `Shape` contains the rectangular area and the `intersects` method returns `true` and the containment calculations would be too expensive to perform.\nSince:\n1.2\n`Area`, `Shape.intersects(double, double, double, double)`\n• #### contains\n\n`public boolean contains​(Rectangle2D r)`\nTests if the interior of the `Shape` entirely contains the specified `Rectangle2D`. The `Shape.contains()` method allows a `Shape` implementation to conservatively return `false` when:\n• the `intersect` method returns `true` and\n• the calculations to determine whether or not the `Shape` entirely contains the `Rectangle2D` are prohibitively expensive.\nThis means that for some `Shapes` this method might return `false` even though the `Shape` contains the `Rectangle2D`. The `Area` class performs more accurate geometric computations than most `Shape` objects and therefore can be used if a more precise answer is required.\nSpecified by:\n`contains` in interface `Shape`\nParameters:\n`r` - The specified `Rectangle2D`\nReturns:\n`true` if the interior of the `Shape` entirely contains the `Rectangle2D`; `false` otherwise or, if the `Shape` contains the `Rectangle2D` and the `intersects` method returns `true` and the containment calculations would be too expensive to perform.\nSince:\n1.2\n`Shape.contains(double, double, double, double)`\n• #### intersects\n\n```public boolean intersects​(double x,\ndouble y,\ndouble w,\ndouble h)```\nTests if the interior of the `Shape` intersects the interior of a specified rectangular area. The rectangular area is considered to intersect the `Shape` if any point is contained in both the interior of the `Shape` and the specified rectangular area.\n\nThe `Shape.intersects()` method allows a `Shape` implementation to conservatively return `true` when:\n\n• there is a high probability that the rectangular area and the `Shape` intersect, but\n• the calculations to accurately determine this intersection are prohibitively expensive.\nThis means that for some `Shapes` this method might return `true` even though the rectangular area does not intersect the `Shape`. The `Area` class performs more accurate computations of geometric intersection than most `Shape` objects and therefore can be used if a more precise answer is required.\nSpecified by:\n`intersects` in interface `Shape`\nParameters:\n`x` - the X coordinate of the upper-left corner of the specified rectangular area\n`y` - the Y coordinate of the upper-left corner of the specified rectangular area\n`w` - the width of the specified rectangular area\n`h` - the height of the specified rectangular area\nReturns:\n`true` if the interior of the `Shape` and the interior of the rectangular area intersect, or are both highly likely to intersect and intersection calculations would be too expensive to perform; `false` otherwise.\nSince:\n1.2\n`Area`\n• #### intersects\n\n`public boolean intersects​(Rectangle2D r)`\nTests if the interior of the `Shape` intersects the interior of a specified `Rectangle2D`. The `Shape.intersects()` method allows a `Shape` implementation to conservatively return `true` when:\n• there is a high probability that the `Rectangle2D` and the `Shape` intersect, but\n• the calculations to accurately determine this intersection are prohibitively expensive.\nThis means that for some `Shapes` this method might return `true` even though the `Rectangle2D` does not intersect the `Shape`. The `Area` class performs more accurate computations of geometric intersection than most `Shape` objects and therefore can be used if a more precise answer is required.\nSpecified by:\n`intersects` in interface `Shape`\nParameters:\n`r` - the specified `Rectangle2D`\nReturns:\n`true` if the interior of the `Shape` and the interior of the specified `Rectangle2D` intersect, or are both highly likely to intersect and intersection calculations would be too expensive to perform; `false` otherwise.\nSince:\n1.2\n`Shape.intersects(double, double, double, double)`\n• #### getPathIterator\n\n`public PathIterator getPathIterator​(AffineTransform at)`\nCreates a `PathIterator` for the outline of this `Area` object. This `Area` object is unchanged.\nSpecified by:\n`getPathIterator` in interface `Shape`\nParameters:\n`at` - an optional `AffineTransform` to be applied to the coordinates as they are returned in the iteration, or `null` if untransformed coordinates are desired\nReturns:\nthe `PathIterator` object that returns the geometry of the outline of this `Area`, one segment at a time.\nSince:\n1.2\n• #### getPathIterator\n\n```public PathIterator getPathIterator​(AffineTransform at,\ndouble flatness)```\nCreates a `PathIterator` for the flattened outline of this `Area` object. Only uncurved path segments represented by the SEG_MOVETO, SEG_LINETO, and SEG_CLOSE point types are returned by the iterator. This `Area` object is unchanged.\nSpecified by:\n`getPathIterator` in interface `Shape`\nParameters:\n`at` - an optional `AffineTransform` to be applied to the coordinates as they are returned in the iteration, or `null` if untransformed coordinates are desired\n`flatness` - the maximum amount that the control points for a given curve can vary from colinear before a subdivided curve is replaced by a straight line connecting the end points\nReturns:\nthe `PathIterator` object that returns the geometry of the outline of this `Area`, one segment at a time.\nSince:\n1.2" ]

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[ "8.2\n\n#### 4.1Equality\n\nEquality is the concept of whether two values are “the same.” Racket supports a few different kinds of equality by default, though equal? is preferred for most use cases.\n\n procedure(equal? v1 v2) → boolean? v1 : any/c v2 : any/c\nTwo values are equal? if and only if they are eqv?, unless otherwise specified for a particular datatype.\n\nDatatypes with further specification of equal? include strings, byte strings, pairs, mutable pairs, vectors, boxes, hash tables, and inspectable structures. In the last six cases, equality is recursively defined; if both v1 and v2 contain reference cycles, they are equal when the infinite unfoldings of the values would be equal. See also gen:equal+hash and prop:impersonator-of.\n\nExamples:\n > (equal? 'yes 'yes) #t > (equal? 'yes 'no) #f > (equal? (* 6 7) 42) #t > (equal? (expt 2 100) (expt 2 100)) #t > (equal? 2 2.0) #f > (let ([v (mcons 1 2)]) (equal? v v)) #t > (equal? (mcons 1 2) (mcons 1 2)) #t > (equal? (integer->char 955) (integer->char 955)) #t > (equal? (make-string 3 #\\z) (make-string 3 #\\z)) #t > (equal? #t #t) #t\n\n procedure(eqv? v1 v2) → boolean? v1 : any/c v2 : any/c\nTwo values are eqv? if and only if they are eq?, unless otherwise specified for a particular datatype.\n\nThe number and character datatypes are the only ones for which eqv? differs from eq?. Two numbers are eqv? when they have the same exactness, precision, and are both equal and non-zero, both +0.0, both +0.0f0, both -0.0, both -0.0f0, both +nan.0, or both +nan.fconsidering real and imaginary components separately in the case of complex numbers. Two characters are eqv? when their char->integer results are equal.\n\nGenerally, eqv? is identical to equal? except that the former cannot recursively compare the contents of compound data types (such as lists and structs) and cannot be customized by user-defined data types. The use of eqv? is lightly discouraged in favor of equal?.\n\nExamples:\n > (eqv? 'yes 'yes) #t > (eqv? 'yes 'no) #f > (eqv? (* 6 7) 42) #t > (eqv? (expt 2 100) (expt 2 100)) #t > (eqv? 2 2.0) #f > (let ([v (mcons 1 2)]) (eqv? v v)) #t > (eqv? (mcons 1 2) (mcons 1 2)) #f > (eqv? (integer->char 955) (integer->char 955)) #t > (eqv? (make-string 3 #\\z) (make-string 3 #\\z)) #f > (eqv? #t #t) #t\n\n procedure(eq? v1 v2) → boolean? v1 : any/c v2 : any/c\nReturn #t if v1 and v2 refer to the same object, #f otherwise. As a special case among numbers, two fixnums that are = are also the same according to eq?. See also Object Identity and Comparisons.\n\nExamples:\n > (eq? 'yes 'yes) #t > (eq? 'yes 'no) #f > (eq? (* 6 7) 42) #t > (eq? (expt 2 100) (expt 2 100)) #f > (eq? 2 2.0) #f > (let ([v (mcons 1 2)]) (eq? v v)) #t > (eq? (mcons 1 2) (mcons 1 2)) #f > (eq? (integer->char 955) (integer->char 955)) #t > (eq? (make-string 3 #\\z) (make-string 3 #\\z)) #f > (eq? #t #t) #t\n\n procedure(equal?/recur v1 v2 recur-proc) → boolean? v1 : any/c v2 : any/c recur-proc : (any/c any/c -> any/c)\nLike equal?, but using recur-proc for recursive comparisons (which means that reference cycles are not handled automatically). Non-#f results from recur-proc are converted to #t before being returned by equal?/recur.\n\nExamples:\n> (equal?/recur 1 1 (lambda (a b) #f))\n\n#t\n\n> (equal?/recur '(1) '(1) (lambda (a b) #f))\n\n#f\n\n > (equal?/recur '#(1 1 1) '#(1 1.2 3/4) (lambda (a b) (<= (abs (- a b)) 0.25)))\n\n#t\n\n##### 4.1.1Object Identity and Comparisons\n\nThe eq? operator compares two values, returning #t when the values refer to the same object. This form of equality is suitable for comparing objects that support imperative update (e.g., to determine that the effect of modifying an object through one reference is visible through another reference). Also, an eq? test evaluates quickly, and eq?-based hashing is more lightweight than equal?-based hashing in hash tables.\n\nIn some cases, however, eq? is unsuitable as a comparison operator, because the generation of objects is not clearly defined. In particular, two applications of + to the same two exact integers may or may not produce results that are eq?, although the results are always equal?. Similarly, evaluation of a lambda form typically generates a new procedure object, but it may re-use a procedure object previously generated by the same source lambda form.\n\nThe behavior of a datatype with respect to eq? is generally specified with the datatype and its associated procedures.\n\n##### 4.1.2Equality and Hashing\n\nAll comparable values have at least one hash code an arbitrary integer (more specifically a fixnum) computed by applying a hash function to the value. The defining property of these hash codes is that equal values have equal hash codes. Note that the reverse is not true: two unequal values can still have equal hash codes. Hash codes are useful for various indexing and comparison operations, especially in the implementation of hash tables. See Hash Tables for more information.\n\n procedure v : any/c\nReturns a hash code consistent with equal?. For any two calls with equal? values, the returned number is the same. A hash code is computed even when v contains a cycle through pairs, vectors, boxes, and/or inspectable structure fields. Additionally, user-defined data types can customize how this hash code is computed by implementing gen:equal+hash.\n\nFor any v that could be produced by read, if v2 is produced by read for the same input characters, the (equal-hash-code v) is the same as (equal-hash-code v2) even if v and v2 do not exist at the same time (and therefore could not be compared by calling equal?).\n\nChanged in version 6.4.0.12 of package base: Strengthened guarantee for readable values.\n\n procedure v : any/c\nLike equal-hash-code, but computes a secondary hash code suitable for use in double hashing.\n\n procedure v : any/c\nReturns a hash code consistent with eq?. For any two calls with eq? values, the returned number is the same.\n\nEqual fixnums are always eq?.\n\n procedure v : any/c\nReturns a hash code consistent with eqv?. For any two calls with eqv? values, the returned number is the same.\n\n##### 4.1.3Implementing Equality for Custom Types\n\n value\nA generic interface (see Generic Interfaces) for types that can be compared for equality using equal?. The following methods must be implemented:\n\n• equal-proc : (-> any/c any/c (-> any/c any/c boolean?) any/c) tests whether the first two arguments are equal, where both values are instances of the structure type to which the generic interface is associated (or a subtype of the structure type).\n\nThe third argument is an equal? predicate to use for recursive equality checks; use the given predicate instead of equal? to ensure that data cycles are handled properly and to work with equal?/recur (but beware that an arbitrary function can be provided to equal?/recur for recursive checks, which means that arguments provided to the predicate might be exposed to arbitrary code).\n\nThe equal-proc is called for a pair of structures only when they are not eq?, and only when they both have a gen:equal+hash value inherited from the same structure type. With this strategy, the order in which equal? receives two structures does not matter. It also means that, by default, a structure sub-type inherits the equality predicate of its parent, if any.\n\n• hash-proc : (-> any/c (-> any/c exact-integer?) exact-integer?) computes a hash code for the given structure, like equal-hash-code. The first argument is an instance of the structure type (or one of its subtypes) to which the generic interface is associated.\n\nThe second argument is an equal-hash-code-like procedure to use for recursive hash-code computation; use the given procedure instead of equal-hash-code to ensure that data cycles are handled properly.\n\nAlthough the result of hash-proc can be any exact integer, it will be truncated for most purposes to a fixnum (e.g., for the result of equal-hash-code). Roughly, truncation uses bitwise-and to take the lower bits of the number. Thus, variation in the hash-code computation should be reflected in the fixnum-compatible bits of hash-proc’s result. Consumers of a hash code are expected to use variation within the fixnum range appropriately, and producers are not responsible to reflect variation in hash codes across the full range of bits that fit within a fixnum.\n\n• hash2-proc : (-> any/c (-> any/c exact-integer?) exact-integer?) computes a secondary hash code for the given structure. This procedure is like hash-proc, but analogous to equal-secondary-hash-code.\n\nTake care to ensure that hash-proc and hash2-proc are consistent with equal-proc. Specifically, hash-proc and hash2-proc should produce the same value for any two structures for which equal-proc produces a true value.\n\nWhen a structure type has no gen:equal+hash implementation, then transparent structures (i.e., structures with an inspector that is controlled by the current inspector) are equal? when they are instances of the same structure type (not counting sub-types), and when they have equal? field values. For transparent structures, equal-hash-code and equal-secondary-hash-code derive hash code using the field values. For opaque structure types, equal? is the same as eq?, and equal-hash-code and equal-secondary-hash-code results are based only on eq-hash-code. If a structure has a prop:impersonator-of property, then the prop:impersonator-of property takes precedence over gen:equal+hash if the property value’s procedure returns a non-#f value when applied to the structure.\n\nExamples:\n (define (farm=? farm1 farm2 recursive-equal?) (and (= (farm-apples farm1) (farm-apples farm2)) (= (farm-oranges farm1) (farm-oranges farm2)) (= (farm-sheep farm1) (farm-sheep farm2)))) (define (farm-hash-code farm recursive-equal-hash) (+ (* 10000 (farm-apples farm)) (* 100 (farm-oranges farm)) (* 1 (farm-sheep farm)))) (define (farm-secondary-hash-code farm recursive-equal-hash) (+ (* 10000 (farm-sheep farm)) (* 100 (farm-apples farm)) (* 1 (farm-oranges farm)))) (struct farm (apples oranges sheep) #:methods gen:equal+hash [(define equal-proc farm=?) (define hash-proc  farm-hash-code) (define hash2-proc farm-secondary-hash-code)]) (define eastern-farm (farm 5 2 20)) (define western-farm (farm 18 6 14)) (define northern-farm (farm 5 20 20)) (define southern-farm (farm 18 6 14))\n\n> (equal? eastern-farm western-farm)\n\n#f\n\n> (equal? eastern-farm northern-farm)\n\n#f\n\n> (equal? western-farm southern-farm)\n\n#t\n\n value\nA structure type property (see Structure Type Properties) that supplies an equality predicate and hashing functions for a structure type. Using the prop:equal+hash property is discouraged; the gen:equal+hash generic interface should be used instead. A prop:equal+hash property value is a list of three procedures that correspond to the methods of gen:equal+hash:" ]

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Essay Writing Services – Essay Editing and Proofreading\n\n2021年11月30日\n\nEssay writing solutions are those hired by universities, colleges and schools to compose essays for students they will be grading. Sometimes these composing services will also be hired by instructors to 続きを読む\n\n\n1 2 3 4 5 29\n\n##", null, "お電話でのご予約・お問い合わせはこちら！\n\nご心配なことやご不明な点など、専門スタッフへお気軽にご相談ください。", null, "" ]

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[ "# Treatment of Trailing Zeros\n\nAll binary not null columns are padded with zeros to the full width of the column. Trailing zeros are truncated in all varbinary data and in binary null columns, since columns that accept null values must be treated as variable-length columns.\n\nThe following example creates a table with all four variations of binary and varbinary datatypes, NULL, and NOT NULL. The same data is inserted in all four columns and is padded or truncated according to the datatype of the column.\n\n```create table zeros (bnot binary(5) not null,\nbnull binary(5) null,\nvnot varbinary(5) not null,\nvnull varbinary(5) null)\n\ninsert zeros values (0x12345000, 0x12345000, 0x12345000, 0x12345000)\ninsert zeros values (0x123, 0x123, 0x123, 0x123)\n\nselect * from zeros```\n```bnot             bnull        vnot        vnull\n------------     ---------    ----------  ---------\n0x1234500000     0x123450     0x123450    0x123450\n0x0123000000     0x0123       0x0123      0x0123```\n\nBecause each byte of storage holds 2 binary digits, the SAP ASE server expects binary entries to consist of the characters “0x” followed by an even number of digits. When the “0x” is followed by an odd number of digits, the SAP ASE server assumes that you omitted the leading 0 and adds it for you.\n\nInput values “0x00” and “0x0” are stored as “0x00” in variable-length binary columns (binary null, image, and varbinary columns). In fixed-length binary (binary not null) columns, the value is padded with zeros to the full length of the field:\n\n```insert zeros values (0x0, 0x0,0x0, 0x0)\nselect * from zeros where bnot = 0x00```\n```bnot           bnull       vnot        vnull\n----------     ------      -----       ------------\n0x0000000000   0x00        0x00        0x00```\n\nIf the input value does not include the “0x”, the SAP ASE server assumes that the value is an ASCII value and converts it. For example:\n\n```create table sample (col_a binary(8))\n\ninsert sample values (’002710000000ae1b’)\n\nselect * from sample```\n```col_a\n------------------\n0x3030323731303030```" ]

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[ "# Java Math.random() Method with Examples\n\nOct 12, 2019 · 3 mins read", null, "The java.lang.Math class that comes bundled with Java contains various methods for performing basic numeric operations such as the elementary exponential, logarithm, square root, and trigonometric functions.\n\nMath.random() method is part of the Math class. It is used for generating random values between 0.0 and 0.1. The random values are chosen pseudorandomly. This means the algorithm relies solely on mathematical equations to generate random numbers. In contrast, true random number generators use unpredictable and external physical criteria (e.g. atmospheric data) to generate random numbers. If you’re looking for true random number generation in Java, take a look at javax.crypto.SecureRandom. It uses your system’s source for randomness (e.g. `/dev/urandom`) to generate random numbers. I digress, but if you’re interested in learning more about pseudorandom vs true random, here’s a good blog post by Bo Allen.\n\nLet’s look at the method signature of the Math.random() method:\n\n``````public static double random()\n``````\n\nNote that just like all other methods of the Math class, Math.random() is a `static` method so you can call it directly on the Math class without needing an object. It returns a value of type `double`. Another thing to keep in mind is that this method creates a new object of type Random which it uses internally to generate random numbers. The same random generator object is used for all calls to this method from anywhere in your code.\n\n``````Math.random(); // e.g. 0.5674616450812238\n``````\n\n## Example: Random Numbers Within a Range\n\nA common use case is finding random numbers between a range. Let’s see an example where we’ll generate a random number within 1 to 10.\n\n``````public class MathEx {\npublic static void main(String[] args) {\nrandom();\n}\n\nprivate static void random() {\n// Generate random number within 1 to 10 (inclusive) 10 times\nfor (int i=0; i<10; i++) {\nSystem.out.println(\n(int)randomWithinRange(1,10)\n);\n}\n\n}\nprivate static double randomWithinRange(int min, int max) {\nassert min < max;\n\nint range = (max - min) + 1;\nreturn ((Math.random() * range) + min);\n}\n}\n``````\n\nHere’s output from the code above generating random number between 1 and 10 when run 10 times.\n\nOutput\n\n``````6\n9\n3\n4\n7\n3\n10\n7\n2\n3\n``````\n\nHere’s a screenshoot of the code above if you’re on mobile and having trouble reading the code above.", null, "### Special cases\n\n• This method is properly synchronized to allow correct use by more than one thread. However, if many threads need to generate pseudorandom numbers at a great rate, it may reduce contention for each thread to have its own pseudorandom-number generator.\n\nIf you need more detailed information, please see Javadocs. It’s worth noting that the `Math` class in Java contains several other useful methods for arithmetic, logarithms and trignometric operations.\n\n### Java Versions Tested\n\nThe examples in this post were compiled and executed using Java 8. Everything contained in this post is accurate to all versions up to Java 13 (which is the latest version.)\n\n#### You May Also Enjoy\n\n• ##### Compare Server-side Rendering vs Client-side Rendering\n\nIf you like this post, please share using the buttons above. It will help CodeAhoy grow and add new content. Thank you!" ]

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[ "Operations research\n\n# Simplex Method\n\nSimplex method is a solving problem analytic method of linear programming, able to resolve complex models than the resolved through graphic method.\n\nSimplex method is an iterative method that improves the solution on each step. The mathematical reason of this improvement is that the method consists in walking through a neighbor vertex in such a way that raises or decreases (according to the context of the objective’s function, whether it is maximize or minimize), given that the number of vertex that a solution polyhedron is finite, there always be a solution.\n\nThis popular method was created in 1947 by the American George Bernard Dantzig and the Russian Leonid Vasilyevich Kantorovich, with the purpose of creating an algorithm able to fix problems of “m” restrictions and “n” variables.\n\n## ¿What is an identity matrix?\n\nA matrix can be defined as a rectangular arrangement of elements, (o finite list of elements), which can be real numbers or complex numbers, disposed in rows and columns.\n\nThe Identity matrix or a sometimes ambiguously called a is square (it has the same number of rows, as it has the same number of columns), of order n that has all the diagonal elements equal to one (1) and all other components equal to zero (0), it’s called Identical matrix or Order “n” Identity, and it is denoted by:", null, "The importance of this matrix theory in the Simplex Method is fundamental, given that the algorithm is based on said theory for itself problem resolution.\n\n## Important considerations when using the simplex method\n\n### Slack and Surplus Variables\n\nThe Simplex Method works based on equations and initial restrictions that are model trough lineal programing are not,, for this you have to convert this inequalities in equations using some type of variables called: Slack and Surplus Variables that are related with the resource on which the restrictions makes reference, and that on the final tabulate represents the “Slack or Surplus”, which many resolution of operation programs make reference, this variables get a great added value in the analysis of sensitivity and play a key role in the creation of the Simplex base Identity Matrix.\n\nThese variables are usually represented by the letter “S”, they add if the restriction is «<= », and they subtract if the restriction is «>=».\n\nFor example:\n\n### Artificial variable / “M” method:\n\nAn artificial variable is a math trick to turn «>=» inequalities, into equations, or when equalities appear in the original problem to resolve, the main characteristic of this variables is that they should not be part of the solution, given that they do not represent any resource. The main objective of this variables is the formation of the identity matrix.\n\nThese variables are represented by the letter “A”, constraints are always added, their coefficient is M (this is why it is called M Method, where M means a number way too big, and very less attractive to the objective’s function), and the sign in the objective’s function goes counter-wise of itself, that is to say, in problems of maximization it sign is minus (-), and on minimization problems it sign is plus (+), repeat with the objective that it’s value on the solution is zero (0).\n\n## Step by step: Simplex Method\n\n### The problem\n\nCompany SAMAN Limited. Dedicated to the manufacture of furniture, it has expanded its production in two more lines. Thus currently manufactures tables, chairs, beds and libraries. Each table requires 2 rectangular pieces of 8 pins each, and two square pieces of 4 pins each. Each chair requires 1 rectangular piece of 8 pins and 2 square pieces of 4 pins each; each bed requires 1 square piece of 8 pins, 1 square piece of 4 pins and 2 trapezoidal bases of 2 pins each; and finally each library requires 2 rectangular pieces of 8 pins each, 2 trapezoidal bases of 2 pins each and 4 rectangular pieces of 2 pins each.\n\nEach table’s manufacture cost is \\$10000, and are sold in \\$30000; each chair’s manufacture cost is \\$8000, and are sold in \\$28000; each bed’s manufacture cost is \\$20000, and are sold in \\$40000; each library’s manufacture cost is \\$40000, and are sold in \\$60000. The main goal of the Company is to maximize profits.", null, "## Step 1: Modeling trough lineal programing:\n\nVariables:\n\nX1 = Quantity of tables to manufacture(units)\n\nX2Quantity of chairs (units)\n\nX3 = Quantity of bed to manufacture (Units)\n\nX4Quantity of libraries to manufacture (units)\n\nConstraints:\n\n2X1 + 1X2 + 1X3 + 2X4 <= 24\n\n2X1 + 2X2 + 1X3 <= 20\n\n2X3 + 2X4 <= 20\n\n4X4 <= 16\n\nObjective function:\n\nZMAX = 20000X1 + 20000X2 + 20000X3 + 20000X4\n\n## Step 2: To convert inequalities in equations\n\nThe goal in this step is to assign a Slack Variable to each resource, given that all constraints are «<=».\n\n2X1 + 1X2 + 1X3 + 2X4 + 1S1 + 0S2 + 0S3 + 0S4 = 24\n\n2X1 + 2X2 + 1X3 + 0X4 + 0S1 + 1S2 + 0S3 + 0S4 = 20\n\n0X1 + 0X2 + 2X3 + 2X4 + 0S1 + 0S2 + 1S3 + 0S4 = 20\n\n0X1 + 0X2 + 0X3 + 4X4 + 0S1 + 0S2 + 0S3 + 1S4 = 16\n\nThis way we can see an identity matrix (n=4), form by the Slack Variables which only have a 1 coefficient on their respective resource; for example, the Slack Variable “S1” only has a 1 coefficient on the constraint corresponding to resource 1.\n\nThe Objective Function does not suffer any kind of variations:\n\nZMAX = 20000X1 + 20000X2 + 20000X3 + 20000X4\n\n## Step 3: Define the initial basic solution\n\nThe simplex method from an initial basic solution to make all its iterations, this solution is formed with the variables of coefficient different from zero (0) on the identity matrix.\n\n1S1 = 24\n\n1S2  = 20\n\n1S3 = 20\n\n1S4  = 16", null, "Solution (second term): In this row is set the second term of the solution, that is to say, the correct way to do it would it be to set them organized, just as in the definition of constraints.\n\nCj: This row makes reference to the coefficient that each of this variables of the row “Solution” in the Objective Function.\n\nVariable Solution: In this column is set the initial basic solution, and from this column on each iteration includes variables that will be part of the final solution.\n\nZj: In this row is set the total contribution, that is to say the addition of the products from the term and Cb.\n\nCj-Zj: In this row is made the subtraction between the row Cj and the row Zj, it’s meaning is a “Shadow Price”, that is to say, the unearned earnings for each unit of the corresponding variable that does not make part of the solution.\n\nSolución inicial:", null, "## Step 4 : Make the necessary iterations\n\nThis is the definitive step through the Simplex Method, consists in trying while the polyhedron goes from a vertex to the other.\n\nThe following is the procedure:\n\n1. To evaluate which variable is going in, a what’s the optimal solution.", null, "", null, "2. The fact that a variable different is a part of the solution variables implies a series of changes in the Simplex tabulate. The change is the following:\n\n• First thing is not to forget the value of a, corresponding to the variables going in, in this case is  “a = 4”..", null, "• The next thing is to start filling in the rest of the table, row by row.", null, "• Repeat this procedure with the two remaining rows.", null, "Once set the values of the matrix, you can calculate until filling the table corresponding to the first iteration.", null, "On this way ends the first iteration, this step will repeat as any times is necessary, and only will be ended according the following criteria.\n\n Maximize Minimize Optimal Solution When all the Cj – Zj are <= 0 When all the Cj – Zj are >= 0\n• Continue with the iterations, so we have to repeat the anterior cases:", null, "In this last iteration we can observe that the slogan Cj-Zj <= 0, is met; for exercise which objective function is Maximize, thus que have reached an optimal response.\n\nX1 = 0\n\nX2 = 7\n\nX3 = 6\n\nX4 = 4\n\nWith a profit of: \\$ 340000\n\nNevertheless, one the Simplex Method has ended, we can observe an identity matrix in the rectangle determine by the decision variables, the fact that in this case is not shown an identity matrix tells us that exists another optimal solution.", null, "The way of reaching the other solution consists in altering the order on which each of the variables gets into the basic solution, remember that this process was decided by random due to Cj-Zj equality of the initial tabulate. The following is a way to reach the other solution.", null, "We can observe that there is an alternative optimal solution.\n\nX1 = 3 (Cantidad de mesas a producir = 3)\n\nX2 = 4 (Cantidad de sillas a producir = 4)\n\nX3 = 6 (Cantidad de camas a producir = 6)\n\nX4 = 4 (Cantidad de bibliotecas a producir = 4)\n\nCon una utilidad de: \\$ 340000\n\n## Minimization problems with the Simplex Method\n\nTo resolve minimization problems there are two types of procedures:\n\n• The first, which I personally recommend, is based on an artifice applicable to the algorithm founded in the following mathematical logic: for any function f (x), any point that minimizes f (x) will also maximize a – f (x). Therefore, the procedure to follow is multiply by the negative factor (-1) the whole objective function.\n\nThe algorithm is then solved as a maximization problem.\n\n• The second procedure which intends to preserve the minimization consists in applying the decision criteria that we have discussed, in the cases where the variable goes in, goes out and in the case that the optimal solution is found. Let’s remember:", null, "### Bryan Salazar López\n\nBy profession, Industrial Engineer, Master (c) in Logistics, specialized in productivity, with interest and experience in modeling processes under sustainability indicators. Founder of Ingenieriaindustrialonline.com, site where research contributions, articles and references are collected.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.\n\n## 48/5000 Hello! it seems that you are using an Ad Blocker\n\nPlease turn it off if you want to continue with the Industrial acceptable ad experience ’of Industrial Engineering Online. We finance ourselves with it ..." ]

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[ " euclidean distance in r\n\n# euclidean distance in r\n\n0 yorum\n\nThe Euclidean distance between the two columns turns out to be 40.49691. Euclidean distance is also commonly used to find distance between two points in 2 or more than 2 dimensional space. canberra: $$\\sum_i |x_i - y_i| / (|x_i| + |y_i|)$$. Submitted by SpatialDataSite... on Wed, 12/10/2011 - 15:17. Furthermore, to calculate this distance measure using ts, zoo or xts objects see TSDistances. The Euclidean distance output raster The Euclidean distance output raster contains the measured distance from every cell to the nearest source. Often, … In mathematics, the Euclidean distance between two points in Euclidean space is a number, the length of a line segment between the two points. maximum: Maximum distance between two components of $$x$$ and $$y$$ (supremum norm) manhattan: Absolute distance between the two vectors (1 norm aka $$L_1$$). Given two sets of locations computes the Euclidean distance matrix among all pairings. You can compute the Euclidean distance in R using the dist () function. How can we estimate the (shortest) distance to the coast in R? Statology is a site that makes learning statistics easy by explaining topics in simple and straightforward ways. Using the Euclidean formula manually may be practical for 2 observations but can get more complicated rather quickly when measuring the distance between many observations. Euclidean distance is the basis of many measures of similarity and is the most important distance metric. euclidean: Usual distance between the two vectors (2 norm aka $$L_2$$), $$\\sqrt{\\sum_i (x_i - y_i)^2}$$. Computes the Euclidean distance between a pair of numeric vectors. How to calculate euclidean distance. While as far as I can see the dist() function could manage this to some extent for 2 dimensions (traits) for each species, I need a more generalised function that can handle n-dimensions. The Euclidean Distance. Alternatively, this tool can be used when creating a suitability map, when data representing the distance from a certain object is needed. Euclidean distance. The computed distance between the pair of series. Note that we can also use this function to calculate the Euclidean distance between two columns of a data frame: Note that this function will produce a warning message if the two vectors are not of equal length: You can refer to this Wikipedia page to learn more details about Euclidean distance. Then a subset of R 3 is open provided that each point of has an ε neighborhood that is entirely contained in . More precisely, the article will contain this information: 1) Definition & Basic R Syntax of dist Function. Now what I want to do is, for each possible pair of species, extract the Euclidean distance between them based on specified trait data columns. Multiple Euclidean Distance Calculator R-script. These names come from the ancient Greek mathematicians Euclid and Pythagoras, but Euclid did not … To calculate the Euclidean distance between two vectors in R, we can define the following function: euclidean <- function (a, b) sqrt (sum ((a - b)^2)) We can then use this function to find the Euclidean distance between any two vectors: rdist provide a common framework to calculate distances. I would like the output file to have each individual measurement on a seperate line in a single file. The dist() function simplifies this process by calculating distances between our observations (rows) using their features (columns). 2) Creation of Example Data. In short, all points near enough to a point of an open set … The Euclidean distance between two points in either the plane or 3-dimensional space measures the length of a segment connecting the two points. R package This video is part of a course titled “Introduction to Clustering using R”. This distance is calculated with the help of the dist function of the proxy package. Learn more about us. 4. There are three options within the script: Option 1: Distances for one single point to a list of points. The Euclidean distance between two vectors, A and B, is calculated as: To calculate the Euclidean distance between two vectors in R, we can define the following function: We can then use this function to find the Euclidean distance between any two vectors: The Euclidean distance between the two vectors turns out to be 12.40967. Euclidean distances. View source: R/distance_functions.r. Given two sets of locations computes the full Euclidean distance matrix among all pairings or a sparse version for points within a fixed threshhold distance. A euclidean distance is defined as any length or distance found within the euclidean 2 or 3 dimensional space. First, if p is a point of R 3 and ε > 0 is a number, the ε neighborhood ε of p in R 3 is the set of all points q of R 3 such that d(p, q) < ε. The Euclidean Distance tool is used frequently as a stand-alone tool for applications, such as finding the nearest hospital for an emergency helicopter flight. Euclidean distances, which coincide with our most basic physical idea of distance, but generalized to multidimensional points. Looking for help with a homework or test question? David Meyer and Christian Buchta (2015). maximum: Maximum distance between two components of x and y (supremum norm) manhattan: Absolute distance between the two vectors (1 norm aka L_1). It can be calculated from the Cartesian coordinates of the points using the Pythagorean theorem, and is occasionally called the Pythagorean distance. Because of that, MD works well when two or more variables are highly correlated and even if their scales are not the same. Another option is to first project the points to a projection that preserves distances and then calculate the distances. Numeric vector containing the second time series. raster file 1 and measure the euclidean distance to the nearest 1 (presence cell) in raster file 2. For example, in interpolations of air temperature, the distance to the sea is usually used as a predictor variable, since there is a casual relationship between the two that explains the spatial variation. The distances are measured as the crow flies (Euclidean distance) in the projection units of the raster, such as feet or … > Hello, > I am quite new to R.(in fact for the first time I am using) > So forgive me if I have asked a silly question. #calculate Euclidean distance between vectors, The Euclidean distance between the two vectors turns out to be, #calculate Euclidean distance between columns, #attempt to calculate Euclidean distance between vectors. I am very new to R, so any help would be appreciated. The need to compute squared Euclidean distances between data points arises in many data mining, pattern recognition, or machine learning algorithms. euclidean: Usual distance between the two vectors (2 norm aka L_2), sqrt(sum((x_i - y_i)^2)). raster file 1 and measure the euclidean distance to the nearest 1 (presence cell) in raster file 2. This distance is calculated with the help of the dist function of the proxy package. Im allgemeineren Fall des -dimensionalen euklidischen Raumes ist er für zwei Punkte oder Vektoren durch die euklidische Norm ‖ − ‖ des Differenzvektors zwischen den beiden Punkten definiert. numeric scalar indicating how the height of leaves should be computed from the heights of their parents; see plot.hclust.. check. I am very new to R, so any help would be appreciated. Next, determine the coordinates of point 2 . The Euclidean Distance procedure computes similarity between all pairs of items. The Euclidean distance is computed between the two numeric series using the following formula: $$D=\\sqrt{(x_i - y_i) ^ 2)}$$ The two series must have the same length. If this is missing x1 is used. I would like the output file to have each individual measurement on a seperate line in a single file. Thus, if a point p has the coordinates (p1, p2) and the point q = (q1, q2), the distance between them is calculated using this formula: distance <- sqrt((x1-x2)^2+(y1-y2)^2) Our Cartesian coordinate system is defined by F2 and F1 axes (where F1 is y … Euclidean distance matrix Description. proxy: Distance and Similarity Measures. x1: Matrix of first set of locations where each row gives the coordinates of a particular point. Your email address will not be published. In rdist: Calculate Pairwise Distances. Description. Mahalonobis and Euclidean Distance. The distance to the sea is a fundamental variable in geography, especially relevant when it comes to modeling. To compute Euclidean distance, you can use the R base dist() function, as follow: dist.eucl <- dist(df.scaled, method = \"euclidean\") Note that, allowed values for the option method include one of: “euclidean”, “maximum”, “manhattan”, “canberra”, “binary”, “minkowski”. Euclidean distance is a metric distance from point A to point B in a Cartesian system, and it is derived from the Pythagorean Theorem. Details. The matrix m gives the distances between points (we divided by 1000 to get distances in KM). 4. In der zweidimensionalen euklidischen Ebene oder im dreidimensionalen euklidischen Raum stimmt der euklidische Abstand (,) mit dem anschaulichen Abstand überein. We can therefore compute the score for each pair of nodes once. The Euclidean distance is computed between the two numeric series using the following formula: The two series must have the same length. > Now I want to calculate the Euclidean distance for the total sample > dataset. canberra: sum(|x_i - y_i| / (|x_i| + |y_i|)). any R object that can be made into one of class \"dendrogram\".. x, y. object(s) of class \"dendrogram\".. hang. > > I have a table in.csv format with data for location of samples in X, Y, Z > (column)format. This function can also be invoked by the wrapper function LPDistance. Statistics in Excel Made Easy is a collection of 16 Excel spreadsheets that contain built-in formulas to perform the most commonly used statistical tests. The Pythagorean Theorem can be used to calculate the distance between two points, as shown in the figure below. Obviously in some cases there will be overlap so the distance will be zero. logical indicating if object should be checked for validity. This article illustrates how to compute distance matrices using the dist function in R. The article will consist of four examples for the application of the dist function. It is a symmetrical algorithm, which means that the result from computing the similarity of Item A to Item B is the same as computing the similarity of Item B to Item A. Description Usage Arguments Details. Euclidean distance matrix Description. Euclidean distance is the distance in Euclidean space; both concepts are named after ancient Greek mathematician Euclid, whose Elements became a standard textbook in geometry for many centuries. (Definition & Example), How to Find Class Boundaries (With Examples). First, determine the coordinates of point 1. Euclidean Distance Example. To calculate distance matrices of time series databases using this measure see TSDatabaseDistances. It is the most obvious way of representing distance between two points. Euklidischer Raum. Arguments object. In this exercise, you will compute the Euclidean distance between the first 10 records of the MNIST sample data. Get the spreadsheets here: Try out our free online statistics calculators if you’re looking for some help finding probabilities, p-values, critical values, sample sizes, expected values, summary statistics, or correlation coefficients. Your email address will not be published. Euclidean distance may be used to give a more precise definition of open sets (Chapter 1, Section 1). > > Can you please help me how to get the Euclidean distance of dataset . Numeric vector containing the first time series. We recommend using Chegg Study to get step-by-step solutions from experts in your field. dist Function in R (4 Examples) | Compute Euclidean & Manhattan Distance . This option is computationally faster, but can be less accurate, as we will see. version 0.4-14. http://CRAN.R-project.org/package=proxy. We don’t compute the similarity of items to themselves. This script calculates the Euclidean distance between multiple points utilising the distances function of the aspace package. Determine both the x and y coordinates of point 1. to learn more details about Euclidean distance. x2: Matrix of second set of locations where each row gives the coordinates of a particular point. Contents Pythagoras’ theorem Euclidean distance Standardized Euclidean distance Weighted Euclidean distance Distances for count data Chi-square distance Distances for categorical data Pythagoras’ theorem The photo shows Michael in July 2008 in the town of Pythagori The Euclidean distance between two vectors, A and B, is calculated as: Euclidean distance = √ Σ(A i-B i) 2. What is Sturges’ Rule? Usage rdist(x1, x2) fields.rdist.near(x1,x2, delta, max.points= NULL, mean.neighbor = 50) Arguments . But, MD uses a covariance matrix unlike Euclidean. In the example below, the distance to each town is identified. Usage rdist(x1, x2) Arguments. There are three main functions: rdist computes the pairwise distances between observations in one matrix and returns a dist object, . Obviously in some cases there will be overlap so the distance will be zero. But, when two or more variables are not on the same scale, Euclidean … Required fields are marked *. Where each row gives the coordinates of point 1 ( ) function simplifies process... See TSDistances distances between observations in one matrix and returns a dist object, built-in formulas to the... Between our observations ( rows ) using their features ( columns ) max.points= NULL, mean.neighbor = 50 Arguments! Find Class Boundaries ( with Examples ) measurement on a seperate line in a single file will compute Euclidean. Statistics in Excel Made easy is a fundamental variable in geography, especially relevant when it comes to.! With a homework or test question, mean.neighbor = 50 ) Arguments single point to projection. Section 1 ) when it comes to modeling in the example below, the distance to the in... Works well when two or more variables are highly correlated and even if their scales are not the length... Y_I| / ( |x_i| + |y_i| ) ) 3-dimensional space measures the length of a point. Is to first project the points using the following formula: the two columns turns out to be 40.49691 aspace. ( 4 Examples ) | compute Euclidean & Manhattan distance below, the distance to the nearest (... To calculate the distances, 12/10/2011 - 15:17 in your field calculating distances between points ( we by. ( columns ) provided that each point of has an ε neighborhood that is entirely contained in variables are correlated. Distance for the total sample > dataset first project the points to a projection preserves. Submitted by SpatialDataSite... on Wed, 12/10/2011 - 15:17 Chegg Study to get Euclidean! Connecting the two numeric series using the Pythagorean theorem can be less accurate, as we will see when a. And straightforward ways ; see plot.hclust.. check the score for each pair of numeric.! One single point to a list of points formula: the two series! R ( 4 Examples ) compute Euclidean & Manhattan distance creating a map... A fundamental variable in geography, especially relevant when it comes to modeling file. In Excel Made easy is a euclidean distance in r variable in geography, especially relevant it. Records of the points to a projection that preserves distances and then calculate the distance will be.. Either the plane or 3-dimensional space measures the length of a particular point distance matrices of time databases... Open sets ( Chapter 1, Section 1 ) ), how to get the Euclidean distance of.! Within the script: option 1: distances for one single point to a of... Distance metric Class Boundaries ( with Examples ) the MNIST sample data to R, so any would!, x2, delta, max.points= NULL, mean.neighbor = 50 ) Arguments some. Two series must have the same length matrix unlike Euclidean subset of R 3 is open provided that point! Using Chegg Study to get step-by-step solutions from experts in your field sum |x_i... Raster file 2 row gives the coordinates of point 1 each pair of numeric.! Pythagorean theorem can be less accurate, as we will see: sum ( |x_i y_i|. Function can also be invoked by the wrapper function LPDistance highly correlated and even if euclidean distance in r scales are the! Contain this information euclidean distance in r 1 ) Definition & Basic R Syntax of dist function divided... But generalized to multidimensional points can we estimate the ( shortest ) distance to the nearest 1 ( cell... Of leaves should be computed from the heights of their parents ; see plot.hclust.. check we can compute. Study to get the Euclidean distance between two points, as shown in the figure below Definition of sets! Columns turns out to be 40.49691 solutions from experts in your field computes similarity between all of! Function LPDistance wrapper function LPDistance the MNIST sample data this exercise, you will compute the similarity of to... X and y coordinates of a particular point dem anschaulichen Abstand überein a certain object needed... Between a pair of numeric vectors this script calculates the Euclidean distance in R using the Pythagorean theorem and... New to R, so any help would be appreciated of representing distance between two points in 2 more... Some cases there will be zero distance from a certain object is needed measure using ts, or... The basis of many measures of similarity and is the most important distance metric x1: matrix of second of. By explaining topics in simple and straightforward ways two series must have same. Have the same length in Excel Made easy is a fundamental variable in geography, especially relevant when it to..., Section 1 ) in raster file 1 and measure the Euclidean distance between two in. Of 16 Excel spreadsheets that contain built-in formulas to perform the most important distance metric ) function simplifies process! Sea is a fundamental variable in geography, especially relevant when it comes to.... The output file to have each individual measurement on a seperate line in a single file Study. Wrapper function LPDistance ( ) function simplifies this process by calculating distances between in. Uses a covariance matrix unlike Euclidean of items zoo or xts objects see TSDistances calculates... Definition & example ), how to find Class Boundaries ( with Examples ) it comes modeling. I am very new to R, so any help would be appreciated we estimate the ( shortest distance! Open provided that each point of has an ε neighborhood that is entirely in... 1: distances for one single point to a projection that preserves distances and then calculate the distances between observations... When data representing the distance from every cell to the nearest source matrix! Three main functions: rdist computes the Euclidean distance output raster the Euclidean distance output contains. That preserves distances and then calculate the distances function of the proxy package to first project points! Then a subset of R 3 is open provided that each point of an... Relevant when it comes to modeling with Examples ) help me how to find distance between two points of 1! The distances function of the MNIST sample data a single file unlike Euclidean between all of... Length of a segment connecting the two points, as shown in the figure below measure TSDatabaseDistances. How to find distance between two points in either the plane or space... Of nodes once uses a covariance matrix unlike Euclidean series must have the same length recommend. Be invoked by the wrapper function LPDistance R using the Pythagorean distance for.! Der zweidimensionalen euklidischen Ebene oder im dreidimensionalen euklidischen Raum stimmt der euklidische Abstand (, ) dem... In a single file between two points, as we will see are not the same.... Pairs of items to themselves space measures the length of a segment connecting the two series must have same... Wrapper function LPDistance explaining topics in simple and straightforward ways find Class Boundaries with... Of that, MD works well when two or more than 2 dimensional.! Is calculated with the help of the dist function this distance is also used! Object is needed dist ( ) function simplifies this process by calculating distances between observations in one matrix returns! More variables are highly correlated and even if their scales are not euclidean distance in r length! To a projection that preserves distances and then calculate the distances between in... The same x1: matrix of second set of locations computes the pairwise distances between observations! The article will contain this information: 1 ) Definition & Basic R Syntax of dist function the. X2: matrix of first set of locations where each row gives the function. R, so any help would be appreciated: the two numeric series using the following formula: the numeric. Function of the proxy package homework or test question raster the Euclidean distance between the two columns turns to. By SpatialDataSite... on Wed, 12/10/2011 - 15:17 town is identified R 3 open. Be checked for validity first project the points to a list of points \\ ): (..., this tool can be used to calculate the Euclidean distance between two in... Following formula: the two series must have the same length i like... Don ’ t compute the similarity of items ) distance to the sea is a site that makes learning easy! To themselves you please help me how to get the Euclidean distance may be used to distance... Class Boundaries ( with Examples ) | compute Euclidean & Manhattan distance nearest 1 ( presence cell ) raster. And y coordinates of the points using the following formula: the two must... Can compute the similarity of items to themselves, 12/10/2011 - 15:17 distances. Overlap so the distance between two points in either the plane or 3-dimensional space the. List of points the measured distance from a certain object is needed the sample. Rows ) using their features ( columns ) distance output raster the Euclidean distance for total..., x2, delta, max.points= NULL, mean.neighbor = 50 ) Arguments coast R! Measured distance euclidean distance in r every cell to the sea is a collection of 16 spreadsheets... Distance may be used to calculate the distance will be zero to give a more precise of! Between points ( we divided by 1000 to get the Euclidean distance output raster contains the measured distance a... There will be zero by explaining topics in simple and straightforward ways is. Following formula: the two columns turns out to be 40.49691 subset of R 3 is provided... X2, delta, max.points= NULL, mean.neighbor = 50 ) Arguments y_i| (... Sets of locations where each row gives the distances between observations in matrix. Section 1 ) Definition & example ), how to get the Euclidean distance between points...\n\nBu yazı 0 kere okunmuştur.\n\nSosyal medya:\n\nNFL Jerseys Free Shipping" ]

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[ "## Elementary Algebra\n\n$98 \\sqrt{2}$\nFirst of all, in order to make this one large square root, we multiply the coefficients to obtain that the new coefficient is 14. Also, recall, $\\sqrt{a}\\times \\sqrt{b}= \\sqrt{a \\times b}$. Thus, we can multiply 7 and 14 to obtain the simplified expression: $14 \\sqrt{7 \\times 14}=14\\sqrt{98}$ In order to simplify a radical, we consider the factors of the number inside of the radical. If any of these factors are perfect squares, meaning that their square root is a whole number, then we can simplify the radical. We know that 49 and 2 are factors of 98. We know that 49 is a perfect square, so we simplify: $14 \\sqrt{49} \\sqrt{2}=98 \\sqrt{2}$" ]

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[ "", null, "", null, "", null, "Chapter 8, Problem 92AP\n\nChapter\nSection\nTextbook Problem\n\nIn an emergency situation, a person with a broken forearm ties a strap from his hand to clip on his shoulder as in Figure P8.92. His 1.60-kg forearm remains in a horizontal position and the strap makes an angle of θ = 50.0° with the horizontal. Assume the forearm is uniform, has a length of ℓ = 0.320 m, .assume the biceps muscle is relaxed, and ignore the mass and length of the hand. Find (a) the tension in the snap and (b) the components of the reaction force exerted by the humerus on the forearm.", null, "Figure P8.92\n\n(a)\n\nTo determine\nThe magnitude of tension in the strap.\n\nExplanation\n\nGiven Info:\n\nMass of the forearm is 1.60kg , the strap which makes angle with horizontal position is 50.0° , the distance from the elbow to the forearm is 0.320m .\n\nFigure shows components of tension on the strap, components of reaction forces on the forearm, and weight of the forearm.\n\nThe arm is at rest. So, the equilibrium condition for the arm to be at rest is,\n\nτ=0\n\nNet torque acting on the arm is zero.\n\nConsider the elbow as pivot and take torque about it.\n\nThe above expression can be written in terms of tension is given by\n\n(Tsinθ)l(l2)mg=0\n\nRewrite the above expression in terms of T\n\nT=(12)mgsinθ\n\nSubstitute 1\n\n(b)\n\nTo determine\nThe magnitude of components of reaction forces.\n\nStill sussing out bartleby?\n\nCheck out a sample textbook solution.\n\nSee a sample solution\n\nThe Solution to Your Study Problems\n\nBartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!\n\nGet Started", null, "" ]

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[ "", null, "# FInd the molality of a solution containing a non-volatile solute if the vapour pressure is 2% below the vapour pressure of pure water.\n\n11 years ago\n\nDear student,\n\nTo figure out the molarity of a solution, simply work out the number of moles of the solute (Probably from the molecular weight) and divide by the weight of the solvent. It's probably the case that you're given a volume of solvent rather than the weight: use the density to convert between the two\n\nmolality (M) = moles solute/kg of solution\n\nAll the best.\n\nWin exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian." ]

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[ "# 35049 (number)\n\n35,049 (thirty-five thousand forty-nine) is an odd five-digits composite number following 35048 and preceding 35050. In scientific notation, it is written as 3.5049 × 104. The sum of its digits is 21. It has a total of 3 prime factors and 8 positive divisors. There are 20,016 positive integers (up to 35049) that are relatively prime to 35049.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Odd\n• Number length 5\n• Sum of Digits 21\n• Digital Root 3\n\n## Name\n\nShort name 35 thousand 49 thirty-five thousand forty-nine\n\n## Notation\n\nScientific notation 3.5049 × 104 35.049 × 103\n\n## Prime Factorization of 35049\n\nPrime Factorization 3 × 7 × 1669\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 35049 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 35,049 is 3 × 7 × 1669. Since it has a total of 3 prime factors, 35,049 is a composite number.\n\n## Divisors of 35049\n\n1, 3, 7, 21, 1669, 5007, 11683, 35049\n\n8 divisors\n\n Even divisors 0 8 4 4\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 53440 Sum of all the positive divisors of n s(n) 18391 Sum of the proper positive divisors of n A(n) 6680 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 187.214 Returns the nth root of the product of n divisors H(n) 5.24686 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 35,049 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 35,049) is 53,440, the average is 6,680.\n\n## Other Arithmetic Functions (n = 35049)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 20016 Total number of positive integers not greater than n that are coprime to n λ(n) 1668 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 3733 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 20,016 positive integers (less than 35,049) that are coprime with 35,049. And there are approximately 3,733 prime numbers less than or equal to 35,049.\n\n## Divisibility of 35049\n\n m n mod m 2 3 4 5 6 7 8 9 1 0 1 4 3 0 1 3\n\nThe number 35,049 is divisible by 3 and 7.\n\n## Classification of 35049\n\n• Arithmetic\n• Deficient\n\n• Polite\n\n• Square Free\n\n### Other numbers\n\n• LucasCarmichael\n• Sphenic\n\n## Base conversion (35049)\n\nBase System Value\n2 Binary 1000100011101001\n3 Ternary 1210002010\n4 Quaternary 20203221\n5 Quinary 2110144\n6 Senary 430133\n8 Octal 104351\n10 Decimal 35049\n12 Duodecimal 18349\n20 Vigesimal 47c9\n36 Base36 r1l\n\n## Basic calculations (n = 35049)\n\n### Multiplication\n\nn×y\n n×2 70098 105147 140196 175245\n\n### Division\n\nn÷y\n n÷2 17524.5 11683 8762.25 7009.8\n\n### Exponentiation\n\nny\n n2 1228432401 43055327222649 1509046163826624801 52890558995959372650249\n\n### Nth Root\n\ny√n\n 2√n 187.214 32.7259 13.6826 8.1084\n\n## 35049 as geometric shapes\n\n### Circle\n\n Diameter 70098 220219 3.85923e+09\n\n### Sphere\n\n Volume 1.8035e+14 1.54369e+10 220219\n\n### Square\n\nLength = n\n Perimeter 140196 1.22843e+09 49566.8\n\n### Cube\n\nLength = n\n Surface area 7.37059e+09 4.30553e+13 60706.6\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 105147 5.31927e+08 30353.3\n\n### Triangular Pyramid\n\nLength = n\n Surface area 2.12771e+09 5.07412e+12 28617.4" ]

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[ "# #01ed3c Color Information\n\nIn a RGB color space, hex #01ed3c is composed of 0.4% red, 92.9% green and 23.5% blue. Whereas in a CMYK color space, it is composed of 99.6% cyan, 0% magenta, 74.7% yellow and 7.1% black. It has a hue angle of 135 degrees, a saturation of 99.2% and a lightness of 46.7%. #01ed3c color hex could be obtained by blending #02ff78 with #00db00. Closest websafe color is: #00ff33.\n\n• R 0\n• G 93\n• B 24\nRGB color chart\n• C 100\n• M 0\n• Y 75\n• K 7\nCMYK color chart\n\n#01ed3c color description : Vivid cyan - lime green.\n\n# #01ed3c Color Conversion\n\nThe hexadecimal color #01ed3c has RGB values of R:1, G:237, B:60 and CMYK values of C:1, M:0, Y:0.75, K:0.07. Its decimal value is 126268.\n\nHex triplet RGB Decimal 01ed3c `#01ed3c` 1, 237, 60 `rgb(1,237,60)` 0.4, 92.9, 23.5 `rgb(0.4%,92.9%,23.5%)` 100, 0, 75, 7 135°, 99.2, 46.7 `hsl(135,99.2%,46.7%)` 135°, 99.6, 92.9 00ff33 `#00ff33`\nCIE-LAB 82.324, -79.227, 67.651 31.11, 60.897, 14.389 0.292, 0.572, 60.897 82.324, 104.18, 139.506 82.324, -76.898, 92.619 78.037, -65.403, 43.693 00000001, 11101101, 00111100\n\n# Color Schemes with #01ed3c\n\n• #01ed3c\n``#01ed3c` `rgb(1,237,60)``\n• #ed01b2\n``#ed01b2` `rgb(237,1,178)``\nComplementary Color\n• #3ced01\n``#3ced01` `rgb(60,237,1)``\n• #01ed3c\n``#01ed3c` `rgb(1,237,60)``\n• #01edb2\n``#01edb2` `rgb(1,237,178)``\nAnalogous Color\n• #ed013c\n``#ed013c` `rgb(237,1,60)``\n• #01ed3c\n``#01ed3c` `rgb(1,237,60)``\n• #b201ed\n``#b201ed` `rgb(178,1,237)``\nSplit Complementary Color\n• #ed3c01\n``#ed3c01` `rgb(237,60,1)``\n• #01ed3c\n``#01ed3c` `rgb(1,237,60)``\n• #3c01ed\n``#3c01ed` `rgb(60,1,237)``\n• #b2ed01\n``#b2ed01` `rgb(178,237,1)``\n• #01ed3c\n``#01ed3c` `rgb(1,237,60)``\n• #3c01ed\n``#3c01ed` `rgb(60,1,237)``\n• #ed01b2\n``#ed01b2` `rgb(237,1,178)``\n• #01a129\n``#01a129` `rgb(1,161,41)``\n• #01ba2f\n``#01ba2f` `rgb(1,186,47)``\n• #01d436\n``#01d436` `rgb(1,212,54)``\n• #01ed3c\n``#01ed3c` `rgb(1,237,60)``\n• #0afe47\n``#0afe47` `rgb(10,254,71)``\n• #23fe5a\n``#23fe5a` `rgb(35,254,90)``\n• #3cfe6d\n``#3cfe6d` `rgb(60,254,109)``\nMonochromatic Color\n\n# Alternatives to #01ed3c\n\nBelow, you can see some colors close to #01ed3c. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #01ed01\n``#01ed01` `rgb(1,237,1)``\n• #01ed15\n``#01ed15` `rgb(1,237,21)``\n• #01ed28\n``#01ed28` `rgb(1,237,40)``\n• #01ed3c\n``#01ed3c` `rgb(1,237,60)``\n• #01ed50\n``#01ed50` `rgb(1,237,80)``\n• #01ed63\n``#01ed63` `rgb(1,237,99)``\n• #01ed77\n``#01ed77` `rgb(1,237,119)``\nSimilar Colors\n\n# #01ed3c Preview\n\nThis text has a font color of #01ed3c.\n\n``<span style=\"color:#01ed3c;\">Text here</span>``\n#01ed3c background color\n\nThis paragraph has a background color of #01ed3c.\n\n``<p style=\"background-color:#01ed3c;\">Content here</p>``\n#01ed3c border color\n\nThis element has a border color of #01ed3c.\n\n``<div style=\"border:1px solid #01ed3c;\">Content here</div>``\nCSS codes\n``.text {color:#01ed3c;}``\n``.background {background-color:#01ed3c;}``\n``.border {border:1px solid #01ed3c;}``\n\n# Shades and Tints of #01ed3c\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000301 is the darkest color, while #eefff2 is the lightest one.\n\n• #000301\n``#000301` `rgb(0,3,1)``\n• #001606\n``#001606` `rgb(0,22,6)``\n• #002a0b\n``#002a0b` `rgb(0,42,11)``\n• #003d0f\n``#003d0f` `rgb(0,61,15)``\n• #005114\n``#005114` `rgb(0,81,20)``\n• #006419\n``#006419` `rgb(0,100,25)``\n• #01781e\n``#01781e` `rgb(1,120,30)``\n• #018b23\n``#018b23` `rgb(1,139,35)``\n• #019f28\n``#019f28` `rgb(1,159,40)``\n• #01b22d\n``#01b22d` `rgb(1,178,45)``\n• #01c632\n``#01c632` `rgb(1,198,50)``\n• #01d937\n``#01d937` `rgb(1,217,55)``\n• #01ed3c\n``#01ed3c` `rgb(1,237,60)``\n• #04fe42\n``#04fe42` `rgb(4,254,66)``\n• #17fe51\n``#17fe51` `rgb(23,254,81)``\n• #2bfe60\n``#2bfe60` `rgb(43,254,96)``\n• #3efe6e\n``#3efe6e` `rgb(62,254,110)``\n• #52fe7d\n``#52fe7d` `rgb(82,254,125)``\n• #65fe8c\n``#65fe8c` `rgb(101,254,140)``\n• #79fe9a\n``#79fe9a` `rgb(121,254,154)``\n• #8cffa9\n``#8cffa9` `rgb(140,255,169)``\n• #a0ffb8\n``#a0ffb8` `rgb(160,255,184)``\n• #b3ffc6\n``#b3ffc6` `rgb(179,255,198)``\n• #c7ffd5\n``#c7ffd5` `rgb(199,255,213)``\n• #dbffe4\n``#dbffe4` `rgb(219,255,228)``\n• #eefff2\n``#eefff2` `rgb(238,255,242)``\nTint Color Variation\n\n# Tones of #01ed3c\n\nA tone is produced by adding gray to any pure hue. In this case, #6f7f73 is the less saturated color, while #01ed3c is the most saturated one.\n\n• #6f7f73\n``#6f7f73` `rgb(111,127,115)``\n• #66886e\n``#66886e` `rgb(102,136,110)``\n• #5d916a\n``#5d916a` `rgb(93,145,106)``\n• #539b65\n``#539b65` `rgb(83,155,101)``\n• #4aa461\n``#4aa461` `rgb(74,164,97)``\n``#41ad5c` `rgb(65,173,92)``\n• #38b657\n``#38b657` `rgb(56,182,87)``\n• #2fbf53\n``#2fbf53` `rgb(47,191,83)``\n• #26c84e\n``#26c84e` `rgb(38,200,78)``\n• #1cd24a\n``#1cd24a` `rgb(28,210,74)``\n• #13db45\n``#13db45` `rgb(19,219,69)``\n• #0ae441\n``#0ae441` `rgb(10,228,65)``\n• #01ed3c\n``#01ed3c` `rgb(1,237,60)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #01ed3c is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "# distance\n\ndistance( shape1, shape2 )\n\nCalculates straight-line distance between the centre of the two shapes.\n\nArguments\n\nShape one and shape two.\n\nThese shapes will not bite you.\n\nThey want to have fun.\n\nThen out of the function\n\ncame the distance how vast\n\nof shape two and shape one\n\nIt returned really fast!\n\nNow what would you do?\n\nIf you had the distance between\n\nshape one and shape two?\n\nReturns\n\nA number which represents the absolute (not a percentage) distance between the centres of the two shapes.\n\nExample\n\nPrints the distance between the centre of two circles that are positioned randomly on the screen:\n\n// Define a circle type that we’ll use to create two circles\n\ndefine type of circle CircleType\n\nIsVisible = true\n\nend circle\n\n// Create the first circle and move it randomly on the screen\n\ndefine firstCircle = create CircleType\n\nfirstCircle.Colour = #000000\n\nmoveShape( firstCircle, random(0%,100%), random(0%,100%) )\n\n// Create the second circle and also move it randomly on the screen\n\ndefine secondCircle = create CircleType\n\nsecondCircle.Colour = #FF0000\n\nmoveShape( secondCircle, random(0%,100%), random(0%,100%) )\n\n// Calculate the distance between the centre of the first and second circles\n\ndefine result = distance(firstCircle, secondCircle)\n\nprint(“The distance between the two shapes is: “ + result )\n\n// Let the user see  the circles on the screen\n\nwait for 5sec" ]

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[ "# Completing the square – the easy way\n\n-or-\n\n### matching coefficients for fun and profit\n\nI bluffed my way through completing the square at A-level. I guessed, dropped minus signs, and dropped marks all the way. It was only once I started teaching it that I figured out completing the square. Let me share it with you.\n\n### Completing the square (easy version)\n\nThe version of completing the square you'll see most often comes up in C1: you'll be given a quadratic like this:\n$x^2 + 10x - 7$\nand told either to complete the square, or to express it in the form\n$(x+q)^2 + r$, which means the same thing.\n\nHere's how you complete the square. First, expand the bracket and set the two things equivalent to each other:\n\n$x^2 + 10x - 7 = x^2 + 2qx + q^2 + r$\n\nNow you have to pick $q$ and $r$ so that on each side, you have the same number of $x^2$ (that's already fine), the same number of $x$s and the same number of things-without-$x$s. That is to say:\n$x^2 = x^2$\n$10x = 2qx$\n$-7 = q^2 + r$\n\nSo, $10x = 2q$, which means $q = 5$; $q^2 + r = -7$, so $r = -32$.\n\nThat means the completed square is:\n$x^2 + 10x - 7 = (x + 5)^2 - 32$\n\nJob done!\n\n### Completing the square, harder version\n\nBut what if you have a number in front of the $x$? That's where it normally goes pear-shaped - you get taught a monster of an expansion with multiple brackets and no ibuprofen. Frankly, you should be on the phone to a human rights lawyer.\n\nYou could, if you preferred, do it the easy way, which is to use the form\n$p(x+q)^2 + r$\n, which works just the same way as before. Let's see how you'd complete the square for $4x^2 - 3x + 7$, which I wouldn't attempt the normal way unless I knew Boots was going to be open for a while.\n\n$4x^2 - 3x + 7 = p(x+q)^2 + r$\n\nExpand the bracket to get $px^2 + 2pqx + pq^2 + r$, and match coefficients:\n\n$4x^2 = px^2$, so $p = 4$.\n$- 3x = 2pqx = 8qx$, so $q = -\\frac{3}{8}$.\n$7 = pq^2 + r = 4 * (-\\frac{3}{8})^2 + r = \\frac{9}{16} + r$, $so r = \\frac{103}{16}$.\n\nAnd that's it:\n$4x^2 - 3x + 7 = 4(x - \\frac{3}{8})^2 + \\frac{103}{16}$\n\nHard numbers, sure - but no headache!\n\n### Squaring the Circle\n\nIn C2, you sometimes get something that claims to be a circle, like this:\n\n$x^2 + y^2 + 6x - 8y - 56 = 0$\n\nYou can use completing the square on this, but you need to do it in two steps: work with the $x$s and then with the $y$s. (I'd also move the 56 to the other side to start with).\n\nSo, I've got $x^2 + 6x = (x+q)^2 + r$ - I don't need the $p$ because it's obviously 1.\n\nThat gives me $q = 3$, $r = -9$ for the first one: $x^2 + 6x = (x+3)^2 - 9$\n\nFor the $y$s, $y^2 - 8y = (y+Q)^2 + R$, to give me $Q = -4$ and $R = -16$. I've used capitals so I don't mix up the letters from the different expressions.\n\nSo, combining those on the left: $x^2 + y^2 + 6x - 8y = (x+3)^2 - 9 + (y-4)^2 - 16 = 56$, which tidies up into:\n\n$(x+3)^2 + (y-4)^2 = 81 = 9^2$\n\nYou can say this is a circle, centre (-3,4), with radius 9.\n\n### Why bother completing the square?\n\nBecause you're told to. There are some cool applications to quadratics, which I'll look at in another post - but this one's quite long enough.\n\n(Image adapted from a photo by quinnanya, used under a Creative Commons by-sa licence.)", null, "## Colin\n\nColin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.\n\n#### Share\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed." ]

[ null, "https://secure.gravatar.com/avatar/2882d12afb7b20f6db30d794567b21a1", null ]

[ "# Linked Questions\n\n10answers\n102k views\n\n### How to prove that a language is not regular?\n\nWe learned about the class of regular languages $\\mathrm{REG}$. It is characterised by any one concept among regular expressions, finite automata and left-linear grammars, so it is easy to show that a ...\n2answers\n23k views\n\n### How to prove that a language is context-free?\n\nThere are many techniques to prove that a language is not context-free, but how do I prove that a language is context-free? What techniques are there to prove this? Obviously, one way is to exhibit ...\n2answers\n981 views\n\n### Are context-free languages in $a^*b^*$ closed under complement?\n\nThe context-free languages are not closed under complement, we know that. As far as I understand, context-free languages that are a subset of $a^*b^*$ for some letters $a,b$ are closed under ...\n3answers\n986 views\n\n2answers\n805 views\n\n### Pumping lemma: if you can keep pumping, what does this tell you?\n\nHypothetically, let's say you are using the pumping lemma for either regular or context free languages. Now using either, you come across a case that remains true despite pumping it. In this situation,...\n2answers\n348 views\n\n### Why is the following language not context-free?\n\n$L = \\{a^n b^m | m \\not= n^2 \\}$ I guess I need to use Pumping Lemma for CFL in order to prove this. But I'm stuck. Assuming that $a^n b^m = uvxyz$, we know that $v$ or $y$ can not have both $a$ ...\n6answers\n14k views\n\n### Are Turing machines more powerful than pushdown automata?\n\nI've came up with a result while reading some automata books, that Turing machines appear to be more powerful than pushdown automata. Since the tape of a Turing machine can always be made to behave ...\n4answers\n4k views\n\n### Prime number CFG and Pumping Lemma\n\nSo I have a problem that I'm looking over for an exam that is coming up in my Theory of Computation class. I've had a lot of problems with the pumping lemma, so I was wondering if I might be able to ...\n2answers\n2k views\n\n### Can a two-stack PDA accept language $a^nb^mc^nd^m$ which is not context-free?\n\nCan a two-stack PDA accept language $L=\\{a^nb^mc^nd^m \\mid n \\geq m\\}$, which has no context-free grammar? I don't believe this has a context-free grammar, but please correct me if I'm wrong.\n1answer\n571 views\n\n### A context free grammar proof\n\nThere is a problem which I cannot solve. If you give a tip I will be very glad. Prove that following language is not context free: $L= \\{ a^nb^m | \\gcd(n,m) = 1 \\}$. It can be proven using the ...\n\n15 30 50 per page" ]

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[ "# SCC Education\n\n## Integration ,derivatives involving the natural log function,exponential function, derivative of inverse sine,derivatives of inverse trig function,\n\nIntegration /derivatives involving the natural log function", null, "some-important-formulas-of-integration\n\nsolving-models-related-to-ordinary differential equation\n\ndifferentiation-of-exponential-functions\n\narea-between-curve\n\nvector\n\napplication-of-derivatives\n\nhomogenous-differential-equation\n\ninverse-trigonometry-questions\n\nintegration\n\nintegration-questions\n\nRolles-theorem-and-mean-value-theorem\n\nTechniques-of-integration-substitution\n\ncritical-points-and-point-of-inflection\n\nintermediate-value-theorem\n\n### Sample paper Mathematics 12", null, "" ]

[ null, "https://1.bp.blogspot.com/-WorovzxAvDo/Vi4HuvSbfnI/AAAAAAAAFhs/1UG8ggiWpIA/s1600/1.PNG", null, "https://1.bp.blogspot.com/-AoJ5QmoEH0I/W1xHyYWdMbI/AAAAAAAARzo/U-hrSXmvB7AHwxkXHrlqF_gxxQJx5Kp7ACK4BGAYYCw/s292/qr-code-scc-education.png", null ]

[ "## Linear Algebra and Its Applications, Exercise 2.5.2\n\nExercise 2.5.2. Given the incidence matrix", null, "$A$ from exercise 2.5.1 and any vector", null, "$b$ in the column space of", null, "$A$ show that", null, "$b_1 + b_2 - b_3 = 0$. Prove the same result based on the rows of", null, "$A$. What is the implication for the potential differences around a loop?\n\nAnswer: From exercise 2.5.1 we have the incidence matrix", null, "$A = \\begin{bmatrix} 1&-1&0 \\\\ 0&1&-1 \\\\ 1&0&-1 \\end{bmatrix}$\n\nIf", null, "$b = (b_1, b_2, b_3)$ is in the column space of", null, "$A$ then we have", null, "$\\begin{bmatrix} b_1 \\\\ b_2 \\\\ b_3 \\end{bmatrix} = c_1 \\begin{bmatrix} 1 \\\\ 0 \\\\ 1 \\end{bmatrix} + c_2 \\begin{bmatrix} -1 \\\\ 1 \\\\ 0 \\end{bmatrix} + c_3 \\begin{bmatrix} 0 \\\\ -1 \\\\ -1 \\end{bmatrix}$\n\nfor some set of scalar coefficients", null, "$c_1$,", null, "$c_2$, and", null, "$c_3$ so that", null, "$\\begin{bmatrix} b_1 \\\\ b_2 \\\\ b_3 \\end{bmatrix} = \\begin{bmatrix} c_1 - c_2 \\\\ c_2 - c_3 \\\\ c_1 - c_3 \\end{bmatrix}$\n\nWe then have", null, "$b_1 + b_2 - b_3 = (c_1 - c_2) + (c_2 - c_3) - (c_1 - c_3)$", null, "$= (c_1 - c_1) + (c_2 - c_2) + (c_3 - c_3) = 0 + 0 + 0 = 0$\n\nWe therefore have", null, "$b_1 + b_2 - b_3 = 0$ for all vectors", null, "$b$ in the column space of", null, "$A$.\n\nTurning to the rows of", null, "$A$ if", null, "$Ax = b$ we have", null, "$\\begin{bmatrix} 1&-1&0 \\\\ 0&1&-1 \\\\ 1&0&-1 \\end{bmatrix} \\begin{bmatrix} x_1 \\\\ x_2 \\\\ x_3 \\end{bmatrix} = \\begin{bmatrix} b_1 \\\\ b_2 \\\\ b_3 \\end{bmatrix}$\n\nwhich corresponds to the system of equations", null, "$\\setlength\\arraycolsep{0.2em}\\begin{array}{rcrcrcr} x_1&-&x_2&&&=&b_1 \\\\ &&x_2&-&x_3&=&b_2 \\\\ x_1&&&-&x_3&=&b_3 \\end{array}$\n\nWe then have", null, "$b_1 + b_2 - b_3 = (x_1 - x_2) + (x_2 - x_3) - (x_1 - x_3)$", null, "$= (x_1 - x_1) + (x_2 - x_2) + (x_3 - x_3) = 0 + 0 + 0 = 0$\n\nWe also have", null, "$(x_1 - x_2) + (x_2 - x_3) - (x_1 - x_3)$", null, "$= (x_1 - x_2) + (x_2 - x_3) + (x_3 - x_1)$\n\nso that", null, "$(x_1 - x_2) + (x_2 - x_3) + (x_3 - x_1) = 0$\n\nThe 3 by 3 incidence matrix", null, "$A$ represents a graph with three nodes and three edges and hence one loop. Each node of the graph is represented by a column of", null, "$A$ and each edge by a row of", null, "$A$. The first row represents edge 1 from node 2 to node 1 (i.e., leaving node 2 and entering node 1). The second row represents edge 2 from node 3 to node 2. The third row represents edge 3 from node 3 to node 1.\n\nIf the vector", null, "$x$ represents potentials at the nodes (", null, "$x_1$ at node 1,", null, "$x_2$ at node 2, and", null, "$x_3$ at node 3) then", null, "$x_1 - x_2$ is the potential difference along edge 1 (from node 2 to node 1),", null, "$x_2 - x_3$ is the potential difference along edge 2 (from node 3 to node 2) and", null, "$x_3 - x_1$ is the potential difference along edge 3 (from node 3 to node 1). From the equations above we see that the sum of the potential differences around the loop is zero (Kirchoff‘s Voltage Law).\n\nNOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition", null, "by Gilbert Strang.\n\nIf you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition", null, ", Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition", null, "and the accompanying free online course, and Dr Strang’s other books", null, ".\n\nThis entry was posted in linear algebra. Bookmark the permalink." ]

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[ "", null, "Run in Google Colab View on GitHub Download notebook\n\n# Tutorial on Diffraction\n\nIn this notebook, you will\n\n• Learn what diffraction is and why it is important\n\n• Make various ray tracing experiments to validate some theoretical results\n\n• Familiarize yourself with the Sionna RT API\n\n• Visualize the impact of diffraction on channel impulse responses and coverage maps\n\n## Background Information", null, "Let’s consider an infinitely long wedge as shown in the figure above. To better visualize this, think of an endless slice of pie. The edge vector is pointed straight out of your screen and the wedge has an opening angle of $$n\\pi$$, where $$n$$ is a real number between 1 and 2.\n\nThe wedge has two faces, the 0- and the n-face. They are labeled this way to indicate from which surface the angle $$\\phi'\\in[0,n]$$ of an incoming locally planar electromagnetic wave is measured. Both faces are made from possibly different materials, each with their own unique properties. These properties are represented by a term known as complex relative permittivity, denoted by $$\\eta_0$$ and $$\\eta_n$$, respectively. Without diving too deep into the specifics, permittivity measures how a material reacts to an applied electric field.\n\nWe can define three distinct regions in this figure: Region $$I$$, in which the incident field as well as the reflected field from the 0-face are present, Region $$II$$, in which the reflected field vanishes, and Region $$III$$, in which the incident field is shadowed by the wedge. The three regions are separated by the reflection shadow boundary (RSB) and the incident shadow boundary (ISB). The former is determined by the angle of specular reflection $$\\pi-\\phi'$$, while the latter is simply the prolongation of the direction of the incoming wave through the edge, i.e., $$\\pi+\\phi'$$.\n\nUsing geometrical optics (GO) alone, the electromagnetic field would abruptly change at each boundary as the reflected and incident field components suddenly disappear. As this is physically not plausible, the geometrical theory of diffraction (GTD) , as developed by Joseph B. Keller in the 1960s, introduces a so-called diffracted field which ensures that the total field is continuous. The diffracted field is hence especially important in the transition regions between the different regions and then rapidly decays to zero. Most importantly, without diffraction, there would be no field beyond the ISB in Region $$III$$.\n\nDiffraction is hence a very important phenomenon which enables wireless coverage behind buildings at positions without a line-of-sight of strong reflected path. As you will see later in this tutorial, the diffracted field is generally much weaker than the incident or reflected field. Moreover, the higher the frequency, the faster the diffracted field decays when moving away from the RSB and ISB.", null, "According to the GTD, when a ray hits a point on the edge, its energy gets spread over a continuum of rays lying on the Keller cone. All rays on this cone make equal angles with the edge of diffraction at the point of diffraction, i.e., $$\\beta_0'=\\beta_0$$. One can think of this phenomenon as an extension of the law of reflection at planar surfaces. The figures above illustrates this concept.\n\nThe GTD was later extended to the uniform theory of diffraction (UTD) [2,3] which overcomes some of its shortcomings.\n\nWe will explore in this notebook these effects in detail and also validate the UTD implementation in Sionna RT as a by-product.\n\n## GPU Configuration and Imports\n\n:\n\nimport os\ngpu_num = 0 # Use \"\" to use the CPU\nos.environ[\"CUDA_VISIBLE_DEVICES\"] = f\"{gpu_num}\"\nos.environ['TF_CPP_MIN_LOG_LEVEL'] = '3'\n\n# Thus, the preview does not work in Colab. However, whenever possible we\n# strongly recommend to use the scene preview mode.\ntry: # detect if the notebook runs in Colab\ncolab_compat = True # deactivate preview\nexcept:\ncolab_compat = False\nresolution = [480,320] # increase for higher quality of renderings\n\n# Allows to exit cell execution in Jupyter\nclass ExitCell(Exception):\ndef _render_traceback_(self):\npass\n\n# Import Sionna\ntry:\nimport sionna\nexcept ImportError as e:\n# Install Sionna if package is not already installed\nimport os\nos.system(\"pip install sionna\")\nimport sionna\n\n# Configure the notebook to use only a single GPU and allocate only as much memory as needed\n# For more details, see https://www.tensorflow.org/guide/gpu\nimport tensorflow as tf\ngpus = tf.config.list_physical_devices('GPU')\nif gpus:\ntry:\ntf.config.experimental.set_memory_growth(gpus, True)\nexcept RuntimeError as e:\nprint(e)\n# Avoid warnings from TensorFlow\ntf.get_logger().setLevel('ERROR')\n\ntf.random.set_seed(1) # Set global random seed for reproducibility\n\n%matplotlib inline\nimport matplotlib.pyplot as plt\nimport numpy as np\nimport sys\n\nfrom sionna.rt.utils import r_hat\nfrom sionna.constants import PI, SPEED_OF_LIGHT\nfrom sionna.utils import expand_to_rank\n\n\n## Experiments with a Simple Wedge\n\n:\n\nscene = load_scene(sionna.rt.scene.simple_wedge)\n\n# Create new camera with different configuration\nmy_cam = Camera(\"my_cam\", position=[10,-100,100], look_at=[10,0,0])\n\n# Render scene\nscene.render(my_cam);\n\n# You can also preview the scene with the following command\n# scene.preview()", null, "The wedge has an opening angle of $$\\frac{3}{2}\\pi=270^\\circ$$, i.e., $$n=1.5$$. The 0-face is aligned with the x axis and the n-face aligned with the negative y axis.\n\nFor the following experiments, we will configure the wedge to be made of metal, an almost perfect conductor, and set the frequency to 1GHz.\n\n:\n\nscene.frequency = 1e9 # 1GHz\nscene.objects[\"wedge\"].radio_material = \"itu_metal\" # Almost perfect reflector\n\n\nWith our scene being set-up, we now need to configure a transmitter and place multiple receivers to measure the field. We assume that the transmitter and all receivers have a single vertically polarized isotropic antenna.\n\n:\n\n# Configure the antenna arrays used by the transmitters and receivers\nscene.tx_array = PlanarArray(num_rows=1,\nnum_cols=1,\nvertical_spacing=0.5,\nhorizontal_spacing=0.5,\npattern=\"iso\",\npolarization=\"V\")\n\nscene.rx_array = scene.tx_array\n\n:\n\n# Transmitter\ntx_angle = 30/180*PI # Angle phi from the 0-face\ntx_dist = 50 # Distance from the edge\ntx_pos = 50*r_hat(PI/2, tx_angle)\nref_boundary = (PI - tx_angle)/PI*180\nlos_boundary = (PI + tx_angle)/PI*180\nposition=tx_pos,\norientation=[0,0,0]))\n\n# We place num_rx receivers uniformly spaced on the segment of a circle around the wedge\nnum_rx = 1000 # Number of receivers\nrx_dist = 5 # Distance from the edge\nphi = tf.linspace(1e-2, 3/2*PI-1e-2, num=num_rx)\ntheta = PI/2*tf.ones_like(phi)\nrx_pos = rx_dist*r_hat(theta, phi)\n\nfor i, pos in enumerate(rx_pos):\nposition=pos,\norientation=[0,0,0]))\n\n:\n\n# Render scene\nmy_cam.position = [-30,100,100]\nmy_cam.look_at([10,0,0])\nscene.render(my_cam);", null, "In the above figure, the blue ball is the transmitter and the green circle corresponds to 1000 receivers uniformly distributed over a segment of a circle around the edge.\n\nNext, we compute the channel impulse response between the transmitter and all of the receivers. We deactivate scattering in this notebook as it would require a prohibitive amount of memory with such a large number of receivers.\n\n:\n\n# Compute paths between the transmitter and all receivers\npaths = scene.compute_paths(num_samples=1e6,\nlos=True,\nreflection=True,\ndiffraction=True,\nscattering=False)\n\n# Obtain channel impulse responses\n# We squeeze irrelevant dimensions\n# [num_rx, max_num_paths]\na, tau = [np.squeeze(t) for t in paths.cir()]\n\n:\n\ndef compute_gain(a, tau):\n\"\"\"Compute $|H(f)|^2\"\"\" a = tf.squeeze(a, axis=-1) tau = expand_to_rank(tau, tf.rank(a), axis=-2) e = a*tf.exp(tf.complex(tf.cast(0, tf.float32), -tf.cast(2*PI, tf.float32)*scene.frequency*tau)) h_f_2 = tf.math.abs(tf.reduce_sum(e, axis=-1))**2 h_f_2 = tf.where(h_f_2==0, 1e-24, h_f_2) g_db = 10*np.log10(h_f_2) return tf.squeeze(g_db) Let’s have a look at the channel impulse response of one of the receivers: : n = 400 plt.figure() plt.stem(tau[n]/1e-9, 10*np.log10(np.abs(a[n])**2)) plt.title(f\"Angle of receiver$\\phi: {int(phi[n]/PI*180)}^\\circ$\"); plt.xlabel(\"Delay (ns)\"); plt.ylabel(\"$|a|^2$(dB)\");", null, "For an angle of around 108 degrees, the receiver is located within Region I, where all propagation effects should be visible. As expected, we can observe three path: line-of-sight, reflected, and diffracted. While the first two have roughly the same strength (as metal is an almost perfect reflector), the diffracted path has significantly lower energy. Next, let us compute the channel frequency response $$H(f)$$ as the sum of all paths multiplied with their complex phase factors: $H(f) = \\sum_{i=1}^N a_i e^{-j2\\pi\\tau_i f}$ : h_f_tot = np.sum(a*np.exp(-1j*2*np.pi*scene.frequency.numpy()*tau), axis=-1) We can now visualize the path gain $$|H(f)|^2$$ for all receivers, i.e., as a function of the angle $$\\phi$$: : fig = plt.figure() plt.plot(phi/PI*180, 20*np.log10(np.abs(h_f_tot))) plt.xlabel(\"Diffraction angle$\\phi$(deg)\"); plt.ylabel(r\"Path gain$|H(f)|^2$(dB)\"); plt.ylim([-100, -59]); plt.xlim([0, phi[-1]/PI*180]);", null, "The most important observation from the figure above is that $$H(f)$$ remains continous over the entire range of $$\\phi$$, especially at the RSB and ISB boundaries at around $$\\phi=150^\\circ$$ and $$\\phi=209^\\circ$$, respectively. To get some more insight, the convenience function in the next cell, computes and visualizes the different components of $$H(f)$$ by their type. : def plot(frequency, material): \"\"\"Plots the path gain$|H(f)|^2 versus $phi$ for a given\nfrequency and RadioMaterial of the wedge.\n\"\"\"\n# Set carrier frequency and material of the wedge\n# You can see a list of available materials by executing\nscene.frequency = frequency\n\n# Recompute paths with the updated material and frequency\npaths = scene.compute_paths(num_samples=1e6,\nlos=True,\nreflection=True,\ndiffraction=True,\nscattering=False)\n\ndef compute_gain(a, tau):\n\"\"\"Compute $|H(f)|^2\"\"\" a = tf.squeeze(a, axis=-1) tau = expand_to_rank(tau, tf.rank(a), axis=-2) e = a*tf.exp(tf.complex(tf.cast(0, tf.float32), -tf.cast(2*PI, tf.float32)*scene.frequency*tau)) h_f_2 = tf.math.abs(tf.reduce_sum(e, axis=-1))**2 h_f_2 = tf.where(h_f_2==0, 1e-24, h_f_2) g_db = 10*np.log10(h_f_2) return tf.squeeze(g_db) # Compute gain for all path types g_tot_db = compute_gain(*paths.cir()) g_los_db = compute_gain(*paths.cir(reflection=False, diffraction=False, scattering=False)) g_ref_db = compute_gain(*paths.cir(los=False, diffraction=False, scattering=False)) g_dif_db = compute_gain(*paths.cir(los=False, reflection=False, scattering=False)) # Make a nice plot fig = plt.figure() phi_deg = phi/PI*180 ymax = np.max(g_tot_db)+5 ymin = ymax - 45 plt.plot(phi_deg, g_tot_db) plt.plot(phi_deg, g_los_db) plt.plot(phi_deg, g_ref_db) plt.plot(phi_deg, g_dif_db) plt.ylim([ymin, ymax]) plt.xlim([phi_deg, phi_deg[-1]]); plt.legend([\"Total\", \"LoS\", \"Reflected\", \"Diffracted\"], loc=\"lower left\") plt.xlabel(\"Diffraction angle$\\phi$(deg)\") plt.ylabel(\"Path gain$|H(f)|^2$(dB)\") ax = fig.axes ax.axvline(x=ref_boundary, ymin=0, ymax=1, color=\"black\", linestyle=\"--\") ax.axvline(x=los_boundary, ymin=0, ymax=1, color=\"black\", linestyle=\"--\") ax.text(ref_boundary-10,ymin+5,'RSB',rotation=90,va='top') ax.text(los_boundary-10,ymin+5,'ISB',rotation=90,va='top') ax.text(ref_boundary/2,ymax-2.5,'Region I', ha='center', va='center', bbox=dict(facecolor='none', edgecolor='black', pad=4.0)) ax.text(los_boundary-(los_boundary-ref_boundary)/2,ymax-2.5,'Region II', ha='center', va='center', bbox=dict(facecolor='none', edgecolor='black', pad=4.0)) ax.text(phi_deg[-1]-(phi_deg[-1]-los_boundary)/2,ymax-2.5,'Region III', ha='center', va='center', bbox=dict(facecolor='none', edgecolor='black', pad=4.0)) plt.title('$f={}\\$ GHz (\"{}\")'.format(frequency/1e9, material))\nplt.tight_layout()\nreturn fig\n\n:\n\nplot(1e9, \"itu_metal\");", null, "The figure above shows the path gain for the total field as well as that for the different path types. In Region $$I$$, the line-of-sight and reflected paths dominate the total field. While their contributions are almost constant over the range of $$\\phi\\in[0,150^\\circ]$$, their combined field exhibits large fluctutations due to constructive and destructive interference. As we approach the RSB, the diffracted field increases to ensure continuity at $$\\phi=150^\\circ$$, where the reflected field immediately drops to zero. A similar observation can be made close to the ISB, where the incident (or line-of-sight) component suddenly vanishes. In Region $$III$$, the only field contribution comes from the diffracted field.\n\nLet us now have a look at what happens when we change the frequency to $$10\\,$$GHz.\n\n:\n\nplot(10e9, \"itu_metal\");", null, "The first observation we can make is that the overall path gain has dropped by around $$20\\,$$dB. This is expected as it is proportional to the square of the wavelength $$\\lambda$$.\n\nThe second noticeable difference is that the path gain fluctuates far more rapidly. This is simply due to the shorter wavelength.\n\nThe third observation we can make is that the diffracted field decays far more radpily when moving away from the boundaries as compared to a frequency of $$1\\,$$GHz. Thus, diffraction is less important at high frequencies.\n\nWe can verify that the same trends continue by plotting the result for a frequency of $$100\\,$$GHz, which is the upper limit for which the ITU Metal material is defined (see the Sionna RT Documentation).\n\n:\n\nplot(100e9, \"itu_metal\");", null, "It is also interesting to change the material of the wedge. The preconfigured materials in Sionna RT can be inspected with the following command:\n\n:\n\nlist(scene.radio_materials.keys())\n\n:\n\n['vacuum',\n'itu_concrete',\n'itu_brick',\n'itu_plasterboard',\n'itu_wood',\n'itu_glass',\n'itu_ceiling_board',\n'itu_chipboard',\n'itu_plywood',\n'itu_marble',\n'itu_floorboard',\n'itu_metal',\n'itu_very_dry_ground',\n'itu_medium_dry_ground',\n'itu_wet_ground']\n\n\nLet’s see what happens when we change the material of the wedge to wood and the frequency back to $$1\\,$$GHz.\n\n:\n\nplot(1e9, \"itu_wood\");", null, "We immediately notice that wood is a bad reflector since the strength of the reflected path has dropped by $$10\\,$$dB compared to the metal. Thanks to the heuristic extension of the diffracted field equations in to non-perfect conductors in (which are implemented in Sionna RT), the total field remains continuous.\n\nYou might now want to try different materials and frequencies for yourself.\n\n## Coverage Maps with Diffraction\n\nSo far, we have obtained a solid microscopic understanding of the effect of scattering. Let us now turn to the macroscopic effects that can be nicely observed through coverage maps.\n\nA coverage map describes the average received power from a specific transmitter at every point on a plane. The effects of fast fading, i.e., constructive/destructive interference between different paths, are averaged out by summing the squared amplitudes of all paths. As we cannot compute coverage maps with infinitely fine resolution, they are approximated by small rectangular tiles for which average values are computed. For a detailed explanation, have a look at the API Documentation.\n\nLet us now load a slightly more interesting scene containing a couple of rectangular buildings and add a transmitter. Note that we do not need to add any receivers to compute a coverage map.\n\n:\n\nscene = load_scene(sionna.rt.scene.simple_street_canyon)\n\n# Set the carrier frequency to 1GHz\nscene.frequency = 1e9\n\nscene.tx_array = PlanarArray(num_rows=1,\nnum_cols=1,\nvertical_spacing=0.5,\nhorizontal_spacing=0.5,\npattern=\"iso\",\npolarization=\"V\")\n\nscene.rx_array = scene.tx_array\n\nposition=[-33,11,32],\norientation=[0,0,0]))\n\n# Render the scene from one of its cameras\n# The blue dot represents the transmitter\nscene.render('scene-cam-1');", null, "Computing a coverage map is as simple as running the following command:\n\n:\n\ncm = scene.coverage_map(cm_cell_size=[1,1], num_samples=10e6)\n\n\nWe can visualizes the coverage map in the scene as follows:\n\n:\n\n# Add a camera looking at the scene from the top\nmy_cam = Camera(\"my_cam\", position=[10,0,300], look_at=[0,0,0])\nmy_cam.look_at([0,0,0])\n\n# Render scene with the new camera and overlay the coverage map\nscene.render(my_cam, coverage_map=cm);", null, "From the figure above, we can see that many regions behind buildings do not receive any signal. The reason for this is that diffraction is by default deactivated. Let us now generate a new coverage map with diffraction enabled:\n\n:\n\ncm_diff = scene.coverage_map(cm_cell_size=[1,1], num_samples=10e6, diffraction=True)\nscene.render(my_cam, coverage_map=cm_diff);", null, "As expected from our experiements above, there is not a single point in the scene that is left blank. In some areas, however, the signal is still very weak and will not enable any form of communication.\n\nLet’s do the same experiments at a higher carrier frequency (30 GHz):\n\n:\n\nscene = load_scene(sionna.rt.scene.simple_street_canyon)\nscene.frequency = 30e9\nscene.tx_array = PlanarArray(num_rows=1,\nnum_cols=1,\nvertical_spacing=0.5,\nhorizontal_spacing=0.5,\npattern=\"iso\",\npolarization=\"V\")\n\nscene.rx_array = scene.tx_array\nposition=[-33,11,32],\norientation=[0,0,0]))\n\ncm = scene.coverage_map(cm_cell_size=[1,1], num_samples=10e6)\ncm_diff = scene.coverage_map(cm_cell_size=[1,1], num_samples=10e6, diffraction=True)\nscene.render(my_cam, coverage_map=cm);\nscene.render(my_cam, coverage_map=cm_diff);", null, "", null, "While the 1 GHz and 30 GHz carrier frequency coverage maps appear similar, key differences exist. The dynamic range for 30 GHz has grown by around 16dB due to the reduced diffracted field in deep shadow areas, such as behind buildings. The diffracted field at this frequency is considerably smaller compared to the incident field than it is at 1 GHz, leading to a significant increase in dynamic range.\n\nIn conclusion, diffraction plays a vital role in maintaining the consistency of the electric field across both reflection and incident shadow boundaries. It generates diffracted rays that form a Keller cone around an edge. As we move away from these boundaries, the diffracted field diminishes rapidly. Importantly, the contributions of the diffracted field become less significant as the carrier frequency increases.\n\nWe hope you enjoyed our dive into diffraction with this Sionna RT tutorial. We really encourage you to get hands-on, conduct your own experiments and deepen your understanding of ray tracing. There’s always more to learn, so do explore our other tutorials as well." ]

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", null, 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[ "WIAS Preprint No. 1106, (2006)\n\n# Random walk on fixed spheres for Laplace and Lamé equations\n\nAuthors\n\n• Sabelfeld, Karl\n• Shalimova, Irina\n• Levykin, Alexander I.\n\n2010 Mathematics Subject Classification\n\n• 65C05 65C20 74B05\n\n2008 Physics and Astronomy Classification Scheme\n\n• 02.70.Lq\n\nKeywords\n\n• Poisson integral formula, random walk on fixed spheres, Lamé equation, successive over relaxation method, divergent Neumann series, discrete random walks\n\nAbstract\n\nThe Random Walk on Fixed Spheres (RWFS) introduced in our previous paper is presented in details for Laplace and Lamé equations governing static elasticity problems. The approach is based on the Poisson type integral formulae written for each disc of a domain consisting of a family of overlapping discs. The original differential boundary value problem is equivalently reformulated in the form of a system of integral equations defined on the intersection surfaces (arches, in 2D, and caps, if generalized to 3D spheres). To solve the obtained system of integral equations, a Random Walk procedure is constructed where the random walks are living on the intersecting surfaces. Since the spheres are fixed, it is convenient to construct also discrete random walk methods for solving the system of linear equations approximating the system of integral equations. We develop here two classes of special Monte Carlo iterative methods for solving these systems of linear algebraic equations which are constructed as a kind of randomized versions of the Chebyshev iteration method and Successive Over Relaxation (SOR) method. It is found that in this class of randomized SOR methods, the Gauss-Seidel method has a minimal variance. In our prevoius paper we have concluded that in the case of classical potential theory, the Random Walk on Fixed Spheres considerably improves the convergence rate of the standard Random Walk on Spheres method. More interesting, we succeeded there to extend the algorithm to the system of Lamé equations which cannot be solved by the conventional Random Walk on Spheres method. We present here a series of numerical experiments for 2D domains consisting of 5, 10, and 17 discs, and analyze the dependence of the variance on the number of discs and elastic constants. Further generalizations to Neumann and Dirichlet-Neumann boundary conditions are also possible.\n\nAppeared in\n\n• Monte Carlo Methods Appl., 12 (2006) pp. 55--93." ]

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[ "Simplification is one of the most common types questions that come in in the IBPS RRB 2020. Most of the candidates understand the basic concepts of simplification but they often worry about solving the questions. The biggest reason is the fact that solving simplification questions take a lot of time to solve. As a result, a lot of candidates try to avoid the questions. However, if you know certain tricks and techniques then solving the questions on simplification really become easy. In this post, we would be taking a look at some of the things that you can do to solve IBPS RRB simplification questions with speed and accuracy.\n\n## IBPS RRB Simplification Questions: Common Types of Questions\n\nLet us start by analyzing the previous year question papers, to see what are the common types of questions in the RRB exam. This would help you to develop the right strategy for bank exam preparation. The following types of questions are most common in IBPS RRB exam. Therefore, you can expect that at least some of the given types of questions would come in IBPS RRB 2020 as well.\n\n• Finding Missing Numbers – You would be given an equation, that would have one missing number on either side. You have to fill them up using simplification methods. Occurrence – Rare. Difficulty – Moderate to difficult\n• Straight Simplification – You would be given a problem which you have to solve an equation using simplification methods. Occurrence – Common. Difficulty – Easy\n\n## FAQs on IBPS RRB Simplification Questions\n\n### Is BODMAS still relevant?\n\nBODMAS is the fundamental method for solving simplification questions. Yes, there are questions that be solved quickly with methods other than BODMAS. However, we would recommend that you start preparing with simple BODMAS questions. This would build a strong foundation that would help you solve trickier questions in the exam.\n\n### How many questions come from simplification on an average?\n\nAs with almost every other topic, the number of questions varies from year to year. However, you can rest assured that you would get a lot of questions on this topic. Every year, there are about 15-20 questions from simplification in IBPS RRB.\n\n### How difficult are they?\n\nAs there are a lot of questions, the difficulty also varies in the same year. In the same question paper, you might get questions that are tough as well as questions that are easy. Generally speaking, the difficulty of questions vary from easy to moderate.\n\n### What to do in order to solve them quicker?\n\nA lot of candidates waste a lot of time in solving simplification questions. There are two things that you need to solve them quickly. One is the right technique and the other is ‘practice’.\n\n### How much do I need to practice?\n\nDepends on how much time you got left for the IBPS Mains exam. If you still have six months or so, and you are relatively strong in maths, then just 10-15 minutes every day would be enough.\n\n## Some Quick Tricks for Solving IBPS RRB Simplification Questions\n\nLet us take a look at some of the common types of questions and some easy tricks that would help you to solve them easily and quickly.\n\nQuestion: Solve (10.8x16x12)x(3.6x56x9.2)\n\nOption:\n\nA) 3928.32\n\nB) 3927.51\n\nC)3941.23\n\nD) 3875.11\n\nSolution: In solving these type of questions, always remember that product of even numbers is always even. If you look at the equation, you would find that all six numbers are even. Therefore, their product must be even. If you look at the options, you would find that there is only one number that is even. That’s the correct answer.\n\nNow, let us look at another example.\n\nQuestion: 203% of 12.3+1110% of 23 = ?\n\nOptions:\n\nA)270.269\n\nB) 250.269\n\nC) 280.269\n\nD) 220.269\n\nSolution: In such cases, all you need to do is just break down the large numbers into smaller numbers like this:\n\n203% of 12.3 = 200% of 12.3+ 3% of 12.3= 24.6+0.369= 24.969\n\n1110% of 23 = 1000% of 23+ 100% of 23+10% of 23=230+23+2.3 =255.3\n\nAdding them both together, we get\n\n24.969+255.3=280.269\n\nThese were some of the tips and tricks that you can use to solve IBPS RRB Simplification Questions. Let us solve some practice questions.\n\n## Practice Questions\n\nQuestion 1: 3.2% of 500 × 2. 4% of ? = 288\n\nOption:\n(a) 650\n(b) 700\n(c) 600\n(d) 750\n\nQuestion 2: (6.5% of 375) – (0.85% of 230) =?\n\nOption:\n(a) 23.42\n(b) 24.24\n(c) 25.76\n(d) 22.42\n\nQuestion 3:  13.141 + 31.417 – 27.118 =?\n\nOption:\n(a) 16.441\n(b) 17.543\n(c) 17.440\n(d) 17.590" ]

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[ "", null, "", null, "Home", null, "What's New", null, "What's Hot", null, "Submit eBook", null, "Search eBooks", null, "Contact Us", null, "FREE JOBS NEWSLETTER 3,67,839[96,218 + 2,71,621] MEMBERS!\n\n Free eBook Categories", null, "Computers and Technologies eBook", null, "Business eBook", null, "Children eBook", null, "Literature eBook", null, "Marketing eBook", null, "Miscellaneous eBook", null, "Publishing eBook", null, "Reference eBook", null, "Self Improvement eBook", null, "Tutorials eBook", null, "Internet eBook", null, "Games eBook", null, "Travel eBook", null, "Health eBook", null, "Cooking eBook", null, "Science & Technology eBooks eBook", null, "Medical & Medicine eBook", null, "Fun & Entertainment eBook", null, "Astrology eBook", null, "Engineering Related Ebooks eBook", null, "Mathematics eBooks eBook", null, "Web Quality eBook", null, "Schema eBook", null, "XQuery eBook", null, "Manuals and Documentations eBook", null, "Operating System eBook", null, "General Tests preparation ebooks eBook", null, "Politics & Government Ebooks eBook", null, "War & Military eBook", null, "Social Issues Ebooks eBook", null, "Arts eBook", null, "Biographies eBook", null, "Education eBook", null, "Languages eBook", null, "Essays eBook", null, "Sports eBook", null, "History eBook", null, "Home & Garden eBook", null, "Mysterious & Supernatural eBook", null, "Philosophy eBook", null, "Religion & Spirituality eBook", null, "Fantasy eBook", null, "Fiction eBook", null, "Drama eBook", null, "Romance eBook", null, "Audio eBook", null, "Exploration eBook", null, "Law & Legals eBook", null, "World eBook", null, "Animals eBook", null, "Ebooks eBook", null, "Computer Tools eBook", null, "Servers eBook", null, "Applications eBook", null, "Microsoft Office eBook", null, "Microsoft Technologies eBook", null, "Mobile Technology eBook", null, "", null, "", null, "# Differential Equations EBooks", null, "Below we have listed all the Free Differential Equations EBooks for download. Feel free to comment on any Differential Equations EBooks for download or answer by the comment feature available on the page.", null, "Mastering Differential Equations:The Visual Method: For centuries, differential equations have been the key to unlocking nature's deepest secrets. Over 300 years ago, Isaac Newton invented differential equations to understand the problem of motion, and he developed calculus in order to solve differential equations.Since then, differential equations have been the essential tool for analyzing the process of change, whether in physics, engineering, biology, or any other field where it's important to predict how something behaves over time.The pinnacle of a mathematics education, differential equations assume a basic knowledge of calculus, and they have traditionally required the rote memorization of a vast \"cookbook\" of formulas and specialized tricks needed to find explicit solutions. Even then, most problems involving differential equations had to be simplified, often in unrealistic ways; and a huge number of equations defied solution at all using these techniques.But that was before computers revolutionized the field, extending the reach of differential equations into previously unexplored areas and allowing solutions to be approximated and displayed in easy-to-grasp computer graphics. For the first time, a method exists that can start a committed learner on the road to mastering this beautiful application of the ideas and techniques of calculus.Mastering Differential Equations: The Visual Method takes you on this amazing mathematical journey in 24 intellectually stimulating and visually engaging half-hour lectures taught by a pioneer of the visual approach, Professor Robert L. Devaney of Boston University, coauthor of one of the most widely used textbooks on ordinary differential equations.Date Added: 9/23/2011 10:48:15 PM | Visits: 31331", null, "MathTutor Differential Equations Vol. 1 First Order Equations: Differential equations is used in all branches of engineering and science. In essence, once a student begins to study more complex problems, nature usually obeys a differential equation which means that the equation involves one or more derivatives of the unknown variable.In other words, a differential equation involves the rate of change of a variable rather than the variable itself. The simplest example of this is F=ma. The \"a\" is acceleration which is the second derivative of the position of the object. Although differential equations may look simple to solve by just integration, they frequently require complex solution methods with many steps.This 10 hour DVD course teaches how to solve first order differential equations using fully worked example problems. All intermediate steps are shown along with graphing methods and applications of differential equations in science and engineering.Date Added: 9/23/2011 10:44:55 PM | Visits: 31638", null, "Mastering Differential Equations The Visual Method: For centuries, differential equations have been the key to unlocking nature's deepest secrets. Over 300 years ago, Isaac Newton invented differential equations to understand the problem of motion, and he developed calculus in order to solve differential equations.Since then, differential equations have been the essential tool for analyzing the process of change, whether in physics, engineering, biology, or any other field where it's important to predict how something behaves over time.The pinnacle of a mathematics education, differential equations assume a basic knowledge of calculus, and they have traditionally required the rote memorization of a vast \"cookbook\" of formulas and specialized tricks needed to find explicit solutions. Even then, most problems involving differential equations had to be simplified, often in unrealistic ways; and a huge number of equations defied solution at all using these techniques.Date Added: 9/23/2011 10:44:02 PM | Visits: 31198", null, "The Differential Equations Tutor Volume 2 - Higher Order Equations: Solving higher order differential equations is challenging for most students simply because the solution methods usually run several pages in length even for the easier problems. The student must identify the type of equation to solve and apply the appropriate solution method, which can lead to valuable lost time on an exam if the wrong solution method is chosen at the outset.We begin by showing the student real life applications of second order and higher ODEs to provide motivation for the material. Next, we show how to solve elemenary second order ODEs, and show the student that all solutions have a similar form.Next, we discuss linear independence of solutions and show the students how to use the wronskian test to determine of a set of functions describe the entire solution space of the ODE.Date Added: 9/23/2011 10:43:12 PM | Visits: 31744", null, "\"Image Processing Based on Partial Differential Equations: Proceedings of the International Conference: This book publishes a collection of original scientific research articles that address the state-of-art in using partial differential equations for image and signal processing. Coverage includes: level set methods for image segmentation and construction, denoising techniques, digital image inpainting, image dejittering, image registration, and fast numerical algorithms for solving these problems.Date Added: 9/23/2011 10:32:38 PM | Visits: 30725", null, "\"Stochastic Differential Equations\": differential equations and applications.- D.W. Stroock, S.R.S. Varadhan: Theory of diffusion processes.- G.C. Papanicolaou: Wave propagation and heat conduction in a random medium.- C. Dewitt Morette: A stochastic problem in Physics.- G.S. Goodman: The embedding problem for stochastic matrices.Date Added: 9/23/2011 10:31:10 PM | Visits: 32515", null, "The Differential Equations Tutor: Solving higher order differential equations is challenging for most students simply because the solution methods usually run several pages in length even for the easier problems. The student must identify the type of equation to solve and apply the appropriate solution method, which can lead to valuable lost time on an exam if the wrong solution method is chosen at the outset.We begin by showing the student real life applications of second order and higher ODEs to provide motivation for the material. Next, we show how to solve elemenary second order ODEs, and show the student that all solutions have a similar form.Next, we discuss linear independence of solutions and show the students how to use the wronskian test to determine of a set of functions describe the entire solution space of the ODE.We then get into the core solution techniques which revolve around constant coefficient differential equations. We examine the case where the roots of the characteristic polynomial are real and complex separately, to give the student a good grounding in what to do in either case.Date Added: 9/23/2011 10:30:30 PM | Visits: 32411", null, "Differential Equations with Boundary � Value Problems: Differential Equations with Boundary-Value Problems (7th Edition) strikes a balance between the analytical, qualitative, and quantitative approaches to the study of differential equations. This proven and accessible text speaks to beginning engineering and math students through a wealth of pedagogical aids, including an abundance of examples, explanations, �Remarks� boxes, definitions, and group projects. Using a straightforward, readable, and helpful style, this book provides a thorough treatment of boundary-value problems and partial differential equations.Date Added: 9/23/2011 10:27:25 PM | Visits: 31735", null, "MIT Open Courseware Differential Equations: * Solution of First-order ODE`s by Analytical, Graphical and Numerical Methods;* Linear ODE`s, Especially Second Order with Constant Coefficients;* Undetermined Coefficients and Variation of Parameters;* Sinusoidal and Exponential Signals: Oscillations, Damping, Resonance;* Complex Numbers and Exponentials;* Fourier Series, Periodic Solutions;* Delta Functions, Convolution, and Laplace Transform Methods;* Matrix and First-order Linear Systems: Eigenvalues and Eigenvectors; and* Non-linear Autonomous Systems: Critical Point Analysis and Phase Plane Diagrams.Date Added: 9/23/2011 10:18:00 PM | Visits: 31379", null, "Fractional Differential Equations: This book is a landmark title in the continuous move from integer to non-integer in mathematics: from integer numbers to real numbers, from factorials to the gamma function, from integer-order models to models of an arbitrary order. For historical reasons, the word 'fractional' is used instead of the word 'arbitrary'.This book is written for readers who are new to the fields of fractional derivatives and fractional-order mathematical models, and feel that they need them for developing more adequate mathematical models.In this book, not only applied scientists, but also pure mathematicians will find fresh motivation for developing new methods and approaches in their fields of research.Date Added: 8/25/2011 2:58:16 AM | Visits: 33330\n\n1| 2 | 3 | 4 | 5 | Next >>\n\nFree Differential Equations eBooks - Download Differential Equations eBooks - List of Differential Equations eBooks\n\n Free EBooks & Online Resources for Download Related Pages Computer And Internet EBooks | Business EBooks | Children EBooks | Literature EBooks | Marketing EBooks | Misc. 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